Intermediate Algebra Third Edition Pdf
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Elementary And Intermediate Algebra 3rd Edition Pdf
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- Pre-Algebra - Integers Objective: Add, Subtract, Multiply and Divide Positive and Negative Numbers. The ability to work comfortably with negative numbers is essential to success in algebra. For this reason we will do a quick review of adding, subtracting, multi-plying and dividing of integers. Integers are all the positive whole numbers, zero.
mm 71
INTERMEDIATE ALGEBRA
THE MACMILLAN COMPANY NEW YORK
BOSTON CHICAGO DALLAS SAN FRANCISCO
ATLANTA
MACMILLAN AND LONDON
CO., LIMITED
BOMBAY CALCUTTA MELBOURNE
MADRAS
THE MACMILLAN COMPANY OF CANADA, TORONTO
LIMITED
INTERMEDIATE ALGEBRA
Ralph
S.
UndcrWOOd,
of Mathematics,
Professor
Texas
Technological College
Thomas R.
Nelson,
Associate Professor of Mathematics,
Agricultural and Mechanical
College of Texas
Samuel Selby,
of
College
Head
the
Department
of
Mathematics,
of Engineering, University of Akron
THE MACMILLAN COMPANY
NEW YORK
1947
COPYKIGHT, 1947, BY THE MACMILLAN COMPANY no part of this book may be reproduced in All rights reserved without any form permission in writing from the publisher, except by a reviewer who wishes to quote brief passages in connection with a review written for inclusion in magazine or newspaper.
PRINTED
IN
THE UNITED STATES OF AMERICA
PREFACE
This text is designed for students who, for one reason or another, such as inadequate previous training or merely the
immaturity of youth, must learn algebra at a somewhat slower pace than that set by the typical college freshman who plans to enter a scientific, engineering, or allied field. In other words, it is intended to be a true intermediate algebra, suitable for college classes catering to those students who in high school had less than the normal amount of mathematics. As such it starts at ' bedrock' in algebra, moves on with a leisurely but accelerating pace, and leaves the student at the
ledge in the
cliff
of learning
and logarithms. There
is
marked roughly by progressions
ample material
for a course of three
or six semester hours, according to the needs
and proficiency
of the class.
Specific features of the text include the following. 1.
The
illustrative
examples are numerous, and they are
carefully selected to cover typical cases. 2. The fact that in algebra the rules of arithmetic are
applied to letters instead of numbers is emphasized by frequent use of two illustrative examples, one of which is purely arithmetic. rules,
and
Thus the student is encouraged to check methods,
results
by use
of simple numbers.
Common
errors are pointed out and discussed before ' they are hidden as traps' in problems. In fact, in certain problems the sole question to be considered is whether a 3.
given operation is or is not correct. 4. In a text of this level the primary aim
is clarity,
to-
PREFACE
vi
gather with as much mathematical rigor as can be preserved in the light of the first requirement. A major goal, therefore,
has been simplicity and directness of style. To this end exceptions and qualifying statements have been for the most part relegated to footnotes. 5.
More than 2600 problems provide ample
drill.
Experience has indicated that the common practice of printing only half of the answers in a text in effect cuts 6.
down
the available problem list by almost one half, since most teachers assign for home work only those problems for
which the answers are to be found
in the text.
On
the other
hand, some problems without answers are needed for correspondence courses, and students acquire self-reliance by learning to check results for themselves. These two conflicting considerations make it seem advisable to include most, but not all, of the answers and so four fifths of them are included in this text. We think it inadvisable to have the ;
remaining answers printed even in pamphlet form, since copies would eventually reach unintended hands and destroy the effectiveness of the book as a text for correspondence courses. 7.
Our guiding
principle has been to maintain, so far as
the subject matter allows, a consistent, upward-sloping level of difficulty. Such a policy dictates, for example, an early treatment of linear equations, and the deferment of the chapter on exponents and radicals as long as seems
expedient.
The
mimeographed form in the appropriate classes at Texas Agricultural and Mechanical College, and the present form embodies changes made in 8.
text has been used in
the light of student reactions. It has since been read critically by several referees, who have contributed many valuable suggestions.
CONTENTS
One
Algebra as a Language and a Tool
Two
Type Products and Factoring
21
Three
Fractions
34
Four
Linear Equations in
Five
Functions and Graphs
87
Six
Simultaneous Linear Equations
99
Seven
Exponents and Radicals
121
Eight
The Number System
143
Nine
Quadratic Equations
150
Ten
Special Equations in
Eleven
Simultaneous Equations
Twelve
Ratios, Proportions,
Thirteen
The Binomial Theorem
216
Fourteen
Progressions
221
Fifteen
Logarithms
233
Tables
254
Answers
259
One Unknown
One Unknown
and Variations
1
54
173 189
201
281
Index vii
INTERMEDIATE ALGEBRA
Chapter One
ALGEBRA AS A LANGUAGE AND A TOOL
The
1.1.
relation between algebra
and English
As
applied, algebra is a method of solving problems by the use of numbers, letters, and symbols. As written and as read, algebra is a brief and useful
international language. One must learn to read it, at least in part, before the calculations can be started. Fortunately for
the beginning student,
familiar ones of arithmetic.
many The
symbols are the may be learned as
of its
others
they occur in the text. For reference they are grouped together in Table 1. This language of algebra is usually much briefer than English. For example, the English phrase, 'the number which is 3 less than 4 times the number x' becomes, in algebra,
simply '4x ber Xj plus
3.' Again, the statement that 'twice a 5, equals one-third of the sum of x and written in algebra as the equation
2x
(1)
When
certain
ner, this fact
is
+
5
=
num4' is
i.
numbers are always related in a given manexpressed conveniently in an algebraic for-
mula. For instance, the distance, d, traveled by a body moving at uniform speed equals the product of its rate, r,
and the time, product of
its
t.
Or the
area, -A, of a rectangle
length, L,
and
its
width,
to the following formulas (2) (3)
= rt; A = LW. d
1
is
equal to the
W. These
facts lead
ALGEBRA AS A LANGUAGE AND A TOOL
2
[Ch. I
The problems in Exercise 1 test the student's ability to translate English phrases or sentences into algebraic language.
EXERCISE
1
// x represents any number which we wish to think about, express in terms of x the numbers described in problems 1-6.
4
1.
The double
2.
The number which
3.
Two more
4.
One
5.
One-half of the
6.
Three times the product of x by
7.
A man
How
less
of x. is
4 less than
than one-half of
than 3 times
x.
x.
x.
number which
is
2 less than its
x.
double.
walks toward a point 30 miles away at 3 m.p.h. he far has traveled, and how far has he to go, after x hours? apples and oranges sell for 3 and 4^f each, respectively, the selling price of x apples and y oranges?
8. If
what
is
9. A and B, 20 miles apart, walk toward each other. After A has walked x miles and B has walked y miles, how far apart arc
they? 10.
Work problem
9
if
A and B
go
in the
same
direction, with
A
behind B. Write as formulas the facts stated in problems 11-14. 11.
base, 12.
The area, P, of a parallelogram B, by its altitude, H. The
base, B, by
equals the product of
its
area, T, of a triangle equals one-half the product of its its altitude,
13.
The
14.
Simple
H.
circumference, C, of a circle equals twice the product of its radius, R, by the constant w (pronounced 'pie'), which is nearly, though not exactly, equal to 3.1416.
the rate,
r,
interest, /, is the
by the
time,
t.
product of the principal, P, by
NUMBERS
1.3]
Change 15.
x and
3
the statements in
Twice the number
problems 15-18
x,
plus
5,
to
equations in algebra.
equals one-third of the
sum
of
4.
16.
Three more than twice x
17.
Half the
sum
of x
is
2 less than 3 times
and twice y
twice the
is
x.
sum
of y
and
thrice x. 18.
the
Two
less
than twice the
sum
number which exceeds x by
of x
and
1 is
3 more than half
4.
Represent in terms of x the pairs of numbers described in problems 19-23. 19.
Their
sum
is 10.
20. Their difference
(Answer: x and 10
x.)
is 5.
21.
One
is
one-third of the other.
22.
One
is
3 more than twice the other.
23.
One
is
2
less
than two-thirds of the other.
24. A boy with x dimes has 3 times as Find the value in cents of what he has.
man is x years old now, old will he be in 10 years?
25. If a
How 1.2.
The
how
relation between algebra
many
old
nickels as dimes.
was he 5 years ago?
and arithmetic
We have seen that algebra is a method of solving problems by the use of letters which stand for numbers. Arithmetic, on the other hand, is the science of computing with particular numbers. The result of an algebraic problem often can be applied to many cases simply by assigning different values to the letters, whereas an arithmetic result applies only to
the one problem concerned.
1.3.
Numbers
The concept
of
what
is
meant by a number develops and
changes as one studies mathematics.
We shall here give illus-
ALGEBRA AS A LANGUAGE AND A TOOL
4
trations of certain types of of numbers in general.
numbers rather than a
[Ch. I
definition
Most
easily understood are the whole numbers, or integers, such as 1, 2, 3, etc. Then there are fractions, such as %, f, f, etc.
Both
integers
and
fractions
may
be
positive, like those
Negative numbers that the temis 20 below or that a river perature (twenty degrees zero) 2 feet, meaning that its surface is 2 feet below level is normal. Numbers are either real, like those mentioned above, or imaginary. The second group will be discussed later. All real numbers may be represented conveniently as points on a line, as shown (in part) in Fig. 1. above, or negative, as 11, 3, are useful in various situations, as
f, etc.
when we say
Negative numbers
Positive
Fig.
numbers
1
This figure also calls attention to the very special number zero, written 0, which is between the positive and negative numbers and may be classed with either group. That is,
=+0 =0.
This fact makes necessary special rules for computation with zero. (Art. 1.6.)
1.4.
Absolute and numerical values
The symbol a
read 'the absolute value of a,' or a , numread 'the absolute value of a,' means the positive ber equal to +a if a is positive, or to a, if a is negative.
,
Thus, 3 = - 3 = 3. Consider two numbers a and 6. If a has a larger absolute value than 6, it is said to be larger than 6 numerically. If a comes to the right of 6 in Fig. 1, it is larger than b algebraically. Or, to state it another way, a exceeds 6 algebraically 3 both b is positive. Thus, 10 is larger than whenever a
:
POSITIVE
1.5]
AND NEGATIVE NUMBERS
5
3 algebrainumerically and algebraically; while 2 exceeds cally but not numerically. The symbols > and < mean ' respectively greater than' and 'less than' in the algebraic sense. Thus, > 2 (zero is greater than minus two) and 5 < 2 (minus five is less than two).
Operations with positive and negative numbers
1.5.
When
negative numbers are taken into account the ordinary operations of arithmetic must be explained anew. The rules for addition and subtraction are as follows :
(1) To add two numbers with like signs (both positive or both negative) add their absolute values and prefix their
common
sign.
To add two numbers with
unlike signs, subtract the smaller absolute value from the larger one and prefix the (2)
sign of the numerically larger number. If the
numerically equal, their (3)
To
sum
numbers are
is zero.
number from another, change number subtracted and then add.
subtract one
sign of the
These operations are examples below.
illustrated in
column form
the
in the
Addition (a)
7-33-7
2
(b)
A
2
(c)
-5
-2
(d)
-2 -5
(h)
-2 ~5
5
Subtraction
2
(e)
(f)
2
-5
5
-3
-2
7-7 (g)
5
3
operations above may be written on single lines by observing the rule that when parentheses are removed the signs enclosed are changed if and only if the sign before the
The
parentheses
is
minus. Thus
(+6)
=
6;
(
6)
= +6
(or
ALGEBRA AS A LANGUAGE AND A TOOL
6
just 6);
and
+(+6) =
(h), for
and +(-6) example, become (-2) (-2)
(d) (h)
6;
+ -
(-5) = -2 (-5) =-2
For multiplication and division numbers a single rule suffices
=-6.
-5 +5
[Ch.
I,
Accordingly, (d)
= -7; = 3.
of positive
and negative
:
The
(4)
product,
with like signs it is
and
two two numbers with unlike signs
also the quotient, of
is positive; of
negative.
-20; and (-4)(+5) =-20. Also, 10 -5- 2 = 5; (-1C) -f(-2) = 5; 10 -f- (- 2) = -5; and (-10) -h 2 = -5. As shown in the above example, multiplication of positi^ numbers may be indicated by dots instead of parentheses, provided that the dots are raised to avoid confusing them with decimal points. In the case of letters representing numbers, the omission of a sign between them indices multiplication, so that, for example, ob means 'a times 6.' This practice, of course, would not do for numbers themselves.
Why? Division
symbol 1.6.
-*-
.
is
indicated
by use
For example, 10
~
of a horizontal bar, or the
5
=
^=
2.
Operations with zero
The
rules for operations with zero are illustrated in equations (1) to (5) below.
+ 6 = 6 + 0 = -0 + 5 5 -
(1) (2)
0-0
=
The value to
it
(3)
6;
0-7= -7 +
=
5;
-3 -
= -7; + = 0. = -0 - 3 = -3;
0.
of a
number
or subtracted
from
is
unchanged when zero
is
it.
0-4 = 4-0 = 0- (-4) = (-4) -0 =
0.,
added
OPERATIONS WITH ZERO
?.6]
The product
of
any number and zero
is zero.
(4)
The
number
quotient obtained by dividing zero by any other than zero is zero. -
5)
That
is, it is
To
impossible
*
6?'
sum
to divide
any number by
zero.
see that the statements in (5) are reasonable, notice
that the value of f, or add up to 6.' Similarly, tv>
- are not sokable.
and
of
definite
-
'the number of 2
'how many
asks
;
s
which will add up
zeros will
Evidently no number meets this condition, since the
any number
but this
3, is
of zeros
number), then
is
times that
impossible, as seen
is
EXERCISE
2
zero. Similarly,
if
$ = (some
number should equal
from equation
6;,
(3).
(ORAL)
Arrange the following numbers so that, as read from left to 1, 15, 0, 6, f, 4, 5, right, they increase algebraically: 5, -2. 1.
Hb
2. Arrange the numbers of problem 1 so that, as read from left to right, they increase numerically, or in absolute values.
Perform
the additions
and subtractions as indicated in problems
3-14. 3. 6.
9.
12.
3
+
-15 6
-
+
(-7).
+
(+5).
4. 7.
11.
10.
(-17).
13.
(-6)
+
3.
-9 - (-2). - (8). (13) - (-32).
5. 8.
11. 14.
-8 +
(-5).
8 - (-4). -7 - 0. -15 - (-0).
* It might be thought that by this test the value of must be one, since 'one up to zero'; but so do two zeros, three zeros, etc. The rule stated in (5)
zero adds
holds in
all cases.
ALGEBRA AS A LANGUAGE AND A TOOL
8
Perform
[Ch. I
the additions as indicated in problems 15-26.
27-38. Subtract the lower from the upper numbers in problems 15-26.
39-50. Multiply the numbers in problems 15-26.
Perform
the indicated multiplications in problems 51-62.
51. 6(-7).
52. (-15)(0).
53. 0(100).
54.
9(-4)(-5).
55. (-0)(28).
56. (-5)(6).
57.
+7(-3).
58.
-(-!)(-!).
59. 4(-l).
60.
-(2)(-3).
61.
-0(-15).
62.
Express as integers 63. 20 *
4.
64.
(+7)(-2).
the quotients indicated in problems 63-71.
(-21)
-f-
(7).
65. 14
-r-
(-2).
66.
Which of the quotients indicated in problems 72-87 Find the values of these numbers.
bers?
(-20)
are true
+
2.
num-
ADDITION AND SUBTRACTION OF POLYNOMIALS
1.8]
1.7.
A
9
Algebraic expressions single
number
or letter, or a group of them,
is
called
an
algebraic expression.
An
2 - x o
1
Examples. 4; x 3x; 2x
1; -;
x
expression with no plus or minus or equality signs be-
tween
its
parts
is
called a term.
2
Examples. 7;
The term
2; a; xy] xyz]
2a6 3 stands for
'
x
2a6 3
;
.
2 times a times 6 3 ,' where
a and 6 are literal numbers, or U ' and 6 3 (read 6 cubed or 'the numbers, third power of 6') means 666, or 6 taken 3 fo'raes as a factor. Here 6 is the base and 3 is the exponent of 6. Thus, if a = 3 and 6 = 2, then -2a6 3 = -2 3 2 2 2 = -48. Terms differing only in numerical coefficients, such as 3# 2 3as 3 are called like terms. An 2x 2 or as 3 4as 3 and and expression of one term is called a monomial; of two terms, a binomial; of three terms, a trinomial; and of two or more 2
is
the numerical
letters
coefficient,
standing for
,
,
,
,
terms, a polynomial. The terms of a polynomial are connected by plus or minus signs, as in the example:
2a
1.8.
+
6.
Addition and subtraction of polynomials
convenient to state in the form of laws certain reasonable assumptions about numbers reprebut unproved sented by letters. Such assumptions are called postulates.
Here
it is
The commutative law of addition: The sum of two more algebraic terms is not changed by changing their
(1)
or
order.
Thus, just as 3 ter
+4=
4
+
3,
so x
+y=
what number values we assign to x and
y
y.
+ x,
no mat-
ALGEBRA AS A LANGUAGE AND A TOOL
10
The number of
The sum
associative law of addition:
(2)
algebraic terms
is
the same, however
[Ch. I
of
any
we group
them.
That
just as
is,
The
(3)
+y+
x
in general,
4+9+5 = (4+9)+5 = 4+(9+5) = 18, z
=
+
(x
distributive law:
y)
+
z
The sum
term by each of several other terms term by the sum of the others.
+
+
=
= x
+
(y
so,
+ z).
of the products of one is the product of this
+ +
c The student a(b d). should verify this equation for various special values of the = 2, b = 3, etc. Once granted, it enables letters, such as a
That
db
is,
ac
ad
+
us to combine the like terms of a polynomial. Thus, 2x 3 5x 3 3 2 3 2 2 2 = x (2+5) = 7x and ax +7ax -2ax = ax (l+7-2) = 6ax 2 .
,
We
are
now
(x
=
=
add or subtract polynomials by
For example, 2x + x + 7) + (3x + 4x 3 - 5 + 6x 2 ) 3 2x 2 + x + 7 + 3x + 4x 3 - 5 + 6x 2 ) (x like terms.
combining 3
in position to
2
+
3
(x
4x
-
3
2z
2
+
6z
2
(by the associative law) t7 - 5) 3z
+x+
+
(by the commutative law)
=
(x
+
=
5x 3
+ 4x + 4x +
3
4x 3 )
+
(-2z
2
+
2
6x 2 )
+
+
+
-
3x) (7 5) (by the associative law) (by the distributive law)
2.
(x
In subtracting one polynomial from another we note that -(a+6-c) = (-l)( a +6-c) = (-l) a +(-l)&+(-l)(-c)
= a 6 c (by the rule of signs (by the distributive law) in multiplication). This indicates that when parentheses preceded by a minus sign are removed, the signs of all terms which
+
had been inside are changed. Thus, 3
(x
-
+ x + 7) - (3x + 4x - 5 + 6x - 2x + x + 7 - 3x - 4x + 5 - 6x
)
=x = -3x 3
is
3
2
3
The work
2
3
2x 2
shown
8x in
2
- 2x+
2
12.
column form below. For convenience
the terms are arranged in the order of descending powers of
x.
REMOVAL AND INTRODUCTION OF PARENTHESES
1.9]
Addition
z
3
-
2x*
+
x 3o:
11
Subtraction
+ -
3
x 4x 3
7
5
-
+ -
2z2 6z 2 Sx 2
+ x+ + 3o: - 2x +
7
5 12
Removal and introduction of symbols of aggregation
1.9.
We
have seen that the signs of terms within parentheses preceded by a minus sign were changed when the parentheses were removed, but were not changed if the preceding sign was plus. Like statements are true of the other symbols of aggregation (Table 1). This enables us to simplify expressions containing such symbols (parentheses, brackets, and braces) by removing them one pair at a time, beginning with the innermost ones. The procedure is then continued on the
new
expressions formed until
removed and
like
Example 2x
such symbols have been
all
terms collected.
- 5(s - 1)]} = 2x+ {3x - [2x - 5z + 5]} = 2x + {3x - 2x + 5x - 5} - 2x + 3x - 2x + 5x - 5 = Sx - 5.
+ [3x~
[2x
When parentheses, or other symbols of aggregation, are introduced into an expression, the signs of the terms within are not changed if the preceding sign is plus, but are changed if the preceding sign is minus. This statement may be verified
by removing the symbols
in accordance with the pre-
ceding rules.
Example 3z
1.
+ 2y + z - w =
Example 3x
(3x
+
2y)
+
(z
-
w)
2.
-
2y
-
z
=
3x
+
(~2y
-
2)
=
to
-
(2y
+ z).
ALGEBRA AS A LANGUAGE AND A TOOL
12
Example 8. 2x - y - (a
+ 6) =
2x
+
= 2s-
-
[-
+
]/
(a
+ 6)]
+
(a
[Ch. I
V)]. f
important to notice that the
It is
same terms can be en-
closed within parentheses preceded
minus
The above examples
sign.
by
either a plus or a
how
illustrate
the last two
terms in each are enclosed with either sign preceding.
EXERCISE
3
terms involving x
and
1.
Write three
2.
Write two examples of monomials; of binomials; of
like
9
y,
z.
tri-
nomials.
Add 3. 4. 5.
6. 7. 8.
9.
10. 11.
12.
the polynomials in problems 3-12.
4x
-
2x
7;
+ 3.
+ 3;4 - 7rr. 2ax* + 3z - 2; ax - 5x + 3. 3az - 2x + 4; 2az - 3x + 5. x - 5z + ax - 2; 3z + 2z - 3ax + 1. x* - 2z - 3az + 1; 2z + 2x - 3az + 1. x - Sx4 + 3z - 2a; 2s - 2z - 1 + 3x. 4z + 3z - x4 + 3; 2z + z + 4a - 3x 2;z - + 7; 3x + 4 - 2. - 2; a - 3z + x - 1; 6 + x 2z
2
2
2
3
2
2
3
2
2
3
2
4
2
4
3
2
.
3
2
a:
2
2
4
.
13-16. Verify the additions in problems 3-6
replaced
by
1
and
by 2 and
are
when x and a
are
respectively.
17-20. Verify the additions in problems 3-6
replaced
when x and a
3 respectively.
21. If the arithmetic results obtained in problems 13-20 do not expose any errors in the algebraic sums obtained in problems 3-6, does this prove that these sums are correct?
22-29. Subtract the second from the
problems 3-10. 30-37. Subtract the of problems 3-10.
first
first
polynomial in each
from the second polynomial
of
in each
MULTIPLICATION AND DIVISION OF POLYNOMIALS
1.10]
In problems 38-47 remove 38.
2a - (b + 2 - 3(s -
-
3a
the symbols of aggregation
39.
d).
- 4z + 1). (2y + 2(a - 26) - 3(2a -
46. 47.
-
42.
44. 45.
2
6).
1.
c]
43.
simplify.
-
3x
41. 4 y) + 2(x + y). 2a + [3a + (26 + 1) 3c + [26 3. (a c) + 4] - 2}. 2 [z (2z {3z 1) + 2] 3a - (26 - [3a - 2(26 - 1)] + 3}. - {2z - [(3ax + 36)x + x
40.
and
13
2
}}.
Insert parentheses preceded by a minus sign which will include all but the first term in each of problems 48-51. 48. 2x 50. a
52.
-
3a
+
+ 26 - c + 2. To
sum
the
- 3y. 51. 3a + 46 - c - 1. + 3a and -6a - 26 + 3c -
49. 4x
6.
-
of 6
2c
+ 4c from 3a 6 from 5x 2z + 7x
result of subtracting 6
53. Subtract
-3x
3
54.
+
4x
2
3
To what -3?
55.
3
add the
2c.
2
2 and add the result to
1.
Take 2a2
mainder to 4a 1?
3x 2
2a
-
+
4a6 from
3a
2a6
expression
1.10. Multiplication
-5a 2
+
+3
6a6
and add the
re-
2 .
nfcist
2a2
3a6
+
6 be added to give 0?
and division of polynomials
In addition to the distributive law already discussed, two other useful laws about multiplication may be mentioned here.
The commutative law of multiplication: The product of two or more factors is not changed by changing their order. (1)
That (2)
three
is,
just as 2
3
=
3
2,
so ab
=
ba in
all cases.
The associative law of multiplication: The product or more factors is not changed by grouping them
of
in
different ways.
Since (2 3)4 = 2(3-4), for example, so, in general, assume that (db}c = a(6c).
we
ALGEBRA AS A LANGUAGE AND A TOOL
14
Two
[Ch.
I
of the so-called laws of exponents, which are discussed 7, will be needed here.
at length in Chapter
2 3 Examples. 2 2
=
25
dm (4)
a
Examples.
is
= am+ n = x aaV = a'aV =
a ma n
(3)
=
^ o
;
=
n
35
~2
.
xW
;
~n
when
33
~
am
=
11
=
m> a7
;
.
$~
w
2
4
'
Chapter
1.
(2az
3
35
=
(3aV)
)
3aa 2z3a; 4 [by
2
= QaW 8
,
Example
a 10a x
2.
4
-
=
/10/a
8
[by
by making
(1)]
(3)].
4
/a: ru
(^(-) 4
[by
/CN1 (5)]
1
[by (4)]
=
5a 5z 3
.
The multiplication of two polynomials, such as is in the expression
3
(x
-
3,
3
5-7
Algebraic terms may be multiplied or divided use of the five laws above.
Exampk
.
n.
Finally, law (5) below, treated more fully in used in the division of polynomials.
Example. (f
a 10
2x4
+ + x
3x2
-
l)(x
-
indicated 3a;
3
+
2),
can be accomplished conveniently by arranging the polynomials in terms of descending powers of x and then employing law (3), as in the following example.
_ -2z4
+ x* +
3x2 3xs
+x- 1 +x+2
(first
factor)
(second factor)
6x 7
a; 4
6z
T
-
3z
6
-
llz
5
-
4s 6z 4
+ 2x + + 8z 3
3
-f-
2
6x2 7x2
- -
2x
+
a;
-
2 2 (product).
1.10]
MULTIPLICATION AND DIVISION OF POLYNOMIALS
A An
,.
.
,.
,
,
.
,
.
indicated division, such as
(5Z
-
3
5x 4
+
+
+
5)L y
6x
(x~
X
15
2)
be carried through as here shown. The explanation of the steps may be found just below the actual division.
may
x*-2x 2 + x-1 (Quotient*) 1 (divisor) x
3x
2 x
r
5o:
4
+5x
+ z+ 5
3
(dividend)
x+5
-
x 2 +3:r+2 3 (remainder)
Explanation. The first step is to arrange the dividend and divisor in the order of the descending powers of the same letter. Here the dividend, divisor, and Quotient, so arranged
5z 4 + with respect to the powers of x, are respectively x 5 3 2 3 2x* x 1. The first 3z - 2, and x 5r + re + 5, z 3 term (x ) of the Quotient is obtained by dividing the first
+
term
5
(#
)
of the dividend
by the
first
term
2
(x
)
of the divisor.
then multiplied by the divisor, and the product is subtracted from the dividend. The remainder thus 2x 4 + 7x 3 + x + 5. To obtained forms the new dividend, obtain the second term, ( 2x 2 ), of the Quotient, the pro2 2x 4 cedure above is repeated, with x being divided into
The
3
result, (x
),
is
.
Succeeding terms of the Quotient are thus obtained until either a zero remainder is reached or one in which the ex-
ponent of the based (in this
letter
case, x) is less
The
in the divisor.
X5
_
5^4
upon which the
^_
^
_
2
than the largest exponent shows that
it
is
has
final result
+ 5X + x + g 3
original arrangement
_
x
_
2x
+
x
3 l+ x2 _ 3x _2
The word 'quotient' is often used with two different meanings. We shall use with a small 'q' to mean the total result of an indicated division. Thus the 'quotient' of 27 by 5 is 5f, while the 'Quotient' is 5. *
it
ALGEBRA AS A LANGUAGE AND A TOOL
16
~ = Quotient
, dividend In general, -37-:
,.
T
divisor
The
result of
,
.
[Ch. I
remainder rr. divisor
H
a long division should always be stated in the
form indicated above.
EXERCISE Perform
4 the multiplications or divisions as indicated in prob-
lems 1-7. 1.
2
3
5z 3 (-36z).
2.
(3z a)(4a z).
6a 7& 6c 3
12oV
3
-6xy*z
-3a26c' 8. Verify that
2 3 (-4z)(2az)(-3a z ).
3.
2
2
'
3
[(a;+l)(a:-l)](2x+l)= (x+l)[(x-l)(2x+l)].
Multiply the pairs of polynomials in problems 9-21. 9.
(6z
7
10. (3z 5 11. (3z
13. 14.
15. 16. 17. 18. 19.
20. 21.
2
4
(3z).
- 9z + 3); (3z - 1). + 5z + 5z - 2); (z - 3z + 2). (2Z - 10z - 5z + 2); (2z + 3z - 1). (6z + 7z - 5z + 5z + 4z - 2); (z - 3z + 1). (z + x - 5z + 4z + 4z + 5z - 2); (-1 + 2z + (6z - z - 4z + 6Z ); (z + 3z - 1). (4 + 2z - 3 + 2z); (z - 1 + 2z). (3z + 8z - 12z + 4z ); (3z + 2z - 2). (8 - 6z - llz + 3z ); (3z - 1 + x (13z - 5z + 4z ); (1 + 4z - 3z). (5z + 1 (-z + 6z - 3 + Hz ); (x - 2 + 3z 4
12.
4-H* -z-5);(2z). - llz + 7z2 - 5z + 2);
4
-
7z
2z2
3
4
5
3
2
2
3
2
3
4
2
3
2
3
2
4
3
8
2
2z2 ).
2
4
2
2
3
2
2
4
3
2
).
2
3
2
2
3
2
).
22-34. Divide the
first
problems 9-21. 35. Divide 3z 5 3
36. Divide
a
37. Divide
a3
polynomial by the second one in each of
+ 2z - 3z + + 6 by a + b.
38. Divide 27az
2
8
1
2 by 2z
-
3z
+
1.
3
63 2
-
by a 25o
3
b.
+ 20a z 2
18z 3 by 6z
-
5a.
REVIEW EXERCISES
A =
//
3z
-
2
17
+ 4,
2x
B=
3z
^nd
the values of the quantities in
39.
2A
-
35.
40.
2
-
x
-
1,
and C
=
-3a:
+
1,
problems 39-41.
EG + AB.
41. (A
+ 25)
-:-
C.
REVIEW EXERCISES // x represents a number, write the numbers indicated in problems 1-7. 1.
Four times the number.
2.
Three-Fourths of
3.
20%
of
4.
The
difference
5. 6.
Eight more than the number. Nine less than three times the number.
7.
The
it.
it.
between the number and
(Two
7.
answers.)
excess of 5 over 2.5 times the number. 9
// x and y represent the tens' digit and the units digit respectively in a number of two digits, write the numbers indicated in problems 8-11. 8.
The number.
9.
The number when the
digits are reversed.
10.
The sum
11.
Twice the number, increased by
of the digits divided
by
3.
36.
// x represents the width in yards of a rectangular field, express in terms of x the length, perimeter, and area of the field under the conditions stated in problems 12 and 13. 12. 13.
//
The length The length
A
is
four times the width.
is
nine yards
less
walks x feet per second and
than twice the width.
B
walks two feet a second faster,
write the quantities described in problems 14-16.
A B
14.
The
distance
15.
The
distance
16.
The
distance traveled
travels in ten minutes. travels in half
by both
an hour. in
one hour.
ALGEBRA AS A LANGUAGE AND A TOOL
18
[Ch. I
Write as formulas the statements in problems 17-26. 17.
The volume,
18.
The
V, of a sphere and the cube of the radius, R. area, A, of a trapezoid
by
altitude, h,
The
19.
one-half the
sum
equal to the product of
is
-J,
TT,
equal to the product of the
is
a and
of the parallel sides,
6.
a telegram of 50 words is comcents for each of the first ten words and
cost, T, in dollars for
puted at the rate of m x cents for each of the others.
C
20.
At p
dollars each,
21.
D is
the difference between twice x and
22.
A man's
2$.
On
x% (x% =
$5000 at
dollars.
(Two
y.
formulas.)
x now and was y ten years ago.
is
age
x books would cost
the yearly interest
y^n)
is
I dollars.
The yearly interest on R dollars at 7% is 7 dollars. 25. The cost in cents of s pounds of nuts at 50jf per pound and y pounds of candy at 75^ per pound is C. 26. The distance in feet traveled in x hours at the rate of five 24.
miles per hour
is s.
Evaluate the numerical expressions in problems 27-34. 27.
-
30. [(-1)
32
-
(~3)][4
+
-20
-(-3)
28.
(-4) (-2) (3).
39.
20
-
20
20
+
20*
+
If a
+
3,
b
=
2,
38. 40.
and c
=
1,
-8 -
-3 2
.
0.
34
36.
(?)
=
-20
20
+
*
missing number in each of problems 35
= -H. (-2) X (?) - 16. - 8 = -10. (?)
35. 7
37.
31.
(-2)].
0(-20) the
29.
.
'
'
Find
2
20
-
20 20'
to 40.
= 10. -25 + (?) = 5. 3 X (?) = -15. (?)
find the values of the expressions
in problems 41-46. 41.
-3a2
44. (a
+
+
2b
b)(b
-
a c.
42.
+J* +
43. 3a 3c
-
Zoc
+ c).
45.
a
+
6(6
+
~~ a c).
46.
7
5.
REVIEW EXERCISES Add 47.
19
the expressions in each of problems 47-51.
-362
48.
3y
262
-2y
-7b2 50.
4x2
2
49.
2
+ 6) + 6) 7(o + 6)
-3(a -2(o
-y2
-
+
2
y
2x
51.
z
-2?/ -32
-3x
+ 2z
x
2
3x2
-
y + 3 + 3*/-l + y-2
- 3a6 + 6c; 4a& - 2a - 3c; -2c - 10 ab + 7mn - 3n2 + 6; 5n2 - 7mn - 2; 6 - ran.
52.
Add: 4a2
53.
Add:
2
6a2
.
Subtract the lower from the upper expression.
4a
-
76
-2a
+
36
54.
-
55.
c
To
y
sum
the
of
+
2z from 3x
-4x
2
+ 3xy -
2 over 2x
the excess of 4xy
+ -
2c
56. Subtract 4x 57.
3x 2
2
Sxy
+
y
2 and
+
4xy 5xy
-
y
+
2y
Remove 59. 60.
61.
62.
-5xy -
65. x
+ 8x
2
add
6.
7c2
2d and
symbols of aggregation and
the
collect like terms.
- 1 - (3x - 1)]. 4 [2x - [-z + (2y - x)]. -[(2x - 3i/ + y] -3(o - 26) + 6{ -4 - [6 - 2(4 + 6)]}. - [(2x - y - 4) - (3y - 2)]. (3x + y + 6) -
2
2
2
)
-3x2 2
7
2
Insert parentheses preceded by a but the first two terms. 63.
2
4.
58. How much must be added to the sum of 3a& + -6 + 3c to give -5a6 - c ? 2
2
+
y
2xy
Carry out
-
2x
+ 3t/
2
+z+ 4.
3w.
minus sign which
- b2 + 4a - 2a 6 -5 + y - x - 3 4
64. a 2
66.
will include all
2
2 .
2
.
the indicated multiplications.
67. (-3x)(2y). 2
2
68.
2 (-3xV)(-zy )(7xy).
3x2 (2xy
-
2 ?/ ).
69.
-(-3mn )(-5m n).
70.
71.
-5mn(2m2n - 3m
72. (3x 2 -i/+6x)(-x+y-x
+ 2n).
2
).
ALGEBRA AS A LANGUAGE AND A TOOL
20 73.
75.
77. 79.
81.
-
4a a
26
74.
+ 7b -
2x2
2xy 3x
-
2
-y (x + 4)(x + 3). - 3). (x + 7)(x -3(a - l)(a +
83. (3a
-
2)(2a
-
3x 78. (x
80. 82.
1).
5).
84.
-
86.
85.
-3(2
87.
(-3x + y)(y + 3x). 2 2 -5(3x - y)(3x + y).
89.
91. (3
-
-
2x)(3
+ 2)
93.
-3(x
95.
-(5x -
Note.
x)(3x
88.
2
+xy - 2xy + 2
-
2y y
-
5)(x
+
-2(4
94. (lOx 2
by
y
2).
-(x + 2)(x - 3). - 4). (2a + 3)(a -2(5a l)(3a + 1). - y). (2x + 2/)(2x -(2x - y)(2x + y).
92.
2x).
.
4j/)
2
90. (2x 2
2
When
carried out
1).
+y
x3
76.
y
-
x2 x
[Ch. I
7)(2x
-
+
2
x)(4
3)
+ 7). -
x).
2 .
96. (4x
.
the multiplications in problems 77-96 have been the methods given in this chapter, the student
should study Arts. 2.2 and 2.3 and then do the problems orally. the indicated divisions.
Perform 97.
=&.
98.
99. 18a 2 6 4 H- 2a6 2
101. (6x
s
102. (9x
2
103. (3x
2
-
100.
.
+ x + 3) -r 24xy + 16?/ 5x + 2x - 1) 2x
2
2
)
3
-r-
2
(-12x y
- 1). (2x - 4y). (3x
-f-
(3x
+
1).
3
2)
-5-
2
(-4X ?/
2 ).
Chapter
Two
TYPE PRODUCTS AND FACTORING
2.1.
The factoring rule
In arithmetic, since 12 = 2-2-3 = 4-3 = 6-2=l-12, we say that the positive integral factors of 12 are 1, 2, 3, 4, 6, and 12. This means that if we divide the integer 12 by any
one of these integers, we get another integer. Of these factors only 2 and 3 are prime, or without positive integral factors other than themselves and 1. In algebra it is possible to make use of a similar factoring rule. To explain it, we shall need some definitions. A polynomial which is the sum of terms such as 2x 2t/3 in which all exponents are positive integers, is said to be integral and rational.* In this chapter we shall deal only with integral ,
rational polynomials, including in this general description 3 not only expressions with two or more terms, such as 2x
+
5xy + 4, but even single terms such as 3x and 5. When a polynomial is expressed as the product of two or more integral rational polynomials, it is said to be factored, and the parts which are multiplied together are the factors. 2 1 = (x Thus, when we write z l)(x + 1), and 3x + 6 =
and 3x 2
3(x
x
+ 1,
except
2),
x
we
+
itself
factor the left sides.
1, 3,
and
Each
of the four factors
+ 2 is prime, since it has no factors In this chapter we shall bar any factoring
and x 1.
which introduces coefficients that are fractions or types of numbers not yet discussed in our text. Thus, while it is correct * The reason for the customary use of these adjectives at the end of Art. 7.5.
21
is
suggested in the note
TYPE PRODUCTS AND FACTORING
22
to writes
+2=
+
2(%x
1),
andz 2 -2 =
where (V2) =
-/2)(z 2 and # 2 2
+
2
(x
[Ch. II
+ V2),
will be 2, the expressions z considered here as prime, or non-factorable. When we completely factor a polynomial, we express it as x = the product of its prime factors. For example, x 3
not completely factored in the second form be1 is not prime.
x(x*
1) is
cause x
2
2.2.
Useful identities
Certain operations in algebra are repeated so often that much time is saved by memorizing them in type forms instead of working them out on each occasion. This is especially true of the following results, which may be considered as multiplications as they stand or as factorizations as read from right to left.
a(b
(1)
+ c) = + b) 2
(a
(2)
(3)
(a
(4)
(a
-
-
+
b)(a
2
b)
b)
= =
ab a2 a
2
a
2
+ ac. + 2ab + - 2ab + -
b
b2
.
b
2 .
The student should check these results by multiplication, and should verify them for various special values of the = 4, 6 = 2, etc. They are examples of letters, such as a algebraic identities.
An
identity is
an equation which
values of the letters involved.
makes each
A
is
true for
value
is
side of the equation a definite
all
permissible
permissible
if
it
number. In the
identity
2x
x the values
1
and
2
1
1
x
1
for
-
1
x
1
+
1
x are not permissible.
Why?
Also useful from the standpoint of factoring are the identities
(5)
(6)
:
a3 a3
-
+
b3 b3
= =
(a
-
(a
+
6) (a
b)(a
2 2
+ -
ab ab
+ +
b 2 ). b 2 ).
2.2]
USEFUL IDENTITIES
Illustrative
A. 1.
23
Examples
Type Products
3(2z
-
2
3x
2.
(2x
+
3t/)
3.
(3z
2)
4.
(5z
-
2
2
+ =
=
4)(5x
= =
1)
2
(2z)
2
(3z)
+
4)
-
2
3(2z ) 3(3x) 6z2 - Qx 3.
+
3-1
=
9z 2
+
+
-
2(2*)
+
2(3z)(2)
-
2
(5x)
42
22
=
25x 2
-
-
12*
+
4.
16.
B. Examples in Factoring 5.
6. 7.
- Qax + 2a = 2a(o; - 3z + 1). 4z - 9 = (2x) 2 - 3 = (2x - 3)(2x + 3). 4x + 4az + a - 2 + 2^/6 - 6 - 2y6 + & = (4z + 4ax + a = (2a; + a) - (y - 6) = [(2a; + a) - (y - 6)][(2s + a) + (y = (2x + a - + 6)(2a; + o + y - 6). 8z - 27 = (2o;) 3 - 3 = (2x - Z)[(2x)* + (2s) (3) + 3 = (2x - 3)(4x + 60; + 9). 2az 2
2
2
2
2
2
2
j/
2
2
2
)
2
(t/
2
)
2
6)]
T/
8.
3
3
2
]
2
9.
27y
10. 9a; 11.
2
4a 2
+
EXERCISE 1.
= = =
+I - (3y)(l) + (3y + l)[(3t/) + l)(9t/ 3y + 1). + 24:n/ + 16^ = (3x + - 20a + 25 = (2o - 5)
3
1
3
3
(Si/)
2
I 2]
2
(3t/
2
2
4j/)
.
2
.
5
State each of the identities
(1), (2), (3)
and
(4), Art. 2.2, in
English.
Write a trinomial involving x which integral in x, (6), not rational and integral in 2.
is
x.
(a), rational
and
TYPE PRODUCTS AND FACTORING
24
By
use of the proper type products perform (orally
when
[Ch. II
possible}
the multiplications indicated in problems 3-29.
- a - 1). 4a 6(6 - ab). 6a(-2x - a).
3.
5x(2x
5.
2
7.
9. (x
11.
3
+
-(3xy 2
13. (2a
-
2
17. (3a c
+ 2)
2
2L
2
b)
14.
.
+ x)
2
16.
.
2
)
2
18.
.
2
aX?
(?
-2x
2xy).
-
(3x
+g + 4) -2(3m +
3p(p
2
2
-
2
).
-
-i/
-
(3x
-(2x
3x)
2n)
2 .
2 .
2 2 .
)
+ 2y). )(2z + y
2y)(3x
3
-
3 2/
+ 0) '
22< [(X
a)
.
(4m
20.
4j/ ).
2).
2
2
(2
+b+
2ax
2
2
12.
.
+ 4y )(2x ' +
19. (2x
6.
2
-
10. (2x 2
.
-2(-3y
15.
-3y(y
8.
2
5)
4.
- 2 - 3t/)(3x - 2 + HINT: 3x - 2 - 3y = [(3x - 2) - 3y]. 24. [(x + y) - (2r - s)][(x + y) + (2r -
3
3
).
2][(X
23. (3x
+ y)(x* -xy + y*). - 3ab + 6 (3a + 6)(9a - Qmp + 9m p (2p + 3mp)(4p -(7/-a)(2/ + ar/ + a -3(2a - 6)(4a + lab + 6 Prove the identity (a + 6 + c) =
)].
25. (x 26. 27. 28. 29.
30.
2ac
+
2
2
).
4
2
2
3
2
).
2
2
).
2
2
).
2
26c both
member
in the
by form
+ b) +
31. State the identity in
+ b + c + 2ab + 2
2
and by writing the
direct multiplication [(a
a2
left
2
c]
.
problem 30
a suitable form for oral
in
problems.
Use
the statement obtained in
problem 31
to
find the squares indi-
cated in problems 32-34.
32. (x
+y+
2z)
2
33. (3x
.
-
2y
+ z)
2
34. (2a
.
Factor the expressions in problems 35-60. 35. 6ax 2 37.
2
2ay
+ 2ox -
y.
4a.
36. x(a 38. 4x
2
+ 6) + x. + 4x +
1.
-
b
-
3c)
2 .
FACTORING A TRINOMIAL
2.3]
z - Qxy + 9y - 1. 4y 1 - (2x - y)*. 2
39.
25
2
40. 4x
.
2
41. 43.
4^.
X
*j i
~i~
49. (x
+ +
6
-
47. (a
51. x
53. 16a
2
yY y
-
(x
+
256
60.
^y
2 .
*
2 .
2
6
.
2
6)
55. (ro
.
-
n)
2
-
2
(x
3
)
+
2
(t/ )
3 .
16.
57.
yY.
5
4
+ d)
27y*.
- (x + - IGy*.
1
(c
+ y HINT: x + y* =
2
(jfc
58.
.
-
52. x 6
.
2
56.
50.
y).
6
- yY ~ (a + (x + I) - (x 12a4 6 - 36. 72a 6 - 9ab2
54.
2
48. 64
8.
6)
9y
ZiiX
TcO.
3
-
44. 8
2
+ 6)
42. (a
JL.
-
2
59.
2
.
2.3. Factoring
The
a trinomial
identities (x
(1)
+
a)(x
+
6)
6) (ex
+
d)
=
x2
=
acx 2
+
(a
+
6)x
(ad
+
+ 06,
and (ox
(2)
+
+
6c)x
+
bd
are not worth memorizing in themselves; but they suggest a practical method of multiplying two monomials and thereby
aid in the factoring of a trinomial. To illustrate, we may find the product (2x 3)(3x 2) mentally by use of the follow-
+
ing scheme.
The and The 6x 2
The
first
term of the product
is
=
2
Go; (2x)(3x) ' second term is the sum of the products of 'inside terms 'outside terms/' or 3(3x) 4x = 5x. 2x( 2) = Qx third term is 3 ( 2) Hence the product is
+
To
,
+
=6.
5x
6. 2
-}- Ilx 2, for example, we with binomial factors whose first terms are Go: experiment and x or 3x and 2x, and whose second terms are factors of 2.
factor the trinomial
Among the possibilities are (6x +
A
2)(x
-
1),
(te
-
Go:
+ i)(x 2), (Qx l)(x + 2), + 1), (3x + l)(2x - 2), etc.
(Qx
2)(x
quick inspection of 'inside plus outside products
' (the
TYPE PRODUCTS AND FACTORING
26
[Ch. II
12x = x + llx, key to the whole method) yields x = I2x llx, and we stop right there, having found that
6x2
+
-
=
-
+
l)(x 2). The student should alcheck the middle term in the product of the trial factors
ways
llx
2
before accepting
(Ox
them
as correct.
unchanged but shorter when the coefficient of a; is 1. In factoring x 2 -- x 12, for example, we seek two 12 and whose sum is 1, the conumbers whose product is efficient of x. These are readily found to be 4 and 3. We check the inside write the factors (x + 4) (x and rapidly 3) and outside products. It is more important for the student to become accustomed to this check as a method than it is
The method
is
2
for huii to learn (2) or (1) as formulas.
EXERCISE
Do
6
Review Exercises of Chapter
orally problems 77-96 in the
1.
Factor the following trinomials. 1.
4. 7.
+ a lOa + 25. 9x 12x + 4. +
x2
13.
16.
x2
19.
2
x
x
25. x 28. x
2
40.
43. 46. 49.
2 .
+ + + 5x - 6. + x - 2.
2.
llx - 12. + 4x - 5. + 12x - 13.
-
2x
2
+
2
-
5x
+
2.
- 2. 5x - 6. 6x2 8x 14x - 15. 2 9x 3x - 2. 2 24x + 7x - 6. 36x2 + Qx - 10.
34. 5x 2
37.
-24rs+4s 3x
2
2
2
8.
3x
2
x2
17.
2
x
2
x
23. x
29.
32. 35.
38.
2
-
+
1.
+
x
9.
4m2
2x
+
6x
12.
x
6.
15.
x2
3x
+ 2.
18.
x
2
21. x
2
24. x
2
2.
llx
-
12.
30.
2
2
36.
+
13x
+ 5.
+15x 27x -6x 2
2
+ 3x
-
6. 8. 2.
-
llx
4x
3. 7.
7.
+ 30.
-
5.
2
2
39. lOx 2 42. 15x 2 2
45. 14x 48.
6x
-
+ 12x + 32. x + lOx - 11. 3x + 5x - 2. 6x + 5x - 6.
27. x
2
33.
1.
-
6.
-
+ 40x + 25. 4x + 4. - 4m +
2
-
x
-
6.
2
50. 27x
2
+
2
2
16x2
5x
x - lOx + 21. x - 12x - 13. 2x - 3x - 2. 4x - x - 3. lOx2 - llx + 3.
44. 9x
3.
5x
2
41. 6x 2
47.
6fc
9.
25x -40x+16.
14.
26.
2
12x -f
2
11. x
20.
+ +
4x2
5. 9A;
2
x2
31.
2.
1.
2
10. 36r
22.
2x
2
20x
-
-
3.
19x
+
x
6.
+ llx - 15. + 16x + 3.
2.4]
FACTORING BY GROUPING TERMS
2.4. Factoring
27
by grouping terms
In general, factoring is more difficult than multiplication because there is no routine method which can always be applied. Yet there is one device which often proves useful that of grouping terms. dx = x(c Since ex
+
it
tiplication,
d),
follows that,
+
c(a
(1)
+
6)
+
if
+
d(a
by the distributive law of mulwe replace x by (a + 6), 6)
=
(a
+
6)(c
+
d).
we can arrange the terms of an expression in connected groups by plus and minus signs in such a way that a certain quantity appears as a factor of each group, then that quantity is a factor of the whole expression. Certain frequently made errors will be avoided if the student will study carefully the examples below. That
is,
if
Illustrative 1.
2ax
= = = 2.
12ac
= = = 3.
3ax
= =
Examples
+ 4bx + Qby (2ax + Say) + (46z + a(2x + Sy) + 2b(2x + (2x + Sy)(a + 26). - 86c - Sad + 2bd
+
Say
- (Sad - 2bd) - d(3a - 26) 26) 4c(3a - d). (3a 26) (4c (12ac
+
x Sax (Sax
-
667;)
Sy)
86c)
+
+
z 3az y 3az Say 3ofe) Say
+
+ +
x
+
no common factor appears
3ay y (x
-
+ y
z
+
z)
after various experimental be possible to arrange them may groupings of the terms, so that the expression to be factored is seen to be the difference of two squares. If
it
TYPE PRODUCTS AND FACTORING
28
Illustrative 4.
4x
2
[Ch. II
Examples
- 1 - Qy + 1) - (3y - I) (2*) - l)][(2x) + (3y (3y [(2x) - 1). (2x -3y + l)(2x + 3y
-
2
+
6y 9y 4z 2 - (9y 2
= = = =
2
+ 4?/ + x + 4xy + 4i/ -
= =
2
2
(z (
=
(x
1
)
2
x 4.
I
2i/)
+
2y
1)]
2
2
1
5. 4xi/
2
-
2
l)(a;
+
+
2y
1).
worth noting that a definite procedure is indicated for factoring, or attempting to factor, a polynomial of four terms. The 'two-two' grouping will aid in exposing any binomial factor. If this fails and there is at least one perfect square term in the expression, the 'three-one' grouping It is
be tried. binomial or trinomial containing two perfect terms can sometimes be changed to a perfect square addition of a term. If the term thus added is a perfect itself, the original expression may be factored, as
may
A
square by the square in the
examples below. Illustrative 6.
7.
x
x
4
4
+
4
+
x
EXERCISE
Examples
= = 2
+
4
(x 2
(z 1
+ + = =
4z 2
-
2 4
(x 2
(x
+
+ +
4)
-
2x)(x
4z 2
+
2
=
+
2
(x
+
-
2
2)
2
+
2
(2x)
2x).
- x = (x + - x)(x +l+x).
2x 1
2
2
2
-
2
1)
I)
2
7
Factor the following expressions. 1.
3. 5.
+ b(x - y). x(a + b) + y(a + 6). h(r + s) + k(r + a(x
y)
s).
2.
+ y} -
b(x
4.
m(k
6.
ax
-
1)
-
c(x
+
n(k
y).
-
1).
+ ay + bx + by.
x2
DEFINITIONS
2.5]
7.
9.
11. 13. 15.
17. 19.
21. 23. 25.
27. 29. 31. 33. 35. 37. 39.
41.
bx
2x
by
+
3
+ ex
3z
2
-
29 8.
cy.
2x
-
10.
3.
- y + x - y. 4z - 9^ + 2x - 3y. 16z - 25z/ + 8az - Way. z + y + 2z?/ - 9. x + y - 2xy - 25. - x + 2y + 1. y z - y - 6x + 9. x - y - c2 - 4x + 2cy -f 4. x4 - 3x2 + 1. x2
2
2
2
2
2
12. 14. 16.
2
2
18.
2
2
20.
2
2
22.
2
2
24.
2
2
+ 9. x 7x + 9. 4 x + 4x + 16. ax + by + bx + ay. 4x - y + y - 4. 25x + x + 9x - 10x + 4
x
+
4
2
4
4
28. 30.
2
32.
2
34. 36.
2
1
38.
40.
1.
2
42.
1-
ax2
2x2
+
-
x2
y
-
+x-
ax 2
2.
bx
-
x
b.
+ y.
-
3x - 4y. 1 - y 4z + 2y 16 + 2xy - x - y 16 - 8x + x - y 4z2 - y - 4x + 1. x +o -6 -2ax+26-l. 2
9x
2
16?/
2
2
.
2
2
2
.
2
2
2
2
+x + x + 3z + 4. x - 5z + 4. x + 2x + 9. 4z + 3z + 1. x + y - x + y 4x + 1. 9X + 2x + 1. 9x - 40z + 16. 4
26. x 9
2
5x
-
x3
1.
4
2
4
2
4
2
4
2
2
2
.
4
4
2
4
2
2.5. Definitions
The degree of an integral rational term in any letters is the sum of the exponents of the letters in that term. Its degree in a given letter is the exponent of that letter. Thus 3x 2y 3
is
of degree 5 in
x and
y, of
degree 2 in
x,
and
of degree
3 iny. its term 3 5 1 is a sixth x of highest degree. For example, y + xy degree polynomial of degree 3 in x and of degree 5 in y.
The degree
of a polynomial
is
the same as that of
polynomial A is a multiple of a polynomial B if B is a factor of A. To illustrate, 5, 10, 15x and 25(z t/) are mul2 1. 1 are multiples of x 1 and x tiples of 5, while x is a called more of two or A multiple of each polynomials common multiple. For instance, 25zV is a common multiple
A
+
of 5,
x and y z
.
TYPE PRODUCTS AND FACTORING
30
2.6.
[Ch. II
Lowest common multiple
The
common multiple (L.C.M.} of two or more polythe common multiple of lowest degree and with the
lowest
nomials
is
We
smallest possible numerical coefficients. find it by multiplying the prime factors of all the polynomials, assigning to each the largest exponent which it has in any one polynomial.
Factors such as x x, which y and y should be considered as the same.
Example 1. The L.C.M. and 2 2 3, is 2 2 3 2 or 36.
of 6, 9,
and
differ
only in sign,
12, or of 2- 3,
3
3,
+
1)
-
,
Example is
36x
2
(y
The L.C.M.
2.
+
+
of 6x, 9(y
3
I)
,
and Ylxy
3
I)
.
2 x 2 is y 2 x 2 or Example 3. The L.C.M. of x y and y 2 2 2 x ). (Note that there are two x y and not (x y)(y correct forms of the L.C.M. one being the negative of the
2
,
other.) 2.7.
common factor
Highest
The highest common factor (H.C.F.) of two nomials
more polycommon factor of highest degree and with the
is the
numerical
largest possible
ing the
common
exponent which
it
Example 1. The H.C.F. 3 2 is 2 3 or 2 3, and 2 ,
2. 3
18x*(y
.r)
EXERCISE Find 1.
the
of 30, 12,
of 30x 3 (x 2 y), or Qx (y
2
Qx (x
and
18, or of
2
3
5,
6.
The H.C.F.
is
it
has in any one polynomial.
2
Example
find
by multiplyassigning to each the smallest
coefficients.
factors,
We
or
-
2 yY, I2x (x
-
y),
and
x}.
8
H.C.F. and L.C.M. of
9, 12, 15.
2.
the
12, 18, 24.
polynomials in each group. 3.
18, 27, 36.
4. 14, 21, 36.
5. 20, 35, 50.
6. 30, 45, 75.
7. 80, 32, 96.
8. 54, 36, 90.
9. 65, 91, 104.
REVIEW EXERCISES
31
10.
3 x Sx ?/, 4zy.
11.
12.
12xV, 15xV, 9xy.
13.
+
2
14.
(x
(x
x?/),
-
+ y),
-
2
(x
5a 3x2 15a2x 3 10a4x4 ,
(x
.
,
-
y), (x
-
2
2
y*), (x
2 ?/ ).
- y (x - x (x 2xy + (x - y (x + x (x + 2xy + y (x - 3x + 2), (x - 4), (x - 5x + 6). (x - 3x + 1), (2x + 5x - 3), (4x - 1). (2x - 2), (3x - 4x + 1), (3x + 2x - 1). (3x + 5x - 5x - 3), (2x - x - 1), (2x + 3x + 1). (2x - 1), (3x - 4x + 1), (6x - 5x + 1). (6x + x - x - 4), (15x + 2x - 8), (10x - 7x - 12). (5x 2
15.
2
2
?/ ),
2
18. 19.
20.
2
?/).
2
2
2
2
2
2
2
2
2
2
2
2
2
2
22.
3
),
2
2
21.
2
?/).
2
),
2
17.
3
),
2
2
16.
2
-
2
- 2yY, (2y x) (x - 2x - yY, (y - x) (xy - y )
23. (x
3
3
2
7/).
,
3
24. (x
2
3 2
,
.
REVIEW EXERCISES Factor by using the type form ab 1.
3.
4y
-
3
2
127/
m?n
-
- 2x 5. 6x 2x - 2o(3a - 6). 7. c(3a 8. -x (2m - n) (2m 3
3
3
i/.
2/
=
ac
+
a(b
c).
4.
+ 6b 3a 6 + 5a6
6.
y(x
-
3(2m
-
3a&
2.
.
mri*.
5
+
2
3
.
3
3y)
2
a6.
+
-
2(x
fc)
2
2
n)
t/
+
n).
Factor by grouping.
+ my +
9. rax
-
x
3
13.
x
3
15.
6m
11.
3
x
-
2
2
4x y
+
x
ex
+
-
12m?n
+ cy.
10.
2xy*
+
3
8j/
3
2o 3
2o
4.
16.
18.
2x2
14.
.
+ 4mn + 8n 2
+
6ab - 2a + 36 - 1. - 9p + 4? - 6. Qpq x - 9y + x - 3y.
12.
1.
a4
2
2
Factor the following trinomials. 17. y 2 19.
21. 23.
-
y
-
12.
+ lOx + 21. x + xy - 12y*. 12a -o-l.
x
2 2
2
+
20. ra n
22. 24.
+ 2.
5x
- 5mn - 15. - 36. p + 16p 25r + lOrs - 3s 2
2
4
2
2
2
.
-
xy).
TYPE PRODUCTS AND FACTORING
32 25. 27. 29. 31.
-
a 5ab - 246 20 + 8x - x 4c - 4cd + d 4 - 20x + 25x 2
2
26. 12x
.
2
2
2
2
Factor by using the type form a2 33. 35. 37.
39. 40.
4z - 121y 16a2 - 256 x2 - (2y + z) 1 - (x + y) 4x - 4xy + y 2
34.
.
36.
2
38.
.
+
6) (a
(a
b).
81 - x4 (three factors). - y) - 9. (3x - 9(w - n) (x + 2j/) 2 2
2
.
.
2
2
-
41. 9
42. 4x
p
2
4xy
a
+
y
2
z2
+
806
-
166
4ax
+
2by
+
-
2
2
-
2
+
z2.
2
q
a
+
=
2
(2x
z2 .
y)
2 .
-
a2
b2
-
y
2 .
2pq.
+
Factor by using the type forms a 3 b3 ab b 2 ). and a 3 - b 3 = (a - 6) (a2
+
44. x 47.
1.
2
2
HINT. 4x2
43.
.
2
25. .
b2
2
[Ch. II
8
4
32.
.
71x
8
2
30.
.
+
-
+ 4xV + 4y 9a + 6a + 9x - Qx yz + j/V.
28. x
.
2
3
x
+ 8. +a
45.
6
50. x
6
48.
.
-
y
=
(a
+
2
06
6) (a
+
6 2 );
+
m - 27. - l) (2x 3
s
46.
49.
8.
ay + 64. 27y + (a + 2h) 3
3 .
6 .
HINT. First treat as difference of squares. Factor as a difference of two squares by adding and subtracting a perfect square.
51.
x4
54.
55. x4
+
52. x4
64.
-
+ 16. - 12xV + 16tA 4
x
24x
2
+ 4y.
HINT, x
4
53. x 4
-
24x
2
56. a 4
=
+
16
-
10a262
+xy +
(x
2
2
+
-
2
2
4)
1664
-
y'.
I6x2
.
.
Factor completely, using the type forms given in this chapter. Refirst any monomial divisor which exists in the given expression
move
it
before factoring
59.
4?V ~ x - a
61.
a 3 -f a2
57.
4
further.
25r
-
4
60.
.
-
4a
~ 8 ^ + W2o - Sy*. 4 y 2aV + a4
58 - ^ 6
2
-
4.
62.
6
.
REVIEW EXERCISES
33
63. m(n - 1) - 2(n - 1). 65. 4z - 1 - 4y - 4y*. 67. 3z a - 2az3/ - aj/ 69. (x + y) - 6 (x + y). 71. pV - 1. 73. x - (a + 2) 75. (a + b)(x + y) - (a + 2
2
64. z
2
66. (2s
2
2
72.
74.
.
76.
Why is it
10 3 that
+
1fl
i
b)(3x
4)
2
Sr mx -
3 .
-
+
(3y
16rm
-
2
2)
48r
3
+ 6) + 5(o + 6) 4* + 28ca: + 49C x (2x + 1) - (4x 2
70. 2(a
3
-
z
68.
.
2
3
+ 27m
3
-
.
m
3 .
-
12.
2
2
.
2
2
1).
y).
73 o can be simplified
by the
illegitimate
operation of canceling the exponents? 77.
Show
that
to problem 76.
-,
,L
,
= 3
^,
,
and apply
this result to
Chapter Three
FRACTIONS
3.1. Definitions
An
an indicated quotient of two algebraic expressions. The dividend and divisor are called respectively the numerator and denominator. algebraic fraction
x v j Examples.-'
A
is
o+l
2
-
3x
;
*
* a
;
l
+ y2
--
2 -:
3
simple if neither its numerator nor its denominator contains a fraction; otherwise it is complex. For fraction
is
2 2x example, and j
2
+- 1 are I[
2tX
simple, while
1
reciprocal of
the only
3.2.
2
Thus the
x for
is
-
J
number which has no
1,
each
reciprocal of
,
+
2 and o
a;
is is
reciprocal.
-;
a
j-
o
which equals
called the
and the
---
Zero
.
34
is
Why?
to its lowest
terms
The value of a fraction is not changed when numerator and denominator are multiplied or divided by same quantity (except zero) 1.
are
^
x is
Reduction of a simple fraction
LAW
j
complex. If the product of two numbers reciprocal of the other.
+
the the
REDUCTION OF A SIMPLE FRACTION
3.2]
v
i
Jbxampies.
By
?
1 -; .
2
23
6
8
__
12
virtue of this law
ax-
.
a
__
;
xy
35
.
a
;
7-
__
o
y
we may
2
reduce
to
xa ~~^r* x 2b lowest terms a
by factoring the numerator and denominator and striking out the common factors. This operation, called canceling, is a short way of dividing both numerator and simple fraction
denominator by a
F 1 example.
ax*
-
common
a
factor.
ft&^~r}(x
+
+ +
A fraction is not reduced factors of the
x
1)
x
1)
to lowest terms until all
common
numerator and the denominator have been
canceled; or in other words, until the numerator and denominator have been divided by their highest common factor.
When
all
j.i
of the
celed, the factor 1
numerator or denominator has been canalways remains.
Example,
1
1
1
jf $
--
=
T
6;
1
1
^-j-l1
CAUTION. Errors due to faulty cancellation are very numerous, and the student who is not sure of his knowledge or ability in algebra should be especially careful when he is tempted to scratch out letters or expressions just because they are above or below a line. The simple rule to be applied is that anything canceled must be a factor of the whole numerator and also a factor of the whole denominator. Hence the student should make sure that this factoring is done correctly before he does
any canceling
at
all.
For example,
if
the
numerator is xy + a, then neither x nor y nor a is a factor of the whole numerator, although the expression as a whole is
a factor of
itself,
as
shown
in
some
of the examples.
FRACTIONS
36
Examples in which canceling
xy
+
a
=
1
=
[Ch. Ill
is permissible.
+4=
2x
.
Examples in which canceling
+ 2)
%(x
is incorrect.
x
(Decide
+2
why
in
each case.)
xy
+ x
a. '
o a
+ +
l.
2(s
+
y]
2' I
-3(x
+
-
In the fraction
+
a
x
yY
+ 6)(c + (a + b)c +
+ y + g. + y + z'
2x
.
(a
the canceling of a
L
is
d)
'
d
of course not
would violate the rule about factors. But why is it not permissible? Note that when we cancel a factor we are actually dividing the numerator and denominator by that factor, whereas if we incorrectly strike
permissible since
it
out a from the fraction
a j_
i
+
2i
a
-
we
are subtracting a from the
numerator and denominator. This changes the value of a 2 -
fraction, as illustrated in the case
+
x
It is permissible to cancel
o
^
-
2 o
1 / 1
2 -
(or o
^
1 5)'
//
2 from the numerator and 1
1
*^
H 5( denominator of 3y(f TjL7 (x
obtaining
Z)
since ^ , x
1
amounts to dividing both members of the fraction by the canceled factor. However, it is a safer procedure to write this
this
fraction as
(x
It will help to full
^
* ,
+
/
2) (3
~~
- ^ 2)
avoid errors
before carrying out cancellation. if
both the
full
numerator and the
denominator of a fraction are completely factored before
cancellation.
3.2)
REDUCTION OF A SIMPLE FRACTION
EXERCISE 1.
9
How many
and how many prime factors, has each of
terms,
+
the following expressions: (a) xyz; (b) x2 2.
37
In which of the two following cases
missible:
The numerator and denominator
common
term, (b) a
common
+
x
is
1; (c)
z2
is
ax
have
-
11 '
13
-
y)(x
15.
3(a?
+ y)
a
In
these
y
(x
'
-
l)(x
-
3(x .
(x >
x
-
+ y) + x-y(x y)' - y (a + b)x - y)' (a + b)(x (x
'
+
+ l)o +1 y)(x
Explain the errors which have been made problems 14-17.
14
(a)
x-y
1)
+ y) + 1 2(x + y) (x l)x + 1 x+l (x + l)a (x + 1) + 6' a + b + c 3(a; + y)(a + b + c) - y)(x + y) (x ax + ay (x
+
x
+
?
canceling always per-
of a fraction
canceling permissible? cases reduce the fractions to lowest terms.
x(b
2
factor?
In which of problems 3-13
*
y
in cancellations for
FRACTIONS
38
+
(a
1
6)
Reduce
terms.
to lowest
^
)
!,f.
30'
36x 54x
-
-
x2 29.
x
2
x2
40.
-
x2
2x
-
15
x2
-
+
3x 3x
+
4 30.
x
2
__L
PL/v
tJJU
x2 2
6x2
+
l'
2x 2
-
-
x
72
-
5x
-
A/
9
^
'
x2
2 A J
x2
- x - 6' - x - 2 -x - 1
_
3
2
'
x2 37.
15x 2
3
-
+
Zx*
-
x
41.
-
x 46.
5x 48.
+
4?/<
47.
+
y)(s
y)
y
+ 8b + a 3
a (3x
'
(3x
2
2
-
46
2y)(3x
2
1
4
2
3x
-f-
(g
-
2
+ x
x3 a3
49.
45.
2
(x
8
-
2
r
-
-
2x
x
'
10
H
5s
rf)
-
+
+
- x - 9x + 9 x - 5x + 2x -
x3
3
3 43.
2
-
+ 7z + 12 + x - 12 +x-6 + 2x - 3'
10x2
15
4
Q^
oj
2x 2
12
43x - 14 -21 + 5z + 6x2
5(c
5)'
Q-i/f/r
-
x3
5)
oyJu
42.
44.
3(x
-
'
4'
-
7(x
y)
+
12x 2
8x
25.
)
+
2x 2 2x 2
3)
2
3x 5x
32.
-
-3)'
5(x
35.
x(y
-y
1
22.
99
26.
-
x3
2L
2
23.
[Ch. Ill
5y(x
-
4b2
2
+ xy
-
2y
y) _
3
2
)
2rs 2
j/) 2
+
+
+s
2
ll(x
+ y) + 24 y
2
-
9
3.3]
3.3.
RULES ABOUT FRACTIONS
39
Rules about fractions
In this and the next two such as - or
fractions,
x
articles
~-
2
-,
we
shall deal
with simple
whose numerators and de-
nominators are polynomials.
RULE 1. The sign before a fraction is changed when the sign of either the numerator or the denominator is changed. For by the rule
of signs [(4), Art. 1.5],
^ = _5 6
~=
~,
and
6'
RULE
2. The sign before a fraction is changed when the sign a of factor of either the numerator or the denominator is changed.
For again by the rule of signs, a( 6) = a&, and hence 6 the sign of the whole when the factor 6 is changed to numerator or denominator in which it appears will be changed, so that Rule 1 will apply.
111
1
Example
1.
Example
2.
(-2)(-3) a ,
,
t
2-3
2(-3)
=x(
x)
1
+
,
or
6
-
xY
x(x
1)
Reminder. It is important to note here that Rule 2, as does Rule 3 below, refers to factors and not terms. For instance, in is
Example 2 above, when the
changed
it
becomes
1
+
x or x
sign of the factor 1 x. 1, and not 1
x
+
Definition. Integers divisible by 2 are even; all other in4, etc., 2, tegers are odd. Even numbers include 2, 4, 6,
as well as 0;
RULE
odd numbers include
1, 3, 5,
1,
3, etc.
The sign before a fraction is or is not changed according as the number of factors whose signs are changed is odd or even. The altered factors may be all in the numerator or in the denominator, or they may be in both. 3.
FRACTIONS
40
[Ch. Ill
Example. (1
(x
-
(2
a?)
l)(x
- sXs - 3) = - 2) (3 -x)
Here the minus sign
is
second step because the sign is 3, which is odd.
RULE nators, the
4.
add
(x(x-
l)(x l)(x
-
2)(x 2)(x
-
_
3) 3)
inserted before the fraction in the
number
of factors there
changed
in
To add
or subtract fractions with the same denomior subtract the numerators and place the result over
common denominator.
(by Rule
EXERCISE
1)
=
10 1-9, write each factor with a positive literal term, 3 should be left sign where necessary. [For instance, x
In problems changing
its
x should be changed to (x unchanged, but 3 and reduce each where fraction permissible factors .
L
x-
Z>
'
3
'
8>
-
-x (1 2) (x l)(x x)' (1 x)(2 x )(x - 2) (3x 1)(4 - x)(x - 4)(3x - 1)' (2 - 1)10 (x - x)10' (1 1
4
(x
3
2
I)
x)
(1
5-
Then
cancel
to lowest terms.
(x
*'
'
2
3).]
-
3
I)
x)
'
3
- 1)(2 - x)(x - x)(x - 2)(3 (1 - 5) (x - 6) (x - x) (6 - x) (5 - 3) (4 - x) (2x - 4) (3 2x) (x (x
2
3
2
3
'
6
6
6
6
'
3) x)'
ADDITION AND SUBTRACTION OF FRACTIONS
3.4]
the operations indicated below.
Perform 3x *
<
55 -
x
1
,
+
~r r
e
3x ., 12.
-+ 5
2
2x
+
3
-2
2s
1f '
e
7
+
2
3x
+3
ie . is.
-
3 2x + 4 + + x + l' 4x + 5 _ 3a: - 2 'x-9 9-x' lOx + 3 3 + 10x
x x
.
-
3.4.
6
''
6
-
5x
6x
5
5x
3s
+
7 1
2x 2z
5
2:r
-
3a: 1
+x +
2
-
6
+
6
-
l'
3 '
x
15z-7 3x2 - 5
'
2
+ g
-
2x
1
3x
+
13.
'
^
I
-
5x
,
.
2a + 3 --
,
5
g
7x
41
7
5
+ -
1
3x2
Addition and subtraction of simple fractions
By use of Law 1, Art. 3.2. two or more simple may be written with a common denominator.
fractions
Write the numbers 4 (or -f), f, %, and ^-f as fractions with a lowest common denominator (L.C.D.).
Example
1.
Solution.
The L.C.D.
r .u j * of the denominators
i
1,
of the fractions
is
4 o * j ir AT 3, 6, and 15. Now,
the L.C.M. (30) 120 4-30 -
? 3
=
2j^lO 3 10
Example
=
20. 30'
5 6
= 5-5 = 6-5
25. 30'
Write the fractions
2.
an
-^
= 16-2 =
15
15
-
^. 3x(x
oO
1
16
,
=
=
1
and ^- ---2)
oO 32
2
6(2
;
30'
-
x)
with a L.C.D. Solution.
-
r x)
-
2)
Qx(x
-
_.
,
G(X
L.C.D. of the fractions 2a a 3x(x
5
=-
2)
is ,
and
by Rule
then 6x(x 5 6(x
1,
Art. 3.3.
The
Z)
-
2)
2).
We now 5x
Qx(x
-
2)
write
FRACTIONS
42
In view of Rule
RULE
mon
[Ch. Ill
we have
4, Art. 3.3,
To add
or subtract fractions, write them with a comdenominator (their L.C.D.), add or subtract the numerators 1.
as indicated, and put the result over the L.C.D.
Example
1.
12
6
66 63
6
+ 3-5
12
2.
Example x
+2
+
2
5
10
6
6
+2_x +2 x -!' '! +1 = (x + 2)(x 1) _ x- + 2 + l)(x - 1) x - 1
2 a:
z+ll-z
= 2
2
x z
a:
1
2
x(x
2
z2
(a;
x2
~ -
x2
From Rule
we
1
-
;
=
4 1
get the simple results: r
But note that these
r^ bd
d
o
- 1) - 1
1
o
and r
2
+
=
d
f^ bd
results are not
formulas when b and d have some factors in common. For example, if we use the method indicated by these efficient as
results
we have
1,1 ^ + T~ '
X
1
X
2
:
(x 1
2
-
(X __ '
1)
+
x(x
(x+
2
1)(X
+ 1
-
-
1)
'
1)
x2
_
(X+
1)
^
1) 2
l)(x
whereas the method of Rule once.
+ (x+
+
x 2
1)(X
-
1)
x
x 2 '-
1'
gives the simplified answer at
3.4]
ADDITION AND SUBTRACTION OF FRACTIONS Find the blunder
Exercise.
43
in the following incorrect addi-
is so common that it should be thoroughly understood and then carefully avoided. A second common blunder is the omission of the denominator. The student should cultivate the habit of writing the denominator first
typical error
to avoid forgetting is
long.
EXERCISE
11
In each problem fraction and reduce
L 4
it to
lowest terms.
1,1 + 3'
2
+
5
'
oil+ 3
k x
7
3
+
+
1
8
3x
- + 2x
j
-
3x
3
+
x
+3 +
4
-
3 _L_i
4 '
5
4
17.
-
6
-
'
~
5x
5
6
Y
5
,
4a;
5x 2x
2 -
'
5*
~
'
5*
+
3*
'
3*
~
'
16.
below, combine the given quantities into a single
2
1
7
when the computation of the numerator
it
4x 5
9
-
+3
3x
1
5 4
_
2x 2x
+ -
2 7
+3
' 3
4*
+
4
9
7^
3 -
18.
2
-
3^ 2
5
3a:
+
2
2x
-
3
2x
+
3
3x
-
2
-
4
5 3x x - 3
9x 4x
+2
+
3'
x-2
3x-5 '
+ 6 ' 3x -
+
5'
'
'
8'
2x
+
4x
3.5. Multiplication
RULE
1.
and
division of simple fractions
TTie product of two or
product of the
more simple fractions
nominators.
2-5-1 3-3 f.
is the
numerators divided by the product of the de-
(a)f^YA =
f?WW = 2f. ceg
10
9
3.5]
MULTIPLICATION AND DIVISION OF FRACTIONS
45
RULE 2. To divide one fraction by another, multiply the one by the reciprocal of the second.
3d.
91L
/*}/7 O / /
*
/
example
^^'^^^iFjljj^
Example
2.
*/
X
9
iC
JL
~7~
/
rfl */
/y.
~~'
f
As
cases.
l/
X
may
and division
=
=
A
4.
2 thirds
a
= = dec
^ By
d
EXERCISE Perform
s
' ;
,
rule 2,
8 2 --=--
'
/-V^ VbAc/
6c
d
the indicated multiplications.
>
-
8 35
JL)
all
33 =
12
24
JL J
arithmetic
where a can be anything
4,
a 6_
of fractions are postu-
'
6
c
and subtraction, the
be observed to be true in
8 thirds
v
!.
For example, since
except zero,
1 ^ ( /
X X
/
in the cases of addition
which
vfr V
1
if
^+
rules for multiplication lates
*
X/'r 2
X
1
= A^ote.
r
**/
first
'
^X V
15
/8V3 =
(-)(-) 3/2/
A
4.
FRACTIONS
46 i3 -
15.
(!-^T>
-
3*
(trf^Vex +
[Ch. Ill
2 >16.
4).
20.
(^F^f)(3x+l)(-
22.
24.
/ 3x
_w
27.
I
6x
on 29
'
2
2
-
+a;-2 V2-3o: /1c , 15 '
A X'K 13a:+6/ 5 2 /o 2 + ,
fa ~ 127T3A
x /)(
)-
,
-
3
/5x
34
-3/
+ 7/2x-3 -
(x
+ 3)
2x
EXERCISE Perform
12a
the indicated divisions.
^~ 4-19 -riH* +
x'
3x
-
o;
'
4
'
'
4,/r
3
^ 5.
6x }JL/
+ ~T~
15 1<-I
^T-T-F ^,
.
-5-
/rk
+3
9r
4-
'
-iM^
' ^ x * 6. 2
4*
+ -j-^ +4
^x f-
5).
v
.
'
x'
-
(a:
/o (2
-
2)
x).
a;
O
* ^ TFX 'T'
+ , i
(2*
4
3 >-
5x
O/v.2 _L_
(x
+
x)
-
8
-
2x2 ^SX
-
Q/v
x a/
O
+2 o -3-^
)-
MULTIPLICATION AND DIVISION OF FRACTIONS
3.5J
-
-
Ix 7x
3x 3x 2
+ -
2s 2x
14. (7x
+
2)
16. (6x
+
5)
2
12
'
v
12
1V'
*'
-
1
47
' .
1
11.
13. (5x /
,15.
-
1-^4Jlx 5
-T-
3)
17. (3
-
5
- 5) ^^ .
(3x
^-_ 6
+
-
(x
23. x
-
-T-
27. (6x 2
2Q
2)
-
1)
2
+2x-l)
-7x-3
+
5x
9x
-T-
6x
2
'
-
^-+ x
zz. '
2
+ 3) _._
7
-J-
2ar
-
5
3y
+
7
2-3x 14 + lOx 3x + 2x 1 + 2x - 3x 3x 3x +
+
-
14
+
'
2x
2
.
'
3x
2
3x2
2x + xy - y2 + xy - y* 2 2x - xy - y2 2y Qxy + 4x 2 6x - 13x?/ + 6?/2 3y 5xy + 2x 3x 6 6 + xy 2y 2y + xy - 3x' 2 2 2x 5xy + 2?y ^ 2y - 3xy - 2x2 - 2x2 2x - xy - y 2?/ + 3xy 2 %y ~ % - 2y + 2 ^ xy + x 2y - 2x + y - 2 xy xy + x + y+l' 2x2 2
2
.
'
'
-
5)
_ '
2
2
-
*
lOx
2
.
2
2
. *
'
'
2
'
5
x2 -3xy-4y 2
2
36
T
+ 2)
2
'
2x2 -x-3
12
28. (5x 2
-
2
32 '
35
g^jx+l
12
26. (3x
-
'
- 9x2 x - 1
I
'
.
fl
-
-f-
-( 3x ~ 2)
33 '
^
-
'
'
-x-5 _^_
6x2
+
-
3x
O
/v.
O/y.2
6x
2
1^ 5
2
(3x .
-T-
12)
-
x
)
iA *'
-
'
2
-
25. (2x
Z
7x
18
.
t
^~
^x
20. (12x 2
6
5
f+ f-- 2 - llx
2x)4- R 2 6x
10 T2r2 4^r iy. -f- ^x
'
3x
-i-
2
+
-
-2y-a?
'
FRACTIONS
48
3.6. Simplification of
[Ch. Ill
complex fractions
We
have seen that a fraction is complex if either the numerator or the denominator contains a fraction. In this case we shall speak of the minor denominators. For example,
in the
are 2
To
complex fraction
and
l+ 26 +3 ,
the minor denominators
62
6.
simplify a complex fraction we reduce
it
to a simple
fraction in its lowest terms.
When
a complex fraction
itself
A
nor the denominator B of contains a complex fraction, there
neither the numerator
are two methods of simplifying
A L>
Method 1 Multiply both the numerator and the denominator by the L.C.M. of the minor denominators. (See Law 1, Art. 3.2.) .
Example 1
1.
!
+
2
+
2
_
3
1
! d+ +2_ +2 3 I 1
12
-
12
+ 6 + 8-1 = 10-8 + 12
12
x
2
Example
2.
2
9 2 X* x)
ax2
-
25 14'
2bx
Method 2. Reduce the numerator and likewise the denominator to a simple fraction, and then apply Rule 2, Art. 3.5.
Example o
.
'*'
1 2
1
.
_
_! 3
1 3
12,3_2 =
66
9_1
33
6
13
=
_6_
8 3
= /13V3 =
U/W
13 16'
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS
Example 1
49
2.
+ ,
x- 1 1 + 1 -
n 2
1
2(x
,
x- 1 1 -x
'
x
- 1) - 1
2
2x- 1 x - 1 2
1
-x
2
-x 1 - x 1 - x - 1 = /2x 1V1 a? = /2x IVx x-l]2-x } x-ljx -2/ '
2
1
2
2
2 2
2
_
(2x
-
a;
2
l)(x 2
-
+
1)
In general, Method 1 is shorter and more efficient when the minor denominators are single terms; otherwise Method 2 is
preferable. When either the
plex fraction
numerator or the denominator of a com-
is itself
must be
simplified fraction and so on.
a complex fraction, the latter fraction
first.
The same
rule applies to the latter
1
Example. 1
+ X
Here we simplify successively the fractions within parentheses, brackets, and braces, thus :
EXERCISE
13
Simplify the following fractions. 1
2.
y
y
FRACTIONS
50
1-1X
+ +
4.
3.
1
1
>+!'
-
1
5.
x-y
+y
x
+
y
x
+
x 1
6.
x
+y+
y
x
7.
x
1
a
_
l
bx
5
o 9>
xy
M
g
5
I
r
10.
y
y
11.
- 13xy 3x + 2y - 6 xy - 5xy + 2x 3y - 3x + xy 6 2y 2
IS.
2
2
2x2 13.
f
ay
x
_
b
y_
8. _
y
y
+
-
2y 2 2y 2x2
+
xy xy xy
y
-
y
17.
5(x
-
2
2y 2x 2
+
14.
2
+ -
3
+
2
x3 '
16.
(x
x2 (x
2xy
y
y
3 '
2
18. 1
3
- y) -y - y) x
y
f-^)'
2
y
y
y)
-'M
2
2xy
-
2
+
x-y x
19.
2
2y 2 2y 2x 2 y
1
+
-
-
'
3x 2 3x 2
2 - 3x2 2y xy 2 2y 3xy + x* - 6y2 6x2 5xy 2 6z Uxy - 2y 2
-3(x
5xy 3xy 3xy xy
2
2x '
2
'
x
2x 2
20.
3z3u x
2-y a;
x
x
2
2
'
[Ch. Ill
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS 2x
21.
y
X
y - xf
(
(x
-
--4 +x z
v
?/)
22.
-
+ x-5
+ 6x -x+1
X 24.
23.
2
25.
2
-
27. 1
-
29.
3x
+
2
3x
-
1
26. 2
1
+
x 28.
30.
2x
2
-
3x
+x
1
~~
<>
r
REVIEW EXERCISES Reduce the fractions in 1-6
4a2 1.
3.
(x (x
+
6o&
-
to lowest terms.
462 2.
+ 2y) + z + 2?/) - z 3
3
2
2
4.
6.
5.
27x 3
3mn 3mn (2x
+ + -
-
%
6fen
+
Qkn
y)
3
3
-
me me
5(2x
+
2fcc
2kc
-
y)
2 j/
Cam/ out the indicated operations and express the answer as a simple fraction in lowest terms.
-
3x
5x
1
7.
8.
+
3y 3x
9.
a
a
2y
% -
'
4
3
.
2xy
-
1
8 4 3
3y
y
x
.
T'
-
1
_
2x + 4x 2
1
1
2x'
to
each
FRACTIONS
-
o3 3x 11.
x
12.
2
3x-5 13.
x 3x
1
14.
15.
-
4x
4
-3 -
4z x - 2
3a-6 + 96 c
20.
a
2
~T~
-
(c
6'
-
a)(6
-
-
'*'
o)
(6
-
c)(o
c)'
- L
+2 ^ +3
x
,
-
2(x
+
1)
ISz
18.
26.
(x
-
36 ab
1
%y
i^$/
+
3 *^
- xy - 6y x2 + llxy + 28y x - x2 + 4x - 4 2x + x - 1 2x2 - x - 262 1 a(a + 26) 06 ^ a + 46 o 46 a 2a6 + 2a x2
2
2
2
3
21.
-
2
16. 1
+ 3a&
a262 19.
a
7/)
2.
-362
ab 17.
1
c)
2x
7^3-x+
+
2
^
-
6)(6
+ ab + 6
-
(x
i/
-
(a
a2
2x
^ .
-
2
'*'
63
'
'
4
4
2
2
:
2
262
*
3x 1
-
x
+
27.
2
1
o
-
1
28. 1
-x
1
-
a
1
+a 2
1
[Ch. Ill
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS x
y
29.
m m
30.
m m
n p
n
1
1 I
y
1
m
V) x
tt
11 y
1' c
53
m
p
n
1
-
a 31.
32.
33. 3
-
+
-
2
2(x
3(a
-
2
+ 2b
-
4) (2s
2
-
3)
-
3)
/
-
2x(2x
3)
1
il
2
+ 4) (a + 5) - (a + 5) (a + 4) 3 2_ _ x + l' 'x + 2 x + 3/ - 2u(u + 2) 2
2
2
2
1
(u
b
2
+
(s
1
^
2
-
a
2
'
38.
34. 3
ab
b
(2x 36.
+
1
a 35.
1
b
(x
+ 2)
(t*
a 39.
+b
a
b
- b) 2 - c2 (a - cY - b - b) - c2 (o + b) 2 - cz (a - c) o - (6 + c) a (b 3 2 b b2 - a a^ + ab + b* ['a a 2ab (6 a) J L o 2
(a
2
40.
'
2
2
'
2
3
2
2
'I
'
42.
a
+&_
a
a
b
a
(m
2
f'V
4
43.
1
'
--
b
+
f l
n 'V
+
'
ab
n ac
2
a
2b
b2
a2 *
-
II
n /m
2
_
b
2
La
_._
2
3
a2
1
I
___
m /
II -f I
*'V
If
-t
II
I
-
3
/
+ ab + acj
mn
3
-
I
^ / '
a6c
m + mn 2
Chapter Four
UNKNOWN
LINEAR EQUATIONS IN ONE
4.1. Identities
An equation
and conditional equations
is
an assertion that two expressions are equal.
a sentence, in the language of algebra, in which the equality sign serves as the verb. The two expressions involved are called the sides or members of the equation. It is
An identity is an equation which the letters involved.* Example
1. (a
Example
2.
+
x1
=
2
6)
y
1
=
a2
+
(x
2a6 y)(x
true for
is
+ +
62
all
values of
.
?/)
When
it is desired to emphasize the fact that an equation an identity, the equality sign may be replaced by the symbol =, read 'is identically equal to.' Thus
is
a
-
1
=_(1 -
x).
A
conditional equation is one which is not an identity. It true for special values only, if any at all, of the letters involved. is
Example. 5x
3
=
+
2x
9.
A
conditional equation may be translated into English as a question. The one in the example above asks, 'What is * More precisely, an identity is true for all values of the letters make each side of the equation a definite number. For example,
3 (x is
-
an identity; but the values x
by
zero
is
_JL
l)(x 1
1__
+ 2)~* and x =
not allowed.
54
involved which
1
*
+2
2 are not permissible, since division
4.3]
LINEAR EQUATIONS
the
number such that 3
55 less
than
its
'
product by 5
is
9 more
than its double? ' Henceforth we shall use the word equation' to mean '
conditional equation.' While, as indicated above, an equation
is
in
one sense an
implied question, it is also, in many cases, a camouflaged answer (like the query, 'Who wrote 'Gray's Elegy'?'). For as will be shown presently, when we apply a routine method to certain equations, the numbers pop very readily out of hiding.
The
which they ask about
numbers or expressions sought from the last part of the alphabet (x, y, are usually chosen z, u, Wj etc.) and are called unknowns. letters representing the
4.2. Integral rational equations
An
integral rational equation
is
one in which two integral
rational polynomials in the unknowns are equated. Thus the unknowns appear either without exponents (since the ex-
ponent
1 is
usually omitted) or with the exponents
3x
4
value sought for
it is
Example Example
1
.
2.
ay
=
by
5x.
Here the unknown
2, as
=
may
is 2,
be verified by
Here the unknown
c.
2, 3, etc.
and the
trial. is y,
which
/
stands for the expression for
-,
a
since this value substituted
b
y makes the equation an identity.
As we
shall see,
many
practical problems lead to integral
rational equations.
4.3.
Linear equations
An
integral rational equation in which the unknowns appear to the first degree in each term containing them is called a first degree, or linear
*
equation.
letters We shall see that a first degree equation in not more than two ' hence the adjective linear.*' represented graphically by a straight line
*
may
be
LINEAR EQUATIONS IN ONE UNKNOWN
56
Example
1.
3x
Example
2.
5a*y
Example
8.
5x
=
7
=
+ 2y
= 4.
c.
shall consider only linear
a linear equation in one unknown an equation we find
or the particusubstituted for the unknown,
solve
lar quantities
make
we
one unknown.
4.4. Solving
When we
(Linear in y.}
(Linear in x and y.)
Hereafter in this chapter
equations in
IV
(Linear in x.)
4.
7b*y
[Ch.
the two
which, when members equal. The
its roots,
root
said to satisfy the
is
equation.
Example.
5-4-3
The
= 2-4
We shall now
root
of
5x
3
=
2x
+9
is
4,
since
+ 9.
solve the equation
(An axiom, as those who have studied geometry may recall, is a statement which seems to accord with experience, though we do not prove it, and which is often used in proofs of other statements. Thus a set of axioms, helps to serve as a foundation upon which one may build a system of mathematics.)
by use
of several axioms.
AXIOM
1.
When
equals are multiplied by equals, the products
are equal.
Application.
x
-
3 6
=
fe
(
+
2)
6,
or
Sx
(2)
AXIOM equal.
2.
When
-
18
=
Zx
+
12.
equals are added to equals, the
sums are
SOLVING A LINEAR EQUATION IN ONE
4.4]
-
(Sx
Application.
+
18)
18
=
(3x
Sx
=
3x
+
UNKNOWN
12)
+
57
18,
or (3)
AXIOM
3.
When
+
equals are subtracted
30. equals, the re-
from
mainders are equal.
Sx
Application.
3x
=
(3x
5x
=
30.
+
30)
3#,
or (4)
AXIOM
4.
When
equals are divided by equals, the quotients
are equal.
5x
.
.
A
7
Application.
=
30 5
o
,
or
x
(5)
Check in It
(1)
:
^
o
-3
=
= ^ 2t
6.
+
or
2,
should be noted that the root,
A
tions (1) to (5). which are satisfied
8-3
=
6, satisfies
+
3
2.
each of equa-
group of equations, such as
(1) to (5),
the same value of the unknown, are
by
called equivalent.
An
finding a succession of equivalent equations, the last one of which exposes the root of all of them.
equation
is
solved, then,
by
The simple rule that a term may be transposed from one member of the equation to the other by changing its sign may be used in practice in place of Axioms 2 and 3; but the reason for the rule should be understood.
The
essential process in solving
be described as follows Step
1.
Simplify
Step
2.
Remove
plications.
all
any
linear equation
may
:
complex fractions involved.
parentheses, carrying out indicated multi-
LINEAR EQUATIONS IN ONE
58
Step
UNKNOWN
[Ch.
IV
Clear the equation of fractions by multiplying both by the L.C.M. of the denominators.
3.
members
Transpose the terms containing the unknown to the side of the equality sign, and all other terms to the right
Step left
4.
side.
Step 5. Factor the left member, expressing it as the product of the unknown by a second factor. The latter is called the
unknown. Divide both members of the equation by the of the unknown.
coefficient of the
Step
6.
coefficient
For example, applying these steps to the equation
we
-
-
(6)
-
=
6*
f+
4,
get in succession the following equivalent equations:
m
3(s
+
_
6)
ftc
,
4 1+4 ~y +
(7)
3s
(X (8)
+
18
_ -
g
21x 21x
(9)
(10) (11)
= = 30x z(-9) =
+
126
-
x
(12)
,
>
- + 4,
6x
SOx 140
.
.
+ -
140;
126;
14
-9
The student can easily avoid blunders which are made if he will keep in mind the following simple Whatever member.
is
done
to the left
member must
often rule:
be done to the right
For example, if 2x = 5, we should be wrong in concluding that x = 5 2 = 3, because the rule does not permit us to divide the left member by 2 and to subtract 2 from the right member, as is done in the foregoing incorrect 'solution.'
4.4]
UNKNOWN
SOLVING A LINEAR EQUATION IN ONE
EXERCISE Remove
59
14
signs of aggregation according to the rules, multiplying factors as indicated, and then determine which of the following equations are identities and which are conditional equations. Solve the all
latter.
11.
= 15. - 1) = x - x. x(x 2z - 3x = x(2z - 3). - I) = x - 2x + 1. (x 5x2 +3x-l = x(5x+3)-l. x + 2 + 3x - 1 = 4x + 1.
15.
x
1.
3. 5. 7.
9.
3x
2
2
2
-
=^-
17. 1.5x 19.
20.
21. 22. 23. 24. 25. 26.
27. 28.
20
2
.7x
=
a
= -
5.
.6x
+
+
o
(5
2x)
=
3x-2 _ ~~ 2
_ 2x-3
3z
12.
20 = 0. Ix 10 = x + 2. 4x - 3 = 5x - 5. 3 - 2x = 7 + 2x. 1 - 3x = 5x + 4. 3 + 4x = 6x - 7.
16.
3x
-
4. 6. 8.
10.
18. .5x
.7.
+
=
10
.07
-
10
=
.8
3x.
+
-06x.
2
o
o
-
-
2
2x
-
4x
x
1
2
^ o
o
7;
3x
+
- 5. 7 - (5x + 3) + 6x = 10 - (3x + 5). 8x - (3 + 4x) + 5 = 7x - (3 + x). - 7x) + (3 + x) = (15 - 8x) + 4. (9 lOx + 5 - (3x - 4) = 4x - 3. -2(5x - 7) + (3x + 2) = 4x - 2(3 + x). - [3(7 - 3x) - 3] = 7x + 5. 5(2 + x) - [3(x - 7) + 4(2x - 3)] = x. 5(2 + x) - x) - 5(3 + x) = -[2(x - 1) + 3x]. 8(2 - 3) - 2(3 + x) = 5[(1 - 3x) - 4]. 7(x 3x
r
31.
5x
2.
=
2x -3 3
4
3
3x
+
3x-2 _ ~~
2
^
+2 o ^
+
A
32. 4
-
+ .
3x
'
2x
x 2
2
2
-
3
4x
K
O
6x+5
5
2(2x-5)
2
3
4x-7
LINEAR EQUATIONS IN ONE UNKNOWN [Ch. IV What number added to 3 is twice as much as the number
60 35.
minus 1?
when
36. If result
is
subtracted from 3 times a certain number, the is that number?
1 is
5 more than the original number, what
number is multiplied by 5, the product is 2 less than sum of the number and 7. Find the number. One child is twice as old as another. Two years ago he was
37. If a
twice the 38.
3 times as 39.
one he
is
old.
Find their
ages.
Fred has 3 times as many marbles as Bob. If he gives Bob will have twice as many. How many has each?
40. The quotient obtained by dividing a certain 5 less than twice the number. Find the number.
A man
41.
invested one
sum
receiving $80.00 interest in
at
6% and
one year.
number by
double that
How much
sum
.3
at 5%, at
was invested
each rate? 42.
An
airplane traveling 200 miles an hour starts at
noon
after
a transport which left at 10 A.M. and travels 150 miles per hour. When will the plane overtake the transport?
4.5. Operations
which yield equivalent equations
When
terms are transposed in an equation, or when both members are multiplied or divided by the same number, excluding zero, the resulting equation is equivalent to the first one. All of the operations required in Exercise 14 were of this nature.
4.6.
Operations which
may
not yield equivalent equations
Multiplication of both members of an equation by zero = 0. While this of course is true, it is yields the identity
not equivalent to the first equation. Division of both bers by zero is not allowable, as we have seen. Again,
when we multiply both members
an equation
expression containing the unknown, the new equation have roots not satisfying the first one. These roots are
by an
may
of
mem-
called extraneous.
OPERATIONS YIELDING UNEQUIVALENT EQUATIONS
4.6]
61
=
1 2 The equation x has the root 2. Mul= both x we members have: 3, tiplying by (x 2)(x 3) = 0. This has the roots 2 and 3; but 3 does not 0(# 3) satisfy the first equation and hence is extraneous.
Example
.
2.
Example
Consider the equation
= _2
1
(x
-
-
!)(
x
2)
1
-
2
Multiplying both members by (x
we
fractions,
root
1.
But
1
1
=
2(x get 1) does not satisfy the
for the non-permissible division
ous,
and the
first
we
if
Finally,
x
-
I
l)(x
(x
2).
2)
to clear
This has the
equation, since it calls zero. Hence 1 is extrane-
first
by
equation has no root.
members by an expression conusually lose one or more roots of
divide both
taining the unknown, the original equation.
we
= 3(x Example. The equation x(x 1) 1) is satisfied = = 1 x as x or be verified 3, may by by substitution. = 3, the root 1 x x we members both by 1, get Dividing being
lost.
Summing
up,
we
find that
when
members of an equation an expression containing
the
are multiplied or divided by zero or the
unknown,
the resulting equation
may
not be equivalent to
the first one.
When operations of this sort
are necessary in solving equa2 in as above, the solution is not complete tions, Example until the roots found have been tested in the original equation, and the extraneous roots have been rejected.
EXERCISE
15
The following equations may be reduced to linear ones and then solved. Where extraneous roots may have been introduced the answers must be tested. .
x-l
i
_2__-_3_ ii x x+1
1* 1
3 *'
~
x
-2o
8
1 i
x
+
2
r 2
4
LINEAR EQUATIONS IN ONE UNKNOWN
62
3
,
.* *'*
g
x+2
3
4
x+3
x-4
x
-x-12'
10 '
x-24-x
~
x 4
'
x s
01
2x 2x
+1 +1^ ~ +3 =
~
x
-
+
5
5
_
20
4
+2
-
3x
,
2
5
.
4x
.
24
-
3
-
5 6
6x
27.
x-3 3x + 1 x + 2 3
28 x
-
x
4x
-
2x
+5^ ~ +7
3x x
+
2x
~
5
'x +
2
1
+9 x
2
x
+
x2
x
5
= 3
-9 2
x
2
-
+7
-x-6 ~ 2x
a:
-
.
6x 2x
3
1
x-3
' -I
5
-
x2
^
-
-6~
+3
,
2x
2'
=
5 4'
1
25.
26.
~
3'
4
3
-
4x
2
___
+3~ _ -
__ _ '
1
+3_
3
6-
'
x
4
3
~
5x 2x
'
O
-
2x 4x
-
'
+ ^.
2
3x 3x
l'
x
-
1
=
2_3x + 2_2-3x x + 1 x + 1
2
2x
2
x2 +5x+6
'
2'
16
-
^
*
+
x
x_+_5
2x 1 x + 2
? x
3x
2^3^
1<^
o*
3
x
+ xx+_3 -2=
4x 2x 3
o
3
~
1
5
lft
4'
x
4
-
5
3
x+3
2x
3
3
3x 3x 17
28 2
x
-
x+2
3
+5~ 6x^7
? x
*'
/->
1
1~
3x
_ 9>
19
+3= -
5x 2x
7
^l
,
o
x 2 -x-6
f*i
i
IV
4
5
1 _. ______
r
x-3
[Ch.
15
5
__
2
=
x x
^
+7 + 3' + 5x 3'
6 '
x
3
+3 + 2:c x + 2
.
LITERAL COEFFICIENTS
4.7]
4
5 29.
x
30.
+
x+l
5
~'
3x
+
10
+
6x
+5
5x
-
-
13x
4
5
2x
-
3
3x
-
2
6x
3x
-
2
^ 4x -
1
12x2
2
3 32.
x2
63
5x
+
1
x
1
2x
5z 5x 2
1
2
3x
4
+
2
^
1
llx
+ 1
10
6x2
-
2
7
4x
,
-
-
7
x
-
~
u>
1
4.7. Literal coefficients If
W, by
P
is
the perimeter of a rectangle of length
the relation between P, L, and the formula:
is
width
evidently expressed
P = 2L + 2W.
(1)
Solving
W
L and
(1) for
L,
we have -2L = 2JT-P; 2L = P-2W;
(2)
L =
-=
=
-=
Similarly,
TT
(3)
Results
and
(1), (2),
(3) are different
versions of the
equation, solved respectively for P, L, and
same
W in terms of the
remaining letters. In such equations any one of the letters may be considered as the unknown. Thus, if the perimeters and widths of many different rectangles were given, and we were asked to get the various lengths, the most efficient way = 6, would be to use (2) as a formula. If P = 100 and
W
AA LT = 100-2-6 =44; etc. Similarly (1)
tively.
.,
if
and
D = nn P 90
1A and TJ7 17=10,
(3) are
,
QK Lj = 90-2-10 = 35,
formulas for
P
and
W respec-
LINEAR EQUATIONS IN ONE UNKNOWN
64
Whenever an equation contains other
letters
known, those letters will usually appear equation. For example, given
Ax
(4)
+B=
[Ch.
IV
than the un-
in the root of the
0,
D the root
is
evidently
A-p
Such equations are said to have literal coefficients. It should be understood that the coefficients in an equation include not only multipliers of the unknown, as A in (4), but also the terms,
such as
B
in (4),
which do not contain the
unknown.
Any linear by
(4).
equation in the
For instance,
2ax
(5)
the
A
unknown x may be
represented
in
of (4) stands for 2a,
-
3y
=
and the
0,
B for
3y. Since the root
T>
of (4)
is
-j, A
which represents a
single
number for any given
set of values for the literal coefficients,
we can conclude
that:
A
linear equation has exactly one root.
Clearly an equation with literal coefficients represents inparticular equations with numerical coeffi2 = 0, cients. Special cases of (1), for example, are 3z 17 = 0, etc. An equation like (4) is said to be more 5x finitely
many
+
general than one with numerical coefficients. In algebra, and in fact in all mathematics, it is often desirable to have problems and solutions as general as possible, thus covering
many
cases in a single operation.
In solving an equation with Art. 4.4 may still be used.
Example. Solve
ax
literal coefficients,
the steps of
LITERAL COEFFICIENTS
4.7]
65
Solution.
Step
2.
Step
3.
Not needed
?-- + x + -
Zaxd
or
Step
4-
Step
5.
in this case.
2bc
+
Qbdx
+ Gbdx = x(3ad + 66d) = 3azd
+
Qbd
=
2bc
-
Qbd
2bc
+
66d.
2bc
I2bd.
+
I2bd.
or 2fe)
A
complete check, of course, would require the substitution of the literal root in the original equation. By way of a
and practical partial check, however, we may substitute = 6 = specific numbers for a, 6, c, and d. For instance, if a brief
c
=
=
d
.
root
8
then
1,
26c -
A1 Also
-.
(6)
T
3aa
9
x becomes ~
+ 66d ^, + ooa ,
.
o?+l = 2orx =
l,
1 O
A
+#+
8 u Or becomes - ^ 9
,
,
let
1
a
with the
=2,
=
c
n 0.
=
^ Then
while the root being tested becomes
Qbd
The
partial check
to replace set a
=
6
EXERCISE
which
is
when
shortest,
permissible,
except the unknown by zero. d = 0in the case above?
all letters
=
c
=
Why
is
not
16
Solve the following equations with literal coefficients. The for is in each case x, y, z, u, v, or w.
unknown
to be solved
,
8 + ,.fe^_ o 3. cy
d
=
a
Z cy.
2
.
4.
c
L_2 a cw
bd
_
<
o
=
be
2, dw.
=
o.
LINEAR EQUATIONS IN ONE UNKNOWN
66
7.
=
dv
5. ac
=
6x
ac
be
u
+n_
u
a
6
a
'
II.
-
2r 13 -
+
y 2u a
ie
_ ^_
n
a
n 12.
T'
+ + 2n b
y
u
+ 2n
^/
2m
=
3n
m+
n
a
x
-
n
=
__
bu
ad.
-
f
m+n
m
v
+b _ m n a + x r + 2n a
;
2b
-
a
26
b
2r
n
~
ad
cy *
m+v
;
=
bd
b
2b __
8.
an
n
-
~'~'
CC/
r
~~
+
17. r
w
b
6.
ax.
+
m+n_ m r w
ad.
cv
26* __
f^
2rw.
18.
-
L^^/
4rf//C/
r-r-
jr
= m+ ,
n.
__ Solve each of the following formulas for the
=
a
+
21.
Z
22.
E = I(R + -
23.
T
=
o
=
24
-
-
+ -, +
jtt
'ET 111
25.
P=
A(l
26.
-
=
v
27.
=
31. 32.
R *>
for jB
A
for
*t2
-
I
H~T>
KmM =
f
f
,
r
f
r
for
m
; J
^5 ^ or
m;
P
r
f
a
for
5
^; '
f
f
r v
r r-
, M.
= VQ + ^<, for t; for VQ; for ^. s = afi, for a. A = /t(a + 6), for h; for 6. F = r (a - 6); for a.
29. v
30.
IT) for ^iJ for
M + m'
a
for r; for /?; for n.
for q] for p; for/.
-^^~^
S=
^ rt 28. ,P
l)d, for a; for n; for d.
(n
2
.
-
-
letters
indicated:
[Ch.
IV
4.7J
LITERAL COEFFICIENTS
w _ w, 33. 34.
What 35.
w
One boy is
many
is
a
'
f
r
W
;
f
^
r
2a years old and a second boy
26 years old.
is
their average age?
One
than the 36.
=
-W
67
child
first.
a years old, and a second one
is
What
is
Henry had a marbles. After buying
How many
as John.
A boy who
is
26 years older
their average age?
10
more he had
half as
had John?
had x dimes
found half a dollar and then gave half of the money he had with him to his mother. What was the value of her share in cents? 37.
in his pocket
A man who owed
a dollars paid x dollars on the debt and then owed f of the original amount. Find x. 38.
39. In
making a trip a man averaged a miles per hour going and hour on his return. If he was c hours on the road, how
6 miles per
far
from home did he go?
40. How many pounds of cream testing x% butter fat must be added to y pounds of milk testing z% butter fat to give milk testing
w%
butter fat?
