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'ALGEBRA
FOR
SECONDARY SCHOOLS
BY
WEBSTER WELLS, S.B.
PROFESSOR OF MATHEMATICS IN THE MASSACHUSETTS
INSTITUTE OF TECHNOLOGY
BOSTON^, U.S.A.
D. C. HEATH & CO., PUBLISHERS
1906
Copyright, 1906,
By WEBSTER WELLS.
All rights reserved.
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PREFACE
The present work is intended to meet the needs of High
Schools and Academies of the highest grade. While in the
main similar to the author's " Essentials of Algebraj" many
additional topics have been introduced, and improvements
made; attention is especially invited to the following:
1. The product by inspection of two binomials of the form
mx + n and px \ q (% 100).
2. In the chapter on Factoring will be found the factoring
of expressions of the forms x^ + ax^y^ f y* and aar' + 6a; + c,
when the factors do not involve surds (§§ 115, 117).
These forms are considered later in §§ 298 and 300.
The solution of equations by factoring is also taken up in
this chapter.
In § 107 will be found many new varieties of examples.
3. In § 176 will be found a set of problems in which the
solutions are negative, fractional, or zero.
4. In the chapter on Evolution will be found the square
root by inspection of polynomials of the form
a^ 4 62 _^ c2 + 2ab + 2ac + 26c,
and the cube root by inspection of polynomials of the form
a' + S a'b + Sab'h b' (§§ 212, 223).
The development of the rules for the square and cube root
of polynomials and arithmetical numbers leaves nothing to be
22l7b8
iv PREFACE
desired from a theoretical point of view. (See §§ 213, 214,
217, 224, 225, 228.)
5. In the solution of quadratic equations by formula (§ 289),
the equation is in the form ax^ \ bx { c = 0.
6. In all the theoretical work in Chapter XXI, the quadratic
equation is in the form ax^ \ bx { c = 0.
7. In the chapter on Ratio and Proportion, in several of the
demonstrations of theorems, fractions are used in place of
ratio symbols.
8. In §§ 386 and 387 will be found the same proof of the
Binomial Theorem for Positive Integral Exponents as is given
in the " Essentials of Algebra " ; those wishing a more com
plete proof, in which the general law of coefficients is proved
for any two consecutive terms, will find it in § 447.
9. The proof of the Theorem of Undetermined Coefficients
given in § 396 is the same as that given in the " Essentials of
Algebra " ; a more rigorous proof is given in § 450.
10. The author has thought it best to omit the proof of the
Binomial Theorem for Fractional and Negative Exponents, as
a rigorous demonstration is beyond the capacity of pupils in
preparatory schools.
11. In Chapter XXXIII will be found Highest Common
Factor and Lowest Common Multiple by Division; and also
the reduction of a fraction to its lowest terms, when the
numerator and denominator cannot be readily factored by
inspection.
Any teacher who so desires can take this work in connection
with Chapters IX and X.
Chapter XXXIII also contains the proof of (1), § 235, for
all values of m and n (§ 445); and the reduction of a frac
tion whose denominator is irrational to an equivalent fraction
PREFACE V
having a rational denominator, when the denominator is the
sum of a rational expression and a surd of the nth degree, or
of two surds of the nth degree (§ 446).
An important feature of the work is the prominence given
to graphical methods; in Chapter XIII will be found the
graph of a linear equation with two unknown numbers,
and also of a linear expression involving one unknown
number.
In §§ 184, 185, and 186 will be found the graphical repre
sentations of the solutions of simultaneous linear equations,
including inconsistent and indeterminate equations.
The subject is taken up for quadratic equations in §§ 303 to
305, and 314 to 316.
To meet the demands of many schools, a number of physi
cal problems have been introduced; these will be found at
the end of Exercises 62, 128, 129, and 145.
At the end of the chapter on Variation will be found a set
of problems in physics in which the principles of variation
are employed ; and also several illustrations of the application
of graphs in physics. All the above work in physics has been
prepared by Professor Eobert A. Milliken, of the University
of Chicago.
In nearly every set of numerical equations, beginning with
Exercise 5S, will be found examples in which other letters
than X, y, and z are used to represent unknown numbers.
The examples and problems are about 4000 in number;
and no example is a duplicate of any in the author's "Aca
demic Algebra" or "Essentials of Algebra."
There is throughout the work a much greater variety of
examples than in the above treatises.
An important and useful feature of the work is the Index,
vi PREFACE
which, contains references to all operations and important
definitions.
To meet the wants of the most advanced schools, the author
has introduced nine additional chapters/ which will cover the
entrance requirements in algebra at any American college or
scientific school.
These additional chapters are:
XXXIII. The Fundamental Laws for Addition and Multipli
cation.
XXXIV. Additional Methods in Factoring.
XXXV. Mathematical Induction.
XXXVI. Equivalent Equations.
XXXVII. Graphical Eepresentation of Imaginary Numbers.
XXXVIII. Indeterminate Forms.
XXXIX. Horner's Synthetic Division.
XL. Permutations and Combinations.
XLI. Exponential and Logarithmic Series.
The author desires to express his thanks to the many
teachers in secondary schools, whose suggestions in the prepa
ration of the work have been of the greatest service.
WEBSTER WELLS.
Boston, 1906.
/
CONTENTS
CHAPTER PAGE
I. Definitions and Notation 1
Symbols 1
Equations . . 2
Axioms 2
Solution of Problems by Algebraic Methods . . .3
Algebraic Expressions 9
11. Positive anb Negative Numbers 11
Addition of Positive and Negative Numbers ... 12
Multiplication of Positive and Negative Numbers . . 14
III. Addition and Subtraction of Algebraic Expressions . 17
Addition of Monomials 18
Addition of Polynomials 21
Subtraction of Monomials . . . ... 24
Subtraction of Polynomials . c .... 26
Parentheses . 28
IV. Multiplication of Algebraic Expressions ... 32
Multiplication of Monomials 33
Multiplication of Polynomials by Monomials . . 84
Multiplication of Polynomials by Polynomials . . 35
V. Division of Algebraic Expressions 42
Division of Monomials . . . . . . .43
Division of Polynomials by Monomials .... 45
Division of Polynomials by Polynomials . . . • 46
VI. Integral Linear Equations . . . . . .61
Principles used in solving Integral Equations . . 52
Solution of Integral Linear Equations .... 54
Problems leading to Integral Linear Equations with One
Unknown Number 57
VII. Special Methods in Multiplication and Division . 63
VIII. Factoring 74
Miscellaneous and Review Examples .... 91
Solution of Equations by Factoring .... 94
vii
Vlll
CONTENTS
CHAPTER
IX.
X.
Highest Common Factor. Lowest Common Multiple
Highest Common Factor
Lowest Common Multiple
j
\ XI.
Fractions
Reduction of Fractions
Addition and Subtraction of Fractions
Multiplication of Fractions .
Division of Fractions .
Complex Fractions
Miscellaneous and Review Examples
Fractional and Literal Linear Equations
Solution of Fractional Linear Equations
Solution of Literal Linear Equations .
Solution of Equations involving Decimals
Problems involving Linear Equations .
Problems involving Literal Equations .
Simultaneous Linear Equations
Elimination by Addition or Subtraction
Elimination by Substitution
Elimination by Comparison .
Simultaneous Linear Equations containing
Two Unknown Numbers .
Problems involving Simultaneous Linear Equati
XIII. Graphical Representation
Graph of a Linear Equation involving Two
Numbers
Intersections of Graphs
XIV. Inequalities
XIL
XV. Involution
XVL
Evolution
Evolution of Monomials
Square Root of a Polynomial
Square Root of an Arithmetical Numbet
Cube Root of a Polynomial .
Cube Root of an Arithmetical Number
XVII.
Theory of Exponents
Miscellaneous Examples
More than
Unknown
CONTENTS
IX
/ CHAPTER PAGE
\j XVIII. Surds 222
Reduction of a Surd to its Simplest Form . . . 222
Addition and Subtraction of Surds .... 226
To reduce Surds of Different Degrees to Equivalent
Surds of the Same Degree 227
Multiplication of Surds 228
Division of Surds 230
Involution of Surds 231
Evolution of Surds 233
Reduction of a Fraction whose Denominator is Irra
tional to an Equivalent Fraction having a Rational
Denominator 233
Properties of Quadratic Surds 237
Imaginary Numbers 242
XIX. Quadratic Equations 248
Pure Quadratic Equations 248
Affected Quadratic Equations 250
Problems involving Quadratic Equations with One
/ Unknown Number . . .. . . . .261
XX. Equations solved like Quadratics .... 268
XXI. Theory of Quadratic Equations 273
Factoring 276
Graphical Representation of Quadratic Expressions
with One Unknown Number 283
XXII. Simultaneous Quadratic Equations .... 286
Graphical Representation of Simultaneous Quadratic
Equations with Two Unknown Numbers . . 300
XXIII. Variables and Limits 304
XXIV. Indeterminate Equations 308
XXV. Ratio and Proportion 312
Properties pf Proportion 313
XXVI. Variation 321
XXVII. Progressions 331
Arithmetic Progression ....... 331
Geometric Progression 338
Harmonic Progression 346
XXVIII. The Binomial Theorem 350
Positive Integral Exponent 350
CONTENTS
CHAPTEK
XXIX.
XXX.
XXXI.
XXXII.
XXXIII.
XXXIV.
XXXV.
XXXVI.
XXXVII.
XXXVIII.
XXXIX.
XL.
XLI.
Index
Undetermined Coefficients
Convergency and Divergency of Series .
The Theorem of Undetermined Coefficients
Expansion of Fractions ....
Expansion of Surds ....
PAGE
. 357
. 358
. 360
. 361
. 363
Partial Fractions 364
Reversion of Series 370
The Binomial Theorem . . . . . . 372
Fractional and Negative Exponents . . . 372
Logarithms ......... 376
Properties of Logarithms ..... 378
Use of the Table 383
Applications ........ 388
Miscellaneous Examples 391
Exponential Equations 393
Miscellaneous Topics ...... 395
Highest Common Factor and Lowest Common
Multiple by Division 395
Proof of (1), § 235, for All Values of m and u . 405
The Binomial Theorem for Positive Integral Expo
nents 407
The Theorem of Undetermined Coefficients . . 408
The Fundamental Laws for Addition and Multi
plication ....
Additional Methods in Factoring
Symmetry ....
Mathematical Induction .
Equivalent Equations
Graphical Representation of Imaginary
Indeterminate Forms
Horner's Synthetic Division .
Permutations and Combinations
Exponential and Logarithmic Series
The Exponential Series .
The Logarithmic Series .
Calculation of Logarithms
Numbers
410
413
416
422
426
434
438
441
445
453
453
455
456
459
ALGEBRA FOR
SECONDARY SCHOOLS
ALGEBRA
I. DEFINITIONS AND NOTATION
1. In Algebra, the operations of Arithmetic are abridged
and generalized by means of Symbols.
SYMBOLS REPRESENTING NUMBERS
2. The symbols usually employed to represent numbers are
the Arabic Numerals and the letters of the Alphabet.
The numerals represent known or determinate numbers.
The letters represent numbers which may have any values
whatever, or numbers whose values are to be found.
SYMBOLS REPRESENTING OPERATIONS
3. The Sign of Addition, +, is read "plus J'
Thus, a{b signifies that the number represented by b is to
be added to the number represented by a ; a \b { c signifies
that the number represented by b is to be, added to the number
represented by a, and then the number represented by c added
to the result ; and so on.
The result of addition is called the Sum.
We shall use the expression " the number a," or simply " a," to signify
" the number represented hy a^^'' etc.
4. The Sign of Subtraction, — , is read " minus.^^
Thus, a — b signifies that the number b is to be subtracted
from the number a ; a— b — c signifies that b is to be sub
tracted from a, and then c subtracted from the result; and
so on. ' * '. ^ ^, = ' ^"^ !"
2 ALGEBRA
5. The Sign of Multiplication, X, is read ^' times, ^' or
" multiplied by."
Thus, a X b signifies that the number a is to be multiplied
by the number b; a xb x c signifies that a is to be multiplied
by b, and the result multiplied by c ; and so on.
The sign of multiplication is usually omitted in Algebra,
except between two numbers expressed in Arabic numerals.
Thus, 2 X signifies 2 multiplied by x ; but the product of
2 by 3 could not be expressed 23.
6. The Sign of Division, ;, is read '^divided by.^^
Thus, a7b signifies that the number a is to be divided by
the number b.
The division of a by 6 is also expressed •
EQUATIONS
7. The Sign of Equality, =, is read " equals. ^^
Thus, a = b signifies that the number a equals the number b.
8. An Equation is a statement that two numbers are equal.
The Jirst member of an equation is the number to the left
of the sign of equality, and the second member is the number
to the right of that sign.
Thus, in the equation 2 ic — 3 = 5, the first member is 2 ic — 3,
and the second member 5.
AXIOMS
9. An Axiom is a truth which is assumed as selfevident.
Algebraic operations are based on the following axioms :
1. Any number equals itself.
2. Any number equals the sum of all its parts.
3. Any number is greater than any of its parts.
4. Two numbers which are equal to the same number, or to
equal ^lumbers, are. equal.
DEFINITIONS AND NOTATION 3
5. If the same number, or equal numbers, be added to equal
numbers, the resulting 7iumbers 2vill be equal.
6. If the same number, or equal numbers, be subtracted from
equal numbers, the resulting numbers imll be equal.