41. If oranges cost c cents per dozen, one buy for a dollars?
EXERCISE
how many
oranges can
17
Solve the following stated problems involving fractions. 1.
If
a certain number
remainder Note.
is
divided by 4 the Quotient
is
3 and the
Find the number.
is 3.
In this and succeeding problems we use the relation
(Art. 1.10)
dividend 77-
:
divisor
=
~
,.
,
,
Quotient H
remainder -TT
.
divisor
Observe the distinction here made between 'Quotient' and 'quotient.' For example, the Quotient of 13 divided by 4 is 3, while the quotient, or total result of the division, 2.
What number
remainder
is
2?
is
divided
by 7
if
is
3 .
the Quotient
is
3 and the
68 3. If
2
is
LINEAR EQUATIONS IN ONE UNKNOWN [Ch. IV added to a certain number and this sum is divided by
3 the Quotient 4.
4 and the remainder
is
One number
is
by
One number
is
2
less
the second the quotient 6.
The
divided
by
7. If
^, the
8.
x
new
difference
is f.
If
the
one
first
is
Find the numbers.
than another. If the first one is f Find the numbers.
is
divided
.
between two numbers is 3. If the larger is ^. Find the numbers.
is
the smaller the quotient is
added to both the numerator and denominator of
fraction formed equals . Find the value of x.
Find the number which must be added to both the numer-
ator and denominator of f to 9.
Find the number.
3 more than another.
divided by the second the quotient 5.
is 2.
What number must be
and denominator 10. If
of
-^
make
the
new
fraction equal to f
.
subtracted from both the numerator
to yield the fraction f ?
a certain number
is
added to the numerator and subnew fraction equals f Find
tracted from the denominator of f the the number.
.
,
The numerator
of a fraction is 2 less than the denominator. added to both numerator and denominator, the resulting fraction is f Find the original fraction. 11.
If 1 is
.
The denominator
of a fraction is 2 more than the numeradded to both numerator and denominator, the new fraction can be reduced to ^-. Find the original fraction. 12.
ator. If 11
13.
The denominator
numerator.
new
is
If
3
is
more than 3 times the both numerator and denominator, the
of a fraction
added to
is 1
fraction equals . Find the original fraction.
4.8. Geometric
Many
problems
problems that are geometric
in character occur in
one's normal experience. They vary widely and include lengths, areas, volumes, relative sizes of angles, etc.
Example 1. A 5-foot string is cut into two pieces, one which is - of the other. Find the length of each piece.
of
4.8]
GEOMETRIC PROBLEMS
69
Solution.
x
Let
Then
=
no.
ft.
in the longer piece.
=
no.
ft.
in the shorter piece,
o
+j=
*
(1)
sum of the parts (1), we have
since the
Solving
x
=
5,
of anything equals the whole of
,
the greater length;
,
the smaller length.
it.
7
4x
20
=
o
y
,
.,
,
Example 2. Two rectangles have the same width. One is 2 inches longer than the other and 5 inches longer than its width. If the difference in their areas is 10 square inches, find the dimensions of each.
Solution.
Let x
x
+5 x + 3
= = =
the width of each rectangle in inches. the length of the longer rectangle, in inches. the length of the shorter rectangle, in inches.
x(x Solving,
+
we have x =
-
5)
5;
x
x(x
+
5
+ =
3)
=
10;
x
10.
+
3
=
8.
Hence the rectangles are 10 by 5 and 8 by 5 inches
re-
spectively.
Note.
The area
The volume height h
is irr
Example
of a circle of radius r
is irr
2
square units.
of a right circular cylinder of radius r h cubic units.
and
2
3.
Each
of
altitude of 10 inches.
two
The
right circular cylinders has an radius of the base of the larger is
two inches greater than that of the other. The difference in volumes is 80?r cubic inches. Find the radius of the
their
base of each cylinder.
LINEAR EQUATIONS IN ONE UNKNOWN
70 .
[Ch.
IV
Solution.
= radius of base of smaller cylinder, in inches. + 2 = radius of base of larger cylinder, in inches. 107TX = volume of the smaller cylinder, in cubic inches. = volume of the larger cylinder, in cubic inches. I0ir(x + 2) - lOTra: = SOvr. 107r(o; + 2) (2) Solving (2), we find that x = 1 and x + 2 = 3. x
Let x
2
2
2
Example
4-
The
second angle and
first is
2
angle of a triangle is equal to -J- of the of the third angle. Find
also equal to
the three angles. Solution.
Let
a:
Then 3x 2x
= = =
no. degrees in the first angle. no. degrees in the second angle. no. degrees in the third angle.
x Solving,
we have x =
EXERCISE Form
+
3z
30,
+ 2x
3x
=
=
90,
180.
2z
=
60.
18
algebraic equations
and
solve the following problems.
A
1. 7-foot cord is cut into two pieces so that one piece as long as the other. Find the length of each piece.
2.
A
form two
is
f
piece of wire 4 feet long is cut and the pieces are bent to circles. If the diameter of one circle is f that of the other,
find the length of each piece. 3.
One
bined length 4.
is
A 26-inch
as the second, is
3 inches longer than another. If their com17 inches, find the length of each.
string
is
cord
is
cut into three pieces.
and the second
first is
f as long as the third.
is
f as long
How
long
each? 5.
Some
6.
two pieces to as long as the other.
12-foot boards are to be cut into
two boxes, one of which should each board be divided? lids for
If
The
The width
of a rectangle
the length of the rectangle
is
is
^
and the
make
How
side of a square are equal.
5 inches
more than
its
width, and
PROBLEMS INVOLVING TIME, RATE, DISTANCE
419]
71
area is 50 square inches more than that of the square, find the dimensions of each.
its
7.
and
The radius of one by
27?r
The diameter
of
one
8.
another, and
its
area
3 inches more than that of another, square inches. Find the radius of each.
circle is
their areas differ
is
circle is
greater
by
4 inches more than that of 12?r
square inches. Find the
diameter of each. 9.
Two
triangle
have equal bases. The two sides of each base and altitude. One triangle is isosceles but
right triangles
form
its
4 inches longer than its base. If their areas differ by 12 square inches, find the dimensions of each. the altitude of the other
is
10. The altitudes of two triangles and the base of one of them are equal. The base of the other is 6 inches more than its altitude, and the difference in the areas is 21 square inches. Find the base
and
altitude of each.
The radius of the base of one right circular cylinder inches more than that of another, and the altitude of each 11.
3
15 inches. If the difference in their volumes
is
is is
2257r cubic inches,
find the radius of the base of each.
One
of the acute angles of a right triangle is 10 the other. Find the number of degrees in each angle. 12.
13.
The
first
14.
The
vertical angle of
less
than
angle of a triangle equals f of the second, and the third equals the sum of the first and second. Find each angle.
an isosceles triangle
is
20 more than
sum of the equal angles. Find each angle. 15. Work problem 14 with '20 more' replaced by '4
the
less.'
16. The first angle of a triangle is as large as the second as large as the third. Find each angle. I
4.9.
Problems involving time,
Many (1)
rate,
and
and distance
problems are solved by means of the formula d
=
rt,
where d represents the number of units of distance, r the number of units of distance traveled in one unit of time, and t the number of units of time. For brevity, d, r, and t are
LINEAR EQUATIONS IN ONE
72
called distance, rate,
and
time. It
UNKNOWN
[Ch.
PV
must be understood, how-
ever, that (1) holds good only if the rate is constant, such as it is, for instance, in a car whose speedometer needle re-
mains
fixed at the 40 m.p.h. (miles per hour) position. Equation (1) is used, of course, to find the distance
the rate and time are known. Solved for r and
two
t, it
when
takes the
alternate forms, r
(2)
,
and
=
(3)
used respectively to find r and
t
when
the other two quanti-
known.
ties are
EXERCISE Form 1.
.
19
equations and solve the following problems.
If
one car runs 40 m.p.h. and another car runs 50 m.p.h., will the sum of their distances be 300 miles?
in
how many hours
x
Let
Then
40:c
40z
+
5Qx 50z 90z ^
= = = = = =
no. hrs. required. no. mi. first car runs. (Here no. mi. second car runs.
rt
=
40.)
300. 300.
W=
2. A car running north at 45 m.p.h. passes point A at 12 o'clock. A second car running north at 50 m.p.h. passes A at 1 :30 o'clock.
When will
the second car overtake the
first
one?
3. A is 450 miles west of B. A car starts east from 40 m.p.h., and at the same time another starts west from 45 m.p.h. In how many hours will they meet?
4.
A B
at
at
Two men are 400 yards apart and walk straight toward each
one walks 80 y.p.m. (yards per minute) and the other 90 y.p.m., in how many minutes will they meet?
other. If
4.9]
PROBLEMS INVOLVING TIME, RATE, DISTANCE 5. Two men run in the same direction around a
73
440-yard
one runs 10
y.p.s. (yards per second) and the other f as in how fast, many seconds will the faster one gain a lap? 6. One man can run 9 y.p.s. and the other can run 8 y.p.s. If they start at the same time and run in opposite directions around a 440-yard track, in how many seconds will they meet?
track. If
A
car starts east at 40 m.p.h. A second car starts east from later at 50 m.p.h. When will the second car be 10 miles ahead of the first one? 7.
same point one hour
the
A
car starts south at 40 m.p.h. One hour later a second car starts south from the same point. If the second car overtakes the 8.
first
one
9.
in three hours, find its speed.
A man
walks toward a certain town at 2 m.p.h. One hour
man starts from the same place at 3 m.p.h. He overtakes the first man just as he reaches the town. How far was it to town and how long did each man require for the trip? later a
second
10. A train that runs 50 m.p.h. passes a station 2 hours behind a slower train and overtakes it in 3 hours. Find the speed of the
slower train.
current in a certain stream flows 3 m.p.h. A crew can row twice as fast downstream as it can row upstream. How fast 11.
can
it
The
row
in still water?
x
Let
Then
x
+3
x
3
x Solving,
+
3
= = = =
The
f times it
-
3).
as fast
downstream as
in still water?
13.
The
it
upstream
still
water?
as
it
can row downstream.
rate of the current
is
downstream in the time required fast can it row in still water?
A
crew can row can row upstream. How fast can
A How
current in a stream flows 4 m.p.h.
fast
The
9.
current in a stream flows 2 m.p.h.
row
14.
2(x
we have
x= 12.
rowed in still water. rowed downstream. no. m.p.h. rowed upstream. no. m.p.h. no. m.p.h.
3 m.p.h. for
crew can row f as fast can it row in
A crew
can row 9 miles
rowing 4 miles upstream.
How
LINEAR EQUATIONS IN ONE UNKNOWN
74
[Ch.
IV
15. Work problem 14 if the current flows 2 m.p.h. and the downstream and upstream distances are respectively 18 and 6 miles. 16. A man can row 4 m.p.h. in still water. He can row 12 miles downstream in the time required for rowing 4 miles upstream. Find the rate of the current. 17.
day
An
its
airplane has a cruising speed of 300 m.p.h. On a certain ground speed when traveling with the wind was twice its
speed against the wind. 18.
How
fast
was the wind blowing?
On a certain day the wind at 5000 feet was blowing 50 m.p.h.,
at 10,000 feet it was blowing 20 m.p.h. in the same direction. in 3 hours, plane flew at 5000 feet with the wind from A to
and
A
B
and returned at the 10,000 feet level in 5 hours. Find still air and the distance from A to B.
4.10. Problems concerning
its
speed in
money
Most
practical problems about money deal with principal, interest, rate of interest, wages, profit and loss. In some other
problems, usually less practical in nature, the object is to find the number or denomination of pieces of money involved.
The The
principal is the sum of money that bears interest. rate is the fraction of the principal paid for its use
during a certain period of time
usually a year. Stated in this fraction multiplied by 100. Thus, the
percentage, it is rate of or .06 becomes
^
'6%.' The time is the interval during which the principal is used. The interest is the total sum paid for use of the principal. The amount is the sum of the principal and interest. In formulas the following notation
is
customary.
A represents the number of dollars in the amount. P represents the number of dollars in the principal. / represents the number of dollars in the t
represents the
number
interest.
of interest periods in the time in-
terval.
r
represents the rate, written as a decimal fraction.
4.10]
PROBLEMS CONCERNING MONEY
We
75
shall use the following formulas.
/
(1)
=
Prt.
Solving (1) for P,
r,
and
t
successively,
we have
P-1
(2)
-
r
(3)
-jre
and
-
.
(4)
A = P+
(5)
From
and
(5)
When
from the
definition.
(1),
A = P
(6)
7,
+
Prt
=
P(l
+
rt).
two or more investments, subscripts may be used to distinguish between them. For example, PI, read 'P sub-one,' means the first principal, P2 the second principal, etc. Similarly, 7i means the interest on P x 7 2 is the interest on P 2 etc. there are
;
,
Example L A man invests $6000, part at 6%. The total interest for one year is $320.
5% and
part at
How much
was
each investment?
two investments be P and 6000 PI. = = r t .05 and for the second, investment, 1;
Solution. Let the
For the r
=
.06
first
and
t
=
1.
Substituting in (1)
we have 7!
and
72
But Hence .05Pi
+
Solving,
and
7i
(360
-
+7
2
.06Pi)
= = =
(6000 -Pi)(.06)(l)= 360 -.06Pi. 320, the total interest.
=
320.
= =
4000, 2000.
Pi(.05)(l)
we have 6000
-
P! PI
=
.05Pi,
LINEAR EQUATIONS IN ONE UNKNOWN
76
What
2.
Example 2 years at
[Ch.
principal will yield $24 interest
IV in
6%?
=
Here /
Solution.
24, r
known. Using these values 24
_ p -
.06,
in (2),
_ -
24
12
i
=
and
2,
P
is
the un-
we have
_ -
2400
~w
Two men work
3.
Example
=
6 days and receive $54 as as much per day as the other, find
wages. If one receives the daily wage of each. Solution. Let
x
=
the daily wage of the second
man
in
dollars. 4:r
Then
=
o
X
_ O_ ^tvC
/*
Solving,
I
-j-
JL
5
first
man
in dollars.
f^ {
U.
6
we have x
EXERCISE 1.
the daily wage of the
=
5,
20
How much interest 5% for 2 years?
will
be earned by an investment of
$300 at 2.
An
investment of $200 yields $8 interest in a year. Find the
rate. 3.
At what
rate will $250 yield $20 interest in 2 years?
4.
At what
rate will $3500 yield $350 in 3 years?
5.
At what
rate will $4700 yield $282 in 2 years?
6.
Find the time required
7.
A sum
of $7500
is
for
$6500 to yield $650 at 5%.
invested, part at
annual interest on the two investments investments.
part at 5%. The $400. Find the two
6% and is
4.10]
PROBLEMS CONCERNING MONEY
77
8. Part of $8000 is invested at 4% and the rest at 5%. The 4% investment yields $50 more interest in a year than the other. How much is invested at each rate?
One part
of $9000 is invested at 4% and the rest at 5%. incomes the two thus yielded are equal, fincj each investment. 9.
If
10.
A certain sum invested at 6%
as an investment of $6000 at 5%.
What
is
the
same yearly income
sum
invested?
one
Two men worked 9 days and together received $144 in wages. man received $7 per day, how much did the other get?
12.
A boy received
11. If
yields the
as
much pay
as his father. If together they
got $48 for 4 days' work, what was the daily wage of each?
two common laborers and four skilled workmen receive $80 per day altogether, and if the wages for skilled labor are twice as much as for common labor, find the wage of each. 14. Ten men were employed, some at $6 per day and some at $8 per day. The total daily wages amounted to $68. Find the num13. If
ber employed at each wage.
Nine men were employed, some at $7 per day and some at $9 per day. If the total daily wages amounted to $75, find the number working at each wage. 16. P = 325, t = 2, r = .06. Find I. 15.
17.
18. 19.
20. 21.
22. 23.
= 3, r = .05. Find A. 450, = = = 100. Find r. 7 3, P 18, 7 = 60, P = 500, r = .06. Find A = 770, = 2, r = 5%. Find P and 7. A = 690, = 3, r = 5%. Find P and 7. 7 = 80, = 2, r = .04. Find P. A man has $1.25 in nickels and dimes.
P=
times as
t
t
t.
t
t
t
many
nickels as dimes, find the
x
Let
Then
3x lOx 5(3x) l(te
Solving,
+
15x
= = = = =
x= =
there are three
of each.
no. dimes. no. nickels.
the value of the dimes in cents.
15x 125.
we have 3z
If
number
5.
15.
=
the value of the nickels in cents.
LINEAR EQUATIONS IN ONE UNKNOWN
78
24.
many 25.
dimes. 26.
dimes
A man
[Ch.
IV
has $3 in nickels and dimes, there being twice as nickels. Find the number of each.
dimes as
The sum
of $1.90
is
made up
of quarters plus f as
many
How many of each are there? A boy has twice as many nickels as dimes and twice as many as quarters. How many of each has he if the total sum is
$1.30? 27. A man has $3.25 in nickels, dimes, and quarters. He has the same number of dimes as of nickels, and twice as many quarters as dimes. Find the number of each. 28.
A man
has f as
quarters as nickels, and as many of the other coins. How many of each
many
dimes as the total number has he if their total value is $2.30? 4.11.
Mixture problems
Many problems arise from the forming of mixtures of various materials. When two or more substances are mixed, an equation may be formed on the basis of the fact that the quantity of a certain substance in one material plus the quantity of this same substance in a second material is equal to the quantity of that substance in the mixture formed by combining the two materials. This same principle will
extend to any mixture containing a given number of
materials.
When
the value of a mixture
is
the matter of chief in-
Example 2 below, we form the equation not on the basis of quantity but on the principle of directly value. When two substances are mixed we say that the value terest, as in
of the first one plus the value of the second equals the value of the mixture.
Example 1. How much metal that is 5% silver must be added to 10 pounds of metal, 2% of which is silver, to form a mixture having
3%
silver?
Solution.
Let
x
=
no. Ibs. of metal containing
5%
silver.
MIXTURE PROBLEMS
4.11]
Then
= = = =
.05z (.02) (10)
(.03) (#
+
.05z
+
10)
(.02)(10)
Solving,
79
no. Ibs. of silver in the first metal; no, Ibs. of silver in the second metal; no. Ibs. of silver in the mixture; (.03) (x
+
10).
we have
=
x
5.
Example 2. How many pounds of coffee worth 28^ per pound must be added to 40 pounds of coffee worth 35^ per pound to form a mixture worth 30 per pound? x 28x
Solution. Let
35(40)
30(x + 40) 28x + 35(40) Solving,
EXERCISE
the value of the 28
the value of the mixture in cents.
30(z
+ 40).
=
100.
21
alloy to form a
How many pounds of
12% copper
be added to 20 pounds of alloy?
cream containing 25% butter
be added to 30 pounds of milk containing mixture containing 15% butter fat? 3.
20%
coffee in cents.
the value of the 40 ji coffee in cents.
How much 10% copper alloy must
15% copper 2.
no. Ibs. of 28 j4 coffee added.
we have x
1.
= = = = =
How much 30%
5%
fat
must
butter fat to form a
gold alloy must be added to 25 ounces of
gold alloy to produce a
27%
gold alloy?
4.
How much
20%
5.
How much
pure copper must be taken from 35 pounds of it to a 25% copper alloy?
pure gold must be taken from 23 ounces of gold alloy to reduce it to a 15% gold alloy?
40%
copper alloy to reduce
6.
35%
How much
silver alloy to
7.
How much
pure silver must be taken from 45 ounces of reduce it to a 20% silver alloy? candy, worth
per pound, must be mixed per pound to form a mixture
25ff
with 20 pounds candy worth 13 of candy worth 20jzf per pound? of
}f
UNKNOWN
LINEAR EQUATIONS IN ONE
80 8.
Two kinds of candy worth
15j
and
25ff
is
is
How much
of
used?
Two
kinds of coffee worth 30 jf and 22 jf per pound are mixed form 100 pounds of 25 per pound coffee. How much of each 9.
to
IV
per pound are mixed
to form 30 pounds of candy worth 18ji per pound.
each
[Ch.
used? 10.
A lady bought
10 pounds of grapes for $1.35.
sold for 10f per pound and pounds of each kind did she
some
for 15^ per
Some of them
pound.
How many
buy?
A
dealer bought 15 cases of fruit for $41.50. Some cases cost $3.00 and some $2.50. How many cases of each did he buy? 11.
4.12. Lever problems
A
a mechanical device by which a force applied at one point is transferred to a second point and there intensified. When we roll over a huge rock with a crowbar we apply this principle. Many problems in algebra are concerned with levers and related mechanical devices such as pulleys. lever is
Some preliminary definitions are necessary. The fulcrum is the non-moving support upon which the lever swings when force is applied to it. The moment of any force, for a given set of units, is equal to the number of units in the force multiplied by the number of units in the distance
from the fulcrum to the point on the
lever where the force
applied.
What
Question.
is
is
the
moment
of a force of 10
pounds
applied 5 feet from the fulcrum of a lever?
Z
I T 10 Ib
Fulcrum
ZL I
1
Weight Fig. 2
Answer.
Moment =
5
10
=
50 foot pounds, for the given
units.
The
shown
above diagram causes the lever to rotate about the fulcrum in a counterclockwise direction force
in the
LEVER PROBLEMS
4.12]
81
the figure. Any force acting downward on the lever to the right of the fulcrum, such as the indicated weight, as
we view
would tend to rotate the lever
in the opposite, or clockwise,
direction. If
the
sum
of the
moments
the lever counterclockwise
moments
of all forces tending to rotate equal to the sum of the
is
of all forces tending to rotate
lever will be balanced
it
clockwise, the lever prob-
and stationary. In most
lems we calculate either the force or distance necessary to
make
the lever balance.
Example
1. If
a force of 50 pounds is applied 5 feet from must a weight of 60 pounds
the fulcrum of a lever, where be placed to balance it? Solution.
x
Let 5
50 QOx 60z z
(A diagram similar to
= = = =
no.
250 the
ft.
Fig. 2
may
be drawn.)
from the fulcrum to the weight.
=
the moment of the 50-lb. force. moment of the 60-lb. weight.
250.
^
s
^at the weight must be 4 ft.,
2
in.
from
the fulcrum.
In the above example the weight of the lever is not considered. This weight may be important, however, in the case of a heavy lever such as a plank. To allow for it the lever is considered as composed of two parts called arms, each of which lies wholly on one side of the fulcrum. If the lever is
uniform, or of the same weight for each unit of length, the
moment
of
an arm
may
be found by multiplying
its
weight
by length, as expressed in the units chosen for the problem. In other words, the moment of a lever arm equals the moment of a force equal to its weight applied at one-half
its
a point halfway from the fulcrum to the end of the arm.
Example 2. The fulcrum and 4 feet from the other.
of a lever If
the
is
10 feet from one end
beam weighs 8 pounds
per
LINEAR EQUATIONS IN ONE
82
what weight must be placed
foot,
arm
at the
UNKNOWN
[Ch.
IV
end of the shorter
to balance the lever?
80
Ib
Fig. 3
The weights of the two arms are 80 and 32 pounds respectively. We must equate the counterclockwise (c.c.) Solution.
and clockwise x
=
80
5
=
the
32
2
= = =
the
Let
moments.
(c.)
ho. Ibs. of force applied at the end of the shorter arm.
moment
of the left
arm
in the
arm
(c.).
diagram
(c.c.).
4#
+
4z
32
Solving,
2
moment moment
the
80
of the right of the force
x
(c.).
5.
we have x
=
84.
two persons are carrying a weight swung on a pole, we may consider the weight as a fulcrum and the supporting If
forces as the acting ones. This
then a lever just like the others we have discussed, except that it is upside down. is
Example 3. Two persons carry a weight of 100 pounds swung on a 9-foot pole. Where should the weight be placed so that one person will carry 60 pounds and the other 40 pounds?
60
100 Ib
Ib
40
Ib
Fig. 4
Solution.
Let 9
x x 60z
= = =
no. of no.
ft.
40(9
from weight to man carrying 60 from wt. to man carrying 40 Ibs.
ft.
-
x).
Ibs.
4.12]
LEVER PROBLEMS
Solving,
we have 9
EXERCISE 1.
A
83
-
x x
= J#, = Y,
or
3f
.
or 5f.
22
90-pound boy and a 60-pound boy balance on a 12-foot
teeter board.
Where
is
the fulcrum?
2. What weight must be placed 6 feet from the fulcrum to balance a 70-pound weight 8 feet on the other side of the fulcrum?
3. How much weight can a 160-pound man raise with a 10-foot beam placed so that the weight is 2 feet from the fulcrum? *
4.
What
force
with a 12-foot
is
beam
necessary to raise a weight of 1000 pounds placed so that the weight is 3 feet from the
fulcrum? 5.
What
6.
If
weight must be placed 5 feet from the fulcrum to balance a 12-pound weight 4 feet on the other side of the fulcrum? the fulcrum
is
beam whose arms and 40 pounds respectively,
4 feet from each end of a
are uniform but unlike, weighing 75
where should a 50-pound weight be placed to balance the beam? 7.
long.
Two men Where
carry a 90-pound weight swung from a pole 8 feet must the weight be placed so that they will carry
40 and 50 pounds respectively? 8.
A
12-foot
beam weighs
9.
A
14-foot
beam weighs 8 pounds per
7 pounds per foot. If the fulcrum is 4 foot from one end, what force must be applied at the end of the shorter arm to balance the beam? foot. If the
fulcrum
is
6 feet from one end, what weight must be placed at the end of the shorter arm to balance the beam? 10.
What
force applied 7 feet
from the fulcrum
will
balance
will
balance
1500 pounds 2 feet from the fulcrum? 11.
What
force applied 8 feet
2000 pounds 2
feet
from the fulcrum
from the fulcrum?
* In this and subsequent problems subject to two interpretations, assume that the fulcrum is between the applied forces.
LINEAR EQUATIONS IN ONE UNKNOWN
84
[Ch.
IV
What weight placed 3 feet from the fulcrum can be balanced a by 170-pound force applied 7 feet on the opposite side of the fulcrum? 12.
13.
lever 14.
What if
weight can a 110-pound boy balance with a 12-foot the weight is 2 feet from the fulcrum?
Two men
and placed 4 15.
carry a 70-pound weight swung on a 10-foot pole from one end. How much does each carry?
feet
Two men
end of a 9-foot
carry a 120-pound weight swung 4 feet from one pole. How much does each carry if the pole itself
weighs 20 pounds? 16.
Two men
The one who
is
carry a weight of 150 pounds swung on a pole. 4 feet from the weight carries 90 pounds. How long
the pole?
is
A weight of 300 pounds rests over the end of a lever 6 inches from the fulcrum. A boy weighing 60 pounds can just lift the weight. How long is the lever? 17.
4.13.
Work problems
problems involve the rate at which an action is being performed. For example, if a man can do a piece of work in 10 days, his rate of work is to do -fa of it per day.
Many
oc
he works x days, he has performed
If
of the total work.
*
The guiding
in problems of this type is the fact principle that the product of the rate of work by the number of units
of time involved
is
equal to the fractional amount of the
work done. Example. If A can do a piece of work in 20 days and B can do the same work in 25 days, how many days will both need to do the job when working together?
if
* Naturally, good judgment must be used in applying the principle. For example, one dentist can fill a tooth in one hour, it does not follow that two dentists could
fill it
in half
an hour.
4.13]
WORK PROBLEMS
85
Solution.
Let
x
=
the no. of days needed for both to do the work.
-
=
the fractional part both do in one day.
=
the fractional part
A
does in one day.
=
the fractional part
B
does in one day.
C
2J
Zo
1 4- 1
=
1
x
25
20
Clearing fractions (multiplying both
members by lOOx)
we have 5x
+
or
EXERCISE 1.
A
4x
=
x
=
100,
100
- - ,
= Hi
,
days.
23
can do a piece of work in 30 days and B can do the same How long will it take for both to complete the job
job in 8 days. if
they work together? 2.
One pipe can
fill
a certain tank in 25 minutes. After this
pipe has been running for 10 minutes, it is shut off and a second pipe is opened. The second pipe finishes the filling in 30 minutes.
How
long would tank alone?
it
have taken the second pipe to have
filled
the
3. A can do a certain amount of work in J the time B requires; can do the same amount in the time C needs. The three together can do the work in 36 days. How long will it take each of them to do the work alone?
B
4. it
A
large pipe
in 18 minutes.
pipes to
fills
How
the tank
a tank in 12 minutes and a small pipe fills long will it take one large and three small
they are
opened simultaneously? John started a job which ordinarily took him 5 hours, and quit after working hours. James finished the job in 8 hours. How long would it have taken James to do the whole job by him5.
self?
fill
if
all
LINEAR EQUATIONS IN ONE UNKNOWN
86
[Ch.
IV
B can do it in 25 hours. been working together for 14 hours, they are joined by C and finish the job in 6 more hours. How long would it have taken C to have done the whole job alone? 6.
After
7.
A can paint a A and B have
One
faucet can
42 minutes. if
house in 40 hours and
How
long
fill
a tank in 18 minutes and a second in
will it
take to
both faucets are opened at the
fill
three-quarters of the tank
same time?
takes John a days to do a piece of work and James b days to do the same work, how long will it take for both to do the 8. If it
work together? 9. A and B together can do a
piece of
works a times as
time each would require alone.
10.
A
list
in 15 hours
of
fast as B, find the
names can be typed
by another.
the work together?
How
in 5 days. If
A
hours by one typist and it take them to complete
in 10
long will
work
Chapter Five
FUNCTIONS AND GRAPHS
5.1.
Functions
Perhaps the most important word in mathematics is 'function.' This being the case, it is certainly advisable to study the meaning of the word as long as necessary to understand it. A function * of x is an expression which has a special value to go
with each value assigned
Examples. 2x in the next x
1;
x
men you
+
see,
to x.
3; x]
who
x
will
+
-;
the
number
5
be more than
of
men,
six feet tall.
a function of x? The reason Why, is that it has an associated value for each value we assign 1 = 1; for x = 2, it beto x. For x = 1, it becomes 2-1 for instance,
comes 2-2
1=3;
is
2x
1
and so on.
3 Similarly, examples of functions of y are 3y, y 3 some functions of z are z 2 -, etc. 1, 2z,
+
z
7,
-;
^
2t
be seen that any mathematical expression containing a letter is a function of that letter. Such quantities are called mathematical functions, and are illustrated by all but It will
one of the examples above. The one non-mathematical function there listed is 'the number of men, in the next x men you see, who will be more than six feet tall.' For every value of x there will be a definite value for this quantity (which *
Some
consider
functions of x have
them
more than one value
here.
87
for a given x;
but we need not
FUNCTIONS AND GRAPHS
88
makes
it
a function of x) but ;
that value will be.
An
we cannot
[Ch.
V
advance what
tell in
important goal in science
to express such related quantities as mathematical functions of each other. This goal has been reached
is
many times,
as for example thrown up vertically and the height it reaches, or the volume of a pound of specified gas and the pressure it exerts on the container. Again, accurate knowledge of the positions of the sun, earth, and moon as functions of the time has enabled astronomers
with regard to the
initial
speed of a stone
to predict eclipses far in the future.
5.2.
Notation concerning functions
A
symbol which can represent any function of x is f(x) read '/ of x.' If in a given problem f(x) stands for some particular function of x, then other symbols needed to represent other particular functions of x in that problem can be written variously as g(x} h(x), F(x) (read 'g of x' 'hoi x^ 9
y
F
of 2,') etc. 'capital = 2x2 3, then /(O), if, in a given problem, f(x) read '/of zero/' means 'the value of f(x) when x = 0,' and
+
Now
equals
2
O2
+
/(-2) = 2(-2)
3
2
+
=
3.
3
=
Similarly, /(I) 11, and so on.
=
2
I2
+3=
5,
A
function of two variables, say x and y, can be represented asf(x, y) and is read as '/ of x and y' Thus /(I, 2) y
would read '/of 1 and 2' and would mean the value = 1 and y = 2. f(x, y) for x
Example Answer.
Example
1.
-
(3
1)
2. If f(x)
Answer. (4
2) (3 3.
Example ,
=
If f(x)
If
-1).
+
=
+
3x
and F(x)
= -1 and g(y) = y
(-2)
4x
7)
1
=
f(x, y)
3
8
10
=
3z
= 2
8
+
=
x3
,
of
evaluate
= -9. 7,
find the value
80.
2xy
+
6,
evaluate
5.2]
NOTATION CONCERNING FUNCTIONS
- 2(1)(3) +6 = 3-6 + -1) = 0-2(0)(-l)+6 = 6; =
Answer. /(I, 3) /(O,
3(1)
2
-1)=3-6 =
,
EXERCISE
89
6
=
3;
18.
24
Given f(x)
=
+
2x2
3z
2,
find the values of the expressions in
problems 1-10. 3.
6.
/(-a).
=
+
3z 2
2 and F(z) expressions in problems 11-20. (ru>en /(z)
11. /(O)
-
12. /(I)
F(0).
15.
14. /(O)F(O).
17.
/(IW).
Gwrcn F(z,
+
2z 2
/(a
-
3,
^2 ^.
*J
10. /(a
6).
-
6).
find the values of the
-
F(l).
16.
19.
=
+
13. /(2)
F(l).
.
18.
?/)
9.
S-/^)
7./(~)-
=
/(-I).
G (x
V)
>
.
=
20.
L _ 3^ / ^ ^ l
w^^
of
(?(-5,
0).
e
the expressions in problems 21-25.
24. F(2,
-3)
23. F(0,
22. (7(0, -1).
21. F(0, -1).
+
-5)
25.
3G(-1, -2).
26. Express the area of a circle as a function of its radius,
r.
27. Express the area of a triangle with a base of 10 inches as a
function of
its altitude, h.
28. Express the area of a triangle
function of
its
base,
whose altitude
is
6 feet as a
b.
29. Express the area of a square as a function of its side,
30. Express as a function of t the in t seconds at 5 feet per second.
31. Express as a function of t
t
seconds at sixty miles per hour.
the
man
s.
number
of feet a
number
of feet a car goes in
walks
FUNCTIONS AND GRAPHS
90
5.3.
[Ch.
V
The coordinate system
The rectangular coordinate system is a simple and widely used device which enables us to show, by means of a picture, or graph, the manner in which the value of a function of x changes along with the value of x. Because it was invented by a seventeenth-century Frenchman named
it is
Descartes,
sometimes called the cartesian coordinate system.
P(*.y)
IV
III
Fig. 5
The device consists of two mutually perpendicular lines, one horizontal and one vertical * (Fig. 5), drawn on a plane. The
horizontal line
called the X-axis; the other, the Y-axis; and, together, they are the coordinate axes. They divide the plane into four parts called quadrants, which are
numbered as shown
is
in the figure. Their intersection, the
called the origin. point 0, is located point such as is
A
P
OA
distances
zero
when
is
P
AP
or y
of
The
two
hori-
is
P
is
This
and
in the figure.
the abscissa, the x-distance, or simply the is on the right of the F-axis, positive when
zontal distance
x of P, and
or x
on the plane by means
on
this axis,
and negative when
P
is
to the
conveniently indicated for memory purposes by the arrow on the right end of the X-axis, which shows the positive horizontal direction. Similarly, the vertical distance left of it.
is
the ordinate, y-distance, or y of P, and is positive, zero, or negative according as P is above, on, or below the X-axis, and also as indicated by the arrow pointing upward on the is
* In this statement as well as in the subsequent discussion, we assume for simarrow on the F-axis points upward, as on a vertical blackboard.
plicity that the
THE GRAPH OF A FUNCTION OF
5.4]
x
91
F-axis. Together, x and y are called the coordinates of P. It is in any quadrant important to note that the coordinates of are plus x and plus y, though x stands for a negative number
P
quadrants II and III, as does y in quadrants III and IV. Beginning at the origin, we mark in advance equal units on each axis on a scale to fit the problem. Then, in locating, or plotting a point with given numerical coordinates, designated here as x and y, we measure x units to the right in
or left according as the #- value
reaching a point on the #-axis.
positive or negative, thus
is
From
this point
we measure
downward to the designated point. Thus, ( 2, 3) we measure 2 units to the left of the origin and then 3 units downward. The point A, along with others, is shown in Fig. 6. Note that a capital letter before
y units upward or to plot
A
the designated coordinates, while not necessary, when the point is to be referred to again.
is
convenient
r -0(0,3)
I
>
*
I
D{-4,2)
I
<
b(o,o) H 1
1
B (3 2 )
*-/
1
(-3,0)
A (-2 ,-3) Fig.
5.4.
6
The graph of a function of x
Using the coordinate system,
let
or
picture of the special function z
may
y
(i) *
designate this function of x
=
+
+
us find the geometric 1-
by the
For convenience we letter y, so that
i.*
The meaning of the word function is here illustrated, thus: Choose any numTake one-half of it. Add 1 to the result. Call the sum y. Thus, the value of
ber x.
y depends upon the original value chosen for
x.
FUNCTIONS AND GRAPHS
92
By means
of (1)
we may
[Ch.
V
find the value of y to pair with x, listing them in a brief
each of several small values of table thus:
x
-2
-3
-1
1
2
3
V
The
points whose coordinates are listed are seen in Fig. 7 to lie in a straight line. It can be shown that all other points obtained from (1) lie also on this line. The line is said to be or the graph of the function -
+
1,
and
also the
graph or locus
of equation (1) as well as of
2y
(2)
obtained from
(1)
by
=
x
+
2,
clearing fractions.
K5)1
f
2 y=2x -3x
J(I,0) Fig. 7
Fig.
2 Similarly, the graph of the function 2x
8 3#, or of the
equation (Q (3)
4,
y
turns out to be the curve
The
O~, OX
9~.2 LX
shown
in Fig. 8.
coordinates of the points plotted are
shown
following table.
x y
^1
1424 20-1-1025
-i
i
in the
VARIABLES AND CONSTANTS
5.5]
EXERCISE
93
25
Plot the points having the following pairs of coordinates. (Remember number is the value of x.)
that the first 1.
2. (1, 4).
(3, 1).
6.
(0, 0).
7.
10. (0,3).
11.
Make a
3. (2, 3).
4. (3, 5).
5.
1).
8. (3,
-4).
9.
(-2,0).
12. (0,
-1).
13.
(-2,
and
table of coordinates
plot the
(2, 6).
(-1, -2). (4,0).
graph of each of the follow-
ing functions. (Hint: Let y equal the function.)
26. x 2 29. 2
-
32. 2x 2
35.
38.
x
-
2
30. 1
3x.
33. x 2
-
36.
x
~x2
39.
x
28. x 2
- 2x - 4x 2
.
x
5.3.
+ 2.
27. x2
.
31. x
.
+
1.
34. x 2 37. -
x
1
^r-
2
40.
1
x
-
2.
+ x - 1. + 2x - 2.
+
1.
-
1.
Variables and constants
When
a
letter
may
take on an unlimited
number
of values
in a problem, as x or y may do in equations (1), (2), and (3) of the preceding article, it is called a variable. If the equation is solved for y thus being in the form y
(1)
y
=
/(*),
called the independent variable, and y, or /(x), the dependent variable. Similarly, in the equation
x
is
(2)
*
=
FUNCTIONS AND GRAPHS
94
[Ch.
V
the independent variable. To find corresponding pairs of values it is convenient to assign values to the independent
y
is
variable.
A letter representing the same number throughout a problem is called a constant. Constants and variables are usually taken respectively from the first and last parts of the alphabet. By common agreement some constants frequently represent the same number
many
in
different problems, as in the case of
TT,
the
ratio of the circumference to the diameter of a circle.
The graph of a linear function of x
5.6.
A
a rational integral function of the where a and 6 are constants and a ^ 0.
linear function of
form ax
+
6,
Examples. 3x It is
shown
a:
2, x,
in the
is
o
,
-
5
2i
branch of mathematics called analytic
geometry that the graph of the function ax
+
6,
or of the
equation
y
(1)
=
ax
+
6,
Evidently it crosses the F-axis at the point (0, 6). The second point needed to determine the line may be found by assigning any value to x at will. Thus, 1 goes through (0, 1) and (2, 3); y = 2x, through y = x a straight
is
line.
+
and
(0,0)
5.7.
(3, 6).
The graph of a linear equation in one or two
The
general linear equation in x or y or both ten in the form
ax
(1)
where zero.
a, 6,
and
c are
+
by
=
variables
may
be writ-
c,
constants and both a and 6 cannot be
THE GRAPH OF A LINEAR EQUATION
5.7]
95
When two
equations are so related that all pairs of values which satisfy one also satisfy the other, they are said to be dependent. Evidently, from the definition, equations have the same graph.
+
Example, x
by
=
y
and 2x
1
+
=
2y
two dependent
2 are both satisfied
-l),etc.
(0, 1), (1, 0), (2,
The operations
leading to dependent equations are essentially the same as those yielding equivalent equations in one variable (Art. 4.5).
That
is,
terms
may
be transposed, or both members
may
be multiplied or divided by any number or constant except zero. If b
^
in (1), the solution for y,
/o (2)
y=-
a
+ lX .
namely
c -,
involves operations of this sort, so that (2) and (1) are dependent and their graphs are the same. Since y is a linear
function of x in If b
=
0,
(2), this
graph
^
then a
by
a straight
is
line.
the statement following
(1),
so
that (1) takes the form
ax
(3)
whose graph
is
=
the vertical line through (-,
For evidently the value of y
etc.
c,
is
(A
(-, 1
J,
(-,
not restricted in any
2),
way
/
by
(3),
while x
Example. 2x allel
is
restricted to the
=
3,
to the F-axis
Similarly,
if
a
or x
a horizontal
f, is the
and f units to the
=
0,
so that
by
(4) is
=
one value,
6^0, =
equation of a
line par-
right.
the graph of
c
line.
Example. The 2 units below it.
line
y
=
2
is
parallel to the X-axis and-
FUNCTIONS AND GRAPHS
96
Our argument,
[Ch.
V
then, leads to this conclusion:
The graph of a linear equation in x or y or both
is
a straight
line.
EXERCISE
26
Plot the graph of each of the following equations. 1.
3. 5. 7.
9.
11. 13.
15. 17.
-
= 0. x + y - I = 0. 2x - y = 2. 4z - y + 4 = 0. 3z + 2y - 6 = 0. 2z + y - 4 = 0. 2x + 3 = 0. = -1. s=0. x
12.
+ y = 0. x - y + 1 = 0. 3x + y - 6 = 0. 2x - 3y + 6 = 0. 2z - y + 4 = 0. 3z + y = 5.
14.
5y
16.
y
18.
y
2.
y
4. 6. 8.
10.
a;
x
- 1 = = -5. = 0.
0.
19-30. Solve each equation in problems 1-12 for y in terms of x,
and then solve
it
for
x
in
terms of
y.
9C =
Given 5F 160, the relation between the Fahrenand centigrade (C) temperature readings, express: (a), F as a function of C; (b), C as a function of F; (c), 5(F + 40) as a 31.
heit (F)
function of C.
Given 7 = Prt, express: (a), P as a function of 7, r, and t a function of 7, P, and t and (c), t as a function of 7, Pj and r. 32.
(b), r as
5.8.
Graphs of non-mathematical functions
*
and suggestive to make a graph showing the relation between two quantities even though it may not yet be possible to express one as a mathematical function of the other. In business, science, war, and practically every Often
field
of
it is
helpful
human
activity,
graphs or charts are used very
extensively. * This article
may
be omitted without loss of continuity.
5.8]
GRAPHS OF NON-MATHEMATICAL FUNCTIONS
97
For example, the meteorologist, or 'weather-man,' might record the Fahrenheit temperature in a given city for every third hour of the day, getting the following result :
Time
A.M.
4
Temperature
3
6
9
P.M. 12
6
6
5
20
25
35
9
12
14
6
40V 30
0)
fw I
i
i
-10 Fig.
The graph
9
with the time as the independent variable shows clearly how the temperature has changed during the day. In this case it is advisable to draw a smooth curve through the plotted points to indicate that the tem(Fig.
9)
perature changes continuously, or from moment to moment. On the other hand, if we record quantities such as the yield per year of an acre of wheat, the smoothly curved line would
be meaningless, and a broken line such as that in Fig. 10 is
better. b 40 o
-ofe '5
a
30
>J2 20 o>
-5
a.
10
o 1930
1935 Year
1940
Fig. 10
The types
of graphs used in practice are
much
too nu-
merous to be shown here. Many of them, however, are based upon some modification of the coordinate system. Perhaps the most frequently used independent variable is time, which
FUNCTIONS AND GRAPHS
98
[Ch.
V
be measured in seconds, minutes, hours, days, years, etc. It is interesting to note that sometimes in this type of graph, the results may be carried more or less reliably into
may
the future.
EXERCISE
27
Graph the relations indicated by the data in the following problems. Use the type of curve (smooth or broken) which seems to be more appropriate for each problem. 1. Given the following readings, graph the temperature as a function of the time. Can you suggest an explanation of the unusual conditions here indicated?
Time (A.M.) Temp. (F)
4
5
6
7
8
9
10
11
12
60
62
65
69
75
75
65
55
55
Given the following data, graph the yield per year tain farm in bushels of corn. 2.
Year Yield 3.
1920
1924
600
750
1932
600
550
Graph the indicated weight
Age (years) Weight (Ibs.) 4.
1928
A river
12
3
25
33
15
of a
1936 510
of a cer-
1940
1944
675
790
boy as a function
456789
39
44
49
54
of his age:
68
60
gauge registered as follows. Graph the water
10
77
level as
a function of the time.
Hour Level
(A.M.) (feet)
4
5
-&
6
f
7
8
f
2
9
10
f
1
11
12
i~A
Chapter Six
SIMULTANEOUS LINEAR EQUATIONS
6.1.
Problems leading
to
simultaneous equations
Sometimes we are confronted by a situation in which two unknowns are related in two separate ways. In this case the rules of algebra enable us to find the values of the unknowns. Example. After a storm, a mixture of water and oil filled a 100-gallon cask. The oil would eventually rise to the top, where it could be skimmed off; but it was so heavy that the process would take longer than the buyer wanted to wait. It was found that the liquid weighed 800 pounds, exclusive of the cask. It was known that the water and oil weighed about 8.4 and 7.4 pounds per gallon respectively. The deal was completed at once after a short algebraic calculation.
How? x and y be the number of gallons of water oil, respectively, in the cask. We have at once, since there were 100 gallons altogether, Solution. Let
and
x
(1)
Also, since
of oil
respectively,
SAx
+ 7 Ay =
Multiplying both sides of (3)
100.
x gallons of water and y gallons
and 7 Ay pounds, (2)
+y=
7.4z
(1)
by
+ 7 Ay = 99
800. 7.4,
740.
we
get
weigh 8Ax
SIMULTANEOUS LINEAR EQUATIONS [Ch. VI Subtracting the two members of (3) from the corresponding ones of (2), we find that x = 60; and then, from (1), that - x = 100 - 60 = 40. Hence there were 60 gallons y = 100 of water and 40 gallons of oil in the cask. Statements (1) and (2), which together carry in disguise the values of the unknowns x and y, are called simultaneous 100
equations. Just as
one unknown may be looked upon as a sentence-question, so a set of two or more simultaneous equations may be considered as a compound question instead of a group of sentences. Taken together, they propose a question and provide the answer, in concealed form, to
an equation
anyone able to
find
in
it.
6.2. Algebraic solutions of
simultaneous linear equations
The
solution of (1) and (2) in the preceding article was ' obtained by use of a combination of the addition and suband ' substitution ' methods. traction shall now illus-
We
7 '
trate these formal procedures as applied to the simultaneous
equations
:
(1)
3z
-
4y
=
2,
4z
+
3y
=
11.
and (2)
A. Addition and Subtraction Method (Note. The meanings of the directions in parentheses below are as here indicated: (1) X 3 means that both members
+
of equation (1) are multiplied by 3; (3) (4) means that of and members equations (3) (4) are added; corresponding
and
so on.)
(4)
(2)
X X
(5)
(3)
+
(3)
(1)
3
4 (4)
Qx 16x
-
+
12y 12y 25x
= = =
6.
44.
50.
SOLUTIONS OF SIMULTANEOUS LINEAR EQUATIONS
6.2]
101
Hence
9y 25y
= = = =
y
=
1.
=
11.
x
(6)
(7)
(1)
(8)
(2)
(9)
(8)
X X -
4
-
12x
IGy
+
I2x
3 (7)
2. 8.
33. 25.
Hence (10)
The
x
=
4x
+
is:
solution, then,
2;
t/
=
1.
B. Substitution Method (11) Solve (1) for y. (12) Substitute in (2).
2 .- 4y
=
y
=
x
(14) Substitute in
3
(1).
Hence
=
2)
4
(13) Solve (12).
Solution: x
~
3(3 *
2,
y
=
2. 2. 1.
I.
C. Combination Method
The method used
in solving the illustrative
Art. 6.1 consists of solving for one
example
in
unknown by method
A,
and then substituting this value in either equation to find the second unknown. This method is a combination of the other two.
should be kept clearly in mind that the solution of two simultaneous equations consists of a pair of values which satisfies both equations. Applying this test to our solution of It
(1)
and
(2),
3-2-4-1 = 4 2 + 3 1 =
(15)
we
2 (correct)
;
11 (correct).
= 2, y = 1) is a solution. Furthermore, pair (x shall show in Art. 6.3, it is the only solution.
Hence the as
we have
SIMULTANEOUS LINEAR EQUATIONS
102
[Ch.
VI
But why, it may be asked, does equation (5) give us the x-value of the equation pair? How can we add the I2y of of when we know from to the the (4) (3) +I2y graphs of (3) and (4) that y can have any value in either equation, so that the two y's may be representing different numbers and hence need not be equal?
The answer may be
stated thus
:
solving simultaneous equations we consider the letters not as variables but as unknown constants. When this is done, all
In
Thus x and y are the same constants in each of equations (1) to (10), and their numerical values are exposed
is
explained.
and
(10).
choice of the
most
in equations (6)
The
a given problem first
efficient of
the three methods for
a matter of judgment. In general, if the is an integer or a very simple algebraic
is
unknown found
expression such as 2a, 36, etc.,
unknown by
quicker to find the second substitution; otherwise, by addition and subit is
some trivial cases, such as 2x = 3, 4y = 7, none the three methods applies; but here, of course, the solu-
traction. In of
= f, y = When some of
tion,
x
,
is
seen
by
inspection.
the coefficients are
literal,
method
usually preferable.
Example. Solve for x and y the following equations
ax dx
(16) (17)
+ by = + ey = /.
c,
Solution. (18) (19) (20)
(16) (17)
(18)
X X -
e
aex
b
bdx
(19)
aex
bdx
= = =
bd)x
=
x
=
+ +
-
bey bey
ce. bf.
ce
-
ce
-
&/,
or (21)
(ae
-
bf.
Hence (22)
^-j*. ae bd
:
A
is
.3]
GRAPHICAL SOLUTION OF LINEAR EQUATIONS
23)
(16)
24)
(17)
25)
(23)
X X -
d a
adx adx bdy
(24)
26)
+ + -
(6d
aey
= = =
ae)y
=
y
=
bdy aey
103
cd. af.
cd
cd
-
a/,
af.
lence 27)
The
-af bd
solution is the pair (22) and (27). It direct substitution in (16) and (17). >y
i.3.
or
ae'
may
af ae
cd
6d
be checked
The graphical solution of simultaneous linear equations
we graph
the lines (1) and (2), Art. 6.2, we find that hey intersect at the point (2, 1), Fig. 11. This point lies on >oth lines, and hence its coordinates must satisfy both If
iquations.
Fig. 11
In general, to solve graphically any two linear equations; vith numerical coefficients, we draw the two straight lines nvolved and then find by inspection of the figure the co>rdinates of their
common
point, or point of intersection,
they have one. These estimated coordinates should be learly the same as the correct solution-pair found algef
SIMULTANEOUS LINEAR EQUATIONS
104
[Ch.
VI
braically; and of course they would be exactly the same if the figure were perfectly accurate. Thus, the graphical solution is a check on the algebraic one, and vice versa.
And now new light is thrown on algebraic appear
in
some attempted
solutions.
coincide or be parallel,
may intersect,
difficulties which For two straight lines and each of these three
possibilities leads to a different type of algebraic result. If they intersect, they do so in a single point, and there is one
and only one
they coincide, there are evidently infinitely many solution-pairs. Finally, if they are parallel there is no common point, and hence no solution. In these respective cases the simultaneous equations are called solution-pair.
If
independent, dependent, and inconsistent. To test these possibilities graphically,
merely draws the
A simple
lines.
of
algebraic test
course, one the follow-
is
ing:
Let any two linear equations in two unknowns be written so that their right members contain all terms not involving the unknowns. Then, if their left members cannot be made identical by multiplying one equation by a constant, they are independent. When the left members can be and are
made
identical,
they are dependent when the right members
are then equal, and inconsistent otherwise.
Example
1.
The equations 2x
(1)
+
y
=
3,
and
x-3y =-2
(2)
are independent, since the tical.
The
solution-pair
Example (3)
2.
left sides
is (1, 1).
The equations 2x
+y=
3,
and (4)
4z
+
2y
=
6
cannot be
made
iden-
6.3]
GRAPHICAL SOLUTION OF LINEAR EQUATIONS
are dependent, since (3) can be
made
105
identical with (4)
common
multiplying its members by 2. Points on the simultaneous algebraic solutions, are (0, 3),
line,
(1, 1), (2,
by or 1),
etc.
Example
3.
The equations 2x
(5)
+y
=
3,
and 4x
(6)
+
2y
=
7
are inconsistent, since when the left members are made identical the right members are unequal. Clearly no pair of values for x and y can make 4# 2y equal to both 6 and 7.
+
EXERCISE
28
Solve by the addition 1.
x
x 4.
7.
10.
13.
+ 3y = -
9x 8z
5y
+
and subtraction method.
11,
2.
3x x
8,
5.
9z 9z
= - 13.
7y
- 9y = -69. - 1, 9o: - 5y = = 5. 10s 8y = 2x + Qy -3, 3x - 5y = 6. 7z - 2y = 15, 60: - y = 10.
8.
11.
14.
+ ty = -1, + 5y = 7. + 8y = 3,
3.
6.
- 8y = -77. 5x - lly = -4, 6x - 8y = -10. 15x + 4y = 7, 6x + Uy =9. x + 3y = 9, 4x + 5y = 22.
9.
12.
15.
-
= 4, = 4x + 53. 7z + 3y = 2, Sx + 7y = -2. 7x + 3y = 4, 8x + 7y = 26. 3x + 5y = -9, 4z - 3y = 17. 2x - 3 = 5y, y + 5 = 3x. llx
y substitution.
= 6, 2 X - y = 7. 5z = -10, 18.' 3x + 4y = 2. 20. 7z + 6y = -11, Sx - 5y = -60. 22. IQx + I2y = -7, 8x + 9 = -8. 16.
3z
17.
3y 2y
0,
6. + Z + 4^=16, 2x + 3y = 17.
3x 19.
= =
21. 6z
-
9z
+
Qy 5y
23. 2x
-
3y
5x
+
8j/
= =
= =
-7, 8.
-29, 5.
Qy 15y
SIMULTANEOUS LINEAR EQUATIONS
106
- 14y = -15, lOx - Qy = -5. 4z = 12, 3z = + 3.
24. I5x
26.
25.
4x 27. 2x
Gx
2/
[Ch.
VI
= -9, = 2. + = -f 9y 13, -7y = -63. 3y 2y
28-39. Graph the pairs of equations in problems 16-27, and estimate the coordinates of the points of intersection. Compare with the algebraic solutions. Solve algebraically.
49.
az
+
bx 51.
53.
ax ax
by
ay
+
to/
by
3x= 5y =
= = = =
50.
ab,
b
2 .
a:
+ =
2a,
x
y
26.
2a,
52. az
26.
ax 54.
4, 7.
2/
=
+ 6y =
2a2
by
ab
=
+ a6 +6 2
.
5az = -10, 3by = 18.
/n problems 55-63, w/iic/& pairs o/ equations have no solutions and which have more than one solution? 55.
58.
3x 6z 5x
6y 61. ax
ax
= =
= 5, = 20. 8t/ 7 3y, - x). 10(1 + 6t/ = 5, + 6y = 10. 4y
56. 2x
4z 59. 6z
3x 62.
-
ax 36y
= 4, = 8. = 10j/, = 5y.
3y 6y
57.
5
60.
2
c
=
3c
by,
3ax.
2x 2y
= =
14
28
-
y,
4z.
2x=l+32/, 6x-9t/=3. 63. ax by =0, = 0. ax by
THREE EQUATIONS AND THREE UNKNOWNS
6.4]
64.
x
+^=
2,'
3
_
2
7
y
x
=
3'
y
HINT. Consider the
become: u Solving,
+ 3v =
we
=
x
y
=
, 4 66. -
~
x
2v
=
x
y
The equations then
and y
where u
-,
3
u=l,v=-, 6
1,
--- =0, L
,.*
x
as -
unknowns
2 and 3u
find that
final solution is:
65.
107
or
=
- and x
-=!,-=y&
v
= y
Hence the
x
3.
o +-=
_
,
^
67.
5,
y
4 -
x
o +-= .
y
-
5,
70.^ + ^=1, x '
y
2a3b_ - ^ x y
6.4.
Three equations and three unknowns
Consider the simultaneous equations
+
2x + y 4* - y
(1)
(2)
60;
(3)
We (2),
seek
and
now a
(3)
as constants
+
2y
2z
-
+
z
3z
= = =
set of three values
l,
3,
2.
which
will satisfy (l) r
simultaneously. Again we consider x, y, and z whose values are fixed in advance, so that we
may combine them
at will.
' guiding principle will be found helpful to avoid ' working in a circle/ to wit: Eliminate one unknown at a time, using all of the equations in which it appears, thus reduc-
Here a
'
'
ing the problem to one with fewer unknowns and fewer equations. For example, to solve equations (1) to (3) we
may
eliminate z
first (if
we
wish), thus:
SIMULTANEOUS LINEAR EQUATIONS
108 (4)
(1)
unchanged
(5)
(2)
X
2
(6)
(4)
+
(5)
Equation
(3)
+ +
2x y 8x 2y
- 2z Ite - y
must now be used,
(7)
(3)
unchanged
(8)
(2)
X
3
(9)
(7)
+
(8)
2z
= = =
[Ch.
VI
1.
6. 7.
either with (1) or with (2).
+
Qx I2x
-
+
3z 2y 3z 3y 18x - y
= = =
2.
9.
11.
There remain two equations and two unknowns, (6) and (9). y = 2. Substitution of these Solving these, we get x = ,
values in (1) yields 2=1. The tentative (or untested) solu= 1. The test is not complete tion is then: x = %, y = 2, z until it is shown that these values satisfy all three original equations. If it
a
is
one of the simultaneous equations, usually advisable to eliminate this letter first. Thus, letter is missing in
given (10) (11) (12)
.
2x - 3y - z x + 2y + 2z
2,
y
2,
= = -3z =
1,
x should be removed from (10) and (11), and the resulting equation can then be used with (12) to get y and z. Thus far we have considered equations with two and three unknowns. In general, if the number of simultaneous equations is the same as the number of unknowns, it is often (though not always) possible to find a set of values which will satisfy all of the equations.
above equation
EXERCISE
The guiding
(4) holds good in
principle stated
all cases.
29
Solve the following systems of equations. 1.
x x x
+ y + z= +y-z= y + z =
G, 0, 2.
2.
x
x x
-y + z= +y-z= y
z
=
Q, 2, 4.
DETERMINANTS
6.5]
x
-
x
+y
x
y
3.
5.
7.
y
+z= - s= z =
2,
+ 4y + - 2y + 3z + y + 2x - 3y + 4z
6.
1,
= 2, = 3. = 4, = 3, z = 5. 32 = 5, 2z = 3, 5z = 4.
8.
10.
3x
12.
7,
A+2B
2,
x 2x
15.
C= -C=
6.5.
A
y
2z
z
-w
2z
+w
+
-
+
+y
3, 4, 5,
1,
2, 1,
2w
0.
=
3,
N=
f,
iAT
L
4
+y+ = z
2x = 2x y = + y 2z =
14.
-2B + 2C = 12. x + y^ x+z_ y 3
x
5y
-
x y
+ 2w = 3, x + 2w = 4. A + 4B + 3A
-
$x
= 5, = 3, = 2. + 3z = 4, - 2z = 5,
+ 3y Zx + 2y + z=2.
z
13.
2,
2x - 3y + z x + 2y - 2z 3x + y - z 5x
2y=l, y + 2z=2,
11.
0,
x-y + z=2.
4.
2x - y - 2z x - 2y + 2z 5z + 3y - z 4z - y + 2z 3z
9.
4.
4,
+ 2y + z =
a;
109
x + y + z = z = x y
2
6.
Determinants
a set of numbers or expressions arranged in rows and columns, having the same number of rows as of determinant
is
columns, and enclosed within two vertical bars.
Example.
2-1 5 3 2-1 1
The
-2
vertical bars
may
4 be considered the mark of a deter-
minant. The numbers between them are called elements.
The
SIMULTANEOUS LINEAR EQUATIONS
110
[Ch.
VI
whole expression has a definite value found by rules which will be explained below. If the determinant has two rows and two columns, it is said to be a second order determinant. If there are three rows and three columns, it is of the third order. In any determinant, the order is the same as the number of rows or of columns. The value of a second order determinant is the product of the upper left and lower right elements minus the product of the lower left and upper right elements. In other words, we find the product of the numbers read diagonally downward and subtract the product of numbers read diagonally upward always from left to right.
= ad
be.
The value of a third order determinant may be found in a simple manner when a fourth column like the first one and a fifth like the second are placed to the right of the determinant. The value is the sum of three products of elements read diagonally downward to the right minus the sum of three similar products read diagonally upward.
=
(aei
+ bfg +
cdh)
(gee
+
hfa
+
idb).
looking at the letters in the answer as read from the original determinant, the student can see easily how to
By
evaluate the determinant without adding the two extra columns, if he so desires. He might try this shorter scheme in
checking through the next example.
6.5]
DETERMINANTS
Example
111
4-
3-12 1
2
3
2
-3
1
-[(2)(2)(2)
+
+
(-3)(3)(3)
= [6 + (-6) + (-6)] - [8 + (-27) = -6 - (-20) = -6 + 20 = 14.
+
(-1)]
These short-cut methods of evaluating second and third order determinants do not apply to those of higher order. The method which does apply to determinants in general is called 'expansion
ing example
Example
by minors.'
It is illustrated
by the
follow-
:
5.
a(ei
=
aei
b(di
hf}
+
bfg
+
gf)
cdh
+
c(dh
bdi
ahf
ge) cge.
In the above expansion the second order determinants are called the minors of the letters before them. c,
for example,
is
the determinant
d h
g
,
The minor
of
found by striking
out the row and column of the original determinant in which c appears; and the other minors are found similarly.
The
sign in front of the letter
minor
in the expansion
is
which
found by
multiplied by its starting with the upper is
element and calling off the signs alternately as 'plus, minus, plus, minus,' each successive element reached being immediately to the right of, or below, the one just called. If
left
an element comes on the minus count, its sign is changed; otherwise not. To illustrate, the determinant in Example 5 may be expanded by minors as above or by using the elements of any row or column as multipliers. If the second column is used, for instance, the expansion is
SIMULTANEOUS LINEAR EQUATIONS
112
[Ch.
VI
-h
-b
The value of the determinant is always the same, regardless of the method of expansion, as the student may verify. EXERCISE
30
Evaluate the following determinants by the 'diagonal multiplying' method.
22-32. Check the results in problems 11-21 by the method of expansion by minors.
6.6. Solving systems of equations
by use of determinants
any unknown in a system of equations may be represented by the ratio of two determinants.
The value
of
Example. Solve the system, 2x x
3
+
1
= =
y,
3y.
SOLVING EQUATIONS BY USE OF DETERMINANTS
6.6]
Solution. First, arrange the equations thus
:
2x ~
JO
Then
x
=
and
y
=
-10 =
-5 = -5
y OU
113
=
Q*.
3, 1
1.
2,
1.
be noted that the determinant in each denominator is composed of the coefficients of x and y arranged in the same order in which they occur in the equations. The determinant in each numerator is the same except that the It will
numbers to the
right of the equality signs replace the coefficients of the letter whose value is sought.
The student may
verify the solutions given for the equa-
~~
^ = in section 6.2 C by now using the dedx ef // terminant method for finding the values of x and y. The tions
(
latter
method
7
+ ,
gives
C
x
= /
e
ce
a
b
ae
d
e
bf bd' /
ni
y
=
a
c
d a d
f
af ae *^
b
dc .
bd
e
method will give the correct results for values of the coefficients a, 6, c, d, e, and /.
It follows that this *
allowable
all
unknowns may be
Systems of three equations in three solved in a similar way. Example. Solve the system 3x 2x
fc
+
2^/
{y
x,- y
+
-
z.
2z
= = 1; =-2. 3*,
bd = 0, the values given for x and y are meaningless, since Note that if ae by zero is not permissible. In this case the equations are inconsistent or dependent, and should be studied by the method explained in Art. 6-3. *
division
SIMULTANEOUS LINEAR EQUATIONS
[Ch.
-18+4+1-6+3-4 = -20 _ -18-2+2+3+3-8 -20 ~~'
-6+3+4+1+6+12 = 20 (evaluated above)
(x
=
1,
?/
=
1,
z
=
i JL
= -1.
-40 = -20
-20
Answer:
20 20
VI
2.
2).
Note that the denominator in each case contains the coefficients arranged in their normal order. The only difference between the numerator and the denominator in each fraction constants on
that, in the numerator, the the right sides of the equations replace the
coefficients of the letter
this
method
is
whose value
to the general system:
+ by + cz = + ey + fz = + hy + mz =
ax dx gx it
follows that
x
=
N,.
D'
z
= D'
k I
n,
is
sought. Applying
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
115
where k
b
c
I
e
f
n
h
m
a d
b
k
e
I
g
h
n
D* =
and
EXERCISE
31
1-27. Solve problems 1-27 of Exercise 28
by use
of determinants.
28-36. Solve problems 1-9 of Exercise 29 by use of determinants.
6.7. Stated
Often
problems with more than one unknown
more
a problem stated in English into a set of simultaneous equations than it is to solve the equations. Therefore, a consideration of the genit is
difficult to translate
eral principles applying to all such
problems should prove
helpful.
The
essential steps in solving
a stated problem are two-
fold. (1)
The unknowns
nated by
must be identified, desigw, etc.) and described clearly.
to be sought
letters (as x, y,
z,
Examples of unsatisfactory Let x
=
starts.
time, or length, or rate, or distance, or amount.
Examples of
satisfactory starts.
Let Let Let Let Let For the
case,
D -
x x x x x
= = = = =
0, see
no. of minutes after 3 P.M. no. of inches in length. no. of feet per second. no. of miles in distance.
no. of bushels of corn. the discussion in the preceding footnote.
SIMULTANEOUS LINEAR EQUATIONS [Ch. VI (2) If there are two unknowns, two equations must be found expressing two relations between the unknowns which 116
should be described in the problem. If there are three unknowns, three relations must be indicated, and so on. Thus equations as there are unknowns will be obtained, and the problem will be 'set up/ or ready for the formal
many
as
7
algebraic solution.
Example. The combined age of John, Bill, and Harry is 36 years. Two years ago Bill was three times as old as John. Eight years hence Harry will be twice as old as Bill. How old is each? Solution. In view of the question at the
end we can
start
confidently thus:
x
Let
y z
At
this point
= = =
no. of years in John's age; no. of years in Bill's age no. of years in Harry's age. ;
we
re-read the
sentence and find that
first
states in English one relation between the goes into algebra thus
it
The statement
x
(1)
+y+z
The second sentence
y
(2)
=
36.
two years ago, which 2. Thus 2, and z y
deals with ages
evidently are represented
by x
-
2
=
2,
3(x
-
Eight years hence the ages will be x According to the third sentence, z
(3)
Solving
y
=
and
(1),
unknowns.
:
(2),
and
+
8
(3)
=
2(y
2).
+
+
8,
y
+
8,
and z
+
8.
get x
=
4,
8).
simultaneously
we
=
24 (answer). Not every problem has a direct question to indicate the 8,
z
unknowns and separate sentences to describe the conditions but these quantities must always be described in one way ;
or another.
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
117
It is interesting to note that impossible simultaneous conditions will be revealed algebraically by inconsistent equations; while conditions which may seem to be inde-
pendent, but which really amount to the same thing, lead to dependent equations with more than one solution.
EXERCISE
32
Three times one number minus twice another equals of twice the first number plus three times the second is Find the numbers. 1.
4.
The sum
7.
5 times one number
2. If
result is 3.
The sum
1.
of the
is
by 3 times another, the
divided
numbers
is 8.
Find the numbers.
sum of two numbers Find the numbers.
One-half the
difference
is 6.
is 6,
and 3 times
their
Three times one number is equal to 4 times another, and half their sum is 7. Find the numbers. 4.
5.
Find two numbers such that 3 times their difference
and the sum 6. If
of 4
and the
first
a certain number
number
is
is
added to the numerator and sub-
tracted from the denominator of ^, the result equals 3. The of half this number and a second one is 7. Find the numbers. 7.
The rowing
rate
is 6,
3 times the second.
down a
certain stream
The
sum
1
m.p.h. less upstream is 3 times is
than twice the upstream rowing rate. the rate of the current. Find the rate of the current and the rate of rowing in Let
still
water.
x y
Then
rate
+y
= = = = =
(1)
x x x
(2)
x-y=
y
+y
The simultaneous
no. m.p.h. rowed in still water. no. m.p.h. the current flows. no. m.p.h. no. m.p.h.
2(x
-
y)
rowed downstream. rowed upstream.
-
1.
3y.
solution of (1)
and
(2) yields:
x
=
4,
y
=
1
(answer).
The
rate of rowing downstream is 2 m.p.h. less than 3 times the rate of the current. The upstream rowing rate is ^ the rate of the current. Find the rate of the current and the rate of rowing in 8.
still
water.
SIMULTANEOUS LINEAR EQUATIONS
118
[Ch.
VI
9. A man can row downstream 5 miles in the same time it would tak him to row 1 mile upstream. The rate of rowing in still water is 1 m.p.h. more than the rate of the current. Find the two
last-mentioned quantities. 10.
The
rate of a boat in
still
water
3 times the rate of the
is
current and also 2 m.p.h. more than this rate. Find the two rates.
A
plane can travel 350 m.p.h. with the wind, and its speed the wind is 320 m.p.h. Find its speed in still air and the against of the wind. velocity 11.
12. A plane goes twice as fast with the wind as against it. If the velocity of the wind were doubled, the speed of the plane when flying with the wind would be 100 m.p.h. more than 3 times its
speed against the wind. Find the speed of the plane in the velocity of the wind. 13.
The sum
of the digits of a 3-digit
digit is twice the
versed, the
hundreds'
new number
is
digit. If
number
is 6.
still air
The
and
units'
the order of the digits
is re-
99 more than the original one. Find the
number.
= = u = + KM + u = h + + u = u = + Wt + h = h
Let
t
100A
Then,
t
(3) (4)
lOOu
(5)
Solving
and u 14.
=
2,
the hundreds' digit; the tens' digit; the units' digit. the number. 6.
2h.
100A
+
10*
+ u + 99.
simultaneously, we get h so that the number is 132 (answer).
(3),
The sum
(4),
and
(5)
=
1, t
=
3,
number is 7. If the order number is 2 more than twice
of the digits of a 2-digit
is reversed, the resulting the original number. Find the number.
of the digits
15.
of the digits of a 3-digit number is 12. The tens' units' digit 1 more, than the hundreds'
less, and the Find the number.
digit digit.
16. digit
The sum is
1
The sum of the digits of a 3-digit number is 14. The units' more than twice the hundreds' digit and 1 less than twice
is 1
the tens' digit. Find the number.
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
a certain 3-digit
17. If the order of the digits of
new number exceeds
versed, the
the old one
number The
396.
by
and the sum of the Find the number.
digit is 3 times the hundreds' digit, 1
more than twice the
tens' digit.
The
length of a rectangle in inches is width. If its length is decreased by 1 inch and 18.
by 3
inches,
19.
The
width. If
creased
it
becomes a square. Find
its
119 is
re-
tens'
digits is
than twice its width is increased
1 less
its
dimensions.
length of a rectangle in inches is 2 less than twice its length is decreased by 2 inches and its width is in-
its
by 3
inches,
it
becomes a square. Find
its
dimensions.
width of a rectangle is increased by 2 inches, its area increased by 14 square inches. If its length is decreased by 2 inches, its area is decreased by 8 square inches. Find its dimen20. If the
is
sions.
by 3 feet, and its width is increased by 2 feet, its area is increased by 2 square feet. If its length is decreased by 2 feet and its width is increased by 21. If the length of a rectangle
1 foot, its
area
22. If the
is
unchanged. Find
width of a rectangle
is
its is
decreased
dimensions.
increased
decreased by 4 inches, its area length becomes a square. Find its dimensions. is
23. If $6000
yearly interest
is is
invested, part at
by 3 inches and its unchanged and it
5% and
amount
$340. Find the
is
part at 6%, the total invested at each rate.
24. If $10,000 is invested, part at 5% and part at 6%, and if the total yearly interest is $540, find the amount invested at each rate.
$8000 is invested, part at* 4% and part at 5%, and if the interest on the 4% investment is $50 more than the other, find the amount invested at each rate. 25. If
The interest received yearly on a 6% investment is twice much as the interest on a 4% investment. If the total yearly
26.
as
interest
is
27. If is
A works 3 A works
$60. If
pay
is
$360, find the
amount
of each investment.
days and B works 5 days, their combined pay 5 days and B works 3 days, their combined
$68. Find the daily
wage
of each.
SIMULTANEOUS LINEAR EQUATIONS
120
28.
A man has $2 in dimes and nickels.
dimes and half as $2.50.
29.
How many
A man
the same
many
nickels as he
If
now
he has twice as has, he
[Ch.
VI
many
would have
of each has he?
has $3.50 in nickels, dimes, and quarters. If he had of dimes but half as many nickels and twice as
number
quarters, he would have $4.50. If he traded his quarters for dimes and his dimes for quarters, he would have $4.10. How many coins of each kind does he have?
many
pounds of bananas of two grades, one selling for 5j per pound and the other for 6^, bring $5.30, how many pounds of 30. If 100
each grade are there?
One
is 5% pure silver and another 15% pure silver. pounds of each must be mixed to form 100 pounds of alloy of which 8% is pure silver? 32. A lady buys 10 pounds of grapes, of which part cost 25?f for 2 pounds and the rest 25 for 3 pounds, l^she paid $1.00 al-
31.
alloy
How many
how many pounds of each kind did she buy? 33. How many pounds each of 20 and 35 coffee must be mixed to make 100 pounds of coffee worth 25ff per pound? together,
34. If corn
meal
how many pounds
is
worth 3
of each does
per pound and flour 4j5 per pound, one buy to get 50 pounds for $1.80?
Chapter Seven
EXPONENTS AND RADICALS
The laws of exponents
7.1.
As we have already seen, the symbol a m when m is a positive integer, means 'the base a taken m times as a ,
factor.' Thus, 2 3 = 2 2 2 powers is called involution.
=
8.
This process of raising to
We shall now examine the laws concerning operations with positive integral exponents. Perhaps even more useful than knowledge of the laws themselves is the realization of how very easily these working
hand by anyone who understands the meaning of the symbol a m There is not a better place to apply the arithmetic tests which are so useful in m n = a m+n or algebra. For example, which is correct: a a can be discovered at
rules
first
.
,
=
a ma n
n =
=
3.
25
a wn ? Try Then,
= 2X with small numbers, say a = 2, = 2 22 3 = (2 2) (2 -2- 2) =2-2-2-2-2
m
it
cw
Evidently the exponents here are added, and not m n = a m+ n multiplied. This suggests that the correct law is a a Once the simplicity of the testing-by-arithmetic method is grasped, the student will be self-reliant when his memory fails. But he can save time by learning the five all-sufficient .
.
:
laws below, which, when supplemented by certain definitions, will be shown to hold even when the exponents are not positive integers. It is helpful to learn
them
in groups of 2, 2,
and
1,
called respectively the repeated base, the repeated exponent, and the single base cases. Easy extensions of the laws are
indicated
by the
illustrative examples. 121
EXPONENTS AND RADICALS
122
[Ch. VII
Repeated base cases
LAW
=
3 4 Examples. 2 2
LAW
= =
a ma n
1.
2 7 3 43 23 am
3 4+2+1
;
=
2.
a
Example. J
a m+ n .
=
27
~3
=
am
n
24
~n
=
37
.
.
.
Thus, if the same number or letter appears as a base in each of two exponential numbers which are multiplied or divided, this same base appears in the result.* Repeated exponent cases
LAW
3
3 5
Examples.
3
2 2 3 25 2a 2
LAW
=
ambm
3.
= =
3
(3
5)
(2
3
am
A
4.
bm
- =
Examples.
(-}
am
a
-
(ab)
=
m
.
(15)
5a)
=
(a (-} b/
=
42
2
=
3 ;
(30a)
2 .
m
;
/
number or letter appears as an extwo each of exponential numbers which are multior divided, this same exponent appears in the result.
That
is,
if
the same
ponent in plied
Single base case
LAW
5.
(a
Examples. (4
Sometimes
3
2
)
it is
=
43
'
2
m
=
n )
46
=
a mn
.
(2aW) = 3
;
23
aW
2 .
convenient to use the laws in reverse order, left. The memorization in the form
or as read from right to *
An
apparent exception to this statement appears in the example:
can be written in the form a by definition of a However, am ~ Law 2 gives: m = am m = a (or 1). a 1
(Art. 7.5).
am
=
1.
Note that
POWERS OF A NEGATIVE NUMBER
7.2]
given, however, is simpler tain common errors.
and more
123
likely to
prevent cer-
5 2 Example. Can 2 3 be simplified by use of a law of exponents?
Answer. No, since neither the base nor the exponent
is
repeated.
While the definitions given later (Art. 7.5) enable us to say that the above five laws are true for all values of the exponents, whether positive, negative, zero, or fractional, the proofs below apply only when the exponents involved are positive integers, and when m > n in Law 2.
Law = aaa
Proof of (n factors)
Proof of = aa
Law
am
a
n
2.
1.
a ma n
(m
=
+
aaaa
n
(m
factors)
(assuming that
m>
(m factors) = aa ^
aa
= n)
,
r
factors)
=
am
~n .
(n factors)
Proof of Law 3. a b m = [aa
=
(m
[(a6)(a6)
(m
factors)][(66
(m
Proof of Law a m = aa
=
(m 7 (m
bb 1,
(ab)
m
.
4-
n
r~
bm
factors)]
factors)]
(by the commutative law of multiplication)
(by Rule
.
:
n
(m
aaa
factors)
am+n
Art. 3.5)
Law 5. m n = a ma m (a ) Law 1) = a mn
factors) F~~I factors)
/a/a = (rKz) bjbj
'
'
'
/ (
m rfactors)
= (-JT.
Proof of
(by 7.2.
A
(n factors)
=
am+m+
*
*
'
(fltennfl>
.
Powers of a negative number a number of the form x n where n is an an even power when n is even and an odd power
power of x
integer. It
when n
is
is
odd.
is
,
EXPONENTS AND RADICALS
124
Since
(
3)
n ,
VII
n negative factors, it is n is even or odd. More
for example, has
positive or negative according as
generally
[Ch.
:
For negative numbers, even powers are
and odd
positive
powers are negative.
EXERCISE Apply
^7 o/.
vii
TZX.
,76
^R oo
-. L
-
n
:J_. o ax 3
4.0 'XA*.
/~rr
^ ^ o x2 y
4^ 1*^.
.
-
x
4fl tv/.
., 6
X*
^
. .
x 3 ^4
44. 4. JL.
2 ?/
2
46.
-(-f)
7.3. Radicals, roots,
The
^.10
/y.5
^O ov.
-
or
2
;
33 (ORAL)
the laws of exponents in the following exercises.
xj.3
X
= (-2)(-2) = 4; = (_ 2 )(-2)(-2)=-8.
(-2)2 (_2)3
Thus,
f
47.
f-
I
and principal
radical fa, read
a
48.
,
x 3y
roots
the rth root of a,'
is
a number
whose rth power is a. That is, (V^) = a. 4 3 5 Thus, (^S) = 8; (A/10) = 10; (A^I?) = -17. The symbol V~is called the radical sign; the quantity below it, or a in the case of Va, is the radicand; and' the '
7
integer r
is
the index of the root.
RATIONAL AND IRRATIONAL NUMBERS
7.4]
125
is 2, it is customarily omitted, and the symbol 'the read square root of a.' In other roots such as v'a, or 'the cube root of a,' the index must be written. There are two numbers whose squares are a given positive number. For instance, (+2) 2 = 4 and also ( 2^2 = 4. Thus, 4 has 2 square roots. However, the symbol V4 stands only for +2, which is called the principal square root of 4. The
If
Va
the index
is
VI. By more advanced methods it can be shown that any number has 3 cube roots, 4 fourth roots, 5 fifth roots, etc. The symbol Va, however, represents only one of the rrth 2
root
designated as
is
namely, the one which is called the principal rth a is positive this root is positive; when a is
roots of a
When
root of a.
negative and r
is
odd,
it is
Thus, V9 = 3; V8 = since (-2) =-8.
negative. 2;
Vl6 =
2;
and
V^8
=-2,
3
In Art. 8.1 we shall consider the case in which a
and
is
negative
r is even.
Rational and irrational numbers
7.4.
From
the definition in Art. 7.3, the
a positive number such that (V2) 2 exactly? It is more than 1.4 and 2 = 1.96 and (1.5) 2 = 2.25. That (1.4)
number V2 must be 2. But what is it
=
than 1.5, since is, V2 is between 1.4 and 1.5, or 1.4
Assume that V.2 =
-,
less
where p and q are integers and the fraction -
to lowest terms. 1,
2
since
=
7)2
V2
is
reduced
Then p and
q have no factors in common, and q cannot be not an integer. Squaring both sides of the equation, we have:
= pp which ,
q*
is
Q
q
is
impossible since there are no
common
factors in
p and q
to
qq
be canceled out. Hence our assumption is impossible. Similarly we can prove that 'V^4, vT, etc., cannot be expressed as quotients of integers.
EXPONENTS AND RADICALS
126
[Ch. VII
7
v !,
v^6, and in general the rth root of a number a perfect rth power. Such roots are called surds, and are included in the large group of irrational numbers. is
true of
which
is not
An
number
irrational
is
one which cannot be expressed
exactly as the quotient of two integers.
Examples. All surds, such as V2 and /5, and many other numbers such as ?r, the ratio of the circumference to the diameter of a
A
rational
circle.
number
is
one which can be expressed as the
quotient of two integers.
Examples, f
,
f,
N/f (or f ), 1.41 (or
^),
3 (or
f ).
As suggested by the alternate forms of 3 and 1.41 in parentheses above, all integers and all decimal numbers are rational. It follows that irrational numbers cannot be expressed exactly in decimal form. Most of the numbers in various tables, such as lists of square and cube roots (Table 2 in this text) are merely decimal approximations, or numbers as close to the actual roots as possible with the given of decimal places.
number
Irrational numbers are connected in an interesting manner with incommensurable quantities in geometry. Consider, for example, the length OP in Fig. 12. It may be computed 0(2,1)
A (1,0)
6(2,0)
Fig. 12
by use
of the
famous Pythagorean Theorem, which states
that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the sides. Since the sides qf the right triangle
OAP
are
1
and
1,
the hypotenuse
OP
is
equal
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
7.5]
to
= Vl + common unit
Vl + 2
had a
m
exactly
OA
I
2
1
= V2.
Geometrically,
if
OA
127
OP OP
and
measure which would go into times, say, and into OA exactly n times, then of
we know this
n
to be impossible, since ^-r
V2,
OA
,
'
and V2 cannot be expressed as the quotient of two integers. Hence OP and OA have no common unit of measure and
V5
are said to be incommensurable. Similarly, since OQ is it is incommensurable with the unit length BQ.
units long,
EXERCISE
34
Give the two square roots of the following numbers, naming first: (a), 1; (b), 4; (c), 1; (d), ; (e), ^; (f), x
1.
the principal root
In each case below state whether the radical
2.
is
rational or
_
irrational. If it is rational, find its exact value; otherwise give its
approximate value
as^
found in_Table
2.
_
V9; (b), VlO; (c), ^8; (d), ^27; (c^O; (f),_Vl2; Vl6; (h), V^T; (i), VI; (j), ^^7; (k), V-8; (1), Vg.
(a), (g),
Find the exact lengths of the diagonals of the rectangles whose dimensions in inches are as follows 3.
:
X 2; (a), (0, 4X6; (g), 1
4.
(b),
8
X
2X3;
(c),
3X4;
(d),
3X7;
(e),
5
X
12;
15.
In which cases in problem 3 are the lengths of the diagonals
rational?
7.5. Negative, zero, It
is
and fractional exponents
desirable that the five laws stated in Art. 7.1 shall
be true even
when
the exponents are not positive integers. It can be so arranged by use of the following definitions.
a
Definition 1.
=
1.
^
(a
0.)
By Law' 1, aa w = a+ m = a m Also, by the definition, &m = 1 while by aa m = 1 a m = a m Or again, by division, .
.
;
a
Law
2 and the definition,
am
=
a m~ m
=
a
=
1.
In other
EXPONENTS AND RADICALS
128
when the exponent
words, the laws remain valid
VII
[Ch.
zero
is
involved, given this definition.
a~ k
Definition 2.
Consider
first
r
a**
a numerical example.
22
By Law
2,
=
22
~5
2i
=
2~ 3 This .
is
when 2~3
correct
22
1
defined as
is
,
since
Zt
2 -2
1
=
=
--
=
<
_!_ ' 3
2-2-2-2-2~2-2-2~2 sim
In general, when n definition,
Two
a~ (n
~ m)
=
>
m,
=
am
~n
=
a~
a
(n
~m
and
by
this
a n-m
convenient rules are consequences of Definition
2.
>
For bj
(2) tion,
^e may
// it
/a m
W
term a~ m
a rn
a a^
a
6^ is
a factor of the denominator of a fraca m and written as a factor of the
be changed to
numerator; if it is a factor of the numerator, it may be changed m and written as a to a factor of the denominator.
Examples.
3- 24
4
5
3
a 6-3 (c
+
d)
a 6(c
+
c
= '
d)-
2
5
a&
3
+
q(c
2
Note, however, that in the case
4 __ 45'
d'
+
2 c?)
6
:
a -i a~ l ,
_
+ .
-1
,;, the terms with 6' 1
negative exponents are not factors of the numerator or de-
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
7.5]
nominator. In such cases the simplification must be l replacing a~
complex
by
b~ l
-,
-
6- 1
+
6-
^ o
more
fraction. Or,
a- 1 a~ l
by
a
1
(a-
i
and then simplifying the
-
1
b~ l )ab
+
1
=
l
b~ )ab
b
-
b
+
a a
= /a.
ar
Definition 8.
made by
directly,
(a-
=
etc.,
129
i
r By Law 5, (a ) = a = a = a. Also, by the definition, r/ = a. For the special case r = 3, for example, = (Va) (a a*aia* = a*+l+* = a = a, while v^ ^a A^a = (by Law 1) (/aY = a. '
r
1
r
r
i
r
r
r
)
1
?
aq
Definition 4-
Q
Q,
,
= (va) p
=va 5
For,
by Law 5 and Definition
also a a
=
Thus, 8*
Here the
= Va p = ( v'g) =
(a
3,
aq
p .*
=
^
(a
p
G
)
=
(
V'/-a)
p ,
and
a
p
.
)
2
final result is
form of Definition
22
=
4,
more
and
also 8*
= v^ = v'64 =
easily calculated from the
This should be used for a q when a
4.
4.
first is
a
perfect gth power.
Again, 7*
-
7
(v
2
?)
,
and
also
7*
-v7^ =^49. 2
In this
P '
case the second definition of a
is
preferable, since 7
is
not
a perfect cube.
and and
The reason
for the use of the adjectives 'rational' 'integral' as applied to letters is suggested in Articles 7.4 1 7.5. Just as is not^ rational, and 2- or ^, is not
Note.
V2
integral, so -, is
x
*
definition
by
,
Vx
is
not rational in x,
and or 1 or ,
not integral in x.
The
75
must be reduced
fraction
to lowest terms.
9
=
1) is
not the same as
(-!
Note that
jL
(
1)*,
or
EXPONENTS AND RADICALS
130
EXERCISE
[Ch. VII
35
Evaluate each of the following.
38.
+
39. (-27)-*.
40. -(5b)
41. l(3o).
42. (-0.027)*.
43.
45.
.
44.
-5-
.
46.
-(-3)-'.
2
-(J)3 (-f)4
47. [2
+
(16)-*]-
(-56).
1 .
1 .
Change each of the following to expressions in which the exponents are positive, and in which the operations indicated in Laws 1 to 5 have been carried out completely.
~ 9r
48.
11
y
*}-)
49.
2 *
3
AD..
2r &*>
2
^7
**'
vJV/
/ff
O
1O*
*^
ej. '
2?>y-i
^
*-*
/
i*'~l'i/*
y
co
3v>
_o*
r^n
OA
/^^
o
OO*
*}
^77-2 &y
M 3V
Ki W> 70/>
1
* X
50.
^ ^.
r^n
*'
re
'
2
^v^ti y /
_i
_i'
CQ *>'
V/^
(200xi/)
65.
68.
70.
v
' .
V.
71.
(30ary)
-1 -
7.5]
72.
74. 76.
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
- y) (x + y)~ (x - y)~ (x + 2y). (2x - 37/)- (x + y). (2x 2
l
.
l
2
-
78. (x
yfi(x
-
82. (a 2
+ z )*(a + rO~i
to
2
x 2 )-*.
6).
+ y)~
(x
-
y)(x
2 .
y)-*.
'4'
2
81. (2z 2
2
(a
131
-
83. (x 2
2
-
a 2 )Kz 2
2
)*.
a2 )~K
Express each of the following as a complex fraction and simplify a simple fraction in lowest terms. 2 84-
a~
+
2
,
2
07,-
,
fc~
volution.
o2
r-
J_
b
+ a~ b
ab~2
rr~-j
2
2
^
+
a(o
2a 2 6 2 ; b + a2
=
L
i
a2
gs
-
-
+ 2/)K^ 2/)~^ + a )^(2x + a
79. (x
y)~^.
-
75.
3
77. (5x
80. (a 2
z 2 )^(a 2
+ 6)~ (3x + y)
73. (a
3
b)-
a-
-
1
1
-
6(a
86>
^_2
00
(a
'
y
_2
^
t
+ x )~ + + (a + x
1
b)'
__
2
2
6-'
2
2
2 2
)
x~* '
x2
89. Select appropriate multipliers for each of the problems 85, 87, and 88 and simplify without first expressing each as a complex fraction.
Reduce
the following to fractions
whose numerators have no frac-
tional exponents.
--
-
90.
a
and denominator by
-
(a
s)i
+
-
a
x).
(a
-
(a
g)~*
-
(a
x
g)*
_
(a
x )*
(a
_
-
2
(a
(a
+ oo -
x2 )^ o
(a
95
-
(a
2
-
x2)-*
*
-^
2
2
(a
^
-
x2 )*
-
2
-
x2 )*
+ x (a - x )'* 2
2
(a
-
+
g
-
+
2
x2 )*
X2 )'*
-
(a
x)
x)'
1 '
x)*
z2 )*
+ (a + x )'* + X2)i - (a + X )'* 2
2
i
-94-
2
'
+ -
(a
2
2
1
2
a;
(a
x)
(a
-
a
(a
V1
numerator
Solution. Multiply r J both
x
2
(a
(a '
2
+ x )* 2
2
2
+ x )* - x (a (a + x )* 2
2
2
2
2
*
EXPONENTS AND RADICALS
132
Reduce
[Ch.
VII
whose denominators have no
the following to fractions
fractional exponents.
-*
2
(a
97 I
CZ
~'~'
X
~~' )
2
)* '~~'
CZ
yCZ
JC J
I
CZ
I
-
100. 2
x (a ftl
2f
u/
O
x
(cz
7.6. La?/;s
)
2
+
/
^
9
I
~T~
x
(.a
9
^
102.
i:
9/9
a (d
)~*
I
~T~
CZ
+
)'*
fj2
x
9 )
* 2 )*
l_
2
a;
/y.2
Z-r-Z i /
9
~~~
/Tt2
?-r^
101.
+
2
ju j
72
fl%
99.
(a
98
_ _ ,
^ -4-
^CZ
X
I
i
->2
X
-
(a
2
+ z )* 2
/>2'2'
X)
/
(cz
-' 9
I
~f~
x
9-i-
)^
concerning radicals
be performed by changing them to exponential form and then applying the laws of Operations with radicals
may
exponents.
Example
Example
L 2.
=
V2
=
= 2*
A/V2 =
4*
= V4 =
2.
2/
=
(2*)*
= /2.
2J
method should be held in reserve in case of doubt, it is more efficient to learn directly the more frequently used rules involving radicals. The student may However, while
this
prove each rule by changing the radicals to exponential form. The laws as stated below are valid when all letters stand for
positive
integers.
(Otherwise
some
exceptions
are
necessary.)
LAW
v^ = a
1.
Examples. /V*
LAW
/ab
2.
From a perfect
= ^2^ =
we see power may
this rule rth
k .
2 4 /3 ;
32
.
= ^fa^/b.
that a factor of the radicand which is be taken from the radicand if its rth
root is written outside the radical sign.
have been removed the radical terms.
=
is
When
all
such factors
said to be reduced to lowest
7.7]
ADDITION AND SUBTRACTION OF RADICALS
Exampk
1.
Vs = V4
Example
2.
v 32a 6 = v 2 a 6
Example
3.
/
(Why This
4
= /4
2 /
6
3
3
=
/2
6
(4a)
=
133
2/2. /
2a6 2 v 4a.
V4a + 166 = V4(a + 46 = 2Va + 46 V4a + 166 = V4o + Vl66 = 2a +46?) 2
2
2
2
2
)
2
2
not
2
2
.
2
common error should be carefully noted and avoided.
Ja
i.
Example
2.
+
/ 3/20
2V2
3V20 = 2V2
= Vg' 2V2
1
V 2
2
3
'^^V
'
2
2 4.
^^a
Examples, v /2
7.7.
=v 2; /
2
3VlQ
. '
LAW
/20
= v' /a = Vfl. v
Vs = V v7^ =v 2. /
Addition and subtraction of radicals
Radicals are of the same order are equal.
the index
Thus cube
roots are
if
all of
the indices of the roots the
same
order, having
3.
Radicals of the same order whose radicands are equal are called like radicals. Only like radicals can be combined by addition and subtraction.
To determine whether two
are like radicals, the radicands should lowest terms, as in the examples below. * Historically, the symbol 'v
and the vinculum 10.
is
first
a union of two symbols:
(treat as a single
number). Thus,
radicals
be reduced to
V (take the root
V64 +
36
= V64
-f
of)
36
EXPONENTS AND RADICALS
134
[Ch. VII
L Vl2 + V27 = 2/3 + 3^3 = 5V3. Example 2. /54a - v'lGa = ^/2a - 2v'2a = /2a. Example
The sums left
or differences of unlike radicals
uncombined or
else
may
may
be approximated
either be
in
decimal
form.
For example, V3 + V2 cannot be further simplified in exact form, but /3 +V2 = 1.732 + 1.414 = 3.146 (nearly).
Note that /3
+ V2 is not
since
V5,
V5 =
2.236 (nearly).
Multiplication of radicals
7.8.
By
the use of
Law
2,
same order can be found
the product of two radicals of the directly.
The law can be extended
to handle cases where the multiplicand or multiplier or both are indicated sums of radicals.
The
Law
of
following illustrative examples 2 :
Example
L
Solution.
V28 V$ =
_
/(28)(f)
VI + 3/2 2V5 - 4V2 2
Thus,
5
of
= /12 =
V5+3/2 and 2 V5-
+ 6VlO
- 4VlO -12-2 10 + 2/10 - 24 = -14 + 2VlO - 4V2) = -14 + 2/10. (/5 + 3V2)(2V5
EXERCISE Reduce
direct use
Find the product of /28 and
Example 2. Find the product Solution.
show the
36
the radicands to lowest terms.
VT6.
1.
V^
2.
5.
^27.
6. '^125.
3.
/36.
4.
A^
7.
Vl2.
8.
V20.
7.8]
9.
MULTIPLICATION OF RADICALS V28.
10.
V2.
11. /18.
14.
V54.
15. /63.
18.
v7^
19.
--v/ 128.
23.
Vl25x2
22.
26.
VlSOx.
30.
VlOSx4 98x~8
49. v/375?/ 9
56.
V/3V27.
57.
60.
V? V63. ^9 ^24.
61.
74. 77.
Vs Vl2 =
V8 /20. v^ v7 !^. 65.
V6 /4 V4 + V4 -
(2/2)(2/3)
=
4V6.
58. /5
V20.
59. v/18 /27.
62. -^5
>^^2ioO.
63. v/16
^16 ^32. =0.
67.
73.
.
54.
Solution for problem 55.
70.
.
.
53.
64.
.
^54.
66. v^25 v'lo.
68.
69.
V2
V3
71.
8x 2
= V4(l -
12x
2
16x2
.
75.
.
78.
= 2/l - 18j/ Vl6 - 4o
76.
Vg +
.
79.
V25
2
92.
.
.
2
81.
2x 2
27x*.
+
lOOx 2
.
EXPONENTS AND RADICALS
136
Combine 93. 95.
97.
the following radicals as indicated.
V27 - Vl2. V32 - VlS. V?5 + 2Vl2.
99. (1 101. (1 103. (5 105. (4 107.
94. 96. 98.
-V8) + (2 + Vl8). +V8) - (2 +VT8). - V50) - (4 -V32). -V75) - (2 - Vl08).
100.
V3(l -A/5x)
109.
V7(
V75 -Vl2. VSO - 2V. V27a3 + Vl2a62 Vl2x3y - V27xy*. .
102. A/98
104. 106.
V5(l -V2) +V2(/5 -
108.
+ ^72.
Vl25x - V45z. - 3V3. 2(/3 + 1)
1).
+/5(V3x -
1).
110. 2/8
111. 112. 113.
114. 115. 116.
117.
(2V6 - 3/3)(3V6 + 4V3). (-4 - 2V7)(-4 + 2V7). (V2x + V50>._ (3V7 - 2)(3V7 + 2). - 2>/3) 2 - 2(3 - 2V3) -2. (3 - V3)(l - /2 +V3). (1 +V2 - 3>/8 + VG). (3 + 5V2)(2V5
Criticise the errors
made in
the following.
ft
1
2c
2c
Simplify the following. 121.
X
122. x
=
=
[Ch. VII
-(66
-
a)
/36&2 rr
+
12a6
4 +/32 ,
123. x
+
a2
= ac
RATIONALIZING DENOMINATORS
7.9]
137
denominators
7.9. Rationalizing
and often desirable to move
It is possible
all
radicals
appearing in a simple fraction into the numerator. This called rationalizing the denominator. comments on the illustrative examples below, see the is
operation
For
paragraph that follows them. 7
y
Example
1.
Example
2.
Example
3.
n Example
4-
Example
5.
/
7
V3 + V3 -
1
3/3
=
jj-
22
3/3
=
1 TF= =
= (V3
-
3 /-
+
2a
l)(/3
(V3 - 1)(V3
1
V/12
=
3/12
^-
jj
+ +
1)
=
1)
3
+_2/3 2 (V3) -
+
1 2
(I)
1
=
4
+
2V3 =
2(2
+ V3) = 1
3 JV /*/'! /vv /y> J0 JliXUTTlpl/e O. 1
/y3' XII
Z
~'
sm:
3
i/^
___ ^
.
2^;
zzz:
o
t> t
^^
2/^
^.o
y*
These illustrations show some of the commonly used methods. In each of Examples 1 and 2 the numerator and denominator of the radicand are multiplied by a factor which makes the denominator a perfect rth power, so that it can be taken from under the radical sign. In Examples 3 and 4,
where the radicals appear
first in
the denominators only,
EXPONENTS AND RADICALS
138
[Ch.
VII
the multipliers must themselves be radicals. Example 5 = a2 makes use of the fact that (a 62 6) (a 6)
+
.
how
6 shows
the laws of exponents can be used to Example rationalize an expression which involves fractional exponents for
some
of its divisors in a denominator.
An
important advantage of the process of rationalizing the denominator appears in numerical cases, where it shortens the work of getting decimal approximations to the values of surds.
Example. Evaluate
Long
solution.
Short solution.
~
^ to V2
four decimal places.
=
=
.7071 (by long division).
~ = ^^ =
-~ = v2
^
While the rationalization
.7071.
*
of
denominators
is
useful at
times, the student should not get the impression that the resulting form is necessarily the 'better' one. For example,
and
^ is shorter than ~
,
.
,
Certainly
10 .
V543
.
is
just as
ul preferable to ,
,
good for
10V543 7:
543
many
purposes.
,
when one
is
pre-
paring to square this fraction.
EXERCISE
37
Rationalize the denominators. Approximate in decimal values of the numerical fractions by use of Table 2.
form
the
7.9]
RATIONALIZING DENOMINATORS
139
21. 24.
#f
2-
3 -
a;
27.
Change '1A o^.
to radical
form and ^^J j*j.
-7*Tfl/V~T <& . */ t/
I
44.
V2 V3 +V2
'
/3 50.
/
A/2
45.
.
and
-V3
5-V3 2
.
y
V3 3
1
54.
/
3a6 56.
62.
,--
V6
,-
Vx
65. -
,--
Vy
x 66.
V
2+V5
V3
1
x2+
simplify.
-V' -
33
3/c o/.
-i
y
Rationalize the denominators .
/!+*
rationalize the denominator.
TO, --~ 6
u/
30.
-
y
i
EXPONENTS AND RADICALS
140
+2 +6
5V6 3V2
,
fi Oo
/-
3V2-V3 2V3-V2
0'
,
VII
2V5-3V2 3V5 + 2V2
70
/
[Ch.
,
/
71-84. Rationalize the numerators in problems 41-7 and 61-7.
Simplify
the following
expressions,
the
leaving
denominators
rationalized.
oe
85.
'Y
/ /2V U/r 3/
o
/
,-
T
Solution
/2
2
V6
2V6
3/
VV3/A/3
3
3
9
2V =
f^l
l)
Lls) J
/2
s
/
/2
2
3
'
/2*
(3)
32 86.
s
87.
.
(V )
(V ). 4
90. (^ )<.
91.
(^f)
94.
95.
V^.
98.
V(i).
v.
7.10.
99.
Changing
When
The
2*
s
9 89.
(V ). 4
92.
(v^)
96.
V(j).
^()
100.
V()*.
93.
.
97.
4
101.
.
the order of radicals is
replaced
said to be reduced. This
is
when
radical v'a' 1.
.
the index of a radical
teger the order
than
88.
'
details
m
by a smaller is
in-
possible hi the
r have a common factor larger be carried through by changing to
and
may
exponential form.
Example
1.
/4
Example
2.
v^o*
= v^ =
=
afa
=
2*
=
a*
=
2*
= V2.
is possible always to equalize the orders of two radicals a fact sometimes useful for comparison purposes.
It
Example. Which Solution.
/2
v3 /
Hence,
vis
is
larger,
V2
or
= 2* = 2* = v^ = /8. / / 2 = 3*_=3*=v 3' =v 9. > /2.
STEPS IN SIMPLIFYING A RADICAL
7.11]
141
When
the orders of two radicals are equalized their product or quotient may be found as a single radical, as illustrated below.
Example
1.
=
/2 /3
2*3*
=
3
2
(2 3 )*
=/2 53~2 =/72.
7.11. Steps in simplifying a radical
When
the following steps are taken, a radical is said by definition to be in simplest form. The order of steps as indi-
cated 1.
is
satisfactory,
The radicand
though not necessary.
made
a simple fraction. Negative and zero exponents are removed is
by use
of
In all cases involving radicals the student should keep mind two fundamental principles
in
2.
the proper definitions. 3. The radicand is reduced to lowest terms. 4. 5.
The order is reduced as much as The denominator is rationalized.
possible.
:
Operations may be verified in cases of doubt by changing the radicals to exponential form. 2. When the processes are long and involved the practical 1.
procedure
may
be to use the decimal approximations for
various surds.
EXERCISE
38
^7 2
'
Change
the orders of the following radicals as directed.
1.
V3
to the 4th order.
2.
/2 to the 6th order.
3.
^2 to the 6th order.
4.
f$ to the 6th order.
5.
/25 to the 2nd order.
6.
v^'to the 2nd
7.
v'2 to the 10th order.
8.
V^32 to the 2nd order.
9.
v^27 to the 3rd order.
10.
v^ to the 2nd order.
order.
EXPONENTS AND RADICALS
142
Find
14. v^2,
12. /5, A^lT.
^5.
Reduce
15.
v^,
the radicals in each
v^,
13.
/3.
v/3.
problem below
to the
same
order,
then perform the indicated operations. 16.
17.
V3
v^.
18.
20.
21.
V3
v'S.
22. -~5-
23.
26.
27.
-
25. -r^-
x/9 =.
VII
the larger quantity in each pair.
11. /3, v/5.
28.
[Ch.
29.
-^
v^Vs.
19.
-^r
30. -7
31.
Express in simplest form. 34.
37
4 '
l&x y
~^7 27^2
and
Chapter Eight
THE NUMBER SYSTEM
8.1.
Imaginary numbers
Thus
numbers we have met, whether rational or have been real, or representable by points on the
far all
irrational,
line of Fig._l. To get the point representing the irrational for instance, we simply measure to the right
number V2,
of the zero-point a length equal to the diagonal of a square with unit sides. But, since 2 2 = 4 and ( 2) 2 = 4, the quantity exist
V
or the
4,
among
number whose square
is
4, just
does not
the real numbers. It turns out to be highly
useful in mathematics to invent a
new type
of
number
in
V
terms of which we can express such quantities as 4. 1 is designated by By definition, then, the quantity
V
the letter root of
That
i.
1.
is,
=
1,
root of
and
i is the
principal square
imaginary unit.
1 is
V
1,
or
i.
the rules for operations with radicals (Art. 7.6)
the rule _that
V^T Va
2
It is also called the
The other square
Among
i
Vab = Va Vb.
= iVa. That
Hence,
V^a = V(
l)a
is
=
is,
(1)
Examples.
V
4
= iVi =
= (+2) 4 =
2z;
16, V-16 cannot be either +2 In or general, it can be shown that all even roots of negative numbers are imaginary, or expressible in terms of
Since (-2) 4 2.
143
THE NUMBER SYSTEM
144
[Ch. VIII
There are still other quantities which can be so expressed, but we shall not consider them here.
the imaginary unit
An
imaginary number is one which can be bi, where a and b are real and 6^0. form a
Definition. written in the
Examples,
i.
+
i,
Z
,
+
4
6
5i,
iy
V
2.
4, *$/
Imaginary numbers should be expressed in terms of i before they are used in algebraic operations. For instance, it is false that =V(-3)(-3) = V9 = 3, since
V^3 V-3
from the definition of the square
we
write
V
3 as
Powers of
(V^3) = 2
3.
i
2
(/3)
2
=
i,
= -l
(by definition).
Note that
t
5
=
i
1 ,
t
6
=
t
2 ,
t
7
=
t
/2w/6 /or finding the value of 4.
The remainder
3 ,
i
= i^' = i = i4 t 9 = t
9
8
3
=
any power of will
be t
0,
=
i
1
2,
1, 1
=
i y etc.
etc.
,
i.
=
Divide the or
3. 2
i,
i
= - 1, i27 =
t
1, i
=
The 1,
i.
Examples, 8.3.
.
,
exponent by corresponding values of the power are i
1,
i
i
and
i,
= -3.
(-1)(3)
of i is equal to one of the four numbers This conclusion is reached as follows:
1.
if
y
Any power and
But
iV3 then
2 ^3) =
8.2.
root,
z
20
=
1, i
25
=
i
1
=
i, i
30
=
i
2
3
=
-i.
Operations with imaginary numbers
The
results of simple operations with
can be expressed in the form a examples below.
+
imaginary numbers
bi y as in the illustrative
8.3]
OPERATIONS WITH IMAGINARY NUMBERS
145
Addition.
+ 3i) +
(2
-
(4
= =
5i)
+ 4) + (3i + (-2)i
(2
6
=
5z)
-
6
2i
Subtraction.
-
(3
-
7t)
-
(4
= =
9i)
-
(3 (3
7f)
4)
+ (-4 + 9i) + (-7z + 9z) = -1 + 2i.
Afwttip/ica^on.
+ 2i)(4 -
(3
= =
i)
+ 8t + 5z -
12 i2
-
3t
2z 2
=
2(-l)
14
+ 5t.
1 JL
^^ 7i' i U
Dmszon. ,/
v
1
+ 3t
=
/o *J
(1
'W1
+ 30(1 -
~ = -1
7t
than a
2
6
EXERCISE
(a
V-25.
3.
V
32. (3 35. (4
4x
-
1
1
-
9t
2
,
10
=
60
a
2
t2*2
=
a2
+6
2
rather
.
39 the following expressions to simplest form.
Answer:
2 .
V
2.
5i.
17.
Answer:
Answer:
+ 20(3 - 20+ 0(4 - 30-
38. 3t(3
30
2
Reduce each of 1.
+ 6t)(a
=
tJt/J
=
1+9
Note that
3-A
7 t/j^.
0-
15.
-v-i
18.
-V-25t/4
21.
-V-23a
24.
z
.
2 .
14 .
27. (1
+O
2
30. (2
-O
2
-
-
+ 40(5 - 40- 50(6 + 50(6
+ 30(3 t(3 + 40-
33. (5
34. (2
36.
37.
39. 2(3
-O
3
2 -
40.
0-
THE NUMBER SYSTEM
146
64.
V-3 +
[Ch. VIII
V^5 V^5. 63. V-4 + V-16. 65. V^2 v^3 V^5.
2V-12.
'^10
/9
+V-9
8.4*
We are now prepared to examine again this thing called a number. To represent all of its types thus far met we shall need to extend Fig. 1 as shown in Fig. 13. The line of Fig. 1, now called the axis of reals, contains all numbers except imaginary ones. Represented by points on this integers (0, 1, 2,
2, etc.), fractions
1,
(,
f,
line
are
J,
ty,
and irrational / 2 +v 16, TT, etc.). But there is no numbers (V2, /7, place on this line for imaginary numbers such as i and 1 2i. The place is provided in a very simple and neat etc.), decimal numbers
* This article
may
(0.5, .33,
be omitted without
Article 7.4 should be re-read
first.
1.21, etc.),
loss of continuity.
If it is studied,
8.5]
NUMBERS
CLASSIFICATIONS OF
147
manner by means of the plane (that of the paper) containing the axis of reals. In Fig. 13 the vertical line is called the axis 2i, and of imaginaries. It contains all numbers, such as i, 'iy
+
which have the form
where b
bi,
is
real
and not
Axis of imaginaries
-*- Axis of reals
Fig. 13
Such numbers are called pure imaginaries. The general number a + bi, where a and b are real, is represented by a point whose coordinates would be (a, 6) if the axes of reals and imaginaries were respectively the X and Y axes of the zero.
rectangular coordinate plane. Thus the point representing 2i has the rectangular coordinates (1, the number 1 2). in The numbers represented their totality by all of the points on the plane are called complex numbers. containing them is known as the complex plane. 8.5. Classifications of
The
numbers
following diagram
Numbers
is
helpful.
in General, or
Complex Numbers
Real Numbers
Imaginary Numbers
L Classifi cation 1
Zero Positive
numbers
Negative numbers
The plane
2 Rational numbers
Classifi cation
Irrational
numbers
THE NUMBER SYSTEM
148
In this diagram Classification represented
by points on
1 divides
[Ch. VIII
the real numbers,
the axis of reals, into three groups
corresponding with: (1), the point the right of 0; and (3), points to the
(zero); (2), points to left of 0.
simply another grouping of real numbers. Here algebra and reason have stepped beyond geometry, since the points representing rational and irrational numbers are mixed together too thickly in any segment of the axis of reals to be separated visually, though we may Classification 2
is
find sample points representing each of the
8.6.
Numbers
two types.
as roots of equations
It is interesting that the
requirements for roots of rational
integral equations in one unknown, which appear in the solutions of very simple problems, have called into use all
the types of numbers thus far discussed. illustrates the point. Root or roots
following table
Type of number represented by the root
A 3
The
type by
itself
Positive integer
2
Negative integer Positive fraction
_5
Negative fraction
+ /2
numbers Pure imaginary numbers Irrational
Imaginary numbers
EXERCISE
40
Show on a complex plane the points representing the follownumbers: i; (c), 3i; (d), 1 i; (a), 2; (b), 2 i;(e), ing 1.
+
(f), (1),
*
-2 + 3t; -3 - 2. The
(g),
4
-
i;
(h),
;
(i),
;
(j),
+
solutions of these equations will be discussed in Chapter 9.
00,
0;
8.6]
NUMBERS AS ROOTS OF EQUATIONS
2.
The numbers a
naries.
What
is
149
+
bi and a bi are called conjugate imagithe relation between the points which represent
them on the complex plane? 3. In the light of the discussion in answer the following questions? (a)
How many Is so,
how would you
real numbers are there between and 1? a smallest positive number, and, if so, what is it? there a largest positive number which is less than 1? If
(b) Is there (c)
this chapter,
what
is it?
Chapter Nine
QUADRATIC EQUATIONS
9.1.
The quadratic in standard form
A quadratic equation, or simply a quadratic, in a given letter is
a rational integral equation of the second degree in that
letter.
Example
1.
2x 2
Example
2.
3y
3x
2
+
2ay
=
1
+
by
(quadratic in x).
+
2
c
=
(quadratic in y).
When
terms of the same degree are grouped together, a quadratic in x may be written in the standard form
ax 2
(1)
where the
+
bx
+c=
coefficients a, 6,
2 Example. Write 5z ard form.
-
a
c
do not involve
and
7z
+3 =
2x*
-
7x
-
2x 2
0,
+ 3x -
#.
4 in stand-
we have
Solution. Grouping, the terms,
-
^
0,
3x
+3+4=
0,
=
0,
=
0.
or (5
-
2)*
2
+
(-7 -
3)x
+
(3
3z 2
-
lOz
+
(1),
we find
+
4)
or (2)
6
Comparing (2) with standard form = - 10, and c = 7. 150
7
that a
=
3,
THE ROOTS OF A QUADRATIC EQUATION
9.2]
9.2.
The
151
a quadratic equation
roots of
We
have seen that a root of an equation is a number or expression which satisfies the equation when substituted for the unknown. Thus, for the quadratic x2
(1)
one root is second root is
The
5,
2
-
4z
5 2
since
since I
1,
+
5)
(
+
4
=
0,
+ 4( 5
1
5)
=
5
=
0;
and a
0.
roots of
x2
(2)
+
4z
-
1
=
+ /5 and 2 /5. Checking the first one, we find that (-2 + /5) + 4(-2 +V5) -1=4- 4/5 + 5-8 + 4V5 -1 = 0. The test of the root -2 - V5 is left to the 2
are
2
student.
Again, the quadratic
x2
(3)
by x =
+
4x
+
4
=
but in this case there is no second root different from the first one. For reasons to be discussed 2 is a double root of (3). later we shall say that satisfied
is
2;
Finally,
x2
(4) is
satisfied
where 4
by the imaginary
8
=
roots
2
+ 2i
the imaginary unit (Art. 8.1). For 4z 2 - 8 Si 8 = 4 - Si 2i)
i is
+ + + 4(-l) =
4(-2
=
+ 4* + +
0. Similarly,
-2 -
+
and
2
2i,
(2 + 2z) + + 8 = 4 + 4i 2
2
2i satisfies (4).
The
roots of equations (l)-(4) represent among them the four different kinds of root-pairs for quadratics rational and unequal, irrational, rational and equal, and
imaginary. In Art. 9.9 we shall discuss a geometric interpretation of the various types. As may be suspected here, and will be proved later, every quadratic equation has two roots,
though
in
some
cases they are equal.
QUADRATIC EQUATIONS
152
EXERCISE it
[Ch.
IX
41
Write each of the followng equations in standard form. Comparing c = 0, find the value of a, b, and c in each case. ax2 bx
+
+
with:
Prove that each of the numbers
3x
1.
-
2
=
2x
1;
(1,
listed is
a
root.
2 -i). Solution. Bx
-
-
2x
1
=
0,
or
+(-2)x+(-l) = 0;a=3, 6=-2, andc=-l;3- ! -2- 1=1; 3(-*) - 2(-i) = 1. 2. 4x = 1 + 4x 3. 6 = 2x 2 + x; (-2, f). ( ). 2 2 4. 1 = 3x + 2x; (-1, ). 5. -3x = 2x + !;(-!, -). 2 6. 3 = x + 2x; (1, -3). 7. 6 = 5x + 4x 2 (-2, ). 8. 3 = x + 2x 2 (1, -f). 9. 3x = -2 - x 2 (-1, -2). 11. 4x = 3 - 4x 2 (, -f). 10. 5x = 2x 2 + 2; (2, ). 12. 8x = -3 - 4x 2 (- J, -f). 13. 2x = x 2 (0, 2). 14. x 2 = 3x; (0, 3). IS. 2x 2 = 3x; (0, f). 16. 2x = -3x 2 (0, -). 17. 3x = -2x (0, -f). 3x
2
2
2
2
;
;
;
;
;
;
;
2
;
18. <>
20.
=x
2 ;
(d,
;
19.
-d).
x2 -(r+s)x+r*=0;
4x2
=
-
9e';
,
21.
(r, *).
22. x2 +(r-s)x-r=0; (-r,s).
23. o 2 x 2
=
16b2
;
,
Write each of the following equations in standard form and find the expressions corresponding with a, b,
24.
4x2 + x - Bx = C + x2 -
+4x+2=4x2 +x+C. 3x +Bx-x=4x2 -Cx-fl. Ax* + B = Cx2 + Dx.
4.
26.
2
28.
29.
9.3.
T/ie solution of ax*
When
6
=
0,
+
30.
c
-
l)x
+ (1 - B)x
Ax*+B+Cx=Bx*-Ax+C. ex2 + 2Bx = 4 - x + Ax.
the general quadratic equation
ox2
2
=
takes the form (1)
c.
Answer. (A
25. 3x 2 27.
and
+c=
0.
(1),
Art. 9.1,
SOLVING QUADRATICS BY FACTORING
9.4]
Since (1)
linear in the
is
unknown x2
(2)
it
,
153
follows that
*__!,
and hence,
-
Example
1. If
If
2.
~~
-
4i!
-
2s 2
=
9
3
=
0,
x1
0,
x2
=
4
=
and x ^,
Ve 2~~~2~' _
/3
and x = f or x 2
?,
=-
-
2
= Vs = ^r
or
=
or
-
Exampk
3. If
Example
4-
4s2
Solve
+5=
:
5x 2
=-
0, a*
f
,
and Z
-
Multiply both members of the equation by 2 ). The new equation becomes 3 !)(!
Solution. 2
(So:
+ 4x
-
2,
= -t.Vio
+
or 2z
2
= -5. Hence x = 2
and x =
f
or
.
9.4. Solving quadratics
When
,
the
left
by factoring
member
of (1), Art. 9.1,
is
factorable,
when
and
the factors can be found readily, the equation can usually be solved most easily by the method illustrated
below.
Example
1.
Solve: x2
Solution. First (1)
we
+ 4x
5
=
factor the left (x
-
1)(*
+
5)
0.
member, thus
=
0.
:
QUADRATIC EQUATIONS
154
This equation
is
by x =
satisfied
since (1
1,
[Ch.
IX
+
5)
1)(1
= 0; and also by x = 5, since 5 5 + 5) = 1)( (_6)(0) = 0. Hence 1 and -5 are roots of (1). We get x = 1 1 equal to zero. Similarly, x + 5 = by setting the factor x = 5. gives x =
6
(
+
Solve: 4z 2
+
12x
Example
2.
Solution.
Noting that the
left
9
=
0.
member
is
a perfect square,
we have
+ 3)
(2z
(2)
2
=
or
0,
(2x
+
3)(2z
+
3)
=
0.
= f This exfactor, placed equal to zero, yields x repeated or double plains the algebraic basis for calling Each
.
fa
root.
The geometric reason
will
appear
later.
CAUTION. The solution of a quadratic by factoring the
member zero.
is
is
left
correct in principle only when the right member test will show, the roots of the
For example, as a
quadratic (x
(3)
are not found
by
is left
explanation which are 6 and
new
left
member
-
2)(x
-
=
3)
12
2 or x 3 equal to 12. The setting x to the student. The correct roots of (3), 1, are found when 12 is transposed and the
is
simplified
and factored.
-
Examples. Solve:
Solution. After clearing fractions
-
by the L.C.D. 4(2z (4)
-
7(2x
which
l)(2x
simplifies,
l)(2x
+ 3)
-
+
3),
+ 3) =
24(2*
after the
by multiplying through we get 4(2x
-
3)(2z
-
1),
indicated operations are per-
formed, to
4z 2
(5)
The
factors of the left
so that x
=
+
-
35
member of
% or f Each .
4z
=
0.
(5) are
2#
of these values for
to satisfy the original equation.
and 2x 5, x will be found
-f 7
9.4]
SOLVING QUADRATICS BY FACTORING
EXERCISE
155
42
Solve by the method of Art. 9.3 where possible; otherwise by the factoring method.
-
58.
= 0. 2. 3x - 12 = 0. 3. 5x - 20 = 0. 2 2x 18 = 0. 5. 2x - 98 = 0. 6. 4x 2 - 100 = 0. 2 3x + 27 = 0. 8. 2x + 8 = 0. 9. 3x 2 + 12 = 0. 11. 2x2 + 98 = 0. 2x + 18 = 0. 12. 4x + 100 = 0. 2 = = 14. 3x 2x 36 0. 15. 2x - 100 = 0. 36 0. 2a x - 64 = 0. 17. 26 2x - 24 = 0. 18. a x - 96c = 0. 2x + 36a2 = 0. 20. 3x 2 + 366 = 0. 21. a x 2 + 1006 = 0. 23. 2x 2 + 48c = 0. 2x + 6462 = 0. 24. a x 2 + 966 = 0. 26. ax 2 + lOOc = 0. ax + 256 = 0. 27. 206 = -ax - 2)(x + 3) = 6. 29. (x + 2)(x - 3) = 6. (x 31. (x - l)(x - 2) = 6. x(x + 5) = 14. = 4. 33. x(x - 3) = 10. (x + l)(x 2) 2x2 + x - 1 = 0. 35. 3x + 2x - 1 = 0. 37. 4x 2 - 3x - 1 = 0. 3x - 2x - 1 = 0. 39. 9x + 6x + 1 = 0. 4x - 4x + 1 = 0. 4x + 126x + 96 = 0. 41. x - Sax + 16a 2 = 0. 43. 6x - x = 0. 9x + 30x + 25 = 0. 45. 8x 2 + lOx = 0. 9x - 3x = 0. 47. ax 2 + 6x = 0. lOx + ax = 0. ex - dx = 0. 49. 6x + x = 2. 2 = 51. lOx - 13x = 3. 8x + lOx 3. 53. 15x + x = 2. lOx2 - x = 2. 55. llx = 6 + 3x 2 x = 2 - 15x2 57. 9x = 2 + lOx 9x = -2 - lOx 2 59. 23x = 6 + 21x 2 23x = -6 - 21x2
60.
25x
1.
4. 7.
10. 13. 16. 19.
22.
2x 2
2
2
8
2
2 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
.'
25.
28. 30. 32. 34. 36. 38. 40. 42.
44. 46. 48. 50. 52. 54. 56.
2
2
.
2
2 2
2
2
2 2
2
2
2
2
2
2
2
.
.
2
.
.
.
- -6 -
5
64.
2
14x2
.
^- = % - -^^2
.
7
2 (2x-3)(2x+3) = 7x -4.
61.
_ 65.
x
5
15
x
2
,
+
12
^
=
0.
QUADRATIC EQUATIONS
166 1 t
5
h
v'
1
*
4z 2 2
,
8z2
=
-
-
16z 4
5
IX
3
=
-
25*
-x +2
1
x- 2^5(2 -*)-
68. 10z 2
70.
4'*'
[Ch.
a:
7*.
3x
=
0.
9.5. Solving quadratics
by completing the squares
The
algebraic methods of solving quadratics thus far discussed, while short and efficient, cannot always be applied.
The method
called
'
completing the square' has the ad-
vantage that it will solve all types of quadratic equations, besides giving practice in an operation which is useful else-
where in mathematics.
To
illustrate, consider
3z 2
(1)
Step
L
the equation
-
2x
-
2
=
0.
Divide both members by 3: ~2 x
/o (2)
^ ^ n _____
o.
Step 2. Transpose the constant term: /o
3.2
/
_
?x o
_
2 o
Add to both sides the number which will make the member a perfect square. Note that in the perfect square
Step 8. left
x - a )2 =a: 2._2aa;+a 2 the third term, or a 2 is [( )(-2a)] 2 Hence we add to both members of the equation the square (
,
of one-half the coefficient of x, or [(!)(
yielding
.
,
2
I)]
i
n
this case,
9.5]
SOLVING QUADRATICS BY COMPLETING SQUARES
157
HH
or
Note that the number added to complete the square is always positive, and that the middle sign of the left member of (5) is the same as the sign before the term of first degree in (4).
Step
4-
Solve for x !
*.
5.
/7
4.
Transpose the term
(7,
Equation (*) (&)
:
s-^+^i+
(R (6)
Step
.-'* (7) is
^:
o,
-!$
a brief way of stating that
--13
x
V?
-
u
,
or
-rx
1
'^ 5
o
Note that the various steps yield equivalent equations, and that the two equations in (6), (7), and (8) are equivalent to the one equation (5). In other words, each of the two values of x shown in (8) satisfies (1), as the student may verify. If, at the stage in the solution represented by equation (5), the right side is negative, the roots will be imaginary. For
example, given (9)
x*
+ 2x + 4 -
0,
then (10)
x2
+ 2x + 1 = 4 + 1; (x + l) =-3; x + l ^V^3 =iV; z=-l tV. 2
(11) (12) (13)
QUADRATIC EQUATIONS
158
The sum and product of
9.6.
[Ch.
IX
the roots
To
get a result which will prove useful for checking purposes, we note that (1) is
ax 2
+
x2
+
equivalent to
(2)
obtained from
(1)
-
satisfied
by x =
r
(r
a
+
a
by dividing the x2
(3) is
+c=
bx
=
coefficients
by
a.
Now
+ s)x + rs =
and by x
=
s,
as
may be verified by trial.
Hence if the general equation (1) is also satisfied by the r and s, it follows, by comparison of (3) with (2), that r
(4)
roots
+ s=-^
and
=
(5)
-
In words, .
the
sum
of the roots of (1) is
,
and
their product
c
is
a
be used in checking the roots found in the solution of any quadratic. The check is often shorter than that obtained by substitution of the roots in the original equation, particularly when the roots are irrational or This result
may
imaginary.
Example L^By use .,
.
that
of the^sum
-+Vll - --Vll 1
2
and
and product formulas, show
1
r
.
,
2t
2x* Solution.
a
-
.,
are roots of the equation,
2x
-
5
=
0.
THE SUM AND PRODUCT OF THE ROOTS
9.6]
159
and
+VTI
1
-Vil _
' 1
_
_
+VTT/i -Vilx =
/i
(2 X^
,
and
Example
2.
Show
-/IP =
44
2
T
)
that r
~ 3 + tv/31
=
are roots of the equation, 2z Sofoilton.
r
i
+s=
2
+
+
3z
10
-T
an d
=
5
s
=
=
5
-2
~ 3 ~ t%/31
0.
636 -=
-= ---
4
2
a
Also,
= /_ 3 Y _
Find
^^ 4
4J
(
EXERCISE
i
t
=
_ ~ 31 =
40
16
16
16
=
5 2
the roots of each of the following equations
x2
5.
2
9.
11. 13. 15.
17. 19.
21. 23.
6z
+8=
0.
- 8 = 0. x - 2z - 8 = 0. 6x - 7* - 3 = 0. 6x + 7x - 3 = 0. 12x2 + 7x - 12 = 0. 3x - 2x - 2 = 0. 5x - 5x - 7 = 0. 2x + 6x = 5. 7x = 14x - 3. 5x = -3x - 4. 4x2 = -6x - 3.
3.
7.
-
+
2x
a
by the method of
sum and
formulas.
x2
c
43
completing the square, and check by use of the
1.
=
2.
4. 6.
2
8.
2
10. 12.
2
14.
2
16.
2
18.
2
20.
2
22. 24.
x2
+ 5x + 6 = -
-
0.
= 0. x + 3z - 10 = 0. 6x2 + x - 2 = 0. 6x2 - x - 2 = 0. 8x2 - 6x - 9 = 0. 5x + 3x - 3 = 0. 3x + 4x = 5. 8x2 + 4x = 3. 9z2 = 6x - 5. 3x2 = -5x - 4. 10x2 = -3z - 2. x
2 2
2 2
3x
10
product
QUADRATIC EQUATIONS
160
25.
3z
27.
3z
29.
6z
31.
(re
33.
= = =
a:
z
2 2
z2
+ 7. + 6. + 2.
+ l)(z (x + 3)(z
34-43. Check
= 4z = 8z = (x
26. 2x 28.
30.
= =
2) 4)
32.
3.
IX
+z z + 6. z + 3. 2)(z + 3) = -7. 2
5
.
2
2
1.
substitution the
direct
by
[Ch.
answers found in
problems 1-10. Solve,
44.
9.7.
z x
and check
+3 -
TTfce
1
3
answers by direct substitution.
the
-
2z
2z
7
+2
2
quadratic formula
When we (1)
solve the general quadratic
ax 2
+
+c=
bx
by completing the square, the various equivalent equations obtained in the successive steps are as follows (2)
x2
++ a x2
(3)
/r
/
--
-
0;
=-
+
v
i
=
a
;
u
:
4dC.
4a (6)
(7)
x
A + 2a
=
x
=
I
y-k
_!_
x->
2a ^
'
formula. In a sense, the infinitely many quadratic equations are here solved collectively once for all. They are seen to have two roots apiece,
Equation
(7) is called the quadratic
THE DISCRIMINANT OF A QUADRATIC EQUATION 161 though these roots are equal when the quantity under the radical sign is zero. To get the roots of a specific quadratic 9.8]
necessary only to identify the coefficients a, 6, and c, and then to substitute their values in (7). If the equation is it is
quadratic in some letter or quantity other than or quantity should replace x in (7).
x, this letter
Example. Solve the quadratic 3z 2
(8)
Here a =
Solution.
these values in (7),
-
3, 6
2x
=
-
4
2,
=
0.
and
6
6 is,
one root
.
,
4.
Substituting
+ V4 - (-48)
_ /13)
6
=
1
6
+VT3 3
and the other
-~r
is
2
-/13 is
o
After getting x to write x
=
we have
-(-2) + V(-2) 2 - 4(3)(-4) _ 2-3 _ 2Vl3 = 2(1 = 2 +V52 = 2 That
c
=
= 4VT3
o
=-,
or x
6
the careless student
= OVT3 Why 6
is
likely
u are these results .,
.
m-
correct?
Note that
if
were solved by completing the square, the would be obtained at once. This is some who work in the field of applied mathe-
(8)
correct simplified root
one reason why matics prefer the square-completing method to the formula. The latter method, on the other hand, is shorter for some problems, especially where literal coefficients are involved; and it also gives other useful information about quadratics,
we
as
9.8.
shall see.
The discriminant of a quadratic equation
The (1)
roots of the quadratic
ax2
+
bx
+c=
0,
QUADRATIC EQUATIONS
162
as given in the quadratic formula, rately thus:
_ -b + Vb - r2
/0 ,
_
(2)
The radicand
62
2
,
(2),
which we
IX
be written sepa-
- s-_ -b -Vb
4ac
4ac in
may
[Ch.
may
4ac
designate
by
called the discriminant of the quadratic equation (1). D, For example, the discriminant of is
2z 2
(3)
for
which a
9 +_32
=
= 2,
41.
6
-
3x
-
4
-
0,
= -3, andc = -4,
Hence, the roots of
V41, and will be irrational. Having the formulas (2)
is
2 (~3) -4(2)(-4) =
(3) will
involve the radical
mind, we can draw certain conclusions about the roots of (1) from the value of D, assuming that a, fr, and c are rational numbers. Premise about (a)
(b)
D> D>
in
D
Then the
and a perfect square and not a perfect
(d)
unequal, and rational.
real,
and
unequal,
irra-
tional.
square (c)
real,
roots of (1) are
D= D<
real, equal,
and
rational.
imaginary.
These results have immediate and practical application as an aid to the solution of quadratic equations. When, for a
D
is negative or not a perfect square, particular equation, it is useless to waste time on the factoring method of solu-
tion, since the roots are
Question.
Can 3x 2
imaginary or irrational.
4x
5 be factored?
60 = 76 Answer. Since D = (-4) 2 - 4 3(-5) = 16 (not a perfect square), the answer, if we bar factors with irrational coefficients,* is 'No.' Hence, the quadratic
+
*
Note that 3z 2
-
4z
-
5
=
/
3
I
x
- -2
--VigW x 11
cover after solving the equation. This type of factoring obtaining the solution.
- -2 + Vl9 -
1,
is
as
we
dis-
of course not helpful in
9.8]
3z
THE DISCRIMINANT OF A QUADRATIC EQUATION
2
4z
must be solved by completing the square the quadratic formula. If the latter method is
5
by use
or
of
used, the test-value
writing
D=
In each
:
6 44
of problems 1-45, determine the nature of the roots by use
of the discriminant.
Then
perfect square; otherwise
1.
3. 5. 7.
9.
11. 13. 15. 17.
19.
21. 23.
25. 27.
29. 31. 33. 35. 37. 39.
41. 43. 45.
76 already found can be used_m = 4 + V76 2 Vl9
down at once the solution x
EXERCISE
zero is
solve the equation
by factoring if
D
- 3x + 5 = 0. 4x2 - 4x + 1 = 0. x2 - 3x + 2 = 0. x2 + x - 3 = 0. x - 3x + 1 = 0. x - x + 3 = 0. x2 + x + 1 = 0. 3x + x - 3 = 0. z - 10* + 1 = 0. 3x2 - 5x + 1 = 0. - 5y + 1 = 0. 4?/ 2 9x 3x + 2 = 0. 2 5x 6* + 1 = 0. 2 $y + 2 = 0. 3y 10x - 3x - 1 - 0. 7x2 - 3x - 2 = 0. % - 12y + 1 = 0. 15x2 - lOx + 1 = 0. ax2 - bx - c = 0. ay + Vy + c2 = 0. 3ax + 26x + c = 0. - l)(x - 2) = 3. (x = 5. y(y + 4) 2x2
3x 2
4.
2
6. 8.
10.
2
12. 14.
2
16.
2
18.
20.
2
2
.
- 2 = 0. 5x + 2x - 3 = 0. x - 6x - 9 = 0. x + 3x - 1 = 0. x - 2x + 2 = 0. 9x2 - 30z + 25 = 0. 2z - 5x + 1 = 0. 4z2 + 3 - 1 = 0. - y + 1 = 0. ICty 3x - 7x + 2 = 0. 7z2 - 8x + 1 = 0. - 5y + 1 = 0. Gy 8z - 7z - 1 = 0. 5x - 4x - 1 = 0. 3/ - Qy + 2 = 0. 12x - Sx + 1 = 0. llx - 12z + 1 = 0. 12^ - lOy + 2 = 0. 6x2 + ex + a = 0. ox2 + 26z + 3c = 0. - by - 2c = 0. 2ay - l)(x + 2) = 2. (x
2.
2
2
is
a
by use of the quadratic formula. (Note that
a perfect square.)
2
163
=
22. 24. 26.
28. 30. 32.
34. 36. 38. 40.
42. 44.
+
2 2 2
2
2
2
2 2 2 2
2 2
2
5x
QUADRATIC EQUATIONS
164
46. Solve s
=
47. Solve s
=
for
vrf
n -^
[2a
+
(n
-
for
t
t>
[Ch.
IX
.
l)d] for n.
t
+ y + 4x + 2y - 1 =
48. Solve x 2
2
49. Solve g
=
^
=
S0 a S lve
9.9.
.
'
C
for y.
R for
- 2c) - c)
R;
for x.
-
f r C'
T'/^ graphical solution of
Four types of results
a quadratic equation
for quadratic equations are illustrated
by the roots of the quadratics (l)-(4) discussed in Art. 9.2. The graphs of the functions of x in the left members of these equations give us a geometric interpretation of their roots. Designating each of these functions as y, we have the four equations:
=
x2
(1)
+
y
+
=
x2
+ 4x
4:X 4, and (4) y y in table the below. analyzed
(3)
5,
(2)
= x2
+
y 4x
=
x2
+ 8,
+ 4#
1,
which are
GRAPHICAL SOLUTION OF A QUADRATIC EQUATION The four graphs are shown in Fig. 14. The curves are members of the family of curves
9.9]
165 all
c.
(5)
y
Each value assigned
to c yields a special curve, as indicated
in the figure.
The graph
(5,
of (1) crosses the X-axis at the points (1, 0) and 0). Another way of saying this is that the rr-intercepts
166
of the
graph are the equation
and
1
5.
+ 4z -
x*
(6)
For evidently at x
=
QUADRATIC EQUATIONS [Ch. IX These numbers are the roots of
=
5
0.
and
5 the curve (1) crosses the and hence (6) X-axis, so that for these values of x, y = is
1
satisfied.
Similarly, the graph of (2) crosses the X-axis in two points; but in this case their abscissas are irrational, namely,
2
+ V5
and
V5. These numbers
2
are the roots of
the quadratic
In the graph of
ward
until the
(2,
0).
(3)
1
=
0.
the curve has in effect been lifted upat the point
two ^-intercepts have coincided Hence the equation
+
z2
(8)
may
+ 4z -
x2
(7)
be said
still
+4=
4*
two roots
to have
2 and
(
2)
which are
identical. The graph of (3) is said to be tangent to the X-axis at (-2,0). Finally, since the graph of (4) does not intersect the X-axis, there are no positive or negative values of x for which y = 0,
and hence there are no z2
(9)
Summing
real roots of
+
8
=
0.
up, if the roots of the quadratic
ax 2
(10)
are real,
+
4x
+ bx + c =
we may find them as accurately as the precision of the from the graph of the corresponding equation
drawing allows (11)
y
=
ax 2
+
bx
Incidentally, the graph of (11),
+
c.
where a
^
0, will
always be a parabola, a very important curve entering into life and mathematics in many ways. Different sets of values for a, b,
and
c give different parabolas;
in general like the ones
shown
but
all
them are shaped They open upward
of
in Fig. 10.
9.9]
GRAPHICAL SOLUTION OF A QUADRATIC EQUATION
as in that figure
a
if
is
positive,
and downward when a
167 is.
negative. The formula for the abscissa (or x- value) of the vertex of each parabola represented by (11) is
,
(12)
Example. Sketch the parabola, y
(13)
Solution.
=-2z + 2
Using (12) and
of the vertex are:
x=-
(13),
4x
-
we
find that the coordinates;
1.
4
1-1 = 1.
2
=1; ?/=-2(l )+4 Tt
The curve opens downward, negative. The coordinates, (0,
since the coefficient of x2 1),
of the point at
is
which the
curve crosses the F-axis are easily found by setting x
=
0*
we know
the vertex, the direction of the axes, and one on the point curve, we can now sketch the parabola shown in Fig. 15. For greater accuracy, we of course use more points whose coordinates satisfy (13). From the graph we might estimate the values of the two roots of Since
-2z2
(14)
as about
.3
and
1.7.
nearly as estimated.
The
+
4*
-
1
=
Actually they are
2
+ /2
^
,
or very
graphical solution of a quadratic equation is interesting not only because it gives geometric meaning to the differ-
QUADRATIC EQUATIONS
168
ent types of roots encountered, but also because
it
[Ch.
IX
illustrates
a very general method of getting approximations to the real roots of an equation in one unknown. This method is very
no shorter way can be found. In the case of quadratic equations, however, the algebraic methods are more efficient, precise and practical. reliable as a last resort
EXERCISE
if
45
Graph the left members of equations 1-15, Exercise 43. From the graphs, estimate to one decimal place the values of the real roots, and compare with the precise values found algebraically. 9.10. Quadratics with given roots
THEOREM
1.
If r
a root of the equation
ax 2
(1)
then x
To say
that r
2
(1)
+
bx
and
+
+
br
0,
member
+c
(1)
=
of (I).
means that
0.
(2),
c)
ar
or, since
left
a root of
is
ar2
(2)
From
+ bx + c =
r is a factor of the
Proof.
(ax
is
2
ax 2
-
(ar
2
+
+ br + c + bx + c
br
= = =
+
=
c)
a(z
2
-
r2 )
+
b(x
-
r),
0,
a(x
r)(x
(x
r)(ax
+ r) + b(x + ar + 6).
r)
2 has the root ^, and, there3 = Example. 2x + 5x 2 5x 3. must be a factor of 2# fore, by the theorem, x 2 = 3 2(x &(x + 3). Actually, 2z + 5x From Theorem 1 there follows, as a corollary,
+
THEOREM the (3)
2.
All quadratics whose roots arer and
form a(x
-
r)(s
-
where a can be any constant except
*)
=
zero.
0,
s,
must take
PROBLEMS LEADING TO QUADRATIC EQUATIONS
9.11]
Example 1. Find a quadratic equation with and cients whose roots are f
169
integral coeffi-
.
Solution.
The equation must have
the form
+ f) =
Letting a
6 to clear fractions,
Qx
Example .
,
cients
2.
2
=0.
we have
5x
6
=
(answer).
Find a quadratic_equation with integral
,
,
whose roots are
- -2
coeffi-
+ -tV3 Zi
Solution.
or,
The equation must have
when a =
2
+ tV3T
2
-
2
the form
-
tV3'
4,
(2z
-
-
2 -f tV_3) 2 2 2) (zV3) (2x 2 4x 8x 7
tV3)(2z
or
+
or
EXERCISE
= = =
0, 0,
0.
46
Find equations with
integral coefficients whose roots are the
num-
bers below. 1.
7. f,
11.
2. 3,
1, 2.
3
IS. 1
-f V2. i.
-1.
-1, -3.
3.
8. 0, f.
4. 0, 2.
9. 1,
12. 1
~
13.
16. 2
3.
17.
-4.
10.
-f.
2
6.
i
f.
+ V3.
3
X
^ 2
5. 0,
-
i/3.
14. 18. 1
t.
+
^-
9.11. Stated problems leading to quadratic equations
The
stated problems calls for the use of quadratic equations. If two different numbers satisfy the conditions of the problem, they will be the roots of the solution of
many
QUADRATIC EQUATIONS [Ch. IX quadratic obtained. If only one number meets the required condition, either this number will appear as a double root or else the second root must be rejected as meaningless. Finally, 170
the conditions as stated are inconsistent, this fact will appear algebraically in the form of imaginary solutions to if
the quadratic.
Example is
1
.
Find two consecutive numbers whose product
56.
x
Solution. Let
Then
x x(x
Solving,
+ + 1)
1
= = =
the smaller
number
(algebraically).
the other number. 56.
we have x
+
x
i
Hence the answers
= =
7 or
Its length
A
building lot has an area of 56 square rods. one rod greater than its width. Find its dimen-
2.
is
and 8; (2), 8 and 7. Here meet the required condition.
are: (1), 7
both roots of the quadratic
Example
8;
8 or -7.
sions.
Here the quadratic obtained is the same as that Example 1 but the root 8 must be rejected as meaning-
Solution. for
;
less in this case.
Example 4 units
less
3.
The
than
area in square feet of a certain square is perimeter in feet. Find the length of its
its
side.
Solution. Let
Then and
Solving,
x x2 4x x2
= = = =
the length of a side, in feet. its area in square feet,
=
2 (answer). Here 2
its
4x
perimeter, in feet. 4.
we have x
is
a repeated root.
PROBLEMS LEADING TO QUADRATIC EQUATIONS 171 Example 4- The area in square feet of a certain square is
9.11]
5 units
less
than
its
perimeter in
Find the length of
feet.
its
side.
Solution. Stating the conditions as in
x
2
=
The quadratic formula x
4#
Example
3,
we have
5.
yields
=
4
+ =
2i
=
2
+
.
,
z,
and hence the conditions are not met by any square.
EXERCISE 1. is
47
Find two numbers whose difference
is
3 and whose product
40.
The base
of a ladder 20 feet long which leans against a barn 4 feet above from the barn. The top of the ladder is x the ground. Find x. 2.
is
x
+
feet
3.
A rectangular lot is
diagonal 4.
A
diagonal
is
250
lot is is
130
f as wide as it dimensions.
its
70 feet longer than it is wide. The length of Find its dimensions.
its
Find
is
long.
its
feet.
5.
Find two numbers whose sum
6.
The
If its
The
length of
feet.
is
20 and whose product
is
99.
length of a rectangle is 3 inches longer than its width. length is increased by 2 inches and its width is increased by
3 inches,
its
area will be doubled. Find
its
dimensions.
The length of a rectangle is twice its width. If its width is increased by 1 inch and its length by 3 inches, its area will be 7.
doubled. Find 8.
its
dimensions.
The diagonal
of a square
for the length of its side (a),
2 feet longer than its side. Solve by use of a linear equation only; is
by use of a quadratic equation. 9. The diagonal of a rectangle is 2 inches longer than length and 9 inches longer than its width. Find its dimensions. (b),
10.
more
The area
its
of a certain square, in square feet, is (a), 5 units than, (b) equal to, and (c) 5 units less than, its perimeter in
172
QUADRATIC EQUATIONS Which
of these conditions, the dimensions in these cases? feet.
[Ch.
IX
any, are possible, and what are
if
11. The length of a rectangle is 1 inch more than and its area in square inches is equal to its perimeter Find its perimeter if the conditions are consistent.
its
width,
in inches.
The length of a rectangle is 1 inch more than its width, and area in square inches is one-half of its perimeter in inches. Find perimeter if the conditions are consistent.
12. its its
13.
One
root of the quadratic, 2x2
bx
1
=
0, is 1.
Find the
other root. 14.
15. 16. 4,
and 17.
One root One root The sum
of 3x 2
of ax 2
+ 2x + c = 4z + 3 =
-1. Find the other
is 2.
Find the other
is
3. if
What must
root.
root.
2k = hx be the values of h and
of the roots of the equation 2x 2
their product Find the value of b
are equal.
is
+
is fc?
the roots of the equation 3x 2 +bx+2=
Chapter Ten
SPECIAL EQUATIONS IN ONE
10.1.
UNKNOWN
The general equation in one unknown
How
can we represent in one algebraic sentence all of the equations in one unknown, say x, which could be written? ' This appears to be a large order' but the answer is simple. ;
It is the
equation
=
/(*)
(1)
0.
Thus far we have learned how to solve when f(x) is a linear function, such as 2x
(1) algebraically 3,
or a quadratic
+
2
2x 5. function, such as 3x have seen further that there
We
is
a graphical method of
approximating the real roots of (1) (as illustrated in Art. 9.9 for the special case of the quadratic) which is so dependable that every student should bear it in mind as a last resort when algebraic methods fail him. The method consists of drawing
the graph of the related curve
y
(2)
=
/(*),
and then
finding, as nearly as the accuracy of the graph the allows, ^-intercepts of (2), which will be the real roots
of (1). In this chapter we shall discuss special equations in one unknown for which the exact values of the roots can be found ^
algebraic means. Linear and quadratic equations, which belong in this group, have been disposed of already
by
(Chapters 4 and
9). 173
UNKNOWN
SPECIAL EQUATIONS IN ONE
174
[Ch.
X
10.2. Solving equations by factoring
The method of member when the
solving a quadratic by factoring the left right member is zero carries over directly to equations of higher degree. The method succeeds when all factors found are linear or quadratic.
Example
Solve the equation
1.
-
x5
(1)
-
x
= =
x(x* x(x*
-
=
1)
+
0.
member, we have
Solution. Factoring the left
x5
=
x
x(x
-
l)(x
2
+
l)(x
l)(z
+
The roots obtained by setting each of these equal to zero are 0, b it follows that i
i
1,
i,
i,
=
and i
i
=
0.
-
1)
1).
factors separately
Since
1.
2
:
i
b
=
i
(Art. 8.2),
The student may check
each of the other four roots. Solve the equation
2.
Example
x*
(2)
=
1.
term and factoring the
Solution. Transposing the constant left
member, we have:
When z-l = 0, Hence
z=
a;
3
1
=
(x
l)(x
2
+#+ '
l;
when z 2 +z+l=0, x=
the_ roots of (2), or J;he three cube roots of
-1 +iV3 -1 -and
=
0.
are
1,
1)
1= i 1,
,
,
example one might easily overlook the two imaginary roots; but he will not do so if he remembers the following simple and important result, proved in more advanced courses: In the
last
THEOREM
1.
has exactly n
Every rational integral equation of the nth degree
roots, not necessarily all different.
For example, the hundredth degree equation, (3)
x(x
-
1)'
=
0,
EQUATIONS WITH GIVEN ROOTS
10.3]
175
has exactly 100 roots, including the single root zero and the multiple or repeated root
10.3.
1,
which
is
of multiplicity 99.
Equations with given roots
As a
corollary of the factoring method, and also as an extension of the method of Art. 9.10 applying to quadratics, it is possible to write immediately the equation in one un-
known having any given Example and -3. Solution.
1.
roots.
Write an equation whose roots are
a(x
- 0)0 -
a can be any constant.
T)(x
we
If
-
-
2)[z
let
a
=
(-3)]
1,
2,
where
0,
and perform the + 6x = 0. An-
1
indicated multiplication, we have: x4 12x = 0. 14x 2 other answer is 2x 4
=
0,
7x 2
+
Example
2.
Write an integral rational equation with
tegral coefficients
numbers)
whose roots are
1
&nd
i
in-
+ /3
2
=r
(four
.
Solution.
An equation
with the desired roots, unsimplified,
is r
a[x
-
/i
(1
-MF + i)][x ,
/i
(1
-
-xif
i)]x
2+V3T x II
2-V31 = A0.
-^
J
is not an answer because the problem calls an equation with integral coefficients. Hence we must let a = 9 (or 18, or 27, etc.) to clear fractions. With a = 9, we have
This, however,
for
(a-
- i)(x - 1 -f i)(3x - 2 -V_3)(3o; - 2 +V3) = [(x - I) - i ][(3x - 2) - (V3) = (x - 2x + 2)(9x - 12x + 1) = 0, 9x - 30x + 43x - 2Qx + 2 = 0. (Answer.) -
1
2
2
2
2
]
2
4
or If
2
3
2
we had chosen a =
coefficients
with the
18 the equation obtained would have
common
factor 2.
When
integral factors of the coefficients different
all
from
common
+1
or
1
UNKNOWN
SPECIAL EQUATIONS IN ONE
176
are divided out, the equation form, as is the answer above.
EXERCISE
is
[Ch.
X
said to be reduced to simplest
48
Solve by the factoring method. 1.
3. 5. 7.
9.
11.
z3
= -l.
z + 2x + 1 = 0. x + x - 6x = 0. x - x - x + 1 = 0. 2z - 4z + 3z - 6 = x4 - 4 = 0. 4
2
3
2
3
2
Find rational reduced
to
IS. 0, 1, 1,
19. 0, 1
6. 8. 0.
integral equations,
23. 0,1,
27.
i, i,
5
+x=
2z 3
-
0.
z 6z = 0. x + x - x = 2z = z - 2. + x + x = 0. 3
2
2
3
0.
4
3
with integral coefficients and sets of numbers.
14. 1,
-2.
16. 0,
-2, -3. 2, -3.
i.
18. 0, 1
iV2.
20.
1,
22. 1
-if.
3z.
V2,
1
^-
24. 1,1,2,2.
-i, -i.
26. 0, 0, 0. 28. 0, 0,
10.4. Equations in quadratic
an equation
is
,
1, 1.
form
quadratic, not in x, but in
of x such as x 2 x3 -, etc., the ,
V2.
2
-
-1, -1, -1.
If
4
1.
simplest form, whose roots are the given
1
,
25.
x5
4.
13. 1, 2, 3.
21.
12.
2
z
17.
10.
4= x5 z z4 z -
2.
x
method
some function
for quadratic equations
be used to find numerical values for the function. The solution is then completed by placing the function equal to each of the roots of the quadratic, and solving the two equa-
may
tions thus obtained.
Example (1)
1.
Solve the equation
x6
-
7x*
-
8
=
0.
EQUATIONS IN QUADRATIC FORM
10.4]
form
Solution. Writing (1) in the
we
-
2
3
(x
(2)
)
7(z
177
3
)
-8
=
0,
a quadratic in x 3 The solution by factoring or quadratic formula yields see that
it is
.
(3)
x3
=
z3
=-l.
8,
or (4)
By
the factoring
and
+ iV3,
1
Thus,
(1)
has
Example
2.
method we
find that the roots of (3) are
while those of (4) are
1
and
=^ A
2
--
:
six distinct roots.
Solve
_
.
x Solution. Let y
=
-4- 3
X
Then
(5)
-
-
4,
=
0,
y
(6)
becomes
jj
or 2
(7)
t/
from which value
is
i/
=
sought,
5 or
we
-
1.
replace
stands, getting (8)
4y
-
5
Remembering that it is x whose y by the function of x for which it
W3
=
5>
and
0) The
^nr roots of (8) are ^
and
1;
1
'
those of (9),
-1
t%/23 .
4 and two two real roots. has When it is imaginary (5) cleared of fractions it is seen to be a fourth degree equation.
Hence
UNKNOWN
SPECIAL EQUATIONS IN ONE
178
EXERCISE
[Ch.
X
49
Solve the following equations in quadratic form.
-
- 6 = 0. 2. 2x + x - 6 = 0. = 1 3. 6x< + x 0. 4. 6x - x - 1 = 0. 5. 2x - x - 1 = 0. 6. 2x + x - 1 = 0. 7. x + 7x - 8 = 0. 8. x - 7x - 8 = 0. 9. x + 26x - 27 = 0. 10. x - 26z - 27 = 0. 11. 8x - 63x -8 = 0. 12. 8x + 63x -8 = 0. 13. x - 28x + 27 = 0. 14. 8x - 19x - 27 = 0. IS. 8x + 19x 27 = 0. 16. 2(x +x) -5(x +x)+3=0. 17. 2(x 3 = 0. (x x) x) x 2x + 2x + 2 = 0. 18. (x I) 19. (x + x - I) - 3(x + x - 1) + 2 = 0. 20. (x + x + I) - x - x - 3 = 0. 21. (2x - x) + 2x - x - 2 = 0. 1.
2x*
xz
4
2
2
4
2
2
4
2
4
6
6
3
6
3
6
3
6
6
2
2
2
2
2
2
x)
2
2
2
2
2
2
2
2
-
3
2
3
25. 2(x 2
3
6
2
10.5.
3
6
3
6
3
2
2
- -^
+1 =
0.
Equations involving radicals
When the unknown appears in one radicand, or in several, the processes necessary to eliminate the radicals may lead to rational integral equations of the types already considered. Example (1)
1
.
Solve the equation
V2x
+ 3 +Vx +
1
=
1
Here the student often makes the
error of lifting off the radicals with the mistaken impression that he is .thus 2 6) squaring both sides. But this is a serious error, since (a Solution.
+
10.5]
=
EQUATIONS INVOLVING RADICALS
179
+ 2a& + 6 and hence (/2x + 3 + Va + I) = Thus the (V2x + 3) + 2V2o: + 3Vx + + (Vx + I) remains. To get a simpler radical 2V(2# + 3)(x + 1) a2
2
2
,
2
2
1
.
still
radicand, however,
it is
(2)
2z
When
members
its
better
+
=
3
first
1
to rewrite (1) thus:
-z +
1.
are squared, (2) becomes
2x+3-l ~2V^Ti + (V^Tl) = l 2
(3)
2
Simplifying and transposing terms so that the one remaining radical is by itself in the left member, we have
2Vz +
(4)
A
= -x -
1
1.
second squaring yields 4(x
(5)
+
1)
=
x2
3
=
0,
+
2x
+
1,
or
x2
(6)
-
2x
-
1. from which x = 3 or jBt^f we /iat>e not finished the problem. For when we squared both sides in steps (3) and (5) we in effect multiplied both sides by functions of x, and hence may have introduced ex-
+V3 +
+
extraneous and must 1 = 3 V2(-l)
3
is
+ +V-1 +
(1)
and
is
member
of (1) we (not 1), so that be rejected. But, for x
traneous j^ot^Testingj^^ 3 in the 1 = 3 3 have: /2 3
left
+ 2-5
1
+
=
=1,
1.
Thus,
-1
satisfies
the only root.
Example
2.
Solve
Vz=-l.
(7)
Solution. Squaring both sides, we get x = 1 ; but this l. Hence (7) has no root. be rejected since ?
must
VI
Note that
not a rational integral equation in x. Thus while, as noted before, it can be proved that every rational integral equation of the nth degree has n roots, other types of equations
(7) is
may
not have any roots.
SPECIAL EQUATIONS IN ONE
180
EXERCISE
UNKNOWN
[Ch.
X
50
Solve the following equations involving radicals. 1.
3.
+
5.
x
7.
x +~2
1
-Vx -3 = +V3 - x =
2.
x
1.
+
2
+V2x + 5=1.
9.
11.
Vx +Vx + 2 = 2. 15. V3x +Vx + 1 = 2 = 16. Vx_+ + /z +
13.
1
17.
Vx
18.
Vx -
1
+Vz + 3 = V2a; +
2.
In the following articles a few theorems and rules are given which are found to be of great assistance in solving equations of a higher degree than the second. For supplementary material covering this work a student
is
referred to the
chapter headed 'Theory of Equations' usually included in
any
college algebra text.*
The remainder theorem
10.6.
The remainder theorem may be // r
is
stated as follows
:
a constant and if any polynomial in x is divided by a remainder is obtained that does not contain x,
r) until
(x that
remainder
is the
value that the polynomial would have if r
were substituted for x. Proof. Let (1)
f(x)
= (x-
r)Q(s)
+
R,
* For example, see Rider's College Algebra, alternate edition, pages 187-227. t A polynomial in x, as used here, means an integral and rational function of x, 3. (See Art. 2.1.) For with integral coefficients, such as 3z* - 2z 4 -f x 2 - 4x 4 3x 1 means that f(x) brevity, it is often designated as f(x). Thus, f(x) = 2x 1. here stands for the particular polynomial, 2x* 4- 3z
+ +
SYNTHETIC DIVISION
10.7]
181
the Quotient and obtained when f(x) is divided
where Q(x)
member
of (1)
is
the constant remainder
by x
=
/(r)
Since the right
r.
the same function of x as the
is
when x =
(2)
R
two
different algebraic form, the
though in equal
is
member,
sides will be
Hence
r.
-
(r
left
+R =
r)Q(r)
Q(r)
+R=
R.
Example. Divide 2xz (2z
-
3
3x
2
-
3z 2
+x~
+x^
1)
i
/y
_ x
-
by x 1)
-
2s3
-I
X __
(x
1
-
3z 2
X2
~~~ ^
L
-
1.
+xi
I
~> X
1
____ ~'~'
-
1
1 1.
-
1
x
1
= 1. The rer is x In this example x 1, so that r 1. Here, according to the mainder, which we shall call R, is 1 in if we for x the polynomial 2x3 substitute theorem, x 1. To 3z 2 1, the value of the polynomial will be that note we check this,
+
on^s 4(L)
_
Qn> o^i;
2
4-1 -fl
1 i
9 z
.
Q 4_ i o-fi
i 1.
i l
10.7. Synthetic division
A
condensation of the operation above which retains only
the essential numbers
Example
1.
is
Divide 2z3
called synthetic division.
-
3z2
+x-
1
by x
-
1.
and place 1 (the value of in the first line below. as in the position of a divisor, First,
copy the
coefficients
2-3 1 - 1 2-1 2-1 0-1
r),
UL
Next, draw a line two spaces below the coefficients and copy the first number below this line. Multiply this number
SPECIAL EQUATIONS IN ONE
182
by the number
(2)
UNKNOWN
[Ch.
X
the divisor's position, place the coefficient (3) and add. Place
(1) in
under the next the result ( 1) under the
result (2)
Multiply this number
line.
1)
(
by the number (
1)
sult
in the divisor's position (1), place result the next coefficient (1), and add. Place the re-
below (0) below the
Multiply this number
line.
(0)
by the
number
(1) in the divisor's position, place the result (0) the next coefficient ( 1), and add. Place the result
below (
1)
below the
The
line.
first
numbers below the
three
line
are the coefficients of the Quotient and the last number ( 1) below the line is R, the remainder. The degree of the Quotient will be one less than the degree of the polynomial divided.
The complete
Example
2.
-
Divide 3x 3
-
4z 2
5x
3-4-5
Solution.
2x 2
is
quotient, then,
+
7
-
x
by x
x
-
1
2.
7 [2
4-2 2-1 5
6 3
The Quotient
is
3x
+ 2x
2
The complete quotient
is
and the remainder (R) 5 1 H 3z2 + 2z 1
x
For our purpose, the remainder
(5) is
part of the result obtained because polynomial with 2 substituted for x.
This
is
it
the most important is the value of the
stated in functional notation as follows
If
}(x)
then
/(2)
Example
3.
3x3
-
4z2
-
+
5z
7,
5.
Find /(2) by synthetic division f(x)
Solution.
= =
=
5z4
-
Zx3
+ 6x + x 2
5-3
6
10
14
40
7
20
41
5
/.
is 5.
/(2)
1-4L2 =
78.
82 78
if
4.
:
10.8]
THE FACTOR THEOREM
183
In the examples given, every power of x from the highest 2 3 power down to the constant occurs. That is: x x x and ,
the constant term
occur. If the constant or
all
,
any power
of
missing, a zero must be placed in its proper position in writing the coefficients in form for synthetic division.
x
is
Example
-
Divide 2x 5
4.
+
2x 3
1
by
+
(x
3).
position,
To determine the number to place in the divisor's r = x + 3, the divisor, and solve for r, we let x
getting r
=
Solution.
3.
Notice that the x 4 term, the x 2 term, and the x term are missing from the expression. We must use zero for the coefficient of each of these.
20-20 18-48
144
2-6
144
1
6
Thus,
if
10-48 /(x)
=
2x 5
-
2x 3
-
-3
432 431
+
1,
-3 =-431. The factor theorem
10.8.
The
factor theorem
///(r)
=
is
0, then (x
We know
usually stated as follows: r) is
=
that /(r)
J?,
a factor off(x).
the remainder. Thus, if /(r) = 0, r), divides the function exactly.
then R = 0, and hence (x r is a factor of the function. Synthetic In other words, x to determine whether or not /(r) = 0. used is division
Example
5. Is
x
/(x)
2 a factor of
=
2x3
-
2-3
Solution.
3x2 5
+
-
5x
14
-
14?
J2
2170 4
/.
/(2)
=
0,
and x
-
2
2
is
14
a factor of /(x).
SPECIAL EQUATIONS IN ONE
184
EXERCISE
UNKNOWN
[Ch.
X
51
Divide, using synthetic division. 1.
2. 3.
4. 5. 6. 7. 8. 9.
10. 11.
12. 13. 14.
By on the 15. 16. 17.
18. 19.
20.
21.
10.9.
(3x
3
-
+ 4x -
2x2
-
5) -^ (x
-
1).
x - x - 2) (5x + 3x (x + 1). - 2). 8x 7x + 7x + 6) (2x (x 4 - 2x - 3x + 2) (x + 2). (3x + 6x - 3x - 9) + (x + 3). (4x + 12x 4 5x x + 5) -Mx - 5). (x 12x 8x + 32) 4- (x - 4). (3x - 12x - 3x + 18) ^- (x - 6). (2x - 5x + 2x + 4) (x - 3). (3x - 3) ^- (x - 2). (5x + 4x - 2x + 5) (x - 1). (3x - 3x + 2x - 5) (x + 1). (4x - 6x + 5x) -^ (x + 4). (7x 6x - 2x + 4x) H- (x - 2). (3x 4
3
4
3
2
-T-
2
3
-f-
2
3
2
4
3
4
3
-r-
5
3
2
3
2
4
2
5
3
3
2
4
3
-=-
-T-
2
-f-
2
use of the factor theorem determine whether or not the expression left is
a factor of
the
polynomial on the
-
+
x
-
2;
x
3
4
3
4
2
3
4
right.
-
+ 2x x 2. x 2. I;x + 2x x + 1; 3X + x + x + 4x + x + 3; 2x + 6x - x - 3. x + 2; 3x + x - lOx. x - 3; 5x - 3x + x - 2. x + 4; x - 3x + 2x - 8. x
4
1.
3
2
3
4
3
3
2
Theorem on rational
roots
If an equation '
+ aiZ'- + 1
a 2x n
~2
+
with integral coefficients, has a
+ a -& + a = n
n
el
duced
to lowest terms,
divisor of
a
.
0,
c c rational root ~, where - is re-
then c is a divisor of a n
a and d
is
a
10.10]
UPPER AND LOWER LIMITS OF ROOTS
Example.
If
the equation
3z 3 has a rational root,
-
4z 2
Xn tircta
+
Any
f
+
& 2X n-2
If
of the following numbers. *.
an equation
----
an
=
+ fc^ +
=
&n
integral divisor of
bn
(^
.
the equation
x3 has a rational root, 1,
10.10.
_
2
i,
rational root of
& ia .-l
integral coefficients, is
Example.
+ 5x +
must be one
it
f,
Corollary.
185
it
3z 2 will
2,
+ 4z +
=
be one of the following numbers.
+4,
3,
Upper and lower
12
12.
6,
limits of roots
The
possible rational roots are checked by synthetic division to determine whether or not they are actual roots. If,
when any
positive
number
tested, the
is
sums beneath the
can be proved * that there is no root larger than the number being tested. This number is therefore an upper limit of the roots. Similarly if, when a negative number is tested, the sums line are all positive, or zero, it
are alternately plus and minus throughout the line, it can be shown that there is no root less than the number being tested. That number is therefore a lower limit.
Example. Test the number 2 as a possible root of 5x
+
2
3-4
5
6
4
18
2
9
20
3
* The proof is not difficult. The student Exercise 51. See also problem 24.
=
0.
2[2
is
challenged to try
it
in
problem 23 of
The
UNKNOWN
SPECIAL EQUATIONS IN ONE
186
[Ch.
X
and hence the roots
signs are all plus in the lower line
of this equation are all less than 2. 1 as a possible root. Test
3-4 -
5
-1
2[
3
7-12
3-7
12-10
The
signs in the lower line are alternately plus and minus, 1. and hence the roots of the equation are greater than ' By this means we may eliminate some of the possible
roots' without testing them by synthetic division. 10.11. Depressed equations
Consider the equation 2
(x
(1)
Any value makes x x
3
=
-
+ 2)(x -
3z
0.
x that makes x 3 = 0, or any value of x that = 3x + 2 0, is a root of (1). The statements, and x 2 3x + 2 = are called depressed equa-
of
2
0,
with relation to equation found by synthetic division.
tions is
=
3)
Example
L
-
Solve: x 3
6z 2
1-6
The depressed equation
(1).
+
-
llx
-
396 1-3
Solution.
11
6
6
=
0.
_3
2
Here the remainder is zero. Use the numbers below the and reduce the degree of the expression 2 3x + 2 = 0. by one. The depressed equation, then, is x
line as coefficients
Since any root of the depressed equation is also a root of the original equation, we solve the depressed equation for the
remaining roots.
z2 or
(x
-
z~2 x
-
3z
2)(x
= =
+2
-
1)
= =
0;
3-1
2;
x
0,
0.
= =
0; 1.
DEPRESSED EQUATIONS
10.11]
But our synthetic
187
division shows that 3
is
a root, since
produces a zero remainder. Therefore, the roots of the nal equation are 3, 2, and 1.
Example
+
4z 3
The
3,
-
8x 2
-
3x
-
6
0.
possible rational roots are 3,
2,
1,
Testing
origi-
Find the roots of the equation
2.
Solution.
it
i,
6,
:
f,
f.
i,
we have 4
-
8
-
3 12
12
4-4
-
6
-3
27
9-33
not a root, and this test shows that all roots are 3, since the sums below the line alternate in greater than sign. Next, -2 4 3
Thus,
is
-
8-3-6
6
8
0-3
4 This test shows that
a root, since the remainder is 2 Ox 3 0. Upon zero. The depressed equation is 4x 2 = = + Vf- = + f /3. solving the latter, we find that x f and x 2
is
+
,
Thus the
roots are
2,
,
and
Zi
EXERCISE
Lt
52
By use of the preceding theory, determine what numbers could possibly be roots of the following equations, and then test them. If the final depressed equation is quadratic, find all the roots.
- x + 3 = 0. 2. x - 3x + 2 = 0. 4. x 4 - 2x 3 + 1 = 0. 3. x + 2z - 3 = 0. 2 6. 2x + 3x 4 - 3x + 2 = 0. 5. x 4 - 3x + 2 = 0. 7. 2x 3 - x 2 - 2x + 1 = 0. 8. 6x 3 + 19x + 15x + 2 = 0. 9. x4 + 3x - 3x2 - 12x - 4 = 0. 10. 6x 3 + 19x + x - 6 = 0. 1.
x3
-
3
3x 2
3
s
2
2
3
2
SPECIAL EQUATIONS IN ONE
188 11. 12. 13. 14. 15. 16. 17. 18. 19.
20. 21. 22.
UNKNOWN
[Ch.
X
- 2x - 6x + 6x + 9 = 0. Sx - 4x - 14x + 5x + 5 = 0. 6X + 3x - llx - 4x + 4 = 0. 2x + 5x + 2 = 0. 3X4 + 2x + 13x + 8a; + 4 = 0. + x - 2 = 0. 4 3s + 2x - 49x - 32rc + 16 = 0. x + 3x - x - 3 = 0. x - 4x - 8x + 32 = 0. 4X - 12x - 25x + 27x + 36 = 0. x + 8x - 12x - 24x + 27 = 0. x4 + 3x + x - 2 = 0. s4
a;
3
2
4
3
2
4
3
2
4
2
3
2
3
2
3
6
3
6
3
4
4
z
2
3
2
2
3
3
2
23. Prove the statement beginning '.. first paragraph of Art. 10.10.
it
can be proved' in
24. Prove the statement beginning '.
it
can be shown' in
the
the second paragraph of Art. 10.10.
.
.
Chapter Eleven
SIMULTANEOUS EQUATIONS
The general problem
11.1.
In this chapter we shall deal with pairs of simultaneous equations in two unknowns, at least one member of each pair being quadratic, or of the second degree, in one or both of the unknowns. Equation-pairs of this sort arise frequently in the solution of simple problems.
As
in the case of linear equations, the solution of simul-
taneous quadratics may be obtained graphically by use of the rectangular coordinate system. In this case, however, it is necessary to plot loci which are not lines. Some typical
ones
be discussed in the next
will
11.2.
Typical
article.
loci
A
quadratic equation in two variables may not have a locus. For example, there is no pair of values for x and y which satisfies the equation
z2
(1)
since the tive.
sum of two
But
if
it
+
2 2/
=-l,
squares of real numbers cannot be nega-
exists at all the locus is
always one of
five
things: (a), an ellipse (Fig. 16), including the circle as a special (Fig.
17); (c), a hyperbola straight lines which may or may not
case; (6), a parabola (Fig.
18); (d),
two
coincide (Fig. 19), or (e), a single point. These curves are studied in analytic geometry. For our purposes we may note simply that if the student has in mind the general appear189
SIMULTANEOUS EQUATIONS
190
[Ch.
XI
ance of each of the three main curves, he can usually sketch one in roughly when he has located a few points on it.
Example. Sketch the curve 9x 2
(2)
+
2
25t/
Solution. Solving (2) for y,
y
(3)
=
=
225.
we have
f V25
-
x
5, we find from 3, and Assigning to x the values 0, 3, 5, + ^, 0, ^, (3) that the corresponding values of y are +3, of the on are and 0. These points Fig. 16. Note that ellipse
y
(0,0)
x +y-l =
[(x-y)(x+y-l)]=0
Fig. 18.
Hyperbola
Fig. 19.
Two
straight lines
x exceeds 5 numerically, y is imaginary. This means that the curve does not extend to the right of the line x = 5, nor 5. to the left of the line x = if
LINEAR AND QUADRATIC EQUATIONS
11.3]
EXERCISE
191
53
The graphs of the equations in problems 1 and 2 below are ellipses; and 4, parabolas; in 5 and 6, hyperbolas. Find at least 6 points on each, and sketch the curve. in 3
1.
3. 5.
x2 y z2
+
y
=
2
25.
= 2z + 3 - y2 = 9. 2
Draw
4.
= 36. +3= y 4y = 4z 36. 9y
2.
4z 2
4.
2
+
Qy
2
a;
2
6.
0.
2
graphs indicated for problems 7-15. They are straight
the
lines. 7.
9.
(x
11.
2
13.
15.
-
(x
x y
2 2
?/
y)(x y)
= ~
Graph
=
4
+
=
2
y
-
1)
=
8.
0.
12.
0.
x
y
2
=
14. (2s
4. 2
(3x-47/-
10. (x
1.
=
-
y
2
2)
=
0.
0.
+ 3y +
2
6)
=
0.
0.
the following equations.
+ 4y = 36. + y = 9. 2
9x 2
-
2
=
16.
9z 2
19.
x
2
21. x
2
-
y
23. x 2
+
y
11.3.
Linear and quadratic equations
17.
2
2 2
= =
4y
20. x
2
0.
22. (x
0.
24. (s
36.
18.
9x 2
- y = 9. - y) = 0. - 2) + (y +
-
4y
2
=
0.
2
2
2
2
I)
=
0.
Suppose, for example, we seek the dimensions of a rectangle whose diagonal is 10 inches long and whose perimeter is 28 inches.
X Fig.
20
Designating the two unknowns, the measures of length and width in inches, by x and y respectively, we have, from Fig. 20 and the Pythagorean relation,
SIMULTANEOUS EQUATIONS
192
+
X2
(1)
2
I/
=
Also, since half of the perimeter
x
(2)
Solving (2) for
y,
+
10 2 is
[Ch.
XI
.
14,
y
=
14.
y
=
14
we have
(3)
-
x.
Substitution of the value of y from (3) in (1) yields
x2
(4)
The
+
(14
- xY =
solutions of (4) are x
=
6 or
100.
8.
Substituting in
(2), or,
we get the solutions: x = 6, y = 8, and x = 8, y = 6. Both of these pairs satisfy (1) and (2); but since algebraic solutions must be examined in the light of the demands of the problem, and since by agreement x stands for the measure of length, the first pair of values must be rejected. The required rectangle is 8 inches long and 6 inches better yet, in (3),
wide.
To
follows, as 1.
and quadratic pair, then, we proceed as illustrated by the example above.
solve a linear
Solve the linear equation for one letter in terms of the
other. 2.
Substitute the obtained literal value of the
first letter
in the second degree equation.
Solve the resulting quadratic in the second letter. Substitute each of the two quadratic roots now found in the linear equation to find the corresponding value of the 3.
4.
first letter.
Consider next the graphical interpretation of the above problem. The circle (1) and the line (2) intersect at the points (8, 6) and (6, 8). (Fig. 21.) It should be noted that there are two points on the circle and only one on the line
which x
=
This suggests the reason for the emphasis upon substitution of the found value in the linear instead of the quadratic equation. for
8.
11.31
LINEAR AND QUADRATIC EQUATIONS
193
Suppose, in this problem, that the perimeter were increased to 20A/2 inches, with the diagonal length unchanged. The algebraic solution of the two equations
=
100,
and X
(5)
+y=
10/2,
= 5/2, y = 5/2. Geoyields the one pair of solutions, x metrically, this means that the^line (5) touches the circle (1) at the single point (5/2, 5V2) (Fig. 21). Finally, with an
assumed perimeter of 40, the algebraic solution is imaginary, as might be suspected from the fact that the line, x + y = 20, in Fig. 21, does not cross the circle.
In general, the physical impossibility of a pair of simultaneous conditions is indicated algebraically whenever i
appears in the solution of the equations stating the condiinconsistent equations; and it is shown the nonintersection of the loci of these equa-
tions, as well as
graphically
by
by
This result applies not only to linear and quadratic pairs, but also to simultaneous equations of many types
tions.
and degrees
in
two unknowns.
SIMULTANEOUS EQUATIONS
194
[Ch.
XI
The number of solution-pairs
11.4.
An
inspection of the sample second degree curves shown in Figs. 16-19 indicates that a straight line cannot cross one of them in more than two points, and also that two of them
cannot intersect in more than four points. The corresponding algebraic results may be stated as follows :
A.* There are at most two distinct pairs of real numbers which satisfy simultaneously a linear and quadratic equation in two unknowns. B.* There are at most four distinct pairs of real numbers which satisfy simultaneously two quadratic equations in two
unknowns.
EXERCISE
54
Solve the following equations graphically, estimating the coordinates of the points of intersection in case they intersect. Then check
by solving algebraically.
= 25, 3x ty = 0. x* + y* = 25, - 2x = L 3y z2 - y2 = 16, 3z - 5y = 0. x2 - 2 = 16, 2x - 3j/ = 1. x2 - y = 0, 2s - y = 1. - y = 0, x2
1.
4.
7.
10.
13.
16.
+
y*
?/
2.
5.
8.
11.
14.
+ y = 25, x + y = 5. x + y = 25, 2
x2
2
2
3x - 4y = 25. x - y = 16, - 3 = 0. y x - y = 16, s = 3. x - y = 0, 2x - y = 2. 2
z
2
2
2
= 25, x y = 0. z + y* = 25, y = 6. x - y* = 16, x + y = 4. s - y = 0, - 4z = 4. 3y x - y2 = 0, x + = 2.
3. x*
6.
9.
12.
15.
+
y*
2
2
2
2/
2
a?
4-3=0. * It
such
assumed here that the equations are independent. That is, we bar pairs, 1 =0 and (x -f y y) =0, whose graphs have a gtraight l)(x common, or such as z2 -f- j/2 = 1 and 2x2 + 2 /2 = 2, whose graphs are
is
asx-hy
line in
identical.
SIMULTANEOUS EQUATIONS IN LINEAR FORM
11.6]
195
Solve algebraically.
+ xy + y = x + y =
17. x 2
2
19. 42/ 2
x2 x
18.
3, 2.
+^+x+22/-l = 0,
+ y = + 2y = 2
-
20. (x 2
2z
xy y
+
= =
2a
2x
22.
a&,
+
6.
z?/
-
y
2
)
(x + y) y = a +
z-42/=4. 21.
*?/
2, 0.
+
ab
=
= =
1, 1.
26,
0.
23. Find the dimensions of a rectangular field whose area 20 square rods and whose perimeter is 24 rods.
is
Find the dimensions of a rectangle whose perimeter 34 inches and whose diagonal is 13 inches.
is
24.
11.5.
Simultaneous quadratics in general
The algebraic solution of simultaneous quadratics in two unknowns is in the general case long and tedious, involving the solution of a fourth degree equation. We shall consider, in the next two articles, two special but important cases.
11.6.
Simultaneous equations in linear form
no more than two of the five quantities, x2 y 2 xy, x, and y appear altogether in two simultaneous quadratics, the method for linear equations may be applied at once to get these two unknowns. After the latter are found, the values of x and y follow at once. Each of the following seven pair2 2 x2 and xy y2 and xy] x2 ings is of interest here: x and y 2 and y; y and x] xy and x] xy and y. If
,
,
]
Example
1.
Solve simultaneously.
x2
(1)
+y = 2
^^^^
(2)
Solution. Solving first for x 2
y
2
=
9.
Hence x =
2,
y
=
13,
and y2 we get x2
3.
,
The
solution
is
=
4 and
not com-
SIMULTANEOUS EQUATIONS
196
[Ch.
XI
however, until the values have been paired properly. Since y = +3 when x = 2 and also when x = 2, the pairs
plete,
are
-3), (-2,
(2, 3), (2,
Example
2.
2x 2 -3*0 s 3z2
+
(4)
x i/
Solution. Treating (3) and xy, we find that x 2
=
-
(Fig. 22).
Solve simultaneously.
(3)
2
and (-2, -3)
3),
= +A/O
or
~^~>
and
=
= =
1,
7.
unknowns Hence x = + V2,
(4) as linear in
2 and xy
=
1.
the
an d the solutions are
f
V2, -
K14)V '(0,12)
-*~x
Fig.
Example (5)
(6)
3.
22
Fig.
23
Solve simultaneously.
2y
-
= 4, =-12.
These equations are linear in re 2 and y. Solving, we get # 2 = 4, y = 4; so that the final solutions are (2i, 4) and ( 2i, 4). The imaginary values for x indicate that the curves do not intersect. This is seen in Fig. 23, to be Solution.
the case.
EQUATIONS REDUCIBLE TO SIMPLER FORMS
11.7]
EXERCISE
55
Solve algebraically. 1.
x2
+
X
2
_
3.
x
2
+
5.
2x -
x
y
+
2 t
y*
2
3x 7.
y
2 2/
= = = =
3y 2y
2
2
2
9.
2.
4.
1,
4.
6.
1,
-
8.
10.
5,
= 1, 3x(y + 2) = 7. x2 = 4xy - 3, = 3x + 4. 2xy 2x + 4y = 3a, 3x - 8?/ = -a.
12.
x)
11.7.
2
2
2
2
14.
2
t/
2
2
5,
2
16.
2 2
Equations reducible
When
2
2
0.
2
15.
2
2
2
8.
2
11. 3(x
= 25, = 50. 2 = 4, = 36. 4x2 - y = 0, 3x - 1 - 2y = 0. 2 Zxy + y = 1, 2 4x/ + y = 2. 3y + 2x - 1 = 0, - 3x + 8 = 0. 2j/ - 3) = 3, t/(2x 2y xy + 1 = 0. 2 y = 3xy Qxy + 2 + y = 0. 2x + 3xy = 2a + 36, 3x - 2xy = 3a - 26.
+y 2x + 3y x + 4x + 9y x2
2
2
13.
1-6, solve graphically also.
4
= = = =
2
In problems
25,
+ xy 6, x + 2zi/ 0. 4x + 3y = - 8x + 4 = 3y 2x
197
to
simpler forms
quadratics in two unknowns are written with the zero, the simultaneous solution of a pair of
right
members
them
is
much
simplified
of the equations
Example
1.
is
if
i)(a;
z2
(2)
Solution. Since
any pair
of at least
one
+ y + l) =0, + y* - 25 = 0.
of values satisfying
3x
(3)
member
left
Solve simultaneously:
(3x-2y-
(1)
the
or can be factored.
-
1
=
0,
+#+
1
=
0,
-
2y
or
z
(4)
will also satisfy (1), it follows that (3)
and
(2)
any common solution
of
or of (4) and (2) will be one of the desired solu-
SIMULTANEOUS EQUATIONS
198
[Ch.
XI
Hence the problem reduces to the solution of two linear-and-quadratic pairs: (3)-(2) and (4)-(2). The solutions of (3) and (2) are (3, 4) and (-- ff, f-f); those of '(4) and (2) are (-4, 3) and (3, -4). The loci of (1) and (2) are shown in Fig. 24. tions of (1)
and
(2).
x'-2y-l)=0
Pig. 24
Example
2.
(5)
(6)
Solve simultaneously (2x (3*
-
3y y
-
+
+
4)(
-
l)(2x
:
y y
-
2) 3)
= =
0,
0.
Equation (5) is satisfied by the coordinates of any point on either one of the lines Solution.
2x
(7)
-
4
=
0,
+y-
2
=
0.
-
3y
and x
(8)
Similarly, (6) (9)
is
satisfied
by
solutions of either
3x
-
y
+
1
=
0,
2x
-
y
-
3
=
0.
or (10)
Hence, the problem reduces to the simultaneous solution of the following four pairs of linear equations: (7)-(9), (7)-(10),
11.7]
EQUATIONS REDUCIBLE TO SIMPLER FORMS
199
and (8)-(10). We must be sure that each pair comfrom (5) and also a factor from (6). The rest a factor prises of the solution is left to the student. (8)-(9),
Evidently the method of Example if
the
left
member
unfactored form. Here the process
Example
3.
1
of either (5) or (6) is
Solve simultaneously 2
(12)
4i/
had been
left in
the
shortened.
:
- xy = - 3xy =
*2
(11)
could have been used
3,
2.
Solution. This system does not come directly under the types illustrated in Examples 1 and 2. However, it can be
reduced to such a type by first producing an equation from the given pair in which the constant term is zero. This new equation is then used to form a system which may be solved
same manner as the above examples. Both members of (11) and (12) are multiplied respectively by 2 and 3, resulting in the equations:
in the
(13)
2x 2
(14)
12y*
-
2xy Qxy
Adding corresponding members 2z
(15)
2
-
+
llxy
= 6, = -6.
of (13)
12y
2
=
and
(14),
we
get
0.
common
solutions of (11) and (12) must also satisfy (15), a new system can be established in which (15) is used as one of the equations, with either (11) or (12) as
Since
all
in this case, (11) The simpler of the two should be given preference. The left member of (15) has the factors 2x 3y and x 4y, so that the new system becomes
the other.
:
z2
(16) (17)
(2x
-
3y)(x
- xy = - 4y) =
3,
0.
This system presents no new difficulty, and by the method of Example 1 the solutions are found to be (3, 2) ( 2) 3, (2, J)
and (-2, -J).
SIMULTANEOUS EQUATIONS
200
EXERCISE
(x
(3x
2
-
-
5y) (y
3)
=
(x
2
2
3x -xy-2/ =l,
9.
11.
2x + xy - y = 0, _ 4y2 _ o - I) - 9 = (x + y -4= (2x + yY 2
2x2
x2
-
2
37/
=
0,
+
+
14.
6x2
- y* = 2, - 3) = 0. + 4^2 = 1, + 4y - 3) = 0.
2xy y)(x + y x2 - 2xy
2y)(x
-
2x
2
7xy
-
2
j/)
+
5xy
(x
+
Example 16.
= -9, = 0.
+
solve algebraically.
0.
-2,
xy + 2y* =3. 2 17. 2x 2 - 13xt/ 2y 2 2 5 x + 3xy */
-
(x
(2x
Solve by the method of illustrative 15.
-
10.
a;2
13.
0.
x2 -2/2 =16,
(2x
12.
2
0,
x2 -y2 =16,
8.
members, and then
left
2
1)
= =
(x-2/-l)(x+y-l) = 0.
(x+y-2)(2x-y-l) = 0. Factor both
+ 4y +
i/)(2x
6.
2
2
y
(x-y-2)(x-2H-l) = 0.
0.
2/
2
-
4.
+ =25, = 0. (x y)(x + y) x + xt/ + 2i/ = 1, - y)(x + y) = 0. (x x2
-
x2
2.
2
-
7.
1-6, solve graphically also.
+ y = 25, = 0. 3)(x + y + 1) x -y =16, 2
3.
5.
In problems
x2
-
XI
56
Solve algebraically. 1.
[Ch.
-
y)
2
2
3y -4 3y
-
= =
0,
= =
0,
2
1
0.
3.
= 5, = 5. + 3xy - 4 = + 2xy - 2 =
3x2 + xy x2 + y2
18. 2y*
3x2
0, 0.
0.
Chapter Twelve
RATIOS, PROPORTIONS,
AND VARIATIONS
12.1. Ratios
A ratio is a fraction which compares two things in terms of the same unit. Its value is determined by the relative sizes of the things
compared, and not by the unit chosen. Thus
the ratio of 6 inches to 3 feet
i alS
since
6
(foot)
'
e
c is -,
^ ,P
.
36 (inches)
!
=
-, 6'
and
3 (foot) 6* Ratios are of great importance in the branch of mathematics known as trigonometry. One of its uses is to measure the distance to an inaccessible object, such as the top of a
mountain or a spot on the moon, by equating the
ratios of
corresponding sides of similar triangles. 12.2. Proportions
A proportion is an equation whose two members are ratios. Thus, to get the height, CD, of a (Fig. 25),
and the
lengths,
OA =
cliff,
a sight
5 feet,
D
Fig. 25
201
is
AB =
taken at 3 feet,
and
202
AND VARIATIONS
RATIOS, PROPORTIONS,
[Ch.
XII
OC =
200 yards, are measured directly. (The representation shows OA and AB too large in comparison with OC and CD. Such a figure is called a diagram rather than a scale drawing.) Then, from similar triangles, in Fig. 25
CD
m (
}
(in yards)
=
200 (yards)
From
(1) it follows
CD =
(2)
3 (feet) 5 (feet)
3 *
=
5*
that
=
( ) (200 yards)
120 yards,
the height of the not-necessarily-climbed
cliff.
The proportion
=
< 3)
* .
is
3
.
sometimes written a
(4)
:
b
=
c
d,
:
'a is to 6 as c is to d.' Here b and c are called a and and d are extremes. There would, however, be means, no point in the new symbol replacing the division sign were and
it
is
read,
not for the gain effected by extending the notation in
(4).
For example, a
(5)
which briefer
is
read
'
a
is
:
to b
b
:
is
c
= d
:
e
to c as d
:
/,
to e
is
to /,' states in
= = beef
form the three proportions 7 :
is
-,
-
-,
and c
=
-
f
Evidently the notation of (5) can be extended indefinitely with as many letters as desired on each side of the equality sign.
Since the equation,
ad
(6)
=
be,
obtained by clearing fractions in (3), is linear in each of the letters involved, the solution for any one of them is simple. ad be 7 ad be j n o = a = and c = -7 Evidently a = b a e a i
,
j.i
,
,
,
* Note that the indicated units merely tell what the numbers represent, and all algebraic operations are upon numbers (as represented by digits or letters).
that
CONSEQUENCES OF A PROPORTION
12.3]
EXERCISE
57
Find in each case
the ratio of the first to the second quantity.
2 yards.
1.
3
3.
2 miles, 50 yards.
5.
1
feet,
2. 7 feet,
ton, 100 ounces.
7. x,
203
X
-
x
8.
2,
-
2
3 miles.
4.
6 ounces, 4 pounds.
6.
4 hours, 13 seconds.
+
a2
9.
x.
1,
a
+ -(t
Solve for the unknowns. 10.
x
11.
x:(x
:
(x
-
+
+ 2) (3x + 1)
12. (x 13.
14. 15.
(x
12.3.
+
5)
-
(x
2)
:
(x
=
(3x
-
3).
+ 4).
-
1)
:
(2x
+ 2).
:
(2x
3)
=
(x
:
:
:
:
x
+ +
:
3 y]
unknowns.
= :
x
:
4
5y.
= 1 - y - 2) (x - 2) (x + y
+
(2x
:
3t/)
:
1
2
:
:
3.
- 1) = 1 2 3. + 1) (2x + 3y - y + 1) = - y + 1) 2 3. (3x (2x = 2y (3y + 2) (5y + 3). (2x + 3) (3x + 5) (x + 1) = 2 3 4. (x + y + 1) z (x + y) (y + z + 1) = 1 2 3 4. (z + 3) (x + y + z) (y + 2) (x + 1)
18. (x
22.
(x
+
:
the following proportions get pairs of simultaneous equations,
17.
21.
(x
1)
:
1
20.
1)=
-
:
16.
y
:
:
:
:
:
:
:
1
:
:
:
:
:
results following
a
_
1
:
c
b'd
are given in equations (2) to (7) below.
:
:
:
:
:
from
:
:
:
Consequences of a proportion
Some fi (1)
:
solve for the
19.
(x
+ 1) (2 - 4x). (-3x) (6x + 1) = (4x + 2) (12x + 9). - 3x) (4x) = (x + 7) (3x + 1). (2
From and
=
1)
:
:
:
RATIOS, PROPORTIONS,
204
When
the
members
of (1) are
AND VARIATIONS [Ch. XII multiplied by bd we learn
that
ad
(2)
=
be,
product of the means equals the product of the extremes. When the members of (2) are divided by dc we find that
or, the
a
,ox
(3)
=
c
b d'
or, referring to (1), the ratio of the
numerators equals
the ratio
of the corresponding denominators. If
the divisor
is ac,
we
get
W
a
(4)
=
c'
or, the reciprocals of equal ratios are equal.
Again, adding
new member
1
to both
as a fraction,
a (5)
+
members we have 6
c
+
of (1)
and writing each
d
d
6
Similarly,
a (6)
b
d
c
d
b
Dividing corresponding members of (5) and (6), (as justified by the axiom: 'When equals are divided by equals, the quotients are equal');
we have a
(7)
The student may
a
+
b
c
b
c
+
d d
find further consequences of
(1)
by
getting equations from (3) and (4) corresponding with (5), (6), and (7) as derived from (1), by inverting all members, and by combining members in various ways. Evidently an unending series of equations of an endless degree of complexity stems from the simple equation (1). Results such as (2) to (7) were formerly given more prominence in algebra texts than at present. Probably one reason
CONSEQUENCES OF A PROPORTION
12.3]
for the decrease in
emphasis
is
205
the fact that such results are
merely what one gets by applying to (1) the guiding prin' ciple for work with equations, namely: Always do to the right side what is done to the left side.' Nevertheless, the processes here suggested may lead to results which are interesting, useful, and far from self-evident. HINT. If, in launching out for himself, the interested student should arrive at a complicated result of (1) such as, say,
a2 a2
v
x
g
+ ab + ac + ac + ab
be
=
be
d2 d2
+
+
bd bd
+ be cd + be
cd
he should not forget the value of frequent arithmetic tests to guard against errors. For instance, when a = 1, 6 = 2, c = 3, and d = 6, (1) is true and so is (8).
EXERCISE
58
(The numbers in 1.
Noting that
this exercise refer to proportions in Art. 12.3.) (5),
(6),
and
three similar consequences of 2.
Get consequences of
(7) are
consequences of
(1),
get
(3).
(4) similar to those of
3. By inverting the members the proportions obtained in problems
of (5), (6), 1
and
and
2,
problem
1.
as well as of
(7),
more conse-
get nine
quences of (1). Test the following proportions with numerical values which satisfy 1, 6 3, d 2, b 2, c 6, or a 3, c 4, (1), such as a 6. // the tests indicate that a given proportion may be a true d
=
=
=
=
=
=
=
-- --
consequence of .
4.
^ ^
a
+b=
c
a
a
b
d
=
d
c
b 7
-
ab
c
+
+b_ c+d -:
^ 2a 7^
c2
=
so.
ad - -a
5.
c
a*
+
+
prove that this is
c
a
a
(1), try to
a
j
cd
+
'6
d
--
b
=
-
d
2c
a
c
2b
=
c
+
2d
d
a2 '
.
7
ab
3
cd
=
206
10 '*
AND VARIATIONS a - b _ c - d ~
RATIOS, PROPORTIONS,
-
a2
b2
_ ~
-
c2
+
b3
c3
+
rf
the following problems
A light is on the
an 8-foot shadow, how In problem 16, will his shadow be? 17.
18. If
casts a 19.
a light
A model,
tain. If
two
the model,
12.4.
is
shadow 10
_ '
(c
a 6
+ C
2
d)*
'
d
by means of proportions.
the
man
is
man
30 feet from the pole,
on top of a pole and a
3 feet high,
known
high
3
&)
2
top of a 20-foot pole. If a 6-foot is he from the pole?
if
is
c
2
casts
far
feet long,
villages
how
(a
15 '
cd
2
2
2
+
3 '
XII
'
a
cd
ab
16.
2
2
13 '
'
afe
a3
d2
[Ch.
how is
is
high
6-foot
man
how long
30 feet away
the pole?
made from photographs
of a
moun-
to be 5 miles apart are 7 feet apart
on
the mountain?
Variation
Two quantities which behave so that their ratio remains fixed may be compared mathematically by use of the symbol oc, read 'varies as.' Thus, if C and D represent and diameter of a circle, then doubled when D is doubled, tripled when D is tripled, and so on. This is expressed in English by saying that 'C respectively the circumference
C
is
varies directly as
D,'
'C
directly proportional to D.' say that 'C varies as D,' or 'C is
or,
is
simply, we may proportional to D.' In symbols,
More
CocD.
(1)
can be put in a form more suitable for algebraic operations as the equation
The statement
(1)
C = kD,
(2)
or
~=
k,
called the 'constant of proportionality,' or the are meas'variation constant.' In this case, when C and
where k
is
D
12.4]
VARIATION
207
ured in terms of the same linear unit, or (nearly) 3.1416, so that
known
that k
=
IT
the formula
C = wD.
(3)
In
we have
it is
problems, however, we must start with an undeThus at a given time of day in a given place the of a man's shadow varies as his height, h. We then
many
termined length,
k.
5,
have s
(4)
=
where k remains to be found. is 9 feet long, then
=
9
(5)
kh, If
or
k$,
the shadow of a 6-foot
man
= f = f
k
Hence, at the time and place in question, s
(6)
=
/i
for people or objects of various heights.
In other words, the
shadow will be as long as the height of the object. Sometimes one quantity varies, not as a second one, but as some function of one or more quantities. Thus, the equations (7)
y
=
(8)
y
=
(9)
y
=
y
=
=
*,
-^
=
k,
xy
=
kx
or
^
kxz,
or
or
k -,
x
fc,
and /IA (10)
2
kx ,
or
2/ 2
w =
?
fc,
a:
state in succession that y varies directly as z 2 , zs, -,
and
/j
-;*
Sometimes
it is
said that, in (8), y varies jointly as x
and z and in (10), y varies directly as x and inversely as 22 and w. More simply, however, if y varies directly as any function of other letters, y is k times that function. Other
common
forms of expression for the variation law
208
RATIOS, PROPORTIONS,
AND VARIATIONS
[Ch.
XII
connecting y with the other variables in (7), (8), (9), and (10) can be used. In order to acquaint the student with some of these phrases, we shall list a few. In (7), it can be said that
'y
varies as x 2 ,' or that
2
to x ,' or simply that
'y
is
11
statements imply that
is
'y
is
directly proportional 2 proportional to x .' All of these
a constant.
X2 can be said that 'y varies directly as
Similarly, in (8), it the product of x and 2,' or that 'y is directly proportional to the product of x and z,' or, more simply, that 'y is proportional to x and z.' All of these statements imply that is
a constant.
xz
can be said that 'y varies inversely as x' or that inversely proportional to x.' These statements imply 'y that xy is a constant. In (10), it can be said that 'y varies directly as x and In
(9), it
is
2 inversely as the product of z and w' or that 'y is directly 2 proportional to x and inversely proportional to z and w.'
1/2 11} is a constant. These statements imply that x It should be observed that the words 'directly' and
'jointly' are frequently omitted, 'directly' being understood when the dependence of one quantity on another is referred to, and 'jointly' quantity involved.
A steps 1.
when
there
is
more than one other
in variation usually consists of the following
problem :
The variation relation is stated as an equation involving
In this step all the quantities involved in the variation law should be described clearly. When any quantity is referred to by name, we shall adopt the plan of using the initial letter of the name as an abbreviation. For example, the word 'pressure' would be described by the letter p or P. When the quantities are not named, the letters z, y, z the constant
k.
y
etc.,
are used.
12.4]
2.
VARIATION
One
209
set of values of all the variables involved is inserted
in the equation, leaving k as the only visable in a problem where step 5 is to
to
make a box scheme
as indicated in
unknown.
It is ad-
be carried out
Example
later,
4, solution 3.
This box should be filled in before going through any numerical work, as it will clearly show what quantities are given, the units being used respectively, and what quantities are to be found.
The resulting equation is solved for k. The value of k is inserted in the original equation. 5. By means of the equation obtained in 4, the value of any specified variable is found when the values of the others are given. However, the same set of units must be used 3.
4.
throughout the problem as were employed in steps 2 and 3 to find the value of k. This point is very important, and is illustrated in the
Example x y
= =
y
2, 1
=
of
Example
y varies directly as
1.
4,
=
z
and w =
Solution.
two versions
The
3,
and
w =
x
4.
2
zw 2.
One Find
set of values is: z
when x =
3,
4.
five steps in order yield the following
equa-
tions.
-
<'> 4
.
(13)
k
=
t*A / (14) ^
y
=
/icN (15)
1
1
sr 6.
zw
=
= -27,(answer). v
-777-,
2(4)
or
z
2
Example 2. State in words the law of variation for x in terms of the variables related to it in (8) and (10).
we use the second forms in (8) and (10), where members include all the variables present, it becomes
Solution. If
the
left
210
AND VARIATIONS
RATIOS, PROPORTIONS,
[Ch.
XII
apparent that in a direct variation between two variables, one appears as a divisor of the numerator while the second appears as a divisor of the denominator. For inverse variation, both variables will appear in the same position either as divisors of the numerator or as divisors of the denominator. Thus, in (8)
'x
varies directly as y
and inversely as z'
Similarly in (10)
w'
'x
varies directly as y, z 2 ,
It
should be noted here that there are other forms of
and
expression as well as those given.
They are left to the student.
Example 3. If the original value of x is 10 units in (8) and (10), what would its new value become if y is doubled, z is tripled, and w is halved? Solution. In (8), since x varies directly as y
and inversely
z, it means that if y is doubled, so is x, but if z is tripled, x becomes one-third of its former measure. Carrying out these changes simultaneously, we get for x,
as
2(1) (10)
In that
(10), since
=
=
6f units.
x varies directly as
y is doubled, so is x] if tripled, x becomes nine times if
w
?/,
z
2 ,
and w,
out these changes simultaneously, we get for 2(i)(9)(10)
Example
4-
Boyle's
=
means
halved, so is x but if z is former measure. Carrying
is
its
it
;
x,
90 units.
Law
temperature the volume
in physics states that at a given of a given quantity of gas is in-
on the walls of the cona given sample the volume is one cubic foot when the pressure is 20 pounds per square inch, what is the
versely proportional to the pressure tainer. If for
volume when the pressure square inch?
is
increased to 60 pounds ,per
12.4]
VARIATION
Solution
1.
211
Let
7 =
(16)
where
F
and
P
or
VP =
represent units of volume and pressure refirst solution let be measured in pounds
P
spectively. In this
per square inch,
Jfc,
and
V
in cubic feet.
=
(18)
k
(19)
V =
(20)
V = Now
Then
20.
r
= j
-
P
be measured in pounds per square foot, with V unchanged in meaning. Then the first and = ^, second values of P in the problem are respectively Solution 2.
and
3%
=
let
-%
3^-.
Equations (17) to (20) are replaced
by the
following. (21)
1
(22)
k
= A.
(23)
F =
(24)
P
g p. p;
1
=
Thus, we see that the value determined for k, the so-called proportionality constant/' depends upon the choice of units; but the same final result is obtained in any case. '
Simple problems in variation may often be easily as problems in proportion; for if one quantity varies as another, corresponding values are proportional. When using the proportion method, it is better to write the variation equation in the second form illustrated IB (7); (8), (9), and (10), where all variables are contained Solution
3.
worked more
212
in one
RATIOS, PROPORTIONS,
member
AND VARIATIONS
F
F P2 = 2
XII
of the equation. Thus, using the second form one set of values for and P,
V
of (16) with Fi, PI indicating
while
[Ch.
that ViPi = FaPa, etc. Arrange the given information in a box
2,
P
and
a second
2
scheme as follows
V
LAW:
set, etc., it follows
:
P
(cu. ft.)
*
per sq.
= yP = FP = = 7 (60). cu. foot. Vz =
V.P, 1 20 .
(Ibs.
2
3
2
3
in.)
etc.
2
5. Here we shall use without proof the following and useful mathematical result important
Example
:
THEOREM
1.
The areas of similar plane figures vary as
the
squares of corresponding dimensions.
The
two
similar rectangles are 15 and 20 square units, respectively. Find the length of the diagonal of the larger rectangle if the diagonal of the smaller one is 10 units
areas of
long.
respectively.
here used (25)
From (26)
A
and diagonal length The mathematical statement of the theorem
Solution. Let
and d stand
is
A =
A
or
i-*-
the smaller rectangle
we have
15
kd
=
This value, substituted in (27)
for area
or
k
=
15
100
=
3 20*
(25), gives
20
= 20, in (27), Inserting the area of the larger rectangle, or A and remembering that in this case d must be positive, we have
12.4]
VARIATION
(28)
d
=
213
Vs =
20 -To
11.5 linear units (answer).
Note. If the student will try to solve this problem by means of simultaneous equations, as is possible, he will
appreciate the simplicity and power of this method. In addition to the required answer he has in (27) a formula applying to any rectangle similar to the first one.
we
In Exercise 59
THEOREM
Theorem
shall use, in addition to
1,
The volumes of similar figures vary as
2.
the
cubes of corresponding dimensions.
EXERCISE In each
59
of the problems 1-8,
express the given relation as an equation; (b) write the equation so that the left member will contain oil the variables; and then
(a)
(c)
write
an equivalent statement in words for
ing x as related ,
I 1.
I/ 2/
to the other variables present.
o J >y rf fit Z ^. OCX?/.
* />o f^ . OCX
5. z oc
6.
w
, OC O. 1/1 U/ (y
oc -^r-
xy
7.
x2
~ y
oc
and
-
A 4X ^k.
,
--
is
4
when x
=
W
Cf C
I/ if
8. ?/ '
z*w
yz*
9. If 2 varies as
the variation involv-
3
.
2
oc
-
t(;
=
2,
y
3,
w=
3,
w?
=
4,
and x
=
4,
and
t^
find the value of (a)
(b) (c)
(d)
= 4; 2, and z y when x = 3, w = x when z = 5, y = 10, -and w = 8; w when x = 3, y = 4, and z = 5; = 4, and w = 2. z when x = 3, y
10. If
y varies as
-
and
is
2
when
z
=
find the value of (a)
(b) (c)
(d)
= z when y = w when y = x when y = y when
z
w = 4a, and x = 6; = 2a, and w = 3c; 5, x = 3, and x = 2a; 2, z = c. = b, and w a, z 3,
2,
214
RATIOS, PROPORTIONS,
AND VARIATIONS
Solve problems 11-15 by use of Theorems 11.
The
12.
A
1
and
2, Art.
[Ch.
XII
12 A.
bases of 6 similar triangles are 2, 3, 4, 7, 8, and 9 inches long respectively. If the area of the smallest triangle is 5 square inches, what are the other areas? sign painter finds that a pint of paint '
is
used in painting
Chicken Dinner. Dine and Dance.' How much paint be used in making a similar sign, with letters of the same type
the words, will
and proportions but three times
as high?
13. The heights of three similarly-proportioned men are 5, 5^, and 6 feet respectively/If the tallest man weighs 200 pounds, about what would be the expected weights of the other two?
14.
The weights of various aerial bombs in pounds
are as follows
:
100, 200, 500, 1000, 2000, 6000, and 12,000. Assuming that all are similarly proportioned and made of the same materials in like proportions, what are the lengths of the other bombs if the
100-pound one 15.
50,
is
2 feet long?
The perimeters
and
of 6 similar plane figures are 4, 5, 10, 20, 100 inches, respectively. If the area of the figure with a
20-inch perimeter
is
10 square inches, what are the other areas?
16. The weight of any given object on a planet varies directly as the mass of the planet and inversely as the square of its radius. For reference let the mass of the earth be one unit and its radius
one unit. If a boy weighs 100 pounds on the earth, how much would he weigh on each of the following bodies?
Name Moon
Radius
Mercury Venus Mars Jupiter
Saturn
Sun 17.
The weight
.27
.012
.39
.04
.97
.81
.53
.11
11.2
317.
9.4
95.
109.
of a
Mass
body within the
330,000 earth, as in a mine, varies
directly as its distance from the center. Assuming the radius of the earth to be 4000 miles, how much would a 200-pound man weigh when 10 miles below the surface?
12.4]
VARIATION
S
18. If
square of
t/,
215
varies directly as the cube of x and inversely as the what change in S results when x is doubled and y is
tripled? 19.
The
force of attraction
between two bodies varies directly
as the product of their masses and inversely as the square of the distance between them. When two masses of 6 units and 24 units
by 192
are separated
tance between them
inches, the force
is
72 units,
(a) If
the dis-
diminished by 24 inches and the smaller mass is doubled, what change in the larger mass will increase the force by 18 units? (b) At what distance will the two given masses is
have twice the force of attraction that they have at 192 inches? 20. Kepler's Law states that the square of the time of a planet's revolution about the sun varies as the cube of its mean distance from the sun. The mean distances of Mars and the earth are in the ratio 3 2. Find in days the time of revolution of Mars (i.e., the Martian year). :
21.
The volume V
T and
of a gas varies directly as the absolute teminversely as the pressure P. If a certain amount of
perature gas occupies 100 cubic feet at a pressure of 16 pounds per square inch and at T = 200, find its volume when the pressure is
20 pounds per square inch and T = 420. 22. The mass of a spherical body varies directly as its density and the cube of the radius. Compare the masses of Jupiter and the earth if the diameter of Jupiter be taken as 11 times that of the earth and 23.
its
density -fo that of the earth.
Given that y
-
HINT. Let y
Show
that
oc x,
=
2 prove that 2x
kx.
*-
=
a constant.
xy
24. If y oc x, prove that 3s4
+ yV oc
+y
2
oc
xy.
Chapter Thirteen
THE BINOMIAL THEOREM
The binomial theorem
13.1. If left
we
carry through the multiplications indicated in the we get the following
side of the expressions below,
formulas.
(F (F (F
n 1 V
/
+ S) = F + 2FS + S = F* + 3F S + 3F + S + + S) = F* + 4F*S + 6F*S + 4FS* + S 2
2
]
2
2
3
)3
;
4
J 1
2
/77 _J_ .QTtn
+
2
= Fn
n(n ,
1
^Z
/xi^^
4 ;
i'*
-4
(n
2)
Q o
F n - /S + 3
3
(to
n
+
1
terms).
The right member in each of equations (1) is said to be the binomial expansion of its corresponding left member. The n S) is called the binomial formula; and expansion of (F the full equation is the algebraic statement of the binomial
+
theorem dealing with a positive integral index, n. The proof of the binomial formula is given in more advanced discussions,
but the student may verify it for various integral values of n. It will be noted that if n represents the exponent of F + S in the left member of any of the equations of (1), then the following statements are true 1. The number of terms in the expansion is n + 1. 2. The first term is F n 3. The exponent of F decreases by one in each succeeding :
.
term of the expansion, while that 216
of
S
increases
by
one.
13.1]
THE BINOMIAL THEOREM
sum
Furthermore, the
term 4.
is
217
of the exponents of
F
and S
in
any
n.
The
term
coefficient of the first
while the coefficient
is 1,
of the second term is n. If we multiply the coefficient of any term by the exponent of F in that term, and then divide the product by one more than the exponent of S in the same term, the result is the coefficient of the next term in the
expansion. 5.
The
coefficients equidistant
from both ends of the ex-
pansion are equal.
Example
1.
Expand
(x
+
7
y)
.
Solution. (x
+
y)i
The
=
x7
+ +
+ +
7xy 7xy
21x*y*
y
+ 35zV + 35xV +
21x*y*
7 .
first coefficient is 1.
Since n
=
7 in this case, the second coefficient
is 7.
1
The
coefficient of the third
term
( ^
is
7
(f
^
'
or 21.
Lt
The
coefficients of the fourth to eighth terms, as
found in
succession, are:
~~~ _~
Example
2.
~~ _~
6 *' (35) (3)
an d
Expand and
binomial formula,
. '
.
7
simplify (x
Comparing the binomial x that x corresponds with F and
Solution. find
'
7;
o
~~ _~ 35(4)
(21) (5)
5
2?/)
.
2y with F 2y with 5.
+ S,
we
By
the
THE BINOMIAL THEOREM
218
Example
8.
Expand and
simplify
f
F =
3/x and S binomial formula, we have
Here
Solution.
13.2.
An
3 Vx
[Ch.
XIII
+ j
= -
Again using the
extension of the binomial formula
It is interesting to
tional in (F
+ S)
n ,
note that
if
n
is
either negative or frac-
the binomial formula
still
holds
when
on the numerical values is not limited to n + 1 made up of an infinite number. It
proper restrictions have been placed of and S. Further, the expansion
F
terms as before, but is is beyond the scope of this text to discuss these restrictions on the binomial expansion when n is not a positive integer. However, it may be stated in general that the expansion is useful when S is numerically smaller than F, and when the successive terms approach zero so rapidly that the sum of the first three or four terms of the expansion serves as a good n S) Example 3 below shows an approximation for (F
+
.
application to the extraction of roots.
Example
1.
Expand
(x
+
2
t/)~
to five terms and simplify.
Solution.
+ s/)- = 2
(x
or2
or 2
2x~*y
+ 3x~
4
2 i/
4z~
V + 5z- V
13.2]
AN EXTENSION OF THE BINOMIAL FORMULA
219
and
simplify.
y}~* to four terms
2.
Expand
Example
8.
Find '^1002 to three decimal
Solution.
Write
Example
(x
Solution.
-
(x
= =
places.
+ 2)* = [1000(1 + .002)]* 10(1 + .002)* _ 10 + 1 (1)^.002) + K-D(ir*(.002)' + [l* = 10(1 + .00067 - .0000004 + = 10.0000 + .0067 - .0000 H
v/1002
(1000
.
)
=
10.0067 or 10.007.
Note that for accuracy to three decimal places we write successive terms to four decimal places until a term is reached which yields zeros in the four places. Then the answer is 'rounded off' to three places. This is a good working rule, though a much more extended discussion would be needed for a rigorous treatment of
EXERCISE
60
Expand and 1.
4.
+ yY(3o + b)
simplify each of the following expressions. 2. (x
(x
2
3 .
7.
(3*-fJ'10.
(SVx
-
problems of this type.
yY-
5.
[2x
yY-
-
-Y-
M* + 5)H
3. (2x
6.
(2Vx
9 '
yY-
+
4 t/)
.
THE BINOMIAL THEOREM
220
[Ch.
XIII
Find the first four terms in each of the following indicated expansions, and simplify the terms obtained. 13. (x*
-
2y)
10
14.
.
Find and simplify
(VE
8
-
-
f)
15.
4/
+ -^Y^ [V^ 2 /
the required terms in the following indicated
expansions. 16. 5th
term of
fx
_
HINT. Let
and
-
x
U
represent the
fifth
x term, with corresponding to
F
y
to 8.
62
I
17. 6th
term of [ db
18. 8th
term of
Assuming
Y
2
j
*J
to
j
x
y
(Vx -
of*)
14 .
that all restrictions have been taken into account,
expand
four terms and simplify.
+ 2/)~ (a* + 26*)3
19. (x
22.
20. (x
.
1
23. (1
.
-
y)*.
+ 3*)^.
+ y)~i
21. (2x 24. (1
-
2s)-
1 .
Bi/ grouping terms, express as binomials and expand by use of the binomial formula.
25. (a 27. (a 29. (a
+ 6 + c) + 6 - c) + 6 + c)
2 ,
or [(a
+
6)
+
2
c]
.
2 .
3 .
26. (a
+6-c-
28. (a
-
30. (a
+
b 6
-
2
d)
.
2
c)
.
3
c)
.
Use the method of illustrative Example 3, Art. 13.2, to find the values of the following roots to three decimal places. 31.
VTOI.
35. A/26.
VUG. 33. V24 (or V25 - 1).
32.
36.
^1.005. 37.
34. v'lOlO.
^10
(or *V
Chapter Fourteen
PROGRESSIONS
14.1. Definitions
A group of numbers arranged in order according to some law of selection is called a sequence. Each number in the sequence is a term. 14.2. Arithmetic progressions
the difference between two adjacent terms is the same, regardless of the pair chosen, the sequence is called an If
arithmetic progression.
Example.
1, 3, 5, 7, 9, 11.
The abbreviation 'A.P.' stands gression. The letters a, d, n,
I,
and
s are
for
an arithmetic pro-
used to represent the
elements of an A.P., where
a represents the first term; d represents the common difference, or what must be added to any term to get the next one; n represents the number of terms / represents the last, or nth, term; and S represents the sum of all the n terms. It may aid the memory to note that the elements are the ' ' letters in the word lands. These elements are so related that if any three of them are known the remaining two can be found by use of the proper formulas.* ;
* In
used,
We
cases, when only the three formulas for an A.P. given in the text are necessary to solve simultaneously two equations with two unknowns.
some
it is
shall
omit examples of
this type.
221
PROGRESSIONS
222
Consider (1)
a,
a
[Ch.
XIV
the progression.
first
+ d, a +
2d,
a
+
3d, a
+ 4d,
a
+
5d, a
+
6d.
Here any term after the first is found by adding d to the term which precedes it. Note that second term is a + Id, the third term is a + 2dt, etc., so that the coefficient of d in any term is one less than the ordinal number of the term. From this we see that the nth term, designated by I, is a + (n l)d. Stated in symbols, this yields an important result which we shall call Formula 1, namely, 1.
=
/
a
+
(n
-
l)d.