7. If equal numbers be multiplied by the same number, or
equal numbers, the resulting numbers will be equal.
8. If equal numbers be divided by the same number, or equal
numbers, the resulting numbers will be equal.
SOLUTION OF PROBLEMS BY ALGEBRAIC METHODS
10. The following examples will illustrate the use of Alge
braic symbols in the solution of problems.
The utility of the process consists in the fact that the un
known numbers are represented by symbols, and that the
various operations are stated in Algebraic language.
1. The sum of two numbers is 30, and the greater exceeds
the less by 4 ; what are the numbers ?
We will represent the less number \)j x.
Then the greater will be represented by x 4 4.
By the conditions of the problem, the sum of the less number and the
greater is 30 ; this is stated in Algebraic language as follows :
x + ic + 4 = 30. (1)
Now, x\x = xx'2\ for to multiply an arithmetical number by 2, we
add it twice.
Again, x x 2 = 2 x aj, or 2 x (§ 5) ; for the product of two arithmeti
cal numbers is the same in whichever order they are multiplied.
Therefore, x + x = 2 x ; and equation (1) can be written
2 X + 4 = 30.
The members of this equation, 2 x + 4 and 30, are equal numbers ; if
from each of them we subtract the number 4, the resulting numbers will
be equal (Ax. 6, § 9).
Therefore, 2 x = 30  4, or 2 x = 26.
Dividing the equal numbers 2 x and 26 by 2 (Ax. 8, § 9), we have
x = 13.
Hence, the less number is 13, and the greater is 13 + 4, or 17.
4 ALGEBRA
The written work will stand as follows :
Let X = the less number.
Then, a; + 4 = the greater number.
By the conditions, jc + x + 4 = 30, or 2 x + 4 = 30.
Whence, 2 a; = 26.
Dividing by 2, x = 13, the less number.
Whence, x + 4 = 17, the greater number.
2. The sum of the ages of A and B is 109 years,. and A is
13 years younger than B; find their ages.
Let X represent the number of years in B's age.
Then, x — 13 will represent the number of years in A's age.
By the conditions of the problem, the sum of the ages of A and B is
109 years.
Whence, a;  13 + x = 109, or 2 cc  13 = 109.
Adding 13 to both members (Ax. 5, § 9),
2x = 122.
Dividing by 2, x = 61, the number of years in B's age.
And, X — 13 = 48, the number of years in A's age.
The written work will stand as follows :
Let X = the number of years in B's age.
Then, x — 13 = the number of years in A's age.
By the conditions, x  13 + x = 109, or 2 x  13 = 109.
Whence, 2 x = 122.
Dividing by 2, x = 61, the number of years in B's age.
Therefore, x — 13 = 48, the number of years in A's age.
It must be carefully borne in mind that x can only represent an abstract
number.
Thus, in Ex. 2, we do. not say "let x represent B's ag^e," but "let x
represent the number of years in B's age."
3. A, B, and C together have $ 66. A has onehalf as much
as B, and C has 3 times as much as A. How much has each ?
DEFINITIONS AND NOTATION 5
Let X = the number of dollars A has.
Then, 2 a; = the number of dollars B has,
and 3 a; = the number of dollars C has.
By the conditions, a; + 2a;43a: = 66.
But the sum of x, twice ic, and 3 times x is 6 times a;, or 6 x.
Whence, Qx = m.
Dividing by 6, a; = 11, the number of dollars A has.
Whence, 2 x = 22, the number of dollars B has,
and 3 a; = 33, the number of dollars C has.
(By letting x represent the number of dollars A has, in Ex. 3, we avoid
fractions.)
EXERCISE I
1. The greater of two numbers is 8 times the less, and
exceeds it by 49 ; find the numbers.
'' 2. The sum of the ages of A and B is 119 years, and A is
17 years older than B ; find their ages.
3.^ Divide $ 74 between A and B so that A may receive $48
more than B.
/ 4. Divide $108 between A and B so that A may receive
5 times as much as B.
* 5. Divide 91 into two parts such that the smaller shall be
onesixth of the greater.
"^ 6. A man travels 112 miles by train and steamer ; he goes
by train 54 miles farther than by steamer. How many miles
does he travel in each way ?
y 7. The sum of three numbers is 69; the first is 14 greater
than the second, and 28 greater than the third. Find the num
bers.
8. The sum of the ages of A, B, and C is 134 years ; B is
13 years younger than A, and 7 years younger than C. Find
their ages.
6 ALGEBRA
9. A cow and sheep together cost f 91, and the sheep cost
onetwelfth as much as the cow ; how much did each cost ?
^ 10. Divide % 6.75 between A and B so that A may receive
onefourth as much as B.
^ 11. A man has % 2. After losing a certain sum, he finds
that he has left 20 cents more than 3 times the sum which he
lost. How much did he lose ?
12. A, B, and C have together % 140 ; A has 4 times as
much as B, and C has as much as A and B together. How
much has each ?
13. A, B, and C have together $ 100 ; A has % 10 less than
C, and C has $ 25 more than B. How much has each ?
»^'14. At an election two candidates, A and B, had together
653 votes, and A was beaten by 395 votes. How many did
each receive ?
■' 15. A field is 7 times as long as it is wide, and the distance
around it is 240 feet. Find its dimensions.
16. My horse, carriage, and harness are worth together
$ 325. The horse is worth 6 times as much as the harness,
and the carriage is worth $ 65 more than the horse. How
much is each worth ?
17. The sum of three numbers is 87 ; the third number is
oneeighth of the first, and the second number 15 less than the
first. Find the numbers.
18. At a certain election, three candidates, A, B, and C,
received together 436 votes; A had a majority over B of 14
votes, and over C of 3 votes. How many did each receive ?
19. The sum of the ages of A, B, and C is 110 years ; B's
age exceeds twice C's by 12 years, and A is 9 years younger
than B. Find their ages.
DEFINITIONS AND NOTATION 7
20. A pole 77 feet long is painted red, white, and black;
the red is onefifth of the white, and the black 21 feet more
than the red. How many feet are there of each color ?
^ 21. Divide 70 into three parts such that the third part shall
be onefifth of the first, and onefourth of the second.
22. Divide $ 7.55 between A, B, and C so that C may
receive onehalf as much as A, and B $ 2.95 less than A and
C together.
23. A, B, and C have together $22.50; B has $1.50 more
than A, and C has $ 8 less than twice the amount that A has.
How much has each ?
y 24. The profits of a shopkeeper in a certain year were one
third as great as in the preceding year, and $ 515 less than in
the following year. If the total profits for the three years
were $ 2615, what were the profits in each year ?
25. The sum of four numbers is 96. The first is 4 times the
fourth, and exceeds the third by 20 ; and the second exceeds
the sum of the first and fourth by 4. Find the numbers.
* 26. Divide $ 468 between A, B, C, and D, so that A may
receive onefifth as much as B, B onefifth as much as C, and
C onefifth as much as D.
DEFINITIONS
11. If a number be multiplied by itself any number of times,
the product is called a Power of the number.
An Exponent is a number written at the right of, and above
another number, to indicate what power of the latter is to be
taken; thus,
a^, read " a square,^' or " a second power,^^ denotes a X a ;
a^, read " a cube,^^ or " a third power,^^ denotes a xaxa\
a!^, read " a fourth/' " a fourth power/' or " a exponeyit 4,"
denotes a x a X ct x a, etc.
8 ALGEBRA
If no exponent is expressed, thejirst power is understood.
Thus, a is the same as a\
12. Symbols of Aggregation.
The parentheses ( ), the brackets [ ], the braces \ j, and the
vinculum , indicate that the numbers enclosed by them are
to be taken collectively ; thus,
{a+b)xc, [a + 6] X c, [a\b\ x c, and a{h x c
all indicate that the result obtained by adding 6 to a is to be
multiplied by c.
EXERCISE 2
What operations are signified by the following ?
1.
2.
2af/.
m{xy).
5.
x(y + z).
8.
\xyj
3.
ab
6.
(m — 7iy.
9.
(2a + 36)(4c
5d).
4.
cd
x + (yz).
7.
a c
b d
10.
eS"
Write the following in symbols :
11. The result of subtracting 6 times n from 5 times m.
12. Three times the product of the eighth power of m and
the ninth power of n.
V 13. The quotient of the sum of a and 6, divided by the sum
of c and d.
* 14. The product of ^x\y and z^.
15. The result of subtracting y — z from x.
16. The product oi a—b and c — d.
17. The result of adding the quotient of m by n, and the
quotient of x by y.
18. The square of m + n.
sj 19. The cube of a  6 + c
DEFINITIONS AND NOTATION 9
20. Tlie fourth power of the quotient of a divided by x.
21. The product of the quotient of 1 by a; and the quotient
of 1 by y.
ALGEBRAIC EXPRESSIONS
' 13. An Algebraic Expression, or simply an Expression, is a
number expressed in algebraic symbols ; as,
2, a, or 2aj2_3a6f 5.
■^14. The Numerical Value of an expression is the result
obtained by substituting particular numerical values for the
letters involved in it, and performing the operations indicated.
1. Find the numerical value of the expression
h
when a = 4, 6 = 3, c = 5, and d = 2.
We have, 4a + — c?3 = 4 x 4 + ^^23 = 16 + 108 =18.
o
If the expression involves parentheses, the operations indi
cated within the parentheses must be performed first.
2. Find the numerical value, when a = 9, & = 7, and c = 4, of
We have, a  6 = 2, & + c = 11, a + & = 16, and &  c = 3.
Then the numerical value of the expression is
2xllM = 22l^ = a
3 3 3
EXERCISE 3 ,, ^P ' '
Find the numerical values of the following when a = 6, & = 3,
c = 4, d = 5, m = 3, and w = 2 :
1. o?hcd\ 2. 2ahcd, 3. 3a6 + 46c5cd.
10 ALGEBRA
4. orh\ g ^_^ I ^.
5. a^ + 6«. 'bed
' 6. «+A. 10. ^ + \l\
he ad abed
7 l + ll. 11. ^_i^. ;/ ^
8 i^ 12 ^4^^
* 28d"* * a^ b^ c"'
Find the numerical values of the following when a = 5, 6 = 2,
c = 3, and d = 4 :
13
2a4
^y. 15. 5a'(ab)2b\c + d).
2b\'
14. (a262d^3. 16. S(aby{3(c + dy
17. (a5)2+(2a3 6)2(6 + c)2.
18. (2a&cHd)(2a + 64cd).
9a — 46 — 3c a + 6 a + c a + ci
Find the numerical values of the following when a = ,
b = ^, c = ^, and a; = 4 :
oi g + c a — c 22 8a + 66 — 15 c
*a — c a + c 'I6a + 10 6 + 9c
23. a^4(2a + 36)ar^(5a4c)a; + a&c.
\a bj \a b cj abc
\
POSITIVE AND NEGATIVE NUMBERS 11
II. POSITIVE AND NEGATIVE NUMBERS
15. There are certain concrete magnitudes which are capa
ble of existing in two opposite states.
Thus, in financial transactions, we may have assets or Ua
hilities, and gains or losses; we may have motion along a
straight line in a certain direction, or in the opposite direc
tion; etc.
In each of these cases, the effect of combining with a mag
nitude of a certain kind another of the opposite kind, is to
diminish the former, destroy it, or reverse its state.
Thus, if to a certain amount of asset we add a certain
amount of liability, the asset is diminished, destroyed, or
changed into liability.
16. The signs + and — , besides denoting addition and sub
traction, are also used, in Algebra, to distinguish between the
opposite states of magnitudes like those of § 15.
Thus, we may indicate assets by the sign +, and liabilities
by the sign — ; for example, the statement that a man's assets
are — $ 100, means that he has liabilities to the amount of $ 100.
EXERCISE 4
1. If a man has assets of $400, and liabilities of $600, how
much is he worth ?
' 2. If gains be taken as positive, and losses as negative, what
does a gain of — $ 100 mean ?
3. In what position is a man who is — 3 miles north of a
certain place ?
4. In what position is a man who is — 50 feet west of a
certain point ?
5. How many miles north of a certain place is a man who
goes 5 miles north, and then 9 miles south ?
12 ALGEBRA
6. How many miles east of a certain place is a man who
goes 11 miles west, and then 6 miles east ?
17. Positive and Negative Numbers.
If the positive and negative states of any concrete magni
tude be expressed witJiout reference to the unit, the results are
called positive and 7iegative numbers, respectively.
Thus, in { $5 and — $3, +5 is a positive number, and —3
is a negative number.
For this reason the sign 4 is called the positive sign, and
the sign — the negative sign.
' If no sign is expressed, the number is understood to be posi
tive ; thus, 5 is the same as f 5.
■ The negative sign must never be omitted before a negative
number.
18. The Absolute Value of a number is the number taken
independently of the sign affecting it.
Thus, the absolute value of — 3 is 3.
ADDITION OF POSITIVE AND NEGATIVE NUMBERS
19. We shall give to addition in Algebra its arithmetical
meaning, so long as the numbers to be added are positive integers
or positive fractions.
We may then attach any meaning we please to addition
involving other forms of numbers, provided the new meanings
are not inconsistent with principles previously established.
20. In adding a positive number and a negative, or two
negative numbers, our methods must be in accordance with
the principles of § 15.
If a man has assets of $ 5, and then incurs liabilities of $ 3,
he will be worth $2.
If he has assets of ^ 3, and then incurs liabilities of $ 5, he
will be in debt to the amount of $2.
POSITIVE AND NEGATIVE NUMBERS 13
If he has liabilities of $5, and then incurs liabilities of f 3,
he will be in debt to the amount of $ 8.
Now with the notation of § 15, incurring liabilities of $3
may be regarded as adding — $ 3 to his property.
Whence, the sum of + f 5 and — $3 is + ^2;
the sum of —$5 and +$3 is — $2;
and the sum of —$5 and — $ 3 is — $8.
Or, omitting reference to the unit,
( + 5) + (3)=:+2;
(_5) + ( + 3)=2;
(_5) + (3)=8.
To indicate the addition of + 5 and — 3, they must be enclosed in
parentheses (§ 12).
We then have the following rules :
^ To add a j^ositive and a negative number, subtract the less
absolute value (§ 18) from the greater, and prefix to the result
the sign of the number having the greater absolute value.
y^ To add two negative numbers, add their absolute values, and
prefix a negative sign to the result.
21. Examples.
1. Find the sum of + 10 and — 3.
Subtracting 3 from 10, the result is 7.
Whence, ( + 10) + (  3) = + 7.
2. Find the sum of — 12 and + 6.
Subtracting 6 from 12, the result is 6.
Whence, (_ 12) + ( + 6) =6.
3. Add _ 9 and  5.
The sum of 9 and 5 is 14.
Whence, (_ 9) + (_ 5) = _„ 14
14 ALGEBRA
EXERCISE 5
Find the values of the following :
</l. (6) + (7).
2. (+8) + (3).
' (IH1
3. (9)+(+5). 9. f+D + fl
»/4. (+4) + (11). V »y V ^,
5. (_13) + (18). 10 (15i) + (+12x).
6. (_42) + (f57). 11. (+17) + (10A).
1^7. (34) + (+82). 12. (14) + (21A).
MULTIPLICATION OF POSITIVE AND NEGATIVE
NUMBERS
22. If two expressions are multiplied together, the first is
called the Multiplicand, and the second the Multiplier.
The result of multiplication is called the Product.
23. We shall retain for multiplication, in Algebra, its arith
metical meaning, so long as the multiplier is a jjositive integer or
a positive fraction.
That is, to multiply a number by a positive integer is to add
the multiplicand as many times as there are units in the mul
tiplier.
For example, to multiply — 4 by 3, we add — 4 three times.
Thus, (4)x(+3) = (4) + (4) + (4) = 12.
24. In Arithmetic, the product of two numbers is the same
in whichever order they are multiplied.
Thus, 3x4 and 4 x 3 are each equal to 12.
If we could assume this law to hold for the product of a
positive number by a negative, we should have
(+3) X (4) = (4) X (+3) = 12 (§ 23) = (3 x4).
POSITIVE AND NEGATIVE NUMBERS 15
Then, if the above law is to hold, we must give the follow
ing meaning to multiplication by a negative number :
A To multiply a number by a negative number is to multiply it by
the absolute value (§ 18) of the multiplier, and change the sign of
the result.
Thus, to multiply + 4 by — 3, we multiply +4 by +3,
giving + 12, and change the sign of the result.
That is, (+ 4) X ( 3) =  12.
Again, to multiply — 4 by — 3, we multiply — 4 by +3,
giving — 12 (§ 23), and change the sign of the result.
That is, (4)x(3) = +12.
25. From §§23 and 24 we derive the following rule :
'^To multiply one iiumber by another, multiply together their
absolute values.
X Make the product plus when the multiplicand and multiplier are
of like sign, and minus whe7i they are of unlike sign.
26. Examples.
1. Multiply +8 by 5.
By the rule, (+ 8) x ( 5) =  (8 x 5) =  40.
2. Multiply  7 by  9.
By the rule, ( 7) x (  9) = + (7 x 9) = + 63.
3. Find the numerical value when a = 4 and b = — 7, of
(a + by.
We have, (a + 6)3 = (4  7)(4  7)(4 7)
= (3)(3)(3) = 27.
EXERCISE 6
Find the values of the following :
1. (+,5)x(4). 2. (11) X (+3).
16
3
ALGEBRA
v'3. (_8)x(7).
v^4. (+9)x(6).
10.
(§)Kii>
5. (12) X (+9).
6. (24) X (5).
^^ll.
(i)Ks>
7. (14)x(+15).
12.
(7)x(«).
8. (+27)x(19).
13.
(6f)x(+6f).
,9. (l)x(l\
14.
(+1H)X(1H)
Find the numerical value when a = 2, b= — 4:^ c = 5, and
d=3 of
15. 6«. 16. d"^. 17. (6(^)2. 18. (abcf.
•19. (2 a 5 6) (4c + 3 d). 20. (ab)(bc)(cd).
v^21. (a + 6)(c + d)(ac)(6d).
22. a263 62c2cU 23. (acY+{b± d}\
/24. 3a6^5 6V + 4c^(^^.
ADDITION AND SUBTRACTION 17
III. ADDITION AND SUBTRACTION OF
ALGEBRAIC EXPRESSIONS. PARENTHESES
*i27. A Monomial, or Term, is an expression (§ 13) whose
parts are not separated by the signs + or — ; as 2 a?^, — 3 ab,
or 5.
2x^, — 3 ah, and + 5 are called the terms of the . expression
2a;2_3a6 + 5.
«^A Positive Term is one preceded by a 4 sign ; as \5a.
If no sign is expressed, the term is understood to be posi
tive.
•''A Negative Term is one preceded by a — sign ; as — 3 ah.
The — sign must never be omitted before a negative term.
/ 28. If two or more numbers are multiplied together, each of
them, or the product of any number of them, is called a Factor
of the product.
Thus, a, h, c, ah, ac, and he are factors of the product ahc.
*" 29. Any factor of a product is called the Coefficient of the
product of the remaining factors.
Thus, in 2 ah, 2 is the coefficient of ah, 2 a of h, a oi 2 h, etc.
V 30. If one factor of a product is expressed in Arabic numer
als, and the other in letters, the former is called the numerical
coefficient of the latter.
Thus, in 2 ah, 2 is the numerical coefficient of ah.
If no numerical coefficient is expressed, the coefficient 1 is
understood ; thus, a is the same as 1 a.
31. By § 25, (3) X a =  (3 X a) = 3a.
That is, — 3 a is the product of — 3 and a.
Then, — 3 is the numerical coefficient of a in — 3 a.
/Thus, in a negative term as in a positive, the numerical coeffir
dent includes the sign.
18 ALGEBRA
^ 32. Similar or Like Terms are those which either do not
differ at all, or differ only in their numerical coefficients ; as
2 x^y and — 7 x^y.
•" Dissimilar or Unlike Terms are those which are not similar ;
as 3 x^y and 3 xy"^.
ADDITION OF MONOMIALS
33. The sum of a and h is expressed a + 6 (§ 3).
V 34. We define Subtraction, in Algebra, as the process of
finding one of two numbers, when their sum and the other
number are given.
The Minuend is the sum of the numbers.
The Subtrahend is the given number.
The Remainder is the required number.
35. The remainder when h is subtracted from a is expressed
a  6 (§ 4).
Since the sum of the remainder and the subtrahend gives
the minuend (§ 34), we have
a—h\h = a.
^ Hence, if the same number he both added to, and subtracted
from, another, the value of the latter is not changed.
^ 36. It follows from § 35 that terms of equal absolute value,
but opposite sign, in an expression, may be cancelled.
37. We will now show how to find the sum of a and — b.
By §35, a + (6) = a + (5) + 65; (1)
for adding and subtracting b does not alter a + (— 6).
But by §20, {b) + b = 0',
for — b and b are numbers of the same absolute value, but
opposite sign.
Therefore, a+(— 6) = a — 6;
for the other terms in the second member of (1) cancel each
other.
ADDITION AND SUBTRACTION 19
*^38. It follows from §§33 and 37 that the addition of mono
mials is effected by uniting their terms with their respective signs
Thus, the sum of a, —b,c, — d, and — e is
a—b\c — d — e.
^39. We assume that the terms can be united in any order j
provided each has its proper sign.
Hence, the result of § 38 can also be expressed
c{a — e — d—bf — d — & + c — efa, etc.
This law is called the Commutative Law for Addition ; compare § 451.
^ 40. To multiply 5 + 3 by 4, we multiply 5 by 4, and then
3 by 4, and add the second result to the first.
Thus, (5 + 3)4=5x4 + 3x4.
We then assume that to multiply a \b by c, we multiply
a by c, and then b by c, and add the second result to the first.
Thus, (a \ b)c = ac \ be.
This law is called the Distributive Law for Multiplication ; its proof
for the various forms of number will be found in § 455.
41. Addition of Similar Terms (§ 32).
1. Required the sum of 5 a and 3 a.
We have, 5 a + 3 a = (5 + 3)a (§40)
= 8 a.
2. Required the sum of — 5 a and — 3 a.
Wehave, (_ 5a) + ( 3a) = ( 5) x a+(3)xa (§31)
= [(5) + (3)]x« (§40)
= (8)xa (§20)
= 8a. (§31)
3. Required the sum of 5 a and — 3 a.
Wehave, 5 a +( 3)a =[5 +(3)] x « (§40)
= 2 a. (§ 20)
20 ALGEBRA
4. Required the sum of —5a and 3a.
We have, (_ 5)a + 3 a =[( 5) + 3] x a (§40)
= (2)x a (§20) =  2 a.
/ Therefore, to add two similar terms, find the sum of their
numerical coefficients (§§ 20, 30, 31), and affix to the result the
common letters.
5. Find the sum of 2 a, — a, 3 a, — 12 a, and 6 a.
Since the additions may be performed in any order, we may add the
positive terms first, and then the negative terms, and finally combine
these two results.
The sum of 2 a, 3 a, and 6 a is 11 a.
The sum of — a and — 12 a is — 13 a.
Hence, the required sum is 11 a + (— 13 a), or —2 a.
6. Add 3{ah), 2(ab), 6(ab), and 4(a6).
The sum of 3(a  b) and 6(a  6) is 9(a  6).
The sum of — 2(a — &) and — 4(a — &) is — 6(a — 6).
Then, the result is [9 +(^ 6)](a  6), or 3(a  b).
If the terms are not all similar, we may combine the similar
terms, and unite the others with their respective signs (§ 38).
7. Required the sum of 12 a, —5x, — 3 /, —5 a, Sx, and
— 3 a;.
The sum of 12 a and — 5 a is 7 a.
The sum of 5x, Sx, and  3 a; is (§ 36).
Then, the required sum is 1 a — Sy^.
EXERCISE 7
Add the following :
^ 1. 11 a and —6 a. 6. — abc and 12 abc.
2. 7x and 10 a;. 7. 8a^2/' and 29 0^^/.
3. 4n and 9n. w8. 9(a + 6) and 2(a + 6).
4. —13ab and 5db. 9. —IWmn^ and 60a%nl
(,
V 5. 17 a^ and loaj^. ^ 10. 8a, la, and ^9a.
ADDITION AND SUBTRACTION 21
11. 15m, —m, —5 771, and —12 m.
12. 16 xyz, — 4 xyz, xyz, and — 6 xyz.
13. Q>{xy), 5(xy), and 10(xy).
14. ISn^, 13 7i2, 2n2,  7^^, and 147il
15. 19a^6, 2a^b, ^a% 17a% and lOa^d.
16. 7 ax, — 9 62/j — 3 aa;, and 2 6?/.
rl7. 8 ic, 2;, — 5 2/, —llz, —2x, and 10 ?/.
18. '8 (m + 7?.), 4 (m — n), — 3 (m 4 w), and — 7 (m — ti).
V 19. 14 a, — 4 d, —8 c, b, —2 a, —3 c, — 15 d, and — c.
•20. 6 a?, —7y,5z,Sy, —4:Z, —Sx, —y, —9z, and 2x.
•/"addition of polynomials
42. A Polynomial is an algebraic expression consisting of
more than one term ; as a + 5, or 2 a:^ — ic?/ — 3 2/^.
A polynomial is also called a multinomial.
A Binomial is a polynomial of two terms ; as a + 6.
A Trinomial is a polynomial of three terms ; as a h 6 — c.
43. A polynomial is said to be arranged according to the
descending powers of any letter, when the term containing the
highest power of that letter is placed first, that having the next
lower immediately after, and so on.
Thus, ic* + 3 ar^?/ 2a^/f 3aj/4 2/^
is arranged according to the descending powers of x.
The term — 4 y^^ which does not involve x at all, is regarded as con
taining the lowest power of x in the above expression.
A polynomial is said to be arranged according to the ascend
ing powers of any letter, when the term containing the lowest
j)Ower of that letter is placed first, that having the next higher
immediately after, and so on.
Thus, x^ + Z o^y 2 Q^y^ \S xf  A: y^
is arranged according to the ascending powers of y.
22 ALGEBRA
44. Addition of Polynomials.
Let it be required to add & + c to a.
Since & + c is the sum of h and c (§ 3), we may add 6  c to
a by adding h and c separately to a.
Then, a + (6 + c) = a f & 4 c.
(To indicate the addition of 6 + c, we enclose it in parentheses.)
The above assumes that, to add the sum of a set of terms, we add the
terms separately.
This is called the Associative Law for Addition; its proof will be found
in § 452.