A
progression having a definite number of terms may be written out in full, but usually some dots are inserted to indicate that
some
of the terms are omitted.
Example
1. 2, 5, 8,
to 8 terms.
Example
2. 1, 3, 5,
19, 21.
14.3. Arithmetic
The
means and extremes
and last terms of a progression are called the and all terms between the extremes are called
first
extremes,
means.
In the special case of an A.P. having only three terms, the middle term is called the arithmetic mean.
Example. Find the arithmetic mean of a and Solution. Let
Since
m
a
m be d,
and
also
m
(1)
Solving for
the desired
m we have,
In words,
the
arithmetic average.
as
m
I
a
mean
=
=
I
Formula
arithmetic
mean
I.
in the A.P.
d, it
:
a,
w,
I.
follows that
m. 2,
of two numbers is their
ARITHMETIC MEANS AND EXTREMES
14.3]
If there is
more than one mean
given extremes,
we
n
and solve
in
Formula
written the
1
until the last
Example
1.
to be inserted between
two and
known values of a, I, The entire set can then be
substitute the for d.
by adding d to each
first,
223
successive term, beginning with is reached.
term (the second extreme)
Find the mean of 2 and
8.
By Formula
2 (here, and in the other examples to follow, the student should first of all write out the formula -Solution.
in question),
^
m=
(2)
+
2
8
10
=
-y-
-2=5.
Find the mean of 7 and
Example
2.
e / , Solution,
m =
7
+
-8 =
(-15) ~ =
-jr-
means between 3 and
three
3. Insert
Solution.
Here a =
3,
(3)
A 4.
2t
2i
Example
15.
/
=
and n =
11,
11
=
3
d
=
2.
+
11.
^ By Formula 5.
1,
4d,
whence (4)
The
progression, then, as obtained by adding 2 to each successive term, is 3, 5, 7, 9, 11; and the required means are 5, 7,
and 9 (answer). Insert four
Example
4-
Solution.
Here a
=
1, 1
means between
=
14,
-14 =
(5)
1
and n
+
1
=
and 6.
14.
By Formula
1,
5d,
whence d
(6)
The progression means are 2,
is
5,
= -3.
2,
1,
8,
and
5,
8,
11,
11 (answer).
14;
and the
PROGRESSIONS
224
Example
d
Solution.
From
n = 7. By Formula 1, = 2 -f (6) (3) = 20
Also
we
the given terms
2, 5, 8,
=
see that a
XIV
2 and
(answer).
l
(7)
2,
Find the 7th term of the progression:
5.
=
3.
[Ch.
Example 6. The first 3 terms of an A.P. having 8 terms are 4. Find the last term. 1, and Solution.
=
Z
(8)
EXERCISE Find
=
Here a 2
+
2,
d
=
3,
7(-3)
and n =
= -19
1.
3 and 19.
2.
2 and -12.
3.
3x and
4.
-3
mean
(answer).
8x.
and -9.
6. 7.
Insert 3
2
.
8. Insert
10.
15.
means between 2 and
5 means between 10 and
22. 2.
Find the 10th term of the progression: If I = 21, n = 7, and d = 3, find a.
11. If
1,
of the numbers given.
and 4y 2 * Insert 2 means between 3 and
9.
By Formula
61
the arithmetic
5. 5t/
8.
n
=
8,
I
=
30,
and a
=
3, 5,
7
.
2, find d.
There are 5 apple trees in a row. Each tree, after the first, produces 10 more apples than the one that precedes it. If the first tree produces 75 apples, how many are obtained from the last 12.
tree? 13.
many 14.
The
first
term of an A.P.
is
21. If
d
= -3
and
I
=
0,
how
terms are there?
Find the 10th term of the progression:
A
3, 5, 7,
.
group of boys, counting their marbles, found that one the next had 24, and so on to the last boy, who had' 12. 27, How many boys were there? 15.
had
THE SUM OF AN 16. A man travels 50
14.4]
on through the month. January?
A man
17.
A.P.
225
miles on Jan.
How
traveled 100 miles on
and so on through the week.
1,
55 miles Jan.
far did he travel
How
on the
Monday, 85
2,
last
and so of
day
miles Tuesday,
far did he travel
on Saturday?
18. A man invested $10 more each month, after the first one, than he invested in the preceding month. His investment for the 10th month was $110. What was the first month's investment?
The sum of an A.P.
14.4.
We
could find the
sum
an A.P. by adding
of the terms of
them; but that would be a long process
in
many
cases. It is
shorter and more convenient to use the formula for the sum derived from the general expressions for S in the two equations below.
The second one
order of the terms in the (1) (2)
first
is
obtained by reversing the
one.
s S
Adding corresponding members
of (1)
and
(2),
we have
(3)
Noting that a
+
I
occurs n times in this sum,
2S = n(a
(4)
and solving
(4) for
S = 5
A
second formula for the
+
Formula mula 4 is
Thus,
4.
S = ~[a
:
S = 5
[2a
:
sum may be obtained by
+ (n + a + (n-
Formula 3 by a
I
1.
3
l).
placing the
in
write
I),
S we have Formula
3.
(a
+
we may
+
(n
-
l)d].
l)d, its
re-
value by
l)d] so that
For-
PROGRESSIONS
226
To
$ we
find
given,
3, of course,
and Formula 4 when we know
Example gression:
1.
Find the sum of the
1, 3, 5,
Solution.
S =
(5)
use Formula
when n,
n, a,
and
a,
[Ch.
and I are
d.
6 terms of the pro-
first
.
Here a
+
f[2(l)
= (6
d
1,
-
=
and n
2,
1)2]
=
3(2
Example 2. Find the sum of the which a = 2 and I = 14.
+
first
=
6.
10)
By Formula 4, = 36 (answer).
5 terms of an A.P. in
By Formula 3, with a = 2, = 14, and n = S = [(2 + 14)] = (16) = 40 (answer).
Solution. (6)
EXERCISE
XIV
I
5,
62
1.
Find the sum of the
2.
Find the sum of the
3.
Find the sum of the
-17,.-. 4. Find the sum
first
first
6 terms of the A.P.:
3, 5, 7,
10 terms of the A.P. 32, 29, 26, :
first
.
.
8 terms of the A.P.: -25, -21,
of the first 12 terms of the A.P.:
7,
4,
-1,..
The sum of
5.
the
8. 9.
10. 11.
12. 13. 14.
15. 16.
the
term of an A.P. first
is
5 and the 7th term
7 terms.
= 4, = 23, and s = 243, find n and d. If a = 5, n = 12, and d = 2, find S and I = 30, find S and n. If a = 12, d = 3, and = -7, find n and d. If a = 101, S = 1457, and = -4, find a and n. If S = 24, d = -2, and = 33, find S and a. If n = 13, d = , and = -25, find S and d. If a = 17, n = 15, and If S = 171, n = 9, and d = 2, find a and I I{ a = 5, d = 3, and S = 98, find n and = 43, and S = 250, find a and d. If n = 10, If a = -6, n = 11, and S = 99, find d and
6. If 7.
first
a
I
I
I
I
Z
I
I.
I
Z.
is 17.
Find
14.5]
GEOMETRIC PROGRESSIONS
227
17. How many boys are required for a triangular formation having 6 boys in the first ro^, 5 in the next, and so on to the last row, in which there is only one boy?
A
and bounces to a height of 25 feet. It continues 5 feet each time. How far does it travel between the less bouncing first and sixth time it strikes the ground? 18.
ball falls
19. Ten bales of hay are lying 4 feet apart in a row. The first 4 feet from a truck, the second is 8 feet, etc. How far must a man travel if he starts at the truck and carries it all, one bale at is
a time, to the truck?
A
boy starts at the first one of 11 marks that are 5 yards apart and touches each of the other ten marks in order. If he returns to his starting point after each touch, what is the total 20.
distance he travels?
boys inherit an estate. Each one after the first receives than the one before him. If the value of the estate was $13,000 how much did each boy receive? 21. Five
$200
less
22. A man invests $100 at 5% simple interest at the beginning of each year. Find the value of his investments at the end of 10 years. 23.
A man
4%
invests $1000 at
simple interest at the beginning of each year. Find the value of his investments at the end of
6 years. 24.
A man
invests $500 at
6%
simple interest at the beginning
of each year. Find the value of his investments at the end of
5 years.
14.5. Geometric progressions
The following sequence as abbreviated, a G.P. a, ar,
(1)
From
is
called a geometric progression, or
ar2 ar3 ,
,
-,?.
we note
that any term after the first is found by term preceding it by r. Or again, if any multiplying the term after the first is divided by the preceding term, the
quotient
(1)
is r.
This quantity,
r, is
called the
common
ratio.
PROGRESSIONS
228
Aside from
[Ch.
XIV
which replaces the d of an A.P. the elements ' of a G.P. are the same as those of an A.P. The word snarl' ' (replacing lands'), here might be used as a memory aid. The exponent of r in any term is one less than the ordinal
number
r,
G.P., the last term 5.
/
= ar-
ar'-
.
if
in
a
.
of the terms of a G.P., called more briefly the may be indicated as follows:
of the G.P.,
S = a
(2) If
is
1
1
The sum S
sum
there are n terms we have Formula 5: Thus,
of the term. It follows that
+
ar
2
we multiply each member rs
(3)
When
=
ar
+
ar*
+
of (2)
ar 3
ar n ~*
+
+ ar +
by
+
r,
1 .
we have
+ ar
-
+ ar*-
n~ l
+
ar
right sides of (3) are subtracted from the corresponding sides of (2), all terms in the right members n drop out except a in (2). and ar in (3). Thus, we get
the
left
and
S - rS = a -
(4)
ar n
,
or
8(1
-
solution of (5) for
S
(5)
The 6
.
s =
ri
Note that value
-,
Since
yields
(r
=
1,
*
I
=
ar n n
r'). :
i).
= na
by Formula by rl in Formula (r
-
Formula 6
~l
S = 5^LL' 1
a(l
Formula 6 gives S the meaningless
whereas actually S
Replacing ar 7.
r
if
=
r)
*
5, 6,
in that case. it
we
follows that
get
rl
Formula 7
=
ar n
.
:
i).
T
any three of the elements a, r, n, I, and S are given, the other two can be found by use of Formulas 5, 6, and 7. If
14.5]
GEOMETRIC PROGRESSIONS
Example
1.
229
Find the 10th term
of the progression: 1, 2,
.-.
4,
Solution.
Here a I
(6)
Example
2.
= 1, r = = 1(2) = 9
(7)
8 -
2( * 1
f
10.
By Formula
5,
first
8 terms of the pro-
.-.
By Formula
~
=
512 (answer).
Find the sum of the
gression: 2, 6, 18, Solution.
and n
2,
=
)
with a
6,
2(
f
o
~
o
1}
3. Insert
Solution.
Here a =
,
I
=
2, r
3*
-
= 1
1
3 geometric
Example
=
=
8,
3,
=
and n =
6560 (answer).
means between and
and n =
5.
8,
By Formula
8.
5,
or
16
(9)
=
r4 .
2 are fourth roots of 16, we have r = 2. The G.P. is then either J, 1, 2, 4, 8 or 1, 2, 4, 8, and hence the geometric means are 1, 2, and 4 or 4. 1, 2, and Since 2
and
,
should be noted that in this chapter we are disregarding any solutions in which the value of r involves the (It
imaginary unit
i.)
Example 4- Find the geometric mean of x and that both are positive or both are negative. Solution. (10)
whence
or (12)
Here a
= y
x,
=
I
=
y and n
x(r}
}
=
3.
y,
assuming
By Formula
5,
PROGRESSIONS
230
[Ch.
XIV
The
G.P., then, is either x, Vxy, # or a, Vxy, y. Thus, we find that there are actually two geometric means of any
two numbers with like signs. 09 we have, as Formula 8,
we denote both
If
M
8.
M
=
by
Vxy. Find the geometric means of 4 and
Example
5.
Solution.
By Formula 8, M g =V(4)(16) =
(13)
of these
Thus, the answers are 8 and
a
=
Example
6. If
Solution.
By Formula
5,
1
=
I
32,
(14)
16.
/64.
8.
=
1,
and
=
r
,
find
n and
5.
l
(32)(J)'~
,
whence
A = (i)-
(15)
But
=
5
( )
^j, so that n
Also, (17)
by Formula
S -
_
and hence
5,
(answer).
?^i
-_y)
Given S
=
Example
7.
Solution.
By Formula
126
.
7,
32 t
=
1
n = 6
(16)
1
126, r
_
=
i^l
2,
and I
7,
= Z
1
Hence a
(19)
Also,
128
by Formula
(20)
64
But 2 6 =
64,
(21)
=
n
-
126
2 (answer).
5,
= (2X2)- = 1
and hence
=
=
6 (answer).
2 1 *-!
= 2.
=
_
63 (answer).
64, find
a and
n.
GEOMETRIC PROGRESSIONS
14.5]
EXERCISE
231
63
1.
Find the 5th term of the G.P.:
2, 4, 8,
2.
Find the 6th term of the G.P.:
1, 3, 9,
3.
Find the 7th term of the G.P.:
4.
Find the 8th term
of the G.P.:
5.
Find the 9th term
of the G.P.: 16,
6. 7.
.
32, 10, 8,
-2,
1,
4,
Insert 4 geometric means between
Find the geometric means of
13. 14.
Find the geometric means of Find the geometric means of
15.
Find the geometric means of 4a and
20. 21.
22. 23.
24.
-.
-.
If a = If a = If a = If a = If a = If S = If S = If a = If a =
1
and
.
2.
1.
9.
3 and
27.
and
20.
5
-
243.
12.
19.
-
and
1
11.
18.
-
2 geometric, means between 16 and
geometric means between 81 and Find the geometric means of 2 and 8.
17.
-
-
10. Insert 3
16.
-
-8, 4, Insert 3 geometric means between 2 and 32. Insert 4 geometric means between 32 and 1.
8. Insert 9.
.
9a.
= 2, and = 16, find n and S. 2, r = -1, and r = - find n and S. 32, = 1, and r 81, J, find n and S. = -2, and = -32, find n and S. 1, r = and = -, find n and S. 16, r = 2, and - 32, find n and a. 63, r = 81, find n and a. 121, r =*3, and = 8, and = 1, find r and S. 128, = 6, and = , find r and S. 16, n I
Z
>,
I
Z
.},
Z
Z
I
?i
I
Z
Find the sum of the areas of 5 squares if the side of the first is 12 inches, the side of the second is 6 inches, the next 3 inches, and so on to the last one. 25.
26.
A man
27.
Four men divide $5625
buys 10 watermelons, paying l for the first, the second, 4^ for the third and so on. Find the total cost. receives half as
one
receives.
much
profit.
Each one
2ff
for
after the first
as the one before him. Find the
amount each
PROGRESSIONS
232
[Ch.
XIV
An
estate of $11,808 is divided among 4 people so that each one after the first receives f as much as the one before him. 28.
How much 29.
does each one receive?
Each person has 2 ancestors
of the first preceding generation
2 (parents), 2 , or 4, of the second (grandparents), etc. ancestors has he in the twelfth preceding generation?
30.
Assuming that a patriarch has 6 children
How many
(first following of these has 6 children that each (second generation), generation) and so on, find the total number of descendants of the patriarch
through and including the fourth generation.
Chapter Fifteen
LOGARITHMS
15.1.
Remarks and
The computation example,
-
'
'J-
definition of a numerical quantity such as, for ' ,
would be a slow and tedious task
not for the invention * called logarithms. This is a device which uses the laws of exponents to simplify calcu-
were
it
lations
involving multiplication, division, involution [or evaluation of powers such as (1.05) 10 ], and evolution (or / extraction of roots such as v 2321). Such laws are applicable because, as the following definition will show, a logarithm is
essentially
an exponent.
N
with respect to Definition. The logarithm of the number the base &, designated as 'logbN,' is the exponent which must be applied to
15.2.
b in order
to
produce N.
Exponential and logarithmic equations
(1)
10 2
=
100
=
100,
and (2)
logic
2,
the latter of which is read 'the logarithm of 100 to the base 10 equals 2,' are two forms of the same statement. The first is an exponential equation, and the second is a logarithmic equation. It should be noted that the exponent * The inventor was John Napier (1550-1617), and the one who applied the invention to practical computation was Henry Briggs (1556-1631).
233
LOGARITHMS
234
of 10
2 and the logarithm of 100
is
=
N
[Ch.
XV
In both equations
is 2.
N
= x. follows that log& Similarly, It is very important to be able to express a logarithmic equation as an exponential one, and vice versa. For example, the base
if
P =
MN 15.3.
=
is 10.
if
P =
then Iog 10
10-,
bx
x
9
it
if
Iog 10
MN
=
x
+
y,
then
10*+*.
Four laws of logarithms
Since logarithms are exponents, the laws applying to both are similar. shall now derive the basic laws for logarithms. 1.
log a
We PQR =
In words, of
log,
P +
the logarithm of
log a
Q +
a product
log. R. is the
sum
of logarithms
73.6
+ Iog
its factors.
For example,
=
logio (58.4) (73.6) (86.4)
P =
Proof. Let log,
Then,
P =
ax
,
58.4
log
Q =
x, log,
Q =
and
a*,
+ Iog
y,
R =
10
and log a a2
R =
10
86.4.
z.
.
Multiplying,
PQ72 = awa' Changing the log,
2Mog a ~ =
1,
Art. 7.1)
equation to the logarithmic form, we have,
last
PQR =
= a^+*. (Law
+
x
y
P -
log a
+
z
=
log,
P+
log,
Q
+
log, R.
logaQ.
In words, the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. For example, by Law 2, ,
10gl
= =
(58.4) (_6L7)
(15^X387)' lo glo [(58.4) (61.7)] (logio 58.4
+
-
lo glo [(15.6) (387)]
log lo 61.7)
-
(logio 15.6
+
Iog 10 387)
(by
=
logio 58.4
+
logio 61.7
logio 15.6
.
logio 387.
Law
1)
15.3]
FOUR LAWS OF LOGARITHMS
Proof. Let
= log a P
x and log a
Then,
P=
ax
Dividing
^
=
^
p =
Hence, 3.
log a
x
log a 7:
v
Pn =
In words,
/ie
Q =
and
-
235
Q =
y. a'.
a-*. (Law
=
y
log a
2, Art. 7.1)
P
log a Q.
n log a P. logarithm of the nth power of any number equals
n times the logarithm of the number. For example, logio (S73.4) 3 = 3 Proof. Let log a
P =
x, so
Then, Hence,
log
tt
P =
that
Pn = Pn =
(a
logic 873.4.
n
x
nx
)
=
a*.
a nx (Law .
= n log a
5,
Art. 7.1)
P.
4.
In words,
/ie
logarithm of the nth root of a number equals
_
- times the logarithm of the number. n For example, logio ^57.6 = ^ logio Proof. Let log
P =
x, so
^P
Then,
lo&
Hence,
The
following law
Corollary, or
For example,
The proof
Law
is
5.
logio
is left
that
=
^P =
P =
57.6.
ax
.
- a.
v^T*
- = n n
(x)
(Def. 4, Art. 7.5)
= - log a P. n
a corollary of Laws 3 and m log a v^Af
=
3 v(O084) = f
to the student.
n
log a
Iog 10 (0.084).
4.
LOGARITHMS
236
EXERCISE Given rithms
to
[Ch.
XV
64
log$
2
=
=
0.30 and logw 3
0.48, use the laws of loga-
numbers in problems 1-16.
find the logarithms of the
Examples. (a)
Iogio6=logio (2-3) logic 2+logio 3= 0.30+0.48= 0.78. = log 3 2 = 2 logio 3 = 2(0.48) = 0.96.
(b) logio 9 (c)
logio
3-2 3 = log
24= log
(d) logio f
=
logio
3-2
3+3
logio
=
0.48
logio 2
2= 0.48+0.90= 1.38.
-
2(0.30)
12.
2.
18.
3. 27.
4. 8.
6. 32.
7.
f
8.
9.
1.
11. 5. HINT.
5
=
.
f
f
-0.12. 5.
13. 125.
15.
16. .12 (or
.2.
16.
10. f.
.
12. 25.
-V*.
14. .3 (or 3^).
.
=
Solve for x the equations in problems 17-34.
Examples. (e) (f)
If
(g)
If
17.
20.
23.
26. 29.
32.
x logs 8 log* 8
If Iog 2
= 3, = x, = f
,
= 2. = 3. logs x = x. Iog 64 = x. Iog ( ) = 2. log, 8 = i log, 8
= 2 = 8. 2 X = 8, and hence x = 3. x% = 8, and hence x = 4.
x
3
18. Iog 4
logsx
a;
21. logs x
= = =
4
24. Iog 9 3
9
27. logs (i) 30. Iog x 9
=
33. log, (J)
19. logs
1.
22. Iog 9 x
3.
= =
25. Iog 4 (f )
x.
=
x
28. Iog 4 (8)
x.
31. log, 3
.
= - 1.
34. log, 4
In each of problems 35-47 state whether the equation is If it is false, change one member so that it will be true.
False.
Change
A
Iog
B
.
x.
x.
.
true or false.
logq^
right side to Iog
f
= =
= - . = -f
Examples. .
f.
logo C.
SYSTEMS OF LOGARITHMS
15.4]
(i)
log a
= t^ VzV
(j)
X y
log a X
=
False.
35.
36. 37. 38.
Change
-
3 logo x
-
~
4 Iog
*.
True.
left side to log a
(-)
a;
=
(1
3(lo ga
( Y fx* <~3 =
--
42. logio
C = =
2108 * 2
15.4.
y
log a y.
4 -
=
logio
(2 log
log a C.
log.*).
3
2 V
46. a
a
^
40. 10g a
44. 10
+ 2 log
= %loga A.s loga VI = ilog.5. V -^loga 5 = I log, A (log, 4)1 = (log. + logn y) log a (xyY
39. log
43.
237
4 -
<
(logio 100) (logs 32).
=
x2
l
45. logio (logio 10) 47. log a a 3
.
=
=
0.
3.
Systems of logarithms
There are two standard systems of logarithms in general use in elementary mathematics. They are: (a) The Briggs or common system, in which 10 is used as is used extensively in numerical calused hereafter in this text. Thus, be culations, whenever the base is omitted, the base is to be understood as 10. For example, 'log 58' will mean 'logio 58.' (b) The Naperian or natural system, in which the base is an irrational number approximately equal to 2.718. This
the base. This system
and
will
so-called natural base, usually designated by the letter is used extensively in advanced mathematics.
e,
LOGARITHMS
238
15.5. Characteristics
[Ch.
XV
and mantissas
In the following table, the two equations on the same line result one from the other, or, in other words, are corre-
sponding exponential and logarithmic statements. 10 10
1
10
2
10 3
= = = =
Note that
logl
1;
log 10
10;
log 100
100;
log 1000
1000;
we apply
= = = =
0. 1.
2.
3 (and so on).
number) exponents 1 and as digits. It is clear that most numbers are omitted from this group. Hence it becomes necessary to use some exponents for 10 that contain decimal fractions in order to write all numbers to 10
if
integral (whole
we produce only numbers having
as powers of 10.
For example, 25 lies between 10 and 100. The exponent that must be applied to 10 to produce 25 is therefore between 1 and 2. Actually 10 1
(1)
-
3979
=
25 (nearly),
=
1.3979.
or log 25
(2)
That
the exponent 1.3979, applied to the base 10, produces the number 25. This exponent consists of the integer 1 is,
fraction .3979. In general, the whole number logarithm is called the characteristic, and the
and the decimal part of any decimal part,
if
and when
For example, while log
it is positive, is
.25
= 1.3979-2= -l+.3979 = is
=
called the mantissa.
(-&) =
-
log 100 -.6021, the mantissa of log .25 log
log 25
.3979 rather than -.6021.
15.6. Operations
When
which
a new number
leave the
mantissa unchanged
obtained from a given number by moving the decimal point to the right or left, the first number is
15.7]
RULES ABOUT THE CHARACTERISTIC
in effect multiplied or divided
is
239
by a power
of 10.
For
example, (1)
2500
=
25.00
X
10 2
.025
=
25
10 3
.
,
and (2)
From
(1)
and Law
log
(3)
From
(2)
2500
and Law
log .025
(4)
= = = =
4-
we have
1,
2 log 25 -f log 10
+ 2 log 1.3979 + 2(1)
log 25
10 (by
Law
3)
3.3979 2, it follows
that
= log 25 - log 10 = 1.3979 - 3(1) = -2 + .3979.
3
In equations (3) and (4) we see why the mantissas of the logarithms of the three numbers: 25, 2500, and .025 are the same. In general, when the decimal point in the number is
moved n
places to the right, the logarithm
the integer n (or log 10 the left the logarithm
n )
;
is
and when
it is
decreased
by
is
increased
moved n n.
Hence
by
places to in either
case the mantissa remains unchanged. This very convenient result makes it possible to have compact tables of logarithms,
example, only one mantissa (.3979) need be recorded to apply to such various numbers as 25, 250, 2500, since, for
2.5, .25, .025, etc.
15.7.
Rules about the characteristic
Since log 10 = 1 and log 100 = 2, it is evident that 1 is the characteristic of the logarithm of 10 and of all numbers
between 10 and 100;
or,
in other words, of all
numbers
LOGARITHMS
240
having two significant point. Similarly, for
*
digits to the left of the
numbers with three
the decimal point, the characteristic
is
[Ch.
XV
decimal
digits to the left of
In general, the
2.
following rule applies.
RULE
1.
The
characteristic of the logarithm of
one
a number
than the number of significant digits
greater than one
is
to the left of the
decimal point.
less
The logarithm
of a number less than one is negative, but a characteristic and a (positive) mantissa. For example, it was noted in (4) Art. 15.6 that log .025 = 2 + .3979, with the characteristic 2 and mantissa .3979. For convenience, this result is usually written in the form
there
is still
log .025
(1)
member
since the right
=
8.3979
of (1)
is
-
10,
easy to use and preserves
the positive mantissa. Continuing the table in Art. 15.5 in the reverse direction,
we have
=
10- 1
^
= =
= -- =
10-*
log.l
.l;
-
01;
log 01
= ~2
log .001
= -3
-
.001;
=-1. -
(and so on).
study of the table above we can arrive at a second rule about characteristics, namely,
With a
RULE
2.
little
The
characteristic of the logarithm of
and
a number
less
one plus the equal numerically number of zeros between the decimal point and the first non-
than one
is negative ,
is
to
zero digit on the right. *
number include the first non-zero digit, as read from the digits that follow it. For example, each of the numbers 10.732 and 00030.21 has two significant digits to the left of the decimal point.
The
significant digits in a
left to right,
plus
all
15.8]
SCIENTIFIC NOTATION
AND CHARACTERISTICS
241
Examples.
THE NUMBER
THE CHARACTERISTIC OF THE LOGARITHM
-(1 + 0) =-1 -(1 + 2) =-3 -(1 + 10) = -11
.8003
.00207
.00000000004
In practice we may count the decimal point as well as the proper zeros, getting at once for the sum the correct negative characteristic. Thus the latter is obtained first as a negative integer, after which it is rewritten in the conventional form when the mantissa is known. log .00025
Example
1.
Example
2. log
- -4
+
.3979
=
6.3979
-
10.
.000000000025= -11 +.3979 = 9.3979 -20.
15.8.* Scientific notation and a second mining characteristics
method for
deter-
It is possible to find the characteristic of the logarithm of
N
a number
to base 10
by taking advantage of what is method for representing N.
called the 'scientific notation'
For example, 4
written as (5.834) (10 2 ); 0.000208 as 3.08 as (3.08)(10); 100,000 as (1)(10 5 ); 0.3141
583.4
if
is
(2.08)(10- ); as (3.141)(10- 1 ) 0.000,000,000,108 as (1.08) (10- 10 ), etc., each of the numbers is said to be written in scientific notation. ;
N
In other words, is in scientific notation if it is represented as (M)(10*), where 1 ^ < 10 (M is equal to or greater than 1 but less than 10). = 2.08, and k = -3. For example, in 0.00208,
M
M
RULE number
3.
To determine
the characteristic of
a logarithm of a
base 10, it is merely necessary to write (or think of) the given number in scientific notation, and the exponent of the 10 in this representation is the characteristic. to
* This article
may
be omitted without
loss of continuity.
LOGARITHMS
242
Example. Since, in (a)
5870
(b) 3.4 (c)
=
[Ch.
XV
scientific notation, 3
(5.87) (10 );
=(3.4) (10);
=
0.0004108
(d) 0.01287
=
4
(4.108) (10- ); 2
(1.287) (10~ ); 6
(f)
1,000,000= (1)(10 ); 15 = (1.5) (10 1 ); and
(g)
0.000,000,000,012
(e)
=
(1.2)(10-);
it follows that the characteristics of the respective logarithms are the exponents of the 10 in the right members of the above equations. Thus the characteristics are (a) 3, (b) 0,
-4,(d) -2,
(c)
EXERCISE
65
(e) 6, (f) 1,
(g)
-11.
(ORAL)
Find the characteristic of numbers by both methods. 1.
and
36.
5. 3.2.
the logarithms of each of the following
2. 407.
3. 8.
4. 2573.
6. 8.05.
7. 0.6.
8. 0.058.
10. 932.5.
11. 0.002.
12. 0.00015.
13. 2.75.
14. 0.235.
IS. 0.056.
16. 0.0058.
17. 283.5.
18. 0.08276.
19.
15.9.
the mantissa.
9. 62.05.
Finding
The mantissa
3258
Four-place tables
of the logarithm of a given use of a table, such as Table 3 at the
by The
20. 865432.
number
is
found
end of the
text.
column on the left with the heading N contains two digits of the number. For example, if the number is 157, the digits 15 are found in the sixth row beneath N. The third digit, 7, is found on the line across the top of the table which contains N and the digits 0, 1, 2, etc. In the sixth row, under 7 and in line with the first two digits 15, we find the entry 1959. This means that the mantissa the
first
first
of the logarithm of 157, or .157, or 15,700, or .000157, etc, is .1959 (the decimal being omitted for brevity).
ANTILOGARITHMS
15.10]
243
number whose logarithm
sought has two digits it is found in the column under AT. Its mantissa will then be on the same line and in the column beneath the digit 0. If the number has one digit, we multiply it by 10 and use the mantissa of the product. For example, the mantissa of = the mantissa of log 70 = ,8451. log 7 If
the
When we its
find the logarithm of a
number we
first
write
by inspection very important to get then find the entry in the table which gives
characteristic
this habit)
is
and
(it is
the digits in the mantissa.
Examples.
=
plus the mantissa. Here the student may be tempted to omit the characteristic; but this is a very bad practice, which often results in errors. The characteristic (a) log 4.3
should always be written first, even when it is zero. In the table, in line with 43 and in the column under the digit 0, is the
= 0.6335. entry 6335. Hence, log 4.3 = 2 plus the mantissa. In line with 56 in the (b) log 564 under the digit 4, is the entry 7513, in column the table, and so that log 564
=
2.7513.
=
2 plus the mantissa. The latter, accordlog 0.0108 in table in line with 10 and under 8, is the the to entry ing 2 in the recommended .0334. Writing the characteristic (c)
form,
we have
log 0.0108
=
8.0334
-
10.
15.10. Antilogarithms If log
A =
and may
A
B
said to be the antilogarithm of be found from the table by reversing the operation J5,
then
is
of finding the logarithm.
Examples. (a)
We
logN = 3.8102. Find N. now look inside the table
found in
line
for the entry 8102. This
with 64 and under the digit
6,
so that
is
N has
LOGARITHMS
244
[Ch.
XV
the digits 646 with the decimal point yet to be fixed. Looking next at the characteristic, 3, we see that there are four digits
N. Hence N = 6460. 10. Find N. 7.3979 (6) log N of the table shows that N has the digits 250. Inspection
to the left of the decimal point in
-
=
Since the characteristic
is
10 or
7
3,
N
is
a decimal point followed by two zeros. Hence (c)
log N
=
0.4133. Find
written with
N
=
.0025.
N.
N
istic
has the digits 259. The charactertable shows that shows that has one place to the left of the decimal
point.
Hence
The
N
EXERCISE Find
N
=
2.59.
66
the logarithm of each of the following numbers.
N in each of problems 21-33. = 1.0607. 22. log N = 1.3541. 24. log N = 1.6444. 26. log N = 9.4609 10. 28. log N = 30. 10. 7.3636 log N = 32. 0.7388. log N = 6.0000- 10. log N
Find 21. 23. 25.
27. 29.
31.
33.
log JV log N log N log N log N log N
= = = = = =
2.1303.
2.6335. 3.5933.
8.4983
8.2455
-
10.
10.
0.0043.
15.11. Interpolation
When a
given number, or a given mantissa, is not found directly in the table, it is necessary to use a process called interpolation, as illustrated in the following examples.
15.11]
INTERPOLATION
245
Find log 2576.
Example
1.
Solution.
Here the
we observe
characteristic
is 3.
To
obtain the
man-
between the numbers 2570 tissa, and 2580, whose associated mantissas are the same as those for 257 and 258 respectively. Schematically the work is that 2576
arranged as follows
lies
:
Number 6 10
Now
4099
Mantissa
T2570
4099'
[2576 2580
4116J
digits
17 (the difference)
+ -&(17) =
4109.2, but since four digits only are mantissa is .4109. Hence, the desired retained,
-
log 2576
3.4109.
Evidently the process of interpolation would not have been changed if the decimal point had been differently placed in the digit sequence 2576. For example, log .0002576
=
6.4109
-
10.
should be realized that interpolation uses the principle proportional parts, which assumes that for a small
It
of
change in the number there corresponds a proportional change in the mantissa. This is only approximately true, but in order to use interpolation this assumption will be accepted when the logarithm table does not contain directly the information desired.
Example
2.
Find
N
if
log
N
=
9.4177
-
10.
4177 are not found among the mantissa digits inside the table but they are seen to lie between the adjacent entries 4166 and 4183. The schematic arrangeSolution.
The
digits
;
ment
follows
:
Number
Mantissa
'2610
41661
L2620
4177J 4183
10
digits
11
17
LOGARITHMS
246
Hence, the significant digits in
2610
+ ii(10) =
[Ch.
XV
N are
2616 (nearest four digit number).
Thus, taking into account the characteristic of logJV,
it
follows that
N
=
0.2616.
When first learning this process, the student should put the work in table form as illustrated in the examples. However, with enough practice and a clear understanding of the principle of proportional parts, he
may
often learn to inter-
polate mentally.
EXERCISE Find 1.
67
the logarithm of each of the following
2536.
4. 35.27. 7.
2. 728.6.
3. 0.08352.
5. 839.6.
6.
0.005706.
9.
0.0007085.
8. 294.3.
5.732.
10. 6.325
X
10 8
11. 9.246
.
numbers.
X
10~ 12
N in each of problems 13-24. = 2.2226. 14. log N
.
X
12. 2.318
1020
.
Find 13. 15. 17.
19.
21. 23.
logN = = log N = log N = log tf = log N
15.12.
3.1145.
9.1463 8.1633
2.3146
16.
-
10.8143.
10.
18.
10.
20.
10.
22. 24.
N= = log N log
logN = = log # = log AT = log N
6.3641.
7.2883
-
10.
0.7714.
7.4161
0.4923
-
20. 10.
15.0129.
Computation with logarithms
should be clearly understood that the usefulness of logarithms lies in the fact that they shorten computations involving multiplication, division, involution, and evolution It
(though not addition and subtraction). Furthermore, these operations could be performed in other ways, so that if time is not saved, the point in the use of logarithms is missed.
15.12]
COMPUTATION WITH LOGARITHMS
247
important, therefore, to take note of the details and practices which do save time. Chief among them are the It is
following. (a)
The quantity
to be
evaluated should be written
directly above the
details of computation,
designated by some
letter, as
(b) left
is
used at (c)
N.
The computation should be
members
and should be
outlined in
full,
with the
of all equations filled in before the logarithm table
all.
The logarithms should be arranged
in neat
column
form, with the decimal points and the preceding equality signs in vertical lines.
The student should study
carefully
the
illustrative
examples below before attempting the problems in the subsequent exercise.
Example
L
Evaluate
N log N
Solution. Let
Then,
log 324 Outline, log 618 log
N N
-.
= $f= log 324 - log 618 '(by Law = 2. = 2. (-) = = (answer).
2).
Note L In what follows we have repeated the outline to show how the mantissas should be entered. The student, however, should understand that after the complete outline is
made, including
write
down by
all
characteristics that
inspection, he should
make
possible to his entries from
it
is
the table in that same outline.
Turning next to the table
in the text,
we
fill
as follows. log 324 log 618 log
N N
= = = =
12.5105
-
10
-
10
2.7910 9.7195
(-)
.5242 (answer).
in the outline
LOGARITHMS
248
[Ch.
XV
Note 2. Since in the details above we are obliged to subtract a larger from a smaller number it is helpful, as soon as this fact becomes apparent, to rewrite the characteristic of 10 instead of 2. Remember that the decimal log 324 as 12 of the part logarithm is not the mantissa unless it is positive.
Note
In actual
3.
life
many numbers
used in computations
are approximations with a limited number of significant digits. In this case the computed result should not have more
than has any of the numbers which enter into the computation, and in the case above should be written as .524. In the subsequent exercise, however, all numbers are assumed to be exact values rather than approxisignificant digits
N
mations.
In the following examples the values found in the table will be included along with the outline, though actually they
should be j
filled in after
Example
the outline
is
made.
^
4
* IT i + -(-02764) (3.268) 2. Evaluate a/
^210,700 Solution. (Here interpolation is in order. The arrangement below suggests a compact, efficient way of recording all logarithms found, so that no details looked up are omitted in the final form of the result.)
Let
Then,
N log N
=
(2764K3.268)*. v/210,700
=
log 2764
+ 4 log 3.268 -
log 2764
4 log 3.268
=
4(0.5142)
log numerator log 210,700
=
1(5.3237) log N N
= = = = = =
log 210,700.
3.4415 2.0568
(+)
5.4983 1.7746
(-)
3.7237
5292 (answer).
the preceding example we found that the antilog of 3.7237 is halfway between 5292 and 5293. In cases
Note
4-
In
15.12]
COMPUTATION WITH LOGARITHMS
of this sort,
we
249
shall agree to use the even
numbers between which the interpolated
Example
3.
Solution. Let
= ^0.01084
N
Evaluate
A = v^.01084
and
result
+
B =
one of the two lies.
4
(0.5136)
.
4
(0.5136)
.
Then,
A = log B = log
$ log 0.0184;
4 log (0.5136);
and
N
= A
+ B.
Details.
A = = log B B = log
(8.0350 4(9.7106 0.06957
A =
0.2213
N
0.2909
=
Note
sum,
Since
5.
10) 10)
- 30) = 9.3450 - 10. (28.0350 38.8424 - 40 = 8.8424 - 10.
= =
(+) (answer).
we have no formula
necessary to find
it is
Note
-
6.
A
and
B
for the logarithm of a
separately.
a second method of finding log
By
-
-
=
10) 3) $(1.0350 $(8.0350 the form indicated in the solution
= is
0.3450
-
A
1.
we have However,
customary.
not necessary to take the final step indicated in the calculation of log/?, since the characteristic, 2, is from 10. 40 as easily as 8 obtained from 38
Note
7. It is
Example
4.
Evaluate
Solution, log
N
N
N
=
(0.008402)*.
= f log (0.008402) = f (7.9244 - 10) = = j(65.8488 - 70) = 9.4070 - 10 = 0.2553.
(15.8488
-
20)
LOGARITHMS
250 5.
Example
Evaluate
N=
(27.84)
-3
log
N= =
3 log (27.84)
i
OT:y
-
3 log (27.84). - 10 (or 0) log 1 = 10.0000 3(1.4446) = 4.3338 (-) log
1
log
EXERCISE
XV
1
= ju
Solution,
[Ch.
N N
= =
-
5.6662
10
0.00004637.
68
use of the four-place table of logarithms compute the value of * that 21 = 21.00, each of the following quantities. (It is agreed
By
.031
1.
=
.03100, etc.)
(25)(76)
2.
(0.283) (41. 6)
388
83 3.
(127.4) (32.6)
4.
(576) (340)
0.0827 5.
(228)
6. (3247) (6.97) (0.0844).
(213.4) (0.529). 3
2
8. (0.732) 2 .
2
7.
3(41.62) (87.2)
9.
(925.6) V387.8.
.
10. (0.00274)*.
11. (0.009082)*.
12. (327)*.
13. (0.0528)1
14. (928)*.
IS.
16 '
'
3
(387J 17.
(576)~
2
18.
.
20. (8.37)-'.
19. '^807.6.
21.
V5298.
22. ^8.362.
^3256.
23. (3. 142) (37.6)
8
24. V(23.8)(47.6)(39.8)(524).
.
1
1
26.
25.
V0.8846
(58.4) (9.836)'
54.8 27.
(-1.06)
3
-4.84
HINT. First find
29.
* See Note 3 in Article 15.12.
N, which
(4.728)(175.3)
(584)(755)
is
30.
positive.
(-4.76) (0.8138) (43.6)(-0,084)
2
'
COMPUTATION WITH LOGARITHMS
15.12]
Given 8
31.
A =
= 3WPA
2 ,
4g
find
S
W=
347.2,
P=
6.7,
and
0.7840.
32.
Given
T = w J-, find T if I =
33.
Given
M=
and
if
251
c
=
48.38.
TO
I
O
=
733,
O
M lia= '
26c~
'
find log
32.2,
and ir
1L20
'
6
=
3.142.
=
43 43 -
'
TABLES
ANSWERS INDEX
Table
+
A LIST OF SYMBOLS
1.
read 'plus.' read 'minus.'
a
X
b
a
-T-
b y or f, or a/6
read 'a multiplied by 6.' read
6
= =
read 'is not equal to.' read 'is less than.'
< >
read 'is greater than.' read 'is less than or equal to.'
^ ^ a! or (
I
a
read 'is greater than or equal to.' a.' read 'factorial a or 1-2-3 -
Parentheses
) ]
{
}
logo
6
6n _
n
-
-
These are signs of aggregation. They are
Brackets
used
Braces
which are to be treated in operations as one algebraic expression,
Vinculum
6.'
read 'is equal to.' read 'is identical with.'
T^
[
'a divided by
to
collect
read 'logarithm of n to base read 'absolute value of 6.'
algebraic
expressions
a.
read 'the nth power of 6' or '6 with the exponent read 'square root of a.'
n'
read 'nth root of a.'
oo
read 'a subscript n' or 'a sub n.' read '/of x' 'g of x,' or 'the/function of x/' etc. read 'infinity'
oo
read 'varies as'
an
/(x), g(x), etc.
264
Table
2.
SQUARES, CUBES, ROOTS
255
Table
3.
COMMON LOGARITHMS FOUR-PLACE LOGARITHMS
256
FOUR-PLACE LOGARITHMS (Continued)
257
ANSWERS EXERCISE
-
x
2.
1. 2z.
Page 2
1.
5
3.
4.
+
3z
-
(20
-
x
/
=
Prt.
4.
2.
6x2
6.
1.
3z miles;
7.
.
2i
(30
-
12.
T=
3x) miles.
-
10
13.
+ 2*/) =
17. (x
3x
+ 4y.
C =
2wR.
8.
2(2/
EXERCISE
14.
+ 3*).
+
18. 2(x
+ 3.
22. x; 2x
21. x; 3x.
x.
9.
1)
-
-
y) miles.
11.
2x
16.
P = ff. = 3x - 2.
+3 + 3.
= ^--^
2
23. x; fx
-
19. x;
24. 25x.
2.
Po0e 7
2.
2. 0, ~i f, -1, -2, -15, -5, -4, -2, -1, -i 0, f, 5, 6. 6. -4. 4. -10. 7. -7. 8. 12. 9. -5. 3. -3. -15. -4, 5, 6, 16. -4. 13. 32. 14. -15. 17. -9. 11. -7. 12. -17. 18. -7. 19. 3. 21. 3. 22. -5. 23. 10. 24. -17. 26. 31. 27. -3. 28. -12. 1.
-7.
29.
58.
-1.
67.
-2.
-42. 59. -4.
76. 0.
86.
84. 2.
EXERCISE
3.
43.
52. 0.
-5.
71. 5.
79.
-30. 64. -3.
63. 5.
72. 0.
Not a number.
73. 0.
74.
81. 0.
82.
38. 31.
47. 0.
56.
54. 180.
-14.
37. 20.
46. 0.
44. 12.
53. 0.
62.
36. -17.
34. 5.
-40.
61. 0.
78. 0.
77. 0.
1.
-8.
69.
68. 6.
33. 3.
32. 4.
42.
51.
49. 0.
83.
-13.
31.
41. 8.
39. 40.
48. 0.
57.
-21.
66.
-10.
Not a number. Not a number.
87. 0.
1.
Page 12
3 2 x*y&. (Sample answers) 2x i/0 5x*y&; Binomials: 3x+y; 4w+3n. Monomials: 2s; 3y.
1.
2.
(Sample answers) Trinomials: 2x 3y
;
:
+
6. 5ox - 5x + 9. 4. 7-5x. 3. 6x - 4. m + 2ran + 5n 9. -3z + z + 4z 8. 3x - Sax + 2. 7. 4s - 3z - 2ax - 1. - 2x + 3. 21. No. 12. x 2x 13. 3x 11. 3x 2a x + 1. + + 26. -2s - 7x + 5. 1. 24. ox + 8x 23. 9x 22. 2x 10. 28. -7x + 5x - 2x - 2a + 1. 4x 27. -x 4ox 3. 32. -ax 31. -9x + 1. 29. -2x + 3x* + 3x + 2x - 4a + 3. 36. 7x 34. 2X + 7x - 4ax + 3. 8x + 5. 33. -ax + x + 1.
+
2
2
2
5;
.
4
3
2
3
3
4
2
2
3
2
3
2
4
2
2
2
.
2
2
4
2
3
259
2
4
ANSWERS 38. -a 37. 2x 4 - 3s3 - 3x2 - 2x + 4a - 3. 5x2 + 2x + 2a - 1. 42. 5a - 26 - c. 6 + 4. 6 + d. 41. -4a 39. 7x-2y- 1. 46. -x2 + 3ax 2 + 36x. 44. -4x + 7. 43. -a + 26 + 4c + 1. 49. 4x - (2a + 3?/). 48. 2x - (3a - 6). 47. -3x2 - 2ax 2 + 2x. 52. -36 53. 2s3 - x 2 + 7x - 3. 5c. 51. 3a (-46 + c + 1). 260
-
54. 4a3
+
10a2
EXERCISE
+ 3.
Page 16
4.
12aV.
1.
12a6
-156x 4
2.
24a3x 5
3.
.
4.
.
-6a^.
-3x.
6.
7. 4x?/.
+ x - 2s 1) l)](2x l)(x - l)(2x + 1)] = (x + l)(2x - x - 1) 2X + x - 2x - 1. (x + l)[(x 11. 9x - 24x + 13X - 29x + 18x - 3. 9. 12x*f+ 22X - 2x - lOx. 13. 12x + 32x 17x + 16x - 4. llx + 19x 5x 12. 2x 14. x - 2x - 7x + 21x - 16X - 9x + 5x - 47X - x + llx - 2. 17. 3x + 16. 18x 6x 13x + 9x3 + 15x + 2x - 4. 2. lOx 18. 8x + 12x - 32X - 20x + 48x - 16. 8x + 3. 7x 14x + ISx 21. 18x + 39x - 4x 19. 3x - 2x - 42x + 6x5 + 45x - 13x. -
+
8. [(x
=
+
(x
2
-
l)(2x
+
= =
1)
2
2
3
6
4
5
4
6
4
55x
x
2
5 .
36.
2x2
-
-21 3x
+
1
rX
31.
.
- ab + b -3x - x + 11.
-
2a 2
2
41.
+ +
-3x
38.
.
4x.
8. lOx
2.
+ y.
4x2
sq. yds.
14.
600x
19.
T=
y
+
28.
45
~
3.
9. l(ty
+
13. (2x
ft.
16.
0.10m
~
-
10.
23. 7 29.
-9.
50z. 31.
-
11.
^
2
3x2
-
17x
+ 48 +
+ 2ax + 5a
-3x2
2 .
^A._.
7 or 7
2(10x
-
Page 17
I.
x.
+ y) +
(6x
9) yds.;
7200(x
=
x
I
~y~
6.
-
3x
-
12.
4x yds.; lOx yds.;
9.
7.
5
2
x.
+ 0.40s.
-9.
4.
'
__ ^^
2
.
2
X~~
+ 4x + 6x + 19 + 29. 2x - 2x ^ +1^ +
2
+ a6 + 6
'
3
(^
27. x3
2.
REVIEW EXERCISES FOR CHAPTER 1.
JL>
.
37. a2
4
__ _L
r4
9-5 &&
f
.
a2
39.
2
3
5
**_.
x
2
3
5
~2 x2 + + 2x ++ 5x 2x - 1 2 32. 2x - 3 + ^ ^ t ^-
g^ _
2S
4
5
4
1
-
26. 3x 2
3.
5
2
f
-
1.
2
2
__
24. x3
6
4
5
5
5
6
2
6
2
3
4
3
3
4
7
6
2
2
3
2
3
2x3
-
36.
18) yds.;
(2x
2
-
9x) sq.yds. b h( a
18. A = V = frR3 21. D = 2x - y D = y - 2z. 26. * = 26,400s. 24. / = .07/2.
+ 1)
-20.
ft.
32.
17.
.
Not a number.
33. 0.
+
-
j
22. x
=
27. 24. 34.
Not a
ANSWERS
261
-. + 26. 51. 3y. 52. 8a - 9a6 + 53. -8n - mn + 10. 54. 6a - 106 + 56. -x + 2y - 62. 57. 2x + lOxt/ - 17. 58. -8a6 - lie + 2d + 6. 59. x + 4. 61. -3a + 126 + 24. 62. x + 5y + 8. 63. -3x - y s + 3w)i 64. a 6 4a 66. y - 5 - (x + 3z (2x (2a 6 67. - 6x0. 68. 21x <#. -15m n 71. -10m n + ISm^/i lOmrc 72. -3x -' 9x - 6x + 4x y + 7xy - y 73. 4a + 26a6 2 2 - Gx - 2x ?/ + 4x0 + 146 74. x x + x0 76. 3x + x - 7x + 10. 79. x + 77. x 7x x 12. 78. + + 2t/ x*y + xV 21. 4x 81. -3a + 3. 82. 2a 12. 5a 83. 6a - 19a + 10. 4a + 2. 86. 4x 87. -9x + 84. -30a 88. ?/ - 4x ?/. 12x 91. 9 -45x 4x 92. -32 89. + + 50 + 16x - 2x 12x 12. 94. lOOx + 60x + 9. 96. 16x + 40x0 + 250 93. -3x 101. 3x + 2x + 97. -80. 98. 25m. 99: 9a6 ^ 102. 3x - 4y. number.
-18. -1. 46.
36.
44.
43. 76.
-8.
37.
38. 30.
-862
47.
9.
2
-2.
39.
-32.
41.
49. 2(a
48. 0.
.
42.
+ 6) =
2a
2
c.
c.
2
2
2
2
2
2
2
4
2
).
6
7
?/
4
2
).
3
3
.
2
3
.
3
2
2
2
.
2
.
2
2
3
.
8
?/
5
4
.
j/
?/
2
2
2
3
2
2
.
2
2
2
2
2
2
2
2
.
2
4
.
2
.
2
.
2
.
2
2
2
.
2
2
.
'
'
27
9
3
EXERCISE 2.
5x.
4.
-
-12ax
12x0 14. 18.
-
9x
2
2
6a
-
26. 27a3
40
2
+4-
12x
+
63
9x
4x
19.
.
27.
.
The square
9. x
.
-
4
90
8p
160
-
4
16.
.
4
21.
.
2
.
+ 27m p 3
+ 4ax + lOx + 25. 13.
.
4
~-
~
b2
a2
3
of a trinomial equals the
22. x 2
+ 2x0 +
28. a
.
17.
.
3
sum
-
4xz
+ 402.
9c 2
-
4a6
33. 9x
-
12ac
+
+
40
+2
2
2
-
c
2
2
.
-
- 4a 6 + 6 - 3a c + . 2
2
.
2
+ a - 4. - 4r + 4rs s 29. -24a + 36 +
2
2ax 2
2
3
.
3
.
of the squares of the terms
+ 6x2 x(a + 6 + 1). 12x0
36.
66c.
9a
2
4x 2
-9x
11.
4a 4
-
.
3
+
2 plus twice the product of each pair of terms. 32. x 2
2
26x 2
3
4
10x 2
3.
3.
6X
+
24. x2
2 6
40
2
-
2x* 3
8n 2
2
2
(6)
6.
.
2
4
- 24m x +
-
3
-18m - 24m n -
12. 2
4
+ 6x0 + 3pg
2
2
3p
+ x + 5.
2x 2
(a)
30 8.
.
4.
16m
23. 9x 2
31.
Page 23
5.
(Sample answers)
5ax 7.
27(3x+l)
+ 4z + 2x0 + 34. 4a + 6 +
2
2
2
402.
2
- 1). c - d) (a + 6 44. (2 - 30) 47. (a + b - 2) 37. 0(2a0
-
-
+ I) 39. (x 3?y) 41. (20 1)(20 +1). 42. 43. (1 - 2x + 0)(1 + 2x - y). 6 c + d). + + (a 46. (3x - 20 )(9x + 6x + 40 (4 + 60 + 90 48. (8 - x - 20) (8 + x + 20). (a + 2a6 + 6 + 2a + 26 + 4). - x0 + 49. (x + 0)(x + 0-1). 51. (x - 0)(x + 0)(x + x0 + )(x -x + 53. (4a - 56) (4a + 56). 54. (x 52. (x + )(x 56. (k + 1 x + y)(k + 1 + x a + a + 6). 0). 6)(x 2
38. (2x
.
2
.
2
2
2
4
2
2
4
).
).
2
2
2
2
2
2
).
2
2
4
2
2
4
).
ANSWERS - 1) 16xy*(2xy
262
-
57. 25x*y(2xy
+
(2xy
7.
(x
(3x
12. (x
+
-
I)
2)
2.
.
2
8.
.
3)(x
(2x
+
+ 3) -
(5x
27. 31.
34. 38.
46. 49.
EXERCISE 1.
-
(a
7.
2
9.
.
+
+
4.
.
-
(2m
2)(x
2
(a
I)
-
5)
2
6.
.
- 2) - 2). 2
(x
.
11. (x - 3)(x 14. (x - 6)(x 18. (x - 7)(x 22. (x - 12)(x 26. (x - 7)(x 29. (x - 13) 33. (3x - l)(x
2 .
1).
- 2)(x - 1). - 5). 21. (x 6)(x 24. (x - 5)(x + 1). - 1). 28. (x + 13) 32. (2x + l)(x - 2). 36. (3x - 2)(2x + 3). 39. (5x - 3)(2x + 1). 43. (3x - 2)(3x + 1). 47. (9x + 4)(3x - 2). (a;
+ + +
1). 1). 1).
-
3).
+ 1). + 2).
(a;
+ 2)(2x - 3). 41. (3x + 5)(2x - 1). 44. 3(x + 2)(3x - 1). 48. (lOx + 3)(2z + 1). 37. (3x
Page 28
-
+ 6)(x
4)
+ 5)
3. (4x
.
17. (x
(5x - 3)(2x - 1). (5x - 3)(3x - 2). - 3)(3x + 2). (8x (12x - 5)(3x + 2).
42.
2
13. (x
1).
+ 6)(x - 1). - 1). (x + 2)(x - 1). (x + 12) (x (x + 8)(x + 4). (2x + l)(x + 2). (5x + 2)(x - 1).
23.
59.
3).
Page 26
6. 2
16. (x 19.
-
58. 12(a 4 6 5
1).
1).
EXERCISE 1.
+
l)(2xy
y).
2.
(b
-
c)(x
+ y).
3. (x
+ y)(a +
4.
b).
(m
-
n)
7. (x - y)(b + c). 8. (x - 2)(x + 1). (k (x + y)(a + b). 1). 11. (x + y + l)(x 12. (x - y) 9. (x y). l)(z + l)(2x + 3). - 4y - 1). (x + y- 1). 13. (2x + 3y + l)(2x 3y). 14. (3x + 4/)(3x 17. (x + y - 3)(x + y + 3). 16. (2x - y + l)(2x + y - 1). - y - 5). 21. (y + x + i/)(4 + x 19. (x 18. (4 y). y + 5)(x 22. (2x 1 1 + y). 23. (x - 3 - y) 1 y)(2x x)(y + 1 + x). (x-3 + y). 24. (x - a - b + l)(x a + b - 1). 26. (x - x + 1) 27. (x + x - l)(x - x - 1). 28. (x (x -x + l)(x + x + 1). x + 2)(x + x + 2). 29. (x + 3 31. (x + x - 3) x)(x + 3 + x). - x - 3). 32. (x - 2x + 3)(x + 2x + 3). 33. (x + 4 - 2x) (x 4 34. x x 36. (x + y) + l)(2x + + 1). + 2x). (x + (2x x + y). 38. (2x - 2x + 1) 37. (2x (1 2). y + 2)(2x + y 3x + l)(5x + 3x + 1). 41. (3x - 1) 39. (5x (2x + 2x + 1). - 1). 42. (3x - 2)(3x + 2)(x - 2)(x + 2). (3x + l)(x + l)(x 2
6.
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
EXERCISE 1.
2
2
3; 180.
8. 18; 540.
8.
Page SO
2. 6; 72.
3. 9; 108.
9. 13; 3640.
11.
4.1; 252. 6. 4 4 5aV; 30a * .
15; 450. 12.
7.-
16; 480.
3xV; 180xy.
ANSWERS x
13.
+
(x
y;
263 2
x3
xy 17. x
2 .
i/)
-
-
14.
.
-
2; (x
+ y x xy 2)(x + 2)(x
2
3
x
-
16.
.
-
l)(x
x
+ y;
x2 (x 2x
18.
3).
y)
-
- l)(x + 2)(x l)(x 1). - 1). 22. 5x + 21. 3x - 1; (3x - l)(2x + l)(x (x + 1). l)(2x - x) 24. (x (5x + 4)(x l)(3x 2)(2x x; x (2y 3). 23. 2y (x y) (2x
l)(x + 3)(2x +
3x
19.
-
1; (3x
2
1)
4; 2
3
.
4
1;
?/)
;
3
.
2/
REVIEW EXERCISES FOR CHAPTER II. Page 81 - 3). 2. 36 (a + 26). 3. mn(m - n). 4. a6(3a + 56 - 1). 1. 47/ 6. (x 7. (c - 2a)(3a - 6). 8. (2m - n)(3 - x - y*). 3y)(y + 2). 9. (m + c)(x + y). 11. (x 12. (36 - l)(2a + 1). l) (x + 1). 13. (x - 4z/)(x - 2?/ 14. (2? - 3)(3p + 2). 16. (x + 3t/)(x - 3y + 1). 17. (y 18. (2x + l)(x + 2). 19. (x + 7)(x + 3). 4)(y + 3). 21. (x + 4y)(x - 3y). 22. (p + 18)(p - 2). 23. (4a + l)(3a - 1). 26. (3x - l)(4x + 25). 24. (5r + 3s) (5r 27. (10 - x)(2 + x). 28. (x + 2?/) 29. (2c 32. (3x - yz) 31. (2 5x) d) 34. (3 33. (2x x)(3 + x)(9 + x lly)(2x + lly). 37. (x - 2y - z)(x + 2y + z). 36. (3x - y - 3)(3x - y + 3). 39. (I x 38. (x + 2y - 3m + 3n)(x + 2y + 3m 3n). y) 42. (2x - a - b + y) 41. (3 - a + 46) (3 + a - 46). (1 + x + y). - a)(p + + a). - a + 6 - y). 44. (x + 2) 43. (p + (2x 2
2
2
(i/
2
2
2
2
).
2
2
).
2
4
2
2
2
.
.
.
2
).
ff
+ 4)(ay-4ay+16). 47. (x + a )(x -a x + a - Say - 66y + a + 49. (3y + a + 26) 48. (2x 3)(4x + 3). 51. (x - 4x + 8)(x + 4x + 8). 52. (x + 2?/ - 2xy) 4a6 + 46 - 4 - 4x) 53. + 2x7/). (x (x + xy + )(x + xy + ?/). 54. (x 57. r*(2y 4 + 4x). 56. (a 26 86 )(a (x 5)(2y + 5). - 2) (y + 2) 61. (a + 2) 59. (x - a)(x + a)(x + a 58. - 2) (a + 1). 62. (y - a)y + a) 63. (m - 2)(n + l)(n - 1). (a 66. (2x - 3y - 6)(2x + 3y - 2). 64. (x + 3m) (x 3mx + 9m 2 - 6r m 69. (1 - 6)(1 + 6) 68. 8rm(rx 67. a(3x + y)(x y). 73. (x - a - 2) 71. (pq - l)(pg + 1). 72. (2x + 7c) (x + y). 74. (2x + l)(x (x + ax + 2x + a + 4a + 4). I) (x
2
-2x + 4).
2
46. (ay
4
2
2
2
4
)
2
2
2
(9?/
2
2
2
2
).
2
2
2
2
2?/
2
2
2/
2
2
2
2
2
).
2
i/
2
2
2
2
.
(7/
).
2
.
2
2
).
2
2
).
2
.
2
2
2
.
EXERCISE 1.
(a)
9.
1; 3.
Page 37 (6)
3; 1.
(c)
2; 2.
Canceling not permissible in problems
2.
In case
4, 5, 7, 9, 10,
a 3.
(6).
and
12.
6.
6
+
1
x
-
1
+ xy +
x2
EXERCISE -1.
1.
2
10.
-
a
?/
Page 40
2. 1.
-1.
3.
4. 1.
-2
6.
-
2x
EXERCISE 5
1
i
11 11.
7 A
lj.
11.
2
2x 30
18.
20
3x
5x
9*
x
5x 2
+ x
2x 2 29.
2
2x
-
/i
14x 12 15. T~r
7 ~*
8x 33.
7'
6JL
v>
-
^o*
3x+
ig
3
+
21x
1A ~' ITT.
Q * 41
6x
3
40
-
12
-
-
5x
+
20x
+ 31x - 43 + x - 12 - 7x + 12
14x 2 31.
'
2x 2
+
9 21.
34.
3x
7
6x
24.
-
6x
-
2
6x
1
14x
+
28x
6x
36.
37.
x3
3x
+
3x 2
-
+ 4x + 3 + 3x - 41
x2 2
6x 2
-
13x
4x?y
+
2
8i/
y
lOx
6x
+
10
-
47.
-
15x 49.
26x
-
15
12x
-
8x2
5
49x?/ ~
9x
-
-
-
+ 9x*
+3
2
32.
*n 23x~ ~ oV. 6x 2
48.
2x 2
- 19 + 5x - 6 x - 13x
2
:
- 1 - 18x - 28
41.
-
5x 2
28.
2
3
15
6x
4x 2
2
'TB'*
18
5
-
+
47 '
7x
+
7x 27.
2x
x
l
-
2Qx
16
60
23.
15
Q v
TTT'
6x 2
9
44.
1
+
x
4x
+
-1.
9.
tx
^ 1 ISlTTf'
12
3
6x 2 2
1
0.
L.
+r
5x 26.
-1.
8.
-
8
10
-
2
1X
4x
8
-
420 * Tpjf-
9
6-x
.
17.
22.
3
^o-
-
4s ~
19
-
9x
1
^J
l^S.
-x
-1.
Page 43
i4 i^.
g T~
+
-
x
7.
5
6
7x
1.
-5
8x
10
12.
-
26
15x
19
'
ANSWERS EXERCISE 2x
-
x 6x
2
16.
+
12.
-
9.
17.
21.
2x
-
.
-
4x
-3.
;.
4
17x 2
+ 9.
39x
2x
18.
-
24. 4.
26. 3.
3x-
28.
+
-
3x
1
29.
2
-
6x 2 33.
x
12a.
13x
- x - 21 + 2x-35'
_
10x 2
_.
34
-9
2
'
x
2
36
2.
3.
+
5x
4.
+
7x
3
' -
6.
+
2x
10
3
-
12x 2 8.
-
2
2x
x
)'
-
3
36x
-
-6 i.
19>
26.
9x
31.
x
6x
1A
+
2
6
+ -
(y
6x 2x
1.2-
-
xy xy
-1. 2x
4y
9.
x
+
-
-
2)
-1.
-
-
9x
2
+2 +4
7x
-
-
43x
12
+
12*
-31x
10x2
2
x
10 '
2
+4 + 2x +
3x 2
2 23. x2
-2.
-
+ 2x -
2x
3?/
24.
.
33.
x)(x
1
-
1
t 3y
-
29. 2.
1).
+
3x)(x
+ 3x)(y y-1 T/
-
28. 7(x 2
-
(y
'
+
. J.
1
-
1
+
2x y 4x - 2tf'
2*
x
-
1
x
+
1
-
;
(3x
+ y.
-
4.
2y)(x
(x~y)(x
-
(^
+ -
!/)!
y)
.
-
1
'
(x 19.
llx
3x
+ 4.
-
(x
(2x
2)
a2
+ xy + y x2
y 21. 2y
2x
-
x
y '
+
lfi
6.
x
2)
2
'
18.
x x2
Page
14.
2x
-3
x
22.
2~x
12x
2)
13. ~'
-
2
(y
+
13.
6x
lly
+
3
2
i
2L
1
2y)(x
^5.
6
10
1
32.
+
x)(x
-
5x
-
27. 8x 2
EXERCISE
13 '
x
(3x
8.
2
+ y-te + 2. 2x - 3
fl*
-
3x
2
5-
x '
Page 46
4
-
5x
2x
15
7.
x
-
7 31.
1
-
13x 2
19.
23. 2x.
30
3
EXERCISE 3x
-
14.
7
22. lOx.
5x
1.
-
6x
14.
- 1 - 3x
f
9.
1
+
4x
13.
3.
A-
7.
2
-2x -
9x
+
M
-A15x -
<*
3
3x
45x2 15x 2x2 - 13x
32.
-
4.
3. f.
2x 2
21x2
27.
Page 45
*.
2.
1. 5.
1L
265 12.
'
*2
+x
2
-
+ -
2
-1.
-
+
y)
)(*-2y)
3y 17 17.
-
3x
10
i/
2xy
-
y)(2s
2
2x2/
7. 1.
'
y2
+y
2
22.
x x
-
1
-4
ANSWERS
266 23>
24>
'
+
x
+ 4x + 1 - 2x)(z + 1)'
6x2 (1
-
2x 2
i
2Q
x
26>
-
27 *
7+T
1'
r^T
+3 2x + l' x
'
REVIEW EXERCISES FOR CHAPTER III. Page 51 3x - 2y x + 4xy + 4y2 - xz - 2yz + z2 4a - 2b 2 2 2 9x + fay + 4y a 36 + 2y - z 2 7a - 15y , 4x 5 3x + 23 3n + c 4xy + y 2
.
'
'
'
'
'
a;
2
'
'
'
3n -c'
2z
-
'
24
-
4xy
g
'
+y -
12xj/*
2
3y
12y.
- xy + x - (x +
5x*
. '
'
2
(x
17 Z'
+6
a
-
2
2x
, 3'
- c)(6 - c) 3a - Qab - 62 3a - 6 2
z(x
+ 3a6 + 26 a + 46* (x + l)(x -4) 2
a2
33 ' ,_ 37
'
+ xy -
x)(3
-x+ y + 56 - 3 + 26 - l'
5a
~
(x
x2 )
'
l
2~'
-
6a6
..
36
2a6
3a
6
-
a
+ 3)(s + 5) (x + 2)
-
,
(6
m
3
a
_
c )[( a
mn
2
-
2n
-
2
c)
MR ,
-4.
26.
^.
34.
-9.
42. 6 P.M.
5) (2a
2
+ 4)
2
+
. '
_
(a
+ by
3
'
3
a2
a6
]'
+6 + 7)
2
2
2
'
2
-
62
.
Page 59
14.
in problems 3, 5, 7, 8.
6. 2.
4. 2.
.
a6
6M + 2 + 2) 4o6 + a - 26
3M '
'
a
The equations 2.
4
1)
3
n3
EXERCISE
+
+
(u
2
6
(a
2
5) (a 2
(a
'
2ac
_
+ 2xy ^
-3a-a a2
'
1 '
4x2
,2
'
-y - ab
x
l
'
'
-
(2o '
6
+y
x
,.
'
-
?R 28
x
Z4
2
>
xy x2
l'
+ 2)
8) (a
2
__ 2a
711 *
-
36
+
2x
_,_
27
^Ti
-
'
12
8
2
(y
1O iy<
-
3
&
y)'
9 , (a 23>
4
2
-
j/)
.,
12x2
+ y)
2(
18<
'
+2
8x 3 - *
(a
2/)
-
-1.
9 and
12. 5.
11 are identities. 13.
~ O
18.
27.
H-
^O
36. 3.
19-
28. f.
37. 4.
21. f. 29.
-1.
38. 4; 8.
22. 31.
.
-1.
39. 9; 3.
1. 5.
14. -^.
16.
-4.
24.
23.
32. -ft.
-1
/.
33. f.
41. $500; $1000.
ANSWERS
267
EXERCISE
No
1.
-
- fj. - T8T
8.
No root. 13 No root.
2. 1.
root.
n
-V-
.
Page 61
15.
9
12 -
-
-
19.
!
21.
f.
29.
1.
EXERCISE
16.
Page 65
CXERCIE
17.
Page 67
15.
2. 23.
.
1.
11.
12.
1. ;
c
in.; in. in.
10;
-2
L
5
12
-V
ft.
6.
in.
8.
4
in.;
12.50; 100.
4. 6;
13.
f
V
16.
T
23. 2.
6. 9;
9.
24.
32. J.
.
- fV
7.
17 26. 10.
-J.
33.
12.
.
No
root.
7. 3.
8. 5.
f
Pa0c 70
18. 4-
ft.;
12.
3.
-
31.
-^.
12. 17.
.
CXERCISE
-
-2.
22.
1.
6. 4.
-
14.
-
28.
7.
4. 2.
3.
8
40.
in.
2.
10
in.
X
3.
ft.
ft.;
10
15
in.;
9. 10 in.
13.30; 60;
X
6
in.
in.;
90.
7
6
10
in.;
X
10
in.
in.
X
6
14.40;
4.
6
in.;
7. 3 in.;
in. in.
100.
11. 1 in.;
16.20;
ANSWERS
268
EXERCISE
Page 72
19.
15 hrs. after 12 o'clock.
2.
after 1st car starts.
20 mph.
13.
EXERCISE
12. $4.00;
11. $9.00 per day.
5%. $6; 4
$8.
$90.
22. $1000.
24. 12 nks.; 24
30
1.
@
8. 21 Ibs.
11. 7 cases
EXERCISE 4.8
1.
6.
I
17. 3
12.
Ibs.
6
6T
5: (5a
+
8.
(a
2
-
12. 4. 19.
-
units.
31.
4%; $4000
19. 2 yrs.
[email protected]
14.
$16.00.
21. $600;
26. 2 qtrs.; 4 dms.; 8 iiks.
4.
oz
5. .1^5
oz>
9. 62.5 Ibs.
28
7.
220; 37.5
Ibs.
Ibs.
300.
$3.
<$
2-
7. Ibs.
93 1
2.
boy.
%
ft.
13.
550
640
3.
Ibs.
from end. Ibs.
8.
28
14.
4.
Ibs.
42
Ibs.;
-5
^
9.
Ibs.
42
Ibs.
333^
Ibs.
16. 10
Ibs.
ft.
Page 85
20 hrs.
7.
A: 48 days; B: 192 days; C: 576 days.
3.
9 min.
-~
8.
'
+
6.
2a 2
-
4a6
+
A
5a
days.
9.
3a
-
7.
+
3a
+ 36-2.
1.
17. 25.
+
5
days;
:
a
24.
-2.
3a
-
3.
21.
f.
27.
-
A =
8&.
-3.
5A
.
22. sq.
12.
2a
-6.
14.
f
4. 9.
4).
13. 15.
#=
75 min.
2.
5) days.
2.
3.
23.
6.
EXERCISE 1.
Ib.
396f
days.
4 min.
4.
$5000
9.
ft.
EXERCISE 1.
$5000 @ 5%;
7.
Page 83
from fulcrum.
ft.
500
11.
22.
2 yrs.
18.6%. dms.
.
250.
$2.50; 8 cases
from 90
ft.
581 oz
3.
150; 9 Ibs.
<&
mph.
525 mph.
Page 79
2. 30.
Ibs.
12. 10
18.
28. 6 nks.; 4 qtrs.; 10 dms.
27. 5 nks.; 5 dms.; 10 qtrs.
21.
6 hrs.
7.
*&*
13. $8.00;
$8.00.
17. $517.50.
16. $39.
EXERCISE
6.
4. -V%. 3. 4%. 4%. 8. $5000 @ 4%; $3000 5%.
$2500 @ 6%.
6.
Page 76
20. 2.
$30.
1.
^
ff
4.
f mph. 9. 6 mi.; 3 hrs.; 2 hrs. 16. 2 mph. 17. 100 mph. mph.
8.
-3
14.
ff
3.
A
+
2
15.
- i
in.
28.
-
-^.
23.
A =
V
T
2fe
2
16.
24.
36 sq.
ft.
2.
-Vtf-.
29.
(a
26.
A =
2
+
-
3a
4).
11. 5. 18.
A = s2
sq.
-
Trr
.
2
sq.
units.
ANSWERS
269
EXERCISE 19.
=
y
-
x
=
y
, *
~
6-g
x
x;
f(7
+
L
(6) r
2.
(-f, - ).
8.
21.
19. (4, 3).
26.
(3, 6).
/in 49.
(6, 0).
3.
.
(-3, -1).
(i
29.
27 y Z7
=
T/
-
4
-
'
+
4.
(-2,
6).
17.
-y-, 4).
23.
41. (V-, f).
42.
6
+ ^a /
i-o
52. j-
i
solutions (inconsistent)
:
5).
(i
!).
9(7
1.
6.
(1,
8
-T T, 14.
2,
-3,
(],
29.
(-7,
-3).
43. 48.
a2 ~~ ^ 2
(*, -*) eo
53.
j-
and
55, 58, 59,
11. (^,
5.
1.
9.
30.
2;1.
(fl,
f,
i
-I,
-16.
3.
32.
4.
14. 57.
More
61.
66. (1,3).
H). 12.
&)
(3,
3,
4.
4).
(1,
(i, -,V, -A).
8.
(A, if,
i
)-
-1, 9.
0).
(tf,
13. (3, , 2).
10.
6.
-1.
16. 43.
7. 31.
17.
-12.
8.
18.
18. 0.
9. 0.
19. 0.
11.
13.
21. 0.
Page 117
2. 3; 5.
2 mph.; 3 mph.
14. 25.
7 J).
Page 112
13. 5.
EXERCISE
/4 ( -,
16. (f, f, 18).
2. 8.
-11.
3.
-1, -2).
(1,
7.
^-).
EXERCISE 1.
2.
3).
(1, 3, 4).
12.
2).
(-ff, -*).
&,
^
(-2,
24. (f, f).
5).
f).
-V-). 7
-f).
Page 108
--
3
360.
69. (, 2).
68. ( , ).
T r)-
+
12. (2,
than one solution (dependent): 56, 57, 60, 62, and 63.
EXERCISE
i^& Z
6. (f,
18.
(2, 0).
-
('/-,
(-^,
47.
-if).
6 a
(-3, 11.
-3).
,
No
67. ( , -f).
=
2z; x
+ 40) =
5(F
(c)
(2, 3).
(-
22.
(-&,
a bl
6
Z
16. (2,
1).
46.
-- ,7-)-
(
= x + 1; = 6 - 3x; ~ ~ 6 3a;
y
24. y '
IL=_1
9.
^
54
22.
y.
t
(2, 3).
Cl /a 51.
t
/i,
L
.
~
3?y
C = f(F - 32); = (c) ^.
(6)
1
+ 2).
(j/
~
x X
-
+ 4; x =
27. (-7, 3).
-M)-
44. (f ,
6
=
=
x
x;
z '
14. (3, 2).
-4).
1
2;
+
2*
~
2x
32;
-
2x
Page 105
28.
(2, 3).
13. (1,
y
=
2g y
EXERCISE 1.
=
26 *
P=
=
21. y
y.
23.
= 6_^2y o 31. (a) F =
7.
=
1.
x
32. (a)
Page 96
26.
16. 347.
3. 7; 5.
11.
4. 8; 6.
3 mph.; 4 mph. 12. 150 mph.; 50 mph.
6. 4; 5.
335 mph.; 15 mph.
17. 397.
18.
9
in.
X
5
in.
8.
19. 12 in.
X
7
in.
ANSWERS
270 21. 10
X
ft.
4
@ 5%; $4000 @ 6%.
24. $6000
$10; B: $6. 31. 70 Ibs.
5%; 30
33.
(/) x, 3.
/5
in.
(/) 2/13
in.
(a)
2.
1.
17
-16.
10.
-8.
21.
-4.
4. 32.
(d)
*
~
^)
+
23. 9.
22. 9.
32. 2.
41.
-
42.
.
V
33.
-.3.
7/
a'
y
~
a
73 '
6
+ 6)
(a
81.
1.
74 3
(a
3.2'
+ *' + a + x _
'
-
52 52.
^2 /.
'
-
-+-
-
93
r2
a
(*)
-2.
(e)
13
in.
~
a2
^ x^2 }
2
2
x*
-
-
x2
53 53.
-
-^.
36.
.
44.
-.
46.
^
1.
J.
54 ^. 54.
2
76
! '
(2x 83. (x2
X2 )i
)
'
-
-
2
77
_1__ -y
'
a;
3?/)
9/7 2 /) 2
i
a2 )i
84.
j~-;*2 + '
1 '
'
2
l
94
'
2
T 5L_L.
x
_1_ (V,2 (a
18. 8.
.
x,2 _(_
98.
2
9. 25.
27.
-
y
-
-
+ 2a6 - a 6) (o + 6)
_
a
?-.
1
10 2.
X
2
-
fft
97. 5
in.
26. 4.
-16.
-
2x
82.
1.
'
2
-.
8. 16.
34.
T
43.
2
_
f;
58.
-
(a -
1.
17. 16.
-27.
24.
^-
51 51.
3
2
79.
1.
Ibs.
(0).
16. 3.
1
02
/58
-
(*
^
(e)
;
-6.
7.
-4.
'
78.
-f. -1. (i)
f
and
(c), (e),
125.
6.
14.
49 -?49.
57.
'
72
(h)
in.
In cases
13. 4.
y
5,.
(d) 4.
fo)
5
(c)
4.
in.
^T
39.
48 I48.
.
(d)
31. .
29. 3.
38. f.
8 47 47. ~ 1?
-3.
in.
3. 27.
12.
.
33.
35.
-9.
37.
^.
@
Ibs.
- .
;
(c)
2.
/13
(6)
11.
28. .
-2.
3.'(c)
(a)
2. 2.
2.
6
44. (-} yl
.
2;
(6)
(g)
EXERCISE
@ -%5^;
34. 20 Ibs. meal; 30 Ibs. flour.
35?f.
-1.
1;
29. 6 qtrs.; 10 dms.; 20 nks.
Page 127
34.
-x.
5%; $4000 @ 6%. 27. A: 4%; $4000 @ 6%.
$3000
32. 4 Ibs.
15%.
29. (2a) 4
.
EXERCISE (a)
23. $2000
in.
Page 124
x
26. (abW)
Ibs.
@
Ibs.
^XERCISE
1.
9 26.
28. 10 dms.; 20 iiks.
-Mp
20^;
X
22. 16 in.
ft.
99.
x2
1
2a2
x2
ANSWERS
271
EXERCISE 1.
36.
2. 4.
3.
11. 3A/2.
4. 2.
3. 6.
12.
3V3. _
19.
-4^2.
-
18.
Page 134
13. 3/5.
26.
27. 5z 2 V7.
32.
33. 6z 8 /
38. 2xv/3.
39.
44.
46.
57. 4 VlO.
78.
2Vl 2V4 -
83.
1.
73.
a
79.
.
84.
31. 6z3 V2. 37.
43.
42.
49.
Vt/2
4^2.
61.
68.14.
69. 4x.
Vl + 3x2 5Vl + 4x2 _ x^ 86.
56. 9.
54.
53.
74. 2
.
2
3/ia
24.
48.
59. 9V/6.
2V?.
9.
17.
5zV5.
36.
34. 41.
6^5. 2x 2
23.
29.
47.
58. 10.
66.10.
8. 2/5. 16. 6/2.
22. 8ar/2&.
28.
S2.
51.
2V.
SV^
14.
4V6.
21.
7.
6. 5.
62. 10.
63.
6^4.
71. 3x/x.
64.
J3.
72. 3x/x*.
76.
.
81.
.
V3.
xy
^6.
91.
+
98. (3a 103. 1
108.
V5.
92.
99. 3
26)/3a.
-V2.
104.
V3-V5.
122. x
+
-
1
^m
EXERCISE 1.
2^
=
+V2 'I'
114. 59.
10 VlO
x
or
=-
-
60.
1
- /2.
-4 +
107.
V5 - /2. -12.
112.
2/6.
121. x
9V^.
102. 13/2.
-3V2.
111.
116.
97.
96. /2.
101. -1 - V2. 106. 2 - Vs.
-V7+V2.
50
= ~~
+/2. 2V5x.
109.
+ 20Vz~. 18 V2 + 6 V5 + 3 V6 +
113. 2z
94. 3/3.
93. /3.
117.
=
or
10V3
-
=
2.
x
P
37.
Pape
0.707.
2.
^5(9
~=
0.577.
2V5
3.
^=
1.224.
4.
2
o A
^,
V30 =
~=
0.866.
2
Q 913
g
V6 =
Q 612
4
=
1.661.
14.
-~^ 5 -
22.
~= 2 27.
= 15
0.908.
11.
V3 >
16.
^32.
23.
^5.
V5 > v'll.
12.
17.
23 24.
18.
_v/243.
-^-
26.
1.
a/3a 33.
32.
EXERCISE 4. 6t. 13.
-3i.
39.
6. 3i.
14.
27.
^
34.
v'3
13.
^108.
^3.
19.
^5 > ^2.
14.
^^2. v/8.
22.
21. /6.
29.
28.
31.
1.
v^.
x
x>^2a2x 2a
i/v^^x 3(J
37.
Page 7. 7t.
-tvTl.
8. fv/47. 16. 3io.
9. 8t.
17. 4tx.
18.
11. 12i.
-5ij/
2 .
^2. 19.
-2i.
NSWERS
+
3
>.
44
i*'. o
>.
38. 3
3i.
59.
V2.
67.
+ 9i. + IQf
61.
^-
_.
'
3.
40.
XERCISE
53.
-6.
62.
5.
-
r
1,
A -
r.
s,
3, 1
~
12
-
Hi
-
3
'
lit
10
~
54.
-10+
KM.
63. 61.
i-=^-
71.
56.
t.
57.
o
69.
-
3l
-
17
o
151
7
68.
and
a, 6,
3, 2, -1. 4, 4, -3.
3
5iV3.
64.
-
-
1
i.
O
(6)
No.
No.
(c)
c are listed in that order.) 6.
-B-
C,
1, 2,
7. 4, 5,
1.
13.
-of.
1, 0,
23. a 2 ,
-rs.
-3.
12. 4, 8, 3.
18.
17. 2, 3, 0.
3, 2, 0.
).
42
Page 152
41.
11.
36. 61.
Page 148
(The values of 4. 2, 1, -6. 1, 3, 2.
-2i.
28.
7i.
10
Unlimited number.
(o)
+
^-^'
+
3
7'
'
lo
XERCISE
34. 9
41.
12i.
27. 2i.
i.
5
10
iV2.
26.
33. 41.
-
39. 16
52. -=^'. Z
f.
-1.
24.
32. 13.
4i.
'13
51.
*.-3i.
>.
15
-t.
23.
-3 -
31.
f
'13
-1.
22.
4t.
-4 + 6 - 4
r.
i
273
-aiV23.
L.
-9e 4
19. 4, 0,
0,
-1662
28.
A -
-
26. 2
.
A +
B,
C.
B, 29.
1.
-3.
-3,
0.
-e2
21. 4d*, 0,
-A -
C,
4,
8. 2, 1,
.
B-
C,
4,
14. 1,
0.
-2,
1,
2.
-6.
.
A + 4. A - C,
D,B.
XERCISE 1.
-2.
2;
2t;
-2x.
Page 155
2. 2; 9. 2i;
-2.
19.
16.
-2.
4. 3;
-7i.
11. 7t;
-3.
a
bo
3aiV2; -3a;/2.
' I0bl
21.
24.
iu
4^.
22. 46f
.
;
-3^/2.
9/x
^;
17.
-3i.
7. 3t;
13. 3v/2;
5i.
9/^
4./9
^-^;
-5.
6. 5;
12. 5i;
^. a
Av/9
2V3; -2V/3. -
3. 2;
-2i.
^-
,
I.
42.
a
v^ ;
25.
-
;
a .
a 1.4; r.
I.
L.
1;
a
a
-1.
32.3;
- .
38.
0;
i
;
- .
44. 0;
;
i
52.
-2.
i
33.5; 39.
46. 0;
- ;
.
-
t,
a
-2.
34. ;
-1.
-^;
-i
41. 4a; 4a.
^rIU
47. 0;
-
53. $;
- .
a
O.
48. 0; -
54. ;
36.
42.
-J; 1. -f; -f.
49.
;
c
-f.
56.
-f
;
-f
.
-f.
ANSWERS
274 57. ;
i
63. 20;
-f
58.
-8.
f
59. ;
-f.
;
61. 4;
-f.
64.^;-^- 66.^;-^-
^
-
62.
^p;
67.
f
68. 0;
-5;
2.
7.
A
.
69. J*L.
EXERCISE 1.4;
-3
14 10 19.
' O
^
H-
.
44.
-2.
4. 5;
2.
4 7-
4>
12 x ^'
-2v 19
6.
&
12
-3
11 1<5>
4-
Vl9
f
-^;
v7
1
o
-1
+V?
>
-3 +
21.
26.
.
11 '
3
tSi
24.
29 3
-4;
3.
/
2V?
7
Page
-3; -2.
2.
2.
2-1 f> 2-
a
<*
43.
i
-, 22.
-5
iV23
3
l2i.
-3
i
28. 2
t
23.
27.
V7.
i
33.
3 6
EXERCISE
44.
Page 163
a._ L
.
7 7.
.
.
26. 1;
-f
33.-
32. J; .
V6 +
4oc
,
23.
1;f
28. 1;
27.
29. J;
-J.
34.1;
A-
-
J
26
-&
V62 -3ac
16ac
.-
.
3
'
3a
-1
Vl7
4a
Vi^ -
-J.
36. J;
VC -
-c
5
2^6.
17. 5
16.
14.
13.
6
33 V2.
.
3V5
-1VJ3
24. }; .
6
2ag.
2a
4a
.
2 2
c
ANSWERS
Vl -4g x2 2
1
A
275
20
EXERCISE 46. Page 1. x - 3x + 2 = 0. 2. x* - 2x - 3 = 0. 2 = 4. 2x 6. 8x - lOx + 3 = 0. 0. 8. 5x - 2x = 0. 9. 5x - 2x - 3 = 0. 12. 2x - 4x - 1 = 0. 13. 2x - 2x - 1 = = 16. x 4x + 13 17. x - 4x + 7 = 0. 0. 2
a:
2
2
-8, -5;
1.
8
6.
47.
11.
X
in.
+
(8
5
-3.
16. 8;
V2; x
-
16.
x
3
+x
19.
x3
13.
16x3
+
24. x
3x
-
4
6x
2
2x 2
13x 2
-
+1=
6x
3
0.
EXERCISE
X
2.
+
8.
ix/2;
_!
4.
50
ft.
X
ft.
15
9.
ft.
in.
-i
13.
in.
1; 1;
13x
28. x
49.
120
X
14.
4. 0; 1; 1;
t;<;-t;-t.
0.
0. 0.
ft.
8
in.
-
*
-1.
8. 0; 1; i;
-i.
-1; -1.
9. 2;
12.
^ 6
= 0. 3x + = 0. + 6x - 1 = 2
3.
i.
-tV5.
-
llx
=
4
x
14.
0.
x
17.
4
21. 8x 3
-
=
0.
20x2
+
1
=
+ 4 0. - 2x + x = 0.
+
18.
46
4x 3
x
2
+x
-
- 5 = 0. 6x - 7x -
2x
26. x
4
3
=
3
x2
27.
0.
2
-
6.,i.
lVS; -1;
11
;iV2. 7.
9.
3.
-2;
-3;
1
i
i
x
= =
0. 0.
-
+ 2x = 0. x + 3x + 2
3
2
3
6
22. 4x 4
18x
23.
0.
12x
3
Page 178
;. 2;
200
+ 2V2
2
2/5)
2.
4.
X
ft.
8.
in.
1;
6x
+
+ +5
0.
Page 176
7.
+
-
2
8x
0.
2V6.
-2.
3
-
1
3
*
- V2;
150
3.
3
12. (4
.
6. 0; 0; 3;
12.
in.
in.
48.
l
11.
6
7.
-1;
1.
2.
5, 8.
17.
EXERCISE
7
= = = =
Page 171
in.
2/17)
18.
2
4x2
14. x 2
0.
2
EXERCISE
2
11.
2
2
6x - x x - 6x + 2
7.
2
+ 4x + 3 =
x2
3.
2
;
ANSWERS
276 11. 2;
*--
-J; -1
iVS;
-1
iv^ -3
, i 13. 1; 6;
-1
-
12. J;
- Stx/S
. 14.
;
V5
V7. -1
zlI - --
,V;
1
-3
3
-1; 7
+V7.
1
.
,
-2;
- ZiVZ
;
-iv/3
1
;
tV3
1
V5
1
2 .
-
-A .
l,
.
24.
2. 20.
14.
1.
50. Pojfe
9. .
-a; -6.
17.
EXERCISE 3z
3
3x2
9.
12. 14.
51.
3X3
16.
3
4
-2.
8.
-1; -2.
-
-
4x 3
+ x* + x -
1.
~
1
x+
3
13.
2.
-1 16. i;
-l;- ;-2.
V
(4, 3);
-
34x
18. Yes.
+
141
-
19. Yes.
17 -
-1
4
V3.
22.
No
54.
;
;
4.
x+4
21. No.
.'-
2;~-
9.
-1;
^
EXERCISE
7x 2
1
564
1;
13. i;
2
-9;
3.
l;l;-2.
~
~^;
3.
Page 187
52.
8.
3.
2
1
17. Yes.
16. Yes.
2x.
2
2
3
1; .
1.
root.
14. 1.
2
_1
21. 1;
4
Page 184
1.
1;3.
12. 1;
No
7.
13. 2.
1.
3
EXERCISE
7.
Li
3. 2x - 3z + 2x 2. 5z - 2x + x - 2. + x + 5. - 2z + 7. 6. x 3x 8. 8. 2x ^ 11. 3x + 3z + x + 1 + + 4x + 14 + -^-x x - 3
4x 4
1.
{
-3.
3z2
1.
4.
12.
-1.
6.
4. 1;2.
3. 0.
18. 1;
1
180
-3.
11. 3;
.
-2,
V2;
1
i
^
EXERCISE
!
^1. l,
^-r-5
14.
-
11.
-1;3;
No
rational roots.
o
i
18 -
L
19 - 2 2 ;
;
~2 -1 ;
i
rational roots.
Page 194
(-4, -3).
2.
3.
(5, 0); (0, 5).
-
i
(No
real intersection8 -)
ANSWERS
277
(-tVH,
(iVll, 6);
11.
13. (1, 1).
14. (1
18.
17. (1, 1).
(I
5
/V58
7. (2,
2
1.
V
23. 10 rods
X
2 rods.
2
'
8
/-V58 V42
/V58 -VJ2
/'
'
2
-4);
(-5,
'
'
2
V
'
2
2
2 '
/'
'
2
V
I
'
'
2
.(-^ ^);(^,- V2).
(f f);
9.
i
)
2
(3,
-4); (-4,
3); (5, 3);
2. (0,0);(4,4);
3).
(-5, -3).
(-1, J);(-l,
(-^,
4. (5, 3); (5, 3);
J).
-f).
(
)
(-i
!
(-1,
I)!
(i
8. (1,2);
(-1, -2); (1,2);
11. (0, 0)
)
;
(0, 0)
13-
(-!, -I)-
);
(-i
;
(0, 0)
(-2,
-f); ( ,_-1); (- , (-JV10, VlO).
is 18.
EXERCISE
-1.
9. a.
16. (2, 2);
21. ($,
57.
,
2).
(0, 0).
-6);
(0,
,
-f
17.
(M);
)
;
(f 1) ,
-2); (-6,
16. (1, 2);
I).
--,
12. (f
;
10).
(-1, -2);
(-1, -1);
(1, 1);
-
-,
Page 203
11. 2.
(-2, -2).
-2,
;
9. (1, 1); (1, 1);
-8).
(11,
6); (4,
;
8.
).
-2).
21. (a, 6);
-J).
(2,
3).
Page 200
56.
(3, 4); (3,
-f);
-a).
(6,
-2&);
2
3
-1); (-2,1).
(-*,-*)
14.
/'
(
3. (5, 3);
19. (1,
-J).
(- ,
4);
16. (4, 2); (4,
-2t).
t,
/-V58 -V42
/42
EXERCISE
-
(1
(-5,
Page 197
(-550). 0^
2 (5.U), (5 0V 2.
12. (2,
in.
55.
'
2
V
2);
i,
); (1,
( ,
EXERCISE
(I,
(-1,
+
22.
X
8. (5, 3);
(-5, -3).
(3,iV7);(3, -iV?).
2aj-
24. 12 in.
6.
7. (5, 3);
6).
9. (4,0).
12.
17.
22. (2, 4, 6).
-9; 1. (i 0).
13. 18.
f -$. ;
(-4,
0).
14.
-^; 19.
-f. (i 2).
ANSWERS
278
EXERCISE
+c_b+d
q
i
a
d
__
c c
a
d
b __ c
b
+a_
d
a
d
a
t
19.
ft.
-
'a
+
6'~c
+
c
6
-
6
+
b
6'~c
d'
d
c,
a
b
16.
_
17.
ft.
18f
d
_
;
d a
c
^.
c
12f
ft.
Page 213 /
varies directly as v?/.
kx
=
kxy]
d
c
?/
2. z
__
_
'
a
b
a
c
+ a '~__ d + c.
mis.
V-
EXERCISE 59. 1. y = kx?] ~ =
c
b
2
d
6
c
c
6 a ~ 6-d'a +
a~c
;
d
+
_
C
-
+
6
c
24
18.
d
-
C
+
t
d
6
+c_6+d
d a 'a
c __ b
c
'
a
a
t
a
t
'
d
c
6
Page 205
58.
x varies directly as
k
z
and inversely as
y.
xy
w =
3.
y
X
x^yz
Z
x2
8.
3
=
~^~ = ;
-- =
;
ft.
Moon:
t(;,
y, /~~
x varies directly as
k:
Vy
and
2
3
(or as .
and inversely as zVw.
iV3;
^;
(c)
11.
(d) 24.
in.
2 v'5
;
x varies directly as wy* and inversely as
fc;
xz
^- sq.
sq. in.;
16.
=
-
^2
i
y
= w
14. 2
i
Vt^ and inversely as
tr^ifiiD
n*?y
---;
9. (a) f ; (6)
80
^/~'
,!
varies directly as
x varies directly as
fc;
x
z*w 7/
,.
,
/c; a;
ID
yz 7.
=
;
w =
6.
x varies directly as /wz and inversely as
k
XfJ
= kw
.
4.
=
-;
Z
-^
_
12._9 pts.
ft.
2 ^10
;
ft.
2 ^20
;
20
sq. in.;
13. ft.
2 ^60
;
3
re
+
60.
3r4/
+
240a?V 3 .
6.
*x
19. (a) It is decreased to
Mass
22.
ft.
EXERCISE 1.
in.;
115.74_lbs.; 154.05jbs. ft.
4 v/15
;
ft.
16.46 Ibs.; Mercury: 26.3 Ibs.; Venus: 86.1 Ibs.; Mars: 43.1 Ibs.;
S is decreased by f y
21. 168 cu.
*& sq.
sq. in.;
Jupiter: 252.8 Ibs.; Saturn: 107.5 Ibs.; Sun: 2777.5 Ibs. 18.
z.
of Jupiter
=
17. 199.5 Ibs.
11 units;
(6)
96 V2
in.
64x 8
-
302.5 (mass of earth).
Page 219 3x?/
-
+
160z
16x2
V-
+
2. x*
y 3
3 ?/
+
60x
2
4 t/
- 3x*y + 3xy* - 12xy + y 4.
3 t/
5
6
.
27a
32xyVx + 24xy* + 8y*Vx + y 8. + ^x^ - TV2/ +
fW
4
B
3.
.
6
+ 7.
a8
4
27a 6
+
ANSWERS
279
~ - 20x% + ISOxV - 960xV +
x6
13.
H----
-70.
16.
.
x8 18. 3432.
.
_
_
22
-792a26 17
17.
a
a2
a
-
14.
-.
W
23.1-2* +
a2
-
x~3
19.
.
_
3x~*y
+
+
*
*4
3
- 2ac + + + + + 2ad 26c 26d + 2cd. 2ac - 26c. 27. a + & + c + 2o6 28. a + 6 + c - 2a6 - 2ac + 26c. 29. a + 3a 6 + Safe + 6 + 3a c + 6a6c + 36 c + Sac + 36c + c
+
24. 1
+
2x
-
8x 3 H----
4x2
-
2
c2
26. a2
.
2
2
2
d2
2
2
2
3
2
2
2
2
2a6
2
3
3
.
EXERCISE 1.
61.
-5.
2.
11.
Pa0e 224
8. 8, 6, 4, 2, 0.
200 mis.
16.
6
17.
EXERCISE 1.
8.
48.
147;
7.
62.
4
Y*-
3.
-6.
7.
6. 7, 11.
12. 115.
11. 4.
25 mis.
18. $20.
-4, -10, -16. 14. 21.
13. 8.
Page 226
2. 185.
3.
9. 31;
-3.6.
-88.
4. 114.
6.
18.
17. 21.
18;
f
150
13. 11; 27. 19.
ft.
440
ft.
23. $6840.
22. $1275.
21. $3000; $2800; $2600; $2400; $2200.
192; 27.
7.
-60; -3.
12.
11. 312; 15.
16. 3; 24.
14. 7; 23.
-
9. 21.
24. $2950.
EXERCISE 1.
7.
32.
63.
2. 243.
8.
16, 8, 4, 2.
13.
9.
21. 6;
Page 231
14.
-8,
9.
4.
-128.
-3,
16. 4; 30.
10.
22. 5;
1.
4.
3. .
9,
27. $3000; $1500; $750; $375.
-27,
81.
or
11.
18. 5; 121.
17. 6; 21.
23. J; 255.
1.
6. 4, 8, 16,
24. J; 31.5.
-4,
8,
12.
4.
19. 6;
-16. 3.
-21.
26. $10.23.
28. $4000; $3200; $2560; $2048.
29. 4096.
EXERCISE 1.
1.08.
8. 0.18.
64. 2.
Page 2S6 1.26.
9. 0.42.
3. 1.44.
11. 0.70.
4. 0.90.
12. 1.4.
6. 1.5.
13. 2.1.
7.
-0.18.
14.
-0.52.
ANSWERS
280 16.
-0.92.
24.
i
33.
.
17. 25.
27.
- .
34.
VA-
36.
Change
side to loga A%.
38. 41.
Change
right side to
44. True.
1.
66.
-
18. 7.7664
EXERCISE
X
24. 1.030
67.
X
13. 0.1109. 18. 72.79.
10 9
.
left side
16.
6. 2.6232.
6.9590
-
12. 0.9180.
17. 6.7782
10.
22. 135.
21. 11.5.
-
10.
23. 22.6.
29. 0.00231.
28. 0.0315.
33. 0.000100.
Page 246 3.
8.9218
6.8503
9.
X
-
-
10.
10.
4. 1.5474.
10~ 8
22. 3.107
.
-
11. 8.9659
16. 0.001942.
6.
20.
17. 0.1401.
X
lO' 10
.
7.7563
-
10.
12. 20.3651. 18. 5.908.
23. 6.521
X
10 10
.
.
Page 250 2. 0.03199.
22.89.
1.644
a y).
logio 2).
Change
11. 0.5705.
27. 0.289.
21. 2.063
68.
43.
+ log
y
logio
4. 2.3711.
1.5119.
- 10. 6.8069 - 10.
2. 2.8625.
10 15
EXERCISE 1.
19.
10.
19. 0.01456.
+
left
47. True.
3. 2.7210.
9.
14. 2,313,000.
13. 167.0.
Change
right side to 10(log x
logio B).
9.8663
8. 2.4688.
7. 0.7583.
7.
14.
10.
3.4041.
37.
Page 244
32. 1.01.
31. 5.48.
A
23. 3. 32. 512.
.
-
46. True.
26. 3920.
24. 430.
1.
(2 logio
8. 2.9657.
13. 8.7924
31.
^B.
left side to log a
Change
2. 2.5403.
2.2625.
7. 2.5563.
29. 2/2.
right side to (logio x
Change
to loga v^C.
EXERCISE
28. f.
i
22.
21. 125.
19. 4.
-1.
39. True. 42.
i
18.
26.
14. 2.624 19. 9.314.
24. 4861.
26. 1.063.
31. 99.75.
32. 15.01.
4. 3.768.
3. 50,220.
8. 0.5358.
X
11. 0.09530.
9. 18,220.
10 7
.
16. 1.725
-46.01.
33. 9.9894
-
1Q- 8
.
22. 1.529.
21. 7.555.
27.
X
28. 10.
-16.05.
34. 0.977.
6.
1910.
12. 47.47.
17. 0.0314.
23. 227,000. 29. 0.04336.
INDEX
refer to pages)
(Numbers
Common Common
Abscissa, 90 Absolute value, 4
Addition Addition Addition Addition
of fractions, 41, 42 of numbers, 5 of radicals, 134
Aggregation, symbols of, 11 Algebra, definition of, 1 Algebraic expression, 9 Algebraic formula, 1 Amount, at simple interest, 74 Analytic geometry, 94, 189 Antilogarithm, 243 Arithmetic, definition of, 1 Arithmetic extremes, 222 Arithmetic means, 222 Arithmetic progression, 221 Associative law of addition, 10 Associative law of multiplication, 13
227
Conditional equation, 54 Constant, 94
Constant of proportionality, 211 Coordinate system, 90 Coordinates, 91 Cube roots, table
of,
255
Decimal approximation, 126 Decimal numbers, 146 Degree, 29 Denominator, 34 Dependent equations, 95, 104, 113 Dependent variable, 93
Axiom, 56 Axis, of coordinate system, 90 Axis of imaginaries, 147 Axis of reals, 146
Determinant, 109 Diagram, 202 Discriminant, 161 Distributive law, 10 Dividend, 15 Division of fractions, 45 Division of numbers, 6 Division of polynomials, 13, 15 Divisor, 15 Double roots, 151, 154
Base, of an exponential expression, 9 Base of a logarithm, 233 Binomial, 9 Binomial formula, 239
Binomial theorem, 239 Braces, 254 Brackets, 254 Briggs* system of logarithms, 237
Ellipse, 189
Equation, 1, 54 Equation in quadratic form, 176 Equations reduced to simplest form,
Cancellation, 35 Characteristic, 238
9
Coefficient, literal, 64 Coefficient of the unknown,
Combination method, 101
Common difference,
ratio,
Commutative law of addition, 9 Commutative law of multiplication, 13 Complex fractions, 34 Complex numbers, 147 Complex plane, 147
and subtraction method, 100
Coefficient,
logarithms, 237
221
58
176 Equations reducible to simpler forms, 197 Equations with given roots, 175 Equivalent equations, 57, 60, 157
281
INDEX
282
Even
integers,
39
Exponential equations, 233 Exponents, 9 laws of, 14, 121 negative, 128 zero, 127 Extraneous roots, 60, 61, 179 Extremes of a proportion, 202 Factor, 21 Fractions,
4, 34, 146 algebraic, 34
complex, 34, 48 rules concerning,
39
simple, 34
Fulcrum, 80 Function, 87 General equation in one unknown, 173 Geometric means, 229 Geometric progression, 227 Graph of a function, 90, 91 of a linear equation, 94 of a linear function, 94 of a non-mathematical function,
96,97 solution of a quadratic equation, 164 Graphical solution of a system of
Graphical
equations, 103, 189-200
Highest common factor, 30 Hyperbola, 189 Identity, 22, 54
Imaginary numbers, 143 Imaginary roots, 157 Imaginary unit, 143 Incommensurable lengths, 126 Inconsistent equations, 104, 113
Independent equations, 104 Independent variable, 93 Index of a root, 124 Integral rational equation, 55 Integral rational polynomial, 21
of logarithms, 234 of multiplication, 13 of radicals, 132
Lever, 80 Lever arms, 80 Like radicals, 133
Like terms, 9 Limits of roots, 185 Linear equation, 55, 56 Linear function, 94 Literal coefficient, 9, 63 Literal number, 9, 64 Literal root, 64 Loci, 189 Locus, 92 Logarithm, 233 Logarithmic equation, 233
Lowest common denominator, 41 Lowest common multiple, 30 Lowest terms of a fraction, 35 Mantissa, 238
Mathematical function, 87 Means of a proportion, 202 Member of an equation, 54 Minor denominators, 48 Minor of a determinant, 111 Moment of a force, 80 Monomial, 9 Multiple, 29 Multiple roots, 175 Multiplication of fractions, 44 of polynomials, 13, 14 of radicals, 134
Napierian system of logarithms, 237 Negative exponents, 128 Negative numbers, 4 Notation concerning functions, 88 Numbers, 3, 143, 146, 148 Numerator, 34 Numerical coefficient, 9
Odd
integers,
39
Interest, simple,
Operations which equations, 60
Involution, 121 Irrational number, 126, 146
Operations which do not yield equivalent equations, 60, 61 Operations with imaginary numbers, 145
74 Interpolation, 244 Inverse variation, 208
Laws
of addition, 9, 10 of exponents, 14, 121
yield
equivalent
Operations with positive and negative numbers, 5, 6
INDEX
283
Operations with zero, 6, 7 Order of a radical, 133 Ordinate, 90 Origin, 90
Rational number, 126 Rationalizing denominators, 137 Real number, 4, 143 Reciprocal, 34 Remainder, 15 Remainder theorem, 180 Repeated base case, 122 Repeated exponent case, 122 Repeated root, 154 Root of a number, 124 Roots of an equation, 56, 151 Roots, table of, 255
Parabola, 166, 189 Parentheses, 5, 10, 11, 264 Plotting a point, 91 Polynomials, 9 11 adoption of, 9, 10, division of, 13, 15 multiplication of, 13, 14 subtraction of, 9, 10, 11 Positive numbers, 4 Postulates, 9, 45
Satisfy an equation, 56 Scale drawing, 202
Power, 123
Scientific notation,
Powers of a negative number, 123 of
t,
144
Prime factors, 21 Prime numbers, 21 Principal in interest problems, 74 Principal rth root, 125 Product of the roots of a quadratic,
158 Progression, arithmetic, 221 geometric, 227 Proportion, 201
Pure imaginary numbers, 147
241 Second order determinant, 110 Side of an equation, 54 Simultaneous equations, 100-109, 189-200 Single base case, 121 Square roots, table of, 255
Stated problems, 115 Substitution method, 101 Subtraction of fractions, 41, 42 of numbers, 5 of radicals, 133 Sum of an arithmetic progression, 225
Sum Sum
Quadrant, 90 Quadratic equations, 150 algebraic solution of, 152-7, 160 of, 161 graphical solution of, 164
of a geometric progression, 228
a quadratic, 158 Surd, 126 Synthetic division, 181 of the roots of
discriminant
Table of logarithms, 256, 257 of roots, 255 of symbols, 254 Tangent, 166 Term, 9 Third order determinant, 110
standard form of, 150 system of, 195-200 with given roots, 168 Quadratic formula, 160 Quotient, 15
Time Time
as an independent variable, 97 in interest problems, 74 Transposition of terms, 57
Radical sign, 124 Radicals, 124 addition of, 133
Trigonometry, 201
equalization of the order of, 140
laws
of,
132
multiplication of, 134 reduction of the order simplest form of, 141 subtraction of, 133
Radicand, 124
Rate
Trinomial, 9
Unknowns, 55 of,
140
Variable, 93 Variation, 206 Vertex of parabola, 167
Vinculum, 254
of interest, 74
Ratio, 201
Zero, operations with, 6, 7
INTERMEDIATE ALGEBRA
THE MACMILLAN COMPANY NEW YORK
BOSTON CHICAGO DALLAS SAN FRANCISCO
ATLANTA
MACMILLAN AND LONDON
CO., LIMITED
BOMBAY CALCUTTA MELBOURNE
MADRAS
THE MACMILLAN COMPANY OF CANADA, TORONTO
LIMITED
INTERMEDIATE ALGEBRA
Ralph
S.