45. Let it be required to add 6 — c to a.
By § 37/ 5 — c is the sum of h and — c.
Then, to add h — c to a, we add h and — c separately to a
(§ 44).
Whence, a + (6 — c) = a + 6 — c.
46. From §§44 and 45 we have the following rule :
*' To add a polynomial, add its terms with their signs unchanged.
1. Add QaTx", 3a^2a + 3/, smd 2 x^  a  mn.
We set the expressions down one underneath the other, similar terms
being in the same vertical column.
We then find the sum of the terms in each column, and write the
results with their respective signs ; thus,
6 a  7 a;2
 2 a + 3 a;2 + 3 ?/3
— a + 2aj2 — mn
a — 2x'^ + Sy^ — mn
2. Add 4:x3a^ll + 5x% 12a^ 78 a^15a;, and
U\6a^\10x9x'.
It is convenient to arrange each expression in descending powers of x
(§ 43) ; thus,
5a;3_ 3a;2+ 4a; 11
 8 cc3 + 12 5c2  15 X  7
6x^ 9a;2 + i0a;+14
Sx^  X A
ADDITION AND SUBTRACTION 23
3. Add 9(a + &)  8(6 + c), _ 3(6 + c)  7(c + a), and
4(c + a)5(a + 6).
9(a + &) 8(& + c)
 3(& + c)7(c + a)
5(« + 6) +4(c + a)
4(a + 6)  11(6 + c)  3(c + a)
4. Add fa + 6 — ic and ia6 + fc.
HaHb + ^\c
EXERCISE 8
Add the following :
1.
2.
«^.
Sa7b
 6(^Wy'
— 17 am 4 4 6n
— 5 « f 4 5
9a^+ 3/
6 am — 11 bn
a26
 12 a^ 4 10 2/'
9 am + 19 bn
4. 7 X + 62/ — 9 2 and 4:X — Sy{5z.
5. 4 m^ — 4 mii + n^, m^ + 4 m?i + 4 n^, and — 5 m^ + 5 n^.
6. 5 a — 7 b, 4 69 c, and 6 c — 2 a.
1^7. 3aj22a;2/ + 7?/2, 5 i^\9 xy10 y% and S x'S xy4: y\
8. a98a2 + 16a^ 5 + 15 a=^ 12 a2 a^,
and Ga^lOa^ + llaia.
9. 5(^a + b)4.x(xy), 6{a\b)\Sx(xy)y
and 8 (a + 6) — 7 a? (a; — y).
10. fa — 16 — y\c and fa + i6 — fc.
Vll. 5m49w + 4(», _3a;72/6n,  10 2/ + 8a; + 2m,
and n\lly — 7m.
12. 3iV«^ + t2/ + H2 and 3\a;^i2;.
24 ALGEBRA
13. U(x{y)17(y + z), 4(y + z)9(z{x\
and —3(x\y) — 7{z\x).
14. 6c + 2a3b, 4.d7c^12a, Sb5d + c,
and 10 a 11 6 + 9 d
n5. 7(a5)2 + 8(a6)+2, 4(a  6)2  5 (a  6),
and 3(a 6)29.
16. 8a3lla7a2, 2a6a2 + 10, 5 + 4a3 + 9a,
and 13a'512^
17. a;22, + 2a;/ + 3ic^ 3 a;/ + 4 ^/^ _ 5 ^^^^ 6a^ \5f 7 xy',
and _82/' + 9i^2/7a^.
'^IS. lla^13 + 4a;3 + 5a;, 14a; + 2a^47 + 12a^,
Sx'3x10{6x% and 1 15cc2 + 9a; 16 a^l
19. a2a, ia^ + a + , and Ja2_ e ^_^.,
•^20. 5mhin^4.m^\2m7i% 7 mn^ lSm^n + 2m^97i^,
 15 mw^ + 3 m^n + 16 71^ + 8 m^
and — 5 m^ + 3 mii^ — 6 ii^ + 10 m^w.
i^l. 6n3 + 2711215 7^2, 14 + 7nn297i3,
6?i2_l3n3^3_ll^^ and 8167i + 107i2 + 4wl
i
SUBTRACTION OF MONOMIALS
47. The remainder when 6 is subtracted from a is expressed
a6(§4).
We will now show how to subtract — 6 from a.
By § 34, the sum of the remainder and the subtrahend equals
the minuend.
Then, the required remainder must be an expression such
that, when it is added to — 6V^he sum shall equal a.
But if a + 6 is added to — 6, the sum is a (§ 35).
Therefore, the required remainder is a + 6.
That is, a(6) = a + 6.
48. From § 47, we have the following rule :
To subtract a monomial, change its sign, and add the result to
the minuend.
ADDITION AND SUBTRACTION 25
1. Subtract 5 a from 2 a.
Changing the sign of the subtrahend, and adding the result to the min
uend,
2 a  5 rt = 2 a + ( 5 a) = 3 a (§ 41).
2. Subtract — 2 a from 5 a,
5«^(2a) = 5a + 2a = 7a.
3. Subtract —b a from —2 a.
2a(5a)=2af5« = 3a.
4. Subtract 5{x\y) from —2(x\y).
 2(a + ?/) 5(x + ?/) =  7(x + ?/).
The pupil should endeavor to put down the results, in examples like the
above, without writing the intermediate step ; changing the sign of the
subtrahend mentally, and adding the result to the minuend.
5. From — 23 a take the sum of 19 a and —5 a.
It is convenient to change the sign of each expression which is to he
subtracted, and then add the results.
We then have — 23 a — 19 a + 5 a, or — 37 a.
EXERCISE 9
Subtract the following :
1. 9 from 3. 4. 5 from 12. 1, 3 from 3^.
V2. 2 from  6. 5. 42 from 15. 8.  f from  f .
3.  16 from  10. v'6. _ 28 from  61. 9. lOf from  3f
10. 11. vl2. 13. ^li.
14 a 4 a; —^a^ — 15 mn — 7 x'y
8a 11a; 4a^  1 mn — 12 x'y
15. 5 he from he. 19. 19 {ah) from 17 {ah).
16. a;2/2 from  8 a;?/2. 20.  18 a%(^ from  45 a^ftc^.
17. 25 aV from 13 a'x^ ' 21. From 7 x take  11 .y.
18.  40 ahc from  23 ahc. 22. From 2a^ take 5 w^.
26 ALGEBRA
23. From the sum of 18 ab and —9ab take the sum of
— 21 ab and 11 ab.
^'24. From the sum of —13n^ and 24: n^ take the sum of
46 n^ and — 19 n^.
V^ 25. From the sum of 16 xy^ and — 37 xy^ take the sum of
— 29 iC2/^, 34 a;?/^, and — 47 iC2/^.
J\ SUBTRACTION OF POLYNOMIALS
49, Since a polynomial may be regarded as the sum of its
separate terms (§ 38), we have the following rule :
To subtract a polynomial, change the sign of each of its terms,
and add the result to the minuend.
1. Subtract 7 a^^ _ 9 ab + 8 6^ from 5 a^  2 a^^ + 4 ab\
It is convenient to place the subtrahend under tlie minuend, so that
similar terms shall be in the same vertical column.
We then mentally change the sign of each term of the subtrahend, and
add the result to the minuend ; thus,
, 5 a3 _ 2 a26 + 4 a62
 9 a% + 7 a62 ^_ 8 53
5a3 + 1 a^b  3 a&2 _ g fts
2. Subtract the sum of 9x^ — %x\x^ and ^ — a?\x from
Q,^lx4.
We change the sign of each expression which is to be subtracted, and
add the results. ^ „ „ m
Qx^ — 7 a; — 4
 x^  9 a;2 + 8 x
+ a;2  a;  5
6x38x2 Tq
EXERCISE iO
Subtract the following:
1. V 2. 3.
a^ 4 13 oj — 11 — 2 m^ — 4 mn 4 9 n^ a6 } 6c + ca
— 3ar' Q>x — 5 8m^ — 7 m?i +. 14 n^ ah — bc{ ca
ADDITION AND SUBTRACTION 27
4. From Sx + 2y — 7z subtract 8x — 2y + 7z.
5. From Aa^ 5a^ 15a6 take a^ 12a^ 3a + ll.
6. From 7a — 9c — 6 subtract — 5c + 12a — 8 6.
7. Subtract —5(x+y)\9(x—y) from 7 (x+y)—6{x—y).
8. Take 49 x^ + 16 m^  56 mx from 25 m^ + 36 a^  60 mx.
9. By how much does 15 ar^ + 6 a^y — 4 a??/^ — 11 2/^
exceed Sa^9x^y iUxy' Sf?
•lO. Take 8a^12a26 46a6263
from a''  6 a'b + 12 a^^ _ 8 b^
V 11. What expression must be added to 3 a^— a; + 5 to give ?
12. By how much does 2 m — 4 m^ — 15 + 17 m^
exceed — 9 + 6 m^ — 11 m — 14 m^ ?
13. From a; + 15 ar^ 18 subtract  2 ar^ 13 + 41 a^.
14. Take 3&16d + 7a10c from 13chl4 a5d^9 6.
15. Subtract 12x — 7 7i — 6y from 11 n + 3 m — 8 x.
^16. From 7 7r5+20n3+13n take 914 n^ +16n+5 Til
17. From . ^^ _ _i_ 5 .f 1 9 c subtract ia + & — c.
18. Subtract 15a21a2 + 17 from  12a^ + 22a39a
19. Take a^ea^ i5a2_8a + 4
from 7a^ + 3a''5a2lla9.
20. From im — Jn + ^p take fm — J?i + Jp.
^1. From 71^  10 A'  nV + 8 nar5 + 3 a;^
take 5n^ + 4n3a;9n2aj2 + 2wa.'312a;*.
22. Take 18a;^8a. + 6a^4128a.'3
from 10a;3^215a;2 + llar'4a;.
23. Take a^  1 a%^ + 13 a'b^  7 a?/  5 6^
from 9 a^ + 3 a^6 H 6 cfb'  arb^  16 6«.
4. From the sum of 2x^—bxy y'^ and 7 y? — 3xy\9y^
subtract 4 a^ — 6 a;?/ + 8 ?/^. .V \
V 25. From subtract the sum of 4 (jl and 3 a — 5 a^ — 1.
V
28 ALGEBRA
26. From 7 x — 5z — 3y subtract the sum of S y \ 2 x —11 z
and 6 z — 12 y^4:X.
27. From 671^ — 671 — 11 subtract the sum of 2 n^ — 4 n — 3,
7w210?i + 4, and 3n2 + 8n12.
• 28. From the sum of 36 + 2 a — 4c and 9G\3b — 5d
subtract the sum of —6 d — 7 a
and 8a — 7d + 96 + 5c.
29. From the sum of 4.al\5a^Sa% 119 a" +3 a^ 7 a,
and 3a^7+10aa^
\ subtract 4.a^{9 aQa' + 2.
30. From the sum of 7 a^4:X^\6x and 3x^10x5
take the sum of— 5a^f4a;412
and 8a;3_ii^2_2.
PARENTHESES
50. Removal of Parentheses.
By § 45, a + (6 — c) = a + & — c.
Hence, jparentlieses preceded by a \ sign may he removed
without changing the signs of the te7'ms enclosed.
Again, by § 49, a — {h — c) = a — h{G.
Hence, pa7entheses preceded by a — sign may be I'emoved if
the sign of each term enclosed be changed, from 4 to — , andfi^om
— to +.
The above rules apply equally to the removal of the brackets,
braces, or vinculum (§ 12).
It should be noticed in, the case of the latter that the sign
apparently prefixed to the first term underneath is in reality
prefixed to the vinculum ; thus, ja — b means the same as
4 (a — b), and —a—b the same as — (a — b).
61. 1. Remove the parentheses from
2 a 3 6 (5 a 4 6) + (4a 6).
By the rules of § 50, the expression becomes
2a365a + 46 + 4a6 = a.
ADDITION AND SUBTRACTION 29
Parentheses sometimes enclose others ; in this case they may
be removed in succession by the rules of § 50.
Beginners should remove one at a time, commencing with the
innermost psiiv, but after a little practice, they should be able
to remove several signs of aggregation at one operation, in which
case they should commence with the outermost pair.
2. Simplify 4a; S3ajf(2a;a;a)S.
We remove the vinculum first, then the parentheses, and finally the
braces.
Thus, 4:X{Sx\ {2xxa)}
= 4x{Sx + {~2xx + a)}
= 4:X — {Sx — 2x — gc{a]
= 4:X — Sx + 2x + x — a = 4:X — a.
EXERCISE II
Simplify the following by removing the signs of aggregation,
and then uniting similar terms :
1. 9m + (— 4?M46n) — (3 m — n).
2. 2xSyl5x\y^{lSx~7y\.
v^3. a62cf 2abc fa  2 6  c.
4. 4:y^23^l4.x^7xy{5y^^ + (Sa^9xy).
5. 3a^5ab{^a'{2ab9b^7a'6ab + b\
6. 5a(7a[9a + 4]).
^7. 7xlSy10xlly}.
♦8. 6 m?i 4 5 — ([ — 7 m7i — 3] — f — 5 mTi — 11 J)
9. 8a29(5a23a + 2) + (6a24a7).
10. 2x(Sy\5x5xy)(9y^3x).
»^11. 25 (8 [341647]).
12. 7a:(5a;[12a; + 6xll]).
1/13. 2a(36 + c Ja6S)(3a + 2c[r26 + 3c]).