UndcrWOOd,
of Mathematics,
Professor
Texas
Technological College
Thomas R.
Nelson,
Associate Professor of Mathematics,
Agricultural and Mechanical
College of Texas
Samuel Selby,
of
College
Head
the
Department
of
Mathematics,
of Engineering, University of Akron
THE MACMILLAN COMPANY
NEW YORK
1947
COPYKIGHT, 1947, BY THE MACMILLAN COMPANY no part of this book may be reproduced in All rights reserved without any form permission in writing from the publisher, except by a reviewer who wishes to quote brief passages in connection with a review written for inclusion in magazine or newspaper.
PRINTED
IN
THE UNITED STATES OF AMERICA
PREFACE
This text is designed for students who, for one reason or another, such as inadequate previous training or merely the
immaturity of youth, must learn algebra at a somewhat slower pace than that set by the typical college freshman who plans to enter a scientific, engineering, or allied field. In other words, it is intended to be a true intermediate algebra, suitable for college classes catering to those students who in high school had less than the normal amount of mathematics. As such it starts at ' bedrock' in algebra, moves on with a leisurely but accelerating pace, and leaves the student at the
ledge in the
cliff
of learning
and logarithms. There
is
marked roughly by progressions
ample material
for a course of three
or six semester hours, according to the needs
and proficiency
of the class.
Specific features of the text include the following. 1.
The
illustrative
examples are numerous, and they are
carefully selected to cover typical cases. 2. The fact that in algebra the rules of arithmetic are
applied to letters instead of numbers is emphasized by frequent use of two illustrative examples, one of which is purely arithmetic. rules,
and
Thus the student is encouraged to check methods,
results
by use
of simple numbers.
Common
errors are pointed out and discussed before ' they are hidden as traps' in problems. In fact, in certain problems the sole question to be considered is whether a 3.
given operation is or is not correct. 4. In a text of this level the primary aim
is clarity,
to-
PREFACE
vi
gather with as much mathematical rigor as can be preserved in the light of the first requirement. A major goal, therefore,
has been simplicity and directness of style. To this end exceptions and qualifying statements have been for the most part relegated to footnotes. 5.
More than 2600 problems provide ample
drill.
Experience has indicated that the common practice of printing only half of the answers in a text in effect cuts 6.
down
the available problem list by almost one half, since most teachers assign for home work only those problems for
which the answers are to be found
in the text.
On
the other
hand, some problems without answers are needed for correspondence courses, and students acquire self-reliance by learning to check results for themselves. These two conflicting considerations make it seem advisable to include most, but not all, of the answers and so four fifths of them are included in this text. We think it inadvisable to have the ;
remaining answers printed even in pamphlet form, since copies would eventually reach unintended hands and destroy the effectiveness of the book as a text for correspondence courses. 7.
Our guiding
principle has been to maintain, so far as
the subject matter allows, a consistent, upward-sloping level of difficulty. Such a policy dictates, for example, an early treatment of linear equations, and the deferment of the chapter on exponents and radicals as long as seems
expedient.
The
mimeographed form in the appropriate classes at Texas Agricultural and Mechanical College, and the present form embodies changes made in 8.
text has been used in
the light of student reactions. It has since been read critically by several referees, who have contributed many valuable suggestions.
CONTENTS
One
Algebra as a Language and a Tool
Two
Type Products and Factoring
21
Three
Fractions
34
Four
Linear Equations in
Five
Functions and Graphs
87
Six
Simultaneous Linear Equations
99
Seven
Exponents and Radicals
121
Eight
The Number System
143
Nine
Quadratic Equations
150
Ten
Special Equations in
Eleven
Simultaneous Equations
Twelve
Ratios, Proportions,
Thirteen
The Binomial Theorem
216
Fourteen
Progressions
221
Fifteen
Logarithms
233
Tables
254
Answers
259
One Unknown
One Unknown
and Variations
1
54
173 189
201
281
Index vii
INTERMEDIATE ALGEBRA
Chapter One
ALGEBRA AS A LANGUAGE AND A TOOL
The
1.1.
relation between algebra
and English
As
applied, algebra is a method of solving problems by the use of numbers, letters, and symbols. As written and as read, algebra is a brief and useful
international language. One must learn to read it, at least in part, before the calculations can be started. Fortunately for
the beginning student,
familiar ones of arithmetic.
many The
symbols are the may be learned as
of its
others
they occur in the text. For reference they are grouped together in Table 1. This language of algebra is usually much briefer than English. For example, the English phrase, 'the number which is 3 less than 4 times the number x' becomes, in algebra,
simply '4x ber Xj plus
3.' Again, the statement that 'twice a 5, equals one-third of the sum of x and written in algebra as the equation
2x
(1)
When
certain
ner, this fact
is
+
5
=
num4' is
i.
numbers are always related in a given manexpressed conveniently in an algebraic for-
mula. For instance, the distance, d, traveled by a body moving at uniform speed equals the product of its rate, r,
and the time, product of
its
t.
Or the
area, -A, of a rectangle
length, L,
and
its
width,
to the following formulas (2) (3)
= rt; A = LW. d
1
is
equal to the
W. These
facts lead
ALGEBRA AS A LANGUAGE AND A TOOL
2
[Ch. I
The problems in Exercise 1 test the student's ability to translate English phrases or sentences into algebraic language.
EXERCISE
1
// x represents any number which we wish to think about, express in terms of x the numbers described in problems 1-6.
4
1.
The double
2.
The number which
3.
Two more
4.
One
5.
One-half of the
6.
Three times the product of x by
7.
A man
How
less
of x. is
4 less than
than one-half of
than 3 times
x.
x.
x.
number which
is
2 less than its
x.
double.
walks toward a point 30 miles away at 3 m.p.h. he far has traveled, and how far has he to go, after x hours? apples and oranges sell for 3 and 4^f each, respectively, the selling price of x apples and y oranges?
8. If
what
is
9. A and B, 20 miles apart, walk toward each other. After A has walked x miles and B has walked y miles, how far apart arc
they? 10.
Work problem
9
if
A and B
go
in the
same
direction, with
A
behind B. Write as formulas the facts stated in problems 11-14. 11.
base, 12.
The area, P, of a parallelogram B, by its altitude, H. The
base, B, by
equals the product of
its
area, T, of a triangle equals one-half the product of its its altitude,
13.
The
14.
Simple
H.
circumference, C, of a circle equals twice the product of its radius, R, by the constant w (pronounced 'pie'), which is nearly, though not exactly, equal to 3.1416.
the rate,
r,
interest, /, is the
by the
time,
t.
product of the principal, P, by
NUMBERS
1.3]
Change 15.
x and
3
the statements in
Twice the number
problems 15-18
x,
plus
5,
to
equations in algebra.
equals one-third of the
sum
of
4.
16.
Three more than twice x
17.
Half the
sum
of x
is
2 less than 3 times
and twice y
twice the
is
x.
sum
of y
and
thrice x. 18.
the
Two
less
than twice the
sum
number which exceeds x by
of x
and
1 is
3 more than half
4.
Represent in terms of x the pairs of numbers described in problems 19-23. 19.
Their
sum
is 10.
20. Their difference
(Answer: x and 10
x.)
is 5.
21.
One
is
one-third of the other.
22.
One
is
3 more than twice the other.
23.
One
is
2
less
than two-thirds of the other.
24. A boy with x dimes has 3 times as Find the value in cents of what he has.
man is x years old now, old will he be in 10 years?
25. If a
How 1.2.
The
how
relation between algebra
many
old
nickels as dimes.
was he 5 years ago?
and arithmetic
We have seen that algebra is a method of solving problems by the use of letters which stand for numbers. Arithmetic, on the other hand, is the science of computing with particular numbers. The result of an algebraic problem often can be applied to many cases simply by assigning different values to the letters, whereas an arithmetic result applies only to
the one problem concerned.
1.3.
Numbers
The concept
of
what
is
meant by a number develops and
changes as one studies mathematics.
We shall here give illus-
ALGEBRA AS A LANGUAGE AND A TOOL
4
trations of certain types of of numbers in general.
numbers rather than a
[Ch. I
definition
Most
easily understood are the whole numbers, or integers, such as 1, 2, 3, etc. Then there are fractions, such as %, f, f, etc.
Both
integers
and
fractions
may
be
positive, like those
Negative numbers that the temis 20 below or that a river perature (twenty degrees zero) 2 feet, meaning that its surface is 2 feet below level is normal. Numbers are either real, like those mentioned above, or imaginary. The second group will be discussed later. All real numbers may be represented conveniently as points on a line, as shown (in part) in Fig. 1. above, or negative, as 11, 3, are useful in various situations, as
f, etc.
when we say
Negative numbers
Positive
Fig.
numbers
1
This figure also calls attention to the very special number zero, written 0, which is between the positive and negative numbers and may be classed with either group. That is,
=+0 =0.
This fact makes necessary special rules for computation with zero. (Art. 1.6.)
1.4.
Absolute and numerical values
The symbol a
read 'the absolute value of a,' or a , numread 'the absolute value of a,' means the positive ber equal to +a if a is positive, or to a, if a is negative.
,
Thus, 3 = - 3 = 3. Consider two numbers a and 6. If a has a larger absolute value than 6, it is said to be larger than 6 numerically. If a comes to the right of 6 in Fig. 1, it is larger than b algebraically. Or, to state it another way, a exceeds 6 algebraically 3 both b is positive. Thus, 10 is larger than whenever a
:
POSITIVE
1.5]
AND NEGATIVE NUMBERS
5
3 algebrainumerically and algebraically; while 2 exceeds cally but not numerically. The symbols > and < mean ' respectively greater than' and 'less than' in the algebraic sense. Thus, > 2 (zero is greater than minus two) and 5 < 2 (minus five is less than two).
Operations with positive and negative numbers
1.5.
When
negative numbers are taken into account the ordinary operations of arithmetic must be explained anew. The rules for addition and subtraction are as follows :
(1) To add two numbers with like signs (both positive or both negative) add their absolute values and prefix their
common
sign.
To add two numbers with
unlike signs, subtract the smaller absolute value from the larger one and prefix the (2)
sign of the numerically larger number. If the
numerically equal, their (3)
To
sum
numbers are
is zero.
number from another, change number subtracted and then add.
subtract one
sign of the
These operations are examples below.
illustrated in
column form
the
in the
Addition (a)
7-33-7
2
(b)
A
2
(c)
-5
-2
(d)
-2 -5
(h)
-2 ~5
5
Subtraction
2
(e)
(f)
2
-5
5
-3
-2
7-7 (g)
5
3
operations above may be written on single lines by observing the rule that when parentheses are removed the signs enclosed are changed if and only if the sign before the
The
parentheses
is
minus. Thus
(+6)
=
6;
(
6)
= +6
(or
ALGEBRA AS A LANGUAGE AND A TOOL
6
just 6);
and
+(+6) =
(h), for
and +(-6) example, become (-2) (-2)
(d) (h)
6;
+ -
(-5) = -2 (-5) =-2
For multiplication and division numbers a single rule suffices
=-6.
-5 +5
[Ch.
I,
Accordingly, (d)
= -7; = 3.
of positive
and negative
:
The
(4)
product,
with like signs it is
and
two two numbers with unlike signs
also the quotient, of
is positive; of
negative.
-20; and (-4)(+5) =-20. Also, 10 -5- 2 = 5; (-1C) -f(-2) = 5; 10 -f- (- 2) = -5; and (-10) -h 2 = -5. As shown in the above example, multiplication of positi^ numbers may be indicated by dots instead of parentheses, provided that the dots are raised to avoid confusing them with decimal points. In the case of letters representing numbers, the omission of a sign between them indices multiplication, so that, for example, ob means 'a times 6.' This practice, of course, would not do for numbers themselves.
Why? Division
symbol 1.6.
-*-
.
is
indicated
by use
For example, 10
~
of a horizontal bar, or the
5
=
^=
2.
Operations with zero
The
rules for operations with zero are illustrated in equations (1) to (5) below.
+ 6 = 6 + 0 = -0 + 5 5 -
(1) (2)
0-0
=
The value to
it
(3)
6;
0-7= -7 +
=
5;
-3 -
= -7; + = 0. = -0 - 3 = -3;
0.
of a
number
or subtracted
from
is
unchanged when zero
is
it.
0-4 = 4-0 = 0- (-4) = (-4) -0 =
0.,
added
OPERATIONS WITH ZERO
?.6]
The product
of
any number and zero
is zero.
(4)
The
number
quotient obtained by dividing zero by any other than zero is zero. -
5)
That
is, it is
To
impossible
*
6?'
sum
to divide
any number by
zero.
see that the statements in (5) are reasonable, notice
that the value of f, or add up to 6.' Similarly, tv>
- are not sokable.
and
of
definite
-
'the number of 2
'how many
asks
;
s
which will add up
zeros will
Evidently no number meets this condition, since the
any number
but this
3, is
of zeros
number), then
is
times that
impossible, as seen
is
EXERCISE
2
zero. Similarly,
if
$ = (some
number should equal
from equation
6;,
(3).
(ORAL)
Arrange the following numbers so that, as read from left to 1, 15, 0, 6, f, 4, 5, right, they increase algebraically: 5, -2. 1.
Hb
2. Arrange the numbers of problem 1 so that, as read from left to right, they increase numerically, or in absolute values.
Perform
the additions
and subtractions as indicated in problems
3-14. 3. 6.
9.
12.
3
+
-15 6
-
+
(-7).
+
(+5).
4. 7.
11.
10.
(-17).
13.
(-6)
+
3.
-9 - (-2). - (8). (13) - (-32).
5. 8.
11. 14.
-8 +
(-5).
8 - (-4). -7 - 0. -15 - (-0).
* It might be thought that by this test the value of must be one, since 'one up to zero'; but so do two zeros, three zeros, etc. The rule stated in (5)
zero adds
holds in
all cases.
ALGEBRA AS A LANGUAGE AND A TOOL
8
Perform
[Ch. I
the additions as indicated in problems 15-26.
27-38. Subtract the lower from the upper numbers in problems 15-26.
39-50. Multiply the numbers in problems 15-26.
Perform
the indicated multiplications in problems 51-62.
51. 6(-7).
52. (-15)(0).
53. 0(100).
54.
9(-4)(-5).
55. (-0)(28).
56. (-5)(6).
57.
+7(-3).
58.
-(-!)(-!).
59. 4(-l).
60.
-(2)(-3).
61.
-0(-15).
62.
Express as integers 63. 20 *
4.
64.
(+7)(-2).
the quotients indicated in problems 63-71.
(-21)
-f-
(7).
65. 14
-r-
(-2).
66.
Which of the quotients indicated in problems 72-87 Find the values of these numbers.
bers?
(-20)
are true
+
2.
num-
ADDITION AND SUBTRACTION OF POLYNOMIALS
1.8]
1.7.
A
9
Algebraic expressions single
number
or letter, or a group of them,
is
called
an
algebraic expression.
An
2 - x o
1
Examples. 4; x 3x; 2x
1; -;
x
expression with no plus or minus or equality signs be-
tween
its
parts
is
called a term.
2
Examples. 7;
The term
2; a; xy] xyz]
2a6 3 stands for
'
x
2a6 3
;
.
2 times a times 6 3 ,' where
a and 6 are literal numbers, or U ' and 6 3 (read 6 cubed or 'the numbers, third power of 6') means 666, or 6 taken 3 fo'raes as a factor. Here 6 is the base and 3 is the exponent of 6. Thus, if a = 3 and 6 = 2, then -2a6 3 = -2 3 2 2 2 = -48. Terms differing only in numerical coefficients, such as 3# 2 3as 3 are called like terms. An 2x 2 or as 3 4as 3 and and expression of one term is called a monomial; of two terms, a binomial; of three terms, a trinomial; and of two or more 2
is
the numerical
letters
coefficient,
standing for
,
,
,
,
terms, a polynomial. The terms of a polynomial are connected by plus or minus signs, as in the example:
2a
1.8.
+
6.
Addition and subtraction of polynomials
convenient to state in the form of laws certain reasonable assumptions about numbers reprebut unproved sented by letters. Such assumptions are called postulates.
Here
it is
The commutative law of addition: The sum of two more algebraic terms is not changed by changing their
(1)
or
order.
Thus, just as 3 ter
+4=
4
+
3,
so x
+y=
what number values we assign to x and
y
y.
+ x,
no mat-
ALGEBRA AS A LANGUAGE AND A TOOL
10
The number of
The sum
associative law of addition:
(2)
algebraic terms
is
the same, however
[Ch. I
of
any
we group
them.
That
just as
is,
The
(3)
+y+
x
in general,
4+9+5 = (4+9)+5 = 4+(9+5) = 18, z
=
+
(x
distributive law:
y)
+
z
The sum
term by each of several other terms term by the sum of the others.
+
+
=
= x
+
(y
so,
+ z).
of the products of one is the product of this
+ +
c The student a(b d). should verify this equation for various special values of the = 2, b = 3, etc. Once granted, it enables letters, such as a
That
db
is,
ac
ad
+
us to combine the like terms of a polynomial. Thus, 2x 3 5x 3 3 2 3 2 2 2 = x (2+5) = 7x and ax +7ax -2ax = ax (l+7-2) = 6ax 2 .
,
We
are
now
(x
=
=
add or subtract polynomials by
For example, 2x + x + 7) + (3x + 4x 3 - 5 + 6x 2 ) 3 2x 2 + x + 7 + 3x + 4x 3 - 5 + 6x 2 ) (x like terms.
combining 3
in position to
2
+
3
(x
4x
-
3
2z
2
+
6z
2
(by the associative law) t7 - 5) 3z
+x+
+
(by the commutative law)
=
(x
+
=
5x 3
+ 4x + 4x +
3
4x 3 )
+
(-2z
2
+
2
6x 2 )
+
+
+
-
3x) (7 5) (by the associative law) (by the distributive law)
2.
(x
In subtracting one polynomial from another we note that -(a+6-c) = (-l)( a +6-c) = (-l) a +(-l)&+(-l)(-c)
= a 6 c (by the rule of signs (by the distributive law) in multiplication). This indicates that when parentheses preceded by a minus sign are removed, the signs of all terms which
+
had been inside are changed. Thus, 3
(x
-
+ x + 7) - (3x + 4x - 5 + 6x - 2x + x + 7 - 3x - 4x + 5 - 6x
)
=x = -3x 3
is
3
2
3
The work
2
3
2x 2
shown
8x in
2
- 2x+
2
12.
column form below. For convenience
the terms are arranged in the order of descending powers of
x.
REMOVAL AND INTRODUCTION OF PARENTHESES
1.9]
Addition
z
3
-
2x*
+
x 3o:
11
Subtraction
+ -
3
x 4x 3
7
5
-
+ -
2z2 6z 2 Sx 2
+ x+ + 3o: - 2x +
7
5 12
Removal and introduction of symbols of aggregation
1.9.
We
have seen that the signs of terms within parentheses preceded by a minus sign were changed when the parentheses were removed, but were not changed if the preceding sign was plus. Like statements are true of the other symbols of aggregation (Table 1). This enables us to simplify expressions containing such symbols (parentheses, brackets, and braces) by removing them one pair at a time, beginning with the innermost ones. The procedure is then continued on the
new
expressions formed until
removed and
like
Example 2x
such symbols have been
all
terms collected.
- 5(s - 1)]} = 2x+ {3x - [2x - 5z + 5]} = 2x + {3x - 2x + 5x - 5} - 2x + 3x - 2x + 5x - 5 = Sx - 5.
+ [3x~
[2x
When parentheses, or other symbols of aggregation, are introduced into an expression, the signs of the terms within are not changed if the preceding sign is plus, but are changed if the preceding sign is minus. This statement may be verified
by removing the symbols
in accordance with the pre-
ceding rules.
Example 3z
1.
+ 2y + z - w =
Example 3x
(3x
+
2y)
+
(z
-
w)
2.
-
2y
-
z
=
3x
+
(~2y
-
2)
=
to
-
(2y
+ z).
ALGEBRA AS A LANGUAGE AND A TOOL
12
Example 8. 2x - y - (a
+ 6) =
2x
+
= 2s-
-
[-
+
]/
(a
+ 6)]
+
(a
[Ch. I
V)]. f
important to notice that the
It is
same terms can be en-
closed within parentheses preceded
minus
The above examples
sign.
by
either a plus or a
how
illustrate
the last two
terms in each are enclosed with either sign preceding.
EXERCISE
3
terms involving x
and
1.
Write three
2.
Write two examples of monomials; of binomials; of
like
9
y,
z.
tri-
nomials.
Add 3. 4. 5.
6. 7. 8.
9.
10. 11.
12.
the polynomials in problems 3-12.
4x
-
2x
7;
+ 3.
+ 3;4 - 7rr. 2ax* + 3z - 2; ax - 5x + 3. 3az - 2x + 4; 2az - 3x + 5. x - 5z + ax - 2; 3z + 2z - 3ax + 1. x* - 2z - 3az + 1; 2z + 2x - 3az + 1. x - Sx4 + 3z - 2a; 2s - 2z - 1 + 3x. 4z + 3z - x4 + 3; 2z + z + 4a - 3x 2;z - + 7; 3x + 4 - 2. - 2; a - 3z + x - 1; 6 + x 2z
2
2
2
3
2
2
3
2
2
3
2
4
2
4
3
2
.
3
2
a:
2
2
4
.
13-16. Verify the additions in problems 3-6
replaced
by
1
and
by 2 and
are
when x and a
are
respectively.
17-20. Verify the additions in problems 3-6
replaced
when x and a
3 respectively.
21. If the arithmetic results obtained in problems 13-20 do not expose any errors in the algebraic sums obtained in problems 3-6, does this prove that these sums are correct?
22-29. Subtract the second from the
problems 3-10. 30-37. Subtract the of problems 3-10.
first
first
polynomial in each
from the second polynomial
of
in each
MULTIPLICATION AND DIVISION OF POLYNOMIALS
1.10]
In problems 38-47 remove 38.
2a - (b + 2 - 3(s -
-
3a
the symbols of aggregation
39.
d).
- 4z + 1). (2y + 2(a - 26) - 3(2a -
46. 47.
-
42.
44. 45.
2
6).
1.
c]
43.
simplify.
-
3x
41. 4 y) + 2(x + y). 2a + [3a + (26 + 1) 3c + [26 3. (a c) + 4] - 2}. 2 [z (2z {3z 1) + 2] 3a - (26 - [3a - 2(26 - 1)] + 3}. - {2z - [(3ax + 36)x + x
40.
and
13
2
}}.
Insert parentheses preceded by a minus sign which will include all but the first term in each of problems 48-51. 48. 2x 50. a
52.
-
3a
+
+ 26 - c + 2. To
sum
the
- 3y. 51. 3a + 46 - c - 1. + 3a and -6a - 26 + 3c -
49. 4x
6.
-
of 6
2c
+ 4c from 3a 6 from 5x 2z + 7x
result of subtracting 6
53. Subtract
-3x
3
54.
+
4x
2
3
To what -3?
55.
3
add the
2c.
2
2 and add the result to
1.
Take 2a2
mainder to 4a 1?
3x 2
2a
-
+
4a6 from
3a
2a6
expression
1.10. Multiplication
-5a 2
+
+3
6a6
and add the
re-
2 .
nfcist
2a2
3a6
+
6 be added to give 0?
and division of polynomials
In addition to the distributive law already discussed, two other useful laws about multiplication may be mentioned here.
The commutative law of multiplication: The product of two or more factors is not changed by changing their order. (1)
That (2)
three
is,
just as 2
3
=
3
2,
so ab
=
ba in
all cases.
The associative law of multiplication: The product or more factors is not changed by grouping them
of
in
different ways.
Since (2 3)4 = 2(3-4), for example, so, in general, assume that (db}c = a(6c).
we
ALGEBRA AS A LANGUAGE AND A TOOL
14
Two
[Ch.
I
of the so-called laws of exponents, which are discussed 7, will be needed here.
at length in Chapter
2 3 Examples. 2 2
=
25
dm (4)
a
Examples.
is
= am+ n = x aaV = a'aV =
a ma n
(3)
=
^ o
;
=
n
35
~2
.
xW
;
~n
when
33
~
am
=
11
=
m> a7
;
.
$~
w
2
4
'
Chapter
1.
(2az
3
35
=
(3aV)
)
3aa 2z3a; 4 [by
2
= QaW 8
,
Example
a 10a x
2.
4
-
=
/10/a
8
[by
by making
(1)]
(3)].
4
/a: ru
(^(-) 4
[by
/CN1 (5)]
1
[by (4)]
=
5a 5z 3
.
The multiplication of two polynomials, such as is in the expression
3
(x
-
3,
3
5-7
Algebraic terms may be multiplied or divided use of the five laws above.
Exampk
.
n.
Finally, law (5) below, treated more fully in used in the division of polynomials.
Example. (f
a 10
2x4
+ + x
3x2
-
l)(x
-
indicated 3a;
3
+
2),
can be accomplished conveniently by arranging the polynomials in terms of descending powers of x and then employing law (3), as in the following example.
_ -2z4
+ x* +
3x2 3xs
+x- 1 +x+2
(first
factor)
(second factor)
6x 7
a; 4
6z
T
-
3z
6
-
llz
5
-
4s 6z 4
+ 2x + + 8z 3
3
-f-
2
6x2 7x2
- -
2x
+
a;
-
2 2 (product).
1.10]
MULTIPLICATION AND DIVISION OF POLYNOMIALS
A An
,.
.
,.
,
,
.
,
.
indicated division, such as
(5Z
-
3
5x 4
+
+
+
5)L y
6x
(x~
X
15
2)
be carried through as here shown. The explanation of the steps may be found just below the actual division.
may
x*-2x 2 + x-1 (Quotient*) 1 (divisor) x
3x
2 x
r
5o:
4
+5x
+ z+ 5
3
(dividend)
x+5
-
x 2 +3:r+2 3 (remainder)
Explanation. The first step is to arrange the dividend and divisor in the order of the descending powers of the same letter. Here the dividend, divisor, and Quotient, so arranged
5z 4 + with respect to the powers of x, are respectively x 5 3 2 3 2x* x 1. The first 3z - 2, and x 5r + re + 5, z 3 term (x ) of the Quotient is obtained by dividing the first
+
term
5
(#
)
of the dividend
by the
first
term
2
(x
)
of the divisor.
then multiplied by the divisor, and the product is subtracted from the dividend. The remainder thus 2x 4 + 7x 3 + x + 5. To obtained forms the new dividend, obtain the second term, ( 2x 2 ), of the Quotient, the pro2 2x 4 cedure above is repeated, with x being divided into
The
3
result, (x
),
is
.
Succeeding terms of the Quotient are thus obtained until either a zero remainder is reached or one in which the ex-
ponent of the based (in this
letter
case, x) is less
The
in the divisor.
X5
_
5^4
upon which the
^_
^
_
2
than the largest exponent shows that
it
is
has
final result
+ 5X + x + g 3
original arrangement
_
x
_
2x
+
x
3 l+ x2 _ 3x _2
The word 'quotient' is often used with two different meanings. We shall use with a small 'q' to mean the total result of an indicated division. Thus the 'quotient' of 27 by 5 is 5f, while the 'Quotient' is 5. *
it
ALGEBRA AS A LANGUAGE AND A TOOL
16
~ = Quotient
, dividend In general, -37-:
,.
T
divisor
The
result of
,
.
[Ch. I
remainder rr. divisor
H
a long division should always be stated in the
form indicated above.
EXERCISE Perform
4 the multiplications or divisions as indicated in prob-
lems 1-7. 1.
2
3
5z 3 (-36z).
2.
(3z a)(4a z).
6a 7& 6c 3
12oV
3
-6xy*z
-3a26c' 8. Verify that
2 3 (-4z)(2az)(-3a z ).
3.
2
2
'
3
[(a;+l)(a:-l)](2x+l)= (x+l)[(x-l)(2x+l)].
Multiply the pairs of polynomials in problems 9-21. 9.
(6z
7
10. (3z 5 11. (3z
13. 14.
15. 16. 17. 18. 19.
20. 21.
2
4
(3z).
- 9z + 3); (3z - 1). + 5z + 5z - 2); (z - 3z + 2). (2Z - 10z - 5z + 2); (2z + 3z - 1). (6z + 7z - 5z + 5z + 4z - 2); (z - 3z + 1). (z + x - 5z + 4z + 4z + 5z - 2); (-1 + 2z + (6z - z - 4z + 6Z ); (z + 3z - 1). (4 + 2z - 3 + 2z); (z - 1 + 2z). (3z + 8z - 12z + 4z ); (3z + 2z - 2). (8 - 6z - llz + 3z ); (3z - 1 + x (13z - 5z + 4z ); (1 + 4z - 3z). (5z + 1 (-z + 6z - 3 + Hz ); (x - 2 + 3z 4
12.
4-H* -z-5);(2z). - llz + 7z2 - 5z + 2);
4
-
7z
2z2
3
4
5
3
2
2
3
2
3
4
2
3
2
3
2
4
3
8
2
2z2 ).
2
4
2
2
3
2
2
4
3
2
).
2
3
2
2
3
2
).
22-34. Divide the
first
problems 9-21. 35. Divide 3z 5 3
36. Divide
a
37. Divide
a3
polynomial by the second one in each of
+ 2z - 3z + + 6 by a + b.
38. Divide 27az
2
8
1
2 by 2z
-
3z
+
1.
3
63 2
-
by a 25o
3
b.
+ 20a z 2
18z 3 by 6z
-
5a.
REVIEW EXERCISES
A =
//
3z
-
2
17
+ 4,
2x
B=
3z
^nd
the values of the quantities in
39.
2A
-
35.
40.
2
-
x
-
1,
and C
=
-3a:
+
1,
problems 39-41.
EG + AB.
41. (A
+ 25)
-:-
C.
REVIEW EXERCISES // x represents a number, write the numbers indicated in problems 1-7. 1.
Four times the number.
2.
Three-Fourths of
3.
20%
of
4.
The
difference
5. 6.
Eight more than the number. Nine less than three times the number.
7.
The
it.
it.
between the number and
(Two
7.
answers.)
excess of 5 over 2.5 times the number. 9
// x and y represent the tens' digit and the units digit respectively in a number of two digits, write the numbers indicated in problems 8-11. 8.
The number.
9.
The number when the
digits are reversed.
10.
The sum
11.
Twice the number, increased by
of the digits divided
by
3.
36.
// x represents the width in yards of a rectangular field, express in terms of x the length, perimeter, and area of the field under the conditions stated in problems 12 and 13. 12. 13.
//
The length The length
A
is
four times the width.
is
nine yards
less
walks x feet per second and
than twice the width.
B
walks two feet a second faster,
write the quantities described in problems 14-16.
A B
14.
The
distance
15.
The
distance
16.
The
distance traveled
travels in ten minutes. travels in half
by both
an hour. in
one hour.
ALGEBRA AS A LANGUAGE AND A TOOL
18
[Ch. I
Write as formulas the statements in problems 17-26. 17.
The volume,
18.
The
V, of a sphere and the cube of the radius, R. area, A, of a trapezoid
by
altitude, h,
The
19.
one-half the
sum
equal to the product of
is
-J,
TT,
equal to the product of the
is
a and
of the parallel sides,
6.
a telegram of 50 words is comcents for each of the first ten words and
cost, T, in dollars for
puted at the rate of m x cents for each of the others.
C
20.
At p
dollars each,
21.
D is
the difference between twice x and
22.
A man's
2$.
On
x% (x% =
$5000 at
dollars.
(Two
y.
formulas.)
x now and was y ten years ago.
is
age
x books would cost
the yearly interest
y^n)
is
I dollars.
The yearly interest on R dollars at 7% is 7 dollars. 25. The cost in cents of s pounds of nuts at 50jf per pound and y pounds of candy at 75^ per pound is C. 26. The distance in feet traveled in x hours at the rate of five 24.
miles per hour
is s.
Evaluate the numerical expressions in problems 27-34. 27.
-
30. [(-1)
32
-
(~3)][4
+
-20
-(-3)
28.
(-4) (-2) (3).
39.
20
-
20
20
+
20*
+
If a
+
3,
b
=
2,
38. 40.
and c
=
1,
-8 -
-3 2
.
0.
34
36.
(?)
=
-20
20
+
*
missing number in each of problems 35
= -H. (-2) X (?) - 16. - 8 = -10. (?)
35. 7
37.
31.
(-2)].
0(-20) the
29.
.
'
'
Find
2
20
-
20 20'
to 40.
= 10. -25 + (?) = 5. 3 X (?) = -15. (?)
find the values of the expressions
in problems 41-46. 41.
-3a2
44. (a
+
+
2b
b)(b
-
a c.
42.
+J* +
43. 3a 3c
-
Zoc
+ c).
45.
a
+
6(6
+
~~ a c).
46.
7
5.
REVIEW EXERCISES Add 47.
19
the expressions in each of problems 47-51.
-362
48.
3y
262
-2y
-7b2 50.
4x2
2
49.
2
+ 6) + 6) 7(o + 6)
-3(a -2(o
-y2
-
+
2
y
2x
51.
z
-2?/ -32
-3x
+ 2z
x
2
3x2
-
y + 3 + 3*/-l + y-2
- 3a6 + 6c; 4a& - 2a - 3c; -2c - 10 ab + 7mn - 3n2 + 6; 5n2 - 7mn - 2; 6 - ran.
52.
Add: 4a2
53.
Add:
2
6a2
.
Subtract the lower from the upper expression.
4a
-
76
-2a
+
36
54.
-
55.
c
To
y
sum
the
of
+
2z from 3x
-4x
2
+ 3xy -
2 over 2x
the excess of 4xy
+ -
2c
56. Subtract 4x 57.
3x 2
2
Sxy
+
y
2 and
+
4xy 5xy
-
y
+
2y
Remove 59. 60.
61.
62.
-5xy -
65. x
+ 8x
2
add
6.
7c2
2d and
symbols of aggregation and
the
collect like terms.
- 1 - (3x - 1)]. 4 [2x - [-z + (2y - x)]. -[(2x - 3i/ + y] -3(o - 26) + 6{ -4 - [6 - 2(4 + 6)]}. - [(2x - y - 4) - (3y - 2)]. (3x + y + 6) -
2
2
2
)
-3x2 2
7
2
Insert parentheses preceded by a but the first two terms. 63.
2
4.
58. How much must be added to the sum of 3a& + -6 + 3c to give -5a6 - c ? 2
2
+
y
2xy
Carry out
-
2x
+ 3t/
2
+z+ 4.
3w.
minus sign which
- b2 + 4a - 2a 6 -5 + y - x - 3 4
64. a 2
66.
will include all
2
2 .
2
.
the indicated multiplications.
67. (-3x)(2y). 2
2
68.
2 (-3xV)(-zy )(7xy).
3x2 (2xy
-
2 ?/ ).
69.
-(-3mn )(-5m n).
70.
71.
-5mn(2m2n - 3m
72. (3x 2 -i/+6x)(-x+y-x
+ 2n).
2
).
ALGEBRA AS A LANGUAGE AND A TOOL
20 73.
75.
77. 79.
81.
-
4a a
26
74.
+ 7b -
2x2
2xy 3x
-
2
-y (x + 4)(x + 3). - 3). (x + 7)(x -3(a - l)(a +
83. (3a
-
2)(2a
-
3x 78. (x
80. 82.
1).
5).
84.
-
86.
85.
-3(2
87.
(-3x + y)(y + 3x). 2 2 -5(3x - y)(3x + y).
89.
91. (3
-
-
2x)(3
+ 2)
93.
-3(x
95.
-(5x -
Note.
x)(3x
88.
2
+xy - 2xy + 2
-
2y y
-
5)(x
+
-2(4
94. (lOx 2
by
y
2).
-(x + 2)(x - 3). - 4). (2a + 3)(a -2(5a l)(3a + 1). - y). (2x + 2/)(2x -(2x - y)(2x + y).
92.
2x).
.
4j/)
2
90. (2x 2
2
When
carried out
1).
+y
x3
76.
y
-
x2 x
[Ch. I
7)(2x
-
+
2
x)(4
3)
+ 7). -
x).
2 .
96. (4x
.
the multiplications in problems 77-96 have been the methods given in this chapter, the student
should study Arts. 2.2 and 2.3 and then do the problems orally. the indicated divisions.
Perform 97.
=&.
98.
99. 18a 2 6 4 H- 2a6 2
101. (6x
s
102. (9x
2
103. (3x
2
-
100.
.
+ x + 3) -r 24xy + 16?/ 5x + 2x - 1) 2x
2
2
)
3
-r-
2
(-12x y
- 1). (2x - 4y). (3x
-f-
(3x
+
1).
3
2)
-5-
2
(-4X ?/
2 ).
Chapter
Two
TYPE PRODUCTS AND FACTORING
2.1.
The factoring rule
In arithmetic, since 12 = 2-2-3 = 4-3 = 6-2=l-12, we say that the positive integral factors of 12 are 1, 2, 3, 4, 6, and 12. This means that if we divide the integer 12 by any
one of these integers, we get another integer. Of these factors only 2 and 3 are prime, or without positive integral factors other than themselves and 1. In algebra it is possible to make use of a similar factoring rule. To explain it, we shall need some definitions. A polynomial which is the sum of terms such as 2x 2t/3 in which all exponents are positive integers, is said to be integral and rational.* In this chapter we shall deal only with integral ,
rational polynomials, including in this general description 3 not only expressions with two or more terms, such as 2x
+
5xy + 4, but even single terms such as 3x and 5. When a polynomial is expressed as the product of two or more integral rational polynomials, it is said to be factored, and the parts which are multiplied together are the factors. 2 1 = (x Thus, when we write z l)(x + 1), and 3x + 6 =
and 3x 2
3(x
x
+ 1,
except
2),
x
we
+
itself
factor the left sides.
1, 3,
and
Each
of the four factors
+ 2 is prime, since it has no factors In this chapter we shall bar any factoring
and x 1.
which introduces coefficients that are fractions or types of numbers not yet discussed in our text. Thus, while it is correct * The reason for the customary use of these adjectives at the end of Art. 7.5.
21
is
suggested in the note
TYPE PRODUCTS AND FACTORING
22
to writes
+2=
+
2(%x
1),
andz 2 -2 =
where (V2) =
-/2)(z 2 and # 2 2
+
2
(x
[Ch. II
+ V2),
will be 2, the expressions z considered here as prime, or non-factorable. When we completely factor a polynomial, we express it as x = the product of its prime factors. For example, x 3
not completely factored in the second form be1 is not prime.
x(x*
1) is
cause x
2
2.2.
Useful identities
Certain operations in algebra are repeated so often that much time is saved by memorizing them in type forms instead of working them out on each occasion. This is especially true of the following results, which may be considered as multiplications as they stand or as factorizations as read from right to left.
a(b
(1)
+ c) = + b) 2
(a
(2)
(3)
(a
(4)
(a
-
-
+
b)(a
2
b)
b)
= =
ab a2 a
2
a
2
+ ac. + 2ab + - 2ab + -
b
b2
.
b
2 .
The student should check these results by multiplication, and should verify them for various special values of the = 4, 6 = 2, etc. They are examples of letters, such as a algebraic identities.
An
identity is
an equation which
values of the letters involved.
makes each
A
is
true for
value
is
side of the equation a definite
all
permissible
permissible
if
it
number. In the
identity
2x
x the values
1
and
2
1
1
x
1
for
-
1
x
1
+
1
x are not permissible.
Why?
Also useful from the standpoint of factoring are the identities
(5)
(6)
:
a3 a3
-
+
b3 b3
= =
(a
-
(a
+
6) (a
b)(a
2 2
+ -
ab ab
+ +
b 2 ). b 2 ).
2.2]
USEFUL IDENTITIES
Illustrative
A. 1.
23
Examples
Type Products
3(2z
-
2
3x
2.
(2x
+
3t/)
3.
(3z
2)
4.
(5z
-
2
2
+ =
=
4)(5x
= =
1)
2
(2z)
2
(3z)
+
4)
-
2
3(2z ) 3(3x) 6z2 - Qx 3.
+
3-1
=
9z 2
+
+
-
2(2*)
+
2(3z)(2)
-
2
(5x)
42
22
=
25x 2
-
-
12*
+
4.
16.
B. Examples in Factoring 5.
6. 7.
- Qax + 2a = 2a(o; - 3z + 1). 4z - 9 = (2x) 2 - 3 = (2x - 3)(2x + 3). 4x + 4az + a - 2 + 2^/6 - 6 - 2y6 + & = (4z + 4ax + a = (2a; + a) - (y - 6) = [(2a; + a) - (y - 6)][(2s + a) + (y = (2x + a - + 6)(2a; + o + y - 6). 8z - 27 = (2o;) 3 - 3 = (2x - Z)[(2x)* + (2s) (3) + 3 = (2x - 3)(4x + 60; + 9). 2az 2
2
2
2
2
2
2
j/
2
2
2
)
2
(t/
2
)
2
6)]
T/
8.
3
3
2
]
2
9.
27y
10. 9a; 11.
2
4a 2
+
EXERCISE 1.
= = =
+I - (3y)(l) + (3y + l)[(3t/) + l)(9t/ 3y + 1). + 24:n/ + 16^ = (3x + - 20a + 25 = (2o - 5)
3
1
3
3
(Si/)
2
I 2]
2
(3t/
2
2
4j/)
.
2
.
5
State each of the identities
(1), (2), (3)
and
(4), Art. 2.2, in
English.
Write a trinomial involving x which integral in x, (6), not rational and integral in 2.
is
x.
(a), rational
and
TYPE PRODUCTS AND FACTORING
24
By
use of the proper type products perform (orally
when
[Ch. II
possible}
the multiplications indicated in problems 3-29.
- a - 1). 4a 6(6 - ab). 6a(-2x - a).
3.
5x(2x
5.
2
7.
9. (x
11.
3
+
-(3xy 2
13. (2a
-
2
17. (3a c
+ 2)
2
2L
2
b)
14.
.
+ x)
2
16.
.
2
)
2
18.
.
2
aX?
(?
-2x
2xy).
-
(3x
+g + 4) -2(3m +
3p(p
2
2
-
2
).
-
-i/
-
(3x
-(2x
3x)
2n)
2 .
2 .
2 2 .
)
+ 2y). )(2z + y
2y)(3x
3
-
3 2/
+ 0) '
22< [(X
a)
.
(4m
20.
4j/ ).
2).
2
2
(2
+b+
2ax
2
2
12.
.
+ 4y )(2x ' +
19. (2x
6.
2
-
10. (2x 2
.
-2(-3y
15.
-3y(y
8.
2
5)
4.
- 2 - 3t/)(3x - 2 + HINT: 3x - 2 - 3y = [(3x - 2) - 3y]. 24. [(x + y) - (2r - s)][(x + y) + (2r -
3
3
).
2][(X
23. (3x
+ y)(x* -xy + y*). - 3ab + 6 (3a + 6)(9a - Qmp + 9m p (2p + 3mp)(4p -(7/-a)(2/ + ar/ + a -3(2a - 6)(4a + lab + 6 Prove the identity (a + 6 + c) =
)].
25. (x 26. 27. 28. 29.
30.
2ac
+
2
2
).
4
2
2
3
2
).
2
2
).
2
2
).
2
26c both
member
in the
by form
+ b) +
31. State the identity in
+ b + c + 2ab + 2
2
and by writing the
direct multiplication [(a
a2
left
2
c]
.
problem 30
a suitable form for oral
in
problems.
Use
the statement obtained in
problem 31
to
find the squares indi-
cated in problems 32-34.
32. (x
+y+
2z)
2
33. (3x
.
-
2y
+ z)
2
34. (2a
.
Factor the expressions in problems 35-60. 35. 6ax 2 37.
2
2ay
+ 2ox -
y.
4a.
36. x(a 38. 4x
2
+ 6) + x. + 4x +
1.
-
b
-
3c)
2 .
FACTORING A TRINOMIAL
2.3]
z - Qxy + 9y - 1. 4y 1 - (2x - y)*. 2
39.
25
2
40. 4x
.
2
41. 43.
4^.
X
*j i
~i~
49. (x
+ +
6
-
47. (a
51. x
53. 16a
2
yY y
-
(x
+
256
60.
^y
2 .
*
2 .
2
6
.
2
6)
55. (ro
.
-
n)
2
-
2
(x
3
)
+
2
(t/ )
3 .
16.
57.
yY.
5
4
+ d)
27y*.
- (x + - IGy*.
1
(c
+ y HINT: x + y* =
2
(jfc
58.
.
-
52. x 6
.
2
56.
50.
y).
6
- yY ~ (a + (x + I) - (x 12a4 6 - 36. 72a 6 - 9ab2
54.
2
48. 64
8.
6)
9y
ZiiX
TcO.
3
-
44. 8
2
+ 6)
42. (a
JL.
-
2
59.
2
.
2.3. Factoring
The
a trinomial
identities (x
(1)
+
a)(x
+
6)
6) (ex
+
d)
=
x2
=
acx 2
+
(a
+
6)x
(ad
+
+ 06,
and (ox
(2)
+
+
6c)x
+
bd
are not worth memorizing in themselves; but they suggest a practical method of multiplying two monomials and thereby
aid in the factoring of a trinomial. To illustrate, we may find the product (2x 3)(3x 2) mentally by use of the follow-
+
ing scheme.
The and The 6x 2
The
first
term of the product
is
=
2
Go; (2x)(3x) ' second term is the sum of the products of 'inside terms 'outside terms/' or 3(3x) 4x = 5x. 2x( 2) = Qx third term is 3 ( 2) Hence the product is
+
To
,
+
=6.
5x
6. 2
-}- Ilx 2, for example, we with binomial factors whose first terms are Go: experiment and x or 3x and 2x, and whose second terms are factors of 2.
factor the trinomial
Among the possibilities are (6x +
A
2)(x
-
1),
(te
-
Go:
+ i)(x 2), (Qx l)(x + 2), + 1), (3x + l)(2x - 2), etc.
(Qx
2)(x
quick inspection of 'inside plus outside products
' (the
TYPE PRODUCTS AND FACTORING
26
[Ch. II
12x = x + llx, key to the whole method) yields x = I2x llx, and we stop right there, having found that
6x2
+
-
=
-
+
l)(x 2). The student should alcheck the middle term in the product of the trial factors
ways
llx
2
before accepting
(Ox
them
as correct.
unchanged but shorter when the coefficient of a; is 1. In factoring x 2 -- x 12, for example, we seek two 12 and whose sum is 1, the conumbers whose product is efficient of x. These are readily found to be 4 and 3. We check the inside write the factors (x + 4) (x and rapidly 3) and outside products. It is more important for the student to become accustomed to this check as a method than it is
The method
is
2
for huii to learn (2) or (1) as formulas.
EXERCISE
Do
6
Review Exercises of Chapter
orally problems 77-96 in the
1.
Factor the following trinomials. 1.
4. 7.
+ a lOa + 25. 9x 12x + 4. +
x2
13.
16.
x2
19.
2
x
x
25. x 28. x
2
40.
43. 46. 49.
2 .
+ + + 5x - 6. + x - 2.
2.
llx - 12. + 4x - 5. + 12x - 13.
-
2x
2
+
2
-
5x
+
2.
- 2. 5x - 6. 6x2 8x 14x - 15. 2 9x 3x - 2. 2 24x + 7x - 6. 36x2 + Qx - 10.
34. 5x 2
37.
-24rs+4s 3x
2
2
2
8.
3x
2
x2
17.
2
x
2
x
23. x
29.
32. 35.
38.
2
-
+
1.
+
x
9.
4m2
2x
+
6x
12.
x
6.
15.
x2
3x
+ 2.
18.
x
2
21. x
2
24. x
2
2.
llx
-
12.
30.
2
2
36.
+
13x
+ 5.
+15x 27x -6x 2
2
+ 3x
-
6. 8. 2.
-
llx
4x
3. 7.
7.
+ 30.
-
5.
2
2
39. lOx 2 42. 15x 2 2
45. 14x 48.
6x
-
+ 12x + 32. x + lOx - 11. 3x + 5x - 2. 6x + 5x - 6.
27. x
2
33.
1.
-
6.
-
+ 40x + 25. 4x + 4. - 4m +
2
-
x
-
6.
2
50. 27x
2
+
2
2
16x2
5x
x - lOx + 21. x - 12x - 13. 2x - 3x - 2. 4x - x - 3. lOx2 - llx + 3.
44. 9x
3.
5x
2
41. 6x 2
47.
6fc
9.
25x -40x+16.
14.
26.
2
12x -f
2
11. x
20.
+ +
4x2
5. 9A;
2
x2
31.
2.
1.
2
10. 36r
22.
2x
2
20x
-
-
3.
19x
+
x
6.
+ llx - 15. + 16x + 3.
2.4]
FACTORING BY GROUPING TERMS
2.4. Factoring
27
by grouping terms
In general, factoring is more difficult than multiplication because there is no routine method which can always be applied. Yet there is one device which often proves useful that of grouping terms. dx = x(c Since ex
+
it
tiplication,
d),
follows that,
+
c(a
(1)
+
6)
+
if
+
d(a
by the distributive law of mulwe replace x by (a + 6), 6)
=
(a
+
6)(c
+
d).
we can arrange the terms of an expression in connected groups by plus and minus signs in such a way that a certain quantity appears as a factor of each group, then that quantity is a factor of the whole expression. Certain frequently made errors will be avoided if the student will study carefully the examples below. That
is,
if
Illustrative 1.
2ax
= = = 2.
12ac
= = = 3.
3ax
= =
Examples
+ 4bx + Qby (2ax + Say) + (46z + a(2x + Sy) + 2b(2x + (2x + Sy)(a + 26). - 86c - Sad + 2bd
+
Say
- (Sad - 2bd) - d(3a - 26) 26) 4c(3a - d). (3a 26) (4c (12ac
+
x Sax (Sax
-
667;)
Sy)
86c)
+
+
z 3az y 3az Say 3ofe) Say
+
+ +
x
+
no common factor appears
3ay y (x
-
+ y
z
+
z)
after various experimental be possible to arrange them may groupings of the terms, so that the expression to be factored is seen to be the difference of two squares. If
it
TYPE PRODUCTS AND FACTORING
28
Illustrative 4.
4x
2
[Ch. II
Examples
- 1 - Qy + 1) - (3y - I) (2*) - l)][(2x) + (3y (3y [(2x) - 1). (2x -3y + l)(2x + 3y
-
2
+
6y 9y 4z 2 - (9y 2
= = = =
2
+ 4?/ + x + 4xy + 4i/ -
= =
2
2
(z (
=
(x
1
)
2
x 4.
I
2i/)
+
2y
1)]
2
2
1
5. 4xi/
2
-
2
l)(a;
+
+
2y
1).
worth noting that a definite procedure is indicated for factoring, or attempting to factor, a polynomial of four terms. The 'two-two' grouping will aid in exposing any binomial factor. If this fails and there is at least one perfect square term in the expression, the 'three-one' grouping It is
be tried. binomial or trinomial containing two perfect terms can sometimes be changed to a perfect square addition of a term. If the term thus added is a perfect itself, the original expression may be factored, as
may
A
square by the square in the
examples below. Illustrative 6.
7.
x
x
4
4
+
4
+
x
EXERCISE
Examples
= = 2
+
4
(x 2
(z 1
+ + = =
4z 2
-
2 4
(x 2
(x
+
+ +
4)
-
2x)(x
4z 2
+
2
=
+
2
(x
+
-
2
2)
2
+
2
(2x)
2x).
- x = (x + - x)(x +l+x).
2x 1
2
2
2
-
2
1)
I)
2
7
Factor the following expressions. 1.
3. 5.
+ b(x - y). x(a + b) + y(a + 6). h(r + s) + k(r + a(x
y)
s).
2.
+ y} -
b(x
4.
m(k
6.
ax
-
1)
-
c(x
+
n(k
y).
-
1).
+ ay + bx + by.
x2
DEFINITIONS
2.5]
7.
9.
11. 13. 15.
17. 19.
21. 23. 25.
27. 29. 31. 33. 35. 37. 39.
41.
bx
2x
by
+
3
+ ex
3z
2
-
29 8.
cy.
2x
-
10.
3.
- y + x - y. 4z - 9^ + 2x - 3y. 16z - 25z/ + 8az - Way. z + y + 2z?/ - 9. x + y - 2xy - 25. - x + 2y + 1. y z - y - 6x + 9. x - y - c2 - 4x + 2cy -f 4. x4 - 3x2 + 1. x2
2
2
2
2
2
12. 14. 16.
2
2
18.
2
2
20.
2
2
22.
2
2
24.
2
2
+ 9. x 7x + 9. 4 x + 4x + 16. ax + by + bx + ay. 4x - y + y - 4. 25x + x + 9x - 10x + 4
x
+
4
2
4
4
28. 30.
2
32.
2
34. 36.
2
1
38.
40.
1.
2
42.
1-
ax2
2x2
+
-
x2
y
-
+x-
ax 2
2.
bx
-
x
b.
+ y.
-
3x - 4y. 1 - y 4z + 2y 16 + 2xy - x - y 16 - 8x + x - y 4z2 - y - 4x + 1. x +o -6 -2ax+26-l. 2
9x
2
16?/
2
2
.
2
2
2
.
2
2
2
2
+x + x + 3z + 4. x - 5z + 4. x + 2x + 9. 4z + 3z + 1. x + y - x + y 4x + 1. 9X + 2x + 1. 9x - 40z + 16. 4
26. x 9
2
5x
-
x3
1.
4
2
4
2
4
2
4
2
2
2
.
4
4
2
4
2
2.5. Definitions
The degree of an integral rational term in any letters is the sum of the exponents of the letters in that term. Its degree in a given letter is the exponent of that letter. Thus 3x 2y 3
is
of degree 5 in
x and
y, of
degree 2 in
x,
and
of degree
3 iny. its term 3 5 1 is a sixth x of highest degree. For example, y + xy degree polynomial of degree 3 in x and of degree 5 in y.
The degree
of a polynomial
is
the same as that of
polynomial A is a multiple of a polynomial B if B is a factor of A. To illustrate, 5, 10, 15x and 25(z t/) are mul2 1. 1 are multiples of x 1 and x tiples of 5, while x is a called more of two or A multiple of each polynomials common multiple. For instance, 25zV is a common multiple
A
+
of 5,
x and y z
.
TYPE PRODUCTS AND FACTORING
30
2.6.
[Ch. II
Lowest common multiple
The
common multiple (L.C.M.} of two or more polythe common multiple of lowest degree and with the
lowest
nomials
is
We
smallest possible numerical coefficients. find it by multiplying the prime factors of all the polynomials, assigning to each the largest exponent which it has in any one polynomial.
Factors such as x x, which y and y should be considered as the same.
Example 1. The L.C.M. and 2 2 3, is 2 2 3 2 or 36.
of 6, 9,
and
differ
only in sign,
12, or of 2- 3,
3
3,
+
1)
-
,
Example is
36x
2
(y
The L.C.M.
2.
+
+
of 6x, 9(y
3
I)
,
and Ylxy
3
I)
.
2 x 2 is y 2 x 2 or Example 3. The L.C.M. of x y and y 2 2 2 x ). (Note that there are two x y and not (x y)(y correct forms of the L.C.M. one being the negative of the
2
,
other.) 2.7.
common factor
Highest
The highest common factor (H.C.F.) of two nomials
more polycommon factor of highest degree and with the
is the
numerical
largest possible
ing the
common
exponent which
it
Example 1. The H.C.F. 3 2 is 2 3 or 2 3, and 2 ,
2. 3
18x*(y
.r)
EXERCISE Find 1.
the
of 30, 12,
of 30x 3 (x 2 y), or Qx (y
2
Qx (x
and
18, or of
2
3
5,
6.
The H.C.F.
is
it
has in any one polynomial.
2
Example
find
by multiplyassigning to each the smallest
coefficients.
factors,
We
or
-
2 yY, I2x (x
-
y),
and
x}.
8
H.C.F. and L.C.M. of
9, 12, 15.
2.
the
12, 18, 24.
polynomials in each group. 3.
18, 27, 36.
4. 14, 21, 36.
5. 20, 35, 50.
6. 30, 45, 75.
7. 80, 32, 96.
8. 54, 36, 90.
9. 65, 91, 104.
REVIEW EXERCISES
31
10.
3 x Sx ?/, 4zy.
11.
12.
12xV, 15xV, 9xy.
13.
+
2
14.
(x
(x
x?/),
-
+ y),
-
2
(x
5a 3x2 15a2x 3 10a4x4 ,
(x
.
,
-
y), (x
-
2
2
y*), (x
2 ?/ ).
- y (x - x (x 2xy + (x - y (x + x (x + 2xy + y (x - 3x + 2), (x - 4), (x - 5x + 6). (x - 3x + 1), (2x + 5x - 3), (4x - 1). (2x - 2), (3x - 4x + 1), (3x + 2x - 1). (3x + 5x - 5x - 3), (2x - x - 1), (2x + 3x + 1). (2x - 1), (3x - 4x + 1), (6x - 5x + 1). (6x + x - x - 4), (15x + 2x - 8), (10x - 7x - 12). (5x 2
15.
2
2
?/ ),
2
18. 19.
20.
2
?/).
2
2
2
2
2
2
2
2
2
2
2
2
2
2
22.
3
),
2
2
21.
2
?/).
2
),
2
17.
3
),
2
2
16.
2
-
2
- 2yY, (2y x) (x - 2x - yY, (y - x) (xy - y )
23. (x
3
3
2
7/).
,
3
24. (x
2
3 2
,
.
REVIEW EXERCISES Factor by using the type form ab 1.
3.
4y
-
3
2
127/
m?n
-
- 2x 5. 6x 2x - 2o(3a - 6). 7. c(3a 8. -x (2m - n) (2m 3
3
3
i/.
2/
=
ac
+
a(b
c).
4.
+ 6b 3a 6 + 5a6
6.
y(x
-
3(2m
-
3a&
2.
.
mri*.
5
+
2
3
.
3
3y)
2
a6.
+
-
2(x
fc)
2
2
n)
t/
+
n).
Factor by grouping.
+ my +
9. rax
-
x
3
13.
x
3
15.
6m
11.
3
x
-
2
2
4x y
+
x
ex
+
-
12m?n
+ cy.
10.
2xy*
+
3
8j/
3
2o 3
2o
4.
16.
18.
2x2
14.
.
+ 4mn + 8n 2
+
6ab - 2a + 36 - 1. - 9p + 4? - 6. Qpq x - 9y + x - 3y.
12.
1.
a4
2
2
Factor the following trinomials. 17. y 2 19.
21. 23.
-
y
-
12.
+ lOx + 21. x + xy - 12y*. 12a -o-l.
x
2 2
2
+
20. ra n
22. 24.
+ 2.
5x
- 5mn - 15. - 36. p + 16p 25r + lOrs - 3s 2
2
4
2
2
2
.
-
xy).
TYPE PRODUCTS AND FACTORING
32 25. 27. 29. 31.
-
a 5ab - 246 20 + 8x - x 4c - 4cd + d 4 - 20x + 25x 2
2
26. 12x
.
2
2
2
2
Factor by using the type form a2 33. 35. 37.
39. 40.
4z - 121y 16a2 - 256 x2 - (2y + z) 1 - (x + y) 4x - 4xy + y 2
34.
.
36.
2
38.
.
+
6) (a
(a
b).
81 - x4 (three factors). - y) - 9. (3x - 9(w - n) (x + 2j/) 2 2
2
.
.
2
2
-
41. 9
42. 4x
p
2
4xy
a
+
y
2
z2
+
806
-
166
4ax
+
2by
+
-
2
2
-
2
+
z2.
2
q
a
+
=
2
(2x
z2 .
y)
2 .
-
a2
b2
-
y
2 .
2pq.
+
Factor by using the type forms a 3 b3 ab b 2 ). and a 3 - b 3 = (a - 6) (a2
+
44. x 47.
1.
2
2
HINT. 4x2
43.
.
2
25. .
b2
2
[Ch. II
8
4
32.
.
71x
8
2
30.
.
+
-
+ 4xV + 4y 9a + 6a + 9x - Qx yz + j/V.
28. x
.
2
3
x
+ 8. +a
45.
6
50. x
6
48.
.
-
y
=
(a
+
2
06
6) (a
+
6 2 );
+
m - 27. - l) (2x 3
s
46.
49.
8.
ay + 64. 27y + (a + 2h) 3
3 .
6 .
HINT. First treat as difference of squares. Factor as a difference of two squares by adding and subtracting a perfect square.
51.
x4
54.
55. x4
+
52. x4
64.
-
+ 16. - 12xV + 16tA 4
x
24x
2
+ 4y.
HINT, x
4
53. x 4
-
24x
2
56. a 4
=
+
16
-
10a262
+xy +
(x
2
2
+
-
2
2
4)
1664
-
y'.
I6x2
.
.
Factor completely, using the type forms given in this chapter. Refirst any monomial divisor which exists in the given expression
move
it
before factoring
59.
4?V ~ x - a
61.
a 3 -f a2
57.
4
further.
25r
-
4
60.
.
-
4a
~ 8 ^ + W2o - Sy*. 4 y 2aV + a4
58 - ^ 6
2
-
4.
62.
6
.
REVIEW EXERCISES
33
63. m(n - 1) - 2(n - 1). 65. 4z - 1 - 4y - 4y*. 67. 3z a - 2az3/ - aj/ 69. (x + y) - 6 (x + y). 71. pV - 1. 73. x - (a + 2) 75. (a + b)(x + y) - (a + 2
2
64. z
2
66. (2s
2
2
72.
74.
.
76.
Why is it
10 3 that
+
1fl
i
b)(3x
4)
2
Sr mx -
3 .
-
+
(3y
16rm
-
2
2)
48r
3
+ 6) + 5(o + 6) 4* + 28ca: + 49C x (2x + 1) - (4x 2
70. 2(a
3
-
z
68.
.
2
3
+ 27m
3
-
.
m
3 .
-
12.
2
2
.
2
2
1).
y).
73 o can be simplified
by the
illegitimate
operation of canceling the exponents? 77.
Show
that
to problem 76.
-,
,L
,
= 3
^,
,
and apply
this result to
Chapter Three
FRACTIONS
3.1. Definitions
An
an indicated quotient of two algebraic expressions. The dividend and divisor are called respectively the numerator and denominator. algebraic fraction
x v j Examples.-'
A
is
o+l
2
-
3x
;
*
* a
;
l
+ y2
--
2 -:
3
simple if neither its numerator nor its denominator contains a fraction; otherwise it is complex. For fraction
is
2 2x example, and j
2
+- 1 are I[
2tX
simple, while
1
reciprocal of
the only
3.2.
2
Thus the
x for
is
-
J
number which has no
1,
each
reciprocal of
,
+
2 and o
a;
is is
reciprocal.
-;
a
j-
o
which equals
called the
and the
---
Zero
.
34
is
Why?
to its lowest
terms
The value of a fraction is not changed when numerator and denominator are multiplied or divided by same quantity (except zero) 1.
are
^
x is
Reduction of a simple fraction
LAW
j
complex. If the product of two numbers reciprocal of the other.
+
the the
REDUCTION OF A SIMPLE FRACTION
3.2]
v
i
Jbxampies.
By
?
1 -; .
2
23
6
8
__
12
virtue of this law
ax-
.
a
__
;
xy
35
.
a
;
7-
__
o
y
we may
2
reduce
to
xa ~~^r* x 2b lowest terms a
by factoring the numerator and denominator and striking out the common factors. This operation, called canceling, is a short way of dividing both numerator and simple fraction
denominator by a
F 1 example.
ax*
-
common
a
factor.
ft&^~r}(x
+
+ +
A fraction is not reduced factors of the
x
1)
x
1)
to lowest terms until all
common
numerator and the denominator have been
canceled; or in other words, until the numerator and denominator have been divided by their highest common factor.
When
all
j.i
of the
celed, the factor 1
numerator or denominator has been canalways remains.
Example,
1
1
1
jf $
--
=
T
6;
1
1
^-j-l1
CAUTION. Errors due to faulty cancellation are very numerous, and the student who is not sure of his knowledge or ability in algebra should be especially careful when he is tempted to scratch out letters or expressions just because they are above or below a line. The simple rule to be applied is that anything canceled must be a factor of the whole numerator and also a factor of the whole denominator. Hence the student should make sure that this factoring is done correctly before he does
any canceling
at
all.
For example,
if
the
numerator is xy + a, then neither x nor y nor a is a factor of the whole numerator, although the expression as a whole is
a factor of
itself,
as
shown
in
some
of the examples.
FRACTIONS
36
Examples in which canceling
xy
+
a
=
1
=
[Ch. Ill
is permissible.
+4=
2x
.
Examples in which canceling
+ 2)
%(x
is incorrect.
x
(Decide
+2
why
in
each case.)
xy
+ x
a. '
o a
+ +
l.
2(s
+
y]
2' I
-3(x
+
-
In the fraction
+
a
x
yY
+ 6)(c + (a + b)c +
+ y + g. + y + z'
2x
.
(a
the canceling of a
L
is
d)
'
d
of course not
would violate the rule about factors. But why is it not permissible? Note that when we cancel a factor we are actually dividing the numerator and denominator by that factor, whereas if we incorrectly strike
permissible since
it
out a from the fraction
a j_
i
+
2i
a
-
we
are subtracting a from the
numerator and denominator. This changes the value of a 2 -
fraction, as illustrated in the case
+
x
It is permissible to cancel
o
^
-
2 o
1 / 1
2 -
(or o
^
1 5)'
//
2 from the numerator and 1
1
*^
H 5( denominator of 3y(f TjL7 (x
obtaining
Z)
since ^ , x
1
amounts to dividing both members of the fraction by the canceled factor. However, it is a safer procedure to write this
this
fraction as
(x
It will help to full
^
* ,
+
/
2) (3
~~
- ^ 2)
avoid errors
before carrying out cancellation. if
both the
full
numerator and the
denominator of a fraction are completely factored before
cancellation.
3.2)
REDUCTION OF A SIMPLE FRACTION
EXERCISE 1.
9
How many
and how many prime factors, has each of
terms,
+
the following expressions: (a) xyz; (b) x2 2.
37
In which of the two following cases
missible:
The numerator and denominator
common
term, (b) a
common
+
x
is
1; (c)
z2
is
ax
have
-
11 '
13
-
y)(x
15.
3(a?
+ y)
a
In
these
y
(x
'
-
l)(x
-
3(x .
(x >
x
-
+ y) + x-y(x y)' - y (a + b)x - y)' (a + b)(x (x
'
+
+ l)o +1 y)(x
Explain the errors which have been made problems 14-17.
14
(a)
x-y
1)
+ y) + 1 2(x + y) (x l)x + 1 x+l (x + l)a (x + 1) + 6' a + b + c 3(a; + y)(a + b + c) - y)(x + y) (x ax + ay (x
+
x
+
?
canceling always per-
of a fraction
canceling permissible? cases reduce the fractions to lowest terms.
x(b
2
factor?
In which of problems 3-13
*
y
in cancellations for
FRACTIONS
38
+
(a
1
6)
Reduce
terms.
to lowest
^
)
!,f.
30'
36x 54x
-
-
x2 29.
x
2
x2
40.
-
x2
2x
-
15
x2
-
+
3x 3x
+
4 30.
x
2
__L
PL/v
tJJU
x2 2
6x2
+
l'
2x 2
-
-
x
72
-
5x
-
A/
9
^
'
x2
2 A J
x2
- x - 6' - x - 2 -x - 1
_
3
2
'
x2 37.
15x 2
3
-
+
Zx*
-
x
41.
-
x 46.
5x 48.
+
4?/<
47.
+
y)(s
y)
y
+ 8b + a 3
a (3x
'
(3x
2
2
-
46
2y)(3x
2
1
4
2
3x
-f-
(g
-
2
+ x
x3 a3
49.
45.
2
(x
8
-
2
r
-
-
2x
x
'
10
H
5s
rf)
-
+
+
- x - 9x + 9 x - 5x + 2x -
x3
3
3 43.
2
-
+ 7z + 12 + x - 12 +x-6 + 2x - 3'
10x2
15
4
Q^
oj
2x 2
12
43x - 14 -21 + 5z + 6x2
5(c
5)'
Q-i/f/r
-
x3
5)
oyJu
42.
44.
3(x
-
'
4'
-
7(x
y)
+
12x 2
8x
25.
)
+
2x 2 2x 2
3)
2
3x 5x
32.
-
-3)'
5(x
35.
x(y
-y
1
22.
99
26.
-
x3
2L
2
23.
[Ch. Ill
5y(x
-
4b2
2
+ xy
-
2y
y) _
3
2
)
2rs 2
j/) 2
+
+
+s
2
ll(x
+ y) + 24 y
2
-
9
3.3]
3.3.
RULES ABOUT FRACTIONS
39
Rules about fractions
In this and the next two such as - or
fractions,
x
articles
~-
2
-,
we
shall deal
with simple
whose numerators and de-
nominators are polynomials.
RULE 1. The sign before a fraction is changed when the sign of either the numerator or the denominator is changed. For by the rule
of signs [(4), Art. 1.5],
^ = _5 6
~=
~,
and
6'
RULE
2. The sign before a fraction is changed when the sign a of factor of either the numerator or the denominator is changed.
For again by the rule of signs, a( 6) = a&, and hence 6 the sign of the whole when the factor 6 is changed to numerator or denominator in which it appears will be changed, so that Rule 1 will apply.
111
1
Example
1.
Example
2.
(-2)(-3) a ,
,
t
2-3
2(-3)
=x(
x)
1
+
,
or
6
-
xY
x(x
1)
Reminder. It is important to note here that Rule 2, as does Rule 3 below, refers to factors and not terms. For instance, in is
Example 2 above, when the
changed
it
becomes
1
+
x or x
sign of the factor 1 x. 1, and not 1
x
+
Definition. Integers divisible by 2 are even; all other in4, etc., 2, tegers are odd. Even numbers include 2, 4, 6,
as well as 0;
RULE
odd numbers include
1, 3, 5,
1,
3, etc.
The sign before a fraction is or is not changed according as the number of factors whose signs are changed is odd or even. The altered factors may be all in the numerator or in the denominator, or they may be in both. 3.
FRACTIONS
40
[Ch. Ill
Example. (1
(x
-
(2
a?)
l)(x
- sXs - 3) = - 2) (3 -x)
Here the minus sign
is
second step because the sign is 3, which is odd.
RULE nators, the
4.
add
(x(x-
l)(x l)(x
-
2)(x 2)(x
-
_
3) 3)
inserted before the fraction in the
number
of factors there
changed
in
To add
or subtract fractions with the same denomior subtract the numerators and place the result over
common denominator.
(by Rule
EXERCISE
1)
=
10 1-9, write each factor with a positive literal term, 3 should be left sign where necessary. [For instance, x
In problems changing
its
x should be changed to (x unchanged, but 3 and reduce each where fraction permissible factors .
L
x-
Z>
'
3
'
8>
-
-x (1 2) (x l)(x x)' (1 x)(2 x )(x - 2) (3x 1)(4 - x)(x - 4)(3x - 1)' (2 - 1)10 (x - x)10' (1 1
4
(x
3
2
I)
x)
(1
5-
Then
cancel
to lowest terms.
(x
*'
'
2
3).]
-
3
I)
x)
'
3
- 1)(2 - x)(x - x)(x - 2)(3 (1 - 5) (x - 6) (x - x) (6 - x) (5 - 3) (4 - x) (2x - 4) (3 2x) (x (x
2
3
2
3
'
6
6
6
6
'
3) x)'
ADDITION AND SUBTRACTION OF FRACTIONS
3.4]
the operations indicated below.
Perform 3x *
<
55 -
x
1
,
+
~r r
e
3x ., 12.
-+ 5
2
2x
+
3
-2
2s
1f '
e
7
+
2
3x
+3
ie . is.
-
3 2x + 4 + + x + l' 4x + 5 _ 3a: - 2 'x-9 9-x' lOx + 3 3 + 10x
x x
.
-
3.4.
6
''
6
-
5x
6x
5
5x
3s
+
7 1
2x 2z
5
2:r
-
3a: 1
+x +
2
-
6
+
6
-
l'
3 '
x
15z-7 3x2 - 5
'
2
+ g
-
2x
1
3x
+
13.
'
^
I
-
5x
,
.
2a + 3 --
,
5
g
7x
41
7
5
+ -
1
3x2
Addition and subtraction of simple fractions
By use of Law 1, Art. 3.2. two or more simple may be written with a common denominator.
fractions
Write the numbers 4 (or -f), f, %, and ^-f as fractions with a lowest common denominator (L.C.D.).
Example
1.
Solution.
The L.C.D.
r .u j * of the denominators
i
1,
of the fractions
is
4 o * j ir AT 3, 6, and 15. Now,
the L.C.M. (30) 120 4-30 -
? 3
=
2j^lO 3 10
Example
=
20. 30'
5 6
= 5-5 = 6-5
25. 30'
Write the fractions
2.
an
-^
= 16-2 =
15
15
-
^. 3x(x
oO
1
16
,
=
=
1
and ^- ---2)
oO 32
2
6(2
;
30'
-
x)
with a L.C.D. Solution.
-
r x)
-
2)
Qx(x
-
_.
,
G(X
L.C.D. of the fractions 2a a 3x(x
5
=-
2)
is ,
and
by Rule
then 6x(x 5 6(x
1,
Art. 3.3.
The
Z)
-
2)
2).
We now 5x
Qx(x
-
2)
write
FRACTIONS
42
In view of Rule
RULE
mon
[Ch. Ill
we have
4, Art. 3.3,
To add
or subtract fractions, write them with a comdenominator (their L.C.D.), add or subtract the numerators 1.
as indicated, and put the result over the L.C.D.
Example
1.
12
6
66 63
6
+ 3-5
12
2.
Example x
+2
+
2
5
10
6
6
+2_x +2 x -!' '! +1 = (x + 2)(x 1) _ x- + 2 + l)(x - 1) x - 1
2 a:
z+ll-z
= 2
2
x z
a:
1
2
x(x
2
z2
(a;
x2
~ -
x2
From Rule
we
1
-
;
=
4 1
get the simple results: r
But note that these
r^ bd
d
o
- 1) - 1
1
o
and r
2
+
=
d
f^ bd
results are not
formulas when b and d have some factors in common. For example, if we use the method indicated by these efficient as
results
we have
1,1 ^ + T~ '
X
1
X
2
:
(x 1
2
-
(X __ '
1)
+
x(x
(x+
2
1)(X
+ 1
-
-
1)
'
1)
x2
_
(X+
1)
^
1) 2
l)(x
whereas the method of Rule once.
+ (x+
+
x 2
1)(X
-
1)
x
x 2 '-
1'
gives the simplified answer at
3.4]
ADDITION AND SUBTRACTION OF FRACTIONS Find the blunder
Exercise.
43
in the following incorrect addi-
is so common that it should be thoroughly understood and then carefully avoided. A second common blunder is the omission of the denominator. The student should cultivate the habit of writing the denominator first
typical error
to avoid forgetting is
long.
EXERCISE
11
In each problem fraction and reduce
L 4
it to
lowest terms.
1,1 + 3'
2
+
5
'
oil+ 3
k x
7
3
+
+
1
8
3x
- + 2x
j
-
3x
3
+
x
+3 +
4
-
3 _L_i
4 '
5
4
17.
-
6
-
'
~
5x
5
6
Y
5
,
4a;
5x 2x
2 -
'
5*
~
'
5*
+
3*
'
3*
~
'
16.
below, combine the given quantities into a single
2
1
7
when the computation of the numerator
it
4x 5
9
-
+3
3x
1
5 4
_
2x 2x
+ -
2 7
+3
' 3
4*
+
4
9
7^
3 -
18.
2
-
3^ 2
5
3a:
+
2
2x
-
3
2x
+
3
3x
-
2
-
4
5 3x x - 3
9x 4x
+2
+
3'
x-2
3x-5 '
+ 6 ' 3x -
+
5'
'
'
8'
2x
+
4x
3.5. Multiplication
RULE
1.
and
division of simple fractions
TTie product of two or
product of the
more simple fractions
nominators.
2-5-1 3-3 f.
is the
numerators divided by the product of the de-
(a)f^YA =
f?WW = 2f. ceg
10
9
3.5]
MULTIPLICATION AND DIVISION OF FRACTIONS
45
RULE 2. To divide one fraction by another, multiply the one by the reciprocal of the second.
3d.
91L
/*}/7 O / /
*
/
example
^^'^^^iFjljj^
Example
2.
*/
X
9
iC
JL
~7~
/
rfl */
/y.
~~'
f
As
cases.
l/
X
may
and division
=
=
A
4.
2 thirds
a
= = dec
^ By
d
EXERCISE Perform
s
' ;
,
rule 2,
8 2 --=--
'
/-V^ VbAc/
6c
d
the indicated multiplications.
>
-
8 35
JL)
all
33 =
12
24
JL J
arithmetic
where a can be anything
4,
a 6_
of fractions are postu-
'
6
c
and subtraction, the
be observed to be true in
8 thirds
v
!.
For example, since
except zero,
1 ^ ( /
X X
/
in the cases of addition
which
vfr V
1
if
^+
rules for multiplication lates
*
X/'r 2
X
1
= A^ote.
r
**/
first
'
^X V
15
/8V3 =
(-)(-) 3/2/
A
4.
FRACTIONS
46 i3 -
15.
(!-^T>
-
3*
(trf^Vex +
[Ch. Ill
2 >16.
4).
20.
(^F^f)(3x+l)(-
22.
24.
/ 3x
_w
27.
I
6x
on 29
'
2
2
-
+a;-2 V2-3o: /1c , 15 '
A X'K 13a:+6/ 5 2 /o 2 + ,
fa ~ 127T3A
x /)(
)-
,
-
3
/5x
34
-3/
+ 7/2x-3 -
(x
+ 3)
2x
EXERCISE Perform
12a
the indicated divisions.
^~ 4-19 -riH* +
x'
3x
-
o;
'
4
'
'
4,/r
3
^ 5.
6x }JL/
+ ~T~
15 1<-I
^T-T-F ^,
.
-5-
/rk
+3
9r
4-
'
-iM^
' ^ x * 6. 2
4*
+ -j-^ +4
^x f-
5).
v
.
'
x'
-
(a:
/o (2
-
2)
x).
a;
O
* ^ TFX 'T'
+ , i
(2*
4
3 >-
5x
O/v.2 _L_
(x
+
x)
-
8
-
2x2 ^SX
-
Q/v
x a/
O
+2 o -3-^
)-
MULTIPLICATION AND DIVISION OF FRACTIONS
3.5J
-
-
Ix 7x
3x 3x 2
+ -
2s 2x
14. (7x
+
2)
16. (6x
+
5)
2
12
'
v
12
1V'
*'
-
1
47
' .
1
11.
13. (5x /
,15.
-
1-^4Jlx 5
-T-
3)
17. (3
-
5
- 5) ^^ .
(3x
^-_ 6
+
-
(x
23. x
-
-T-
27. (6x 2
2Q
2)
-
1)
2
+2x-l)
-7x-3
+
5x
9x
-T-
6x
2
'
-
^-+ x
zz. '
2
+ 3) _._
7
-J-
2ar
-
5
3y
+
7
2-3x 14 + lOx 3x + 2x 1 + 2x - 3x 3x 3x +
+
-
14
+
'
2x
2
.
'
3x
2
3x2
2x + xy - y2 + xy - y* 2 2x - xy - y2 2y Qxy + 4x 2 6x - 13x?/ + 6?/2 3y 5xy + 2x 3x 6 6 + xy 2y 2y + xy - 3x' 2 2 2x 5xy + 2?y ^ 2y - 3xy - 2x2 - 2x2 2x - xy - y 2?/ + 3xy 2 %y ~ % - 2y + 2 ^ xy + x 2y - 2x + y - 2 xy xy + x + y+l' 2x2 2
2
.
'
'
-
5)
_ '
2
2
-
*
lOx
2
.
2
2
. *
'
'
2
'
5
x2 -3xy-4y 2
2
36
T
+ 2)
2
'
2x2 -x-3
12
28. (5x 2
-
2
32 '
35
g^jx+l
12
26. (3x
-
'
- 9x2 x - 1
I
'
.
fl
-
-f-
-( 3x ~ 2)
33 '
^
-
'
'
-x-5 _^_
6x2
+
-
3x
O
/v.
O/y.2
6x
2
1^ 5
2
(3x .
-T-
12)
-
x
)
iA *'
-
'
2
-
25. (2x
Z
7x
18
.
t
^~
^x
20. (12x 2
6
5
f+ f-- 2 - llx
2x)4- R 2 6x
10 T2r2 4^r iy. -f- ^x
'
3x
-i-
2
+
-
-2y-a?
'
FRACTIONS
48
3.6. Simplification of
[Ch. Ill
complex fractions
We
have seen that a fraction is complex if either the numerator or the denominator contains a fraction. In this case we shall speak of the minor denominators. For example,
in the
are 2
To
complex fraction
and
l+ 26 +3 ,
the minor denominators
62
6.
simplify a complex fraction we reduce
it
to a simple
fraction in its lowest terms.
When
a complex fraction
itself
A
nor the denominator B of contains a complex fraction, there
neither the numerator
are two methods of simplifying
A L>
Method 1 Multiply both the numerator and the denominator by the L.C.M. of the minor denominators. (See Law 1, Art. 3.2.) .
Example 1
1.
!
+
2
+
2
_
3
1
! d+ +2_ +2 3 I 1
12
-
12
+ 6 + 8-1 = 10-8 + 12
12
x
2
Example
2.
2
9 2 X* x)
ax2
-
25 14'
2bx
Method 2. Reduce the numerator and likewise the denominator to a simple fraction, and then apply Rule 2, Art. 3.5.
Example o
.
'*'
1 2
1
.
_
_! 3
1 3
12,3_2 =
66
9_1
33
6
13
=
_6_
8 3
= /13V3 =
U/W
13 16'
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS
Example 1
49
2.
+ ,
x- 1 1 + 1 -
n 2
1
2(x
,
x- 1 1 -x
'
x
- 1) - 1
2
2x- 1 x - 1 2
1
-x
2
-x 1 - x 1 - x - 1 = /2x 1V1 a? = /2x IVx x-l]2-x } x-ljx -2/ '
2
1
2
2
2 2
2
_
(2x
-
a;
2
l)(x 2
-
+
1)
In general, Method 1 is shorter and more efficient when the minor denominators are single terms; otherwise Method 2 is
preferable. When either the
plex fraction
numerator or the denominator of a com-
is itself
must be
simplified fraction and so on.
a complex fraction, the latter fraction
first.
The same
rule applies to the latter
1
Example. 1
+ X
Here we simplify successively the fractions within parentheses, brackets, and braces, thus :
EXERCISE
13
Simplify the following fractions. 1
2.
y
y
FRACTIONS
50
1-1X
+ +
4.
3.
1
1
>+!'
-
1
5.
x-y
+y
x
+
y
x
+
x 1
6.
x
+y+
y
x
7.
x
1
a
_
l
bx
5
o 9>
xy
M
g
5
I
r
10.
y
y
11.
- 13xy 3x + 2y - 6 xy - 5xy + 2x 3y - 3x + xy 6 2y 2
IS.
2
2
2x2 13.
f
ay
x
_
b
y_
8. _
y
y
+
-
2y 2 2y 2x2
+
xy xy xy
y
-
y
17.
5(x
-
2
2y 2x 2
+
14.
2
+ -
3
+
2
x3 '
16.
(x
x2 (x
2xy
y
y
3 '
2
18. 1
3
- y) -y - y) x
y
f-^)'
2
y
y
y)
-'M
2
2xy
-
2
+
x-y x
19.
2
2y 2 2y 2x 2 y
1
+
-
-
'
3x 2 3x 2
2 - 3x2 2y xy 2 2y 3xy + x* - 6y2 6x2 5xy 2 6z Uxy - 2y 2
-3(x
5xy 3xy 3xy xy
2
2x '
2
'
x
2x 2
20.
3z3u x
2-y a;
x
x
2
2
'
[Ch. Ill
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS 2x
21.
y
X
y - xf
(
(x
-
--4 +x z
v
?/)
22.
-
+ x-5
+ 6x -x+1
X 24.
23.
2
25.
2
-
27. 1
-
29.
3x
+
2
3x
-
1
26. 2
1
+
x 28.
30.
2x
2
-
3x
+x
1
~~
<>
r
REVIEW EXERCISES Reduce the fractions in 1-6
4a2 1.
3.
(x (x
+
6o&
-
to lowest terms.
462 2.
+ 2y) + z + 2?/) - z 3
3
2
2
4.
6.
5.
27x 3
3mn 3mn (2x
+ + -
-
%
6fen
+
Qkn
y)
3
3
-
me me
5(2x
+
2fcc
2kc
-
y)
2 j/
Cam/ out the indicated operations and express the answer as a simple fraction in lowest terms.
-
3x
5x
1
7.
8.
+
3y 3x
9.
a
a
2y
% -
'
4
3
.
2xy
-
1
8 4 3
3y
y
x
.
T'
-
1
_
2x + 4x 2
1
1
2x'
to
each
FRACTIONS
-
o3 3x 11.
x
12.
2
3x-5 13.
x 3x
1
14.
15.
-
4x
4
-3 -
4z x - 2
3a-6 + 96 c
20.
a
2
~T~
-
(c
6'
-
a)(6
-
-
'*'
o)
(6
-
c)(o
c)'
- L
+2 ^ +3
x
,
-
2(x
+
1)
ISz
18.
26.
(x
-
36 ab
1
%y
i^$/
+
3 *^
- xy - 6y x2 + llxy + 28y x - x2 + 4x - 4 2x + x - 1 2x2 - x - 262 1 a(a + 26) 06 ^ a + 46 o 46 a 2a6 + 2a x2
2
2
2
3
21.
-
2
16. 1
+ 3a&
a262 19.
a
7/)
2.
-362
ab 17.
1
c)
2x
7^3-x+
+
2
^
-
6)(6
+ ab + 6
-
(x
i/
-
(a
a2
2x
^ .
-
2
'*'
63
'
'
4
4
2
2
:
2
262
*
3x 1
-
x
+
27.
2
1
o
-
1
28. 1
-x
1
-
a
1
+a 2
1
[Ch. Ill
3.6]
SIMPLIFICATION OF COMPLEX FRACTIONS x
y
29.
m m
30.
m m
n p
n
1
1 I
y
1
m
V) x
tt
11 y
1' c
53
m
p
n
1
-
a 31.
32.
33. 3
-
+
-
2
2(x
3(a
-
2
+ 2b
-
4) (2s
2
-
3)
-
3)
/
-
2x(2x
3)
1
il
2
+ 4) (a + 5) - (a + 5) (a + 4) 3 2_ _ x + l' 'x + 2 x + 3/ - 2u(u + 2) 2
2
2
2
1
(u
b
2
+
(s
1
^
2
-
a
2
'
38.
34. 3
ab
b
(2x 36.
+
1
a 35.
1
b
(x
+ 2)
(t*
a 39.
+b
a
b
- b) 2 - c2 (a - cY - b - b) - c2 (o + b) 2 - cz (a - c) o - (6 + c) a (b 3 2 b b2 - a a^ + ab + b* ['a a 2ab (6 a) J L o 2
(a
2
40.
'
2
2
'
2
3
2
2
'I
'
42.
a
+&_
a
a
b
a
(m
2
f'V
4
43.
1
'
--
b
+
f l
n 'V
+
'
ab
n ac
2
a
2b
b2
a2 *
-
II
n /m
2
_
b
2
La
_._
2
3
a2
1
I
___
m /
II -f I
*'V
If
-t
II
I
-
3
/
+ ab + acj
mn
3
-
I
^ / '
a6c
m + mn 2
Chapter Four
UNKNOWN
LINEAR EQUATIONS IN ONE
4.1. Identities
An equation
and conditional equations
is
an assertion that two expressions are equal.
a sentence, in the language of algebra, in which the equality sign serves as the verb. The two expressions involved are called the sides or members of the equation. It is
An identity is an equation which the letters involved.* Example
1. (a
Example
2.
+
x1
=
2
6)
y
1
=
a2
+
(x
2a6 y)(x
true for
is
+ +
62
all
values of
.
?/)
When
it is desired to emphasize the fact that an equation an identity, the equality sign may be replaced by the symbol =, read 'is identically equal to.' Thus
is
a
-
1
=_(1 -
x).
A
conditional equation is one which is not an identity. It true for special values only, if any at all, of the letters involved. is
Example. 5x
3
=
+
2x
9.
A
conditional equation may be translated into English as a question. The one in the example above asks, 'What is * More precisely, an identity is true for all values of the letters make each side of the equation a definite number. For example,
3 (x is
-
an identity; but the values x
by
zero
is
_JL
l)(x 1
1__
+ 2)~* and x =
not allowed.
54
involved which
1
*
+2
2 are not permissible, since division
4.3]
LINEAR EQUATIONS
the
number such that 3
55 less
than
its
'
product by 5
is
9 more
than its double? ' Henceforth we shall use the word equation' to mean '
conditional equation.' While, as indicated above, an equation
is
in
one sense an
implied question, it is also, in many cases, a camouflaged answer (like the query, 'Who wrote 'Gray's Elegy'?'). For as will be shown presently, when we apply a routine method to certain equations, the numbers pop very readily out of hiding.
The
which they ask about
numbers or expressions sought from the last part of the alphabet (x, y, are usually chosen z, u, Wj etc.) and are called unknowns. letters representing the
4.2. Integral rational equations
An
integral rational equation
is
one in which two integral
rational polynomials in the unknowns are equated. Thus the unknowns appear either without exponents (since the ex-
ponent
1 is
usually omitted) or with the exponents
3x
4
value sought for
it is
Example Example
1
.
2.
ay
=
by
5x.
Here the unknown
2, as
=
may
is 2,
be verified by
Here the unknown
c.
2, 3, etc.
and the
trial. is y,
which
/
stands for the expression for
-,
a
since this value substituted
b
y makes the equation an identity.
As we
shall see,
many
practical problems lead to integral
rational equations.
4.3.
Linear equations
An
integral rational equation in which the unknowns appear to the first degree in each term containing them is called a first degree, or linear
*
equation.
letters We shall see that a first degree equation in not more than two ' hence the adjective linear.*' represented graphically by a straight line
*
may
be
LINEAR EQUATIONS IN ONE UNKNOWN
56
Example
1.
3x
Example
2.
5a*y
Example
8.
5x
=
7
=
+ 2y
= 4.
c.
shall consider only linear
a linear equation in one unknown an equation we find
or the particusubstituted for the unknown,
solve
lar quantities
make
we
one unknown.
4.4. Solving
When we
(Linear in y.}
(Linear in x and y.)
Hereafter in this chapter
equations in
IV
(Linear in x.)
4.
7b*y
[Ch.
the two
which, when members equal. The
its roots,
root
said to satisfy the
is
equation.
Example.
5-4-3
The
= 2-4
We shall now
root
of
5x
3
=
2x
+9
is
4,
since
+ 9.
solve the equation
(An axiom, as those who have studied geometry may recall, is a statement which seems to accord with experience, though we do not prove it, and which is often used in proofs of other statements. Thus a set of axioms, helps to serve as a foundation upon which one may build a system of mathematics.)
by use
of several axioms.
AXIOM
1.
When
equals are multiplied by equals, the products
are equal.
Application.
x
-
3 6
=
fe
(
+
2)
6,
or
Sx
(2)
AXIOM equal.
2.
When
-
18
=
Zx
+
12.
equals are added to equals, the
sums are
SOLVING A LINEAR EQUATION IN ONE
4.4]
-
(Sx
Application.
+
18)
18
=
(3x
Sx
=
3x
+
UNKNOWN
12)
+
57
18,
or (3)
AXIOM
3.
When
+
equals are subtracted
30. equals, the re-
from
mainders are equal.
Sx
Application.
3x
=
(3x
5x
=
30.
+
30)
3#,
or (4)
AXIOM
4.
When
equals are divided by equals, the quotients
are equal.
5x
.
.
A
7
Application.
=
30 5
o
,
or
x
(5)
Check in It
(1)
:
^
o
-3
=
= ^ 2t
6.
+
or
2,
should be noted that the root,
A
tions (1) to (5). which are satisfied
8-3
=
6, satisfies
+
3
2.
each of equa-
group of equations, such as
(1) to (5),
the same value of the unknown, are
by
called equivalent.
An
finding a succession of equivalent equations, the last one of which exposes the root of all of them.
equation
is
solved, then,
by
The simple rule that a term may be transposed from one member of the equation to the other by changing its sign may be used in practice in place of Axioms 2 and 3; but the reason for the rule should be understood.
The
essential process in solving
be described as follows Step
1.
Simplify
Step
2.
Remove
plications.
all
any
linear equation
may
:
complex fractions involved.
parentheses, carrying out indicated multi-
LINEAR EQUATIONS IN ONE
58
Step
UNKNOWN
[Ch.
IV
Clear the equation of fractions by multiplying both by the L.C.M. of the denominators.
3.
members
Transpose the terms containing the unknown to the side of the equality sign, and all other terms to the right
Step left
4.
side.
Step 5. Factor the left member, expressing it as the product of the unknown by a second factor. The latter is called the
unknown. Divide both members of the equation by the of the unknown.
coefficient of the
Step
6.
coefficient
For example, applying these steps to the equation
we
-
-
(6)
-
=
6*
f+
4,
get in succession the following equivalent equations:
m
3(s
+
_
6)
ftc
,
4 1+4 ~y +
(7)
3s
(X (8)
+
18
_ -
g
21x 21x
(9)
(10) (11)
= = 30x z(-9) =
+
126
-
x
(12)
,
>
- + 4,
6x
SOx 140
.
.
+ -
140;
126;
14
-9
The student can easily avoid blunders which are made if he will keep in mind the following simple Whatever member.
is
done
to the left
member must
often rule:
be done to the right
For example, if 2x = 5, we should be wrong in concluding that x = 5 2 = 3, because the rule does not permit us to divide the left member by 2 and to subtract 2 from the right member, as is done in the foregoing incorrect 'solution.'
4.4]
UNKNOWN
SOLVING A LINEAR EQUATION IN ONE
EXERCISE Remove
59
14
signs of aggregation according to the rules, multiplying factors as indicated, and then determine which of the following equations are identities and which are conditional equations. Solve the all
latter.
11.
= 15. - 1) = x - x. x(x 2z - 3x = x(2z - 3). - I) = x - 2x + 1. (x 5x2 +3x-l = x(5x+3)-l. x + 2 + 3x - 1 = 4x + 1.
15.
x
1.
3. 5. 7.
9.
3x
2
2
2
-
=^-
17. 1.5x 19.
20.
21. 22. 23. 24. 25. 26.
27. 28.
20
2
.7x
=
a
= -
5.
.6x
+
+
o
(5
2x)
=
3x-2 _ ~~ 2
_ 2x-3
3z
12.
20 = 0. Ix 10 = x + 2. 4x - 3 = 5x - 5. 3 - 2x = 7 + 2x. 1 - 3x = 5x + 4. 3 + 4x = 6x - 7.
16.
3x
-
4. 6. 8.
10.
18. .5x
.7.
+
=
10
.07
-
10
=
.8
3x.
+
-06x.
2
o
o
-
-
2
2x
-
4x
x
1
2
^ o
o
7;
3x
+
- 5. 7 - (5x + 3) + 6x = 10 - (3x + 5). 8x - (3 + 4x) + 5 = 7x - (3 + x). - 7x) + (3 + x) = (15 - 8x) + 4. (9 lOx + 5 - (3x - 4) = 4x - 3. -2(5x - 7) + (3x + 2) = 4x - 2(3 + x). - [3(7 - 3x) - 3] = 7x + 5. 5(2 + x) - [3(x - 7) + 4(2x - 3)] = x. 5(2 + x) - x) - 5(3 + x) = -[2(x - 1) + 3x]. 8(2 - 3) - 2(3 + x) = 5[(1 - 3x) - 4]. 7(x 3x
r
31.
5x
2.
=
2x -3 3
4
3
3x
+
3x-2 _ ~~
2
^
+2 o ^
+
A
32. 4
-
+ .
3x
'
2x
x 2
2
2
-
3
4x
K
O
6x+5
5
2(2x-5)
2
3
4x-7
LINEAR EQUATIONS IN ONE UNKNOWN [Ch. IV What number added to 3 is twice as much as the number
60 35.
minus 1?
when
36. If result
is
subtracted from 3 times a certain number, the is that number?
1 is
5 more than the original number, what
number is multiplied by 5, the product is 2 less than sum of the number and 7. Find the number. One child is twice as old as another. Two years ago he was
37. If a
twice the 38.
3 times as 39.
one he
is
old.
Find their
ages.
Fred has 3 times as many marbles as Bob. If he gives Bob will have twice as many. How many has each?
40. The quotient obtained by dividing a certain 5 less than twice the number. Find the number.
A man
41.
invested one
sum
receiving $80.00 interest in
at
6% and
one year.
number by
double that
How much
sum
.3
at 5%, at
was invested
each rate? 42.
An
airplane traveling 200 miles an hour starts at
noon
after
a transport which left at 10 A.M. and travels 150 miles per hour. When will the plane overtake the transport?
4.5. Operations
which yield equivalent equations
When
terms are transposed in an equation, or when both members are multiplied or divided by the same number, excluding zero, the resulting equation is equivalent to the first one. All of the operations required in Exercise 14 were of this nature.
4.6.
Operations which
may
not yield equivalent equations
Multiplication of both members of an equation by zero = 0. While this of course is true, it is yields the identity
not equivalent to the first equation. Division of both bers by zero is not allowable, as we have seen. Again,
when we multiply both members
an equation
expression containing the unknown, the new equation have roots not satisfying the first one. These roots are
by an
may
of
mem-
called extraneous.
OPERATIONS YIELDING UNEQUIVALENT EQUATIONS
4.6]
61
=
1 2 The equation x has the root 2. Mul= both x we members have: 3, tiplying by (x 2)(x 3) = 0. This has the roots 2 and 3; but 3 does not 0(# 3) satisfy the first equation and hence is extraneous.
Example
.
2.
Example
Consider the equation
= _2
1
(x
-
-
!)(
x
2)
1
-
2
Multiplying both members by (x
we
fractions,
root
1.
But
1
1
=
2(x get 1) does not satisfy the
for the non-permissible division
ous,
and the
first
we
if
Finally,
x
-
I
l)(x
(x
2).
2)
to clear
This has the
equation, since it calls zero. Hence 1 is extrane-
first
by
equation has no root.
members by an expression conusually lose one or more roots of
divide both
taining the unknown, the original equation.
we
= 3(x Example. The equation x(x 1) 1) is satisfied = = 1 x as x or be verified 3, may by by substitution. = 3, the root 1 x x we members both by 1, get Dividing being
lost.
Summing
up,
we
find that
when
members of an equation an expression containing
the
are multiplied or divided by zero or the
unknown,
the resulting equation
may
not be equivalent to
the first one.
When operations of this sort
are necessary in solving equa2 in as above, the solution is not complete tions, Example until the roots found have been tested in the original equation, and the extraneous roots have been rejected.
EXERCISE
15
The following equations may be reduced to linear ones and then solved. Where extraneous roots may have been introduced the answers must be tested. .
x-l
i
_2__-_3_ ii x x+1
1* 1
3 *'
~
x
-2o
8
1 i
x
+
2
r 2
4
LINEAR EQUATIONS IN ONE UNKNOWN
62
3
,
.* *'*
g
x+2
3
4
x+3
x-4
x
-x-12'
10 '
x-24-x
~
x 4
'
x s
01
2x 2x
+1 +1^ ~ +3 =
~
x
-
+
5
5
_
20
4
+2
-
3x
,
2
5
.
4x
.
24
-
3
-
5 6
6x
27.
x-3 3x + 1 x + 2 3
28 x
-
x
4x
-
2x
+5^ ~ +7
3x x
+
2x
~
5
'x +
2
1
+9 x
2
x
+
x2
x
5
= 3
-9 2
x
2
-
+7
-x-6 ~ 2x
a:
-
.
6x 2x
3
1
x-3
' -I
5
-
x2
^
-
-6~
+3
,
2x
2'
=
5 4'
1
25.
26.
~
3'
4
3
-
4x
2
___
+3~ _ -
__ _ '
1
+3_
3
6-
'
x
4
3
~
5x 2x
'
O
-
2x 4x
-
'
+ ^.
2
3x 3x
l'
x
-
1
=
2_3x + 2_2-3x x + 1 x + 1
2
2x
2
x2 +5x+6
'
2'
16
-
^
*
+
x
x_+_5
2x 1 x + 2
? x
3x
2^3^
1<^
o*
3
x
+ xx+_3 -2=
4x 2x 3
o
3
~
1
5
lft
4'
x
4
-
5
3
x+3
2x
3
3
3x 3x 17
28 2
x
-
x+2
3
+5~ 6x^7
? x
*'
/->
1
1~
3x
_ 9>
19
+3= -
5x 2x
7
^l
,
o
x 2 -x-6
f*i
i
IV
4
5
1 _. ______
r
x-3
[Ch.
15
5
__
2
=
x x
^
+7 + 3' + 5x 3'
6 '
x
3
+3 + 2:c x + 2
.
LITERAL COEFFICIENTS
4.7]
4
5 29.
x
30.
+
x+l
5
~'
3x
+
10
+
6x
+5
5x
-
-
13x
4
5
2x
-
3
3x
-
2
6x
3x
-
2
^ 4x -
1
12x2
2
3 32.
x2
63
5x
+
1
x
1
2x
5z 5x 2
1
2
3x
4
+
2
^
1
llx
+ 1
10
6x2
-
2
7
4x
,
-
-
7
x
-
~
u>
1
4.7. Literal coefficients If
W, by
P
is
the perimeter of a rectangle of length
the relation between P, L, and the formula:
is
width
evidently expressed
P = 2L + 2W.
(1)
Solving
W
L and
(1) for
L,
we have -2L = 2JT-P; 2L = P-2W;
(2)
L =
-=
=
-=
Similarly,
TT
(3)
Results
and
(1), (2),
(3) are different
versions of the
equation, solved respectively for P, L, and
same
W in terms of the
remaining letters. In such equations any one of the letters may be considered as the unknown. Thus, if the perimeters and widths of many different rectangles were given, and we were asked to get the various lengths, the most efficient way = 6, would be to use (2) as a formula. If P = 100 and
W
AA LT = 100-2-6 =44; etc. Similarly (1)
tively.
.,
if
and
D = nn P 90
1A and TJ7 17=10,
(3) are
,
QK Lj = 90-2-10 = 35,
formulas for
P
and
W respec-
LINEAR EQUATIONS IN ONE UNKNOWN
64
Whenever an equation contains other
letters
known, those letters will usually appear equation. For example, given
Ax
(4)
+B=
[Ch.
IV
than the un-
in the root of the
0,
D the root
is
evidently
A-p
Such equations are said to have literal coefficients. It should be understood that the coefficients in an equation include not only multipliers of the unknown, as A in (4), but also the terms,
such as
B
in (4),
which do not contain the
unknown.
Any linear by
(4).
equation in the
For instance,
2ax
(5)
the
A
unknown x may be
represented
in
of (4) stands for 2a,
-
3y
=
and the
0,
B for
3y. Since the root
T>
of (4)
is
-j, A
which represents a
single
number for any given
set of values for the literal coefficients,
we can conclude
that:
A
linear equation has exactly one root.
Clearly an equation with literal coefficients represents inparticular equations with numerical coeffi2 = 0, cients. Special cases of (1), for example, are 3z 17 = 0, etc. An equation like (4) is said to be more 5x finitely
many
+
general than one with numerical coefficients. In algebra, and in fact in all mathematics, it is often desirable to have problems and solutions as general as possible, thus covering
many
cases in a single operation.
In solving an equation with Art. 4.4 may still be used.
Example. Solve
ax
literal coefficients,
the steps of
LITERAL COEFFICIENTS
4.7]
65
Solution.
Step
2.
Step
3.
Not needed
?-- + x + -
Zaxd
or
Step
4-
Step
5.
in this case.
2bc
+
Qbdx
+ Gbdx = x(3ad + 66d) = 3azd
+
Qbd
=
2bc
-
Qbd
2bc
+
66d.
2bc
I2bd.
+
I2bd.
or 2fe)
A
complete check, of course, would require the substitution of the literal root in the original equation. By way of a
and practical partial check, however, we may substitute = 6 = specific numbers for a, 6, c, and d. For instance, if a brief
c
=
=
d
.
root
8
then
1,
26c -
A1 Also
-.
(6)
T
3aa
9
x becomes ~
+ 66d ^, + ooa ,
.
o?+l = 2orx =
l,
1 O
A
+#+
8 u Or becomes - ^ 9
,
,
let
1
a
with the
=2,
=
c
n 0.
=
^ Then
while the root being tested becomes
Qbd
The
partial check
to replace set a
=
6
EXERCISE
which
is
when
shortest,
permissible,
except the unknown by zero. d = 0in the case above?
all letters
=
c
=
Why
is
not
16
Solve the following equations with literal coefficients. The for is in each case x, y, z, u, v, or w.
unknown
to be solved
,
8 + ,.fe^_ o 3. cy
d
=
a
Z cy.
2
.
4.
c
L_2 a cw
bd
_
<
o
=
be
2, dw.
=
o.
LINEAR EQUATIONS IN ONE UNKNOWN
66
7.
=
dv
5. ac
=
6x
ac
be
u
+n_
u
a
6
a
'
II.
-
2r 13 -
+
y 2u a
ie
_ ^_
n
a
n 12.
T'
+ + 2n b
y
u
+ 2n
^/
2m
=
3n
m+
n
a
x
-
n
=
__
bu
ad.
-
f
m+n
m
v
+b _ m n a + x r + 2n a
;
2b
-
a
26
b
2r
n
~
ad
cy *
m+v
;
=
bd
b
2b __
8.
an
n
-
~'~'
CC/
r
~~
+
17. r
w
b
6.
ax.
+
m+n_ m r w
ad.
cv
26* __
f^
2rw.
18.
-
L^^/
4rf//C/
r-r-
jr
= m+ ,
n.
__ Solve each of the following formulas for the
=
a
+
21.
Z
22.
E = I(R + -
23.
T
=
o
=
24
-
-
+ -, +
jtt
'ET 111
25.
P=
A(l
26.
-
=
v
27.
=
31. 32.
R *>
for jB
A
for
*t2
-
I
H~T>
KmM =
f
f
,
r
f
r
for
m
; J
^5 ^ or
m;
P
r
f
a
for
5
^; '
f
f
r v
r r-
, M.
= VQ + ^<, for t; for VQ; for ^. s = afi, for a. A = /t(a + 6), for h; for 6. F = r (a - 6); for a.
29. v
30.
IT) for ^iJ for
M + m'
a
for r; for /?; for n.
for q] for p; for/.
-^^~^
S=
^ rt 28. ,P
l)d, for a; for n; for d.
(n
2
.
-
-
letters
indicated:
[Ch.
IV
4.7J
LITERAL COEFFICIENTS
w _ w, 33. 34.
What 35.
w
One boy is
many
is
a
'
f
r
W
;
f
^
r
2a years old and a second boy
26 years old.
is
their average age?
One
than the 36.
=
-W
67
child
first.
a years old, and a second one
is
What
is
Henry had a marbles. After buying
How many
as John.
A boy who
is
26 years older
their average age?
10
more he had
half as
had John?
had x dimes
found half a dollar and then gave half of the money he had with him to his mother. What was the value of her share in cents? 37.
in his pocket
A man who owed
a dollars paid x dollars on the debt and then owed f of the original amount. Find x. 38.
39. In
making a trip a man averaged a miles per hour going and hour on his return. If he was c hours on the road, how
6 miles per
far
from home did he go?
40. How many pounds of cream testing x% butter fat must be added to y pounds of milk testing z% butter fat to give milk testing
w%
butter fat?
41. If oranges cost c cents per dozen, one buy for a dollars?
EXERCISE
how many
oranges can
17
Solve the following stated problems involving fractions. 1.
If
a certain number
remainder Note.
is
divided by 4 the Quotient
is
3 and the
Find the number.
is 3.