30 ALGEBRA
14. 5m[7m J3 m 4 wi + 9 6 m8],
15. 37 [41 513 (5628 + 7) j].
16. 9 m — (3 ?2 + j 4 m — [71 — 6 ??i] \ — [m + 7 n]).
17. 2a+[6& S3c + (466c + a)].
18. 7x(6xl5x[4.x3x2^l).
19. 5 7i[8?i(37i + 6) J6 7i + 7 7i5J].
20. 4a[a 57a(8a5a + 3)(6a2a9)].
21. a;Sll2/[2.T(42/{7aj52/J6aj92/)]i.
22. 3a[5(457c)S2a(365c)66 + cn. ■
^23. 2a;[4a^{5aj(aj7i» + 6)j+(3ic8a;9)].
52. Insertion of Parentheses.
To enclose terms in parentheses, we take the converse of the
rules of § 50.
Any number of terms may be enclosed in parentheses preceded
by a \ sign, without changing their signs.
Any member of terms may be enclosed in parentheses preceded
by a — sign, if the sign of each tei^m be changed, from { to —,
or from — to ■^.
Ex. Enclose the last three terms of a — b + c — d\e\n
parentheses preceded by a — sign.
Result, a — h — {— c \ d — e).
EXERCISE 12
In each of the following expressions, enclose the last three
terms in parentheses preceded by a — sign :
1. a — b — c\d. \,5.4:Q^ — y^ — 2yz — z\
9^2. m3 + 2m_2f 3m + 4. 6. a^ { b  (r + d\
3. X' h x^y — xy^ — y\ 7. x^ — 2xy\y^\3x~4: y.
\/8. ?2^5w38?i2 + 6m + 7«
ADDITION AND SUBTRACTION 31
9. In each of the above results, enclose the last two terms in
parentheses in brackets preceded by a — sign.
53. Addition and Subtraction of Terms having Literal Coeffi
cients.
To add two or more terms involving the same power of a
certain letter, with literal, or numerical and literal, coefficients,
it is convenient to put the coefficient of this letter in paren
theses.
1. Add ax and 2 x.
Bj § 40, ax\2x = (a + 2)x.
2. Add (2 m + n)y and (m — 3 n)y. ' "^
{2 m h n)y + (m — 3 n)y = [(2 m + w) + (w — 3 n)'\y
= (2 m + w + TO  3 w)?/(§ 50) = (3 m  2 w)y.
(The pupil should endeavor to put down the result in one operation. )
3. Subtract (h — g)oi? from a^.
By § 48, ax2  (h  'c)x^ = [a{b c)]x2
= (ah + c)x'' (§ 50).
EXERCISE 13
Add the following :
1. ax and bx. 4. mx, —nx, and —px.
5. aV and (ab — b^)x^.
V6. (3 a + 4 b)n and (5 c7 d)n.
V2.
mx^ and —2a^.
3.
— mny and —pqy.
Subtract the following :
^^.
2 bx from 3 ax.
8.
— mny from aby.
9. — 7ixy from — axy.
10. {p + q)x from mx.
V 11. (2 a  3 6)2/2 from (5 a  4 b^f.
32 ALGEBRA
IV. MULTIPLICATION OF ALGEBRAIC
EXPRESSIONS
54. The Rule of Signs.
If a and h are any two positive numbers, we have by § 25,
(+a)x(+6)= + a5, (.^a) x (6) = a6,
{—a)x{\h) = — ab, (_«,) x (— &) = + a6.
From these results we may state what is called the Rule of
Signs in multiplication, as follows :
V TJie product of two terms of like sign is positive; the product
of two terms of unlike sign is negative.
55. We have by § 54,
(— a) x{—b)x (— c) = (ab) x (— c)
= — abc ; (1)
(—a)x{—b)x(—c)x(—d)=(—abc)x(—d), by (1),
= abcd; etc.
That is, the product of three negative terms is negative ; the
product of four negative terms is positive ; and so on.
^ In general, the product of any number of terms is positive or
negative according as the number of negative terms is even or odd.
56. The Law of Exponents.
Let it be required to multiply a^ by al
By § 11, a?=zaxax a,
and a^ = axa.
Whence, a^ x a^ = a X a x a x a x a — a\
We will now consider the general case.
Let it be required to multiply a"* by a", where 7n and n are
any positive integers.
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 33
We have a™ = a x a X • • • to m factors,
and a'' = axax " to n factors.
Then, a"* X a'' = a x a x • •• to m 4 71 factors = a*"+**.
(The Sign of Continuation^ •••, is read '■'•and so on^)
^^ence, the exponent of a letter in the product is equal to its
exponent in the multiplicand plus its exponent in the multiplier.
This is called the Law of Exponents for Multiplication.
A similar result holds for the product of three or more
powers of the same letter.
Thus, a^xa''xa' = d^+^+^ = a^.
MULTIPLICATION OF MONOMIALS
57. 1. Let it be required to multiply 7 a by —2 b.
By §31, ^2b=(2)xb.
Then, 7 a x ( 2 &)= 7 a x ( 2) x 6
= 7 x(2)x a X 6=14a6. (§54)
In the above solution, we assume that the factors of a product can be
loritten in any order.
This is called the Commutative Laic for Multiplication ; its proof for
the various forms of number will be found in § 453,
2. Eequired the product of — 2 a^b^, 6 ab^, and — 7 a^c.
( 2 a263) X 6 a?>5 X ( 7 a'^c)
= (2)a253x 6a&5x(7)«4c
= (— 2) X 6 X (  7) X a2 X Of X a* X 6^ X 6^ X c
= 84 aWc, by §§ 55 and 56.
We then have the following rule for the product of any
number of monomials :
^^ To the product of the numerical coefficients (§§ 30, 31, 55, 56)
annex the letters ; giving to each an exponent equal to the sum of
its exponents in the factors.
34 ALGEBRA
3. Multiply 5a%\)y S ah\
(  5 a%) X (  8 ah^) = 40 a^+i^i+a ^ 40 a*^*.
4. Find the product of 4 n^, — 3 71^, and 2 n^.
4 w2 X ( 3 7z6) X 2 w* =  24 ^2+6+*.=  24 w".
5. Multiply — ic"* by 7 a;^.
6. Multiply 6 (?)i + ny by 7 (m + nf
6 (m + ny X 7 (?n + w)^ = 42 (m + n)"^.
c
EXERCISE 14
Multiply the following:
1. 9 aj"^ by 4 ie2. v" 9. 9(a + &)' by 6(a + 6f.
2. 8a^6by7a6^ 10.  6 a*a^2/^ by 11 ccY^^
V 3. 11 aa; by  3 61/. •^l.  2 a^^sn by  5 a^h'''.
4.7 0^2/^ by — 9 aj^?/ 1^12. 14 a^^?/"* by — 8 aj*?/*"
^ 5. 15 6V by 2 a^^a, 13^ 4 ^^3^ _ 7 ^5^ ^nd  3 m^
V 6.  a;'"?/"^' by ^yh. ^ 14. 2 a^ 6 6^ and  8 c«.
7. 13(a; — 2/) by —(x — yf. 15. a^5', 6*c"», and cV^.
^8.  5 a^6V by  12 a%^(?. 16.  5 a^y,  9y^V, and ^'a;
17. 2 a)^, — aj^, 6 x'^, and 4 a;^.
V 18.  3 a%  5 h%  2 c^a, and  a^&V.
/ 19. 3 m^TiV, — 4 m^7iy, — 5 m^y^y'^, and 6 A'V.
20. a^m^?^ _ ^3p^r^ _ ^n^2^ and  ft^c.
21. m^n^^ — 2 ma;^, 3 mhf, — 5 7iV, and — 4 n^?/^
MULTIPLICATION OF POLYNOMIALS BY MONOMIALS
58. In § 40, we assumed that the product of a + 6 by c was
ao, H he.
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 35
We then have the following rule for the product of a poly
nomial by a monomial :
V Multiply each term of the multiplicand by the multiplier ^ and
add the partial products.
Ex. Multiply 23?bx + l by 8a^.
(2a;2_5x+7)x(8a;3)
= (2:*:2)x(8x3)4(5a;)x(8a;3) + (7)x(8x8)
The student should put down the final result in one operation.
EXERCISE 15
Multiply the following :
/I. 5a;12 by 1 x. 5. Sx^ by Gar^ + SicIT.
2. lOa^h + 1 ah' by Qab\ ^ 6. 4.a%^ by 3d'2ab4.h\
w/3. a;* — 4 x^y^ + 4 2/^ by  xh/ 1 7. 7 x'^y^ — 8 x^y^ by — 3 xhf.
,4. 8m*m23 by 5m^ ^8. 6a^4a^5a« by 9a^
k 9. — m^n + 8 ri^ — 3 m^ by — 12 m^n^.
^ 10. 2 a^54 ^j ^s _Qa'b\ 12 a^^ _ g 53^
HI. 97i2_62n^5?i + 7i^ by 3n^
»^12. 5ar^2i»27/__4ir?/232/^ by lla^?/
4muLTIPLICATI0N of POLYNOMIALS BY POLYNOMIALS
59. Let it be required to multiply a{b by c + d.
As in § 40, we multiply a{b by c, and then a\b by d,
and add the second result to the first; that is,
(a + 6) (c + d) = (a + b)c + (a + &)d
= ac \ be ^ ad \ bd.
We then have the following rule :
Multiply each term of the midtiplicand by each term of the
multiplier, and add the partial pi'oducts.
36 ALGEBRA
60. 1. Multiply 3a 4 6 by 2 a 5b,
In accordance with the rule, we multiply 3 a — ib by 2 a, and then by
— 5 &, and add the partial products.
A convenient arrangement of the work is shown below, similar terms
being in the same vertical column.
3a 4&
2a 5&
6a^ Sab
 15 a6 + 20 62
6 a2 _ 23 a& + 20 &2
The work may be verified by performing the example with the multi
plicand and multiplier interchanged.
2. Multiply ^a3(^ + a^8x^2a^x by 2x + a.
It is convenient to arrange the multiplicand and multiplier in the same
order of powers of some common letter (§ 43), and write the partial
products in the same order.
Arranging the expressions according to the descending powers of a, we
^^^® a32a2x + 4aa:28a;3
a +2x
a*  2 a^x + 4 a'^x^  8 ax^
2a^xi a2.r2 + Sax^ 16 a;*
a* 16x*
EXERCISE 16
Multiply the following :
VI. 5ir7 by 3a; + 2. V4. lOxy + ShjSxyA.
2. Sm\n hj 8m\n, 5. m^ — m — 3 by m + 3.
3. 2aS hj 6a7, v 6. a^ a12 hj a7.
7. 4(a_6)_3by4(a&)+3.
^S. x^2xij\3y^hj x3y.
9. 4m2+ 9n26mwby 37i42m.
10. ia_i6by iai6.
»/ 11. a? — 4 y by a;2 + 4 a;?/ 4 16 2/".
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 37
12. a + 6 + c by a — 5 — c.
13. 5 m^ + 3 m — 4 by 6 r/i^ + 5 m^.
14. 8  4 ?i + 2 71^  ^3 by 2 + 7i.
15. 2a'8a\5hy a\a2.
16. 6(m + ^)  5(m + w) + 1 by 7(m + n)2.
17. 2a^3i»25ajlby 3a;5.
18. 6ic + 2a;2 + 8by 4 + a^2_3^_
19. 2 71^ + m^ + 3 mn by 2?^^ — 3 m7i + m^.
120. ^x'ix + ^\ by fa^ + .
21. 4 a^ + 6 a  10 by 2 a  3 a + 5.
v22. 9a7 + 2a^5by4 + 3x'7ic.
23. 10 n^ f 3 n  4 by 9 71^ _ 5 n  6.
''24. x^p+^y — x^y'i by x^p'^ { y'^\
25. a'^ + 2 a^d + 2 ab' + 6^ by a^ 2abj h\
26. m^3m''' + 97^2_27 ^^2,^81 by m + 3.
\ 27. 3(a4^)'2(a + ?>)+l by 4(a + 5)'(a + 6)45.
28. 3 + aV7a4a2by a + a27.
29. 8 m^ + 12 mhi + 18 m^i^ + 27 r^^ by 2 ^^i^^i  3 mnK
30. 4 a'"+^&2 _ 3 ^4^„ i^y ^m+25 _ 2 «^,"i.
V31. a32a2 + 6a5by a2a + 10.
32. 5i«*6a^4a^ + 2i«3by 3a.'2.
V33. 4 m^ + 6 in'n  5 m^i^  3 n^ by 3 ^ti^ + 2 mw  wl
^ 35. mx + m?/ — nx — ny by ma? — my + nx — ny.
36. a»3a2aj + 3a£c2a^by a3 + 3a2a; + 3aic2_^aj3.
37. a^6a;2/ + 9/ by ar^9a^2^ + 27 0^2/227 2/3.
^ 38. a"* + ^'^ — e by a"" — 6" + e.
38 ALGEBRA
39. 2n33n2n + 4b.y 2n33?i2 + n4
40. 5a^7 + 2x'Sxhj 4:^3x^5x,
41. 5a* + a^2a26a + 3by2aa6.
42. Im'^^ml by \m + \m^'
43. a + 3, a + 4, and a — 5.
44. a;6, 3aj2, and4a; + l.
45. m{2n, w? — 2 mn + 4 w^, and m^ — 8 v?,
46. 4cm — 7, 5m — S, and 6 7?^ — 5.