In this and succeeding problems we use the relation
(Art. 1.10)
dividend 77-
:
divisor
=
~
,.
,
,
Quotient H
remainder -TT
.
divisor
Observe the distinction here made between 'Quotient' and 'quotient.' For example, the Quotient of 13 divided by 4 is 3, while the quotient, or total result of the division, 2.
What number
remainder
is
2?
is
divided
by 7
if
is
3 .
the Quotient
is
3 and the
68 3. If
2
is
LINEAR EQUATIONS IN ONE UNKNOWN [Ch. IV added to a certain number and this sum is divided by
3 the Quotient 4.
4 and the remainder
is
One number
is
by
One number
is
2
less
the second the quotient 6.
The
divided
by
7. If
^, the
8.
x
new
difference
is f.
If
the
one
first
is
Find the numbers.
than another. If the first one is f Find the numbers.
is
divided
.
between two numbers is 3. If the larger is ^. Find the numbers.
is
the smaller the quotient is
added to both the numerator and denominator of
fraction formed equals . Find the value of x.
Find the number which must be added to both the numer-
ator and denominator of f to 9.
Find the number.
3 more than another.
divided by the second the quotient 5.
is 2.
What number must be
and denominator 10. If
of
-^
make
the
new
fraction equal to f
.
subtracted from both the numerator
to yield the fraction f ?
a certain number
is
added to the numerator and subnew fraction equals f Find
tracted from the denominator of f the the number.
.
,
The numerator
of a fraction is 2 less than the denominator. added to both numerator and denominator, the resulting fraction is f Find the original fraction. 11.
If 1 is
.
The denominator
of a fraction is 2 more than the numeradded to both numerator and denominator, the new fraction can be reduced to ^-. Find the original fraction. 12.
ator. If 11
13.
The denominator
numerator.
new
is
If
3
is
more than 3 times the both numerator and denominator, the
of a fraction
added to
is 1
fraction equals . Find the original fraction.
4.8. Geometric
Many
problems
problems that are geometric
in character occur in
one's normal experience. They vary widely and include lengths, areas, volumes, relative sizes of angles, etc.
Example 1. A 5-foot string is cut into two pieces, one which is - of the other. Find the length of each piece.
of
4.8]
GEOMETRIC PROBLEMS
69
Solution.
x
Let
Then
=
no.
ft.
in the longer piece.
=
no.
ft.
in the shorter piece,
o
+j=
*
(1)
sum of the parts (1), we have
since the
Solving
x
=
5,
of anything equals the whole of
,
the greater length;
,
the smaller length.
it.
7
4x
20
=
o
y
,
.,
,
Example 2. Two rectangles have the same width. One is 2 inches longer than the other and 5 inches longer than its width. If the difference in their areas is 10 square inches, find the dimensions of each.
Solution.
Let x
x
+5 x + 3
= = =
the width of each rectangle in inches. the length of the longer rectangle, in inches. the length of the shorter rectangle, in inches.
x(x Solving,
+
we have x =
-
5)
5;
x
x(x
+
5
+ =
3)
=
10;
x
10.
+
3
=
8.
Hence the rectangles are 10 by 5 and 8 by 5 inches
re-
spectively.
Note.
The area
The volume height h
is irr
Example
of a circle of radius r
is irr
2
square units.
of a right circular cylinder of radius r h cubic units.
and
2
3.
Each
of
altitude of 10 inches.
two
The
right circular cylinders has an radius of the base of the larger is
two inches greater than that of the other. The difference in volumes is 80?r cubic inches. Find the radius of the
their
base of each cylinder.
LINEAR EQUATIONS IN ONE UNKNOWN
70 .
[Ch.
IV
Solution.
= radius of base of smaller cylinder, in inches. + 2 = radius of base of larger cylinder, in inches. 107TX = volume of the smaller cylinder, in cubic inches. = volume of the larger cylinder, in cubic inches. I0ir(x + 2) - lOTra: = SOvr. 107r(o; + 2) (2) Solving (2), we find that x = 1 and x + 2 = 3. x
Let x
2
2
2
Example
4-
The
second angle and
first is
2
angle of a triangle is equal to -J- of the of the third angle. Find
also equal to
the three angles. Solution.
Let
a:
Then 3x 2x
= = =
no. degrees in the first angle. no. degrees in the second angle. no. degrees in the third angle.
x Solving,
we have x =
EXERCISE Form
+
3z
30,
+ 2x
3x
=
=
90,
180.
2z
=
60.
18
algebraic equations
and
solve the following problems.
A
1. 7-foot cord is cut into two pieces so that one piece as long as the other. Find the length of each piece.
2.
A
form two
is
f
piece of wire 4 feet long is cut and the pieces are bent to circles. If the diameter of one circle is f that of the other,
find the length of each piece. 3.
One
bined length 4.
is
A 26-inch
as the second, is
3 inches longer than another. If their com17 inches, find the length of each.
string
is
cord
is
cut into three pieces.
and the second
first is
f as long as the third.
is
f as long
How
long
each? 5.
Some
6.
two pieces to as long as the other.
12-foot boards are to be cut into
two boxes, one of which should each board be divided? lids for
If
The
The width
of a rectangle
the length of the rectangle
is
is
^
and the
make
How
side of a square are equal.
5 inches
more than
its
width, and
PROBLEMS INVOLVING TIME, RATE, DISTANCE
419]
71
area is 50 square inches more than that of the square, find the dimensions of each.
its
7.
and
The radius of one by
27?r
The diameter
of
one
8.
another, and
its
area
3 inches more than that of another, square inches. Find the radius of each.
circle is
their areas differ
is
circle is
greater
by
4 inches more than that of 12?r
square inches. Find the
diameter of each. 9.
Two
triangle
have equal bases. The two sides of each base and altitude. One triangle is isosceles but
right triangles
form
its
4 inches longer than its base. If their areas differ by 12 square inches, find the dimensions of each. the altitude of the other
is
10. The altitudes of two triangles and the base of one of them are equal. The base of the other is 6 inches more than its altitude, and the difference in the areas is 21 square inches. Find the base
and
altitude of each.
The radius of the base of one right circular cylinder inches more than that of another, and the altitude of each 11.
3
15 inches. If the difference in their volumes
is
is is
2257r cubic inches,
find the radius of the base of each.
One
of the acute angles of a right triangle is 10 the other. Find the number of degrees in each angle. 12.
13.
The
first
14.
The
vertical angle of
less
than
angle of a triangle equals f of the second, and the third equals the sum of the first and second. Find each angle.
an isosceles triangle
is
20 more than
sum of the equal angles. Find each angle. 15. Work problem 14 with '20 more' replaced by '4
the
less.'
16. The first angle of a triangle is as large as the second as large as the third. Find each angle. I
4.9.
Problems involving time,
Many (1)
rate,
and
and distance
problems are solved by means of the formula d
=
rt,
where d represents the number of units of distance, r the number of units of distance traveled in one unit of time, and t the number of units of time. For brevity, d, r, and t are
LINEAR EQUATIONS IN ONE
72
called distance, rate,
and
time. It
UNKNOWN
[Ch.
PV
must be understood, how-
ever, that (1) holds good only if the rate is constant, such as it is, for instance, in a car whose speedometer needle re-
mains
fixed at the 40 m.p.h. (miles per hour) position. Equation (1) is used, of course, to find the distance
the rate and time are known. Solved for r and
two
t, it
when
takes the
alternate forms, r
(2)
,
and
=
(3)
used respectively to find r and
t
when
the other two quanti-
known.
ties are
EXERCISE Form 1.
.
19
equations and solve the following problems.
If
one car runs 40 m.p.h. and another car runs 50 m.p.h., will the sum of their distances be 300 miles?
in
how many hours
x
Let
Then
40:c
40z
+
5Qx 50z 90z ^
= = = = = =
no. hrs. required. no. mi. first car runs. (Here no. mi. second car runs.
rt
=
40.)
300. 300.
W=
2. A car running north at 45 m.p.h. passes point A at 12 o'clock. A second car running north at 50 m.p.h. passes A at 1 :30 o'clock.
When will
the second car overtake the
first
one?
3. A is 450 miles west of B. A car starts east from 40 m.p.h., and at the same time another starts west from 45 m.p.h. In how many hours will they meet?
4.
A B
at
at
Two men are 400 yards apart and walk straight toward each
one walks 80 y.p.m. (yards per minute) and the other 90 y.p.m., in how many minutes will they meet?
other. If
4.9]
PROBLEMS INVOLVING TIME, RATE, DISTANCE 5. Two men run in the same direction around a
73
440-yard
one runs 10
y.p.s. (yards per second) and the other f as in how fast, many seconds will the faster one gain a lap? 6. One man can run 9 y.p.s. and the other can run 8 y.p.s. If they start at the same time and run in opposite directions around a 440-yard track, in how many seconds will they meet?
track. If
A
car starts east at 40 m.p.h. A second car starts east from later at 50 m.p.h. When will the second car be 10 miles ahead of the first one? 7.
same point one hour
the
A
car starts south at 40 m.p.h. One hour later a second car starts south from the same point. If the second car overtakes the 8.
first
one
9.
in three hours, find its speed.
A man
walks toward a certain town at 2 m.p.h. One hour
man starts from the same place at 3 m.p.h. He overtakes the first man just as he reaches the town. How far was it to town and how long did each man require for the trip? later a
second
10. A train that runs 50 m.p.h. passes a station 2 hours behind a slower train and overtakes it in 3 hours. Find the speed of the
slower train.
current in a certain stream flows 3 m.p.h. A crew can row twice as fast downstream as it can row upstream. How fast 11.
can
it
The
row
in still water?
x
Let
Then
x
+3
x
3
x Solving,
+
3
= = = =
The
f times it
-
3).
as fast
downstream as
in still water?
13.
The
it
upstream
still
water?
as
it
can row downstream.
rate of the current
is
downstream in the time required fast can it row in still water?
A
crew can row can row upstream. How fast can
A How
current in a stream flows 4 m.p.h.
fast
The
9.
current in a stream flows 2 m.p.h.
row
14.
2(x
we have
x= 12.
rowed in still water. rowed downstream. no. m.p.h. rowed upstream. no. m.p.h. no. m.p.h.
3 m.p.h. for
crew can row f as fast can it row in
A crew
can row 9 miles
rowing 4 miles upstream.
How
LINEAR EQUATIONS IN ONE UNKNOWN
74
[Ch.
IV
15. Work problem 14 if the current flows 2 m.p.h. and the downstream and upstream distances are respectively 18 and 6 miles. 16. A man can row 4 m.p.h. in still water. He can row 12 miles downstream in the time required for rowing 4 miles upstream. Find the rate of the current. 17.
day
An
its
airplane has a cruising speed of 300 m.p.h. On a certain ground speed when traveling with the wind was twice its
speed against the wind. 18.
How
fast
was the wind blowing?
On a certain day the wind at 5000 feet was blowing 50 m.p.h.,
at 10,000 feet it was blowing 20 m.p.h. in the same direction. in 3 hours, plane flew at 5000 feet with the wind from A to
and
A
B
and returned at the 10,000 feet level in 5 hours. Find still air and the distance from A to B.
4.10. Problems concerning
its
speed in
money
Most
practical problems about money deal with principal, interest, rate of interest, wages, profit and loss. In some other
problems, usually less practical in nature, the object is to find the number or denomination of pieces of money involved.
The The
principal is the sum of money that bears interest. rate is the fraction of the principal paid for its use
during a certain period of time
usually a year. Stated in this fraction multiplied by 100. Thus, the
percentage, it is rate of or .06 becomes
^
'6%.' The time is the interval during which the principal is used. The interest is the total sum paid for use of the principal. The amount is the sum of the principal and interest. In formulas the following notation
is
customary.
A represents the number of dollars in the amount. P represents the number of dollars in the principal. / represents the number of dollars in the t
represents the
number
interest.
of interest periods in the time in-
terval.
r
represents the rate, written as a decimal fraction.
4.10]
PROBLEMS CONCERNING MONEY
We
75
shall use the following formulas.
/
(1)
=
Prt.
Solving (1) for P,
r,
and
t
successively,
we have
P-1
(2)
-
r
(3)
-jre
and
-
.
(4)
A = P+
(5)
From
and
(5)
When
from the
definition.
(1),
A = P
(6)
7,
+
Prt
=
P(l
+
rt).
two or more investments, subscripts may be used to distinguish between them. For example, PI, read 'P sub-one,' means the first principal, P2 the second principal, etc. Similarly, 7i means the interest on P x 7 2 is the interest on P 2 etc. there are
;
,
Example L A man invests $6000, part at 6%. The total interest for one year is $320.
5% and
part at
How much
was
each investment?
two investments be P and 6000 PI. = = r t .05 and for the second, investment, 1;
Solution. Let the
For the r
=
.06
first
and
t
=
1.
Substituting in (1)
we have 7!
and
72
But Hence .05Pi
+
Solving,
and
7i
(360
-
+7
2
.06Pi)
= = =
(6000 -Pi)(.06)(l)= 360 -.06Pi. 320, the total interest.
=
320.
= =
4000, 2000.
Pi(.05)(l)
we have 6000
-
P! PI
=
.05Pi,
LINEAR EQUATIONS IN ONE UNKNOWN
76
What
2.
Example 2 years at
[Ch.
principal will yield $24 interest
IV in
6%?
=
Here /
Solution.
24, r
known. Using these values 24
_ p -
.06,
in (2),
_ -
24
12
i
=
and
2,
P
is
the un-
we have
_ -
2400
~w
Two men work
3.
Example
=
6 days and receive $54 as as much per day as the other, find
wages. If one receives the daily wage of each. Solution. Let
x
=
the daily wage of the second
man
in
dollars. 4:r
Then
=
o
X
_ O_ ^tvC
/*
Solving,
I
-j-
JL
5
first
man
in dollars.
f^ {
U.
6
we have x
EXERCISE 1.
the daily wage of the
=
5,
20
How much interest 5% for 2 years?
will
be earned by an investment of
$300 at 2.
An
investment of $200 yields $8 interest in a year. Find the
rate. 3.
At what
rate will $250 yield $20 interest in 2 years?
4.
At what
rate will $3500 yield $350 in 3 years?
5.
At what
rate will $4700 yield $282 in 2 years?
6.
Find the time required
7.
A sum
of $7500
is
for
$6500 to yield $650 at 5%.
invested, part at
annual interest on the two investments investments.
part at 5%. The $400. Find the two
6% and is
4.10]
PROBLEMS CONCERNING MONEY
77
8. Part of $8000 is invested at 4% and the rest at 5%. The 4% investment yields $50 more interest in a year than the other. How much is invested at each rate?
One part
of $9000 is invested at 4% and the rest at 5%. incomes the two thus yielded are equal, fincj each investment. 9.
If
10.
A certain sum invested at 6%
as an investment of $6000 at 5%.
What
is
the
same yearly income
sum
invested?
one
Two men worked 9 days and together received $144 in wages. man received $7 per day, how much did the other get?
12.
A boy received
11. If
yields the
as
much pay
as his father. If together they
got $48 for 4 days' work, what was the daily wage of each?
two common laborers and four skilled workmen receive $80 per day altogether, and if the wages for skilled labor are twice as much as for common labor, find the wage of each. 14. Ten men were employed, some at $6 per day and some at $8 per day. The total daily wages amounted to $68. Find the num13. If
ber employed at each wage.
Nine men were employed, some at $7 per day and some at $9 per day. If the total daily wages amounted to $75, find the number working at each wage. 16. P = 325, t = 2, r = .06. Find I. 15.
17.
18. 19.
20. 21.
22. 23.
= 3, r = .05. Find A. 450, = = = 100. Find r. 7 3, P 18, 7 = 60, P = 500, r = .06. Find A = 770, = 2, r = 5%. Find P and 7. A = 690, = 3, r = 5%. Find P and 7. 7 = 80, = 2, r = .04. Find P. A man has $1.25 in nickels and dimes.
P=
times as
t
t
t.
t
t
t
many
nickels as dimes, find the
x
Let
Then
3x lOx 5(3x) l(te
Solving,
+
15x
= = = = =
x= =
there are three
of each.
no. dimes. no. nickels.
the value of the dimes in cents.
15x 125.
we have 3z
If
number
5.
15.
=
the value of the nickels in cents.
LINEAR EQUATIONS IN ONE UNKNOWN
78
24.
many 25.
dimes. 26.
dimes
A man
[Ch.
IV
has $3 in nickels and dimes, there being twice as nickels. Find the number of each.
dimes as
The sum
of $1.90
is
made up
of quarters plus f as
many
How many of each are there? A boy has twice as many nickels as dimes and twice as many as quarters. How many of each has he if the total sum is
$1.30? 27. A man has $3.25 in nickels, dimes, and quarters. He has the same number of dimes as of nickels, and twice as many quarters as dimes. Find the number of each. 28.
A man
has f as
quarters as nickels, and as many of the other coins. How many of each
many
dimes as the total number has he if their total value is $2.30? 4.11.
Mixture problems
Many problems arise from the forming of mixtures of various materials. When two or more substances are mixed, an equation may be formed on the basis of the fact that the quantity of a certain substance in one material plus the quantity of this same substance in a second material is equal to the quantity of that substance in the mixture formed by combining the two materials. This same principle will
extend to any mixture containing a given number of
materials.
When
the value of a mixture
is
the matter of chief in-
Example 2 below, we form the equation not on the basis of quantity but on the principle of directly value. When two substances are mixed we say that the value terest, as in
of the first one plus the value of the second equals the value of the mixture.
Example 1. How much metal that is 5% silver must be added to 10 pounds of metal, 2% of which is silver, to form a mixture having
3%
silver?
Solution.
Let
x
=
no. Ibs. of metal containing
5%
silver.
MIXTURE PROBLEMS
4.11]
Then
= = = =
.05z (.02) (10)
(.03) (#
+
.05z
+
10)
(.02)(10)
Solving,
79
no. Ibs. of silver in the first metal; no, Ibs. of silver in the second metal; no. Ibs. of silver in the mixture; (.03) (x
+
10).
we have
=
x
5.
Example 2. How many pounds of coffee worth 28^ per pound must be added to 40 pounds of coffee worth 35^ per pound to form a mixture worth 30 per pound? x 28x
Solution. Let
35(40)
30(x + 40) 28x + 35(40) Solving,
EXERCISE
the value of the 28
the value of the mixture in cents.
30(z
+ 40).
=
100.
21
alloy to form a
How many pounds of
12% copper
be added to 20 pounds of alloy?
cream containing 25% butter
be added to 30 pounds of milk containing mixture containing 15% butter fat? 3.
20%
coffee in cents.
the value of the 40 ji coffee in cents.
How much 10% copper alloy must
15% copper 2.
no. Ibs. of 28 j4 coffee added.
we have x
1.
= = = = =
How much 30%
5%
fat
must
butter fat to form a
gold alloy must be added to 25 ounces of
gold alloy to produce a
27%
gold alloy?
4.
How much
20%
5.
How much
pure copper must be taken from 35 pounds of it to a 25% copper alloy?
pure gold must be taken from 23 ounces of gold alloy to reduce it to a 15% gold alloy?
40%
copper alloy to reduce
6.
35%
How much
silver alloy to
7.
How much
pure silver must be taken from 45 ounces of reduce it to a 20% silver alloy? candy, worth
per pound, must be mixed per pound to form a mixture
25ff
with 20 pounds candy worth 13 of candy worth 20jzf per pound? of
}f
UNKNOWN
LINEAR EQUATIONS IN ONE
80 8.
Two kinds of candy worth
15j
and
25ff
is
is
How much
of
used?
Two
kinds of coffee worth 30 jf and 22 jf per pound are mixed form 100 pounds of 25 per pound coffee. How much of each 9.
to
IV
per pound are mixed
to form 30 pounds of candy worth 18ji per pound.
each
[Ch.
used? 10.
A lady bought
10 pounds of grapes for $1.35.
sold for 10f per pound and pounds of each kind did she
some
for 15^ per
Some of them
pound.
How many
buy?
A
dealer bought 15 cases of fruit for $41.50. Some cases cost $3.00 and some $2.50. How many cases of each did he buy? 11.
4.12. Lever problems
A
a mechanical device by which a force applied at one point is transferred to a second point and there intensified. When we roll over a huge rock with a crowbar we apply this principle. Many problems in algebra are concerned with levers and related mechanical devices such as pulleys. lever is
Some preliminary definitions are necessary. The fulcrum is the non-moving support upon which the lever swings when force is applied to it. The moment of any force, for a given set of units, is equal to the number of units in the force multiplied by the number of units in the distance
from the fulcrum to the point on the
lever where the force
applied.
What
Question.
is
is
the
moment
of a force of 10
pounds
applied 5 feet from the fulcrum of a lever?
Z
I T 10 Ib
Fulcrum
ZL I
1
Weight Fig. 2
Answer.
Moment =
5
10
=
50 foot pounds, for the given
units.
The
shown
above diagram causes the lever to rotate about the fulcrum in a counterclockwise direction force
in the
LEVER PROBLEMS
4.12]
81
the figure. Any force acting downward on the lever to the right of the fulcrum, such as the indicated weight, as
we view
would tend to rotate the lever
in the opposite, or clockwise,
direction. If
the
sum
of the
moments
the lever counterclockwise
moments
of all forces tending to rotate equal to the sum of the
is
of all forces tending to rotate
lever will be balanced
it
clockwise, the lever prob-
and stationary. In most
lems we calculate either the force or distance necessary to
make
the lever balance.
Example
1. If
a force of 50 pounds is applied 5 feet from must a weight of 60 pounds
the fulcrum of a lever, where be placed to balance it? Solution.
x
Let 5
50 QOx 60z z
(A diagram similar to
= = = =
no.
250 the
ft.
Fig. 2
may
be drawn.)
from the fulcrum to the weight.
=
the moment of the 50-lb. force. moment of the 60-lb. weight.
250.
^
s
^at the weight must be 4 ft.,
2
in.
from
the fulcrum.
In the above example the weight of the lever is not considered. This weight may be important, however, in the case of a heavy lever such as a plank. To allow for it the lever is considered as composed of two parts called arms, each of which lies wholly on one side of the fulcrum. If the lever is
uniform, or of the same weight for each unit of length, the
moment
of
an arm
may
be found by multiplying
its
weight
by length, as expressed in the units chosen for the problem. In other words, the moment of a lever arm equals the moment of a force equal to its weight applied at one-half
its
a point halfway from the fulcrum to the end of the arm.
Example 2. The fulcrum and 4 feet from the other.
of a lever If
the
is
10 feet from one end
beam weighs 8 pounds
per
LINEAR EQUATIONS IN ONE
82
what weight must be placed
foot,
arm
at the
UNKNOWN
[Ch.
IV
end of the shorter
to balance the lever?
80
Ib
Fig. 3
The weights of the two arms are 80 and 32 pounds respectively. We must equate the counterclockwise (c.c.) Solution.
and clockwise x
=
80
5
=
the
32
2
= = =
the
Let
moments.
(c.)
ho. Ibs. of force applied at the end of the shorter arm.
moment
of the left
arm
in the
arm
(c.).
diagram
(c.c.).
4#
+
4z
32
Solving,
2
moment moment
the
80
of the right of the force
x
(c.).
5.
we have x
=
84.
two persons are carrying a weight swung on a pole, we may consider the weight as a fulcrum and the supporting If
forces as the acting ones. This
then a lever just like the others we have discussed, except that it is upside down. is
Example 3. Two persons carry a weight of 100 pounds swung on a 9-foot pole. Where should the weight be placed so that one person will carry 60 pounds and the other 40 pounds?
60
100 Ib
Ib
40
Ib
Fig. 4
Solution.
Let 9
x x 60z
= = =
no. of no.
ft.
40(9
from weight to man carrying 60 from wt. to man carrying 40 Ibs.
ft.
-
x).
Ibs.
4.12]
LEVER PROBLEMS
Solving,
we have 9
EXERCISE 1.
A
83
-
x x
= J#, = Y,
or
3f
.
or 5f.
22
90-pound boy and a 60-pound boy balance on a 12-foot
teeter board.
Where
is
the fulcrum?
2. What weight must be placed 6 feet from the fulcrum to balance a 70-pound weight 8 feet on the other side of the fulcrum?
3. How much weight can a 160-pound man raise with a 10-foot beam placed so that the weight is 2 feet from the fulcrum? *
4.
What
force
with a 12-foot
is
beam
necessary to raise a weight of 1000 pounds placed so that the weight is 3 feet from the
fulcrum? 5.
What
6.
If
weight must be placed 5 feet from the fulcrum to balance a 12-pound weight 4 feet on the other side of the fulcrum? the fulcrum
is
beam whose arms and 40 pounds respectively,
4 feet from each end of a
are uniform but unlike, weighing 75
where should a 50-pound weight be placed to balance the beam? 7.
long.
Two men Where
carry a 90-pound weight swung from a pole 8 feet must the weight be placed so that they will carry
40 and 50 pounds respectively? 8.
A
12-foot
beam weighs
9.
A
14-foot
beam weighs 8 pounds per
7 pounds per foot. If the fulcrum is 4 foot from one end, what force must be applied at the end of the shorter arm to balance the beam? foot. If the
fulcrum
is
6 feet from one end, what weight must be placed at the end of the shorter arm to balance the beam? 10.
What
force applied 7 feet
from the fulcrum
will
balance
will
balance
1500 pounds 2 feet from the fulcrum? 11.
What
force applied 8 feet
2000 pounds 2
feet
from the fulcrum
from the fulcrum?
* In this and subsequent problems subject to two interpretations, assume that the fulcrum is between the applied forces.
LINEAR EQUATIONS IN ONE UNKNOWN
84
[Ch.
IV
What weight placed 3 feet from the fulcrum can be balanced a by 170-pound force applied 7 feet on the opposite side of the fulcrum? 12.
13.
lever 14.
What if
weight can a 110-pound boy balance with a 12-foot the weight is 2 feet from the fulcrum?
Two men
and placed 4 15.
carry a 70-pound weight swung on a 10-foot pole from one end. How much does each carry?
feet
Two men
end of a 9-foot
carry a 120-pound weight swung 4 feet from one pole. How much does each carry if the pole itself
weighs 20 pounds? 16.
Two men
The one who
is
carry a weight of 150 pounds swung on a pole. 4 feet from the weight carries 90 pounds. How long
the pole?
is
A weight of 300 pounds rests over the end of a lever 6 inches from the fulcrum. A boy weighing 60 pounds can just lift the weight. How long is the lever? 17.
4.13.
Work problems
problems involve the rate at which an action is being performed. For example, if a man can do a piece of work in 10 days, his rate of work is to do -fa of it per day.
Many
oc
he works x days, he has performed
If
of the total work.
*
The guiding
in problems of this type is the fact principle that the product of the rate of work by the number of units
of time involved
is
equal to the fractional amount of the
work done. Example. If A can do a piece of work in 20 days and B can do the same work in 25 days, how many days will both need to do the job when working together?
if
* Naturally, good judgment must be used in applying the principle. For example, one dentist can fill a tooth in one hour, it does not follow that two dentists could
fill it
in half
an hour.
4.13]
WORK PROBLEMS
85
Solution.
Let
x
=
the no. of days needed for both to do the work.
-
=
the fractional part both do in one day.
=
the fractional part
A
does in one day.
=
the fractional part
B
does in one day.
C
2J
Zo
1 4- 1
=
1
x
25
20
Clearing fractions (multiplying both
members by lOOx)
we have 5x
+
or
EXERCISE 1.
A
4x
=
x
=
100,
100
- - ,
= Hi
,
days.
23
can do a piece of work in 30 days and B can do the same How long will it take for both to complete the job
job in 8 days. if
they work together? 2.
One pipe can
fill
a certain tank in 25 minutes. After this
pipe has been running for 10 minutes, it is shut off and a second pipe is opened. The second pipe finishes the filling in 30 minutes.
How
long would tank alone?
it
have taken the second pipe to have
filled
the
3. A can do a certain amount of work in J the time B requires; can do the same amount in the time C needs. The three together can do the work in 36 days. How long will it take each of them to do the work alone?
B
4. it
A
large pipe
in 18 minutes.
pipes to
fills
How
the tank
a tank in 12 minutes and a small pipe fills long will it take one large and three small
they are
opened simultaneously? John started a job which ordinarily took him 5 hours, and quit after working hours. James finished the job in 8 hours. How long would it have taken James to do the whole job by him5.
self?
fill
if
all
LINEAR EQUATIONS IN ONE UNKNOWN
86
[Ch.
IV
B can do it in 25 hours. been working together for 14 hours, they are joined by C and finish the job in 6 more hours. How long would it have taken C to have done the whole job alone? 6.
After
7.
A can paint a A and B have
One
faucet can
42 minutes. if
house in 40 hours and
How
long
fill
a tank in 18 minutes and a second in
will it
take to
both faucets are opened at the
fill
three-quarters of the tank
same time?
takes John a days to do a piece of work and James b days to do the same work, how long will it take for both to do the 8. If it
work together? 9. A and B together can do a
piece of
works a times as
time each would require alone.
10.
A
list
in 15 hours
of
fast as B, find the
names can be typed
by another.
the work together?
How
in 5 days. If
A
hours by one typist and it take them to complete
in 10
long will
work
Chapter Five
FUNCTIONS AND GRAPHS
5.1.
Functions
Perhaps the most important word in mathematics is 'function.' This being the case, it is certainly advisable to study the meaning of the word as long as necessary to understand it. A function * of x is an expression which has a special value to go
with each value assigned
Examples. 2x in the next x
1;
x
men you
+
see,
to x.
3; x]
who
x
will
+
-;
the
number
5
be more than
of
men,
six feet tall.
a function of x? The reason Why, is that it has an associated value for each value we assign 1 = 1; for x = 2, it beto x. For x = 1, it becomes 2-1 for instance,
comes 2-2
1=3;
is
2x
1
and so on.
3 Similarly, examples of functions of y are 3y, y 3 some functions of z are z 2 -, etc. 1, 2z,
+
z
7,
-;
^
2t
be seen that any mathematical expression containing a letter is a function of that letter. Such quantities are called mathematical functions, and are illustrated by all but It will
one of the examples above. The one non-mathematical function there listed is 'the number of men, in the next x men you see, who will be more than six feet tall.' For every value of x there will be a definite value for this quantity (which *
Some
consider
functions of x have
them
more than one value
here.
87
for a given x;
but we need not
FUNCTIONS AND GRAPHS
88
makes
it
a function of x) but ;
that value will be.
An
we cannot
[Ch.
V
advance what
tell in
important goal in science
to express such related quantities as mathematical functions of each other. This goal has been reached
is
many times,
as for example thrown up vertically and the height it reaches, or the volume of a pound of specified gas and the pressure it exerts on the container. Again, accurate knowledge of the positions of the sun, earth, and moon as functions of the time has enabled astronomers
with regard to the
initial
speed of a stone
to predict eclipses far in the future.
5.2.
Notation concerning functions
A
symbol which can represent any function of x is f(x) read '/ of x.' If in a given problem f(x) stands for some particular function of x, then other symbols needed to represent other particular functions of x in that problem can be written variously as g(x} h(x), F(x) (read 'g of x' 'hoi x^ 9
y
F
of 2,') etc. 'capital = 2x2 3, then /(O), if, in a given problem, f(x) read '/of zero/' means 'the value of f(x) when x = 0,' and
+
Now
equals
2
O2
+
/(-2) = 2(-2)
3
2
+
=
3.
3
=
Similarly, /(I) 11, and so on.
=
2
I2
+3=
5,
A
function of two variables, say x and y, can be represented asf(x, y) and is read as '/ of x and y' Thus /(I, 2) y
would read '/of 1 and 2' and would mean the value = 1 and y = 2. f(x, y) for x
Example Answer.
Example
1.
-
(3
1)
2. If f(x)
Answer. (4
2) (3 3.
Example ,
=
If f(x)
If
-1).
+
=
+
3x
and F(x)
= -1 and g(y) = y
(-2)
4x
7)
1
=
f(x, y)
3
8
10
=
3z
= 2
8
+
=
x3
,
of
evaluate
= -9. 7,
find the value
80.
2xy
+
6,
evaluate
5.2]
NOTATION CONCERNING FUNCTIONS
- 2(1)(3) +6 = 3-6 + -1) = 0-2(0)(-l)+6 = 6; =
Answer. /(I, 3) /(O,
3(1)
2
-1)=3-6 =
,
EXERCISE
89
6
=
3;
18.
24
Given f(x)
=
+
2x2
3z
2,
find the values of the expressions in
problems 1-10. 3.
6.
/(-a).
=
+
3z 2
2 and F(z) expressions in problems 11-20. (ru>en /(z)
11. /(O)
-
12. /(I)
F(0).
15.
14. /(O)F(O).
17.
/(IW).
Gwrcn F(z,
+
2z 2
/(a
-
3,
^2 ^.
*J
10. /(a
6).
-
6).
find the values of the
-
F(l).
16.
19.
=
+
13. /(2)
F(l).
.
18.
?/)
9.
S-/^)
7./(~)-
=
/(-I).
G (x
V)
>
.
=
20.
L _ 3^ / ^ ^ l
w^^
of
(?(-5,
0).
e
the expressions in problems 21-25.
24. F(2,
-3)
23. F(0,
22. (7(0, -1).
21. F(0, -1).
+
-5)
25.
3G(-1, -2).
26. Express the area of a circle as a function of its radius,
r.
27. Express the area of a triangle with a base of 10 inches as a
function of
its altitude, h.
28. Express the area of a triangle
function of
its
base,
whose altitude
is
6 feet as a
b.
29. Express the area of a square as a function of its side,
30. Express as a function of t the in t seconds at 5 feet per second.
31. Express as a function of t
t
seconds at sixty miles per hour.
the
man
s.
number
of feet a
number
of feet a car goes in
walks
FUNCTIONS AND GRAPHS
90
5.3.
[Ch.
V
The coordinate system
The rectangular coordinate system is a simple and widely used device which enables us to show, by means of a picture, or graph, the manner in which the value of a function of x changes along with the value of x. Because it was invented by a seventeenth-century Frenchman named
it is
Descartes,
sometimes called the cartesian coordinate system.
P(*.y)
IV
III
Fig. 5
The device consists of two mutually perpendicular lines, one horizontal and one vertical * (Fig. 5), drawn on a plane. The
horizontal line
called the X-axis; the other, the Y-axis; and, together, they are the coordinate axes. They divide the plane into four parts called quadrants, which are
numbered as shown
is
in the figure. Their intersection, the
called the origin. point 0, is located point such as is
A
P
OA
distances
zero
when
is
P
AP
or y
of
The
two
hori-
is
P
is
This
and
in the figure.
the abscissa, the x-distance, or simply the is on the right of the F-axis, positive when
zontal distance
x of P, and
or x
on the plane by means
on
this axis,
and negative when
P
is
to the
conveniently indicated for memory purposes by the arrow on the right end of the X-axis, which shows the positive horizontal direction. Similarly, the vertical distance left of it.
is
the ordinate, y-distance, or y of P, and is positive, zero, or negative according as P is above, on, or below the X-axis, and also as indicated by the arrow pointing upward on the is
* In this statement as well as in the subsequent discussion, we assume for simarrow on the F-axis points upward, as on a vertical blackboard.
plicity that the
THE GRAPH OF A FUNCTION OF
5.4]
x
91
F-axis. Together, x and y are called the coordinates of P. It is in any quadrant important to note that the coordinates of are plus x and plus y, though x stands for a negative number
P
quadrants II and III, as does y in quadrants III and IV. Beginning at the origin, we mark in advance equal units on each axis on a scale to fit the problem. Then, in locating, or plotting a point with given numerical coordinates, designated here as x and y, we measure x units to the right in
or left according as the #- value
reaching a point on the #-axis.
positive or negative, thus
is
From
this point
we measure
downward to the designated point. Thus, ( 2, 3) we measure 2 units to the left of the origin and then 3 units downward. The point A, along with others, is shown in Fig. 6. Note that a capital letter before
y units upward or to plot
A
the designated coordinates, while not necessary, when the point is to be referred to again.
is
convenient
r -0(0,3)
I
>
*
I
D{-4,2)
I
<
b(o,o) H 1
1
B (3 2 )
*-/
1
(-3,0)
A (-2 ,-3) Fig.
5.4.
6
The graph of a function of x
Using the coordinate system,
let
or
picture of the special function z
may
y
(i) *
designate this function of x
=
+
+
us find the geometric 1-
by the
For convenience we letter y, so that
i.*
The meaning of the word function is here illustrated, thus: Choose any numTake one-half of it. Add 1 to the result. Call the sum y. Thus, the value of
ber x.
y depends upon the original value chosen for
x.
FUNCTIONS AND GRAPHS
92
By means
of (1)
we may
[Ch.
V
find the value of y to pair with x, listing them in a brief
each of several small values of table thus:
x
-2
-3
-1
1
2
3
V
The
points whose coordinates are listed are seen in Fig. 7 to lie in a straight line. It can be shown that all other points obtained from (1) lie also on this line. The line is said to be or the graph of the function -
+
1,
and
also the
graph or locus
of equation (1) as well as of
2y
(2)
obtained from
(1)
by
=
x
+
2,
clearing fractions.
K5)1
f
2 y=2x -3x
J(I,0) Fig. 7
Fig.
2 Similarly, the graph of the function 2x
8 3#, or of the
equation (Q (3)
4,
y
turns out to be the curve
The
O~, OX
9~.2 LX
shown
in Fig. 8.
coordinates of the points plotted are
shown
following table.
x y
^1
1424 20-1-1025
-i
i
in the
VARIABLES AND CONSTANTS
5.5]
EXERCISE
93
25
Plot the points having the following pairs of coordinates. (Remember number is the value of x.)
that the first 1.
2. (1, 4).
(3, 1).
6.
(0, 0).
7.
10. (0,3).
11.
Make a
3. (2, 3).
4. (3, 5).
5.
1).
8. (3,
-4).
9.
(-2,0).
12. (0,
-1).
13.
(-2,
and
table of coordinates
plot the
(2, 6).
(-1, -2). (4,0).
graph of each of the follow-
ing functions. (Hint: Let y equal the function.)
26. x 2 29. 2
-
32. 2x 2
35.
38.
x
-
2
30. 1
3x.
33. x 2
-
36.
x
~x2
39.
x
28. x 2
- 2x - 4x 2
.
x
5.3.
+ 2.
27. x2
.
31. x
.
+
1.
34. x 2 37. -
x
1
^r-
2
40.
1
x
-
2.
+ x - 1. + 2x - 2.
+
1.
-
1.
Variables and constants
When
a
letter
may
take on an unlimited
number
of values
in a problem, as x or y may do in equations (1), (2), and (3) of the preceding article, it is called a variable. If the equation is solved for y thus being in the form y
(1)
y
=
/(*),
called the independent variable, and y, or /(x), the dependent variable. Similarly, in the equation
x
is
(2)
*
=
FUNCTIONS AND GRAPHS
94
[Ch.
V
the independent variable. To find corresponding pairs of values it is convenient to assign values to the independent
y
is
variable.
A letter representing the same number throughout a problem is called a constant. Constants and variables are usually taken respectively from the first and last parts of the alphabet. By common agreement some constants frequently represent the same number
many
in
different problems, as in the case of
TT,
the
ratio of the circumference to the diameter of a circle.
The graph of a linear function of x
5.6.
A
a rational integral function of the where a and 6 are constants and a ^ 0.
linear function of
form ax
+
6,
Examples. 3x It is
shown
a:
2, x,
in the
is
o
,
-
5
2i
branch of mathematics called analytic
geometry that the graph of the function ax
+
6,
or of the
equation
y
(1)
=
ax
+
6,
Evidently it crosses the F-axis at the point (0, 6). The second point needed to determine the line may be found by assigning any value to x at will. Thus, 1 goes through (0, 1) and (2, 3); y = 2x, through y = x a straight
is
line.
+
and
(0,0)
5.7.
(3, 6).
The graph of a linear equation in one or two
The
general linear equation in x or y or both ten in the form
ax
(1)
where zero.
a, 6,
and
c are
+
by
=
variables
may
be writ-
c,
constants and both a and 6 cannot be
THE GRAPH OF A LINEAR EQUATION
5.7]
95
When two
equations are so related that all pairs of values which satisfy one also satisfy the other, they are said to be dependent. Evidently, from the definition, equations have the same graph.
+
Example, x
by
=
y
and 2x
1
+
=
2y
two dependent
2 are both satisfied
-l),etc.
(0, 1), (1, 0), (2,
The operations
leading to dependent equations are essentially the same as those yielding equivalent equations in one variable (Art. 4.5).
That
is,
terms
may
be transposed, or both members
may
be multiplied or divided by any number or constant except zero. If b
^
in (1), the solution for y,
/o (2)
y=-
a
+ lX .
namely
c -,
involves operations of this sort, so that (2) and (1) are dependent and their graphs are the same. Since y is a linear
function of x in If b
=
0,
(2), this
graph
^
then a
by
a straight
is
line.
the statement following
(1),
so
that (1) takes the form
ax
(3)
whose graph
is
=
the vertical line through (-,
For evidently the value of y
etc.
c,
is
(A
(-, 1
J,
(-,
not restricted in any
2),
way
/
by
(3),
while x
Example. 2x allel
is
restricted to the
=
3,
to the F-axis
Similarly,
if
a
or x
a horizontal
f, is the
and f units to the
=
0,
so that
by
(4) is
=
one value,
6^0, =
equation of a
line par-
right.
the graph of
c
line.
Example. The 2 units below it.
line
y
=
2
is
parallel to the X-axis and-
FUNCTIONS AND GRAPHS
96
Our argument,
[Ch.
V
then, leads to this conclusion:
The graph of a linear equation in x or y or both
is
a straight
line.
EXERCISE
26
Plot the graph of each of the following equations. 1.
3. 5. 7.
9.
11. 13.
15. 17.
-
= 0. x + y - I = 0. 2x - y = 2. 4z - y + 4 = 0. 3z + 2y - 6 = 0. 2z + y - 4 = 0. 2x + 3 = 0. = -1. s=0. x
12.
+ y = 0. x - y + 1 = 0. 3x + y - 6 = 0. 2x - 3y + 6 = 0. 2z - y + 4 = 0. 3z + y = 5.
14.
5y
16.
y
18.
y
2.
y
4. 6. 8.
10.
a;
x
- 1 = = -5. = 0.
0.
19-30. Solve each equation in problems 1-12 for y in terms of x,
and then solve
it
for
x
in
terms of
y.
9C =
Given 5F 160, the relation between the Fahrenand centigrade (C) temperature readings, express: (a), F as a function of C; (b), C as a function of F; (c), 5(F + 40) as a 31.
heit (F)
function of C.
Given 7 = Prt, express: (a), P as a function of 7, r, and t a function of 7, P, and t and (c), t as a function of 7, Pj and r. 32.
(b), r as
5.8.
Graphs of non-mathematical functions
*
and suggestive to make a graph showing the relation between two quantities even though it may not yet be possible to express one as a mathematical function of the other. In business, science, war, and practically every Often
field
of
it is
helpful
human
activity,
graphs or charts are used very
extensively. * This article
may
be omitted without loss of continuity.
5.8]
GRAPHS OF NON-MATHEMATICAL FUNCTIONS
97
For example, the meteorologist, or 'weather-man,' might record the Fahrenheit temperature in a given city for every third hour of the day, getting the following result :
Time
A.M.
4
Temperature
3
6
9
P.M. 12
6
6
5
20
25
35
9
12
14
6
40V 30
0)
fw I
i
i
-10 Fig.
The graph
9
with the time as the independent variable shows clearly how the temperature has changed during the day. In this case it is advisable to draw a smooth curve through the plotted points to indicate that the tem(Fig.
9)
perature changes continuously, or from moment to moment. On the other hand, if we record quantities such as the yield per year of an acre of wheat, the smoothly curved line would
be meaningless, and a broken line such as that in Fig. 10 is
better. b 40 o
-ofe '5
a
30
>J2 20 o>
-5
a.
10
o 1930
1935 Year
1940
Fig. 10
The types
of graphs used in practice are
much
too nu-
merous to be shown here. Many of them, however, are based upon some modification of the coordinate system. Perhaps the most frequently used independent variable is time, which
FUNCTIONS AND GRAPHS
98
[Ch.
V
be measured in seconds, minutes, hours, days, years, etc. It is interesting to note that sometimes in this type of graph, the results may be carried more or less reliably into
may
the future.
EXERCISE
27
Graph the relations indicated by the data in the following problems. Use the type of curve (smooth or broken) which seems to be more appropriate for each problem. 1. Given the following readings, graph the temperature as a function of the time. Can you suggest an explanation of the unusual conditions here indicated?
Time (A.M.) Temp. (F)
4
5
6
7
8
9
10
11
12
60
62
65
69
75
75
65
55
55
Given the following data, graph the yield per year tain farm in bushels of corn. 2.
Year Yield 3.
1920
1924
600
750
1932
600
550
Graph the indicated weight
Age (years) Weight (Ibs.) 4.
1928
A river
12
3
25
33
15
of a
1936 510
of a cer-
1940
1944
675
790
boy as a function
456789
39
44
49
54
of his age:
68
60
gauge registered as follows. Graph the water
10
77
level as
a function of the time.
Hour Level
(A.M.) (feet)
4
5
-&
6
f
7
8
f
2
9
10
f
1
11
12
i~A
Chapter Six
SIMULTANEOUS LINEAR EQUATIONS
6.1.
Problems leading
to
simultaneous equations
Sometimes we are confronted by a situation in which two unknowns are related in two separate ways. In this case the rules of algebra enable us to find the values of the unknowns. Example. After a storm, a mixture of water and oil filled a 100-gallon cask. The oil would eventually rise to the top, where it could be skimmed off; but it was so heavy that the process would take longer than the buyer wanted to wait. It was found that the liquid weighed 800 pounds, exclusive of the cask. It was known that the water and oil weighed about 8.4 and 7.4 pounds per gallon respectively. The deal was completed at once after a short algebraic calculation.
How? x and y be the number of gallons of water oil, respectively, in the cask. We have at once, since there were 100 gallons altogether, Solution. Let
and
x
(1)
Also, since
of oil
respectively,
SAx
+ 7 Ay =
Multiplying both sides of (3)
100.
x gallons of water and y gallons
and 7 Ay pounds, (2)
+y=
7.4z
(1)
by
+ 7 Ay = 99
800. 7.4,
740.
we
get
weigh 8Ax
SIMULTANEOUS LINEAR EQUATIONS [Ch. VI Subtracting the two members of (3) from the corresponding ones of (2), we find that x = 60; and then, from (1), that - x = 100 - 60 = 40. Hence there were 60 gallons y = 100 of water and 40 gallons of oil in the cask. Statements (1) and (2), which together carry in disguise the values of the unknowns x and y, are called simultaneous 100
equations. Just as
one unknown may be looked upon as a sentence-question, so a set of two or more simultaneous equations may be considered as a compound question instead of a group of sentences. Taken together, they propose a question and provide the answer, in concealed form, to
an equation
anyone able to
find
in
it.
6.2. Algebraic solutions of
simultaneous linear equations
The
solution of (1) and (2) in the preceding article was ' obtained by use of a combination of the addition and suband ' substitution ' methods. traction shall now illus-
We
7 '
trate these formal procedures as applied to the simultaneous
equations
:
(1)
3z
-
4y
=
2,
4z
+
3y
=
11.
and (2)
A. Addition and Subtraction Method (Note. The meanings of the directions in parentheses below are as here indicated: (1) X 3 means that both members
+
of equation (1) are multiplied by 3; (3) (4) means that of and members equations (3) (4) are added; corresponding
and
so on.)
(4)
(2)
X X
(5)
(3)
+
(3)
(1)
3
4 (4)
Qx 16x
-
+
12y 12y 25x
= = =
6.
44.
50.
SOLUTIONS OF SIMULTANEOUS LINEAR EQUATIONS
6.2]
101
Hence
9y 25y
= = = =
y
=
1.
=
11.
x
(6)
(7)
(1)
(8)
(2)
(9)
(8)
X X -
4
-
12x
IGy
+
I2x
3 (7)
2. 8.
33. 25.
Hence (10)
The
x
=
4x
+
is:
solution, then,
2;
t/
=
1.
B. Substitution Method (11) Solve (1) for y. (12) Substitute in (2).
2 .- 4y
=
y
=
x
(14) Substitute in
3
(1).
Hence
=
2)
4
(13) Solve (12).
Solution: x
~
3(3 *
2,
y
=
2. 2. 1.
I.
C. Combination Method
The method used
in solving the illustrative
Art. 6.1 consists of solving for one
example
in
unknown by method
A,
and then substituting this value in either equation to find the second unknown. This method is a combination of the other two.
should be kept clearly in mind that the solution of two simultaneous equations consists of a pair of values which satisfies both equations. Applying this test to our solution of It
(1)
and
(2),
3-2-4-1 = 4 2 + 3 1 =
(15)
we
2 (correct)
;
11 (correct).
= 2, y = 1) is a solution. Furthermore, pair (x shall show in Art. 6.3, it is the only solution.
Hence the as
we have
SIMULTANEOUS LINEAR EQUATIONS
102
[Ch.
VI
But why, it may be asked, does equation (5) give us the x-value of the equation pair? How can we add the I2y of of when we know from to the the (4) (3) +I2y graphs of (3) and (4) that y can have any value in either equation, so that the two y's may be representing different numbers and hence need not be equal?
The answer may be
stated thus
:
solving simultaneous equations we consider the letters not as variables but as unknown constants. When this is done, all
In
Thus x and y are the same constants in each of equations (1) to (10), and their numerical values are exposed
is
explained.
and
(10).
choice of the
most
in equations (6)
The
a given problem first
efficient of
the three methods for
a matter of judgment. In general, if the is an integer or a very simple algebraic
is
unknown found
expression such as 2a, 36, etc.,
unknown by
quicker to find the second substitution; otherwise, by addition and subit is
some trivial cases, such as 2x = 3, 4y = 7, none the three methods applies; but here, of course, the solu-
traction. In of
= f, y = When some of
tion,
x
,
is
seen
by
inspection.
the coefficients are
literal,
method
usually preferable.
Example. Solve for x and y the following equations
ax dx
(16) (17)
+ by = + ey = /.
c,
Solution. (18) (19) (20)
(16) (17)
(18)
X X -
e
aex
b
bdx
(19)
aex
bdx
= = =
bd)x
=
x
=
+ +
-
bey bey
ce. bf.
ce
-
ce
-
&/,
or (21)
(ae
-
bf.
Hence (22)
^-j*. ae bd
:
A
is
.3]
GRAPHICAL SOLUTION OF LINEAR EQUATIONS
23)
(16)
24)
(17)
25)
(23)
X X -
d a
adx adx bdy
(24)
26)
+ + -
(6d
aey
= = =
ae)y
=
y
=
bdy aey
103
cd. af.
cd
cd
-
a/,
af.
lence 27)
The
-af bd
solution is the pair (22) and (27). It direct substitution in (16) and (17). >y
i.3.
or
ae'
may
af ae
cd
6d
be checked
The graphical solution of simultaneous linear equations
we graph
the lines (1) and (2), Art. 6.2, we find that hey intersect at the point (2, 1), Fig. 11. This point lies on >oth lines, and hence its coordinates must satisfy both If
iquations.
Fig. 11
In general, to solve graphically any two linear equations; vith numerical coefficients, we draw the two straight lines nvolved and then find by inspection of the figure the co>rdinates of their
common
point, or point of intersection,
they have one. These estimated coordinates should be learly the same as the correct solution-pair found algef
SIMULTANEOUS LINEAR EQUATIONS
104
[Ch.
VI
braically; and of course they would be exactly the same if the figure were perfectly accurate. Thus, the graphical solution is a check on the algebraic one, and vice versa.
And now new light is thrown on algebraic appear
in
some attempted
solutions.
coincide or be parallel,
may intersect,
difficulties which For two straight lines and each of these three
possibilities leads to a different type of algebraic result. If they intersect, they do so in a single point, and there is one
and only one
they coincide, there are evidently infinitely many solution-pairs. Finally, if they are parallel there is no common point, and hence no solution. In these respective cases the simultaneous equations are called solution-pair.
If
independent, dependent, and inconsistent. To test these possibilities graphically,
merely draws the
A simple
lines.
of
algebraic test
course, one the follow-
is
ing:
Let any two linear equations in two unknowns be written so that their right members contain all terms not involving the unknowns. Then, if their left members cannot be made identical by multiplying one equation by a constant, they are independent. When the left members can be and are
made
identical,
they are dependent when the right members
are then equal, and inconsistent otherwise.
Example
1.
The equations 2x
(1)
+
y
=
3,
and
x-3y =-2
(2)
are independent, since the tical.
The
solution-pair
Example (3)
2.
left sides
is (1, 1).
The equations 2x
+y=
3,
and (4)
4z
+
2y
=
6
cannot be
made
iden-
6.3]
GRAPHICAL SOLUTION OF LINEAR EQUATIONS
are dependent, since (3) can be
made
105
identical with (4)
common
multiplying its members by 2. Points on the simultaneous algebraic solutions, are (0, 3),
line,
(1, 1), (2,
by or 1),
etc.
Example
3.
The equations 2x
(5)
+y
=
3,
and 4x
(6)
+
2y
=
7
are inconsistent, since when the left members are made identical the right members are unequal. Clearly no pair of values for x and y can make 4# 2y equal to both 6 and 7.
+
EXERCISE
28
Solve by the addition 1.
x
x 4.
7.
10.
13.
+ 3y = -
9x 8z
5y
+
and subtraction method.
11,
2.
3x x
8,
5.
9z 9z
= - 13.
7y
- 9y = -69. - 1, 9o: - 5y = = 5. 10s 8y = 2x + Qy -3, 3x - 5y = 6. 7z - 2y = 15, 60: - y = 10.
8.
11.
14.
+ ty = -1, + 5y = 7. + 8y = 3,
3.
6.
- 8y = -77. 5x - lly = -4, 6x - 8y = -10. 15x + 4y = 7, 6x + Uy =9. x + 3y = 9, 4x + 5y = 22.
9.
12.
15.
-
= 4, = 4x + 53. 7z + 3y = 2, Sx + 7y = -2. 7x + 3y = 4, 8x + 7y = 26. 3x + 5y = -9, 4z - 3y = 17. 2x - 3 = 5y, y + 5 = 3x. llx
y substitution.
= 6, 2 X - y = 7. 5z = -10, 18.' 3x + 4y = 2. 20. 7z + 6y = -11, Sx - 5y = -60. 22. IQx + I2y = -7, 8x + 9 = -8. 16.
3z
17.
3y 2y
0,
6. + Z + 4^=16, 2x + 3y = 17.
3x 19.
= =
21. 6z
-
9z
+
Qy 5y
23. 2x
-
3y
5x
+
8j/
= =
= =
-7, 8.
-29, 5.
Qy 15y
SIMULTANEOUS LINEAR EQUATIONS
106
- 14y = -15, lOx - Qy = -5. 4z = 12, 3z = + 3.
24. I5x
26.
25.
4x 27. 2x
Gx
2/
[Ch.
VI
= -9, = 2. + = -f 9y 13, -7y = -63. 3y 2y
28-39. Graph the pairs of equations in problems 16-27, and estimate the coordinates of the points of intersection. Compare with the algebraic solutions. Solve algebraically.
49.
az
+
bx 51.
53.
ax ax
by
ay
+
to/
by
3x= 5y =
= = = =
50.
ab,
b
2 .
a:
+ =
2a,
x
y
26.
2a,
52. az
26.
ax 54.
4, 7.
2/
=
+ 6y =
2a2
by
ab
=
+ a6 +6 2
.
5az = -10, 3by = 18.
/n problems 55-63, w/iic/& pairs o/ equations have no solutions and which have more than one solution? 55.
58.
3x 6z 5x
6y 61. ax
ax
= =
= 5, = 20. 8t/ 7 3y, - x). 10(1 + 6t/ = 5, + 6y = 10. 4y
56. 2x
4z 59. 6z
3x 62.
-
ax 36y
= 4, = 8. = 10j/, = 5y.
3y 6y
57.
5
60.
2
c
=
3c
by,
3ax.
2x 2y
= =
14
28
-
y,
4z.
2x=l+32/, 6x-9t/=3. 63. ax by =0, = 0. ax by
THREE EQUATIONS AND THREE UNKNOWNS
6.4]
64.
x
+^=
2,'
3
_
2
7
y
x
=
3'
y
HINT. Consider the
become: u Solving,
+ 3v =
we
=
x
y
=
, 4 66. -
~
x
2v
=
x
y
The equations then
and y
where u
-,
3
u=l,v=-, 6
1,
--- =0, L
,.*
x
as -
unknowns
2 and 3u
find that
final solution is:
65.
107
or
=
- and x
-=!,-=y&
v
= y
Hence the
x
3.
o +-=
_
,
^
67.
5,
y
4 -
x
o +-= .
y
-
5,
70.^ + ^=1, x '
y
2a3b_ - ^ x y
6.4.
Three equations and three unknowns
Consider the simultaneous equations
+
2x + y 4* - y
(1)
(2)
60;
(3)
We (2),
seek
and
now a
(3)
as constants
+
2y
2z
-
+
z
3z
= = =
set of three values
l,
3,
2.
which
will satisfy (l) r
simultaneously. Again we consider x, y, and z whose values are fixed in advance, so that we
may combine them
at will.
' guiding principle will be found helpful to avoid ' working in a circle/ to wit: Eliminate one unknown at a time, using all of the equations in which it appears, thus reduc-
Here a
'
'
ing the problem to one with fewer unknowns and fewer equations. For example, to solve equations (1) to (3) we
may
eliminate z
first (if
we
wish), thus:
SIMULTANEOUS LINEAR EQUATIONS
108 (4)
(1)
unchanged
(5)
(2)
X
2
(6)
(4)
+
(5)
Equation
(3)
+ +
2x y 8x 2y
- 2z Ite - y
must now be used,
(7)
(3)
unchanged
(8)
(2)
X
3
(9)
(7)
+
(8)
2z
= = =
[Ch.
VI
1.
6. 7.
either with (1) or with (2).
+
Qx I2x
-
+
3z 2y 3z 3y 18x - y
= = =
2.
9.
11.
There remain two equations and two unknowns, (6) and (9). y = 2. Substitution of these Solving these, we get x = ,
values in (1) yields 2=1. The tentative (or untested) solu= 1. The test is not complete tion is then: x = %, y = 2, z until it is shown that these values satisfy all three original equations. If it
a
is
one of the simultaneous equations, usually advisable to eliminate this letter first. Thus, letter is missing in
given (10) (11) (12)
.
2x - 3y - z x + 2y + 2z
2,
y
2,
= = -3z =
1,
x should be removed from (10) and (11), and the resulting equation can then be used with (12) to get y and z. Thus far we have considered equations with two and three unknowns. In general, if the number of simultaneous equations is the same as the number of unknowns, it is often (though not always) possible to find a set of values which will satisfy all of the equations.
above equation
EXERCISE
The guiding
(4) holds good in
principle stated
all cases.
29
Solve the following systems of equations. 1.
x x x
+ y + z= +y-z= y + z =
G, 0, 2.
2.
x
x x
-y + z= +y-z= y
z
=
Q, 2, 4.
DETERMINANTS
6.5]
x
-
x
+y
x
y
3.
5.
7.
y
+z= - s= z =
2,
+ 4y + - 2y + 3z + y + 2x - 3y + 4z
6.
1,
= 2, = 3. = 4, = 3, z = 5. 32 = 5, 2z = 3, 5z = 4.
8.
10.
3x
12.
7,
A+2B
2,
x 2x
15.
C= -C=
6.5.
A
y
2z
z
-w
2z
+w
+
-
+
+y
3, 4, 5,
1,
2, 1,
2w
0.
=
3,
N=
f,
iAT
L
4
+y+ = z
2x = 2x y = + y 2z =
14.
-2B + 2C = 12. x + y^ x+z_ y 3
x
5y
-
x y
+ 2w = 3, x + 2w = 4. A + 4B + 3A
-
$x
= 5, = 3, = 2. + 3z = 4, - 2z = 5,
+ 3y Zx + 2y + z=2.
z
13.
2,
2x - 3y + z x + 2y - 2z 3x + y - z 5x
2y=l, y + 2z=2,
11.
0,
x-y + z=2.
4.
2x - y - 2z x - 2y + 2z 5z + 3y - z 4z - y + 2z 3z
9.
4.
4,
+ 2y + z =
a;
109
x + y + z = z = x y
2
6.
Determinants
a set of numbers or expressions arranged in rows and columns, having the same number of rows as of determinant
is
columns, and enclosed within two vertical bars.
Example.
2-1 5 3 2-1 1
The
-2
vertical bars
may
4 be considered the mark of a deter-
minant. The numbers between them are called elements.
The
SIMULTANEOUS LINEAR EQUATIONS
110
[Ch.
VI
whole expression has a definite value found by rules which will be explained below. If the determinant has two rows and two columns, it is said to be a second order determinant. If there are three rows and three columns, it is of the third order. In any determinant, the order is the same as the number of rows or of columns. The value of a second order determinant is the product of the upper left and lower right elements minus the product of the lower left and upper right elements. In other words, we find the product of the numbers read diagonally downward and subtract the product of numbers read diagonally upward always from left to right.
= ad
be.
The value of a third order determinant may be found in a simple manner when a fourth column like the first one and a fifth like the second are placed to the right of the determinant. The value is the sum of three products of elements read diagonally downward to the right minus the sum of three similar products read diagonally upward.
=
(aei
+ bfg +
cdh)
(gee
+
hfa
+
idb).
looking at the letters in the answer as read from the original determinant, the student can see easily how to
By
evaluate the determinant without adding the two extra columns, if he so desires. He might try this shorter scheme in
checking through the next example.
6.5]
DETERMINANTS
Example
111
4-
3-12 1
2
3
2
-3
1
-[(2)(2)(2)
+
+
(-3)(3)(3)
= [6 + (-6) + (-6)] - [8 + (-27) = -6 - (-20) = -6 + 20 = 14.
+
(-1)]
These short-cut methods of evaluating second and third order determinants do not apply to those of higher order. The method which does apply to determinants in general is called 'expansion
ing example
Example
by minors.'
It is illustrated
by the
follow-
:
5.
a(ei
=
aei
b(di
hf}
+
bfg
+
gf)
cdh
+
c(dh
bdi
ahf
ge) cge.
In the above expansion the second order determinants are called the minors of the letters before them. c,
for example,
is
the determinant
d h
g
,
The minor
of
found by striking
out the row and column of the original determinant in which c appears; and the other minors are found similarly.
The
sign in front of the letter
minor
in the expansion
is
which
found by
multiplied by its starting with the upper is
element and calling off the signs alternately as 'plus, minus, plus, minus,' each successive element reached being immediately to the right of, or below, the one just called. If
left
an element comes on the minus count, its sign is changed; otherwise not. To illustrate, the determinant in Example 5 may be expanded by minors as above or by using the elements of any row or column as multipliers. If the second column is used, for instance, the expansion is
SIMULTANEOUS LINEAR EQUATIONS
112
[Ch.
VI
-h
-b
The value of the determinant is always the same, regardless of the method of expansion, as the student may verify. EXERCISE
30
Evaluate the following determinants by the 'diagonal multiplying' method.
22-32. Check the results in problems 11-21 by the method of expansion by minors.
6.6. Solving systems of equations
by use of determinants
any unknown in a system of equations may be represented by the ratio of two determinants.
The value
of
Example. Solve the system, 2x x
3
+
1
= =
y,
3y.
SOLVING EQUATIONS BY USE OF DETERMINANTS
6.6]
Solution. First, arrange the equations thus
:
2x ~
JO
Then
x
=
and
y
=
-10 =
-5 = -5
y OU
113
=
Q*.
3, 1
1.
2,
1.
be noted that the determinant in each denominator is composed of the coefficients of x and y arranged in the same order in which they occur in the equations. The determinant in each numerator is the same except that the It will
numbers to the
right of the equality signs replace the coefficients of the letter whose value is sought.
The student may
verify the solutions given for the equa-
~~
^ = in section 6.2 C by now using the dedx ef // terminant method for finding the values of x and y. The tions
(
latter
method
7
+ ,
gives
C
x
= /
e
ce
a
b
ae
d
e
bf bd' /
ni
y
=
a
c
d a d
f
af ae *^
b
dc .
bd
e
method will give the correct results for values of the coefficients a, 6, c, d, e, and /.
It follows that this *
allowable
all
unknowns may be
Systems of three equations in three solved in a similar way. Example. Solve the system 3x 2x
fc
+
2^/
{y
x,- y
+
-
z.
2z
= = 1; =-2. 3*,
bd = 0, the values given for x and y are meaningless, since Note that if ae by zero is not permissible. In this case the equations are inconsistent or dependent, and should be studied by the method explained in Art. 6-3. *
division
SIMULTANEOUS LINEAR EQUATIONS
[Ch.
-18+4+1-6+3-4 = -20 _ -18-2+2+3+3-8 -20 ~~'
-6+3+4+1+6+12 = 20 (evaluated above)
(x
=
1,
?/
=
1,
z
=
i JL
= -1.
-40 = -20
-20
Answer:
20 20
VI
2.
2).
Note that the denominator in each case contains the coefficients arranged in their normal order. The only difference between the numerator and the denominator in each fraction constants on
that, in the numerator, the the right sides of the equations replace the
coefficients of the letter
this
method
is
whose value
to the general system:
+ by + cz = + ey + fz = + hy + mz =
ax dx gx it
follows that
x
=
N,.
D'
z
= D'
k I
n,
is
sought. Applying
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
115
where k
b
c
I
e
f
n
h
m
a d
b
k
e
I
g
h
n
D* =
and
EXERCISE
31
1-27. Solve problems 1-27 of Exercise 28
by use
of determinants.
28-36. Solve problems 1-9 of Exercise 29 by use of determinants.
6.7. Stated
Often
problems with more than one unknown
more
a problem stated in English into a set of simultaneous equations than it is to solve the equations. Therefore, a consideration of the genit is
difficult to translate
eral principles applying to all such
problems should prove
helpful.
The
essential steps in solving
a stated problem are two-
fold. (1)
The unknowns
nated by
must be identified, desigw, etc.) and described clearly.
to be sought
letters (as x, y,
z,
Examples of unsatisfactory Let x
=
starts.
time, or length, or rate, or distance, or amount.
Examples of
satisfactory starts.
Let Let Let Let Let For the
case,
D -
x x x x x
= = = = =
0, see
no. of minutes after 3 P.M. no. of inches in length. no. of feet per second. no. of miles in distance.
no. of bushels of corn. the discussion in the preceding footnote.
SIMULTANEOUS LINEAR EQUATIONS [Ch. VI (2) If there are two unknowns, two equations must be found expressing two relations between the unknowns which 116
should be described in the problem. If there are three unknowns, three relations must be indicated, and so on. Thus equations as there are unknowns will be obtained, and the problem will be 'set up/ or ready for the formal
many
as
7
algebraic solution.
Example. The combined age of John, Bill, and Harry is 36 years. Two years ago Bill was three times as old as John. Eight years hence Harry will be twice as old as Bill. How old is each? Solution. In view of the question at the
end we can
start
confidently thus:
x
Let
y z
At
this point
= = =
no. of years in John's age; no. of years in Bill's age no. of years in Harry's age. ;
we
re-read the
sentence and find that
first
states in English one relation between the goes into algebra thus
it
The statement
x
(1)
+y+z
The second sentence
y
(2)
=
36.
two years ago, which 2. Thus 2, and z y
deals with ages
evidently are represented
by x
-
2
=
2,
3(x
-
Eight years hence the ages will be x According to the third sentence, z
(3)
Solving
y
=
and
(1),
unknowns.
:
(2),
and
+
8
(3)
=
2(y
2).
+
+
8,
y
+
8,
and z
+
8.
get x
=
4,
8).
simultaneously
we
=
24 (answer). Not every problem has a direct question to indicate the 8,
z
unknowns and separate sentences to describe the conditions but these quantities must always be described in one way ;
or another.
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
117
It is interesting to note that impossible simultaneous conditions will be revealed algebraically by inconsistent equations; while conditions which may seem to be inde-
pendent, but which really amount to the same thing, lead to dependent equations with more than one solution.
EXERCISE
32
Three times one number minus twice another equals of twice the first number plus three times the second is Find the numbers. 1.
4.
The sum
7.
5 times one number
2. If
result is 3.
The sum
1.
of the
is
by 3 times another, the
divided
numbers
is 8.
Find the numbers.
sum of two numbers Find the numbers.
One-half the
difference
is 6.
is 6,
and 3 times
their
Three times one number is equal to 4 times another, and half their sum is 7. Find the numbers. 4.
5.
Find two numbers such that 3 times their difference
and the sum 6. If
of 4
and the
first
a certain number
number
is
is
added to the numerator and sub-
tracted from the denominator of ^, the result equals 3. The of half this number and a second one is 7. Find the numbers. 7.
The rowing
rate
is 6,
3 times the second.
down a
certain stream
The
sum
1
m.p.h. less upstream is 3 times is
than twice the upstream rowing rate. the rate of the current. Find the rate of the current and the rate of rowing in Let
still
water.
x y
Then
rate
+y
= = = = =
(1)
x x x
(2)
x-y=
y
+y
The simultaneous
no. m.p.h. rowed in still water. no. m.p.h. the current flows. no. m.p.h. no. m.p.h.
2(x
-
y)
rowed downstream. rowed upstream.
-
1.
3y.
solution of (1)
and
(2) yields:
x
=
4,
y
=
1
(answer).
The
rate of rowing downstream is 2 m.p.h. less than 3 times the rate of the current. The upstream rowing rate is ^ the rate of the current. Find the rate of the current and the rate of rowing in 8.
still
water.
SIMULTANEOUS LINEAR EQUATIONS
118
[Ch.
VI
9. A man can row downstream 5 miles in the same time it would tak him to row 1 mile upstream. The rate of rowing in still water is 1 m.p.h. more than the rate of the current. Find the two
last-mentioned quantities. 10.
The
rate of a boat in
still
water
3 times the rate of the
is
current and also 2 m.p.h. more than this rate. Find the two rates.
A
plane can travel 350 m.p.h. with the wind, and its speed the wind is 320 m.p.h. Find its speed in still air and the against of the wind. velocity 11.
12. A plane goes twice as fast with the wind as against it. If the velocity of the wind were doubled, the speed of the plane when flying with the wind would be 100 m.p.h. more than 3 times its
speed against the wind. Find the speed of the plane in the velocity of the wind. 13.
The sum
of the digits of a 3-digit
digit is twice the
versed, the
hundreds'
new number
is
digit. If
number
is 6.
still air
The
and
units'
the order of the digits
is re-
99 more than the original one. Find the
number.
= = u = + KM + u = h + + u = u = + Wt + h = h
Let
t
100A
Then,
t
(3) (4)
lOOu
(5)
Solving
and u 14.
=
2,
the hundreds' digit; the tens' digit; the units' digit. the number. 6.
2h.
100A
+
10*
+ u + 99.
simultaneously, we get h so that the number is 132 (answer).
(3),
The sum
(4),
and
(5)
=
1, t
=
3,
number is 7. If the order number is 2 more than twice
of the digits of a 2-digit
is reversed, the resulting the original number. Find the number.
of the digits
15.
of the digits of a 3-digit number is 12. The tens' units' digit 1 more, than the hundreds'
less, and the Find the number.
digit digit.
16. digit
The sum is
1
The sum of the digits of a 3-digit number is 14. The units' more than twice the hundreds' digit and 1 less than twice
is 1
the tens' digit. Find the number.
PROBLEMS WITH MORE THAN ONE UNKNOWN
6.7]
a certain 3-digit
17. If the order of the digits of
new number exceeds
versed, the
the old one
number The
396.
by
and the sum of the Find the number.
digit is 3 times the hundreds' digit, 1
more than twice the
tens' digit.
The
length of a rectangle in inches is width. If its length is decreased by 1 inch and 18.
by 3
inches,
19.
The
width. If
creased
it
becomes a square. Find
its
119 is
re-
tens'
digits is
than twice its width is increased
1 less
its
dimensions.
length of a rectangle in inches is 2 less than twice its length is decreased by 2 inches and its width is in-
its
by 3
inches,
it
becomes a square. Find
its
dimensions.
width of a rectangle is increased by 2 inches, its area increased by 14 square inches. If its length is decreased by 2 inches, its area is decreased by 8 square inches. Find its dimen20. If the
is
sions.
by 3 feet, and its width is increased by 2 feet, its area is increased by 2 square feet. If its length is decreased by 2 feet and its width is increased by 21. If the length of a rectangle
1 foot, its
area
22. If the
is
unchanged. Find
width of a rectangle
is
its is
decreased
dimensions.
increased
decreased by 4 inches, its area length becomes a square. Find its dimensions. is
23. If $6000
yearly interest
is is
invested, part at
by 3 inches and its unchanged and it
5% and
amount
$340. Find the
is
part at 6%, the total invested at each rate.
24. If $10,000 is invested, part at 5% and part at 6%, and if the total yearly interest is $540, find the amount invested at each rate.
$8000 is invested, part at* 4% and part at 5%, and if the interest on the 4% investment is $50 more than the other, find the amount invested at each rate. 25. If
The interest received yearly on a 6% investment is twice much as the interest on a 4% investment. If the total yearly
26.
as
interest
is
27. If is
A works 3 A works
$60. If
pay
is
$360, find the
amount
of each investment.
days and B works 5 days, their combined pay 5 days and B works 3 days, their combined
$68. Find the daily
wage
of each.
SIMULTANEOUS LINEAR EQUATIONS
120
28.
A man has $2 in dimes and nickels.
dimes and half as $2.50.
29.
How many
A man
the same
many
nickels as he
If
now
he has twice as has, he
[Ch.
VI
many
would have
of each has he?
has $3.50 in nickels, dimes, and quarters. If he had of dimes but half as many nickels and twice as
number
quarters, he would have $4.50. If he traded his quarters for dimes and his dimes for quarters, he would have $4.10. How many coins of each kind does he have?
many
pounds of bananas of two grades, one selling for 5j per pound and the other for 6^, bring $5.30, how many pounds of 30. If 100
each grade are there?
One
is 5% pure silver and another 15% pure silver. pounds of each must be mixed to form 100 pounds of alloy of which 8% is pure silver? 32. A lady buys 10 pounds of grapes, of which part cost 25?f for 2 pounds and the rest 25 for 3 pounds, l^she paid $1.00 al-
31.
alloy
How many
how many pounds of each kind did she buy? 33. How many pounds each of 20 and 35 coffee must be mixed to make 100 pounds of coffee worth 25ff per pound? together,
34. If corn
meal
how many pounds
is
worth 3
of each does
per pound and flour 4j5 per pound, one buy to get 50 pounds for $1.80?
Chapter Seven
EXPONENTS AND RADICALS
The laws of exponents
7.1.
As we have already seen, the symbol a m when m is a positive integer, means 'the base a taken m times as a ,
factor.' Thus, 2 3 = 2 2 2 powers is called involution.
=
8.
This process of raising to
We shall now examine the laws concerning operations with positive integral exponents. Perhaps even more useful than knowledge of the laws themselves is the realization of how very easily these working
hand by anyone who understands the meaning of the symbol a m There is not a better place to apply the arithmetic tests which are so useful in m n = a m+n or algebra. For example, which is correct: a a can be discovered at
rules
first
.
,
=
a ma n
n =
=
3.
25
a wn ? Try Then,
= 2X with small numbers, say a = 2, = 2 22 3 = (2 2) (2 -2- 2) =2-2-2-2-2
m
it
cw
Evidently the exponents here are added, and not m n = a m+ n multiplied. This suggests that the correct law is a a Once the simplicity of the testing-by-arithmetic method is grasped, the student will be self-reliant when his memory fails. But he can save time by learning the five all-sufficient .
.
:
laws below, which, when supplemented by certain definitions, will be shown to hold even when the exponents are not positive integers. It is helpful to learn
them
in groups of 2, 2,
and
1,
called respectively the repeated base, the repeated exponent, and the single base cases. Easy extensions of the laws are
indicated
by the
illustrative examples. 121
EXPONENTS AND RADICALS
122
[Ch. VII
Repeated base cases
LAW
=
3 4 Examples. 2 2
LAW
= =
a ma n
1.
2 7 3 43 23 am
3 4+2+1
;
=
2.
a
Example. J
a m+ n .
=
27
~3
=
am
n
24
~n
=
37
.
.
.
Thus, if the same number or letter appears as a base in each of two exponential numbers which are multiplied or divided, this same base appears in the result.* Repeated exponent cases
LAW
3
3 5
Examples.
3
2 2 3 25 2a 2
LAW
=
ambm
3.
= =
3
(3
5)
(2
3
am
A
4.
bm
- =
Examples.
(-}
am
a
-
(ab)
=
m
.
(15)
5a)
=
(a (-} b/
=
42
2
=
3 ;
(30a)
2 .
m
;
/
number or letter appears as an extwo each of exponential numbers which are multior divided, this same exponent appears in the result.
That
is,
if
the same
ponent in plied
Single base case
LAW
5.
(a
Examples. (4
Sometimes
3
2
)
it is
=
43
'
2
m
=
n )
46
=
a mn
.
(2aW) = 3
;
23
aW
2 .
convenient to use the laws in reverse order, left. The memorization in the form
or as read from right to *
An
apparent exception to this statement appears in the example:
can be written in the form a by definition of a However, am ~ Law 2 gives: m = am m = a (or 1). a 1
(Art. 7.5).
am
=
1.
Note that
POWERS OF A NEGATIVE NUMBER
7.2]
given, however, is simpler tain common errors.
and more
123
likely to
prevent cer-
5 2 Example. Can 2 3 be simplified by use of a law of exponents?
Answer. No, since neither the base nor the exponent
is
repeated.
While the definitions given later (Art. 7.5) enable us to say that the above five laws are true for all values of the exponents, whether positive, negative, zero, or fractional, the proofs below apply only when the exponents involved are positive integers, and when m > n in Law 2.
Law = aaa
Proof of (n factors)
Proof of = aa
Law
am
a
n
2.
1.
a ma n
(m
=
+
aaaa
n
(m
factors)
(assuming that
m>
(m factors) = aa ^
aa
= n)
,
r
factors)
=
am
~n .
(n factors)
Proof of Law 3. a b m = [aa
=
(m
[(a6)(a6)
(m
factors)][(66
(m
Proof of Law a m = aa
=
(m 7 (m
bb 1,
(ab)
m
.
4-
n
r~
bm
factors)]
factors)]
(by the commutative law of multiplication)
(by Rule
.
:
n
(m
aaa
factors)
am+n
Art. 3.5)
Law 5. m n = a ma m (a ) Law 1) = a mn
factors) F~~I factors)
/a/a = (rKz) bjbj
'
'
'
/ (
m rfactors)
= (-JT.
Proof of
(by 7.2.
A
(n factors)
=
am+m+
*
*
'
(fltennfl>
.
Powers of a negative number a number of the form x n where n is an an even power when n is even and an odd power
power of x
integer. It
when n
is
is
odd.
is
,
EXPONENTS AND RADICALS
124
Since
(
3)
n ,
VII
n negative factors, it is n is even or odd. More
for example, has
positive or negative according as
generally
[Ch.
:
For negative numbers, even powers are
and odd
positive
powers are negative.
EXERCISE Apply
^7 o/.
vii
TZX.
,76
^R oo
-. L
-
n
:J_. o ax 3
4.0 'XA*.
/~rr
^ ^ o x2 y
4^ 1*^.
.
-
x
4fl tv/.
., 6
X*
^
. .
x 3 ^4
44. 4. JL.
2 ?/
2
46.
-(-f)
7.3. Radicals, roots,
The
^.10
/y.5
^O ov.
-
or
2
;
33 (ORAL)
the laws of exponents in the following exercises.
xj.3
X
= (-2)(-2) = 4; = (_ 2 )(-2)(-2)=-8.
(-2)2 (_2)3
Thus,
f
47.
f-
I
and principal
radical fa, read
a
48.
,
x 3y
roots
the rth root of a,'
is
a number
whose rth power is a. That is, (V^) = a. 4 3 5 Thus, (^S) = 8; (A/10) = 10; (A^I?) = -17. The symbol V~is called the radical sign; the quantity below it, or a in the case of Va, is the radicand; and' the '
7
integer r
is
the index of the root.
RATIONAL AND IRRATIONAL NUMBERS
7.4]
125
is 2, it is customarily omitted, and the symbol 'the read square root of a.' In other roots such as v'a, or 'the cube root of a,' the index must be written. There are two numbers whose squares are a given positive number. For instance, (+2) 2 = 4 and also ( 2^2 = 4. Thus, 4 has 2 square roots. However, the symbol V4 stands only for +2, which is called the principal square root of 4. The
If
Va
the index
is
VI. By more advanced methods it can be shown that any number has 3 cube roots, 4 fourth roots, 5 fifth roots, etc. The symbol Va, however, represents only one of the rrth 2
root
designated as
is
namely, the one which is called the principal rth a is positive this root is positive; when a is
roots of a
When
root of a.
negative and r
is
odd,
it is
Thus, V9 = 3; V8 = since (-2) =-8.
negative. 2;
Vl6 =
2;
and
V^8
=-2,
3
In Art. 8.1 we shall consider the case in which a
and
is
negative
r is even.
Rational and irrational numbers
7.4.
From
the definition in Art. 7.3, the
a positive number such that (V2) 2 exactly? It is more than 1.4 and 2 = 1.96 and (1.5) 2 = 2.25. That (1.4)
number V2 must be 2. But what is it
=
than 1.5, since is, V2 is between 1.4 and 1.5, or 1.4
Assume that V.2 =
-,
less
where p and q are integers and the fraction -
to lowest terms. 1,
2
since
=
7)2
V2
is
reduced
Then p and
q have no factors in common, and q cannot be not an integer. Squaring both sides of the equation, we have:
= pp which ,
q*
is
Q
q
is
impossible since there are no
common
factors in
p and q
to
be canceled out. Hence our assumption is impossible. Similarly we can prove that 'V^4, vT, etc., cannot be expressed as quotients of integers.
EXPONENTS AND RADICALS
126
[Ch. VII
7
v !,
v^6, and in general the rth root of a number a perfect rth power. Such roots are called surds, and are included in the large group of irrational numbers. is
true of
which
is not
An
number
irrational
is
one which cannot be expressed
exactly as the quotient of two integers.
Examples. All surds, such as V2 and /5, and many other numbers such as ?r, the ratio of the circumference to the diameter of a
A
rational
circle.
number
is
one which can be expressed as the
quotient of two integers.
Examples, f
,
f,
N/f (or f ), 1.41 (or
^),
3 (or
f ).
As suggested by the alternate forms of 3 and 1.41 in parentheses above, all integers and all decimal numbers are rational. It follows that irrational numbers cannot be expressed exactly in decimal form. Most of the numbers in various tables, such as lists of square and cube roots (Table 2 in this text) are merely decimal approximations, or numbers as close to the actual roots as possible with the given of decimal places.
number
Irrational numbers are connected in an interesting manner with incommensurable quantities in geometry. Consider, for example, the length OP in Fig. 12. It may be computed 0(2,1)
A (1,0)
6(2,0)
Fig. 12
by use
of the
famous Pythagorean Theorem, which states
that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the sides. Since the sides qf the right triangle
OAP
are
1
and
1,
the hypotenuse
OP
is
equal
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
7.5]
to
= Vl + common unit
Vl + 2
had a
m
exactly
OA
I
2
1
= V2.
Geometrically,
if
OA
127
OP OP
and
measure which would go into times, say, and into OA exactly n times, then of
we know this
n
to be impossible, since ^-r
V2,
OA
,
'
and V2 cannot be expressed as the quotient of two integers. Hence OP and OA have no common unit of measure and
V5
are said to be incommensurable. Similarly, since OQ is it is incommensurable with the unit length BQ.
units long,
EXERCISE
34
Give the two square roots of the following numbers, naming first: (a), 1; (b), 4; (c), 1; (d), ; (e), ^; (f), x
1.
the principal root
In each case below state whether the radical
2.
is
rational or
_
irrational. If it is rational, find its exact value; otherwise give its
approximate value
as^
found in_Table
2.
_
V9; (b), VlO; (c), ^8; (d), ^27; (c^O; (f),_Vl2; Vl6; (h), V^T; (i), VI; (j), ^^7; (k), V-8; (1), Vg.
(a), (g),
Find the exact lengths of the diagonals of the rectangles whose dimensions in inches are as follows 3.
:
X 2; (a), (0, 4X6; (g), 1
4.
(b),
8
X
2X3;
(c),
3X4;
(d),
3X7;
(e),
5
X
12;
15.
In which cases in problem 3 are the lengths of the diagonals
rational?
7.5. Negative, zero, It
is
and fractional exponents
desirable that the five laws stated in Art. 7.1 shall
be true even
when
the exponents are not positive integers. It can be so arranged by use of the following definitions.
a
Definition 1.
=
1.
^
(a
0.)
By Law' 1, aa w = a+ m = a m Also, by the definition, &m = 1 while by aa m = 1 a m = a m Or again, by division, .
.
;
a
Law
2 and the definition,
am
=
a m~ m
=
a
=
1.
In other
EXPONENTS AND RADICALS
128
when the exponent
words, the laws remain valid
VII
[Ch.
zero
is
involved, given this definition.
a~ k
Definition 2.
Consider
first
r
a**
a numerical example.
22
By Law
2,
=
22
~5
2i
=
2~ 3 This .
is
when 2~3
correct
22
1
defined as
is
,
since
Zt
2 -2
1
=
=
--
=
<
_!_ ' 3
2-2-2-2-2~2-2-2~2 sim
In general, when n definition,
Two
a~ (n
~ m)
=
>
m,
=
am
~n
=
a~
a
(n
~m
and
by
this
a n-m
convenient rules are consequences of Definition
2.
>
For bj
(2) tion,
^e may
// it
/a m
W
term a~ m
a rn
a a^
a
6^ is
a factor of the denominator of a fraca m and written as a factor of the
be changed to
numerator; if it is a factor of the numerator, it may be changed m and written as a to a factor of the denominator.
Examples.
3- 24
4
5
3
a 6-3 (c
+
d)
a 6(c
+
c
= '
d)-
2
5
a&
3
+
q(c
2
Note, however, that in the case
4 __ 45'
d'
+
2 c?)
6
:
a -i a~ l ,
_
+ .
-1
,;, the terms with 6' 1
negative exponents are not factors of the numerator or de-
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
7.5]
nominator. In such cases the simplification must be l replacing a~
complex
by
b~ l
-,
-
6- 1
+
6-
^ o
more
fraction. Or,
a- 1 a~ l
by
a
1
(a-
i
and then simplifying the
-
1
b~ l )ab
+
1
=
l
b~ )ab
b
-
b
+
a a
= /a.
ar
Definition 8.
made by
directly,
(a-
=
etc.,
129
i
r By Law 5, (a ) = a = a = a. Also, by the definition, r/ = a. For the special case r = 3, for example, = (Va) (a a*aia* = a*+l+* = a = a, while v^ ^a A^a = (by Law 1) (/aY = a. '
r
1
r
r
i
r
r
r
)
1
?
aq
Definition 4-
Q
Q,
,
= (va) p
=va 5
For,
by Law 5 and Definition
also a a
=
Thus, 8*
Here the
= Va p = ( v'g) =
(a
3,
aq
p .*
=
^
(a
p
G
)
=
(
V'/-a)
p ,
and
a
p
.
)
2
final result is
form of Definition
22
=
4,
more
and
also 8*
= v^ = v'64 =
easily calculated from the
This should be used for a q when a
4.
4.
first is
a
perfect gth power.
Again, 7*
-
7
(v
2
?)
,
and
also
7*
-v7^ =^49. 2
In this
P '
case the second definition of a
is
preferable, since 7
is
not
a perfect cube.
and and
The reason
for the use of the adjectives 'rational' 'integral' as applied to letters is suggested in Articles 7.4 1 7.5. Just as is not^ rational, and 2- or ^, is not
Note.
V2
integral, so -, is
x
*
definition
by
,
Vx
is
not rational in x,
and or 1 or ,
not integral in x.
The
75
must be reduced
fraction
to lowest terms.
9
=
1) is
not the same as
(-!
Note that
jL
(
1)*,
or
EXPONENTS AND RADICALS
130
EXERCISE
[Ch. VII
35
Evaluate each of the following.
38.
+
39. (-27)-*.
40. -(5b)
41. l(3o).
42. (-0.027)*.
43.
45.
.
44.
-5-
.
46.
-(-3)-'.
2
-(J)3 (-f)4
47. [2
+
(16)-*]-
(-56).
1 .
1 .
Change each of the following to expressions in which the exponents are positive, and in which the operations indicated in Laws 1 to 5 have been carried out completely.
~ 9r
48.
11
y
*}-)
49.
2 *
3
AD..
2r &*>
2
^7
**'
vJV/
/ff
O
1O*
*^
ej. '
2?>y-i
^
*-*
/
i*'~l'i/*
y
co
3v>
_o*
r^n
OA
/^^
o
OO*
*}
^77-2 &y
M 3V
Ki W> 70/>
1
* X
50.
^ ^.
r^n
*'
re
'
2
^v^ti y /
_i
_i'
CQ *>'
V/^
(200xi/)
65.
68.
70.
v
' .
V.
71.
(30ary)
-1 -
7.5]
72.
74. 76.
NEGATIVE, ZERO, AND FRACTIONAL EXPONENTS
- y) (x + y)~ (x - y)~ (x + 2y). (2x - 37/)- (x + y). (2x 2
l
.
l
2
-
78. (x
yfi(x
-
82. (a 2
+ z )*(a + rO~i
to
2
x 2 )-*.
6).
+ y)~
(x
-
y)(x
2 .
y)-*.
'4'
2
81. (2z 2
2
(a
131
-
83. (x 2
2
-
a 2 )Kz 2
2
)*.
a2 )~K
Express each of the following as a complex fraction and simplify a simple fraction in lowest terms. 2 84-
a~
+
2
,
2
07,-
,
fc~
volution.
o2
r-
J_
b
+ a~ b
ab~2
rr~-j
2
2
^
+
a(o
2a 2 6 2 ; b + a2
=
L
i
a2
gs
-
-
+ 2/)K^ 2/)~^ + a )^(2x + a
79. (x
y)~^.
-
75.
3
77. (5x
80. (a 2
z 2 )^(a 2
+ 6)~ (3x + y)
73. (a
3
b)-
a-
-
1
1
-
6(a
86>
^_2
00
(a
'
y
_2
^
t
+ x )~ + + (a + x
1
b)'
__
2
2
6-'
2
2
2 2
)
x~* '
x2
89. Select appropriate multipliers for each of the problems 85, 87, and 88 and simplify without first expressing each as a complex fraction.
Reduce
the following to fractions
whose numerators have no frac-
tional exponents.
--
-
90.
a
and denominator by
-
(a
s)i
+
-
a
x).
(a
-
(a
g)~*
-
(a
x
g)*
_
(a
x )*
(a
_
-
2
(a
(a
+ oo -
x2 )^ o
(a
95
-
(a
2
-
x2)-*
*
-^
2
2
(a
^
-
x2 )*
-
2
-
x2 )*
+ x (a - x )'* 2
2
(a
-
+
g
-
+
2
x2 )*
X2 )'*
-
(a
x)
x)'
1 '
x)*
z2 )*
+ (a + x )'* + X2)i - (a + X )'* 2
2
i
-94-
2
'
+ -
(a
2
2
1
2
a;
(a
x)
(a
-
a
(a
V1
numerator
Solution. Multiply r J both
x
2
(a
(a '
2
+ x )* 2
2
2
+ x )* - x (a (a + x )* 2
2
2
2
2
*
EXPONENTS AND RADICALS
132
Reduce
[Ch.
VII
whose denominators have no
the following to fractions
fractional exponents.
-*
2
(a
97 I
CZ
~'~'
X
~~' )
2
)* '~~'
CZ
yCZ
JC J
I
CZ
I
-
100. 2
x (a ftl
2f
u/
O
x
(cz
7.6. La?/;s
)
2
+
/
^
9
I
~T~
x
(.a
9
^
102.
i:
9/9
a (d
)~*
I
~T~
CZ
+
)'*
fj2
x
9 )
* 2 )*
l_
2
a;
/y.2
Z-r-Z i /
9
~~~
/Tt2
?-r^
101.
+
2
ju j
72
fl%
99.
(a
98
_ _ ,
^ -4-
^CZ
X
I
i
->2
X
-
(a
2
+ z )* 2
/>2'2'
X)
/
(cz
-' 9
I
~f~
x
9-i-
)^
concerning radicals
be performed by changing them to exponential form and then applying the laws of Operations with radicals
may
exponents.
Example
Example
L 2.
=
V2
=
= 2*
A/V2 =
4*
= V4 =
2.
2/
=
(2*)*
= /2.
2J
method should be held in reserve in case of doubt, it is more efficient to learn directly the more frequently used rules involving radicals. The student may However, while
this
prove each rule by changing the radicals to exponential form. The laws as stated below are valid when all letters stand for
positive
integers.
(Otherwise
some
exceptions
are
necessary.)
LAW
v^ = a
1.
Examples. /V*
LAW
/ab
2.
From a perfect
= ^2^ =
we see power may
this rule rth
k .
2 4 /3 ;
32
.
= ^fa^/b.
that a factor of the radicand which is be taken from the radicand if its rth
root is written outside the radical sign.
have been removed the radical terms.
=
is
When
all
such factors
said to be reduced to lowest
7.7]
ADDITION AND SUBTRACTION OF RADICALS
Exampk
1.
Vs = V4
Example
2.
v 32a 6 = v 2 a 6
Example
3.
/
(Why This
4
= /4
2 /
6
3
3
=
/2
6
(4a)
=
133
2/2. /
2a6 2 v 4a.
V4a + 166 = V4(a + 46 = 2Va + 46 V4a + 166 = V4o + Vl66 = 2a +46?) 2
2
2
2
2
)
2
2
not
2
2
.
2
common error should be carefully noted and avoided.
Ja
i.
Example
2.
+
/ 3/20
2V2
3V20 = 2V2
= Vg' 2V2
1
V 2
2
3
'^^V
'
2
2 4.
^^a
Examples, v /2
7.7.
=v 2; /
2
3VlQ
. '
LAW
/20
= v' /a = Vfl. v
Vs = V v7^ =v 2. /
Addition and subtraction of radicals
Radicals are of the same order are equal.
the index
Thus cube
roots are
if
all of
the indices of the roots the
same
order, having
3.
Radicals of the same order whose radicands are equal are called like radicals. Only like radicals can be combined by addition and subtraction.
To determine whether two
are like radicals, the radicands should lowest terms, as in the examples below. * Historically, the symbol 'v
and the vinculum 10.
is
first
a union of two symbols:
(treat as a single
number). Thus,
radicals
be reduced to
V (take the root
V64 +
36
= V64
-f
of)
36
EXPONENTS AND RADICALS
134
[Ch. VII
L Vl2 + V27 = 2/3 + 3^3 = 5V3. Example 2. /54a - v'lGa = ^/2a - 2v'2a = /2a. Example
The sums left
or differences of unlike radicals
uncombined or
else
may
may
be approximated
either be
in
decimal
form.
For example, V3 + V2 cannot be further simplified in exact form, but /3 +V2 = 1.732 + 1.414 = 3.146 (nearly).
Note that /3
+ V2 is not
since
V5,
V5 =
2.236 (nearly).
Multiplication of radicals
7.8.
By
the use of
Law
2,
same order can be found
the product of two radicals of the directly.
The law can be extended
to handle cases where the multiplicand or multiplier or both are indicated sums of radicals.
The
Law
of
following illustrative examples 2 :
Example
L
Solution.
V28 V$ =
_
/(28)(f)
VI + 3/2 2V5 - 4V2 2
Thus,
5
of
= /12 =
V5+3/2 and 2 V5-
+ 6VlO
- 4VlO -12-2 10 + 2/10 - 24 = -14 + 2VlO - 4V2) = -14 + 2/10. (/5 + 3V2)(2V5
EXERCISE Reduce
direct use
Find the product of /28 and
Example 2. Find the product Solution.
show the
36
the radicands to lowest terms.
VT6.
1.
V^
2.
5.
^27.
6. '^125.
3.
/36.
4.
A^
7.
Vl2.
8.
V20.
7.8]
9.
MULTIPLICATION OF RADICALS V28.
10.
V2.
11. /18.
14.
V54.
15. /63.
18.
v7^
19.
--v/ 128.
23.
Vl25x2
22.
26.
VlSOx.
30.
VlOSx4 98x~8
49. v/375?/ 9
56.
V/3V27.
57.
60.
V? V63. ^9 ^24.
61.
74. 77.
Vs Vl2 =
V8 /20. v^ v7 !^. 65.
V6 /4 V4 + V4 -
(2/2)(2/3)
=
4V6.
58. /5
V20.
59. v/18 /27.
62. -^5
>^^2ioO.
63. v/16
^16 ^32. =0.
67.
73.
.
54.
Solution for problem 55.
70.
.
.
53.
64.
.
^54.
66. v^25 v'lo.
68.
69.
V2
V3
71.
8x 2
= V4(l -
12x
2
16x2
.
75.
.
78.
= 2/l - 18j/ Vl6 - 4o
76.
Vg +
.
79.
V25
2
92.
.
.
2
81.
2x 2
27x*.
+
lOOx 2
.
EXPONENTS AND RADICALS
136
Combine 93. 95.
97.
the following radicals as indicated.
V27 - Vl2. V32 - VlS. V?5 + 2Vl2.
99. (1 101. (1 103. (5 105. (4 107.
94. 96. 98.
-V8) + (2 + Vl8). +V8) - (2 +VT8). - V50) - (4 -V32). -V75) - (2 - Vl08).
100.
V3(l -A/5x)
109.
V7(
V75 -Vl2. VSO - 2V. V27a3 + Vl2a62 Vl2x3y - V27xy*. .
102. A/98
104. 106.
V5(l -V2) +V2(/5 -
108.
+ ^72.
Vl25x - V45z. - 3V3. 2(/3 + 1)
1).
+/5(V3x -
1).
110. 2/8
111. 112. 113.
114. 115. 116.
117.
(2V6 - 3/3)(3V6 + 4V3). (-4 - 2V7)(-4 + 2V7). (V2x + V50>._ (3V7 - 2)(3V7 + 2). - 2>/3) 2 - 2(3 - 2V3) -2. (3 - V3)(l - /2 +V3). (1 +V2 - 3>/8 + VG). (3 + 5V2)(2V5
Criticise the errors
made in
the following.
ft
1
2c
2c
Simplify the following. 121.
X
122. x
=
=
[Ch. VII
-(66
-
a)
/36&2 rr
+
12a6
4 +/32 ,
123. x
+
a2
= ac
RATIONALIZING DENOMINATORS
7.9]
137
denominators
7.9. Rationalizing
and often desirable to move
It is possible
all
radicals
appearing in a simple fraction into the numerator. This called rationalizing the denominator. comments on the illustrative examples below, see the is
operation
For
paragraph that follows them. 7
y
Example
1.
Example
2.
Example
3.
n Example
4-
Example
5.
/
7
V3 + V3 -
1
3/3
=
jj-
22
3/3
=
1 TF= =
= (V3
-
3 /-
+
2a
l)(/3
(V3 - 1)(V3
1
V/12
=
3/12
^-
jj
+ +
1)
=
1)
3
+_2/3 2 (V3) -
+
1 2
(I)
1
=
4
+
2V3 =
2(2
+ V3) = 1
3 JV /*/'! /vv /y> J0 JliXUTTlpl/e O. 1
/y3' XII
Z
~'
sm:
3
i/^
___ ^
.
2^;
zzz:
o
t> t
^^
2/^
^.o
y*
These illustrations show some of the commonly used methods. In each of Examples 1 and 2 the numerator and denominator of the radicand are multiplied by a factor which makes the denominator a perfect rth power, so that it can be taken from under the radical sign. In Examples 3 and 4,
where the radicals appear
first in
the denominators only,
EXPONENTS AND RADICALS
138
[Ch.
VII
the multipliers must themselves be radicals. Example 5 = a2 makes use of the fact that (a 62 6) (a 6)
+
.
how
6 shows
the laws of exponents can be used to Example rationalize an expression which involves fractional exponents for
some
of its divisors in a denominator.
An
important advantage of the process of rationalizing the denominator appears in numerical cases, where it shortens the work of getting decimal approximations to the values of surds.
Example. Evaluate
Long
solution.
Short solution.
~
^ to V2
four decimal places.
=
=
.7071 (by long division).
~ = ^^ =
-~ = v2
^
While the rationalization
.7071.
*
of
denominators
is
useful at
times, the student should not get the impression that the resulting form is necessarily the 'better' one. For example,
and
^ is shorter than ~
,
.
,
Certainly
10 .
V543
.
is
just as
ul preferable to ,
,
good for
10V543 7:
543
many
purposes.
,
when one
is
pre-
paring to square this fraction.
EXERCISE
37
Rationalize the denominators. Approximate in decimal values of the numerical fractions by use of Table 2.
form
the
7.9]
RATIONALIZING DENOMINATORS
139
21. 24.
#f
2-
3 -
a;
27.
Change '1A o^.
to radical
form and ^^J j*j.
-7*Tfl/V~T <& . */ t/
I
44.
V2 V3 +V2
'
/3 50.
/
A/2
45.
.
and
-V3
5-V3 2
.
y
V3 3
1
54.
/
3a6 56.
62.
,--
V6
,-
Vx
65. -
,--
Vy
x 66.
V
2+V5
V3
1
x2+
simplify.
-V' -
33
3/c o/.
-i
y
Rationalize the denominators .
/!+*
rationalize the denominator.
TO, --~ 6
u/
30.
-
y
i
EXPONENTS AND RADICALS
140
+2 +6
5V6 3V2
,
fi Oo
/-
3V2-V3 2V3-V2
0'
,
VII
2V5-3V2 3V5 + 2V2
70
/
[Ch.
,
/
71-84. Rationalize the numerators in problems 41-7 and 61-7.
Simplify
the following
expressions,
the
leaving
denominators
rationalized.
oe
85.
'Y
/ /2V U/r 3/
o
/
,-
T
Solution
/2
2
V6
2V6
3/
VV3/A/3
3
3
9
2V =
f^l
l)
Lls) J
/2
s
/
/2
2
3
'
/2*
(3)
32 86.
s
87.
.
(V )
(V ). 4
90. (^ )<.
91.
(^f)
94.
95.
V^.
98.
V(i).
v.
7.10.
99.
Changing
When
The
2*
s
9 89.
(V ). 4
92.
(v^)
96.
V(j).
^()
100.
V()*.
93.
.
97.
4
101.
.
the order of radicals is
replaced
said to be reduced. This
is
when
radical v'a' 1.
.
the index of a radical
teger the order
than
88.
'
details
m
by a smaller is
in-
possible hi the
r have a common factor larger be carried through by changing to
and
may
exponential form.
Example
1.
/4
Example
2.
v^o*
= v^ =
=
afa
=
2*
=
a*
=
2*
= V2.
is possible always to equalize the orders of two radicals a fact sometimes useful for comparison purposes.
It
Example. Which Solution.
/2
v3 /
Hence,
vis
is
larger,
V2
or
= 2* = 2* = v^ = /8. / / 2 = 3*_=3*=v 3' =v 9. > /2.
STEPS IN SIMPLIFYING A RADICAL
7.11]
141
When
the orders of two radicals are equalized their product or quotient may be found as a single radical, as illustrated below.
Example
1.
=
/2 /3
2*3*
=
3
2
(2 3 )*
=/2 53~2 =/72.
7.11. Steps in simplifying a radical
When
the following steps are taken, a radical is said by definition to be in simplest form. The order of steps as indi-
cated 1.
is
satisfactory,
The radicand
though not necessary.
made
a simple fraction. Negative and zero exponents are removed is
by use
of
In all cases involving radicals the student should keep mind two fundamental principles
in
2.
the proper definitions. 3. The radicand is reduced to lowest terms. 4. 5.
The order is reduced as much as The denominator is rationalized.
possible.
:
Operations may be verified in cases of doubt by changing the radicals to exponential form. 2. When the processes are long and involved the practical 1.
procedure
may
be to use the decimal approximations for
various surds.
EXERCISE
38
^7 2
'
Change
the orders of the following radicals as directed.
1.
V3
to the 4th order.
2.
/2 to the 6th order.
3.
^2 to the 6th order.
4.
f$ to the 6th order.
5.
/25 to the 2nd order.
6.
v^'to the 2nd
7.
v'2 to the 10th order.
8.
V^32 to the 2nd order.
9.
v^27 to the 3rd order.
10.
v^ to the 2nd order.
order.
EXPONENTS AND RADICALS
142
Find
14. v^2,
12. /5, A^lT.
^5.
Reduce
15.
v^,
the radicals in each
v^,
13.
/3.
v/3.
problem below
to the
same
order,
then perform the indicated operations. 16.
17.
V3
v^.
18.
20.
21.
V3
v'S.
22. -~5-
23.
26.
27.
-
25. -r^-
x/9 =.
VII
the larger quantity in each pair.
11. /3, v/5.
28.
[Ch.
29.
-^
v^Vs.
19.
-^r
30. -7
31.
Express in simplest form. 34.
37
4 '
l&x y
~^7 27^2
and
Chapter Eight
THE NUMBER SYSTEM
8.1.
Imaginary numbers
Thus
numbers we have met, whether rational or have been real, or representable by points on the
far all
irrational,
line of Fig._l. To get the point representing the irrational for instance, we simply measure to the right
number V2,
of the zero-point a length equal to the diagonal of a square with unit sides. But, since 2 2 = 4 and ( 2) 2 = 4, the quantity exist
V
or the
4,
among
number whose square
is
4, just
does not
the real numbers. It turns out to be highly
useful in mathematics to invent a
new type
of
number
in
V
terms of which we can express such quantities as 4. 1 is designated by By definition, then, the quantity
V
the letter root of
That
i.
1.
is,
=
1,
root of
and
i is the
principal square
imaginary unit.
1 is
V
1,
or
i.
the rules for operations with radicals (Art. 7.6)
the rule _that
V^T Va
2
It is also called the
The other square
Among
i
Vab = Va Vb.
= iVa. That
Hence,
V^a = V(
l)a
is
=
is,
(1)
Examples.
V
4
= iVi =
= (+2) 4 =
2z;
16, V-16 cannot be either +2 In or general, it can be shown that all even roots of negative numbers are imaginary, or expressible in terms of
Since (-2) 4 2.
143
THE NUMBER SYSTEM
144
[Ch. VIII
There are still other quantities which can be so expressed, but we shall not consider them here.
the imaginary unit
An
imaginary number is one which can be bi, where a and b are real and 6^0. form a
Definition. written in the
Examples,
i.
+
i,
Z
,
+
4
6
5i,
iy
V
2.
4, *$/
Imaginary numbers should be expressed in terms of i before they are used in algebraic operations. For instance, it is false that =V(-3)(-3) = V9 = 3, since
V^3 V-3
from the definition of the square
we
write
V
3 as
Powers of
(V^3) = 2
3.
i
2
(/3)
2
=
i,
= -l
(by definition).
Note that
t
5
=
i
1 ,
t
6
=
t
2 ,
t
7
=
t
/2w/6 /or finding the value of 4.
The remainder
3 ,
i
= i^' = i = i4 t 9 = t
9
8
3
=
any power of will
be t
0,
=
i
1
2,
1, 1
=
i y etc.
etc.
,
i.
=
Divide the or
3. 2
i,
i
= - 1, i27 =
t
1, i
=
The 1,
i.
Examples, 8.3.
.
,
exponent by corresponding values of the power are i
1,
i
i
and
i,
= -3.
(-1)(3)
of i is equal to one of the four numbers This conclusion is reached as follows:
1.
if
y
Any power and
But
iV3 then
2 ^3) =
8.2.
root,
z
20
=
1, i
25
=
i
1
=
i, i
30
=
i
2
3
=
-i.
Operations with imaginary numbers
The
results of simple operations with
can be expressed in the form a examples below.
+
imaginary numbers
bi y as in the illustrative
8.3]
OPERATIONS WITH IMAGINARY NUMBERS
145
Addition.
+ 3i) +
(2
-
(4
= =
5i)
+ 4) + (3i + (-2)i
(2
6
=
5z)
-
6
2i
Subtraction.
-
(3
-
7t)
-
(4
= =
9i)
-
(3 (3
7f)
4)
+ (-4 + 9i) + (-7z + 9z) = -1 + 2i.
Afwttip/ica^on.
+ 2i)(4 -
(3
= =
i)
+ 8t + 5z -
12 i2
-
3t
2z 2
=
2(-l)
14
+ 5t.
1 JL
^^ 7i' i U
Dmszon. ,/
v
1
+ 3t
=
/o *J
(1
'W1
+ 30(1 -
~ = -1
7t
than a
2
6
EXERCISE
(a
V-25.
3.
V
32. (3 35. (4
4x
-
1
1
-
9t
2
,
10
=
60
a
2
t2*2
=
a2
+6
2
rather
.
39 the following expressions to simplest form.
Answer:
2 .
V
2.
5i.
17.
Answer:
Answer:
+ 20(3 - 20+ 0(4 - 30-
38. 3t(3
30
2
Reduce each of 1.
+ 6t)(a
=
tJt/J
=
1+9
Note that
3-A
7 t/j^.
0-
15.
-v-i
18.
-V-25t/4
21.
-V-23a
24.
z
.
2 .
14 .
27. (1
+O
2
30. (2
-O
2
-
-
+ 40(5 - 40- 50(6 + 50(6
+ 30(3 t(3 + 40-
33. (5
34. (2
36.
37.
39. 2(3
-O
3
2 -
40.
0-
THE NUMBER SYSTEM
146
64.
V-3 +
[Ch. VIII
V^5 V^5. 63. V-4 + V-16. 65. V^2 v^3 V^5.
2V-12.
'^10
/9
+V-9
8.4*
We are now prepared to examine again this thing called a number. To represent all of its types thus far met we shall need to extend Fig. 1 as shown in Fig. 13. The line of Fig. 1, now called the axis of reals, contains all numbers except imaginary ones. Represented by points on this integers (0, 1, 2,
2, etc.), fractions
1,
(,
f,
line
are
J,
ty,
and irrational / 2 +v 16, TT, etc.). But there is no numbers (V2, /7, place on this line for imaginary numbers such as i and 1 2i. The place is provided in a very simple and neat etc.), decimal numbers
* This article
may
(0.5, .33,
be omitted without
Article 7.4 should be re-read
first.
1.21, etc.),
loss of continuity.
If it is studied,
8.5]
NUMBERS
CLASSIFICATIONS OF
147
manner by means of the plane (that of the paper) containing the axis of reals. In Fig. 13 the vertical line is called the axis 2i, and of imaginaries. It contains all numbers, such as i, 'iy
+
which have the form
where b
bi,
is
real
and not
Axis of imaginaries
-*- Axis of reals
Fig. 13
Such numbers are called pure imaginaries. The general number a + bi, where a and b are real, is represented by a point whose coordinates would be (a, 6) if the axes of reals and imaginaries were respectively the X and Y axes of the zero.
rectangular coordinate plane. Thus the point representing 2i has the rectangular coordinates (1, the number 1 2). in The numbers represented their totality by all of the points on the plane are called complex numbers. containing them is known as the complex plane. 8.5. Classifications of
The
numbers
following diagram
Numbers
is
helpful.
in General, or
Complex Numbers
Real Numbers
Imaginary Numbers
L Classifi cation 1
Zero Positive
numbers
Negative numbers
The plane
2 Rational numbers
Classifi cation
Irrational
numbers
THE NUMBER SYSTEM
148
In this diagram Classification represented
by points on
1 divides
[Ch. VIII
the real numbers,
the axis of reals, into three groups
corresponding with: (1), the point the right of 0; and (3), points to the
(zero); (2), points to left of 0.
simply another grouping of real numbers. Here algebra and reason have stepped beyond geometry, since the points representing rational and irrational numbers are mixed together too thickly in any segment of the axis of reals to be separated visually, though we may Classification 2
is
find sample points representing each of the
8.6.
Numbers
two types.
as roots of equations
It is interesting that the
requirements for roots of rational
integral equations in one unknown, which appear in the solutions of very simple problems, have called into use all
the types of numbers thus far discussed. illustrates the point. Root or roots
following table
Type of number represented by the root
A 3
The
type by
itself
Positive integer
2
Negative integer Positive fraction
_5
Negative fraction
+ /2
numbers Pure imaginary numbers Irrational
Imaginary numbers
EXERCISE
40
Show on a complex plane the points representing the follownumbers: i; (c), 3i; (d), 1 i; (a), 2; (b), 2 i;(e), ing 1.
+
(f), (1),
*
-2 + 3t; -3 - 2. The
(g),
4
-
i;
(h),
;
(i),
;
(j),
+
solutions of these equations will be discussed in Chapter 9.
00,
0;
8.6]
NUMBERS AS ROOTS OF EQUATIONS
2.
The numbers a
naries.
What
is
149
+
bi and a bi are called conjugate imagithe relation between the points which represent
them on the complex plane? 3. In the light of the discussion in answer the following questions? (a)
How many Is so,
how would you
real numbers are there between and 1? a smallest positive number, and, if so, what is it? there a largest positive number which is less than 1? If
(b) Is there (c)
this chapter,
what
is it?
Chapter Nine
QUADRATIC EQUATIONS
9.1.
The quadratic in standard form
A quadratic equation, or simply a quadratic, in a given letter is
a rational integral equation of the second degree in that
letter.
Example
1.
2x 2
Example
2.
3y
3x
2
+
2ay
=
1
+
by
(quadratic in x).
+
2
c
=
(quadratic in y).
When
terms of the same degree are grouped together, a quadratic in x may be written in the standard form
ax 2
(1)
where the
+
bx
+c=
coefficients a, 6,
2 Example. Write 5z ard form.
-
a
c
do not involve
and
7z
+3 =
2x*
-
7x
-
2x 2
0,
+ 3x -
#.
4 in stand-
we have
Solution. Grouping, the terms,
-
^
0,
3x
+3+4=
0,
=
0,
=
0.
or (5
-
2)*
2
+
(-7 -
3)x
+
(3
3z 2
-
lOz
+
(1),
we find
+
4)
or (2)
6
Comparing (2) with standard form = - 10, and c = 7. 150
7
that a
=
3,
THE ROOTS OF A QUADRATIC EQUATION
9.2]
9.2.
The
151
a quadratic equation
roots of
We
have seen that a root of an equation is a number or expression which satisfies the equation when substituted for the unknown. Thus, for the quadratic x2
(1)
one root is second root is
The
5,
2
-
4z
5 2
since
since I
1,
+
5)
(
+
4
=
0,
+ 4( 5
1
5)
=
5
=
0;
and a
0.
roots of
x2
(2)
+
4z
-
1
=
+ /5 and 2 /5. Checking the first one, we find that (-2 + /5) + 4(-2 +V5) -1=4- 4/5 + 5-8 + 4V5 -1 = 0. The test of the root -2 - V5 is left to the 2
are
2
student.
Again, the quadratic
x2
(3)
by x =
+
4x
+
4
=
but in this case there is no second root different from the first one. For reasons to be discussed 2 is a double root of (3). later we shall say that satisfied
is
2;
Finally,
x2
(4) is
satisfied
where 4
by the imaginary
8
=
roots
2
+ 2i
the imaginary unit (Art. 8.1). For 4z 2 - 8 Si 8 = 4 - Si 2i)
i is
+ + + 4(-l) =
4(-2
=
+ 4* + +
0. Similarly,
-2 -
+
and
2
2i,
(2 + 2z) + + 8 = 4 + 4i 2
2
2i satisfies (4).
The
roots of equations (l)-(4) represent among them the four different kinds of root-pairs for quadratics rational and unequal, irrational, rational and equal, and
imaginary. In Art. 9.9 we shall discuss a geometric interpretation of the various types. As may be suspected here, and will be proved later, every quadratic equation has two roots,
though
in
some
cases they are equal.
QUADRATIC EQUATIONS
152
EXERCISE it
[Ch.
IX
41
Write each of the followng equations in standard form. Comparing c = 0, find the value of a, b, and c in each case. ax2 bx
+
+
with:
Prove that each of the numbers
3x
1.
-
2
=
2x
1;
(1,
listed is
a
root.
2 -i). Solution. Bx
-
-
2x
1
=
0,
or
+(-2)x+(-l) = 0;a=3, 6=-2, andc=-l;3- ! -2- 1=1; 3(-*) - 2(-i) = 1. 2. 4x = 1 + 4x 3. 6 = 2x 2 + x; (-2, f). ( ). 2 2 4. 1 = 3x + 2x; (-1, ). 5. -3x = 2x + !;(-!, -). 2 6. 3 = x + 2x; (1, -3). 7. 6 = 5x + 4x 2 (-2, ). 8. 3 = x + 2x 2 (1, -f). 9. 3x = -2 - x 2 (-1, -2). 11. 4x = 3 - 4x 2 (, -f). 10. 5x = 2x 2 + 2; (2, ). 12. 8x = -3 - 4x 2 (- J, -f). 13. 2x = x 2 (0, 2). 14. x 2 = 3x; (0, 3). IS. 2x 2 = 3x; (0, f). 16. 2x = -3x 2 (0, -). 17. 3x = -2x (0, -f). 3x
2
2
2
2
;
;
;
;
;
;
;
2
;
18. <>
20.
=x
2 ;
(d,
;
19.
-d).
x2 -(r+s)x+r*=0;
4x2
=
-
9e';
,
21.
(r, *).
22. x2 +(r-s)x-r=0; (-r,s).
23. o 2 x 2
=
16b2
;
,
Write each of the following equations in standard form and find the expressions corresponding with a, b,
24.
4x2 + x - Bx = C + x2 -
+4x+2=4x2 +x+C. 3x +Bx-x=4x2 -Cx-fl. Ax* + B = Cx2 + Dx.
4.
26.
2
28.
29.
9.3.
T/ie solution of ax*
When
6
=
0,
+
30.
c
-
l)x
+ (1 - B)x
Ax*+B+Cx=Bx*-Ax+C. ex2 + 2Bx = 4 - x + Ax.
the general quadratic equation
ox2
2
=
takes the form (1)
c.
Answer. (A
25. 3x 2 27.
and
+c=
0.
(1),
Art. 9.1,
SOLVING QUADRATICS BY FACTORING
9.4]
Since (1)
linear in the
is
unknown x2
(2)
it
,
153
follows that
*__!,
and hence,
-
Example
1. If
If
2.
~~
-
4i!
-
2s 2
=
9
3
=
0,
x1
0,
x2
=
4
=
and x ^,
Ve 2~~~2~' _
/3
and x = f or x 2
?,
=-
-
2
= Vs = ^r
or
=
or
-
Exampk
3. If
Example
4-
4s2
Solve
+5=
:
5x 2
=-
0, a*
f
,
and Z
-
Multiply both members of the equation by 2 ). The new equation becomes 3 !)(!
Solution. 2
(So:
+ 4x
-
2,
= -t.Vio
+
or 2z
2
= -5. Hence x = 2
and x =
f
or
.
9.4. Solving quadratics
When
,
the
left
by factoring
member
of (1), Art. 9.1,
is
factorable,
when
and
the factors can be found readily, the equation can usually be solved most easily by the method illustrated
below.
Example
1.
Solve: x2
Solution. First (1)
we
+ 4x
5
=
factor the left (x
-
1)(*
+
5)
0.
member, thus
=
0.
:
QUADRATIC EQUATIONS
154
This equation
is
by x =
satisfied
since (1
1,
[Ch.
IX
+
5)
1)(1
= 0; and also by x = 5, since 5 5 + 5) = 1)( (_6)(0) = 0. Hence 1 and -5 are roots of (1). We get x = 1 1 equal to zero. Similarly, x + 5 = by setting the factor x = 5. gives x =
6
(
+
Solve: 4z 2
+
12x
Example
2.
Solution.
Noting that the
left
9
=
0.
member
is
a perfect square,
we have
+ 3)
(2z
(2)
2
=
or
0,
(2x
+
3)(2z
+
3)
=
0.
= f This exfactor, placed equal to zero, yields x repeated or double plains the algebraic basis for calling Each
.
fa
root.
The geometric reason
will
appear
later.
CAUTION. The solution of a quadratic by factoring the
member zero.
is
is
left
correct in principle only when the right member test will show, the roots of the
For example, as a
quadratic (x
(3)
are not found
by
is left
explanation which are 6 and
new
left
member
-
2)(x
-
=
3)
12
2 or x 3 equal to 12. The setting x to the student. The correct roots of (3), 1, are found when 12 is transposed and the
is
simplified
and factored.
-
Examples. Solve:
Solution. After clearing fractions
-
by the L.C.D. 4(2z (4)
-
7(2x
which
l)(2x
simplifies,
l)(2x
+ 3)
-
+
3),
+ 3) =
24(2*
after the
by multiplying through we get 4(2x
-
3)(2z
-
1),
indicated operations are per-
formed, to
4z 2
(5)
The
factors of the left
so that x
=
+
-
35
member of
% or f Each .
4z
=
0.
(5) are
2#
of these values for
to satisfy the original equation.
and 2x 5, x will be found
-f 7
9.4]
SOLVING QUADRATICS BY FACTORING
EXERCISE
155
42
Solve by the method of Art. 9.3 where possible; otherwise by the factoring method.
-
58.
= 0. 2. 3x - 12 = 0. 3. 5x - 20 = 0. 2 2x 18 = 0. 5. 2x - 98 = 0. 6. 4x 2 - 100 = 0. 2 3x + 27 = 0. 8. 2x + 8 = 0. 9. 3x 2 + 12 = 0. 11. 2x2 + 98 = 0. 2x + 18 = 0. 12. 4x + 100 = 0. 2 = = 14. 3x 2x 36 0. 15. 2x - 100 = 0. 36 0. 2a x - 64 = 0. 17. 26 2x - 24 = 0. 18. a x - 96c = 0. 2x + 36a2 = 0. 20. 3x 2 + 366 = 0. 21. a x 2 + 1006 = 0. 23. 2x 2 + 48c = 0. 2x + 6462 = 0. 24. a x 2 + 966 = 0. 26. ax 2 + lOOc = 0. ax + 256 = 0. 27. 206 = -ax - 2)(x + 3) = 6. 29. (x + 2)(x - 3) = 6. (x 31. (x - l)(x - 2) = 6. x(x + 5) = 14. = 4. 33. x(x - 3) = 10. (x + l)(x 2) 2x2 + x - 1 = 0. 35. 3x + 2x - 1 = 0. 37. 4x 2 - 3x - 1 = 0. 3x - 2x - 1 = 0. 39. 9x + 6x + 1 = 0. 4x - 4x + 1 = 0. 4x + 126x + 96 = 0. 41. x - Sax + 16a 2 = 0. 43. 6x - x = 0. 9x + 30x + 25 = 0. 45. 8x 2 + lOx = 0. 9x - 3x = 0. 47. ax 2 + 6x = 0. lOx + ax = 0. ex - dx = 0. 49. 6x + x = 2. 2 = 51. lOx - 13x = 3. 8x + lOx 3. 53. 15x + x = 2. lOx2 - x = 2. 55. llx = 6 + 3x 2 x = 2 - 15x2 57. 9x = 2 + lOx 9x = -2 - lOx 2 59. 23x = 6 + 21x 2 23x = -6 - 21x2
60.
25x
1.
4. 7.
10. 13. 16. 19.
22.
2x 2
2
2
8
2
2 2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
.'
25.
28. 30. 32. 34. 36. 38. 40. 42.
44. 46. 48. 50. 52. 54. 56.
2
2
.
2
2 2
2
2
2 2
2
2
2
2
2
2
2
.
.
2
.
.
.
- -6 -
5
64.
2
14x2
.
^- = % - -^^2
.
7
2 (2x-3)(2x+3) = 7x -4.
61.
_ 65.
x
5
15
x
2
,
+
12
^
=
0.
QUADRATIC EQUATIONS
166 1 t
5
h
v'
1
*
4z 2 2
,
8z2
=
-
-
16z 4
5
IX
3
=
-
25*
-x +2
1
x- 2^5(2 -*)-
68. 10z 2
70.
4'*'
[Ch.
a:
7*.
3x
=
0.
9.5. Solving quadratics
by completing the squares
The
algebraic methods of solving quadratics thus far discussed, while short and efficient, cannot always be applied.
The method
called
'
completing the square' has the ad-
vantage that it will solve all types of quadratic equations, besides giving practice in an operation which is useful else-
where in mathematics.
To
illustrate, consider
3z 2
(1)
Step
L
the equation
-
2x
-
2
=
0.
Divide both members by 3: ~2 x
/o (2)
^ ^ n _____
o.
Step 2. Transpose the constant term: /o
3.2
/
_
?x o
_
2 o
Add to both sides the number which will make the member a perfect square. Note that in the perfect square
Step 8. left
x - a )2 =a: 2._2aa;+a 2 the third term, or a 2 is [( )(-2a)] 2 Hence we add to both members of the equation the square (
,
of one-half the coefficient of x, or [(!)(
yielding
.
,
2
I)]
i
n
this case,
9.5]
SOLVING QUADRATICS BY COMPLETING SQUARES
157
HH
or
Note that the number added to complete the square is always positive, and that the middle sign of the left member of (5) is the same as the sign before the term of first degree in (4).
Step
4-
Solve for x !
*.
5.
/7
4.
Transpose the term
(7,
Equation (*) (&)
:
s-^+^i+
(R (6)
Step
.-'* (7) is
^:
o,
-!$
a brief way of stating that
--13
x
V?
-
u
,
or
-rx
1
'^ 5
o
Note that the various steps yield equivalent equations, and that the two equations in (6), (7), and (8) are equivalent to the one equation (5). In other words, each of the two values of x shown in (8) satisfies (1), as the student may verify. If, at the stage in the solution represented by equation (5), the right side is negative, the roots will be imaginary. For
example, given (9)
x*
+ 2x + 4 -
0,
then (10)
x2
+ 2x + 1 = 4 + 1; (x + l) =-3; x + l ^V^3 =iV; z=-l tV. 2
(11) (12) (13)
QUADRATIC EQUATIONS
158
The sum and product of
9.6.
[Ch.
IX
the roots
To
get a result which will prove useful for checking purposes, we note that (1) is
ax 2
+
x2
+
equivalent to
(2)
obtained from
(1)
-
satisfied
by x =
r
(r
a
+
a
by dividing the x2
(3) is
+c=
bx
=
coefficients
by
a.
Now
+ s)x + rs =
and by x
=
s,
as
may be verified by trial.
Hence if the general equation (1) is also satisfied by the r and s, it follows, by comparison of (3) with (2), that r
(4)
roots
+ s=-^
and
=
(5)
-
In words, .
the
sum
of the roots of (1) is
,
and
their product
c
is
a
be used in checking the roots found in the solution of any quadratic. The check is often shorter than that obtained by substitution of the roots in the original equation, particularly when the roots are irrational or This result
may
imaginary.
Example L^By use .,
.
that
of the^sum
-+Vll - --Vll 1
2
and
and product formulas, show
1
r
.
,
2t
2x* Solution.
a
-
.,
are roots of the equation,
2x
-
5
=
0.
THE SUM AND PRODUCT OF THE ROOTS
9.6]
159
and
+VTI
1
-Vil _
' 1
_
_
+VTT/i -Vilx =
/i
(2 X^
,
and
Example
2.
Show
-/IP =
44
2
T
)
that r
~ 3 + tv/31
=
are roots of the equation, 2z Sofoilton.
r
i
+s=
2
+
+
3z
10
-T
an d
=
5
s
=
=
5
-2
~ 3 ~ t%/31
0.
636 -=
-= ---
4
2
a
Also,
= /_ 3 Y _
Find
^^ 4
4J
(
EXERCISE
i
t
=
_ ~ 31 =
40
16
16
16
=
5 2
the roots of each of the following equations
x2
5.
2
9.
11. 13. 15.
17. 19.
21. 23.
6z
+8=
0.
- 8 = 0. x - 2z - 8 = 0. 6x - 7* - 3 = 0. 6x + 7x - 3 = 0. 12x2 + 7x - 12 = 0. 3x - 2x - 2 = 0. 5x - 5x - 7 = 0. 2x + 6x = 5. 7x = 14x - 3. 5x = -3x - 4. 4x2 = -6x - 3.
3.
7.
-
+
2x
a
by the method of
sum and
formulas.
x2
c
43
completing the square, and check by use of the
1.
=
2.
4. 6.
2
8.
2
10. 12.
2
14.
2
16.
2
18.
2
20.
2
22. 24.
x2
+ 5x + 6 = -
-
0.
= 0. x + 3z - 10 = 0. 6x2 + x - 2 = 0. 6x2 - x - 2 = 0. 8x2 - 6x - 9 = 0. 5x + 3x - 3 = 0. 3x + 4x = 5. 8x2 + 4x = 3. 9z2 = 6x - 5. 3x2 = -5x - 4. 10x2 = -3z - 2. x
2 2
2 2
3x
10
product
QUADRATIC EQUATIONS
160
25.
3z
27.
3z
29.
6z
31.
(re
33.
= = =
a:
z
2 2
z2
+ 7. + 6. + 2.
+ l)(z (x + 3)(z
34-43. Check
= 4z = 8z = (x
26. 2x 28.
30.
= =
2) 4)
32.
3.
IX
+z z + 6. z + 3. 2)(z + 3) = -7. 2
5
.
2
2
1.
substitution the
direct
by
[Ch.
answers found in
problems 1-10. Solve,
44.
9.7.
z x
and check
+3 -
TTfce
1
3
answers by direct substitution.
the
-
2z
2z
7
+2
2
quadratic formula
When we (1)
solve the general quadratic
ax 2
+
+c=
bx
by completing the square, the various equivalent equations obtained in the successive steps are as follows (2)
x2
++ a x2
(3)
/r
/
--
-
0;
=-
+
v
i
=
a
;
u
:
4dC.
4a (6)
(7)
x
A + 2a
=
x
=
I
y-k
_!_
x->
2a ^
'
formula. In a sense, the infinitely many quadratic equations are here solved collectively once for all. They are seen to have two roots apiece,
Equation
(7) is called the quadratic
THE DISCRIMINANT OF A QUADRATIC EQUATION 161 though these roots are equal when the quantity under the radical sign is zero. To get the roots of a specific quadratic 9.8]
necessary only to identify the coefficients a, 6, and c, and then to substitute their values in (7). If the equation is it is
quadratic in some letter or quantity other than or quantity should replace x in (7).
x, this letter
Example. Solve the quadratic 3z 2
(8)
Here a =
Solution.
these values in (7),
-
3, 6
2x
=
-
4
2,
=
0.
and
6
6 is,
one root
.
,
4.
Substituting
+ V4 - (-48)
_ /13)
6
=
1
6
+VT3 3
and the other
-~r
is
2
-/13 is
o
After getting x to write x
=
we have
-(-2) + V(-2) 2 - 4(3)(-4) _ 2-3 _ 2Vl3 = 2(1 = 2 +V52 = 2 That
c
=
= 4VT3
o
=-,
or x
6
the careless student
= OVT3 Why 6
is
likely
u are these results .,
.
m-
correct?
Note that
if
were solved by completing the square, the would be obtained at once. This is some who work in the field of applied mathe-
(8)
correct simplified root
one reason why matics prefer the square-completing method to the formula. The latter method, on the other hand, is shorter for some problems, especially where literal coefficients are involved; and it also gives other useful information about quadratics,
we
as
9.8.
shall see.
The discriminant of a quadratic equation
The (1)
roots of the quadratic
ax2
+
bx
+c=
0,
QUADRATIC EQUATIONS
162
as given in the quadratic formula, rately thus:
_ -b + Vb - r2
/0 ,
_
(2)
The radicand
62
2
,
(2),
which we
IX
be written sepa-
- s-_ -b -Vb
4ac
4ac in
may
[Ch.
may
4ac
designate
by
called the discriminant of the quadratic equation (1). D, For example, the discriminant of is
2z 2
(3)
for
which a
9 +_32
=
= 2,
41.
6
-
3x
-
4
-
0,
= -3, andc = -4,
Hence, the roots of
V41, and will be irrational. Having the formulas (2)
is
2 (~3) -4(2)(-4) =
(3) will
involve the radical
mind, we can draw certain conclusions about the roots of (1) from the value of D, assuming that a, fr, and c are rational numbers. Premise about (a)
(b)
D> D>
in
D
Then the
and a perfect square and not a perfect
(d)
unequal, and rational.
real,
and
unequal,
irra-
tional.
square (c)
real,
roots of (1) are
D= D<
real, equal,
and
rational.
imaginary.
These results have immediate and practical application as an aid to the solution of quadratic equations. When, for a
D
is negative or not a perfect square, particular equation, it is useless to waste time on the factoring method of solu-
tion, since the roots are
Question.
Can 3x 2
imaginary or irrational.
4x
5 be factored?
60 = 76 Answer. Since D = (-4) 2 - 4 3(-5) = 16 (not a perfect square), the answer, if we bar factors with irrational coefficients,* is 'No.' Hence, the quadratic
+
*
Note that 3z 2
-
4z
-
5
=
/
3
I
x
- -2
--VigW x 11
cover after solving the equation. This type of factoring obtaining the solution.
- -2 + Vl9 -
1,
is
as
we
dis-
of course not helpful in
9.8]
3z
THE DISCRIMINANT OF A QUADRATIC EQUATION
2
4z
must be solved by completing the square the quadratic formula. If the latter method is
5
by use
or
of
used, the test-value
writing
D=
In each
:
6 44
of problems 1-45, determine the nature of the roots by use
of the discriminant.
Then
perfect square; otherwise
1.
3. 5. 7.
9.
11. 13. 15. 17.
19.
21. 23.
25. 27.
29. 31. 33. 35. 37. 39.
41. 43. 45.
76 already found can be used_m = 4 + V76 2 Vl9
down at once the solution x
EXERCISE
zero is
solve the equation
by factoring if
D
- 3x + 5 = 0. 4x2 - 4x + 1 = 0. x2 - 3x + 2 = 0. x2 + x - 3 = 0. x - 3x + 1 = 0. x - x + 3 = 0. x2 + x + 1 = 0. 3x + x - 3 = 0. z - 10* + 1 = 0. 3x2 - 5x + 1 = 0. - 5y + 1 = 0. 4?/ 2 9x 3x + 2 = 0. 2 5x 6* + 1 = 0. 2 $y + 2 = 0. 3y 10x - 3x - 1 - 0. 7x2 - 3x - 2 = 0. % - 12y + 1 = 0. 15x2 - lOx + 1 = 0. ax2 - bx - c = 0. ay + Vy + c2 = 0. 3ax + 26x + c = 0. - l)(x - 2) = 3. (x = 5. y(y + 4) 2x2
3x 2
4.
2
6. 8.
10.
2
12. 14.
2
16.
2
18.
20.
2
2
.
- 2 = 0. 5x + 2x - 3 = 0. x - 6x - 9 = 0. x + 3x - 1 = 0. x - 2x + 2 = 0. 9x2 - 30z + 25 = 0. 2z - 5x + 1 = 0. 4z2 + 3 - 1 = 0. - y + 1 = 0. ICty 3x - 7x + 2 = 0. 7z2 - 8x + 1 = 0. - 5y + 1 = 0. Gy 8z - 7z - 1 = 0. 5x - 4x - 1 = 0. 3/ - Qy + 2 = 0. 12x - Sx + 1 = 0. llx - 12z + 1 = 0. 12^ - lOy + 2 = 0. 6x2 + ex + a = 0. ox2 + 26z + 3c = 0. - by - 2c = 0. 2ay - l)(x + 2) = 2. (x
2.
2
2
is
a
by use of the quadratic formula. (Note that
a perfect square.)
2
163
=
22. 24. 26.
28. 30. 32.
34. 36. 38. 40.
42. 44.
+
2 2 2
2
2
2
2 2 2 2
2 2
2
5x
QUADRATIC EQUATIONS
164
46. Solve s
=
47. Solve s
=
for
vrf
n -^
[2a
+
(n
-
for
t
t>
[Ch.
IX
.
l)d] for n.
t
+ y + 4x + 2y - 1 =
48. Solve x 2
2
49. Solve g
=
^
=
S0 a S lve
9.9.
.
'
C
for y.
R for
- 2c) - c)
R;
for x.
-
f r C'
T'/^ graphical solution of
Four types of results
a quadratic equation
for quadratic equations are illustrated
by the roots of the quadratics (l)-(4) discussed in Art. 9.2. The graphs of the functions of x in the left members of these equations give us a geometric interpretation of their roots. Designating each of these functions as y, we have the four equations:
=
x2
(1)
+
y
+
=
x2
+ 4x
4:X 4, and (4) y y in table the below. analyzed
(3)
5,
(2)
= x2
+
y 4x
=
x2
+ 8,
+ 4#
1,
which are
GRAPHICAL SOLUTION OF A QUADRATIC EQUATION The four graphs are shown in Fig. 14. The curves are members of the family of curves
9.9]
165 all
c.
(5)
y
Each value assigned
to c yields a special curve, as indicated
in the figure.
The graph
(5,
of (1) crosses the X-axis at the points (1, 0) and 0). Another way of saying this is that the rr-intercepts
166
of the
graph are the equation
and
1
5.
+ 4z -
x*
(6)
For evidently at x
=
QUADRATIC EQUATIONS [Ch. IX These numbers are the roots of
=
5
0.
and
5 the curve (1) crosses the and hence (6) X-axis, so that for these values of x, y = is
1
satisfied.
Similarly, the graph of (2) crosses the X-axis in two points; but in this case their abscissas are irrational, namely,
2
+ V5
and
V5. These numbers
2
are the roots of
the quadratic
In the graph of
ward
until the
(2,
0).
(3)
1
=
0.
the curve has in effect been lifted upat the point
two ^-intercepts have coincided Hence the equation
+
z2
(8)
may
+ 4z -
x2
(7)
be said
still
+4=
4*
two roots
to have
2 and
(
2)
which are
identical. The graph of (3) is said to be tangent to the X-axis at (-2,0). Finally, since the graph of (4) does not intersect the X-axis, there are no positive or negative values of x for which y = 0,
and hence there are no z2
(9)
Summing
real roots of
+
8
=
0.
up, if the roots of the quadratic
ax 2
(10)
are real,
+
4x
+ bx + c =
we may find them as accurately as the precision of the from the graph of the corresponding equation
drawing allows (11)
y
=
ax 2
+
bx
Incidentally, the graph of (11),
+
c.
where a
^
0, will
always be a parabola, a very important curve entering into life and mathematics in many ways. Different sets of values for a, b,
and
c give different parabolas;
in general like the ones
shown
but
all
them are shaped They open upward
of
in Fig. 10.
9.9]
GRAPHICAL SOLUTION OF A QUADRATIC EQUATION
as in that figure
a
if
is
positive,
and downward when a
167 is.
negative. The formula for the abscissa (or x- value) of the vertex of each parabola represented by (11) is
,
(12)
Example. Sketch the parabola, y
(13)
Solution.
=-2z + 2
Using (12) and
of the vertex are:
x=-
(13),
4x
-
we
find that the coordinates;
1.
4
1-1 = 1.
2
=1; ?/=-2(l )+4 Tt
The curve opens downward, negative. The coordinates, (0,
since the coefficient of x2 1),
of the point at
is
which the
curve crosses the F-axis are easily found by setting x
=
0*
we know
the vertex, the direction of the axes, and one on the point curve, we can now sketch the parabola shown in Fig. 15. For greater accuracy, we of course use more points whose coordinates satisfy (13). From the graph we might estimate the values of the two roots of Since
-2z2
(14)
as about
.3
and
1.7.
nearly as estimated.
The
+
4*
-
1
=
Actually they are
2
+ /2
^
,
or very
graphical solution of a quadratic equation is interesting not only because it gives geometric meaning to the differ-
QUADRATIC EQUATIONS
168
ent types of roots encountered, but also because
it
[Ch.
IX
illustrates
a very general method of getting approximations to the real roots of an equation in one unknown. This method is very
no shorter way can be found. In the case of quadratic equations, however, the algebraic methods are more efficient, precise and practical. reliable as a last resort
EXERCISE
if
45
Graph the left members of equations 1-15, Exercise 43. From the graphs, estimate to one decimal place the values of the real roots, and compare with the precise values found algebraically. 9.10. Quadratics with given roots
THEOREM
1.
If r
a root of the equation
ax 2
(1)
then x
To say
that r
2
(1)
+
bx
and
+
+
br
0,
member
+c
(1)
=
of (I).
means that
0.
(2),
c)
ar
or, since
left
a root of
is
ar2
(2)
From
+ bx + c =
r is a factor of the
Proof.
(ax
is
2
ax 2
-
(ar
2
+
+ br + c + bx + c
br
= = =
+
=
c)
a(z
2
-
r2 )
+
b(x
-
r),
0,
a(x
r)(x
(x
r)(ax
+ r) + b(x + ar + 6).
r)
2 has the root ^, and, there3 = Example. 2x + 5x 2 5x 3. must be a factor of 2# fore, by the theorem, x 2 = 3 2(x &(x + 3). Actually, 2z + 5x From Theorem 1 there follows, as a corollary,
+
THEOREM the (3)
2.
All quadratics whose roots arer and
form a(x
-
r)(s
-
where a can be any constant except
*)
=
zero.
0,
s,
must take
PROBLEMS LEADING TO QUADRATIC EQUATIONS
9.11]
Example 1. Find a quadratic equation with and cients whose roots are f
169
integral coeffi-
.
Solution.
The equation must have
the form
+ f) =
Letting a
6 to clear fractions,
Qx
Example .
,
cients
2.
2
=0.
we have
5x
6
=
(answer).
Find a quadratic_equation with integral
,
,
whose roots are
- -2
coeffi-
+ -tV3 Zi
Solution.
or,
The equation must have
when a =
2
+ tV3T
2
-
2
the form
-
tV3'
4,
(2z
-
-
2 -f tV_3) 2 2 2) (zV3) (2x 2 4x 8x 7
tV3)(2z
or
+
or
EXERCISE
= = =
0, 0,
0.
46
Find equations with
integral coefficients whose roots are the
num-
bers below. 1.
7. f,
11.
2. 3,
1, 2.
3
IS. 1
-f V2. i.
-1.
-1, -3.
3.
8. 0, f.
4. 0, 2.
9. 1,
12. 1
~
13.
16. 2
3.
17.
-4.
10.
-f.
2
6.
i
f.
+ V3.
3
X
^ 2
5. 0,
-
i/3.
14. 18. 1
t.
+
^-
9.11. Stated problems leading to quadratic equations
The
stated problems calls for the use of quadratic equations. If two different numbers satisfy the conditions of the problem, they will be the roots of the solution of
many
QUADRATIC EQUATIONS [Ch. IX quadratic obtained. If only one number meets the required condition, either this number will appear as a double root or else the second root must be rejected as meaningless. Finally, 170
the conditions as stated are inconsistent, this fact will appear algebraically in the form of imaginary solutions to if
the quadratic.
Example is
1
.
Find two consecutive numbers whose product
56.
x
Solution. Let
Then
x x(x
Solving,
+ + 1)
1
= = =
the smaller
number
(algebraically).
the other number. 56.
we have x
+
x
i
Hence the answers
= =
7 or
Its length
A
building lot has an area of 56 square rods. one rod greater than its width. Find its dimen-
2.
is
and 8; (2), 8 and 7. Here meet the required condition.
are: (1), 7
both roots of the quadratic
Example
8;
8 or -7.
sions.
Here the quadratic obtained is the same as that Example 1 but the root 8 must be rejected as meaning-
Solution. for
;
less in this case.
Example 4 units
less
3.
The
than
area in square feet of a certain square is perimeter in feet. Find the length of its
its
side.
Solution. Let
Then and
Solving,
x x2 4x x2
= = = =
the length of a side, in feet. its area in square feet,
=
2 (answer). Here 2
its
4x
perimeter, in feet. 4.
we have x
is
a repeated root.
PROBLEMS LEADING TO QUADRATIC EQUATIONS 171 Example 4- The area in square feet of a certain square is
9.11]
5 units
less
than
its
perimeter in
Find the length of
feet.
its
side.
Solution. Stating the conditions as in
x
2
=
The quadratic formula x
4#
Example
3,
we have
5.
yields
=
4
+ =
2i
=
2
+
.
,
z,
and hence the conditions are not met by any square.
EXERCISE 1. is
47
Find two numbers whose difference
is
3 and whose product
40.
The base
of a ladder 20 feet long which leans against a barn 4 feet above from the barn. The top of the ladder is x the ground. Find x. 2.
is
x
+
feet
3.
A rectangular lot is
diagonal 4.
A
diagonal
is
250
lot is is
130
f as wide as it dimensions.
its
70 feet longer than it is wide. The length of Find its dimensions.
its
Find
is
long.
its
feet.
5.
Find two numbers whose sum
6.
The
If its
The
length of
feet.
is
20 and whose product
is
99.
length of a rectangle is 3 inches longer than its width. length is increased by 2 inches and its width is increased by
3 inches,
its
area will be doubled. Find
its
dimensions.
The length of a rectangle is twice its width. If its width is increased by 1 inch and its length by 3 inches, its area will be 7.
doubled. Find 8.
its
dimensions.
The diagonal
of a square
for the length of its side (a),
2 feet longer than its side. Solve by use of a linear equation only; is
by use of a quadratic equation. 9. The diagonal of a rectangle is 2 inches longer than length and 9 inches longer than its width. Find its dimensions. (b),
10.
more
The area
its
of a certain square, in square feet, is (a), 5 units than, (b) equal to, and (c) 5 units less than, its perimeter in
172
QUADRATIC EQUATIONS Which
of these conditions, the dimensions in these cases? feet.
[Ch.
IX
any, are possible, and what are
if
11. The length of a rectangle is 1 inch more than and its area in square inches is equal to its perimeter Find its perimeter if the conditions are consistent.
its
width,
in inches.
The length of a rectangle is 1 inch more than its width, and area in square inches is one-half of its perimeter in inches. Find perimeter if the conditions are consistent.
12. its its
13.
One
root of the quadratic, 2x2
bx
1
=
0, is 1.
Find the
other root. 14.
15. 16. 4,
and 17.
One root One root The sum
of 3x 2
of ax 2
+ 2x + c = 4z + 3 =
-1. Find the other
is 2.
Find the other
is
3. if
What must
root.
root.
2k = hx be the values of h and
of the roots of the equation 2x 2
their product Find the value of b
are equal.
is
+
is fc?
the roots of the equation 3x 2 +bx+2=
Chapter Ten
SPECIAL EQUATIONS IN ONE
10.1.
UNKNOWN
The general equation in one unknown
How
can we represent in one algebraic sentence all of the equations in one unknown, say x, which could be written? ' This appears to be a large order' but the answer is simple. ;
It is the
equation
=
/(*)
(1)
0.
Thus far we have learned how to solve when f(x) is a linear function, such as 2x
(1) algebraically 3,
or a quadratic
+
2
2x 5. function, such as 3x have seen further that there
We
is
a graphical method of
approximating the real roots of (1) (as illustrated in Art. 9.9 for the special case of the quadratic) which is so dependable that every student should bear it in mind as a last resort when algebraic methods fail him. The method consists of drawing
the graph of the related curve
y
(2)
=
/(*),
and then
finding, as nearly as the accuracy of the graph the allows, ^-intercepts of (2), which will be the real roots
of (1). In this chapter we shall discuss special equations in one unknown for which the exact values of the roots can be found ^
algebraic means. Linear and quadratic equations, which belong in this group, have been disposed of already
by
(Chapters 4 and
9). 173
UNKNOWN
SPECIAL EQUATIONS IN ONE
174
[Ch.
X
10.2. Solving equations by factoring
The method of member when the
solving a quadratic by factoring the left right member is zero carries over directly to equations of higher degree. The method succeeds when all factors found are linear or quadratic.
Example
Solve the equation
1.
-
x5
(1)
-
x
= =
x(x* x(x*
-
=
1)
+
0.
member, we have
Solution. Factoring the left
x5
=
x
x(x
-
l)(x
2
+
l)(x
l)(z
+
The roots obtained by setting each of these equal to zero are 0, b it follows that i
i
1,
i,
i,
=
and i
i
=
0.
-
1)
1).
factors separately
Since
1.
2
:
i
b
=
i
(Art. 8.2),
The student may check
each of the other four roots. Solve the equation
2.
Example
x*
(2)
=
1.
term and factoring the
Solution. Transposing the constant left
member, we have:
When z-l = 0, Hence
z=
a;
3
1
=
(x
l)(x
2
+#+ '
l;
when z 2 +z+l=0, x=
the_ roots of (2), or J;he three cube roots of
-1 +iV3 -1 -and
=
0.
are
1,
1)
1= i 1,
,
,
example one might easily overlook the two imaginary roots; but he will not do so if he remembers the following simple and important result, proved in more advanced courses: In the
last
THEOREM
1.
has exactly n
Every rational integral equation of the nth degree
roots, not necessarily all different.
For example, the hundredth degree equation, (3)
x(x
-
1)'
=
0,
EQUATIONS WITH GIVEN ROOTS
10.3]
175
has exactly 100 roots, including the single root zero and the multiple or repeated root
10.3.
1,
which
is
of multiplicity 99.
Equations with given roots
As a
corollary of the factoring method, and also as an extension of the method of Art. 9.10 applying to quadratics, it is possible to write immediately the equation in one un-
known having any given Example and -3. Solution.
1.
roots.
Write an equation whose roots are
a(x
- 0)0 -
a can be any constant.
T)(x
we
If
-
-
2)[z
let
a
=
(-3)]
1,
2,
where
0,
and perform the + 6x = 0. An-
1
indicated multiplication, we have: x4 12x = 0. 14x 2 other answer is 2x 4
=
0,
7x 2
+
Example
2.
Write an integral rational equation with
tegral coefficients
numbers)
whose roots are
1
&nd
i
in-
+ /3
2
=r
(four
.
Solution.
An equation
with the desired roots, unsimplified,
is r
a[x
-
/i
(1
-MF + i)][x ,
/i
(1
-
-xif
i)]x
2+V3T x II
2-V31 = A0.
-^
J
is not an answer because the problem calls an equation with integral coefficients. Hence we must let a = 9 (or 18, or 27, etc.) to clear fractions. With a = 9, we have
This, however,
for
(a-
- i)(x - 1 -f i)(3x - 2 -V_3)(3o; - 2 +V3) = [(x - I) - i ][(3x - 2) - (V3) = (x - 2x + 2)(9x - 12x + 1) = 0, 9x - 30x + 43x - 2Qx + 2 = 0. (Answer.) -
1
2
2
2
2
]
2
4
or If
2
3
2
we had chosen a =
coefficients
with the
18 the equation obtained would have
common
factor 2.
When
integral factors of the coefficients different
all
from
common
+1
or
1
UNKNOWN
SPECIAL EQUATIONS IN ONE
176
are divided out, the equation form, as is the answer above.
EXERCISE
is
[Ch.
X
said to be reduced to simplest
48
Solve by the factoring method. 1.
3. 5. 7.
9.
11.
z3
= -l.
z + 2x + 1 = 0. x + x - 6x = 0. x - x - x + 1 = 0. 2z - 4z + 3z - 6 = x4 - 4 = 0. 4
2
3
2
3
2
Find rational reduced
to
IS. 0, 1, 1,
19. 0, 1
6. 8. 0.
integral equations,
23. 0,1,
27.
i, i,
5
+x=
2z 3
-
0.
z 6z = 0. x + x - x = 2z = z - 2. + x + x = 0. 3
2
2
3
0.
4
3
with integral coefficients and sets of numbers.
14. 1,
-2.
16. 0,
-2, -3. 2, -3.
i.
18. 0, 1
iV2.
20.
1,
22. 1
-if.
3z.
V2,
1
^-
24. 1,1,2,2.
-i, -i.
26. 0, 0, 0. 28. 0, 0,
10.4. Equations in quadratic
an equation
is
,
1, 1.
form
quadratic, not in x, but in
of x such as x 2 x3 -, etc., the ,
V2.
2
-
-1, -1, -1.
If
4
1.
simplest form, whose roots are the given
1
,
25.
x5
4.
13. 1, 2, 3.
21.
12.
2
z
17.
10.
4= x5 z z4 z -
2.
x
method
some function
for quadratic equations
be used to find numerical values for the function. The solution is then completed by placing the function equal to each of the roots of the quadratic, and solving the two equa-
may
tions thus obtained.
Example (1)
1.
Solve the equation
x6
-
7x*
-
8
=
0.
EQUATIONS IN QUADRATIC FORM
10.4]
form
Solution. Writing (1) in the
we
-
2
3
(x
(2)
)
7(z
177
3
)
-8
=
0,
a quadratic in x 3 The solution by factoring or quadratic formula yields see that
it is
.
(3)
x3
=
z3
=-l.
8,
or (4)
By
the factoring
and
+ iV3,
1
Thus,
(1)
has
Example
2.
method we
find that the roots of (3) are
while those of (4) are
1
and
=^ A
2
--
:
six distinct roots.
Solve
_
.
x Solution. Let y
=
-4- 3
X
Then
(5)
-
-
4,
=
0,
y
(6)
becomes
jj
or 2
(7)
t/
from which value
is
i/
=
sought,
5 or
we
-
1.
replace
stands, getting (8)
4y
-
5
Remembering that it is x whose y by the function of x for which it
W3
=
5>
and
0) The
^nr roots of (8) are ^
and
1;
1
'
those of (9),
-1
t%/23 .
4 and two two real roots. has When it is imaginary (5) cleared of fractions it is seen to be a fourth degree equation.
Hence
UNKNOWN
SPECIAL EQUATIONS IN ONE
178
EXERCISE
[Ch.
X
49
Solve the following equations in quadratic form.
-
- 6 = 0. 2. 2x + x - 6 = 0. = 1 3. 6x< + x 0. 4. 6x - x - 1 = 0. 5. 2x - x - 1 = 0. 6. 2x + x - 1 = 0. 7. x + 7x - 8 = 0. 8. x - 7x - 8 = 0. 9. x + 26x - 27 = 0. 10. x - 26z - 27 = 0. 11. 8x - 63x -8 = 0. 12. 8x + 63x -8 = 0. 13. x - 28x + 27 = 0. 14. 8x - 19x - 27 = 0. IS. 8x + 19x 27 = 0. 16. 2(x +x) -5(x +x)+3=0. 17. 2(x 3 = 0. (x x) x) x 2x + 2x + 2 = 0. 18. (x I) 19. (x + x - I) - 3(x + x - 1) + 2 = 0. 20. (x + x + I) - x - x - 3 = 0. 21. (2x - x) + 2x - x - 2 = 0. 1.
2x*
xz
4
2
2
4
2
2
4
2
4
6
6
3
6
3
6
3
6
6
2
2
2
2
2
2
x)
2
2
2
2
2
2
2
2
-
3
2
3
25. 2(x 2
3
6
2
10.5.
3
6
3
6
3
2
2
- -^
+1 =
0.
Equations involving radicals
When the unknown appears in one radicand, or in several, the processes necessary to eliminate the radicals may lead to rational integral equations of the types already considered. Example (1)
1
.
Solve the equation
V2x
+ 3 +Vx +
1
=
1
Here the student often makes the
error of lifting off the radicals with the mistaken impression that he is .thus 2 6) squaring both sides. But this is a serious error, since (a Solution.
+
10.5]
=
EQUATIONS INVOLVING RADICALS
179
+ 2a& + 6 and hence (/2x + 3 + Va + I) = Thus the (V2x + 3) + 2V2o: + 3Vx + + (Vx + I) remains. To get a simpler radical 2V(2# + 3)(x + 1) a2
2
2
,
2
2
1
.
still
radicand, however,
it is
(2)
2z
When
members
its
better
+
=
3
first
1
to rewrite (1) thus:
-z +
1.
are squared, (2) becomes
2x+3-l ~2V^Ti + (V^Tl) = l 2
(3)
2
Simplifying and transposing terms so that the one remaining radical is by itself in the left member, we have
2Vz +
(4)
A
= -x -
1
1.
second squaring yields 4(x
(5)
+
1)
=
x2
3
=
0,
+
2x
+
1,
or
x2
(6)
-
2x
-
1. from which x = 3 or jBt^f we /iat>e not finished the problem. For when we squared both sides in steps (3) and (5) we in effect multiplied both sides by functions of x, and hence may have introduced ex-
+V3 +
+
extraneous and must 1 = 3 V2(-l)
3
is
+ +V-1 +
(1)
and
is
member
of (1) we (not 1), so that be rejected. But, for x
traneous j^ot^Testingj^^ 3 in the 1 = 3 3 have: /2 3
left
+ 2-5
1
+
=
=1,
1.
Thus,
-1
satisfies
the only root.
Example
2.
Solve
Vz=-l.
(7)
Solution. Squaring both sides, we get x = 1 ; but this l. Hence (7) has no root. be rejected since ?
must
VI
Note that
not a rational integral equation in x. Thus while, as noted before, it can be proved that every rational integral equation of the nth degree has n roots, other types of equations
(7) is
may
not have any roots.
SPECIAL EQUATIONS IN ONE
180
EXERCISE
UNKNOWN
[Ch.
X
50
Solve the following equations involving radicals. 1.
3.
+
5.
x
7.
x +~2
1
-Vx -3 = +V3 - x =
2.
x
1.
+
2
+V2x + 5=1.
9.
11.
Vx +Vx + 2 = 2. 15. V3x +Vx + 1 = 2 = 16. Vx_+ + /z +
13.
1
17.
Vx
18.
Vx -
1
+Vz + 3 = V2a; +
2.
In the following articles a few theorems and rules are given which are found to be of great assistance in solving equations of a higher degree than the second. For supplementary material covering this work a student
is
referred to the
chapter headed 'Theory of Equations' usually included in
any
college algebra text.*
The remainder theorem
10.6.
The remainder theorem may be // r
is
stated as follows
:
a constant and if any polynomial in x is divided by a remainder is obtained that does not contain x,
r) until
(x that
remainder
is the
value that the polynomial would have if r
were substituted for x. Proof. Let (1)
f(x)
= (x-
r)Q(s)
+
R,
* For example, see Rider's College Algebra, alternate edition, pages 187-227. t A polynomial in x, as used here, means an integral and rational function of x, 3. (See Art. 2.1.) For with integral coefficients, such as 3z* - 2z 4 -f x 2 - 4x 4 3x 1 means that f(x) brevity, it is often designated as f(x). Thus, f(x) = 2x 1. here stands for the particular polynomial, 2x* 4- 3z
+ +
SYNTHETIC DIVISION
10.7]
181
the Quotient and obtained when f(x) is divided
where Q(x)
member
of (1)
is
the constant remainder
by x
=
/(r)
Since the right
r.
the same function of x as the
is
when x =
(2)
R
two
different algebraic form, the
though in equal
is
member,
sides will be
Hence
r.
-
(r
left
+R =
r)Q(r)
Q(r)
+R=
R.
Example. Divide 2xz (2z
-
3
3x
2
-
3z 2
+x~
+x^
1)
i
/y
_ x
-
by x 1)
-
2s3
-I
X __
(x
1
-
3z 2
X2
~~~ ^
L
-
1.
+xi
I
~> X
1
____ ~'~'
-
1
1 1.
-
1
x
1
= 1. The rer is x In this example x 1, so that r 1. Here, according to the mainder, which we shall call R, is 1 in if we for x the polynomial 2x3 substitute theorem, x 1. To 3z 2 1, the value of the polynomial will be that note we check this,
+
on^s 4(L)
_
Qn> o^i;
2
4-1 -fl
1 i
9 z
.
Q 4_ i o-fi
i 1.
i l
10.7. Synthetic division
A
condensation of the operation above which retains only
the essential numbers
Example
1.
is
Divide 2z3
called synthetic division.
-
3z2
+x-
1
by x
-
1.
and place 1 (the value of in the first line below. as in the position of a divisor, First,
copy the
coefficients
2-3 1 - 1 2-1 2-1 0-1
r),
UL
Next, draw a line two spaces below the coefficients and copy the first number below this line. Multiply this number
SPECIAL EQUATIONS IN ONE
182
by the number
(2)
UNKNOWN
[Ch.
X
the divisor's position, place the coefficient (3) and add. Place
(1) in
under the next the result ( 1) under the
result (2)
Multiply this number
line.
1)
(
by the number (
1)
sult
in the divisor's position (1), place result the next coefficient (1), and add. Place the re-
below (0) below the
Multiply this number
line.
(0)
by the
number
(1) in the divisor's position, place the result (0) the next coefficient ( 1), and add. Place the result
below (
1)
below the
The
line.
first
numbers below the
three
line
are the coefficients of the Quotient and the last number ( 1) below the line is R, the remainder. The degree of the Quotient will be one less than the degree of the polynomial divided.
The complete
Example
2.
-
Divide 3x 3
-
4z 2
5x
3-4-5
Solution.
2x 2
is
quotient, then,
+
7
-
x
by x
x
-
1
2.
7 [2
4-2 2-1 5
6 3
The Quotient
is
3x
+ 2x
2
The complete quotient
is
and the remainder (R) 5 1 H 3z2 + 2z 1
x
For our purpose, the remainder
(5) is
part of the result obtained because polynomial with 2 substituted for x.
This
is
it
the most important is the value of the
stated in functional notation as follows
If
}(x)
then
/(2)
Example
3.
3x3
-
4z2
-
+
5z
7,
5.
Find /(2) by synthetic division f(x)
Solution.
= =
=
5z4
-
Zx3
+ 6x + x 2
5-3
6
10
14
40
7
20
41
5
/.
is 5.
/(2)
1-4L2 =
78.
82 78
if
4.
:
10.8]
THE FACTOR THEOREM
183
In the examples given, every power of x from the highest 2 3 power down to the constant occurs. That is: x x x and ,
the constant term
occur. If the constant or
all
,
any power
of
missing, a zero must be placed in its proper position in writing the coefficients in form for synthetic division.
x
is
Example
-
Divide 2x 5
4.
+
2x 3
1
by
+
(x
3).
position,
To determine the number to place in the divisor's r = x + 3, the divisor, and solve for r, we let x
getting r
=
Solution.
3.
Notice that the x 4 term, the x 2 term, and the x term are missing from the expression. We must use zero for the coefficient of each of these.
20-20 18-48
144
2-6
144
1
6
Thus,
if
10-48 /(x)
=
2x 5
-
2x 3
-
-3
432 431
+
1,
-3 =-431. The factor theorem
10.8.
The
factor theorem
///(r)
=
is
0, then (x
We know
usually stated as follows: r) is
=
that /(r)
J?,
a factor off(x).
the remainder. Thus, if /(r) = 0, r), divides the function exactly.
then R = 0, and hence (x r is a factor of the function. Synthetic In other words, x to determine whether or not /(r) = 0. used is division
Example
5. Is
x
/(x)
2 a factor of
=
2x3
-
2-3
Solution.
3x2 5
+
-
5x
14
-
14?
J2
2170 4
/.
/(2)
=
0,
and x
-
2
2
is
14
a factor of /(x).
SPECIAL EQUATIONS IN ONE
184
EXERCISE
UNKNOWN
[Ch.
X
51
Divide, using synthetic division. 1.
2. 3.
4. 5. 6. 7. 8. 9.
10. 11.
12. 13. 14.
By on the 15. 16. 17.
18. 19.
20.
21.
10.9.
(3x
3
-
+ 4x -
2x2
-
5) -^ (x
-
1).
x - x - 2) (5x + 3x (x + 1). - 2). 8x 7x + 7x + 6) (2x (x 4 - 2x - 3x + 2) (x + 2). (3x + 6x - 3x - 9) + (x + 3). (4x + 12x 4 5x x + 5) -Mx - 5). (x 12x 8x + 32) 4- (x - 4). (3x - 12x - 3x + 18) ^- (x - 6). (2x - 5x + 2x + 4) (x - 3). (3x - 3) ^- (x - 2). (5x + 4x - 2x + 5) (x - 1). (3x - 3x + 2x - 5) (x + 1). (4x - 6x + 5x) -^ (x + 4). (7x 6x - 2x + 4x) H- (x - 2). (3x 4
3
4
3
2
-T-
2
3
-f-
2
3
2
4
3
4
3
-r-
5
3
2
3
2
4
2
5
3
3
2
4
3
-=-
-T-
2
-f-
2
use of the factor theorem determine whether or not the expression left is
a factor of
the
polynomial on the
-
+
x
-
2;
x
3
4
3
4
2
3
4
right.
-
+ 2x x 2. x 2. I;x + 2x x + 1; 3X + x + x + 4x + x + 3; 2x + 6x - x - 3. x + 2; 3x + x - lOx. x - 3; 5x - 3x + x - 2. x + 4; x - 3x + 2x - 8. x
4
1.
3
2
3
4
3
3
2
Theorem on rational
roots
If an equation '
+ aiZ'- + 1
a 2x n
~2
+
with integral coefficients, has a
+ a -& + a = n
n
el
duced
to lowest terms,
divisor of
a
.
0,
c c rational root ~, where - is re-
then c is a divisor of a n
a and d
is
a
10.10]
UPPER AND LOWER LIMITS OF ROOTS
Example.
If
the equation
3z 3 has a rational root,
-
4z 2
Xn tircta
+
Any
f
+
& 2X n-2
If
of the following numbers. *.
an equation
----
an
=
+ fc^ +
=
&n
integral divisor of
bn
(^
.
the equation
x3 has a rational root, 1,
10.10.
_
2
i,
rational root of
& ia .-l
integral coefficients, is
Example.
+ 5x +
must be one
it
f,
Corollary.
185
it
3z 2 will
2,
+ 4z +
=
be one of the following numbers.
+4,
3,
Upper and lower
12
12.
6,
limits of roots
The
possible rational roots are checked by synthetic division to determine whether or not they are actual roots. If,
when any
positive
number
tested, the
is
sums beneath the
can be proved * that there is no root larger than the number being tested. This number is therefore an upper limit of the roots. Similarly if, when a negative number is tested, the sums line are all positive, or zero, it
are alternately plus and minus throughout the line, it can be shown that there is no root less than the number being tested. That number is therefore a lower limit.
Example. Test the number 2 as a possible root of 5x
+
2
3-4
5
6
4
18
2
9
20
3
* The proof is not difficult. The student Exercise 51. See also problem 24.
=
0.
2[2
is
challenged to try
it
in
problem 23 of
The
UNKNOWN
SPECIAL EQUATIONS IN ONE
186
[Ch.
X
and hence the roots
signs are all plus in the lower line
of this equation are all less than 2. 1 as a possible root. Test
3-4 -
5
-1
2[
3
7-12
3-7
12-10
The
signs in the lower line are alternately plus and minus, 1. and hence the roots of the equation are greater than ' By this means we may eliminate some of the possible
roots' without testing them by synthetic division. 10.11. Depressed equations
Consider the equation 2
(x
(1)
Any value makes x x
3
=
-
+ 2)(x -
3z
0.
x that makes x 3 = 0, or any value of x that = 3x + 2 0, is a root of (1). The statements, and x 2 3x + 2 = are called depressed equa-
of
2
0,
with relation to equation found by synthetic division.
tions is
=
3)
Example
L
-
Solve: x 3
6z 2
1-6
The depressed equation
(1).
+
-
llx
-
396 1-3
Solution.
11
6
6
=
0.
_3
2
Here the remainder is zero. Use the numbers below the and reduce the degree of the expression 2 3x + 2 = 0. by one. The depressed equation, then, is x
line as coefficients
Since any root of the depressed equation is also a root of the original equation, we solve the depressed equation for the
remaining roots.
z2 or
(x
-
z~2 x
-
3z
2)(x
= =
+2
-
1)
= =
0;
3-1
2;
x
0,
0.
= =
0; 1.
DEPRESSED EQUATIONS
10.11]
But our synthetic
187
division shows that 3
is
a root, since
produces a zero remainder. Therefore, the roots of the nal equation are 3, 2, and 1.
Example
+
4z 3
The
3,
-
8x 2
-
3x
-
6
0.
possible rational roots are 3,
2,
1,
Testing
origi-
Find the roots of the equation
2.
Solution.
it
i,
6,
:
f,
f.
i,
we have 4
-
8
-
3 12
12
4-4
-
6
-3
27
9-33
not a root, and this test shows that all roots are 3, since the sums below the line alternate in greater than sign. Next, -2 4 3
Thus,
is
-
8-3-6
6
8
0-3
4 This test shows that
a root, since the remainder is 2 Ox 3 0. Upon zero. The depressed equation is 4x 2 = = + Vf- = + f /3. solving the latter, we find that x f and x 2
is
+
,
Thus the
roots are
2,
,
and
Zi
EXERCISE
Lt
52
By use of the preceding theory, determine what numbers could possibly be roots of the following equations, and then test them. If the final depressed equation is quadratic, find all the roots.
- x + 3 = 0. 2. x - 3x + 2 = 0. 4. x 4 - 2x 3 + 1 = 0. 3. x + 2z - 3 = 0. 2 6. 2x + 3x 4 - 3x + 2 = 0. 5. x 4 - 3x + 2 = 0. 7. 2x 3 - x 2 - 2x + 1 = 0. 8. 6x 3 + 19x + 15x + 2 = 0. 9. x4 + 3x - 3x2 - 12x - 4 = 0. 10. 6x 3 + 19x + x - 6 = 0. 1.
x3
-
3
3x 2
3
s
2
2
3
2
SPECIAL EQUATIONS IN ONE
188 11. 12. 13. 14. 15. 16. 17. 18. 19.
20. 21. 22.
UNKNOWN
[Ch.
X
- 2x - 6x + 6x + 9 = 0. Sx - 4x - 14x + 5x + 5 = 0. 6X + 3x - llx - 4x + 4 = 0. 2x + 5x + 2 = 0. 3X4 + 2x + 13x + 8a; + 4 = 0. + x - 2 = 0. 4 3s + 2x - 49x - 32rc + 16 = 0. x + 3x - x - 3 = 0. x - 4x - 8x + 32 = 0. 4X - 12x - 25x + 27x + 36 = 0. x + 8x - 12x - 24x + 27 = 0. x4 + 3x + x - 2 = 0. s4
a;
3
2
4
3
2
4
3
2
4
2
3
2
3
2
3
6
3
6
3
4
4
z
2
3
2
2
3
3
2
23. Prove the statement beginning '.. first paragraph of Art. 10.10.
it
can be proved' in
24. Prove the statement beginning '.
it
can be shown' in
the
the second paragraph of Art. 10.10.
.
.
Chapter Eleven
SIMULTANEOUS EQUATIONS
The general problem
11.1.
In this chapter we shall deal with pairs of simultaneous equations in two unknowns, at least one member of each pair being quadratic, or of the second degree, in one or both of the unknowns. Equation-pairs of this sort arise frequently in the solution of simple problems.
As
in the case of linear equations, the solution of simul-
taneous quadratics may be obtained graphically by use of the rectangular coordinate system. In this case, however, it is necessary to plot loci which are not lines. Some typical
ones
be discussed in the next
will
11.2.
Typical
article.
loci
A
quadratic equation in two variables may not have a locus. For example, there is no pair of values for x and y which satisfies the equation
z2
(1)
since the tive.
sum of two
But
if
it
+
2 2/
=-l,
squares of real numbers cannot be nega-
exists at all the locus is
always one of
five
things: (a), an ellipse (Fig. 16), including the circle as a special (Fig.
17); (c), a hyperbola straight lines which may or may not
case; (6), a parabola (Fig.
18); (d),
two
coincide (Fig. 19), or (e), a single point. These curves are studied in analytic geometry. For our purposes we may note simply that if the student has in mind the general appear189
SIMULTANEOUS EQUATIONS
190
[Ch.
XI
ance of each of the three main curves, he can usually sketch one in roughly when he has located a few points on it.
Example. Sketch the curve 9x 2
(2)
+
2
25t/
Solution. Solving (2) for y,
y
(3)
=
=
225.
we have
f V25
-
x
5, we find from 3, and Assigning to x the values 0, 3, 5, + ^, 0, ^, (3) that the corresponding values of y are +3, of the on are and 0. These points Fig. 16. Note that ellipse
y
(0,0)
x +y-l =
[(x-y)(x+y-l)]=0
Fig. 18.
Hyperbola
Fig. 19.
Two
straight lines
x exceeds 5 numerically, y is imaginary. This means that the curve does not extend to the right of the line x = 5, nor 5. to the left of the line x = if
LINEAR AND QUADRATIC EQUATIONS
11.3]
EXERCISE
191
53
The graphs of the equations in problems 1 and 2 below are ellipses; and 4, parabolas; in 5 and 6, hyperbolas. Find at least 6 points on each, and sketch the curve. in 3
1.
3. 5.
x2 y z2
+
y
=
2
25.
= 2z + 3 - y2 = 9. 2
Draw
4.
= 36. +3= y 4y = 4z 36. 9y
2.
4z 2
4.
2
+
Qy
2
a;
2
6.
0.
2
graphs indicated for problems 7-15. They are straight
the
lines. 7.
9.
(x
11.
2
13.
15.
-
(x
x y
2 2
?/
y)(x y)
= ~
Graph
=
4
+
=
2
y
-
1)
=
8.
0.
12.
0.
x
y
2
=
14. (2s
4. 2
(3x-47/-
10. (x
1.
=
-
y
2
2)
=
0.
0.
+ 3y +
2
6)
=
0.
0.
the following equations.
+ 4y = 36. + y = 9. 2
9x 2
-
2
=
16.
9z 2
19.
x
2
21. x
2
-
y
23. x 2
+
y
11.3.
Linear and quadratic equations
17.
2
2 2
= =
4y
20. x
2
0.
22. (x
0.
24. (s
36.
18.
9x 2
- y = 9. - y) = 0. - 2) + (y +
-
4y
2
=
0.
2
2
2
2
I)
=
0.
Suppose, for example, we seek the dimensions of a rectangle whose diagonal is 10 inches long and whose perimeter is 28 inches.
X Fig.
20
Designating the two unknowns, the measures of length and width in inches, by x and y respectively, we have, from Fig. 20 and the Pythagorean relation,
SIMULTANEOUS EQUATIONS
192
+
X2
(1)
2
I/
=
Also, since half of the perimeter
x
(2)
Solving (2) for
y,
+
10 2 is
[Ch.
XI
.
14,
y
=
14.
y
=
14
we have
(3)
-
x.
Substitution of the value of y from (3) in (1) yields
x2
(4)
The
+
(14
- xY =
solutions of (4) are x
=
6 or
100.
8.
Substituting in
(2), or,
we get the solutions: x = 6, y = 8, and x = 8, y = 6. Both of these pairs satisfy (1) and (2); but since algebraic solutions must be examined in the light of the demands of the problem, and since by agreement x stands for the measure of length, the first pair of values must be rejected. The required rectangle is 8 inches long and 6 inches better yet, in (3),
wide.
To
follows, as 1.
and quadratic pair, then, we proceed as illustrated by the example above.
solve a linear
Solve the linear equation for one letter in terms of the
other. 2.
Substitute the obtained literal value of the
first letter
in the second degree equation.
Solve the resulting quadratic in the second letter. Substitute each of the two quadratic roots now found in the linear equation to find the corresponding value of the 3.
4.
first letter.
Consider next the graphical interpretation of the above problem. The circle (1) and the line (2) intersect at the points (8, 6) and (6, 8). (Fig. 21.) It should be noted that there are two points on the circle and only one on the line
which x
=
This suggests the reason for the emphasis upon substitution of the found value in the linear instead of the quadratic equation. for
8.
11.31
LINEAR AND QUADRATIC EQUATIONS
193
Suppose, in this problem, that the perimeter were increased to 20A/2 inches, with the diagonal length unchanged. The algebraic solution of the two equations
=
100,
and X
(5)
+y=
10/2,
= 5/2, y = 5/2. Geoyields the one pair of solutions, x metrically, this means that the^line (5) touches the circle (1) at the single point (5/2, 5V2) (Fig. 21). Finally, with an
assumed perimeter of 40, the algebraic solution is imaginary, as might be suspected from the fact that the line, x + y = 20, in Fig. 21, does not cross the circle.
In general, the physical impossibility of a pair of simultaneous conditions is indicated algebraically whenever i
appears in the solution of the equations stating the condiinconsistent equations; and it is shown the nonintersection of the loci of these equa-
tions, as well as
graphically
by
by
This result applies not only to linear and quadratic pairs, but also to simultaneous equations of many types
tions.
and degrees
in
two unknowns.
SIMULTANEOUS EQUATIONS
194
[Ch.
XI
The number of solution-pairs
11.4.
An
inspection of the sample second degree curves shown in Figs. 16-19 indicates that a straight line cannot cross one of them in more than two points, and also that two of them
cannot intersect in more than four points. The corresponding algebraic results may be stated as follows :
A.* There are at most two distinct pairs of real numbers which satisfy simultaneously a linear and quadratic equation in two unknowns. B.* There are at most four distinct pairs of real numbers which satisfy simultaneously two quadratic equations in two
unknowns.
EXERCISE
54
Solve the following equations graphically, estimating the coordinates of the points of intersection in case they intersect. Then check
by solving algebraically.
= 25, 3x ty = 0. x* + y* = 25, - 2x = L 3y z2 - y2 = 16, 3z - 5y = 0. x2 - 2 = 16, 2x - 3j/ = 1. x2 - y = 0, 2s - y = 1. - y = 0, x2
1.
4.
7.
10.
13.
16.
+
y*
?/
2.
5.
8.
11.
14.
+ y = 25, x + y = 5. x + y = 25, 2
x2
2
2
3x - 4y = 25. x - y = 16, - 3 = 0. y x - y = 16, s = 3. x - y = 0, 2x - y = 2. 2
z
2
2
2
= 25, x y = 0. z + y* = 25, y = 6. x - y* = 16, x + y = 4. s - y = 0, - 4z = 4. 3y x - y2 = 0, x + = 2.
3. x*
6.
9.
12.
15.
+
y*
2
2
2
2/
2
a?
4-3=0. * It
such
assumed here that the equations are independent. That is, we bar pairs, 1 =0 and (x -f y y) =0, whose graphs have a gtraight l)(x common, or such as z2 -f- j/2 = 1 and 2x2 + 2 /2 = 2, whose graphs are
is
asx-hy
line in
identical.
SIMULTANEOUS EQUATIONS IN LINEAR FORM
11.6]
195
Solve algebraically.
+ xy + y = x + y =
17. x 2
2
19. 42/ 2
x2 x
18.
3, 2.
+^+x+22/-l = 0,
+ y = + 2y = 2
-
20. (x 2
2z
xy y
+
= =
2a
2x
22.
a&,
+
6.
z?/
-
y
2
)
(x + y) y = a +
z-42/=4. 21.
*?/
2, 0.
+
ab
=
= =
1, 1.
26,
0.
23. Find the dimensions of a rectangular field whose area 20 square rods and whose perimeter is 24 rods.
is
Find the dimensions of a rectangle whose perimeter 34 inches and whose diagonal is 13 inches.
is
24.
11.5.
Simultaneous quadratics in general
The algebraic solution of simultaneous quadratics in two unknowns is in the general case long and tedious, involving the solution of a fourth degree equation. We shall consider, in the next two articles, two special but important cases.
11.6.
Simultaneous equations in linear form
no more than two of the five quantities, x2 y 2 xy, x, and y appear altogether in two simultaneous quadratics, the method for linear equations may be applied at once to get these two unknowns. After the latter are found, the values of x and y follow at once. Each of the following seven pair2 2 x2 and xy y2 and xy] x2 ings is of interest here: x and y 2 and y; y and x] xy and x] xy and y. If
,
,
]
Example
1.
Solve simultaneously.
x2
(1)
+y = 2
^^^^
(2)
Solution. Solving first for x 2
y
2
=
9.
Hence x =
2,
y
=
13,
and y2 we get x2
3.
,
The
solution
is
=
4 and
not com-
SIMULTANEOUS EQUATIONS
196
[Ch.
XI
however, until the values have been paired properly. Since y = +3 when x = 2 and also when x = 2, the pairs
plete,
are
-3), (-2,
(2, 3), (2,
Example
2.
2x 2 -3*0 s 3z2
+
(4)
x i/
Solution. Treating (3) and xy, we find that x 2
=
-
(Fig. 22).
Solve simultaneously.
(3)
2
and (-2, -3)
3),
= +A/O
or
~^~>
and
=
= =
1,
7.
unknowns Hence x = + V2,
(4) as linear in
2 and xy
=
1.
the
an d the solutions are
f
V2, -
K14)V '(0,12)
-*~x
Fig.
Example (5)
(6)
3.
22
Fig.
23
Solve simultaneously.
2y
-
= 4, =-12.
These equations are linear in re 2 and y. Solving, we get # 2 = 4, y = 4; so that the final solutions are (2i, 4) and ( 2i, 4). The imaginary values for x indicate that the curves do not intersect. This is seen in Fig. 23, to be Solution.
the case.
EQUATIONS REDUCIBLE TO SIMPLER FORMS
11.7]
EXERCISE
55
Solve algebraically. 1.
x2
+
X
2
_
3.
x
2
+
5.
2x -
x
y
+
2 t
y*
2
3x 7.
y
2 2/
= = = =
3y 2y
2
2
2
9.
2.
4.
1,
4.
6.
1,
-
8.
10.
5,
= 1, 3x(y + 2) = 7. x2 = 4xy - 3, = 3x + 4. 2xy 2x + 4y = 3a, 3x - 8?/ = -a.
12.
x)
11.7.
2
2
2
2
14.
2
t/
2
2
5,
2
16.
2 2
Equations reducible
When
2
2
0.
2
15.
2
2
2
8.
2
11. 3(x
= 25, = 50. 2 = 4, = 36. 4x2 - y = 0, 3x - 1 - 2y = 0. 2 Zxy + y = 1, 2 4x/ + y = 2. 3y + 2x - 1 = 0, - 3x + 8 = 0. 2j/ - 3) = 3, t/(2x 2y xy + 1 = 0. 2 y = 3xy Qxy + 2 + y = 0. 2x + 3xy = 2a + 36, 3x - 2xy = 3a - 26.
+y 2x + 3y x + 4x + 9y x2
2
2
13.
1-6, solve graphically also.
4
= = = =
2
In problems
25,
+ xy 6, x + 2zi/ 0. 4x + 3y = - 8x + 4 = 3y 2x
197
to
simpler forms
quadratics in two unknowns are written with the zero, the simultaneous solution of a pair of
right
members
them
is
much
simplified
of the equations
Example
1.
is
if
i)(a;
z2
(2)
Solution. Since
any pair
of at least
one
+ y + l) =0, + y* - 25 = 0.
of values satisfying
3x
(3)
member
left
Solve simultaneously:
(3x-2y-
(1)
the
or can be factored.
-
1
=
0,
+#+
1
=
0,
-
2y
or
z
(4)
will also satisfy (1), it follows that (3)
and
(2)
any common solution
of
or of (4) and (2) will be one of the desired solu-
SIMULTANEOUS EQUATIONS
198
[Ch.
XI
Hence the problem reduces to the solution of two linear-and-quadratic pairs: (3)-(2) and (4)-(2). The solutions of (3) and (2) are (3, 4) and (-- ff, f-f); those of '(4) and (2) are (-4, 3) and (3, -4). The loci of (1) and (2) are shown in Fig. 24. tions of (1)
and
(2).
x'-2y-l)=0
Pig. 24
Example
2.
(5)
(6)
Solve simultaneously (2x (3*
-
3y y
-
+
+
4)(
-
l)(2x
:
y y
-
2) 3)
= =
0,
0.
Equation (5) is satisfied by the coordinates of any point on either one of the lines Solution.
2x
(7)
-
4
=
0,
+y-
2
=
0.
-
3y
and x
(8)
Similarly, (6) (9)
is
satisfied
by
solutions of either
3x
-
y
+
1
=
0,
2x
-
y
-
3
=
0.
or (10)
Hence, the problem reduces to the simultaneous solution of the following four pairs of linear equations: (7)-(9), (7)-(10),
11.7]
EQUATIONS REDUCIBLE TO SIMPLER FORMS
199
and (8)-(10). We must be sure that each pair comfrom (5) and also a factor from (6). The rest a factor prises of the solution is left to the student. (8)-(9),
Evidently the method of Example if
the
left
member
unfactored form. Here the process
Example
3.
1
of either (5) or (6) is
Solve simultaneously 2
(12)
4i/
had been
left in
the
shortened.
:
- xy = - 3xy =
*2
(11)
could have been used
3,
2.
Solution. This system does not come directly under the types illustrated in Examples 1 and 2. However, it can be
reduced to such a type by first producing an equation from the given pair in which the constant term is zero. This new equation is then used to form a system which may be solved
same manner as the above examples. Both members of (11) and (12) are multiplied respectively by 2 and 3, resulting in the equations:
in the
(13)
2x 2
(14)
12y*
-
2xy Qxy
Adding corresponding members 2z
(15)
2
-
+
llxy
= 6, = -6.
of (13)
12y
2
=
and
(14),
we
get
0.
common
solutions of (11) and (12) must also satisfy (15), a new system can be established in which (15) is used as one of the equations, with either (11) or (12) as
Since
all
in this case, (11) The simpler of the two should be given preference. The left member of (15) has the factors 2x 3y and x 4y, so that the new system becomes
the other.
:
z2
(16) (17)
(2x
-
3y)(x
- xy = - 4y) =
3,
0.
This system presents no new difficulty, and by the method of Example 1 the solutions are found to be (3, 2) ( 2) 3, (2, J)
and (-2, -J).
SIMULTANEOUS EQUATIONS
200
EXERCISE
(x
(3x
2
-
-
5y) (y
3)
=
(x
2
2
3x -xy-2/ =l,
9.
11.
2x + xy - y = 0, _ 4y2 _ o - I) - 9 = (x + y -4= (2x + yY 2
2x2
x2
-
2
37/
=
0,
+
+
14.
6x2
- y* = 2, - 3) = 0. + 4^2 = 1, + 4y - 3) = 0.
2xy y)(x + y x2 - 2xy
2y)(x
-
2x
2
7xy
-
2
j/)
+
5xy
(x
+
Example 16.
= -9, = 0.
+
solve algebraically.
0.
-2,
xy + 2y* =3. 2 17. 2x 2 - 13xt/ 2y 2 2 5 x + 3xy */
-
(x
(2x
Solve by the method of illustrative 15.
-
10.
a;2
13.
0.
x2 -2/2 =16,
(2x
12.
2
0,
x2 -y2 =16,
8.
members, and then
left
2
1)
= =
(x-2/-l)(x+y-l) = 0.
(x+y-2)(2x-y-l) = 0. Factor both
+ 4y +
i/)(2x
6.
2
2
y
(x-y-2)(x-2H-l) = 0.
0.
2/
2
-
4.
+ =25, = 0. (x y)(x + y) x + xt/ + 2i/ = 1, - y)(x + y) = 0. (x x2
-
x2
2.
2
-
7.
1-6, solve graphically also.
+ y = 25, = 0. 3)(x + y + 1) x -y =16, 2
3.
5.
In problems
x2
-
XI
56
Solve algebraically. 1.
[Ch.
-
y)
2
2
3y -4 3y
-
= =
0,
= =
0,
2
1
0.
3.
= 5, = 5. + 3xy - 4 = + 2xy - 2 =
3x2 + xy x2 + y2
18. 2y*
3x2
0, 0.
0.
Chapter Twelve
RATIOS, PROPORTIONS,
AND VARIATIONS
12.1. Ratios
A ratio is a fraction which compares two things in terms of the same unit. Its value is determined by the relative sizes of the things
compared, and not by the unit chosen. Thus
the ratio of 6 inches to 3 feet
i alS
since
6
(foot)
'
e
c is -,
^ ,P
.
36 (inches)
!
=
-, 6'
and
3 (foot) 6* Ratios are of great importance in the branch of mathematics known as trigonometry. One of its uses is to measure the distance to an inaccessible object, such as the top of a
mountain or a spot on the moon, by equating the
ratios of
corresponding sides of similar triangles. 12.2. Proportions
A proportion is an equation whose two members are ratios. Thus, to get the height, CD, of a (Fig. 25),
and the
lengths,
OA =
cliff,
a sight
5 feet,
D
Fig. 25
201
is
AB =
taken at 3 feet,
and
202
AND VARIATIONS
RATIOS, PROPORTIONS,
[Ch.
XII
OC =
200 yards, are measured directly. (The representation shows OA and AB too large in comparison with OC and CD. Such a figure is called a diagram rather than a scale drawing.) Then, from similar triangles, in Fig. 25
CD
m (
}
(in yards)
=
200 (yards)
From
(1) it follows
CD =
(2)
3 (feet) 5 (feet)
3 *
=
5*
that
=
( ) (200 yards)
120 yards,
the height of the not-necessarily-climbed
cliff.
The proportion
=
< 3)
* .
is
3
.
sometimes written a
(4)
:
b
=
c
d,
:
'a is to 6 as c is to d.' Here b and c are called a and and d are extremes. There would, however, be means, no point in the new symbol replacing the division sign were and
it
is
read,
not for the gain effected by extending the notation in
(4).
For example, a
(5)
which briefer
is
read
'
a
is
:
to b
b
:
is
c
= d
:
e
to c as d
:
/,
to e
is
to /,' states in
= = beef
form the three proportions 7 :
is
-,
-
-,
and c
=
-
f
Evidently the notation of (5) can be extended indefinitely with as many letters as desired on each side of the equality sign.
Since the equation,
ad
(6)
=
be,
obtained by clearing fractions in (3), is linear in each of the letters involved, the solution for any one of them is simple. ad be 7 ad be j n o = a = and c = -7 Evidently a = b a e a i
,
j.i
,
,
,
* Note that the indicated units merely tell what the numbers represent, and all algebraic operations are upon numbers (as represented by digits or letters).
that
CONSEQUENCES OF A PROPORTION
12.3]
EXERCISE
57
Find in each case
the ratio of the first to the second quantity.
2 yards.
1.
3
3.
2 miles, 50 yards.
5.
1
feet,
2. 7 feet,
ton, 100 ounces.
7. x,
203
X
-
x
8.
2,
-
2
3 miles.
4.
6 ounces, 4 pounds.
6.
4 hours, 13 seconds.
+
a2
9.
x.
1,
a
+ -(t
Solve for the unknowns. 10.
x
11.
x:(x
:
(x
-
+
+ 2) (3x + 1)
12. (x 13.
14. 15.
(x
12.3.
+
5)
-
(x
2)
:
(x
=
(3x
-
3).
+ 4).
-
1)
:
(2x
+ 2).
:
(2x
3)
=
(x
:
:
:
:
x
+ +
:
3 y]
unknowns.
= :
x
:
4
5y.
= 1 - y - 2) (x - 2) (x + y
+
(2x
:
3t/)
:
1
2
:
:
3.
- 1) = 1 2 3. + 1) (2x + 3y - y + 1) = - y + 1) 2 3. (3x (2x = 2y (3y + 2) (5y + 3). (2x + 3) (3x + 5) (x + 1) = 2 3 4. (x + y + 1) z (x + y) (y + z + 1) = 1 2 3 4. (z + 3) (x + y + z) (y + 2) (x + 1)
18. (x
22.
(x
+
:
the following proportions get pairs of simultaneous equations,
17.
21.
(x
1)
:
1
20.
1)=
-
:
16.
y
:
:
:
:
:
:
:
1
:
:
:
:
:
results following
a
_
1
:
c
b'd
are given in equations (2) to (7) below.
:
:
:
:
:
from
:
:
:
Consequences of a proportion
Some fi (1)
:
solve for the
19.
(x
+ 1) (2 - 4x). (-3x) (6x + 1) = (4x + 2) (12x + 9). - 3x) (4x) = (x + 7) (3x + 1). (2
From and
=
1)
:
:
:
RATIOS, PROPORTIONS,
204
When
the
members
of (1) are
AND VARIATIONS [Ch. XII multiplied by bd we learn
that
ad
(2)
=
be,
product of the means equals the product of the extremes. When the members of (2) are divided by dc we find that
or, the
a
,ox
(3)
=
c
b d'
or, referring to (1), the ratio of the
numerators equals
the ratio
of the corresponding denominators. If
the divisor
is ac,
we
get
W
a
(4)
=
c'
or, the reciprocals of equal ratios are equal.
Again, adding
new member
1
to both
as a fraction,
a (5)
+
members we have 6
c
+
of (1)
and writing each
d
d
6
Similarly,
a (6)
b
d
c
d
b
Dividing corresponding members of (5) and (6), (as justified by the axiom: 'When equals are divided by equals, the quotients are equal');
we have a
(7)
The student may
a
+
b
c
b
c
+
d d
find further consequences of
(1)
by
getting equations from (3) and (4) corresponding with (5), (6), and (7) as derived from (1), by inverting all members, and by combining members in various ways. Evidently an unending series of equations of an endless degree of complexity stems from the simple equation (1). Results such as (2) to (7) were formerly given more prominence in algebra texts than at present. Probably one reason
CONSEQUENCES OF A PROPORTION
12.3]
for the decrease in
emphasis
is
205
the fact that such results are
merely what one gets by applying to (1) the guiding prin' ciple for work with equations, namely: Always do to the right side what is done to the left side.' Nevertheless, the processes here suggested may lead to results which are interesting, useful, and far from self-evident. HINT. If, in launching out for himself, the interested student should arrive at a complicated result of (1) such as, say,
a2 a2
v
x
g
+ ab + ac + ac + ab
be
=
be
d2 d2
+
+
bd bd
+ be cd + be
cd
he should not forget the value of frequent arithmetic tests to guard against errors. For instance, when a = 1, 6 = 2, c = 3, and d = 6, (1) is true and so is (8).
EXERCISE
58
(The numbers in 1.
Noting that
this exercise refer to proportions in Art. 12.3.) (5),
(6),
and
three similar consequences of 2.
Get consequences of
(7) are
consequences of
(1),
get
(3).
(4) similar to those of
3. By inverting the members the proportions obtained in problems
of (5), (6), 1
and
and
2,
problem
1.
as well as of
(7),
more conse-
get nine
quences of (1). Test the following proportions with numerical values which satisfy 1, 6 3, d 2, b 2, c 6, or a 3, c 4, (1), such as a 6. // the tests indicate that a given proportion may be a true d
=
=
=
=
=
=
=
-- --
consequence of .
4.
^ ^
a
+b=
c
a
a
b
d
=
d
c
b 7
-
ab
c
+
+b_ c+d -:
^ 2a 7^
c2
=
so.
ad - -a
5.
c
a*
+
+
prove that this is
c
a
a
(1), try to
a
j
cd
+
'6
d
--
b
=
-
d
2c
a
c
2b
=
c
+
2d
d
a2 '
.
7
ab
3
cd
=
206
10 '*
AND VARIATIONS a - b _ c - d ~
RATIOS, PROPORTIONS,
-
a2
b2
_ ~
-
c2
+
b3
c3
+
rf
the following problems
A light is on the
an 8-foot shadow, how In problem 16, will his shadow be? 17.
18. If
casts a 19.
a light
A model,
tain. If
two
the model,
12.4.
is
shadow 10
_ '
(c
a 6
+ C
2
d)*
'
d
by means of proportions.
the
man
is
man
30 feet from the pole,
on top of a pole and a
3 feet high,
known
high
3
&)
2
top of a 20-foot pole. If a 6-foot is he from the pole?
if
is
c
2
casts
far
feet long,
villages
how
(a
15 '
cd
2
2
2
+
3 '
XII
'
a
cd
ab
16.
2
2
13 '
'
afe
a3
d2
[Ch.
how is
is
high
6-foot
man
how long
30 feet away
the pole?
made from photographs
of a
moun-
to be 5 miles apart are 7 feet apart
on
the mountain?
Variation
Two quantities which behave so that their ratio remains fixed may be compared mathematically by use of the symbol oc, read 'varies as.' Thus, if C and D represent and diameter of a circle, then doubled when D is doubled, tripled when D is tripled, and so on. This is expressed in English by saying that 'C respectively the circumference
C
is
varies directly as
D,'
'C
directly proportional to D.' say that 'C varies as D,' or 'C is
or,
is
simply, we may proportional to D.' In symbols,
More
CocD.
(1)
can be put in a form more suitable for algebraic operations as the equation
The statement
(1)
C = kD,
(2)
or
~=
k,
called the 'constant of proportionality,' or the are meas'variation constant.' In this case, when C and
where k
is
D
12.4]
VARIATION
207
ured in terms of the same linear unit, or (nearly) 3.1416, so that
known
that k
=
IT
the formula
C = wD.
(3)
In
we have
it is
problems, however, we must start with an undeThus at a given time of day in a given place the of a man's shadow varies as his height, h. We then
many
termined length,
k.
5,
have s
(4)
=
where k remains to be found. is 9 feet long, then
=
9
(5)
kh, If
or
k$,
the shadow of a 6-foot
man
= f = f
k
Hence, at the time and place in question, s
(6)
=
/i
for people or objects of various heights.
In other words, the
shadow will be as long as the height of the object. Sometimes one quantity varies, not as a second one, but as some function of one or more quantities. Thus, the equations (7)
y
=
(8)
y
=
(9)
y
=
y
=
=
*,
-^
=
k,
xy
=
kx
or
^
kxz,
or
or
k -,
x
fc,
and /IA (10)
2
kx ,
or
2/ 2
w =
?
fc,
a:
state in succession that y varies directly as z 2 , zs, -,
and
/j
-;*
Sometimes
it is
said that, in (8), y varies jointly as x
and z and in (10), y varies directly as x and inversely as 22 and w. More simply, however, if y varies directly as any function of other letters, y is k times that function. Other
common
forms of expression for the variation law
208
RATIOS, PROPORTIONS,
AND VARIATIONS
[Ch.
XII
connecting y with the other variables in (7), (8), (9), and (10) can be used. In order to acquaint the student with some of these phrases, we shall list a few. In (7), it can be said that
'y
varies as x 2 ,' or that
2
to x ,' or simply that
'y
is
11
statements imply that
is
'y
is
directly proportional 2 proportional to x .' All of these
a constant.
X2 can be said that 'y varies directly as
Similarly, in (8), it the product of x and 2,' or that 'y is directly proportional to the product of x and z,' or, more simply, that 'y is proportional to x and z.' All of these statements imply that is
a constant.
xz
can be said that 'y varies inversely as x' or that inversely proportional to x.' These statements imply 'y that xy is a constant. In (10), it can be said that 'y varies directly as x and In
(9), it
is
2 inversely as the product of z and w' or that 'y is directly 2 proportional to x and inversely proportional to z and w.'
1/2 11} is a constant. These statements imply that x It should be observed that the words 'directly' and
'jointly' are frequently omitted, 'directly' being understood when the dependence of one quantity on another is referred to, and 'jointly' quantity involved.
A steps 1.
when
there
is
more than one other
in variation usually consists of the following
problem :
The variation relation is stated as an equation involving
In this step all the quantities involved in the variation law should be described clearly. When any quantity is referred to by name, we shall adopt the plan of using the initial letter of the name as an abbreviation. For example, the word 'pressure' would be described by the letter p or P. When the quantities are not named, the letters z, y, z the constant
k.
y
etc.,
are used.
12.4]
2.
VARIATION
One
209
set of values of all the variables involved is inserted
in the equation, leaving k as the only visable in a problem where step 5 is to
to
make a box scheme
as indicated in
unknown.
It is ad-
be carried out
Example
later,
4, solution 3.
This box should be filled in before going through any numerical work, as it will clearly show what quantities are given, the units being used respectively, and what quantities are to be found.
The resulting equation is solved for k. The value of k is inserted in the original equation. 5. By means of the equation obtained in 4, the value of any specified variable is found when the values of the others are given. However, the same set of units must be used 3.
4.
throughout the problem as were employed in steps 2 and 3 to find the value of k. This point is very important, and is illustrated in the
Example x y
= =
y
2, 1
=
of
Example
y varies directly as
1.
4,
=
z
and w =
Solution.
two versions
The
3,
and
w =
x
4.
2
zw 2.
One Find
set of values is: z
when x =
3,
4.
five steps in order yield the following
equa-
tions.
-
<'> 4
.
(13)
k
=
t*A / (14) ^
y
=
/icN (15)
1
1
sr 6.
zw
=
= -27,(answer). v
-777-,
2(4)
or
z
2
Example 2. State in words the law of variation for x in terms of the variables related to it in (8) and (10).
we use the second forms in (8) and (10), where members include all the variables present, it becomes
Solution. If
the
left
210
AND VARIATIONS
RATIOS, PROPORTIONS,
[Ch.
XII
apparent that in a direct variation between two variables, one appears as a divisor of the numerator while the second appears as a divisor of the denominator. For inverse variation, both variables will appear in the same position either as divisors of the numerator or as divisors of the denominator. Thus, in (8)
'x
varies directly as y
and inversely as z'
Similarly in (10)
w'
'x
varies directly as y, z 2 ,
It
should be noted here that there are other forms of
and
expression as well as those given.
They are left to the student.
Example 3. If the original value of x is 10 units in (8) and (10), what would its new value become if y is doubled, z is tripled, and w is halved? Solution. In (8), since x varies directly as y
and inversely
z, it means that if y is doubled, so is x, but if z is tripled, x becomes one-third of its former measure. Carrying out these changes simultaneously, we get for x,
as
2(1) (10)
In that
(10), since
=
=
6f units.
x varies directly as
y is doubled, so is x] if tripled, x becomes nine times if
w
?/,
z
2 ,
and w,
out these changes simultaneously, we get for 2(i)(9)(10)
Example
4-
Boyle's
=
means
halved, so is x but if z is former measure. Carrying
is
its
it
;
x,
90 units.
Law
temperature the volume
in physics states that at a given of a given quantity of gas is in-
on the walls of the cona given sample the volume is one cubic foot when the pressure is 20 pounds per square inch, what is the
versely proportional to the pressure tainer. If for
volume when the pressure square inch?
is
increased to 60 pounds ,per
12.4]
VARIATION
Solution
1.
211
Let
7 =
(16)
where
F
and
P
or
VP =
represent units of volume and pressure refirst solution let be measured in pounds
P
spectively. In this
per square inch,
Jfc,
and
V
in cubic feet.
=
(18)
k
(19)
V =
(20)
V = Now
Then
20.
r
= j
-
P
be measured in pounds per square foot, with V unchanged in meaning. Then the first and = ^, second values of P in the problem are respectively Solution 2.
and
3%
=
let
-%
3^-.
Equations (17) to (20) are replaced
by the
following. (21)
1
(22)
k
= A.
(23)
F =
(24)
P
g p. p;
1
=
Thus, we see that the value determined for k, the so-called proportionality constant/' depends upon the choice of units; but the same final result is obtained in any case. '
Simple problems in variation may often be easily as problems in proportion; for if one quantity varies as another, corresponding values are proportional. When using the proportion method, it is better to write the variation equation in the second form illustrated IB (7); (8), (9), and (10), where all variables are contained Solution
3.
worked more
212
in one
RATIOS, PROPORTIONS,
member
AND VARIATIONS
F
F P2 = 2
XII
of the equation. Thus, using the second form one set of values for and P,
V
of (16) with Fi, PI indicating
while
[Ch.
that ViPi = FaPa, etc. Arrange the given information in a box
2,
P
and
a second
2
scheme as follows
V
LAW:
set, etc., it follows
:
P
(cu. ft.)
*
per sq.
= yP = FP = = 7 (60). cu. foot. Vz =
V.P, 1 20 .
(Ibs.
2
3
2
3
in.)
etc.
2
5. Here we shall use without proof the following and useful mathematical result important
Example
:
THEOREM
1.
The areas of similar plane figures vary as
the
squares of corresponding dimensions.
The
two
similar rectangles are 15 and 20 square units, respectively. Find the length of the diagonal of the larger rectangle if the diagonal of the smaller one is 10 units
areas of
long.
respectively.
here used (25)
From (26)
A
and diagonal length The mathematical statement of the theorem
Solution. Let
and d stand
is
A =
A
or
i-*-
the smaller rectangle
we have
15
kd
=
This value, substituted in (27)
for area
or
k
=
15
100
=
3 20*
(25), gives
20
= 20, in (27), Inserting the area of the larger rectangle, or A and remembering that in this case d must be positive, we have
12.4]
VARIATION
(28)
d
=
213
Vs =
20 -To
11.5 linear units (answer).
Note. If the student will try to solve this problem by means of simultaneous equations, as is possible, he will
appreciate the simplicity and power of this method. In addition to the required answer he has in (27) a formula applying to any rectangle similar to the first one.
we
In Exercise 59
THEOREM
Theorem
shall use, in addition to
1,
The volumes of similar figures vary as
2.
the
cubes of corresponding dimensions.
EXERCISE In each
59
of the problems 1-8,
express the given relation as an equation; (b) write the equation so that the left member will contain oil the variables; and then
(a)
(c)
write
an equivalent statement in words for
ing x as related ,
I 1.
I/ 2/
to the other variables present.
o J >y rf fit Z ^. OCX?/.
* />o f^ . OCX
5. z oc
6.
w
, OC O. 1/1 U/ (y
oc -^r-
xy
7.
x2
~ y
oc
and
-
A 4X ^k.
,
--
is
4
when x
=
W
Cf C
I/ if
8. ?/ '
z*w
yz*
9. If 2 varies as
the variation involv-
3
.
2
oc
-
t(;
=
2,
y
3,
w=
3,
w?
=
4,
and x
=
4,
and
t^
find the value of (a)
(b) (c)
(d)
= 4; 2, and z y when x = 3, w = x when z = 5, y = 10, -and w = 8; w when x = 3, y = 4, and z = 5; = 4, and w = 2. z when x = 3, y
10. If
y varies as
-
and
is
2
when
z
=
find the value of (a)
(b) (c)
(d)
= z when y = w when y = x when y = y when
z
w = 4a, and x = 6; = 2a, and w = 3c; 5, x = 3, and x = 2a; 2, z = c. = b, and w a, z 3,
2,
214
RATIOS, PROPORTIONS,
AND VARIATIONS
Solve problems 11-15 by use of Theorems 11.
The
12.
A
1
and
2, Art.
[Ch.
XII
12 A.
bases of 6 similar triangles are 2, 3, 4, 7, 8, and 9 inches long respectively. If the area of the smallest triangle is 5 square inches, what are the other areas? sign painter finds that a pint of paint '
is
used in painting
Chicken Dinner. Dine and Dance.' How much paint be used in making a similar sign, with letters of the same type
the words, will
and proportions but three times
as high?
13. The heights of three similarly-proportioned men are 5, 5^, and 6 feet respectively/If the tallest man weighs 200 pounds, about what would be the expected weights of the other two?
14.
The weights of various aerial bombs in pounds
are as follows
:
100, 200, 500, 1000, 2000, 6000, and 12,000. Assuming that all are similarly proportioned and made of the same materials in like proportions, what are the lengths of the other bombs if the
100-pound one 15.
50,
is
2 feet long?
The perimeters
and
of 6 similar plane figures are 4, 5, 10, 20, 100 inches, respectively. If the area of the figure with a
20-inch perimeter
is
10 square inches, what are the other areas?
16. The weight of any given object on a planet varies directly as the mass of the planet and inversely as the square of its radius. For reference let the mass of the earth be one unit and its radius
one unit. If a boy weighs 100 pounds on the earth, how much would he weigh on each of the following bodies?
Name Moon
Radius
Mercury Venus Mars Jupiter
Saturn
Sun 17.
The weight
.27
.012
.39
.04
.97
.81
.53
.11
11.2
317.
9.4
95.
109.
of a
Mass
body within the
330,000 earth, as in a mine, varies
directly as its distance from the center. Assuming the radius of the earth to be 4000 miles, how much would a 200-pound man weigh when 10 miles below the surface?
12.4]
VARIATION
S
18. If
square of
t/,
215
varies directly as the cube of x and inversely as the what change in S results when x is doubled and y is
tripled? 19.
The
force of attraction
between two bodies varies directly
as the product of their masses and inversely as the square of the distance between them. When two masses of 6 units and 24 units
by 192
are separated
tance between them
inches, the force
is
72 units,
(a) If
the dis-
diminished by 24 inches and the smaller mass is doubled, what change in the larger mass will increase the force by 18 units? (b) At what distance will the two given masses is
have twice the force of attraction that they have at 192 inches? 20. Kepler's Law states that the square of the time of a planet's revolution about the sun varies as the cube of its mean distance from the sun. The mean distances of Mars and the earth are in the ratio 3 2. Find in days the time of revolution of Mars (i.e., the Martian year). :
21.
The volume V
T and
of a gas varies directly as the absolute teminversely as the pressure P. If a certain amount of
perature gas occupies 100 cubic feet at a pressure of 16 pounds per square inch and at T = 200, find its volume when the pressure is
20 pounds per square inch and T = 420. 22. The mass of a spherical body varies directly as its density and the cube of the radius. Compare the masses of Jupiter and the earth if the diameter of Jupiter be taken as 11 times that of the earth and 23.
its
density -fo that of the earth.
Given that y
-
HINT. Let y
Show
that
oc x,
=
2 prove that 2x
kx.
*-
=
a constant.
xy
24. If y oc x, prove that 3s4
+ yV oc
+y
2
oc
xy.
Chapter Thirteen
THE BINOMIAL THEOREM
The binomial theorem
13.1. If left
we
carry through the multiplications indicated in the we get the following
side of the expressions below,
formulas.
(F (F (F
n 1 V
/
+ S) = F + 2FS + S = F* + 3F S + 3F + S + + S) = F* + 4F*S + 6F*S + 4FS* + S 2
2
]
2
2
3
)3
;
4
J 1
2
/77 _J_ .QTtn
+
2
= Fn
n(n ,
1
^Z
/xi^^
4 ;
i'*
-4
(n
2)
Q o
F n - /S + 3
3
(to
n
+
1
terms).
The right member in each of equations (1) is said to be the binomial expansion of its corresponding left member. The n S) is called the binomial formula; and expansion of (F the full equation is the algebraic statement of the binomial
+
theorem dealing with a positive integral index, n. The proof of the binomial formula is given in more advanced discussions,
but the student may verify it for various integral values of n. It will be noted that if n represents the exponent of F + S in the left member of any of the equations of (1), then the following statements are true 1. The number of terms in the expansion is n + 1. 2. The first term is F n 3. The exponent of F decreases by one in each succeeding :
.
term of the expansion, while that 216
of
S
increases
by
one.
13.1]
THE BINOMIAL THEOREM
sum
Furthermore, the
term 4.
is
217
of the exponents of
F
and S
in
any
n.
The
term
coefficient of the first
while the coefficient
is 1,
of the second term is n. If we multiply the coefficient of any term by the exponent of F in that term, and then divide the product by one more than the exponent of S in the same term, the result is the coefficient of the next term in the
expansion. 5.
The
coefficients equidistant
from both ends of the ex-
pansion are equal.
Example
1.
Expand
(x
+
7
y)
.
Solution. (x
+
y)i
The
=
x7
+ +
+ +
7xy 7xy
21x*y*
y
+ 35zV + 35xV +
21x*y*
7 .
first coefficient is 1.
Since n
=
7 in this case, the second coefficient
is 7.
1
The
coefficient of the third
term
( ^
is
7
(f
^
'
or 21.
Lt
The
coefficients of the fourth to eighth terms, as
found in
succession, are:
~~~ _~
Example
2.
~~ _~
6 *' (35) (3)
an d
Expand and
binomial formula,
. '
.
7
simplify (x
Comparing the binomial x that x corresponds with F and
Solution. find
'
7;
o
~~ _~ 35(4)
(21) (5)
5
2?/)
.
2y with F 2y with 5.
+ S,
we
By
the
THE BINOMIAL THEOREM
218
Example
8.
Expand and
simplify
f
F =
3/x and S binomial formula, we have
Here
Solution.
13.2.
An
3 Vx
[Ch.
XIII
+ j
= -
Again using the
extension of the binomial formula
It is interesting to
tional in (F
+ S)
n ,
note that
if
n
is
either negative or frac-
the binomial formula
still
holds
when
on the numerical values is not limited to n + 1 made up of an infinite number. It
proper restrictions have been placed of and S. Further, the expansion
F
terms as before, but is is beyond the scope of this text to discuss these restrictions on the binomial expansion when n is not a positive integer. However, it may be stated in general that the expansion is useful when S is numerically smaller than F, and when the successive terms approach zero so rapidly that the sum of the first three or four terms of the expansion serves as a good n S) Example 3 below shows an approximation for (F
+
.
application to the extraction of roots.
Example
1.
Expand
(x
+
2
t/)~
to five terms and simplify.
Solution.
+ s/)- = 2
(x
or2
or 2
2x~*y
+ 3x~
4
2 i/
4z~
V + 5z- V
13.2]
AN EXTENSION OF THE BINOMIAL FORMULA
219
and
simplify.
y}~* to four terms
2.
Expand
Example
8.
Find '^1002 to three decimal
Solution.
Write
Example
(x
Solution.
-
(x
= =
places.
+ 2)* = [1000(1 + .002)]* 10(1 + .002)* _ 10 + 1 (1)^.002) + K-D(ir*(.002)' + [l* = 10(1 + .00067 - .0000004 + = 10.0000 + .0067 - .0000 H
v/1002
(1000
.
)
=
10.0067 or 10.007.
Note that for accuracy to three decimal places we write successive terms to four decimal places until a term is reached which yields zeros in the four places. Then the answer is 'rounded off' to three places. This is a good working rule, though a much more extended discussion would be needed for a rigorous treatment of
EXERCISE
60
Expand and 1.
4.
+ yY(3o + b)
simplify each of the following expressions. 2. (x
(x
2
3 .
7.
(3*-fJ'10.
(SVx
-
problems of this type.
yY-
5.
[2x
yY-
-
-Y-
M* + 5)H
3. (2x
6.
(2Vx
9 '
yY-
+
4 t/)
.
THE BINOMIAL THEOREM
220
[Ch.
XIII
Find the first four terms in each of the following indicated expansions, and simplify the terms obtained. 13. (x*
-
2y)
10
14.
.
Find and simplify
(VE
8
-
-
f)
15.
4/
+ -^Y^ [V^ 2 /
the required terms in the following indicated
expansions. 16. 5th
term of
fx
_
HINT. Let
and
-
x
U
represent the
fifth
x term, with corresponding to
F
y
to 8.
62
I
17. 6th
term of [ db
18. 8th
term of
Assuming
Y
2
j
*J
to
j
x
y
(Vx -
of*)
14 .
that all restrictions have been taken into account,
expand
four terms and simplify.
+ 2/)~ (a* + 26*)3
19. (x
22.
20. (x
.
1
23. (1
.
-
y)*.
+ 3*)^.
+ y)~i
21. (2x 24. (1
-
2s)-
1 .
Bi/ grouping terms, express as binomials and expand by use of the binomial formula.
25. (a 27. (a 29. (a
+ 6 + c) + 6 - c) + 6 + c)
2 ,
or [(a
+
6)
+
2
c]
.
2 .
3 .
26. (a
+6-c-
28. (a
-
30. (a
+
b 6
-
2
d)
.
2
c)
.
3
c)
.
Use the method of illustrative Example 3, Art. 13.2, to find the values of the following roots to three decimal places. 31.
VTOI.
35. A/26.
VUG. 33. V24 (or V25 - 1).
32.
36.
^1.005. 37.
34. v'lOlO.
^10
(or *V
Chapter Fourteen
PROGRESSIONS
14.1. Definitions
A group of numbers arranged in order according to some law of selection is called a sequence. Each number in the sequence is a term. 14.2. Arithmetic progressions
the difference between two adjacent terms is the same, regardless of the pair chosen, the sequence is called an If
arithmetic progression.
Example.
1, 3, 5, 7, 9, 11.
The abbreviation 'A.P.' stands gression. The letters a, d, n,
I,
and
s are
for
an arithmetic pro-
used to represent the
elements of an A.P., where
a represents the first term; d represents the common difference, or what must be added to any term to get the next one; n represents the number of terms / represents the last, or nth, term; and S represents the sum of all the n terms. It may aid the memory to note that the elements are the ' ' letters in the word lands. These elements are so related that if any three of them are known the remaining two can be found by use of the proper formulas.* ;
* In
used,
We
cases, when only the three formulas for an A.P. given in the text are necessary to solve simultaneously two equations with two unknowns.
some
it is
shall
omit examples of
this type.
221
PROGRESSIONS
222
Consider (1)
a,
a
[Ch.
XIV
the progression.
first
+ d, a +
2d,
a
+
3d, a
+ 4d,
a
+
5d, a
+
6d.
Here any term after the first is found by adding d to the term which precedes it. Note that second term is a + Id, the third term is a + 2dt, etc., so that the coefficient of d in any term is one less than the ordinal number of the term. From this we see that the nth term, designated by I, is a + (n l)d. Stated in symbols, this yields an important result which we shall call Formula 1, namely, 1.
=
/
a
+
(n
-
l)d.
A
progression having a definite number of terms may be written out in full, but usually some dots are inserted to indicate that
some
of the terms are omitted.
Example
1. 2, 5, 8,
to 8 terms.