^47. a;42, .T — 3, i»5, and i» + 6.
48. a + 26, 3a4&, and3a22a6862.
49. 2x + y,2x — y,4.x^ + if, and 16 x^ + ^/^
50. 2 m \ o n, 2 m — 3 n, Z m \2 n, and 3 m — 2 w.
51. ^2 + 71+2, w2n + 2, andn* + 3n24.
752. a2, a + 3, 3al, and 3 a^2 a^ 19 a6.
61. If the product has more than one term involving the
same power of a certain letter, with literal, or numerical and
literal, coefficients, we put the coefficient of this letter in paren
theses, as in § 53.
Ex. Multiply x^ — ax — hx + ah by x — a.
x'^ — ax — hx {■ db
x—a
x^ — ax? — hx'^ + ahx
— ax^ + a'^x + ahx — a'^h
a:3 ^ (2 a + 6)x2 + (a2 + 2 ab^x  cfib
As in §53,2 ax"^  bx^ is equivalent to  (2 a + b)x^, and a^ + 2 abx
to (a2 + 2 ab)x.
EXERCISE 17
Multiply the following :
yl. a^ + ax + bx + ab hj x\c.
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 39
^2. x^ — mx { nx — mn by x—p.
3. x^ — hx — cx\hc by x— a.
4. x^ + ax — hx — Z ah hj X \ h.
y b. x^{ax + 2hx\2 ab hj x — G.
6. x^ {px — 5 qx — 5pq by x — r.
7. 0/'^ — 3ax — bx\3ab by cc + 2 a.
8. x^ — 4: mx \nx — 4: mn by x\Sn.
^d. x^^3ax2bx + 6ab hj x4:C.
10. (a6>3a& by 2a;(a6).
11. x^^ — B ax"" + 4 bx'' — 2ab by x" + c.
V/ 12. (2 a  l)a;2 + (a + 2)x  (a + 3) by (a  2)x  a.
yC»62. Ex. Simplify (a2 xY2(S aix)(ax).
To simplify the expression, we first multiply a — 2x'by itself (§ 11) ; we
then find the product of 2, 3 a + x, and a — x, and subtract the second
result from the first.
a —2x Sa + X
a —2x a — X
2 ax 3 o2 4 Ota;
2 ax + 4 jc2 Sax 
a^4:ax + ix^ Za^2ax x^
2
6a'^4:ax2x^
Subtracting the second result from the first, we have
a"^  iax + ix"^  Q a^ h 4:ax + 2x^ =  6 a'^ ^ 6x^,
EXERCISE 18
Simplify the following :
1. (3a + 5)(2a8)4(4a7)(ah6).
v2. (Sx{2)(4.x + S)(Sx2)(4.xS),
v/3. (a2 x)(b + 3y) + (a{2 x)(b3y).
40 ALGEBRA
4.
5.
6.
^ 7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
^17.
^ 18.
*19.
20.
• 21.
22.
23.
24.
3m + iy{Smlf.
2a + 3 6)24(a6)(a + 5 6).
3x{5yj2z)][_3x~(5y2z)].
711 + 2 n — (2 m — n)] [2 m + n — (m — 2 n)].
a{b){c'(abcy.
a + 2)(a + 3)(a _ 4) + (a  2)(a  3)(a + 4).
i^iy + i^y
2x'+(3xl)(4.xi5)^l5x'(4:X + 3)(x2)'].
a + 2bc3 ay.
a  2)(a + 3)  (a  3)(a + 4)  (a  4)(a + 5).
a; + 2)(2 X  1)(3 05  4)  (oj  2)(2 a.' + 1)(3 x + 4).
^  (2/  ^)] [2/  (^  2=)] [2=  (» 2/)]
a  5)(a3 4_ 6•3^) (^(^ _^ ^) _^ ^^2]^
a^b2cy(b\G2ay+(c + a2by.
x+y+zy+(xyzy+(x+yzy+{xy+zy.
2xiiy\(2xiy.
a\b + c)(ab \bc + ca) — (a + b)(b + c)(c + a).
a + 2&)22(a + 25)(2a + &) + (2a + Z')'.
a; + 2/ + 2;)=^  3 (2/ + 2;) (2; + a^) (oj + 2/) .
a\by + 3(a\by(ab)\3(a + b)(aby + (aby
DEFINITIONS
63. A monomial is said to be rational and integral when it
is either a number expressed in Arabic numerals, or a single
letter with unity for its exponent, or the product of two or
more such numbers or letters.
Thus, 3 a^W, being equivalent to 3 • a • a • 5 • 6 • &, is rational
and integral.
MULTIPLICATION OF ALGEBRAIC EXPRESSIONS 41
A polynomial is said to be rational and integral when each
terra is rational and integral ; as 2 ic^ — a6 + cl
64. If a term has a literal portion which consists of a single
letter with unity for its exponent, the term is said to be of the
fifbt degree.
Thus, 2 a is of the first degree.
The degree of any rational and integral monomial (§ 63) is
the number of terms of the first degree which are multiplied
together to form its literal portion.
Thus, 5 a& is of the second degree ; 3 a^W, being equivalent
to 3 • (X • a • & • 6 • &, is of the fifth degree ; etc.
The degree of a rational and integral monomial equals the
sum of the exponents of the letters involved in it.
Thus, a6V is of the eighth degree.
The degree of a rational and integral polynomial is the
degree of its term of highest degree.
Thus, 2 a^6 — 3 c + d^ is of the third degree.
65. Homogeneity.
Homogeneous terms are terms of the same degree.
Thus, a^^ 3 bh, and — 5 x^y'^ are homogeneous terms.
A polynomial is said to be homogeneous when its terms are
homogeneous ; as a^ + 3 &^c— 4 xyz.
66. If the multiplicand and multiplier are homogeneous, the
product will also be homogeneous, and its degree equal to the
sum of the degrees of the multiplicand and multiplier.
The examples in § 60 are instances of the above law ; thus,
in Ex. 2, the multiplicand, multiplier, are homogeneous, and of
the third, first, and fourth degrees, respectively.
The student should always, when possible, apply the prin
ciples of homogeneity to test the accuracy of algebraic work.
Thus, if two homogeneous expressions be multiplied together,
and the product obtained is not homogeneous, it is evident that
the work is not correct.
42 ALGEBRA
V. DIVISION OF ALGEBRAIC EXPRESSIONS
67. We define Division, in Algebra, as the process of finding
one of two numbers, when their product and the other number
are given.
The Dividend is the product of the numbers.
The Divisor is the given number.
The Quotient is the required number.
68. The Rule of Signs.
Since the dividend is the product of the divisor and quotient,
the equations of § 54 may be written as follows :
±^=+b, =^ = + h, =^ = 6, and +^ = 6.
+ a —a \a —a
Froijd these results, we may state the Rule of Signs in divi
sion, as follows :
Tlie quotient of two terms of like sign is positive; the quotient
of two terms of unlike sign is negative.
69. Let  = aj. (1)
Then, since the dividend is the product of the divisor and
quotient, we have a = bx
Multiply each of these equals by c (Ax. 7, § 9),
ac = hex.
Eegarding ac as the dividend, be as the divisor, and x as the
quotient, this may be written
be
(2)
From (1) and (2),
f5 = ?. (Ax. 4, §9)
be b
(3)
That is, a factor common to the dividend and divisor can be
removed, or cayicelled.
DIVISION OF ALGEBRAIC EXPRESSIONS 43
70. The Law of Exponents for Division.
Let it be required to divide a^ by al ^
By § 11,
w a X a
Cancelling the common factor « x a (§ 69), we have
— = a xaxa = a^.
a"
We will now consider the general case.
Let it be required to divide a"* by a", where m and n are any
positive integers such that m is greater than n.
■VKT X. a"* _ ax axa X " to m factors
a'* a X a X a X • • • to ?i factors
Cancelling the common factor axaxax"ton factors,
— = axaxax •••torn — n factors = a"""".
Hence, the exponent of a letter in the quotient is equal to its
exponent in the dividend, minus its exponent in the divisor.
This is called the Laio of Exponents for Division.
DIVISION OF MONOMIALS
71. 1. Let it be required to divide — 14a"6 by 7a^.
Cancelling the common factors 7 and a^ (§ 69), we have
Then to find the quotient of two monomials :
To the quotient of the numerical coefficients annex the letter Sj
giving to each an exponent equal to its exponent in the dividend
minus its exponent in the divisor, and omitting any letter having
the same exponent in the dividend and divisor.
44 ALGEBRA
2. Divide 54 a'hh^ by  9 a'h\
54 a563c2
 9 a*63
3. Divide —2x^'^y''z'' by — a;'"?/V,
=  6 a54c2 =  6 ac2.
4. Divide 35 (a  6)^ by 7 (a  b)\
EXERCISE 19
Divide the following :
1. 30by5. 4. 64by8. 7. ^by^.
2.  42 by 6. 5.  135 by  9. 8. 21 a^' by 3 al .
V 3.  48 by  4. ~ 6. 176 by  11. 9. 63mV by 7mV.
10. 6 a^2/'° by  x^'  18.  28 a'b'c' by 2 6V.
'•11. 9 (a 6/ by 3 (a 6)2. 19. a^+^b^+^ hj  ab\
12. xy'z^ by 'rc^/^. V 20.  55 afyh''' by  11 2/V.
13.  13 ?/!%•' by  13 ?/i5?i^ 21.  70 a^6V by 14 ab^c\
14. 45(a; + 2/y by 5(a; + ?/y. 22.  32 o.'^^^/'"^' by  8 a; V
15. 72a;yby6a;y. 23.  96 x'^'Y ^J ^^ ^""^Y
, 16.  40 a'b'c' by  8 be. 24. 52 a^^^e^ia by _ 4 a'b(f.
17. 90a"a:«by 9aV. 25. 132 a^y^;!^ by 12 a^y^^.
Find the numerical value when a = 2, & = — 4, c = 5, and
(« = 3of:
2Q 10 ab Sac 28 '^« + 14& — 12c
cc? 6d ' ' 13 a  9 6 + 17 c'
rtty 2 ft — 6 _ ct 4 4 & 29 a — & , 6 — c c — (f
5d 3c4ci a + 36 & + 5c c + 4(^
DIVISION OF ALGEBRAIC EXPRESSIONS 45
DIVISION OF POLYNOMIALS BY MONOMLAXS
72. We have, (a + h)c = ac\ he.
Since the dividend is the product of the divisor and quotient
(§ 67), we may regard ac + he as the dividend, c as the divisor,
and a 4 6 as the quotient.
Whence, ^^±i2 = a4&.
c
Hence, to divide a polynomial by a monomial, we divide each
term of the dividend hy the divisor, and add the results,
Ex. Divide 9 a^h^ ^^"0 + 12 a^hc^ by  3 a\
— 3 a2
EXERCISE 20
Divide the following :
V 1. 25a«15a« + 40a^ by 5a^
"> 2. — 24 mV + 33 mn^ by — 3 mn^.
V 3. S6 xhjz^ 9 xYz^ 27 a^ifz'hy 9 a^y.
V ' 4. 54 a'b'c^  60 a^b^c^ by 6 ab'c'.
K 5.  22 x^V + 30 xY + 26 xy by  2 xy.
6. 70ni356w"63w9 + 49?/by7 7i^
* 7. 66 a;'^?/^ + 77 xy^z — 55 xyz^ by — 11 xyz.
8. 36ai^ + 28ai2_4^9_20a6by4a«.
V 9. ic^+?2/"+2 — a;«+y by a^/.
10. 14 m^n2  28 m^7i^ + 28 m^w^  14 wiV by  14 mV.
• 11. 32a;i5 + 24.'»^3_43^n_40^9]^y _8aj8^
12. 84 oi^y^z^  108 icy^s _ 48 xYz' by 12 xyz^
'^IS. a^pfe'c^'  a^&^c^'  a^^6^^c^'' by a^¥c'\
"^ 14. 30 a^^mhi^ — 60 a^m^n^ — 45 ci^ri^ by — 15 a^n^.
46 ALGEBRA
DIVISION OF POLYNOMIALS BY POLYNOMIALS
73. Let it be required to divide 12 + 10 oj^  11 a;  21 a;^ ^^
Arranging each expression according to the descending
powers of x (§ 43), we are to find an expression which, when
multiplied by the divisor, 2 a^ — 3x — 4:, will produce the
dividend, 10 x^  21 cc^  11 a; + 12.
It is evident that the term containing the highest power of
X in the product is the product of the terms containing the
highest powers of x in the multiplicand and multiplier.
Therefore, 10 x^ is the product of 2 x^ and the term contain
ing the highest power of x in the quotient.
Whence, the term containing the highest power of x in the
quotient is 10 x^ divided^ by 2 x", or 5 x.
Multiplying the divisor by 5 x, we have the product
10 0^—15x^ — 20 X', which, when subtracted from the divi
dend, leaves the remainder —6x^\9x\12.
This remainder must be the product of the divisor by the
rest of the quotient ; therefore, to obtain the next term of the
quotient, we regard — 6a^f9aj412 as a new dividend.
Dividing tlie term containing the highest power of x, — 6x%
by the term containing the highest power of x in the divisor,
2 a^, we obtain — 3 as the second term of the quotient.
Multiplying the divisor by — 3, we have the product
— 6 a.'^ + 9 a; + 12 ; which, when subtracted from the second
dividend, leaves no remainder.