Example
2. 1, 3, 5,
19, 21.
14.3. Arithmetic
The
means and extremes
and last terms of a progression are called the and all terms between the extremes are called
first
extremes,
means.
In the special case of an A.P. having only three terms, the middle term is called the arithmetic mean.
Example. Find the arithmetic mean of a and Solution. Let
Since
m
a
m be d,
and
also
m
(1)
Solving for
the desired
m we have,
In words,
the
arithmetic average.
as
m
I
a
mean
=
=
I
Formula
arithmetic
mean
I.
in the A.P.
d, it
:
a,
w,
I.
follows that
m. 2,
of two numbers is their
ARITHMETIC MEANS AND EXTREMES
14.3]
If there is
more than one mean
given extremes,
we
n
and solve
in
Formula
written the
1
until the last
Example
1.
to be inserted between
two and
known values of a, I, The entire set can then be
substitute the for d.
by adding d to each
first,
223
successive term, beginning with is reached.
term (the second extreme)
Find the mean of 2 and
8.
By Formula
2 (here, and in the other examples to follow, the student should first of all write out the formula -Solution.
in question),
^
m=
(2)
+
2
8
10
=
-y-
-2=5.
Find the mean of 7 and
Example
2.
e / , Solution,
m =
7
+
-8 =
(-15) ~ =
-jr-
means between 3 and
three
3. Insert
Solution.
Here a =
3,
(3)
A 4.
2t
2i
Example
15.
/
=
and n =
11,
11
=
3
d
=
2.
+
11.
^ By Formula 5.
1,
4d,
whence (4)
The
progression, then, as obtained by adding 2 to each successive term, is 3, 5, 7, 9, 11; and the required means are 5, 7,
and 9 (answer). Insert four
Example
4-
Solution.
Here a
=
1, 1
means between
=
14,
-14 =
(5)
1
and n
+
1
=
and 6.
14.
By Formula
1,
5d,
whence d
(6)
The progression means are 2,
is
5,
= -3.
2,
1,
8,
and
5,
8,
11,
11 (answer).
14;
and the
PROGRESSIONS
224
Example
d
Solution.
From
n = 7. By Formula 1, = 2 -f (6) (3) = 20
Also
we
the given terms
2, 5, 8,
=
see that a
XIV
2 and
(answer).
l
(7)
2,
Find the 7th term of the progression:
5.
=
3.
[Ch.
Example 6. The first 3 terms of an A.P. having 8 terms are 4. Find the last term. 1, and Solution.
=
Z
(8)
EXERCISE Find
=
Here a 2
+
2,
d
=
3,
7(-3)
and n =
= -19
1.
3 and 19.
2.
2 and -12.
3.
3x and
4.
-3
mean
(answer).
8x.
and -9.
6. 7.
Insert 3
2
.
8. Insert
10.
15.
means between 2 and
5 means between 10 and
22. 2.
Find the 10th term of the progression: If I = 21, n = 7, and d = 3, find a.
11. If
1,
of the numbers given.
and 4y 2 * Insert 2 means between 3 and
9.
By Formula
61
the arithmetic
5. 5t/
8.
n
=
8,
I
=
30,
and a
=
3, 5,
7
.
2, find d.
There are 5 apple trees in a row. Each tree, after the first, produces 10 more apples than the one that precedes it. If the first tree produces 75 apples, how many are obtained from the last 12.
tree? 13.
many 14.
The
first
term of an A.P.
is
21. If
d
= -3
and
I
=
0,
how
terms are there?
Find the 10th term of the progression:
A
3, 5, 7,
.
group of boys, counting their marbles, found that one the next had 24, and so on to the last boy, who had' 12. 27, How many boys were there? 15.
had
THE SUM OF AN 16. A man travels 50
14.4]
on through the month. January?
A man
17.
A.P.
225
miles on Jan.
How
traveled 100 miles on
and so on through the week.
1,
55 miles Jan.
far did he travel
How
on the
Monday, 85
2,
last
and so of
day
miles Tuesday,
far did he travel
on Saturday?
18. A man invested $10 more each month, after the first one, than he invested in the preceding month. His investment for the 10th month was $110. What was the first month's investment?
The sum of an A.P.
14.4.
We
could find the
sum
an A.P. by adding
of the terms of
them; but that would be a long process
in
many
cases. It is
shorter and more convenient to use the formula for the sum derived from the general expressions for S in the two equations below.
The second one
order of the terms in the (1) (2)
first
is
obtained by reversing the
one.
s S
Adding corresponding members
of (1)
and
(2),
we have
(3)
Noting that a
+
I
occurs n times in this sum,
2S = n(a
(4)
and solving
(4) for
S = 5
A
second formula for the
+
Formula mula 4 is
Thus,
4.
S = ~[a
:
S = 5
[2a
:
sum may be obtained by
+ (n + a + (n-
Formula 3 by a
I
1.
3
l).
placing the
in
write
I),
S we have Formula
3.
(a
+
we may
+
(n
-
l)d].
l)d, its
re-
value by
l)d] so that
For-
PROGRESSIONS
226
To
$ we
find
given,
3, of course,
and Formula 4 when we know
Example gression:
1.
Find the sum of the
1, 3, 5,
Solution.
S =
(5)
use Formula
when n,
n, a,
and
a,
[Ch.
and I are
d.
6 terms of the pro-
first
.
Here a
+
f[2(l)
= (6
d
1,
-
=
and n
2,
1)2]
=
3(2
Example 2. Find the sum of the which a = 2 and I = 14.
+
first
=
6.
10)
By Formula 4, = 36 (answer).
5 terms of an A.P. in
By Formula 3, with a = 2, = 14, and n = S = [(2 + 14)] = (16) = 40 (answer).
Solution. (6)
EXERCISE
XIV
I
5,
62
1.
Find the sum of the
2.
Find the sum of the
3.
Find the sum of the
-17,.-. 4. Find the sum
first
first
6 terms of the A.P.:
3, 5, 7,
10 terms of the A.P. 32, 29, 26, :
first
.
.
8 terms of the A.P.: -25, -21,
of the first 12 terms of the A.P.:
7,
4,
-1,..
The sum of
5.
the
8. 9.
10. 11.
12. 13. 14.
15. 16.
the
term of an A.P. first
is
5 and the 7th term
7 terms.
= 4, = 23, and s = 243, find n and d. If a = 5, n = 12, and d = 2, find S and I = 30, find S and n. If a = 12, d = 3, and = -7, find n and d. If a = 101, S = 1457, and = -4, find a and n. If S = 24, d = -2, and = 33, find S and a. If n = 13, d = , and = -25, find S and d. If a = 17, n = 15, and If S = 171, n = 9, and d = 2, find a and I I{ a = 5, d = 3, and S = 98, find n and = 43, and S = 250, find a and d. If n = 10, If a = -6, n = 11, and S = 99, find d and
6. If 7.
first
a
I
I
I
I
Z
I
I.
I
Z.
is 17.
Find
14.5]
GEOMETRIC PROGRESSIONS
227
17. How many boys are required for a triangular formation having 6 boys in the first ro^, 5 in the next, and so on to the last row, in which there is only one boy?
A
and bounces to a height of 25 feet. It continues 5 feet each time. How far does it travel between the less bouncing first and sixth time it strikes the ground? 18.
ball falls
19. Ten bales of hay are lying 4 feet apart in a row. The first 4 feet from a truck, the second is 8 feet, etc. How far must a man travel if he starts at the truck and carries it all, one bale at is
a time, to the truck?
A
boy starts at the first one of 11 marks that are 5 yards apart and touches each of the other ten marks in order. If he returns to his starting point after each touch, what is the total 20.
distance he travels?
boys inherit an estate. Each one after the first receives than the one before him. If the value of the estate was $13,000 how much did each boy receive? 21. Five
$200
less
22. A man invests $100 at 5% simple interest at the beginning of each year. Find the value of his investments at the end of 10 years. 23.
A man
4%
invests $1000 at
simple interest at the beginning of each year. Find the value of his investments at the end of
6 years. 24.
A man
invests $500 at
6%
simple interest at the beginning
of each year. Find the value of his investments at the end of
5 years.
14.5. Geometric progressions
The following sequence as abbreviated, a G.P. a, ar,
(1)
From
is
called a geometric progression, or
ar2 ar3 ,
,
-,?.
we note
that any term after the first is found by term preceding it by r. Or again, if any multiplying the term after the first is divided by the preceding term, the
quotient
(1)
is r.
This quantity,
r, is
called the
common
ratio.
PROGRESSIONS
228
Aside from
[Ch.
XIV
which replaces the d of an A.P. the elements ' of a G.P. are the same as those of an A.P. The word snarl' ' (replacing lands'), here might be used as a memory aid. The exponent of r in any term is one less than the ordinal
number
r,
G.P., the last term 5.
/
= ar-
ar'-
.
if
in
a
.
of the terms of a G.P., called more briefly the may be indicated as follows:
of the G.P.,
S = a
(2) If
is
1
1
The sum S
sum
there are n terms we have Formula 5: Thus,
of the term. It follows that
+
ar
2
we multiply each member rs
(3)
When
=
ar
+
ar*
+
of (2)
ar 3
ar n ~*
+
+ ar +
by
+
r,
1 .
we have
+ ar
-
+ ar*-
n~ l
+
ar
right sides of (3) are subtracted from the corresponding sides of (2), all terms in the right members n drop out except a in (2). and ar in (3). Thus, we get
the
left
and
S - rS = a -
(4)
ar n
,
or
8(1
-
solution of (5) for
S
(5)
The 6
.
s =
ri
Note that value
-,
Since
yields
(r
=
1,
*
I
=
ar n n
r'). :
i).
= na
by Formula by rl in Formula (r
-
Formula 6
~l
S = 5^LL' 1
a(l
Formula 6 gives S the meaningless
whereas actually S
Replacing ar 7.
r
if
=
r)
*
5, 6,
in that case. it
we
follows that
get
rl
Formula 7
=
ar n
.
:
i).
T
any three of the elements a, r, n, I, and S are given, the other two can be found by use of Formulas 5, 6, and 7. If
14.5]
GEOMETRIC PROGRESSIONS
Example
1.
229
Find the 10th term
of the progression: 1, 2,
.-.
4,
Solution.
Here a I
(6)
Example
2.
= 1, r = = 1(2) = 9
(7)
8 -
2( * 1
f
10.
By Formula
5,
first
8 terms of the pro-
.-.
By Formula
~
=
512 (answer).
Find the sum of the
gression: 2, 6, 18, Solution.
and n
2,
=
)
with a
6,
2(
f
o
~
o
1}
3. Insert
Solution.
Here a =
,
I
=
2, r
3*
-
= 1
1
3 geometric
Example
=
=
8,
3,
=
and n =
6560 (answer).
means between and
and n =
5.
8,
By Formula
8.
5,
or
16
(9)
=
r4 .
2 are fourth roots of 16, we have r = 2. The G.P. is then either J, 1, 2, 4, 8 or 1, 2, 4, 8, and hence the geometric means are 1, 2, and 4 or 4. 1, 2, and Since 2
and
,
should be noted that in this chapter we are disregarding any solutions in which the value of r involves the (It
imaginary unit
i.)
Example 4- Find the geometric mean of x and that both are positive or both are negative. Solution. (10)
whence
or (12)
Here a
= y
x,
=
I
=
y and n
x(r}
}
=
3.
y,
assuming
By Formula
5,
PROGRESSIONS
230
[Ch.
XIV
The
G.P., then, is either x, Vxy, # or a, Vxy, y. Thus, we find that there are actually two geometric means of any
two numbers with like signs. 09 we have, as Formula 8,
we denote both
If
M
8.
M
=
by
Vxy. Find the geometric means of 4 and
Example
5.
Solution.
By Formula 8, M g =V(4)(16) =
(13)
of these
Thus, the answers are 8 and
a
=
Example
6. If
Solution.
By Formula
5,
1
=
I
32,
(14)
16.
/64.
8.
=
1,
and
=
r
,
find
n and
5.
l
(32)(J)'~
,
whence
A = (i)-
(15)
But
=
5
( )
^j, so that n
Also, (17)
by Formula
S -
_
and hence
5,
(answer).
?^i
-_y)
Given S
=
Example
7.
Solution.
By Formula
126
.
7,
32 t
=
1
n = 6
(16)
1
126, r
_
=
i^l
2,
and I
7,
= Z
1
Hence a
(19)
Also,
128
by Formula
(20)
64
But 2 6 =
64,
(21)
=
n
-
126
2 (answer).
5,
= (2X2)- = 1
and hence
=
=
6 (answer).
2 1 *-!
= 2.
=
_
63 (answer).
64, find
a and
n.
GEOMETRIC PROGRESSIONS
14.5]
EXERCISE
231
63
1.
Find the 5th term of the G.P.:
2, 4, 8,
2.
Find the 6th term of the G.P.:
1, 3, 9,
3.
Find the 7th term of the G.P.:
4.
Find the 8th term
of the G.P.:
5.
Find the 9th term
of the G.P.: 16,
6. 7.
.
32, 10, 8,
-2,
1,
4,
Insert 4 geometric means between
Find the geometric means of
13. 14.
Find the geometric means of Find the geometric means of
15.
Find the geometric means of 4a and
20. 21.
22. 23.
24.
-.
-.
If a = If a = If a = If a = If a = If S = If S = If a = If a =
1
and
.
2.
1.
9.
3 and
27.
and
20.
5
-
243.
12.
19.
-
and
1
11.
18.
-
2 geometric, means between 16 and
geometric means between 81 and Find the geometric means of 2 and 8.
17.
-
-
10. Insert 3
16.
-
-8, 4, Insert 3 geometric means between 2 and 32. Insert 4 geometric means between 32 and 1.
8. Insert 9.
.
9a.
= 2, and = 16, find n and S. 2, r = -1, and r = - find n and S. 32, = 1, and r 81, J, find n and S. = -2, and = -32, find n and S. 1, r = and = -, find n and S. 16, r = 2, and - 32, find n and a. 63, r = 81, find n and a. 121, r =*3, and = 8, and = 1, find r and S. 128, = 6, and = , find r and S. 16, n I
Z
>,
I
Z
.},
Z
Z
I
?i
I
Z
Find the sum of the areas of 5 squares if the side of the first is 12 inches, the side of the second is 6 inches, the next 3 inches, and so on to the last one. 25.
26.
A man
27.
Four men divide $5625
buys 10 watermelons, paying l for the first, the second, 4^ for the third and so on. Find the total cost. receives half as
one
receives.
much
profit.
Each one
2ff
for
after the first
as the one before him. Find the
amount each
PROGRESSIONS
232
[Ch.
XIV
An
estate of $11,808 is divided among 4 people so that each one after the first receives f as much as the one before him. 28.
How much 29.
does each one receive?
Each person has 2 ancestors
of the first preceding generation
2 (parents), 2 , or 4, of the second (grandparents), etc. ancestors has he in the twelfth preceding generation?
30.
Assuming that a patriarch has 6 children
How many
(first following of these has 6 children that each (second generation), generation) and so on, find the total number of descendants of the patriarch
through and including the fourth generation.
Chapter Fifteen
LOGARITHMS
15.1.
Remarks and
The computation example,
-
'
'J-
definition of a numerical quantity such as, for ' ,
would be a slow and tedious task
not for the invention * called logarithms. This is a device which uses the laws of exponents to simplify calcu-
were
it
lations
involving multiplication, division, involution [or evaluation of powers such as (1.05) 10 ], and evolution (or / extraction of roots such as v 2321). Such laws are applicable because, as the following definition will show, a logarithm is
essentially
an exponent.
N
with respect to Definition. The logarithm of the number the base &, designated as 'logbN,' is the exponent which must be applied to
15.2.
b in order
to
produce N.
Exponential and logarithmic equations
(1)
10 2
=
100
=
100,
and (2)
logic
2,
the latter of which is read 'the logarithm of 100 to the base 10 equals 2,' are two forms of the same statement. The first is an exponential equation, and the second is a logarithmic equation. It should be noted that the exponent * The inventor was John Napier (1550-1617), and the one who applied the invention to practical computation was Henry Briggs (1556-1631).
233
LOGARITHMS
234
of 10
2 and the logarithm of 100
is
=
N
[Ch.
XV
In both equations
is 2.
N
= x. follows that log& Similarly, It is very important to be able to express a logarithmic equation as an exponential one, and vice versa. For example, the base
if
P =
MN 15.3.
=
is 10.
if
P =
then Iog 10
10-,
bx
x
9
it
if
Iog 10
MN
=
x
+
y,
then
10*+*.
Four laws of logarithms
Since logarithms are exponents, the laws applying to both are similar. shall now derive the basic laws for logarithms. 1.
log a
We PQR =
In words, of
log,
P +
the logarithm of
log a
Q +
a product
log. R. is the
sum
of logarithms
73.6
+ Iog
its factors.
For example,
=
logio (58.4) (73.6) (86.4)
P =
Proof. Let log,
Then,
P =
ax
,
58.4
log
Q =
x, log,
Q =
and
a*,
+ Iog
y,
R =
10
and log a a2
R =
10
86.4.
z.
.
Multiplying,
PQ72 = awa' Changing the log,
2Mog a ~ =
1,
Art. 7.1)
equation to the logarithmic form, we have,
last
PQR =
= a^+*. (Law
+
x
y
P -
log a
+
z
=
log,
P+
log,
Q
+
log, R.
logaQ.
In words, the logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. For example, by Law 2, ,
10gl
= =
(58.4) (_6L7)
(15^X387)' lo glo [(58.4) (61.7)] (logio 58.4
+
-
lo glo [(15.6) (387)]
log lo 61.7)
-
(logio 15.6
+
Iog 10 387)
(by
=
logio 58.4
+
logio 61.7
logio 15.6
.
logio 387.
Law
1)
15.3]
FOUR LAWS OF LOGARITHMS
Proof. Let
= log a P
x and log a
Then,
P=
ax
Dividing
^
=
^
p =
Hence, 3.
log a
x
log a 7:
v
Pn =
In words,
/ie
Q =
and
-
235
Q =
y. a'.
a-*. (Law
=
y
log a
2, Art. 7.1)
P
log a Q.
n log a P. logarithm of the nth power of any number equals
n times the logarithm of the number. For example, logio (S73.4) 3 = 3 Proof. Let log a
P =
x, so
Then, Hence,
log
tt
P =
that
Pn = Pn =
(a
logic 873.4.
n
x
nx
)
=
a*.
a nx (Law .
= n log a
5,
Art. 7.1)
P.
4.
In words,
/ie
logarithm of the nth root of a number equals
_
- times the logarithm of the number. n For example, logio ^57.6 = ^ logio Proof. Let log
P =
x, so
^P
Then,
lo&
Hence,
The
following law
Corollary, or
For example,
The proof
Law
is
5.
logio
is left
that
=
^P =
P =
57.6.
ax
.
- a.
v^T*
- = n n
(x)
(Def. 4, Art. 7.5)
= - log a P. n
a corollary of Laws 3 and m log a v^Af
=
3 v(O084) = f
to the student.
n
log a
Iog 10 (0.084).
4.
LOGARITHMS
236
EXERCISE Given rithms
to
[Ch.
XV
64
log$
2
=
=
0.30 and logw 3
0.48, use the laws of loga-
numbers in problems 1-16.
find the logarithms of the
Examples. (a)
Iogio6=logio (2-3) logic 2+logio 3= 0.30+0.48= 0.78. = log 3 2 = 2 logio 3 = 2(0.48) = 0.96.
(b) logio 9 (c)
logio
3-2 3 = log
24= log
(d) logio f
=
logio
3-2
3+3
logio
=
0.48
logio 2
2= 0.48+0.90= 1.38.
-
2(0.30)
12.
2.
18.
3. 27.
4. 8.
6. 32.
7.
f
8.
9.
1.
11. 5. HINT.
5
=
.
f
f
-0.12. 5.
13. 125.
15.
16. .12 (or
.2.
16.
10. f.
.
12. 25.
-V*.
14. .3 (or 3^).
.
=
Solve for x the equations in problems 17-34.
Examples. (e) (f)
If
(g)
If
17.
20.
23.
26. 29.
32.
x logs 8 log* 8
If Iog 2
= 3, = x, = f
,
= 2. = 3. logs x = x. Iog 64 = x. Iog ( ) = 2. log, 8 = i log, 8
= 2 = 8. 2 X = 8, and hence x = 3. x% = 8, and hence x = 4.
x
3
18. Iog 4
logsx
a;
21. logs x
= = =
4
24. Iog 9 3
9
27. logs (i) 30. Iog x 9
=
33. log, (J)
19. logs
1.
22. Iog 9 x
3.
= =
25. Iog 4 (f )
x.
=
x
28. Iog 4 (8)
x.
31. log, 3
.
= - 1.
34. log, 4
In each of problems 35-47 state whether the equation is If it is false, change one member so that it will be true.
False.
Change
A
Iog
B
.
x.
x.
.
true or false.
logq^
right side to Iog
f
= =
= - . = -f
Examples. .
f.
logo C.
SYSTEMS OF LOGARITHMS
15.4]
(i)
log a
= t^ VzV
(j)
X y
log a X
=
False.
35.
36. 37. 38.
Change
-
3 logo x
-
~
4 Iog
*.
True.
left side to log a
(-)
a;
=
(1
3(lo ga
( Y fx* <~3 =
--
42. logio
C = =
2108 * 2
15.4.
y
log a y.
4 -
=
logio
(2 log
log a C.
log.*).
3
2 V
46. a
a
^
40. 10g a
44. 10
+ 2 log
= %loga A.s loga VI = ilog.5. V -^loga 5 = I log, A (log, 4)1 = (log. + logn y) log a (xyY
39. log
43.
237
4 -
<
(logio 100) (logs 32).
=
x2
l
45. logio (logio 10) 47. log a a 3
.
=
=
0.
3.
Systems of logarithms
There are two standard systems of logarithms in general use in elementary mathematics. They are: (a) The Briggs or common system, in which 10 is used as is used extensively in numerical calused hereafter in this text. Thus, be culations, whenever the base is omitted, the base is to be understood as 10. For example, 'log 58' will mean 'logio 58.' (b) The Naperian or natural system, in which the base is an irrational number approximately equal to 2.718. This
the base. This system
and
will
so-called natural base, usually designated by the letter is used extensively in advanced mathematics.
e,
LOGARITHMS
238
15.5. Characteristics
[Ch.
XV
and mantissas
In the following table, the two equations on the same line result one from the other, or, in other words, are corre-
sponding exponential and logarithmic statements. 10 10
1
10
2
10 3
= = = =
Note that
logl
1;
log 10
10;
log 100
100;
log 1000
1000;
we apply
= = = =
0. 1.
2.
3 (and so on).
number) exponents 1 and as digits. It is clear that most numbers are omitted from this group. Hence it becomes necessary to use some exponents for 10 that contain decimal fractions in order to write all numbers to 10
if
integral (whole
we produce only numbers having
as powers of 10.
For example, 25 lies between 10 and 100. The exponent that must be applied to 10 to produce 25 is therefore between 1 and 2. Actually 10 1
(1)
-
3979
=
25 (nearly),
=
1.3979.
or log 25
(2)
That
the exponent 1.3979, applied to the base 10, produces the number 25. This exponent consists of the integer 1 is,
fraction .3979. In general, the whole number logarithm is called the characteristic, and the
and the decimal part of any decimal part,
if
and when
For example, while log
it is positive, is
.25
= 1.3979-2= -l+.3979 = is
=
called the mantissa.
(-&) =
-
log 100 -.6021, the mantissa of log .25 log
log 25
.3979 rather than -.6021.
15.6. Operations
When
which
a new number
leave the
mantissa unchanged
obtained from a given number by moving the decimal point to the right or left, the first number is
15.7]
RULES ABOUT THE CHARACTERISTIC
in effect multiplied or divided
is
239
by a power
of 10.
For
example, (1)
2500
=
25.00
X
10 2
.025
=
25
10 3
.
,
and (2)
From
(1)
and Law
log
(3)
From
(2)
2500
and Law
log .025
(4)
= = = =
4-
we have
1,
2 log 25 -f log 10
+ 2 log 1.3979 + 2(1)
log 25
10 (by
Law
3)
3.3979 2, it follows
that
= log 25 - log 10 = 1.3979 - 3(1) = -2 + .3979.
3
In equations (3) and (4) we see why the mantissas of the logarithms of the three numbers: 25, 2500, and .025 are the same. In general, when the decimal point in the number is
moved n
places to the right, the logarithm
the integer n (or log 10 the left the logarithm
n )
;
is
and when
it is
decreased
by
is
increased
moved n n.
Hence
by
places to in either
case the mantissa remains unchanged. This very convenient result makes it possible to have compact tables of logarithms,
example, only one mantissa (.3979) need be recorded to apply to such various numbers as 25, 250, 2500, since, for
2.5, .25, .025, etc.
15.7.
Rules about the characteristic
Since log 10 = 1 and log 100 = 2, it is evident that 1 is the characteristic of the logarithm of 10 and of all numbers
between 10 and 100;
or,
in other words, of all
numbers
LOGARITHMS
240
having two significant point. Similarly, for
*
digits to the left of the
numbers with three
the decimal point, the characteristic
is
[Ch.
XV
decimal
digits to the left of
In general, the
2.
following rule applies.
RULE
1.
The
characteristic of the logarithm of
one
a number
than the number of significant digits
greater than one
is
to the left of the
decimal point.
less
The logarithm
of a number less than one is negative, but a characteristic and a (positive) mantissa. For example, it was noted in (4) Art. 15.6 that log .025 = 2 + .3979, with the characteristic 2 and mantissa .3979. For convenience, this result is usually written in the form
there
is still
log .025
(1)
member
since the right
=
8.3979
of (1)
is
-
10,
easy to use and preserves
the positive mantissa. Continuing the table in Art. 15.5 in the reverse direction,
we have
=
10- 1
^
= =
= -- =
10-*
log.l
.l;
-
01;
log 01
= ~2
log .001
= -3
-
.001;
=-1. -
(and so on).
study of the table above we can arrive at a second rule about characteristics, namely,
With a
RULE
2.
little
The
characteristic of the logarithm of
and
a number
less
one plus the equal numerically number of zeros between the decimal point and the first non-
than one
is negative ,
is
to
zero digit on the right. *
number include the first non-zero digit, as read from the digits that follow it. For example, each of the numbers 10.732 and 00030.21 has two significant digits to the left of the decimal point.
The
significant digits in a
left to right,
plus
all
15.8]
SCIENTIFIC NOTATION
AND CHARACTERISTICS
241
Examples.
THE NUMBER
THE CHARACTERISTIC OF THE LOGARITHM
-(1 + 0) =-1 -(1 + 2) =-3 -(1 + 10) = -11
.8003
.00207
.00000000004
In practice we may count the decimal point as well as the proper zeros, getting at once for the sum the correct negative characteristic. Thus the latter is obtained first as a negative integer, after which it is rewritten in the conventional form when the mantissa is known. log .00025
Example
1.
Example
2. log
- -4
+
.3979
=
6.3979
-
10.
.000000000025= -11 +.3979 = 9.3979 -20.
15.8.* Scientific notation and a second mining characteristics
method for
deter-
It is possible to find the characteristic of the logarithm of
N
a number
to base 10
by taking advantage of what is method for representing N.
called the 'scientific notation'
For example, 4
written as (5.834) (10 2 ); 0.000208 as 3.08 as (3.08)(10); 100,000 as (1)(10 5 ); 0.3141
583.4
if
is
(2.08)(10- ); as (3.141)(10- 1 ) 0.000,000,000,108 as (1.08) (10- 10 ), etc., each of the numbers is said to be written in scientific notation. ;
N
In other words, is in scientific notation if it is represented as (M)(10*), where 1 ^ < 10 (M is equal to or greater than 1 but less than 10). = 2.08, and k = -3. For example, in 0.00208,
M
M
RULE number
3.
To determine
the characteristic of
a logarithm of a
base 10, it is merely necessary to write (or think of) the given number in scientific notation, and the exponent of the 10 in this representation is the characteristic. to
* This article
may
be omitted without
loss of continuity.
LOGARITHMS
242
Example. Since, in (a)
5870
(b) 3.4 (c)
=
[Ch.
XV
scientific notation, 3
(5.87) (10 );
=(3.4) (10);
=
0.0004108
(d) 0.01287
=
4
(4.108) (10- ); 2
(1.287) (10~ ); 6
(f)
1,000,000= (1)(10 ); 15 = (1.5) (10 1 ); and
(g)
0.000,000,000,012
(e)
=
(1.2)(10-);
it follows that the characteristics of the respective logarithms are the exponents of the 10 in the right members of the above equations. Thus the characteristics are (a) 3, (b) 0,
-4,(d) -2,
(c)
EXERCISE
65
(e) 6, (f) 1,
(g)
-11.
(ORAL)
Find the characteristic of numbers by both methods. 1.
and
36.
5. 3.2.
the logarithms of each of the following
2. 407.
3. 8.
4. 2573.
6. 8.05.
7. 0.6.
8. 0.058.
10. 932.5.
11. 0.002.
12. 0.00015.
13. 2.75.
14. 0.235.
IS. 0.056.
16. 0.0058.
17. 283.5.
18. 0.08276.
19.
15.9.
the mantissa.
9. 62.05.
Finding
The mantissa
3258
Four-place tables
of the logarithm of a given use of a table, such as Table 3 at the
by The
20. 865432.
number
is
found
end of the
text.
column on the left with the heading N contains two digits of the number. For example, if the number is 157, the digits 15 are found in the sixth row beneath N. The third digit, 7, is found on the line across the top of the table which contains N and the digits 0, 1, 2, etc. In the sixth row, under 7 and in line with the first two digits 15, we find the entry 1959. This means that the mantissa the
first
first
of the logarithm of 157, or .157, or 15,700, or .000157, etc, is .1959 (the decimal being omitted for brevity).
ANTILOGARITHMS
15.10]
243
number whose logarithm
sought has two digits it is found in the column under AT. Its mantissa will then be on the same line and in the column beneath the digit 0. If the number has one digit, we multiply it by 10 and use the mantissa of the product. For example, the mantissa of = the mantissa of log 70 = ,8451. log 7 If
the
When we its
find the logarithm of a
number we
first
write
by inspection very important to get then find the entry in the table which gives
characteristic
this habit)
is
and
(it is
the digits in the mantissa.
Examples.
=
plus the mantissa. Here the student may be tempted to omit the characteristic; but this is a very bad practice, which often results in errors. The characteristic (a) log 4.3
should always be written first, even when it is zero. In the table, in line with 43 and in the column under the digit 0, is the
= 0.6335. entry 6335. Hence, log 4.3 = 2 plus the mantissa. In line with 56 in the (b) log 564 under the digit 4, is the entry 7513, in column the table, and so that log 564
=
2.7513.
=
2 plus the mantissa. The latter, accordlog 0.0108 in table in line with 10 and under 8, is the the to entry ing 2 in the recommended .0334. Writing the characteristic (c)
form,
we have
log 0.0108
=
8.0334
-
10.
15.10. Antilogarithms If log
A =
and may
A
B
said to be the antilogarithm of be found from the table by reversing the operation J5,
then
is
of finding the logarithm.
Examples. (a)
We
logN = 3.8102. Find N. now look inside the table
found in
line
for the entry 8102. This
with 64 and under the digit
6,
so that
is
N has
LOGARITHMS
244
[Ch.
XV
the digits 646 with the decimal point yet to be fixed. Looking next at the characteristic, 3, we see that there are four digits
N. Hence N = 6460. 10. Find N. 7.3979 (6) log N of the table shows that N has the digits 250. Inspection
to the left of the decimal point in
-
=
Since the characteristic
is
10 or
7
3,
N
is
a decimal point followed by two zeros. Hence (c)
log N
=
0.4133. Find
written with
N
=
.0025.
N.
N
istic
has the digits 259. The charactertable shows that shows that has one place to the left of the decimal
point.
Hence
The
N
EXERCISE Find
N
=
2.59.
66
the logarithm of each of the following numbers.
N in each of problems 21-33. = 1.0607. 22. log N = 1.3541. 24. log N = 1.6444. 26. log N = 9.4609 10. 28. log N = 30. 10. 7.3636 log N = 32. 0.7388. log N = 6.0000- 10. log N
Find 21. 23. 25.
27. 29.
31.
33.
log JV log N log N log N log N log N
= = = = = =
2.1303.
2.6335. 3.5933.
8.4983
8.2455
-
10.
10.
0.0043.
15.11. Interpolation
When a
given number, or a given mantissa, is not found directly in the table, it is necessary to use a process called interpolation, as illustrated in the following examples.
15.11]
INTERPOLATION
245
Find log 2576.
Example
1.
Solution.
Here the
we observe
characteristic
is 3.
To
obtain the
man-
between the numbers 2570 tissa, and 2580, whose associated mantissas are the same as those for 257 and 258 respectively. Schematically the work is that 2576
arranged as follows
lies
:
Number 6 10
Now
4099
Mantissa
T2570
4099'
[2576 2580
4116J
digits
17 (the difference)
+ -&(17) =
4109.2, but since four digits only are mantissa is .4109. Hence, the desired retained,
-
log 2576
3.4109.
Evidently the process of interpolation would not have been changed if the decimal point had been differently placed in the digit sequence 2576. For example, log .0002576
=
6.4109
-
10.
should be realized that interpolation uses the principle proportional parts, which assumes that for a small
It
of
change in the number there corresponds a proportional change in the mantissa. This is only approximately true, but in order to use interpolation this assumption will be accepted when the logarithm table does not contain directly the information desired.
Example
2.
Find
N
if
log
N
=
9.4177
-
10.
4177 are not found among the mantissa digits inside the table but they are seen to lie between the adjacent entries 4166 and 4183. The schematic arrangeSolution.
The
digits
;
ment
follows
:
Number
Mantissa
'2610
41661
L2620
4177J 4183
10
digits
11
17
LOGARITHMS
246
Hence, the significant digits in
2610
+ ii(10) =
[Ch.
XV
N are
2616 (nearest four digit number).
Thus, taking into account the characteristic of logJV,
it
follows that
N
=
0.2616.
When first learning this process, the student should put the work in table form as illustrated in the examples. However, with enough practice and a clear understanding of the principle of proportional parts, he
may
often learn to inter-
polate mentally.
EXERCISE Find 1.
67
the logarithm of each of the following
2536.
4. 35.27. 7.
2. 728.6.
3. 0.08352.
5. 839.6.
6.
0.005706.
9.
0.0007085.
8. 294.3.
5.732.
10. 6.325
X
10 8
11. 9.246
.
numbers.
X
10~ 12
N in each of problems 13-24. = 2.2226. 14. log N
.
X
12. 2.318
1020
.
Find 13. 15. 17.
19.
21. 23.
logN = = log N = log N = log tf = log N
15.12.
3.1145.
9.1463 8.1633
2.3146
16.
-
10.8143.
10.
18.
10.
20.
10.
22. 24.
N= = log N log
logN = = log # = log AT = log N
6.3641.
7.2883
-
10.
0.7714.
7.4161
0.4923
-
20. 10.
15.0129.
Computation with logarithms
should be clearly understood that the usefulness of logarithms lies in the fact that they shorten computations involving multiplication, division, involution, and evolution It
(though not addition and subtraction). Furthermore, these operations could be performed in other ways, so that if time is not saved, the point in the use of logarithms is missed.
15.12]
COMPUTATION WITH LOGARITHMS
247
important, therefore, to take note of the details and practices which do save time. Chief among them are the It is
following. (a)
The quantity
to be
evaluated should be written
directly above the
details of computation,
designated by some
letter, as
(b) left
is
used at (c)
N.
The computation should be
members
and should be
outlined in
full,
with the
of all equations filled in before the logarithm table
all.
The logarithms should be arranged
in neat
column
form, with the decimal points and the preceding equality signs in vertical lines.
The student should study
carefully
the
illustrative
examples below before attempting the problems in the subsequent exercise.
Example
L
Evaluate
N log N
Solution. Let
Then,
log 324 Outline, log 618 log
N N
-.
= $f= log 324 - log 618 '(by Law = 2. = 2. (-) = = (answer).
2).
Note L In what follows we have repeated the outline to show how the mantissas should be entered. The student, however, should understand that after the complete outline is
made, including
write
down by
all
characteristics that
inspection, he should
make
possible to his entries from
it
is
the table in that same outline.
Turning next to the table
in the text,
we
fill
as follows. log 324 log 618 log
N N
= = = =
12.5105
-
10
-
10
2.7910 9.7195
(-)
.5242 (answer).
in the outline
LOGARITHMS
248
[Ch.
XV
Note 2. Since in the details above we are obliged to subtract a larger from a smaller number it is helpful, as soon as this fact becomes apparent, to rewrite the characteristic of 10 instead of 2. Remember that the decimal log 324 as 12 of the part logarithm is not the mantissa unless it is positive.
Note
In actual
3.
life
many numbers
used in computations
are approximations with a limited number of significant digits. In this case the computed result should not have more
than has any of the numbers which enter into the computation, and in the case above should be written as .524. In the subsequent exercise, however, all numbers are assumed to be exact values rather than approxisignificant digits
N
mations.
In the following examples the values found in the table will be included along with the outline, though actually they
should be j
filled in after
Example
the outline
is
made.
^
4
* IT i + -(-02764) (3.268) 2. Evaluate a/
^210,700 Solution. (Here interpolation is in order. The arrangement below suggests a compact, efficient way of recording all logarithms found, so that no details looked up are omitted in the final form of the result.)
Let
Then,
N log N
=
(2764K3.268)*. v/210,700
=
log 2764
+ 4 log 3.268 -
log 2764
4 log 3.268
=
4(0.5142)
log numerator log 210,700
=
1(5.3237) log N N
= = = = = =
log 210,700.
3.4415 2.0568
(+)
5.4983 1.7746
(-)
3.7237
5292 (answer).
the preceding example we found that the antilog of 3.7237 is halfway between 5292 and 5293. In cases
Note
4-
In
15.12]
COMPUTATION WITH LOGARITHMS
of this sort,
we
249
shall agree to use the even
numbers between which the interpolated
Example
3.
Solution. Let
= ^0.01084
N
Evaluate
A = v^.01084
and
result
+
B =
one of the two lies.
4
(0.5136)
.
4
(0.5136)
.
Then,
A = log B = log
$ log 0.0184;
4 log (0.5136);
and
N
= A
+ B.
Details.
A = = log B B = log
(8.0350 4(9.7106 0.06957
A =
0.2213
N
0.2909
=
Note
sum,
Since
5.
10) 10)
- 30) = 9.3450 - 10. (28.0350 38.8424 - 40 = 8.8424 - 10.
= =
(+) (answer).
we have no formula
necessary to find
it is
Note
-
6.
A
and
B
for the logarithm of a
separately.
a second method of finding log
By
-
-
=
10) 3) $(1.0350 $(8.0350 the form indicated in the solution
= is
0.3450
-
A
1.
we have However,
customary.
not necessary to take the final step indicated in the calculation of log/?, since the characteristic, 2, is from 10. 40 as easily as 8 obtained from 38
Note
7. It is
Example
4.
Evaluate
Solution, log
N
N
N
=
(0.008402)*.
= f log (0.008402) = f (7.9244 - 10) = = j(65.8488 - 70) = 9.4070 - 10 = 0.2553.
(15.8488
-
20)
LOGARITHMS
250 5.
Example
Evaluate
N=
(27.84)
-3
log
N= =
3 log (27.84)
i
OT:y
-
3 log (27.84). - 10 (or 0) log 1 = 10.0000 3(1.4446) = 4.3338 (-) log
1
log
EXERCISE
XV
1
= ju
Solution,
[Ch.
N N
= =
-
5.6662
10
0.00004637.
68
use of the four-place table of logarithms compute the value of * that 21 = 21.00, each of the following quantities. (It is agreed
By
.031
1.
=
.03100, etc.)
(25)(76)
2.
(0.283) (41. 6)
388
83 3.
(127.4) (32.6)
4.
(576) (340)
0.0827 5.
(228)
6. (3247) (6.97) (0.0844).
(213.4) (0.529). 3
2
8. (0.732) 2 .
2
7.
3(41.62) (87.2)
9.
(925.6) V387.8.
.
10. (0.00274)*.
11. (0.009082)*.
12. (327)*.
13. (0.0528)1
14. (928)*.
IS.
16 '
'
3
(387J 17.
(576)~
2
18.
.
20. (8.37)-'.
19. '^807.6.
21.
V5298.
22. ^8.362.
^3256.
23. (3. 142) (37.6)
8
24. V(23.8)(47.6)(39.8)(524).
.
1
1
26.
25.
V0.8846
(58.4) (9.836)'
54.8 27.
(-1.06)
3
-4.84
HINT. First find
29.
* See Note 3 in Article 15.12.
N, which
(4.728)(175.3)
(584)(755)
is
30.
positive.
(-4.76) (0.8138) (43.6)(-0,084)
2
'
COMPUTATION WITH LOGARITHMS
15.12]
Given 8
31.
A =
= 3WPA
2 ,
4g
find
S
W=
347.2,
P=
6.7,
and
0.7840.
32.
Given
T = w J-, find T if I =
33.
Given
M=
and
if
251
c
=
48.38.
TO
I
O
=
733,
O
M lia= '
26c~
'
find log
32.2,
and ir
1L20
'
6
=
3.142.
=
43 43 -
'
TABLES
ANSWERS INDEX
Table
+
A LIST OF SYMBOLS
1.
read 'plus.' read 'minus.'
a
X
b
a
-T-
b y or f, or a/6
read 'a multiplied by 6.' read
6
= =
read 'is not equal to.' read 'is less than.'
< >
read 'is greater than.' read 'is less than or equal to.'
^ ^ a! or (
I
a
read 'is greater than or equal to.' a.' read 'factorial a or 1-2-3 -
Parentheses
) ]
{
}
logo
6
6n _
n
-
-
These are signs of aggregation. They are
Brackets
used
Braces
which are to be treated in operations as one algebraic expression,
Vinculum
6.'
read 'is equal to.' read 'is identical with.'
T^
[
'a divided by
to
collect
read 'logarithm of n to base read 'absolute value of 6.'
algebraic
expressions
a.
read 'the nth power of 6' or '6 with the exponent read 'square root of a.'
n'
read 'nth root of a.'
oo
read 'a subscript n' or 'a sub n.' read '/of x' 'g of x,' or 'the/function of x/' etc. read 'infinity'
oo
read 'varies as'
an
/(x), g(x), etc.
264
Table
2.
SQUARES, CUBES, ROOTS
255
Table
3.
COMMON LOGARITHMS FOUR-PLACE LOGARITHMS
256
FOUR-PLACE LOGARITHMS (Continued)
257
ANSWERS EXERCISE
-
x
2.
1. 2z.
Page 2
1.
5
3.
4.
+
3z
-
(20
-
x
/
=
Prt.
4.
2.
6x2
6.
1.
3z miles;
7.
.
2i
(30
-
12.
T=
3x) miles.
-
10
13.
+ 2*/) =
17. (x
3x
+ 4y.
C =
2wR.
8.
2(2/
EXERCISE
14.
+ 3*).
+
18. 2(x
+ 3.
22. x; 2x
21. x; 3x.
x.
9.
1)
-
-
y) miles.
11.
2x
16.
P = ff. = 3x - 2.
+3 + 3.
= ^--^
2
23. x; fx
-
19. x;
24. 25x.
2.
Po0e 7
2.
2. 0, ~i f, -1, -2, -15, -5, -4, -2, -1, -i 0, f, 5, 6. 6. -4. 4. -10. 7. -7. 8. 12. 9. -5. 3. -3. -15. -4, 5, 6, 16. -4. 13. 32. 14. -15. 17. -9. 11. -7. 12. -17. 18. -7. 19. 3. 21. 3. 22. -5. 23. 10. 24. -17. 26. 31. 27. -3. 28. -12. 1.
-7.
29.
58.
-1.
67.
-2.
-42. 59. -4.
76. 0.
86.
84. 2.
EXERCISE
3.
43.
52. 0.
-5.
71. 5.
79.
-30. 64. -3.
63. 5.
72. 0.
Not a number.
73. 0.
74.
81. 0.
82.
38. 31.
47. 0.
56.
54. 180.
-14.
37. 20.
46. 0.
44. 12.
53. 0.
62.
36. -17.
34. 5.
-40.
61. 0.
78. 0.
77. 0.
1.
-8.
69.
68. 6.
33. 3.
32. 4.
42.
51.
49. 0.
83.
-13.
31.
41. 8.
39. 40.
48. 0.
57.
-21.
66.
-10.
Not a number. Not a number.
87. 0.
1.
Page 12
3 2 x*y&. (Sample answers) 2x i/0 5x*y&; Binomials: 3x+y; 4w+3n. Monomials: 2s; 3y.
1.
2.
(Sample answers) Trinomials: 2x 3y
;
:
+
6. 5ox - 5x + 9. 4. 7-5x. 3. 6x - 4. m + 2ran + 5n 9. -3z + z + 4z 8. 3x - Sax + 2. 7. 4s - 3z - 2ax - 1. - 2x + 3. 21. No. 12. x 2x 13. 3x 11. 3x 2a x + 1. + + 26. -2s - 7x + 5. 1. 24. ox + 8x 23. 9x 22. 2x 10. 28. -7x + 5x - 2x - 2a + 1. 4x 27. -x 4ox 3. 32. -ax 31. -9x + 1. 29. -2x + 3x* + 3x + 2x - 4a + 3. 36. 7x 34. 2X + 7x - 4ax + 3. 8x + 5. 33. -ax + x + 1.
+
2
2
2
5;
.
4
3
2
3
3
4
2
2
3
2
3
2
4
2
2
2
.
2
2
4
2
3
259
2
4
ANSWERS 38. -a 37. 2x 4 - 3s3 - 3x2 - 2x + 4a - 3. 5x2 + 2x + 2a - 1. 42. 5a - 26 - c. 6 + 4. 6 + d. 41. -4a 39. 7x-2y- 1. 46. -x2 + 3ax 2 + 36x. 44. -4x + 7. 43. -a + 26 + 4c + 1. 49. 4x - (2a + 3?/). 48. 2x - (3a - 6). 47. -3x2 - 2ax 2 + 2x. 52. -36 53. 2s3 - x 2 + 7x - 3. 5c. 51. 3a (-46 + c + 1). 260
-
54. 4a3
+
10a2
EXERCISE
+ 3.
Page 16
4.
12aV.
1.
12a6
-156x 4
2.
24a3x 5
3.
.
4.
.
-6a^.
-3x.
6.
7. 4x?/.
+ x - 2s 1) l)](2x l)(x - l)(2x + 1)] = (x + l)(2x - x - 1) 2X + x - 2x - 1. (x + l)[(x 11. 9x - 24x + 13X - 29x + 18x - 3. 9. 12x*f+ 22X - 2x - lOx. 13. 12x + 32x 17x + 16x - 4. llx + 19x 5x 12. 2x 14. x - 2x - 7x + 21x - 16X - 9x + 5x - 47X - x + llx - 2. 17. 3x + 16. 18x 6x 13x + 9x3 + 15x + 2x - 4. 2. lOx 18. 8x + 12x - 32X - 20x + 48x - 16. 8x + 3. 7x 14x + ISx 21. 18x + 39x - 4x 19. 3x - 2x - 42x + 6x5 + 45x - 13x. -
+
8. [(x
=
+
(x
2
-
l)(2x
+
= =
1)
2
2
3
6
4
5
4
6
4
55x
x
2
5 .
36.
2x2
-
-21 3x
+
1
rX
31.
.
- ab + b -3x - x + 11.
-
2a 2
2
41.
+ +
-3x
38.
.
4x.
8. lOx
2.
+ y.
4x2
sq. yds.
14.
600x
19.
T=
y
+
28.
45
~
3.
9. l(ty
+
13. (2x
ft.
16.
0.10m
~
-
10.
23. 7 29.
-9.
50z. 31.
-
11.
^
2
3x2
-
17x
+ 48 +
+ 2ax + 5a
-3x2
2 .
^A._.
7 or 7
2(10x
-
Page 17
I.
x.
+ y) +
(6x
9) yds.;
7200(x
=
x
I
~y~
6.
-
3x
-
12.
4x yds.; lOx yds.;
9.
7.
5
2
x.
+ 0.40s.
-9.
4.
'
__ ^^
2
.
2
X~~
+ 4x + 6x + 19 + 29. 2x - 2x ^ +1^ +
2
+ a6 + 6
'
3
(^
27. x3
2.
REVIEW EXERCISES FOR CHAPTER 1.
JL>
.
37. a2
4
__ _L
r4
9-5 &&
f
.
a2
39.
2
3
5
**_.
x
2
3
5
~2 x2 + + 2x ++ 5x 2x - 1 2 32. 2x - 3 + ^ ^ t ^-
g^ _
2S
4
5
4
1
-
26. 3x 2
3.
5
2
f
-
1.
2
2
__
24. x3
6
4
5
5
5
6
2
6
2
3
4
3
3
4
7
6
2
2
3
2
3
2x3
-
36.
18) yds.;
(2x
2
-
9x) sq.yds. b h( a
18. A = V = frR3 21. D = 2x - y D = y - 2z. 26. * = 26,400s. 24. / = .07/2.
+ 1)
-20.
ft.
32.
17.
.
Not a number.
33. 0.
+
-
j
22. x
=
27. 24. 34.
Not a
ANSWERS
261
-. + 26. 51. 3y. 52. 8a - 9a6 + 53. -8n - mn + 10. 54. 6a - 106 + 56. -x + 2y - 62. 57. 2x + lOxt/ - 17. 58. -8a6 - lie + 2d + 6. 59. x + 4. 61. -3a + 126 + 24. 62. x + 5y + 8. 63. -3x - y s + 3w)i 64. a 6 4a 66. y - 5 - (x + 3z (2x (2a 6 67. - 6x0. 68. 21x <#. -15m n 71. -10m n + ISm^/i lOmrc 72. -3x -' 9x - 6x + 4x y + 7xy - y 73. 4a + 26a6 2 2 - Gx - 2x ?/ + 4x0 + 146 74. x x + x0 76. 3x + x - 7x + 10. 79. x + 77. x 7x x 12. 78. + + 2t/ x*y + xV 21. 4x 81. -3a + 3. 82. 2a 12. 5a 83. 6a - 19a + 10. 4a + 2. 86. 4x 87. -9x + 84. -30a 88. ?/ - 4x ?/. 12x 91. 9 -45x 4x 92. -32 89. + + 50 + 16x - 2x 12x 12. 94. lOOx + 60x + 9. 96. 16x + 40x0 + 250 93. -3x 101. 3x + 2x + 97. -80. 98. 25m. 99: 9a6 ^ 102. 3x - 4y. number.
-18. -1. 46.
36.
44.
43. 76.
-8.
37.
38. 30.
-862
47.
9.
2
-2.
39.
-32.
41.
49. 2(a
48. 0.
.
42.
+ 6) =
2a
2
c.
c.
2
2
2
2
2
2
2
4
2
).
6
7
?/
4
2
).
3
3
.
2
3
.
3
2
2
2
.
2
.
2
2
3
.
8
?/
5
4
.
j/
?/
2
2
2
3
2
2
.
2
2
2
2
2
2
2
2
.
2
4
.
2
.
2
.
2
.
2
2
2
.
2
2
.
'
'
27
9
3
EXERCISE 2.
5x.
4.
-
-12ax
12x0 14. 18.
-
9x
2
2
6a
-
26. 27a3
40
2
+4-
12x
+
63
9x
4x
19.
.
27.
.
The square
9. x
.
-
4
90
8p
160
-
4
16.
.
4
21.
.
2
.
+ 27m p 3
+ 4ax + lOx + 25. 13.
.
4
~-
~
b2
a2
3
of a trinomial equals the
22. x 2
+ 2x0 +
28. a
.
17.
.
3
sum
-
4xz
+ 402.
9c 2
-
4a6
33. 9x
-
12ac
+
+
40
+2
2
2
-
c
2
2
.
-
- 4a 6 + 6 - 3a c + . 2
2
.
2
+ a - 4. - 4r + 4rs s 29. -24a + 36 +
2
2ax 2
2
3
.
3
.
of the squares of the terms
+ 6x2 x(a + 6 + 1). 12x0
36.
66c.
9a
2
4x 2
-9x
11.
4a 4
-
.
3
+
2 plus twice the product of each pair of terms. 32. x 2
2
26x 2
3
4
10x 2
3.
3.
6X
+
24. x2
2 6
40
2
-
2x* 3
8n 2
2
2
(6)
6.
.
2
4
- 24m x +
-
3
-18m - 24m n -
12. 2
4
+ 6x0 + 3pg
2
2
3p
+ x + 5.
2x 2
(a)
30 8.
.
4.
16m
23. 9x 2
31.
Page 23
5.
(Sample answers)
5ax 7.
27(3x+l)
+ 4z + 2x0 + 34. 4a + 6 +
2
2
2
402.
2
- 1). c - d) (a + 6 44. (2 - 30) 47. (a + b - 2) 37. 0(2a0
-
-
+ I) 39. (x 3?y) 41. (20 1)(20 +1). 42. 43. (1 - 2x + 0)(1 + 2x - y). 6 c + d). + + (a 46. (3x - 20 )(9x + 6x + 40 (4 + 60 + 90 48. (8 - x - 20) (8 + x + 20). (a + 2a6 + 6 + 2a + 26 + 4). - x0 + 49. (x + 0)(x + 0-1). 51. (x - 0)(x + 0)(x + x0 + )(x -x + 53. (4a - 56) (4a + 56). 54. (x 52. (x + )(x 56. (k + 1 x + y)(k + 1 + x a + a + 6). 0). 6)(x 2
38. (2x
.
2
.
2
2
2
4
2
2
4
).
).
2
2
2
2
2
2
).
2
2
4
2
2
4
).
ANSWERS - 1) 16xy*(2xy
262
-
57. 25x*y(2xy
+
(2xy
7.
(x
(3x
12. (x
+
-
I)
2)
2.
.
2
8.
.
3)(x
(2x
+
+ 3) -
(5x
27. 31.
34. 38.
46. 49.
EXERCISE 1.
-
(a
7.
2
9.
.
+
+
4.
.
-
(2m
2)(x
2
(a
I)
-
5)
2
6.
.
- 2) - 2). 2
(x
.
11. (x - 3)(x 14. (x - 6)(x 18. (x - 7)(x 22. (x - 12)(x 26. (x - 7)(x 29. (x - 13) 33. (3x - l)(x
2 .
1).
- 2)(x - 1). - 5). 21. (x 6)(x 24. (x - 5)(x + 1). - 1). 28. (x + 13) 32. (2x + l)(x - 2). 36. (3x - 2)(2x + 3). 39. (5x - 3)(2x + 1). 43. (3x - 2)(3x + 1). 47. (9x + 4)(3x - 2). (a;
+ + +
1). 1). 1).
-
3).
+ 1). + 2).
(a;
+ 2)(2x - 3). 41. (3x + 5)(2x - 1). 44. 3(x + 2)(3x - 1). 48. (lOx + 3)(2z + 1). 37. (3x
Page 28
-
+ 6)(x
4)
+ 5)
3. (4x
.
17. (x
(5x - 3)(2x - 1). (5x - 3)(3x - 2). - 3)(3x + 2). (8x (12x - 5)(3x + 2).
42.
2
13. (x
1).
+ 6)(x - 1). - 1). (x + 2)(x - 1). (x + 12) (x (x + 8)(x + 4). (2x + l)(x + 2). (5x + 2)(x - 1).
23.
59.
3).
Page 26
6. 2
16. (x 19.
-
58. 12(a 4 6 5
1).
1).
EXERCISE 1.
+
l)(2xy
y).
2.
(b
-
c)(x
+ y).
3. (x
+ y)(a +
4.
b).
(m
-
n)
7. (x - y)(b + c). 8. (x - 2)(x + 1). (k (x + y)(a + b). 1). 11. (x + y + l)(x 12. (x - y) 9. (x y). l)(z + l)(2x + 3). - 4y - 1). (x + y- 1). 13. (2x + 3y + l)(2x 3y). 14. (3x + 4/)(3x 17. (x + y - 3)(x + y + 3). 16. (2x - y + l)(2x + y - 1). - y - 5). 21. (y + x + i/)(4 + x 19. (x 18. (4 y). y + 5)(x 22. (2x 1 1 + y). 23. (x - 3 - y) 1 y)(2x x)(y + 1 + x). (x-3 + y). 24. (x - a - b + l)(x a + b - 1). 26. (x - x + 1) 27. (x + x - l)(x - x - 1). 28. (x (x -x + l)(x + x + 1). x + 2)(x + x + 2). 29. (x + 3 31. (x + x - 3) x)(x + 3 + x). - x - 3). 32. (x - 2x + 3)(x + 2x + 3). 33. (x + 4 - 2x) (x 4 34. x x 36. (x + y) + l)(2x + + 1). + 2x). (x + (2x x + y). 38. (2x - 2x + 1) 37. (2x (1 2). y + 2)(2x + y 3x + l)(5x + 3x + 1). 41. (3x - 1) 39. (5x (2x + 2x + 1). - 1). 42. (3x - 2)(3x + 2)(x - 2)(x + 2). (3x + l)(x + l)(x 2
6.
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
EXERCISE 1.
2
2
3; 180.
8. 18; 540.
8.
Page SO
2. 6; 72.
3. 9; 108.
9. 13; 3640.
11.
4.1; 252. 6. 4 4 5aV; 30a * .
15; 450. 12.
7.-
16; 480.
3xV; 180xy.
ANSWERS x
13.
+
(x
y;
263 2
x3
xy 17. x
2 .
i/)
-
-
14.
.
-
2; (x
+ y x xy 2)(x + 2)(x
2
3
x
-
16.
.
-
l)(x
x
+ y;
x2 (x 2x
18.
3).
y)
-
- l)(x + 2)(x l)(x 1). - 1). 22. 5x + 21. 3x - 1; (3x - l)(2x + l)(x (x + 1). l)(2x - x) 24. (x (5x + 4)(x l)(3x 2)(2x x; x (2y 3). 23. 2y (x y) (2x
l)(x + 3)(2x +
3x
19.
-
1; (3x
2
1)
4; 2
3
.
4
1;
?/)
;
3
.
2/
REVIEW EXERCISES FOR CHAPTER II. Page 81 - 3). 2. 36 (a + 26). 3. mn(m - n). 4. a6(3a + 56 - 1). 1. 47/ 6. (x 7. (c - 2a)(3a - 6). 8. (2m - n)(3 - x - y*). 3y)(y + 2). 9. (m + c)(x + y). 11. (x 12. (36 - l)(2a + 1). l) (x + 1). 13. (x - 4z/)(x - 2?/ 14. (2? - 3)(3p + 2). 16. (x + 3t/)(x - 3y + 1). 17. (y 18. (2x + l)(x + 2). 19. (x + 7)(x + 3). 4)(y + 3). 21. (x + 4y)(x - 3y). 22. (p + 18)(p - 2). 23. (4a + l)(3a - 1). 26. (3x - l)(4x + 25). 24. (5r + 3s) (5r 27. (10 - x)(2 + x). 28. (x + 2?/) 29. (2c 32. (3x - yz) 31. (2 5x) d) 34. (3 33. (2x x)(3 + x)(9 + x lly)(2x + lly). 37. (x - 2y - z)(x + 2y + z). 36. (3x - y - 3)(3x - y + 3). 39. (I x 38. (x + 2y - 3m + 3n)(x + 2y + 3m 3n). y) 42. (2x - a - b + y) 41. (3 - a + 46) (3 + a - 46). (1 + x + y). - a)(p + + a). - a + 6 - y). 44. (x + 2) 43. (p + (2x 2
2
2
(i/
2
2
2
2
).
2
2
).
2
4
2
2
2
.
.
.
2
).
ff
+ 4)(ay-4ay+16). 47. (x + a )(x -a x + a - Say - 66y + a + 49. (3y + a + 26) 48. (2x 3)(4x + 3). 51. (x - 4x + 8)(x + 4x + 8). 52. (x + 2?/ - 2xy) 4a6 + 46 - 4 - 4x) 53. + 2x7/). (x (x + xy + )(x + xy + ?/). 54. (x 57. r*(2y 4 + 4x). 56. (a 26 86 )(a (x 5)(2y + 5). - 2) (y + 2) 61. (a + 2) 59. (x - a)(x + a)(x + a 58. - 2) (a + 1). 62. (y - a)y + a) 63. (m - 2)(n + l)(n - 1). (a 66. (2x - 3y - 6)(2x + 3y - 2). 64. (x + 3m) (x 3mx + 9m 2 - 6r m 69. (1 - 6)(1 + 6) 68. 8rm(rx 67. a(3x + y)(x y). 73. (x - a - 2) 71. (pq - l)(pg + 1). 72. (2x + 7c) (x + y). 74. (2x + l)(x (x + ax + 2x + a + 4a + 4). I) (x
2
-2x + 4).
2
46. (ay
4
2
2
2
4
)
2
2
2
(9?/
2
2
2
2
).
2
2
2
2
2?/
2
2
2/
2
2
2
2
2
).
2
i/
2
2
2
2
.
(7/
).
2
.
2
2
).
2
2
).
2
.
2
2
2
.
EXERCISE 1.
(a)
9.
1; 3.
Page 37 (6)
3; 1.
(c)
2; 2.
Canceling not permissible in problems
2.
In case
4, 5, 7, 9, 10,
a 3.
(6).
and
12.
6.
6
+
1
x
-
1
+ xy +
x2
EXERCISE -1.
1.
2
10.
-
a
?/
Page 40
2. 1.
-1.
3.
4. 1.
-2
6.
-
2x
EXERCISE 5
1
i
11 11.
7 A
lj.
11.
2
2x 30
18.
20
3x
5x
9*
x
5x 2
+ x
2x 2 29.
2
2x
-
/i
14x 12 15. T~r
7 ~*
8x 33.
7'
6JL
v>
-
^o*
3x+
ig
3
+
21x
1A ~' ITT.
Q * 41
6x
3
40
-
12
-
-
5x
+
20x
+ 31x - 43 + x - 12 - 7x + 12
14x 2 31.
'
2x 2
+
9 21.
34.
3x
7
6x
24.
-
6x
-
2
6x
1
14x
+
28x
6x
36.
37.
x3
3x
+
3x 2
-
+ 4x + 3 + 3x - 41
x2 2
6x 2
-
13x
4x?y
+
2
8i/
y
lOx
6x
+
10
-
47.
-
15x 49.
26x
-
15
12x
-
8x2
5
49x?/ ~
9x
-
-
-
+ 9x*
+3
2
32.
*n 23x~ ~ oV. 6x 2
48.
2x 2
- 19 + 5x - 6 x - 13x
2
:
- 1 - 18x - 28
41.
-
5x 2
28.
2
3
15
6x
4x 2
2
'TB'*
18
5
-
+
47 '
7x
+
7x 27.
2x
x
l
-
2Qx
16
60
23.
15
Q v
TTT'
6x 2
9
44.
1
+
x
4x
+
-1.
9.
tx
^ 1 ISlTTf'
12
3
6x 2 2
1
0.
L.
+r
5x 26.
-1.
8.
-
8
10
-
2
1X
4x
8
-
420 * Tpjf-
9
6-x
.
17.
22.
3
^o-
-
4s ~
19
-
9x
1
^J
l^S.
-x
-1.
Page 43
i4 i^.
g T~
+
-
x
7.
5
6
7x
1.
-5
8x
10
12.
-
26
15x
19
'
ANSWERS EXERCISE 2x
-
x 6x
2
16.
+
12.
-
9.
17.
21.
2x
-
.
-
4x
-3.
;.
4
17x 2
+ 9.
39x
2x
18.
-
24. 4.
26. 3.
3x-
28.
+
-
3x
1
29.
2
-
6x 2 33.
x
12a.
13x
- x - 21 + 2x-35'
_
10x 2
_.
34
-9
2
'
x
2
36
2.
3.
+
5x
4.
+
7x
3
' -
6.
+
2x
10
3
-
12x 2 8.
-
2
2x
x
)'
-
3
36x
-
-6 i.
19>
26.
9x
31.
x
6x
1A
+
2
6
+ -
(y
6x 2x
1.2-
-
xy xy
-1. 2x
4y
9.
x
+
-
-
2)
-1.
-
-
9x
2
+2 +4
7x
-
-
43x
12
+
12*
-31x
10x2
2
x
10 '
2
+4 + 2x +
3x 2
2 23. x2
-2.
-
+ 2x -
2x
3?/
24.
.
33.
x)(x
1
-
1
t 3y
-
29. 2.
1).
+
3x)(x
+ 3x)(y y-1 T/
-
28. 7(x 2
-
(y
'
+
. J.
1
-
1
+
2x y 4x - 2tf'
2*
x
-
1
x
+
1
-
;
(3x
+ y.
-
4.
2y)(x
(x~y)(x
-
(^
+ -
!/)!
y)
.
-
1
'
(x 19.
llx
3x
+ 4.
-
(x
(2x
2)
a2
+ xy + y x2
y 21. 2y
2x
-
x
y '
+
lfi
6.
x
2)
2
'
18.
x x2
Page
14.
2x
-3
x
22.
2~x
12x
2)
13. ~'
-
2
(y
+
13.
6x
lly
+
3
2
i
2L
1
2y)(x
^5.
6
10
1
32.
+
x)(x
-
5x
-
27. 8x 2
EXERCISE
13 '
x
(3x
8.
2
+ y-te + 2. 2x - 3
fl*
-
3x
2
5-
x '
Page 46
4
-
5x
2x
15
7.
x
-
7 31.
1
-
13x 2
19.
23. 2x.
30
3
EXERCISE 3x
-
14.
7
22. lOx.
5x
1.
-
6x
14.
- 1 - 3x
f
9.
1
+
4x
13.
3.
A-
7.
2
-2x -
9x
+
M
-A15x -
<*
3
3x
45x2 15x 2x2 - 13x
32.
-
4.
3. f.
2x 2
21x2
27.
Page 45
*.
2.
1. 5.
1L
265 12.
'
*2
+x
2
-
+ -
2
-1.
-
+
y)
)(*-2y)
3y 17 17.
-
3x
10
i/
2xy
-
y)(2s
2
2x2/
7. 1.
'
y2
+y
2
22.
x x
-
1
-4
ANSWERS
266 23>
24>
'
+
x
+ 4x + 1 - 2x)(z + 1)'
6x2 (1
-
2x 2
i
2Q
x
26>
-
27 *
7+T
1'
r^T
+3 2x + l' x
'
REVIEW EXERCISES FOR CHAPTER III. Page 51 3x - 2y x + 4xy + 4y2 - xz - 2yz + z2 4a - 2b 2 2 2 9x + fay + 4y a 36 + 2y - z 2 7a - 15y , 4x 5 3x + 23 3n + c 4xy + y 2
.
'
'
'
'
'
a;
2
'
'
'
3n -c'
2z
-
'
24
-
4xy
g
'
+y -
12xj/*
2
3y
12y.
- xy + x - (x +
5x*
. '
'
2
(x
17 Z'
+6
a
-
2
2x
, 3'
- c)(6 - c) 3a - Qab - 62 3a - 6 2
z(x
+ 3a6 + 26 a + 46* (x + l)(x -4) 2
a2
33 ' ,_ 37
'
+ xy -
x)(3
-x+ y + 56 - 3 + 26 - l'
5a
~
(x
x2 )
'
l
2~'
-
6a6
..
36
2a6
3a
6
-
a
+ 3)(s + 5) (x + 2)
-
,
(6
m
3
a
_
c )[( a
mn
2
-
2n
-
2
c)
MR ,
-4.
26.
^.
34.
-9.
42. 6 P.M.
5) (2a
2
+ 4)
2
+
. '
_
(a
+ by
3
'
3
a2
a6
]'
+6 + 7)
2
2
2
'
2
-
62
.
Page 59
14.
in problems 3, 5, 7, 8.
6. 2.
4. 2.
.
a6
6M + 2 + 2) 4o6 + a - 26
3M '
'
a
The equations 2.
4
1)
3
n3
EXERCISE
+
+
(u
2
6
(a
2
5) (a 2
(a
'
2ac
_
+ 2xy ^
-3a-a a2
'
1 '
4x2
,2
'
-y - ab
x
l
'
'
-
(2o '
6
+y
x
,.
'
-
?R 28
x
Z4
2
>
xy x2
l'
+ 2)
8) (a
2
__ 2a
711 *
-
36
+
2x
_,_
27
^Ti
-
'
12
8
2
(y
1O iy<
-
3
&
y)'
9 , (a 23>
4
2
-
j/)
.,
12x2
+ y)
2(
18<
'
+2
8x 3 - *
(a
2/)
-
-1.
9 and
12. 5.
11 are identities. 13.
~ O
18.
27.
H-
^O
36. 3.
19-
28. f.
37. 4.
21. f. 29.
-1.
38. 4; 8.
22. 31.
.
-1.
39. 9; 3.
1. 5.
14. -^.
16.
-4.
24.
23.
32. -ft.
-1
/.
33. f.
41. $500; $1000.
ANSWERS
267
EXERCISE
No
1.
-
- fj. - T8T
8.
No root. 13 No root.
2. 1.
root.
n
-V-
.
Page 61
15.
9
12 -
-
-
19.
!
21.
f.
29.
1.
EXERCISE
16.
Page 65
CXERCIE
17.
Page 67
15.
2. 23.
.
1.
11.
12.
1. ;
c
in.; in. in.
10;
-2
L
5
12
-V
ft.
6.
in.
8.
4
in.;
12.50; 100.
4. 6;
13.
f
V
16.
T
23. 2.
6. 9;
9.
24.
32. J.
.
- fV
7.
17 26. 10.
-J.
33.
12.
.
No
root.
7. 3.
8. 5.
f
Pa0c 70
18. 4-
ft.;
12.
3.
-
31.
-^.
12. 17.
.
CXERCISE
-
-2.
22.
1.
6. 4.
-
14.
-
28.
7.
4. 2.
3.
8
40.
in.
2.
10
in.
X
3.
ft.
ft.;
10
15
in.;
9. 10 in.
13.30; 60;
X
6
in.
in.;
90.
7
6
10
in.;
X
10
in.
in.
X
6
14.40;
4.
6
in.;
7. 3 in.;
in. in.
100.
11. 1 in.;
16.20;
ANSWERS
268
EXERCISE
Page 72
19.
15 hrs. after 12 o'clock.
2.
after 1st car starts.
20 mph.
13.
EXERCISE
12. $4.00;
11. $9.00 per day.
5%. $6; 4
$8.
$90.
22. $1000.
24. 12 nks.; 24
30
1.
@
8. 21 Ibs.
11. 7 cases
EXERCISE 4.8
1.
6.
I
17. 3
12.
Ibs.
6
6T
5: (5a
+
8.
(a
2
-
12. 4. 19.
-
units.
31.
4%; $4000
19. 2 yrs.
[email protected]
14.
$16.00.
21. $600;
26. 2 qtrs.; 4 dms.; 8 iiks.
4.
oz
5. .1^5
oz>
9. 62.5 Ibs.
28
7.
220; 37.5
Ibs.
Ibs.
300.
$3.
<$
2-
7. Ibs.
93 1
2.
boy.
%
ft.
13.
550
640
3.
Ibs.
from end. Ibs.
8.
28
14.
4.
Ibs.
42
Ibs.;
-5
^
9.
Ibs.
42
Ibs.
333^
Ibs.
16. 10
Ibs.
ft.
Page 85
20 hrs.
7.
A: 48 days; B: 192 days; C: 576 days.
3.
9 min.
-~
8.
'
+
6.
2a 2
-
4a6
+
A
5a
days.
9.
3a
-
7.
+
3a
+ 36-2.
1.
17. 25.
+
5
days;
:
a
24.
-2.
3a
-
3.
21.
f.
27.
-
A =
8&.
-3.
5A
.
22. sq.
12.
2a
-6.
14.
f
4. 9.
4).
13. 15.
#=
75 min.
2.
5) days.
2.
3.
23.
6.
EXERCISE 1.
Ib.
396f
days.
4 min.
4.
$5000
9.
ft.
EXERCISE 1.
$5000 @ 5%;
7.
Page 83
from fulcrum.
ft.
500
11.
22.
2 yrs.
18.6%. dms.
.
250.
$2.50; 8 cases
from 90
ft.
581 oz
3.
150; 9 Ibs.
<&
mph.
525 mph.
Page 79
2. 30.
Ibs.
12. 10
18.
28. 6 nks.; 4 qtrs.; 10 dms.
27. 5 nks.; 5 dms.; 10 qtrs.
21.
6 hrs.
7.
*&*
13. $8.00;
$8.00.
17. $517.50.
16. $39.
EXERCISE
6.
4. -V%. 3. 4%. 4%. 8. $5000 @ 4%; $3000 5%.
$2500 @ 6%.
6.
Page 76
20. 2.
$30.
1.
^
ff
4.
f mph. 9. 6 mi.; 3 hrs.; 2 hrs. 16. 2 mph. 17. 100 mph. mph.
8.
-3
14.
ff
3.
A
+
2
15.
- i
in.
28.
-
-^.
23.
A =
V
T
2fe
2
16.
24.
36 sq.
ft.
2.
-Vtf-.
29.
(a
26.
A =
2
+
-
3a
4).
11. 5. 18.
A = s2
sq.
-
Trr
.
2
sq.
units.
ANSWERS
269
EXERCISE 19.
=
y
-
x
=
y
, *
~
6-g
x
x;
f(7
+
L
(6) r
2.
(-f, - ).
8.
21.
19. (4, 3).
26.
(3, 6).
/in 49.
(6, 0).
3.
.
(-3, -1).
(i
29.
27 y Z7
=
T/
-
4
-
'
+
4.
(-2,
6).
17.
-y-, 4).
23.
41. (V-, f).
42.
6
+ ^a /
i-o
52. j-
i
solutions (inconsistent)
:
5).
(i
!).
9(7
1.
6.
(1,
8
-T T, 14.
2,
-3,
(],
29.
(-7,
-3).
43. 48.
a2 ~~ ^ 2
(*, -*) eo
53.
j-
and
55, 58, 59,
11. (^,
5.
1.
9.
30.
2;1.
(fl,
f,
i
-I,
-16.
3.
32.
4.
14. 57.
More
61.
66. (1,3).
H). 12.
&)
(3,
3,
4.
4).
(1,
(i, -,V, -A).
8.
(A, if,
i
)-
-1, 9.
0).
(tf,
13. (3, , 2).
10.
6.
-1.
16. 43.
7. 31.
17.
-12.
8.
18.
18. 0.
9. 0.
19. 0.
11.
13.
21. 0.
Page 117
2. 3; 5.
2 mph.; 3 mph.
14. 25.
7 J).
Page 112
13. 5.
EXERCISE
/4 ( -,
16. (f, f, 18).
2. 8.
-11.
3.
-1, -2).
(1,
7.
^-).
EXERCISE 1.
2.
3).
(1, 3, 4).
12.
2).
(-ff, -*).
&,
^
(-2,
24. (f, f).
5).
f).
-V-). 7
-f).
Page 108
--
3
360.
69. (, 2).
68. ( , ).
T r)-
+
12. (2,
than one solution (dependent): 56, 57, 60, 62, and 63.
EXERCISE
i^& Z
6. (f,
18.
(2, 0).
-
('/-,
(-^,
47.
-if).
6 a
(-3, 11.
-3).
,
No
67. ( , -f).
=
2z; x
+ 40) =
5(F
(c)
(2, 3).
(-
22.
(-&,
a bl
6
Z
16. (2,
1).
46.
-- ,7-)-
(
= x + 1; = 6 - 3x; ~ ~ 6 3a;
y
24. y '
IL=_1
9.
^
54
22.
y.
t
(2, 3).
Cl /a 51.
t
/i,
L
.
~
3?y
C = f(F - 32); = (c) ^.
(6)
1
+ 2).
(j/
~
x X
-
+ 4; x =
27. (-7, 3).
-M)-
44. (f ,
6
=
=
x
x;
z '
14. (3, 2).
-4).
1
2;
+
2*
~
2x
32;
-
2x
Page 105
28.
(2, 3).
13. (1,
y
=
2g y
EXERCISE 1.
=
26 *
P=
=
21. y
y.
23.
= 6_^2y o 31. (a) F =
7.
=
1.
x
32. (a)
Page 96
26.
16. 347.
3. 7; 5.
11.
4. 8; 6.
3 mph.; 4 mph. 12. 150 mph.; 50 mph.
6. 4; 5.
335 mph.; 15 mph.
17. 397.
18.
9
in.
X
5
in.
8.
19. 12 in.
X
7
in.
ANSWERS
270 21. 10
X
ft.
4
@ 5%; $4000 @ 6%.
24. $6000
$10; B: $6. 31. 70 Ibs.
5%; 30
33.
(/) x, 3.
/5
in.
(/) 2/13
in.
(a)
2.
1.
17
-16.
10.
-8.
21.
-4.
4. 32.
(d)
*
~
^)
+
23. 9.
22. 9.
32. 2.
41.
-
42.
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V
33.
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7/
a'
y
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a
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6
+ 6)
(a
81.
1.
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(a
3.2'
+ *' + a + x _
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52 52.
^2 /.
'
-
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-
93
r2
a
(*)
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(e)
13
in.
~
a2
^ x^2 }
2
2
x*
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x2
53 53.
-
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36.
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44.
-.
46.
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1.
J.
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2
76
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(2x 83. (x2
X2 )i
)
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2
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a;
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9/7 2 /) 2
i
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1 '
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l
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18. 8.
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x,2 _(_
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2
9. 25.
27.
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y
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+ 2a6 - a 6) (o + 6)
_
a
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1
10 2.
X
2
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fft
97. 5
in.
26. 4.
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-
2x
82.
1.
'
2
-.
8. 16.
34.
T
43.
2
_
f;
58.
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(a -
1.
17. 16.
-27.
24.
^-
51 51.
3
2
79.
1.
Ibs.
(0).
16. 3.
1
02
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(e)
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7.
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'
78.
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f
and
(c), (e),
125.
6.
14.
49 -?49.
57.
'
72
(h)
in.
In cases
13. 4.
y
5,.
(d) 4.
fo)
5
(c)
4.
in.
^T
39.
48 I48.
.
(d)
31. .
29. 3.
38. f.
8 47 47. ~ 1?
-3.
in.
3. 27.
12.
.
33.
35.
-9.
37.
^.
@
Ibs.
- .
;
(c)
2.
/13
(6)
11.
28. .
-2.
3.'(c)
(a)
2. 2.
2.
6
44. (-} yl
.
2;
(6)
(g)
EXERCISE
@ -%5^;
34. 20 Ibs. meal; 30 Ibs. flour.
35?f.
-1.
1;
29. 6 qtrs.; 10 dms.; 20 nks.
Page 127
34.
-x.
5%; $4000 @ 6%. 27. A: 4%; $4000 @ 6%.
$3000
32. 4 Ibs.
15%.
29. (2a) 4
.
EXERCISE (a)
23. $2000
in.
Page 124
x
26. (abW)
Ibs.
@
Ibs.
^XERCISE
1.
9 26.
28. 10 dms.; 20 iiks.
-Mp
20^;
X
22. 16 in.
ft.
99.
x2
1
2a2
x2
ANSWERS
271
EXERCISE 1.
36.
2. 4.
3.
11. 3A/2.
4. 2.
3. 6.
12.
3V3. _
19.
-4^2.
-
18.
Page 134
13. 3/5.
26.
27. 5z 2 V7.
32.
33. 6z 8 /
38. 2xv/3.
39.
44.
46.
57. 4 VlO.
78.
2Vl 2V4 -
83.
1.
73.
a
79.
.
84.
31. 6z3 V2. 37.
43.
42.
49.
Vt/2
4^2.
61.
68.14.
69. 4x.
Vl + 3x2 5Vl + 4x2 _ x^ 86.
56. 9.
54.
53.
74. 2
.
2
3/ia
24.
48.
59. 9V/6.
2V?.
9.
17.
5zV5.
36.
34. 41.
6^5. 2x 2
23.
29.
47.
58. 10.
66.10.
8. 2/5. 16. 6/2.
22. 8ar/2&.
28.
S2.
51.
2V.
SV^
14.
4V6.
21.
7.
6. 5.
62. 10.
63.
6^4.
71. 3x/x.
64.
J3.
72. 3x/x*.
76.
.
81.
.
V3.
xy
^6.
91.
+
98. (3a 103. 1
108.
V5.
92.
99. 3
26)/3a.
-V2.
104.
V3-V5.
122. x
+
-
1
^m
EXERCISE 1.
2^
=
+V2 'I'
114. 59.
10 VlO
x
or
=-
-
60.
1
- /2.
-4 +
107.
V5 - /2. -12.
112.
2/6.
121. x
9V^.
102. 13/2.
-3V2.
111.
116.
97.
96. /2.
101. -1 - V2. 106. 2 - Vs.
-V7+V2.
50
= ~~
+/2. 2V5x.
109.
+ 20Vz~. 18 V2 + 6 V5 + 3 V6 +
113. 2z
94. 3/3.
93. /3.
117.
=
or
10V3
-
=
2.
x
P
37.
Pape
0.707.
2.
^5(9
~=
0.577.
2V5
3.
^=
1.224.
4.
2
o A
^,
V30 =
~=
0.866.
2
Q 913
g
V6 =
Q 612
4
=
1.661.
14.
-~^ 5 -
22.
~= 2 27.
= 15
0.908.
11.
V3 >
16.
^32.
23.
^5.
V5 > v'll.
12.
17.
23 24.
18.
_v/243.
-^-
26.
1.
a/3a 33.
32.
EXERCISE 4. 6t. 13.
-3i.
39.
6. 3i.
14.
27.
^
34.
v'3
13.
^108.
^3.
19.
^5 > ^2.
14.
^^2. v/8.
22.
21. /6.
29.
28.
31.
1.
v^.
x
x>^2a2x 2a
i/v^^x 3(J
37.
Page 7. 7t.
-tvTl.
8. fv/47. 16. 3io.
9. 8t.
17. 4tx.
18.
11. 12i.
-5ij/
2 .
^2. 19.
-2i.
NSWERS
+
3
>.
44
i*'. o
>.
38. 3
3i.
59.
V2.
67.
+ 9i. + IQf
61.
^-
_.
'
3.
40.
XERCISE
53.
-6.
62.
5.
-
r
1,
A -
r.
s,
3, 1
~
12
-
Hi
-
3
'
lit
10
~
54.
-10+
KM.
63. 61.
i-=^-
71.
56.
t.
57.
o
69.
-
3l
-
17
o
151
7
68.
and
a, 6,
3, 2, -1. 4, 4, -3.
3
5iV3.
64.
-
-
1
i.
O
(6)
No.
No.
(c)
c are listed in that order.) 6.
-B-
C,
1, 2,
7. 4, 5,
1.
13.
-of.
1, 0,
23. a 2 ,
-rs.
-3.
12. 4, 8, 3.
18.
17. 2, 3, 0.
3, 2, 0.
).
42
Page 152
41.
11.
36. 61.
Page 148
(The values of 4. 2, 1, -6. 1, 3, 2.
-2i.
28.
7i.
10
Unlimited number.
(o)
+
^-^'
+
3
7'
'
lo
XERCISE
34. 9
41.
12i.
27. 2i.
i.
5
10
iV2.
26.
33. 41.
-
39. 16
52. -=^'. Z
f.
-1.
24.
32. 13.
4i.
'13
51.
*.-3i.
>.
15
-t.
23.
-3 -
31.
f
'13
-1.
22.
4t.
-4 + 6 - 4
r.
i
273
-aiV23.
L.
-9e 4
19. 4, 0,
0,
-1662
28.
A -
-
26. 2
.
A +
B,
C.
B, 29.
1.
-3.
-3,
0.
-e2
21. 4d*, 0,
-A -
C,
4,
8. 2, 1,
.
B-
C,
4,
14. 1,
0.
-2,
1,
2.
-6.
.
A + 4. A - C,
D,B.
XERCISE 1.
-2.
2;
2t;
-2x.
Page 155
2. 2; 9. 2i;
-2.
19.
16.
-2.
4. 3;
-7i.
11. 7t;
-3.
a
bo
3aiV2; -3a;/2.
' I0bl
21.
24.
iu
4^.
22. 46f
.
;
-3^/2.
9/x
^;
17.
-3i.
7. 3t;
13. 3v/2;
5i.
9/^
4./9
^-^;
-5.
6. 5;
12. 5i;
^. a
Av/9
2V3; -2V/3. -
3. 2;
-2i.
^-
,
I.
42.
a
v^ ;
25.
-
;
a .
a 1.4; r.
I.
L.
1;
a
a
-1.
32.3;
- .
38.
0;
i
;
- .
44. 0;
;
i
52.
-2.
i
33.5; 39.
46. 0;
- ;
.
-
t,
a
-2.
34. ;
-1.
-^;
-i
41. 4a; 4a.
^rIU
47. 0;
-
53. $;
- .
a
O.
48. 0; -
54. ;
36.
42.
-J; 1. -f; -f.
49.
;
c
-f.
56.
-f
;
-f
.
-f.
ANSWERS
274 57. ;
i
63. 20;
-f
58.
-8.
f
59. ;
-f.
;
61. 4;
-f.
64.^;-^- 66.^;-^-
^
-
62.
^p;
67.
f
68. 0;
-5;
2.
7.
A
.
69. J*L.
EXERCISE 1.4;
-3
14 10 19.
' O
^
H-
.
44.
-2.
4. 5;
2.
4 7-
4>
12 x ^'
-2v 19
6.
&
12
-3
11 1<5>
4-
Vl9
f
-^;
v7
1
o
-1
+V?
>
-3 +
21.
26.
.
11 '
3
tSi
24.
29 3
-4;
3.
/
2V?
7
Page
-3; -2.
2.
2.
2-1 f> 2-
a
<*
43.
i
-, 22.
-5
iV23
3
l2i.
-3
i
28. 2
t
23.
27.
V7.
i
33.
3 6
EXERCISE
44.
Page 163
a._ L
.
7 7.
.
.
26. 1;
-f
33.-
32. J; .
V6 +
4oc
,
23.
1;f
28. 1;
27.
29. J;
-J.
34.1;
A-
-
J
26
-&
V62 -3ac
16ac
.-
.
3
'
3a
-1
Vl7
4a
Vi^ -
-J.
36. J;
VC -
-c
5
2^6.
17. 5
16.
14.
13.
6
33 V2.
.
3V5
-1VJ3
24. }; .
6
2ag.
2a
4a
.
2 2
c
ANSWERS
Vl -4g x2 2
1
A
275
20
EXERCISE 46. Page 1. x - 3x + 2 = 0. 2. x* - 2x - 3 = 0. 2 = 4. 2x 6. 8x - lOx + 3 = 0. 0. 8. 5x - 2x = 0. 9. 5x - 2x - 3 = 0. 12. 2x - 4x - 1 = 0. 13. 2x - 2x - 1 = = 16. x 4x + 13 17. x - 4x + 7 = 0. 0. 2
a:
2
2
-8, -5;
1.
8
6.
47.
11.
X
in.
+
(8
5
-3.
16. 8;
V2; x
-
16.
x
3
+x
19.
x3
13.
16x3
+
24. x
3x
-
4
6x
2
2x 2
13x 2
-
+1=
6x
3
0.
EXERCISE
X
2.
+
8.
ix/2;
_!
4.
50
ft.
X
ft.
15
9.
ft.
in.
-i
13.
in.
1; 1;
13x
28. x
49.
120
X
14.
4. 0; 1; 1;
t;<;-t;-t.
0.
0. 0.
ft.
8
in.
-
*
-1.
8. 0; 1; i;
-i.
-1; -1.
9. 2;
12.
^ 6
= 0. 3x + = 0. + 6x - 1 = 2
3.
i.
-tV5.
-
llx
=
4
x
14.
0.
x
17.
4
21. 8x 3
-
=
0.
20x2
+
1
=
+ 4 0. - 2x + x = 0.
+
18.
46
4x 3
x
2
+x
-
- 5 = 0. 6x - 7x -
2x
26. x
4
3
=
3
x2
27.
0.
2
-
6.,i.
lVS; -1;
11
;iV2. 7.
9.
3.
-2;
-3;
1
i
i
x
= =
0. 0.
-
+ 2x = 0. x + 3x + 2
3
2
3
6
22. 4x 4
18x
23.
0.
12x
3
Page 178
;. 2;
200
+ 2V2
2
2/5)
2.
4.
X
ft.
8.
in.
1;
6x
+
+ +5
0.
Page 176
7.
+
-
2
8x
0.
2V6.
-2.
3
-
1
3
*
- V2;
150
3.
3
12. (4
.
6. 0; 0; 3;
12.
in.
in.
48.
l
11.
6
7.
-1;
1.
2.
5, 8.
17.
EXERCISE
7
= = = =
Page 171
in.
2/17)
18.
2
4x2
14. x 2
0.
2
EXERCISE
2
11.
2
2
6x - x x - 6x + 2
7.
2
+ 4x + 3 =
x2
3.
2
;
ANSWERS
276 11. 2;
*--
-J; -1
iVS;
-1
iv^ -3
, i 13. 1; 6;
-1
-
12. J;
- Stx/S
. 14.
;
V5
V7. -1
zlI - --
,V;
1
-3
3
-1; 7
+V7.
1
.
,
-2;
- ZiVZ
;
-iv/3
1
;
tV3
1
V5
1
2 .
-
-A .
l,
.
24.
2. 20.
14.
1.
50. Pojfe
9. .
-a; -6.
17.
EXERCISE 3z
3
3x2
9.
12. 14.
51.
3X3
16.
3
4
-2.
8.
-1; -2.
-
-
4x 3
+ x* + x -
1.
~
1
x+
3
13.
2.
-1 16. i;
-l;- ;-2.
V
(4, 3);
-
34x
18. Yes.
+
141
-
19. Yes.
17 -
-1
4
V3.
22.
No
54.
;
;
4.
x+4
21. No.
.'-
2;~-
9.
-1;
^
EXERCISE
7x 2
1
564
1;
13. i;
2
-9;
3.
l;l;-2.
~
~^;
3.
Page 187
52.
8.
3.
2
1
17. Yes.
16. Yes.
2x.
2
2
3
1; .
1.
root.
14. 1.
2
_1
21. 1;
4
Page 184
1.
1;3.
12. 1;
No
7.
13. 2.
1.
3
EXERCISE
7.
Li
3. 2x - 3z + 2x 2. 5z - 2x + x - 2. + x + 5. - 2z + 7. 6. x 3x 8. 8. 2x ^ 11. 3x + 3z + x + 1 + + 4x + 14 + -^-x x - 3
4x 4
1.
{
-3.
3z2
1.
4.
12.
-1.
6.
4. 1;2.
3. 0.
18. 1;
1
180
-3.
11. 3;
.
-2,
V2;
1
i
^
EXERCISE
!
^1. l,
^-r-5
14.
-
11.
-1;3;
No
rational roots.
o
i
18 -
L
19 - 2 2 ;
;
~2 -1 ;
i
rational roots.
Page 194
(-4, -3).
2.
3.
(5, 0); (0, 5).
-
i
(No
real intersection8 -)
ANSWERS
277
(-tVH,
(iVll, 6);
11.
13. (1, 1).
14. (1
18.
17. (1, 1).
(I
5
/V58
7. (2,
2
1.
V
23. 10 rods
X
2 rods.
2
'
8
/-V58 V42
/V58 -VJ2
/'
'
2
-4);
(-5,
'
'
2
V
'
2
2
2 '
/'
'
2
V
I
'
'
2
.(-^ ^);(^,- V2).
(f f);
9.
i
)
2
(3,
-4); (-4,
3); (5, 3);
2. (0,0);(4,4);
3).
(-5, -3).
(-1, J);(-l,
(-^,
4. (5, 3); (5, 3);
J).
-f).
(
)
(-i
!
(-1,
I)!
(i
8. (1,2);
(-1, -2); (1,2);
11. (0, 0)
)
;
(0, 0)
13-
(-!, -I)-
);
(-i
;
(0, 0)
(-2,
-f); ( ,_-1); (- , (-JV10, VlO).
is 18.
EXERCISE
-1.
9. a.
16. (2, 2);
21. ($,
57.
,
2).
(0, 0).
-6);
(0,
,
-f
17.
(M);
)
;
(f 1) ,
-2); (-6,
16. (1, 2);
I).
--,
12. (f
;
10).
(-1, -2);
(-1, -1);
(1, 1);
-
-,
Page 203
11. 2.
(-2, -2).
-2,
;
9. (1, 1); (1, 1);
-8).
(11,
6); (4,
;
8.
).
-2).
21. (a, 6);
-J).
(2,
3).
Page 200
56.
(3, 4); (3,
-f);
-a).
(6,
-2&);
2
3
-1); (-2,1).
(-*,-*)
14.
/'
(
3. (5, 3);
19. (1,
-J).
(- ,
4);
16. (4, 2); (4,
-2t).
t,
/-V58 -V42
/42
EXERCISE
-
(1
(-5,
Page 197
(-550). 0^
2 (5.U), (5 0V 2.
12. (2,
in.
55.
'
2
V
2);
i,
); (1,
( ,
EXERCISE
(I,
(-1,
+
22.
X
8. (5, 3);
(-5, -3).
(3,iV7);(3, -iV?).
2aj-
24. 12 in.
6.
7. (5, 3);
6).
9. (4,0).
12.
17.
22. (2, 4, 6).
-9; 1. (i 0).
13. 18.
f -$. ;
(-4,
0).
14.
-^; 19.
-f. (i 2).
ANSWERS
278
EXERCISE
+c_b+d
q
i
a
d
__
c c
a
d
b __ c
b
+a_
d
a
d
a
t
19.
ft.
-
'a
+
6'~c
+
c
6
-
6
+
b
6'~c
d'
d
c,
a
b
16.
_
17.
ft.
18f
d
_
;
d a
c
^.
c
12f
ft.
Page 213 /
varies directly as v?/.
kx
=
kxy]
d
c
?/
2. z
__
_
'
a
b
a
c
+ a '~__ d + c.
mis.
V-
EXERCISE 59. 1. y = kx?] ~ =
c
b
2
d
6
c
c
6 a ~ 6-d'a +
a~c
;
d
+
_
C
-
+
6
c
24
18.
d
-
C
+
t
d
6
+c_6+d
d a 'a
c __ b
c
'
a
a
t
a
t
'
d
c
6
Page 205
58.
x varies directly as
k
z
and inversely as
y.
xy
w =
3.
y
X
x^yz
Z
x2
8.
3
=
~^~ = ;
-- =
;
ft.
Moon:
t(;,
y, /~~
x varies directly as
k:
Vy
and
2
3
(or as .
and inversely as zVw.
iV3;
^;
(c)
11.
(d) 24.
in.
2 v'5
;
x varies directly as wy* and inversely as
fc;
xz
^- sq.
sq. in.;
16.
=
-
^2
i
y
= w
14. 2
i
Vt^ and inversely as
tr^ifiiD
n*?y
---;
9. (a) f ; (6)
80
^/~'
,!
varies directly as
x varies directly as
fc;
x
z*w 7/
,.
,
/c; a;
ID
yz 7.
=
;
w =
6.
x varies directly as /wz and inversely as
k
XfJ
= kw
.
4.
=
-;
Z
-^
_
12._9 pts.
ft.
2 ^10
;
ft.
2 ^20
;
20
sq. in.;
13. ft.
2 ^60
;
3
re
+
60.
3r4/
+
240a?V 3 .
6.
*x
19. (a) It is decreased to
Mass
22.
ft.
EXERCISE 1.
in.;
115.74_lbs.; 154.05jbs. ft.
4 v/15
;
ft.
16.46 Ibs.; Mercury: 26.3 Ibs.; Venus: 86.1 Ibs.; Mars: 43.1 Ibs.;
S is decreased by f y
21. 168 cu.
*& sq.
sq. in.;
Jupiter: 252.8 Ibs.; Saturn: 107.5 Ibs.; Sun: 2777.5 Ibs. 18.
z.
of Jupiter
=
17. 199.5 Ibs.
11 units;
(6)
96 V2
in.
64x 8
-
302.5 (mass of earth).
Page 219 3x?/
-
+
160z
16x2
V-
+
2. x*
y 3
3 ?/
+
60x
2
4 t/
- 3x*y + 3xy* - 12xy + y 4.
3 t/
5
6
.
27a
32xyVx + 24xy* + 8y*Vx + y 8. + ^x^ - TV2/ +
fW
4
B
3.
.
6
+ 7.
a8
4
27a 6
+
ANSWERS
279
~ - 20x% + ISOxV - 960xV +
x6
13.
H----
-70.
16.
.
x8 18. 3432.
.
_
_
22
-792a26 17
17.
a
a2
a
-
14.
-.
W
23.1-2* +
a2
-
x~3
19.
.
_
3x~*y
+
+
*
*4
3
- 2ac + + + + + 2ad 26c 26d + 2cd. 2ac - 26c. 27. a + & + c + 2o6 28. a + 6 + c - 2a6 - 2ac + 26c. 29. a + 3a 6 + Safe + 6 + 3a c + 6a6c + 36 c + Sac + 36c + c
+
24. 1
+
2x
-
8x 3 H----
4x2
-
2
c2
26. a2
.
2
2
2
d2
2
2
2
3
2
2
2
2
2a6
2
3
3
.
EXERCISE 1.
61.
-5.
2.
11.
Pa0e 224
8. 8, 6, 4, 2, 0.
200 mis.
16.
6
17.
EXERCISE 1.
8.
48.
147;
7.
62.
4
Y*-
3.
-6.
7.
6. 7, 11.
12. 115.
11. 4.
25 mis.
18. $20.
-4, -10, -16. 14. 21.
13. 8.
Page 226
2. 185.
3.
9. 31;
-3.6.
-88.
4. 114.
6.
18.
17. 21.
18;
f
150
13. 11; 27. 19.
ft.
440
ft.
23. $6840.
22. $1275.
21. $3000; $2800; $2600; $2400; $2200.
192; 27.
7.
-60; -3.
12.
11. 312; 15.
16. 3; 24.
14. 7; 23.
-
9. 21.
24. $2950.
EXERCISE 1.
7.
32.
63.
2. 243.
8.
16, 8, 4, 2.
13.
9.
21. 6;
Page 231
14.
-8,
9.
4.
-128.
-3,
16. 4; 30.
10.
22. 5;
1.
4.
3. .
9,
27. $3000; $1500; $750; $375.
-27,
81.
or
11.
18. 5; 121.
17. 6; 21.
23. J; 255.
1.
6. 4, 8, 16,
24. J; 31.5.
-4,
8,
12.
4.
19. 6;
-16. 3.
-21.
26. $10.23.
28. $4000; $3200; $2560; $2048.
29. 4096.
EXERCISE 1.
1.08.
8. 0.18.
64. 2.
Page 2S6 1.26.
9. 0.42.
3. 1.44.
11. 0.70.
4. 0.90.
12. 1.4.
6. 1.5.
13. 2.1.
7.
-0.18.
14.
-0.52.
ANSWERS
280 16.
-0.92.
24.
i
33.
.
17. 25.
27.
- .
34.
VA-
36.
Change
side to loga A%.
38. 41.
Change
right side to
44. True.
1.
66.
-
18. 7.7664
EXERCISE
X
24. 1.030
67.
X
13. 0.1109. 18. 72.79.
10 9
.
left side
16.
6. 2.6232.
6.9590
-
12. 0.9180.
17. 6.7782
10.
22. 135.
21. 11.5.
-
10.
23. 22.6.
29. 0.00231.
28. 0.0315.
33. 0.000100.
Page 246 3.
8.9218
6.8503
9.
X
-
-
10.
10.
4. 1.5474.
10~ 8
22. 3.107
.
-
11. 8.9659
16. 0.001942.
6.
20.
17. 0.1401.
X
lO' 10
.
7.7563
-
10.
12. 20.3651. 18. 5.908.
23. 6.521
X
10 10
.
.
Page 250 2. 0.03199.
22.89.
1.644
a y).
logio 2).
Change
11. 0.5705.
27. 0.289.
21. 2.063
68.
43.
+ log
y
logio
4. 2.3711.
1.5119.
- 10. 6.8069 - 10.
2. 2.8625.
10 15
EXERCISE 1.
19.
10.
19. 0.01456.
+
left
47. True.
3. 2.7210.
9.
14. 2,313,000.
13. 167.0.
Change
right side to 10(log x
logio B).
9.8663
8. 2.4688.
7. 0.7583.
7.
14.
10.
3.4041.
37.
Page 244
32. 1.01.
31. 5.48.
A
23. 3. 32. 512.
.
-
46. True.
26. 3920.
24. 430.
1.
(2 logio
8. 2.9657.
13. 8.7924
31.
^B.
left side to log a
Change
2. 2.5403.
2.2625.
7. 2.5563.
29. 2/2.
right side to (logio x
Change
to loga v^C.
EXERCISE
28. f.
i
22.
21. 125.
19. 4.
-1.
39. True. 42.
i
18.
26.
14. 2.624 19. 9.314.
24. 4861.
26. 1.063.
31. 99.75.
32. 15.01.
4. 3.768.
3. 50,220.
8. 0.5358.
X
11. 0.09530.
9. 18,220.
10 7
.
16. 1.725
-46.01.
33. 9.9894
-
1Q- 8
.
22. 1.529.
21. 7.555.
27.
X
28. 10.
-16.05.
34. 0.977.
6.
1910.
12. 47.47.
17. 0.0314.
23. 227,000. 29. 0.04336.
INDEX
refer to pages)
(Numbers
Common Common
Abscissa, 90 Absolute value, 4
Addition Addition Addition Addition
of fractions, 41, 42 of numbers, 5 of radicals, 134
Aggregation, symbols of, 11 Algebra, definition of, 1 Algebraic expression, 9 Algebraic formula, 1 Amount, at simple interest, 74 Analytic geometry, 94, 189 Antilogarithm, 243 Arithmetic, definition of, 1 Arithmetic extremes, 222 Arithmetic means, 222 Arithmetic progression, 221 Associative law of addition, 10 Associative law of multiplication, 13
227
Conditional equation, 54 Constant, 94
Constant of proportionality, 211 Coordinate system, 90 Coordinates, 91 Cube roots, table
of,
255
Decimal approximation, 126 Decimal numbers, 146 Degree, 29 Denominator, 34 Dependent equations, 95, 104, 113 Dependent variable, 93
Axiom, 56 Axis, of coordinate system, 90 Axis of imaginaries, 147 Axis of reals, 146
Determinant, 109 Diagram, 202 Discriminant, 161 Distributive law, 10 Dividend, 15 Division of fractions, 45 Division of numbers, 6 Division of polynomials, 13, 15 Divisor, 15 Double roots, 151, 154
Base, of an exponential expression, 9 Base of a logarithm, 233 Binomial, 9 Binomial formula, 239
Binomial theorem, 239 Braces, 254 Brackets, 254 Briggs* system of logarithms, 237
Ellipse, 189
Equation, 1, 54 Equation in quadratic form, 176 Equations reduced to simplest form,
Cancellation, 35 Characteristic, 238
9
Coefficient, literal, 64 Coefficient of the unknown,
Combination method, 101
Common difference,
ratio,
Commutative law of addition, 9 Commutative law of multiplication, 13 Complex fractions, 34 Complex numbers, 147 Complex plane, 147
and subtraction method, 100
Coefficient,
logarithms, 237
221
58
176 Equations reducible to simpler forms, 197 Equations with given roots, 175 Equivalent equations, 57, 60, 157
281
INDEX
282
Even
integers,
39
Exponential equations, 233 Exponents, 9 laws of, 14, 121 negative, 128 zero, 127 Extraneous roots, 60, 61, 179 Extremes of a proportion, 202 Factor, 21 Fractions,
4, 34, 146 algebraic, 34
complex, 34, 48 rules concerning,
39
simple, 34
Fulcrum, 80 Function, 87 General equation in one unknown, 173 Geometric means, 229 Geometric progression, 227 Graph of a function, 90, 91 of a linear equation, 94 of a linear function, 94 of a non-mathematical function,
96,97 solution of a quadratic equation, 164 Graphical solution of a system of
Graphical
equations, 103, 189-200
Highest common factor, 30 Hyperbola, 189 Identity, 22, 54
Imaginary numbers, 143 Imaginary roots, 157 Imaginary unit, 143 Incommensurable lengths, 126 Inconsistent equations, 104, 113
Independent equations, 104 Independent variable, 93 Index of a root, 124 Integral rational equation, 55 Integral rational polynomial, 21
of logarithms, 234 of multiplication, 13 of radicals, 132
Lever, 80 Lever arms, 80 Like radicals, 133
Like terms, 9 Limits of roots, 185 Linear equation, 55, 56 Linear function, 94 Literal coefficient, 9, 63 Literal number, 9, 64 Literal root, 64 Loci, 189 Locus, 92 Logarithm, 233 Logarithmic equation, 233
Lowest common denominator, 41 Lowest common multiple, 30 Lowest terms of a fraction, 35 Mantissa, 238
Mathematical function, 87 Means of a proportion, 202 Member of an equation, 54 Minor denominators, 48 Minor of a determinant, 111 Moment of a force, 80 Monomial, 9 Multiple, 29 Multiple roots, 175 Multiplication of fractions, 44 of polynomials, 13, 14 of radicals, 134
Napierian system of logarithms, 237 Negative exponents, 128 Negative numbers, 4 Notation concerning functions, 88 Numbers, 3, 143, 146, 148 Numerator, 34 Numerical coefficient, 9
Odd
integers,
39
Interest, simple,
Operations which equations, 60
Involution, 121 Irrational number, 126, 146
Operations which do not yield equivalent equations, 60, 61 Operations with imaginary numbers, 145
74 Interpolation, 244 Inverse variation, 208
Laws
of addition, 9, 10 of exponents, 14, 121
yield
equivalent
Operations with positive and negative numbers, 5, 6
INDEX
283
Operations with zero, 6, 7 Order of a radical, 133 Ordinate, 90 Origin, 90
Rational number, 126 Rationalizing denominators, 137 Real number, 4, 143 Reciprocal, 34 Remainder, 15 Remainder theorem, 180 Repeated base case, 122 Repeated exponent case, 122 Repeated root, 154 Root of a number, 124 Roots of an equation, 56, 151 Roots, table of, 255
Parabola, 166, 189 Parentheses, 5, 10, 11, 264 Plotting a point, 91 Polynomials, 9 11 adoption of, 9, 10, division of, 13, 15 multiplication of, 13, 14 subtraction of, 9, 10, 11 Positive numbers, 4 Postulates, 9, 45
Satisfy an equation, 56 Scale drawing, 202
Power, 123
Scientific notation,
Powers of a negative number, 123 of
t,
144
Prime factors, 21 Prime numbers, 21 Principal in interest problems, 74 Principal rth root, 125 Product of the roots of a quadratic,
158 Progression, arithmetic, 221 geometric, 227 Proportion, 201
Pure imaginary numbers, 147
241 Second order determinant, 110 Side of an equation, 54 Simultaneous equations, 100-109, 189-200 Single base case, 121 Square roots, table of, 255
Stated problems, 115 Substitution method, 101 Subtraction of fractions, 41, 42 of numbers, 5 of radicals, 133 Sum of an arithmetic progression, 225
Sum Sum
Quadrant, 90 Quadratic equations, 150 algebraic solution of, 152-7, 160 of, 161 graphical solution of, 164
of a geometric progression, 228
a quadratic, 158 Surd, 126 Synthetic division, 181 of the roots of
discriminant
Table of logarithms, 256, 257 of roots, 255 of symbols, 254 Tangent, 166 Term, 9 Third order determinant, 110
standard form of, 150 system of, 195-200 with given roots, 168 Quadratic formula, 160 Quotient, 15
Time Time
as an independent variable, 97 in interest problems, 74 Transposition of terms, 57
Radical sign, 124 Radicals, 124 addition of, 133
Trigonometry, 201
equalization of the order of, 140
laws
of,
132
multiplication of, 134 reduction of the order simplest form of, 141 subtraction of, 133
Radicand, 124
Rate
Trinomial, 9
Unknowns, 55 of,
140
Variable, 93 Variation, 206 Vertex of parabola, 167
Vinculum, 254
of interest, 74
Ratio, 201
Zero, operations with, 6, 7
Intermediate Algebra 4th Edition Pdf
Edition: A Maple and MATLAB Approach, Third Edition (Textbooks in Mathematics) Solutions Manual to Accompany Introduction to Abstract Algebra, Fourth Edition Student Solutions Manual for Strang's Linear Algebra and Its Applications, 4th Edition Student Solutions Manual for Differential.
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