Hence, 5 a; — 3 is the required quotient.
2 a^ — 3 a; — 4, Divisor. ■ 
10aj321aj2lla.'12
10a^15a;220'a;
5 a; — 3, Quotient.
 6a^f 9a; + 12
 6x^{ 9a; + 12
The example might have been solved by arranging the dividend and
divisor according to the ascending powers of x.
From the above example, we derive the following rule.
DIVISION OF ALGEBRAIC EXPRESSIONS 47
Arrange the dividend and divisor in the same order of powers
of some common letter.
Divide the first term of the dividend by the first term of the
divisor, and ivrite the result as the first term of the quotient.
Multiply the ivhole divisor by the first term of the quotient, and
subtract the product from the dividend.
If there be a remainder, regard it as a new dividend, arid,
proceed as before; arranging the remaiyider in the same order
of powers as the divideiid and divisor.
1. Divide ^ab"" + a^^W^ab by ^W{d'2ab.
Arranging according to the descending powers of «,
^3 _ 5 ^25 _,. 9 ^52 _ 9 ^,3 f. 05? _ 2 a& 4 3 62
Sb
3 a^b + 6 a62
Sa^b + 6a&2_9^3
In the above example, the last term of the second dividend is omitted,
as it is merely a repetition of the term directly above.
The work may be verified by multiplying the quotient by the divisor,
which should of course give the dividend.
2. Divide 4. + 9x^2Sa^ by 3x^ + 2+4:X.
Arranging according to the ascending powers of x,
2 4 4 X  3 x2
4  28 a;2 + x4
4+ 8 a;  6x2
2  4 X  3 ic2
 8 X  22 X + 9 X*
 8 X  16 x2 4 12 x3
6 x2  12 x3 + 9 X*
6 x2  12 x3 4 9 x*
EXERCISE 21
Divide the following :
1. 6a^ + 29a435 by 2a{5.
^2. S0x'53x\S by 6xl.
3. 32a.2 + 28i»2/15/ by 4.x{5y.
»^4. 10a3 + 33a2_52a + 9 by 5al,
48 ALGEBRA
5. a^Sb^ by a2b.
■ 6. 2Sn^ + S4:n12 by 7n\2.
7. 64a^ + 27/ by 4.x\3y.
V 8. 6(xyy7(xy)20 by 3(a;2/)+4.
9. 25m^n^ — S6 by 6 + 5mn.
via 25«3^ + 12a;^ + 12a.'2/ by Sx'Axy.
11. 18m317 maj26a^ by 3m42ic.
12. 27v'6\5n^19n hj Sn{5n^3.
13. 12{13x^19x12x' hy Sx^4.\x,
14. x'^ — y — 2yz — z^hjx — y — z.
15. 8(a + &)^c^ by 2(a + 6)c.
16. 8 m'^/i — 24 m^n^ + 12 ?n^ — 31 m^7i^ by 6m^ — 8n^ — 5mn.
17. 8a + 9a^l16a2 by l3a24a.
18. m^ — ^m — I by m + ^«
vl9. 2a^106a^{x^{llx hj 2 + x'^x.
20. a;^81 by a;33 a;2+ 9 a;27.
v'21. a^256 6^ by a 4 6.
22. m^ — 7 mV + w^ by m^ + 3 mw + n^.
23. 81a;^l by l+3aj.
24. 6a3 + 6a^8a23a2 by 2a + 3.
^25. {x^yf9{x\yy + 27(x\y)27 by (x + y)3.
26. 36ic2l+4a;^12aj by 2x^\l + 6x.
27. 10aa225 + 16a* by 5 + 4a2a.
28. ^0^3^21 by aj + /
^29. 37i*lln325w2137i2 by 37i2 + 47i + l.
30. 2a;y + 2/^ + 9a;^ by  2 a;^/ + 3 a;^ + 2/^.
. 31. 73x + 37a^35h20x^15x^ hj 5^4:x'' + 9x.
DIVISION OF ALGEBRAIC EXPRESSIONS 49
32. 243n^ + l by 3w + l.
33. x'^16i^ + 96a^ + 256x{256 by (a; + 4)2.
34. _60w^ + 127n2 + 214ri336 by  12 n^ \ 11 n + 5Q.
35. 324a' by 8a + a4 + 16 + 2a3 + 4(x2.
v36. a^'^+^ft" + 2 a2'«+352«+i _^ ^2m+552n+2 j^^y a"*6«i + a'^+S^n
37. 2V^'tV«'^+tV«^'6V?>' by i«i&.
38. 5aj3 + 2433a;2 + l0a; + 3a;^ by 4 + 3a;.
39. aj^ + 37a^70a; + 50 by a^2aj + 10.
'' 40. 0^+^ + 8 a;^^ by aj2«»+i _ 2 a^"* + 4 a^'»i.
/41. {Sd' + 5a2)(2d'a6) by (3 al)(2a + 3).
42. 63aj^ + 114ar^ + 49a^16a;20 by dx' + QxQ.
v^43. a^+i^^  a&^^+' by a^+'^b^ab'^+K
,44. a'^_65_5(^45_^5^54_j_L()^3^2_jL0a263
by a^633a26 + 3a6l
45. _67i^25n4 477^^ + 817^2 + 37128
by _27i35n2 + 87i + 7.
46. 23a^5aj412 + 12a^ + 8a;14a^ by a;2 + 3a^.
47. 15a34a^15 + 8a«5a2a^+3a2 by a^Aw'S.
48. 52aj3 + 64 + 18a;^200.'»2 + a^ by 6a;2 8 +a;3_i2a;.
, 49. a"^ — 62" _ 2 6"c^ — c^p by a"* — 6" — c^.
50. a^6aV + 9a27i^47i« by a^  2 a^n  an^ \ 2 n^.
51. 3aj^7a;^llic^ + 5a^ + 7a;2 + 5a;2 by 3a^a?2.
52. 57i3+6nri^8 + 67i6+467i238n*
by 257i2 + 3n347io
53. ^m*2m^ + lm^j\ hj ^m^^ml,
54. 9»2_25 2/2_40 2/2_l6.;22 by 3aj4.52/ + 42!.
^ 55. 90 71*  143 n^  102 n^ + 131 ti + 60 by (2 tj  3)(5 n + 4).
^ 56. «*"» + a;2m^4n _. 2^n by x^"" — x"'y^'' + y^.
50 ALGEBRA
^"57. 4.x^21 xy + 21 xY  4 2/6 by (xy)(x2 y)(2x\ y).
V 58. a^ + 32 + 10a(a3 + 8)+40a2(a + 2) by (a2)2 + 8a.
74. By § 66, if the dividend and divisor are homogeneous,
the quotient will be homogeneous, and its degree equal to the
degree of the dividend minus the degree of the divisor.
75. The operation of division is often facilitated by the use
of parentheses.
JEx. Divide x^}(a{b—c)x^+(ab—bc—ca)x—abc by x+a.
ic^ + (a 4 & — c)x2 + (a6 — be — cd)x — ahc I x_ Ya
x^+ [^ I »^ 4 C&  c)x  be
(6  c)x2
(b — c)x'^ 4 {ab — ca)x
— bcx
— bcx — abc
EXERCISE 22
Divide the following :
^ 1. a^ ■\ (a — b — g)x^ + (— ab ibc — ca)x + abc
by x^ + (a — b)x — ab.
2. a^ + (a + 6 — c)a;^ + (a6 — &c — ca)x — a6c by ic — c.
3. :i?—{a\b{c)x^\(ab\bc{ca)x—abG by a;^— (6 + c)a;+6c.
4. a^  (a  2 6  3 c)^?^ f(_2a6 + 6 6c3 ca)x  6 a6c
by ic^ — (a — 3 c)a; — 3 ac.
5. a:3+ (3 a + &42 c)a^+ (3 a&+2 6c46 ca)x+^ abc by a;+3 a.
. 6. a{ab)x?+{ab\b'^\bc)xc{b+c) by (a6)fl?+c.
V 7. m(m + w)flj^ — (m^ + n^)a; + n{m — n) by ma; — n.
8. a:;^ — (m — 2 n)a; — 2 m^ f 11 mw — 15 w^ by a; + m — 3 w.
9. (2 m2 + 10 mn)x^ + (8 m^  9 mn  15 n'^)x  (12 mw  9 ii^
by 2 ma; — 3 n.
V 10. a;3(3a4264c)a;''^+(6a686c+12ca)a;~24a5cbya;26.
INTEGRAL LINEAR EQUATIONS 51
VI. INTEGRAL LINEAR EQUATIONS
76. Any term of either member of an equation is called a
term of the equation.
77. A Numerical Equation is one in which all the known
numbers are represented by Arabic numerals ; as,
2x — l — x\Q>.
An Integral Equation is one each of whose members is a
rational and integral expression (§ 63) ; as,
78. An Identical Equation, or Identity, is one whose members
are equal, whatever values are given to the letters involved ;
as (a \h) (a — h) = o? — 61
The sign =, read "^s identically equal to,'''' is frequently used in place
of the sign of equality in an identity.
79. An equation is said to be satisfied by a set of values of
certain letters involved in it when, on substituting the value
of each letter in place of the . letter wherever it occurs, the
equation becomes identical.
Thus, the equation x — y = 5 is satisfied by the set of values
x = S, y = S; for, on substituting 8 for x, and 3 for y, the equar
tion becomes 8 — 3 = 5, or 5 = 5; which is identical.
80. An Equation of Condition is an equation involving one
or more letters, called Unknown Numbers, which is satisfied
only by particular values of these letters.
Thus, the equation a? + 2 = 5 is not satisfied by every value
of X, but only by the particular value x — S.
An equation of condition is usually called an equation.
Any letter in an equation of condition may represent an
unknown number ; but it is usual to represent unknown num
bers by the last letters of the alphabet.
52 ALGEBRA
81. If an equation contains but one unknown number, any
value of the unknown number which satisfies the equation is
called a Root of the equation.
Thus, 3 is a root of the equation x\2 = 5.
To solve an equation is to find its roots.
82. If a rationaland integral monomial (§ 63) involves a
certain letter, its degree with respect to it is denoted by its
exponent.
If it involves two letters, its degree with respect to them is
denoted by the sum of their exponents ; etc.
Thus, 2 aWx^y^ is of the second degree with respect to x, and
of the fifth with respect to x and y,
83. If an integral equation (§ 77) contains one or more un
known numbers, the degree of the equation is the degree of its
term of highest degree.
Thus, if X and y represent unknown numbers,
ax—by = c is an equation of the first degree ;
a^\4:X= — 2, an equation of the second degree;
2x^—3xy^ = 5, an equation of the third degree ; etc.
A Linear, or Simple, Equation is an equation of the first degree.
PRINCIPLES USED IN SOLVING INTEGRAL EQUATIONS
84. Since the members of an equation are equal numbers,
we may write the last four axioms of § 9 as follows :
1. TJie same number, or equal numbers, may be added to both
members of an equation ivithout destroying the equality.
2. Hie same number, or equal numbers, may be subtracted
from both members of an equation ivithout destroyiyig the equality.
3. Both members of an equation may be multiplied by the same
number, or equal numbers, without destroying the equality.
4. Both members of an equation may be divided by the same
number, or equal numbers, without destroying the equality.
INTEGRAL LINEAR EQUATIONS 53
85. Transposing Terms. ,
Consider the equation x\a — h = c.
Adding — a and + 6 to both members (§ 84, 1), we have
x=c — a\h.
In this case, the terms + a and — b are said to be transposed
from the first member to the second.
Hence, any term may he transposed from one member of an
equation to the other by changing its sign.
86. It follows from § 85 that
If the same term occurs in both members of an equation affected
with the same sign, it may be cancelled.
87. Consider the equation
a — x = b — c. (1)
Multiplying each term by — 1 (§ 84), we have
x — a = c — b',
which is the same as equation (1) with the sign of every term
changed.
Hence, the signs of all the terms of an equation may be changed,
without destroying the equality.
88. Clearing of Fractions.
Consider the equation
2 5 5 9
X = x
3 4 6 8
Multiplying each term by 24, the lowest common multiple of
the denominators (Ax. 7, § 9), we have
16a;30 = 20a;27,
where the denominators have been removed.
Kemoving the fractions from an equation by multiplication
is called clearing the equation of fractions.
54 ALGEBRA
SOLUTION OF INTEGRAL LINEAR, EQUATIONS
89. To solve an equation involving one unknown number,
we put it into a succession of forms, which finally leads to the
value of the root.
This process is called transforming the equation.
Every transformation is effected by means of the principles
of §§ 84 to 88, inclusive.
90. Examples.
1. Solve the equation
5a; — 7 = 3a; + l.
Transposing 3ic to the first member, and — 7 to the second (§ 85), we
^^^^ 5x3x = 7 + L
Uniting similar terms, 2 sc = 8.
Dividing both members by 2 (§ 84, 4),
To verify the result, put cc = 4 in the given equation.
Thus, 20  7 = 12 + 1 ; which is identical.
2. Solve the equation
6 3~6 4
Clearing of fractions by multiplying each term by 60, the L. C. M. of
6, 3, 5, and 4, we have
70 cc  100 = 36 X  15.
Transposing 36 x to the first member, and — 100 to the second,
70a;36x = 10015.
Uniting terms, 34 a; = 85.
Divided by 34, x = — = .
34 2
3. Solve the equation
(5 3 a;) (3 + 4 ic) = 62  (73 a;)(l "4a;).
INTEGRAL LINEAR EQUATIONS 65
Expanding, 15 + 11 x  12 x^ = 62  (7  31 x + 12 x2).
Or, 15 + 11 X  12 x2 = 62  7 + 31 X  12 x\
Cancelling the — 12 x^ terms (§ 86), and transposing,
llx31x = 62715.
Uniting terms, — 20 x = 40.
Dividing by — 20, x = — 2.
To expand an algebraic expression is to perform the operations indicated.
From the above examples, we have the following rule for
solving an integral linear equation with one unknown number :
Clear the equation of fractions, if any, by multiplying each term
by the L. C. M. of the denominators of the fractional coefficients.
Remove the parentheses, if any, by performing all the opera
tions indicated.
Transpose the unknown terms to the first member, and the
known to the second ; cancelling any term which has the same
coefficient in both members.
Unite similar terms, and divide both members by the coefficient
of the unknown number.
The pupil should verify every result.
EXERCISE 23
Solve the following equations, in each case verifying the
answer :
1. 8» + 7 = 95. 8. 6x2S = 15xlS.
2. 9x = 5xS2, 9. 1913a; = 3129a;.
3. 7a; + 15 = 2a; + 45. 10. 14aj51 = 27a;33.
>4. 10a;3 = 3a;38. 11. 13 +12 a; = 37 a; + 43.
5. 6a; + 13 = lla;7. V12. 21a; 23 = 51 16a;.
6. 518a; = 8312a;. 13. lla; + 17 = 65a; + 47.
V7. lla;3 = 4 + 3a;. 14. 9816a; = 2341a;.
56
ALGEBRA
15.
16.
17.
V18. 2x
17a;9 447 = 41a;35a; + 27.
13 a; 39 = 48 a; 29 a; 81.
54 = 26 a;31a; + 19a;9.
?a; + ia; = 10.
3 7
21.
22.
7'" + 4n'' + 28''
23.
2^ 38 8^ 4^
5^i5=9*r
55
o>i 5 7 7 1
^ 6"8" = 9"i8"48
i/25. 4(2a;7) + 5 = 5(a;3) + 16.
V26. 13x(5a;8) = 6a;(3a; + 7).
27. 806(4x + 3) = 7a;3(6a; + l).
28. x2(47a;) = 4a;9(23a)).
29. 8a;(3a; + 2)27 = 4a;(6a;l)147.
30. 455a;(6ajl)=213a;(10a; + 3).
31. 4(a; + 14) 4 (3 a; 32) = 6(a; + 12) 7(a; 12).
./32. (53 a;)(3 + 4aj) = (7 + 3a;)(l 4 a;)  1.
V^33. (l + 3a;)2=(5a;)2 + 4(la;)(32a;).
34. 6(aj + 4)2 = 5(2a; + 3)25(2a;)(7 + 2a;).
V35. (3aj2)29(a;l)(3a;8) = 18a^f 51a;38.
36. (aj + 4)3(x4)3 = 2(3a;2)(4a.' + l).
1
(4x + lJ + .f6.
V:
__J^«9.
)l(5. + 8)
(.3)(...z)X(._5) =
2.
16
23
)
INTEGRAL LINEAR EQUATIONS 57
PROBLEMS LEADING TO INTEGRAL LINEAR EQUATIONS
WITH ONE UNKNOWN NUMBER
91. For the solution of a problem by algebraic methods, the
following suggestions will be found of service :
1. Represent the unknown number, or one of the unknown
numbers if there are several, by some letter, as x.
2. Every problem contains, explicitly or implicitly, just as
many distinct statements as there are unknowri numbers involved.
Use all but one of these to express the other unknown num
bers in terms of x.
3. Use the remaining statement to form an equation.
92. Problems.
1. Divide 45 into two parts such that the less part shall be
onefourth the greater.
Here there are two unknown numbers ; the greater part and the less.
In accordance with the first suggestion of § 91, we represent the greater
part by x.
The first statement of the problem is, implicitly :
The sum of the greater part and the less is 45.
The second statement is :
The less part is onefourth the greater.
In accordance with the second suggestion of § 91, we use ih.Q first state
ment to express the less part in terms of x.
Thus, the less part is represented by 45 — x.
We now, in accordance with the third suggestion, use the second state
ment to form an equation.
Thus,
45a; = x.
Clearing of fractions,
180  4 a; = x.
Transposing,
 4 a;  X =  180, or  5 a; =  180.
Dividing by  6,
x = 36, the greater part.
Then,
45 — X = 9, the less part.
2. A had twice as much money as B; but after giving B
^28, he has f as much as B. How much had each at first?
58 ALGEBRA
Let X represent the number of dollars B had at first.
Then, 2 x will represent the number A had at first.
Now after giving B $28, A has 2 a;  28 dollars, and B re + 28 dollars ;
we then have the equation
2a;28 = ^(x + 28).
o
Clearing of fractions, 6 a;  84 = 2 (a; + 28) .
Expanding, 6 x — 84 = 2 a; + 56.
Transposing, 4 a: = 140.
Dividing by 4, x = 35, the numberof dollars B had at first ;
and 2 X = 70, the number of dollars A had at first.
3. A is 3 times as old as B, and 8 years ago he was 7 times
as old as B. Eequired their ages at present.
Let X = the number of years in B's age.
Then, Sx= the number of years in A's age.
Also, x — S= the number of years in B's age 8 years ago,
and 3ft — 8 = the number of years in A's age 8 years ago.
But A's age 8 years ago was 7 times B's age 8 years ago.
Whence, 3a;_8 = 7(x8).
Expanding, 3x8 = 7x56.
Transposing, _ 4 x = — 48.
Dividing by — 4, x = 12, the number of years in B's age.
Whence, 3 x = 36, the number of years in A's age.
4. A sum of money amounting to $4.32 consists of 108
coins, all dimes and cents ; how many are there of each kind ?
Let X = the number of dimes.
Then, 108 — x = the number of cents.
Also, the X dimes are worth 10 x cents.
But the entire sum amounts to 432 cents.
Whence, 10 x + 108  x = 432.
Transposing, 9 x = 324.
Whence, x = 36, the number of dimes ;
and 108  X = 72, the number of cents.
INTEGRAL LINEAR EQUATIONS 59
EXERCISE 24
sll. The difference of two numbers is 12, and 7 times the
smaller exceeds the greater by 30. Find the numbers.
"^ 2. The sum of two numbers is 29, and the smaller exceeds
their difference by 4. Find the numbers.
3. Find two numbers whose sum is J, and difference ^.
4. The sum of two numbers is 44, and their difference is
threefourths the smaller number. Find the numbers.
^5. A is 4 times as old as B ; and in 22 years he will be twice
as old. ' Find their ages.
6. A is 3 times as old as B ; and 6J years ago he was 5 times
as old. Find their ages.
J 7. A has 3 times as much money as B ; but after B gives
him $9, he has 6 times as much as B. How much had each
at first?
^ 8. A man has 21 , coins, all dimes and twentyfivecent
pieces, valued in all at $3.30. How many has he of each?
9. A is 25 years of age, and B is 16. In how many years
will B be t\^othirds as old as A ?
\i 10. Divide 43 into two parts such that if the greater be
added to 17, and the less to 30, the resulting numbers shall
be equal.
11. Twice a certain number exceeds 35 by the same amount
that onethird the number exceeds 5. Find the number.
12. Divide $280 between A, B, and C so that A's share may
exceed f of B's by $96, and B's share exceed C's by $20.
13. A is 22 years of age, and B is 18. How many years ago
was A's age f of B's ?
14. A man has $4.10, all fivecent and fiftycent pieces;
and he has 5 more fivecent than fiftycent pieces. How many
has he of each ?
60 ALGEBRA
15. The sum of  and f a certain number exceeds  the
number by . Find the number.
16. If A has $ 5.50, and B ^3.50, how much money must A
give B in order that B may have  as much as A ?
V 17. A room is f as long as it is wide; if the width were
increased by 1^ feet, and the length diminished by the same
amount, the room would be square. Find its dimensions.
18. The sum of two numbers is {^ the greater, and their dif
ference is I. Find the numbers.
V 19. A boy buys a certain number of apples at 2 for 5 cents,
and double the number at 3 for 5 cents, and spent in all 35
cents. . How many of each kind did he buy ?
20. Divide $320 between A, B, C, and D so that A may
receive $ 35 more than B, C $ 15 more than B, and D $ 25 less
than C.
21. The sum of the ages of A, B, and C is 52 years ; A's age
is f of B's, and he is 8 years younger than C. Find their ages.
22. In a certain school the boys are 15 fewer than  of the
whole, and the girls are 33 more than ^. How many boys, and
how many girls, are there ?
I 23. The sum of $900 is invested, part at 4%, and the rest
at 5%, per annum, and the total annual income is $42. How
much is invested in each way ?
1^24. In 9 years B will be f as old as A; and 12 years ago he
was I as old. What are their ages ?
Let X represent the number of years in A's age 12 years ago.
25. A has I of a certain sum of money, B has ^, C has ,
and D has the remainder, $8. How much have A, B, and C ?
V 26. A man bought 8 hens, 7 sheep, and 12 pigs for $269;
each sheep cost y as much as each hen, and $ 3 less than each
pig. What did each cost ?
27. Divide 66 into two parts such that f the greater shall
exceed f the less by 21.
INTEGRAL LINEAR EQUATIONS 61
i/ 28. Find two numbers whose sum is 10, such that the square
of the greater exceeds the square of the less by 40.
29. Find two consecutive numbers such that ^ the greater
exceeds I the less by 2.
1^30. A person attempting to arrange a certain number of
counters in a square finds that he has too few by 12; but on
reducing the number in the side of the square by 3, he has
21 left over. How many has he ?
31. A purse contains a certain number of 10shilling pieces,
twice as many 5shilling pieces, and 5 times as many shillings,
the contents of the purse being worth £5. How many are
there of each coin ?
32. The square of the third of three consecutive numbers
exceeds the product of the other two by 13. Find the numbers.
33. Divide 39 into two parts such that 3 times the smaller
shall be as much below 58 as twice the greater exceeds 38.
34. Find two numbers whose difference is 3, and whose
product is less by 33 than the square of the greater.
'' 35. The total tiumber of persons at a certain factory is 196;
the number of women is f the number of men, and  the num
ber of boys. How many of each are there ?
36. A room is twice as long as it is wide, and it is found
that 50 square feet of carpet, 1 foot in width, are required to
make a border around it. Find its dimensions.
37. A purse contains a certain number of dimes,  as many
cents, and ^ as many $ 1 bills, the value of the entire contents
being $ 5.74. How many are there of each ?
38. A starts to walk from P to Q, 12 miles, at the same
time that B starts to walk from Q to P. They meet at the
end of 2 hours. If A walks one mile an hour faster than B,
what are their rates ?
39. Divide ^210 between A, B, C, and D so that B may re
ceive $ 10 less than A, G y as much as B, and D  as much as A.
62 ALGEBRA
40. The sum of $ 32 is divided between 7 men, 8 women,
and 16 children ; each child receiving i as much as each man,
and each woman 75 cents more than each child. How much is
received by each man, each woman, and each child ?
41. A boy had a certain number of marbles. He lost 6 of
them, gave away ^ the remainder, and then found that he
had 5 more than i of his original number. How many had
he at first ?
42. There are two heaps of coins, one containing 5cent
pieces and the other 10cent pieces. The second heap is worth
20 cents more than the first, and has 8 fewer coins. Find the
number in each heap.
43. In an audience of 435 persons, there are 25 more women
than men, and 3 times as many girls as men ; and the number
of boys is less by 195 than twice the number of girls. Find
the number of each.
{. 44. Find four consecutive odd numbers such that the prod
uct of the first and third shall be less than the product of the
second and fourth by 64.
," 45. A sum of money, amounting to $19.30, consists of $2
bills, 25cent pieces, and 5cent pieces. There are 13 more
5cent pieces than $ 2 bills, and  as many 5cent pieces as
25cent pieces. Find the number of each.
46. Two barrels contain 46 aijd 45 gallons of water, respec
tively. A certain number of gallons are drawn from the first,
and I as many from the second, and the second now contains
f as many gallons as the first. How many gallons were drawn
from each ? .
47. A tank containing 150 gallons can be filled by one pipe
in 15 minutes, and emptied by another in 25 minutes. After
the first pipe has been open a certain number of minutes, it is
closed, and the second pipe opened; and the tank is emptied
in 24 minutes from the time the first pipe was opened. How
many minutes is each pipe open ?
SPECIAL METHODS 63
VII. SPECIAL METHODS IN MULTIPLICA
TION AND DIVISION
93. Any Power of a Power.
Required the value of {a^y.
By § 11, {ay = a'xa'xa^ = a\
We will now consider the general case :
Eequired the value of (ccy, where m and n are any positive
integers.
We have, (a™)" = ar xoT X " ton factors
^m+m+" to n terms __ Qj^n
94. Any Power of a Product.
Eequired the value of (abf.
By § 11, (ahf = abxabxab = a%K
We will now consider the general case :
« Eequired the value of (a&)", where n is any positive integer.
We haye, (aby =ab xab x" to n factors = a''b\
In like manner, (abc •.•)" = a^'b^'c^ • • •,
whatever the number of factors in abC":
95. Any Power of a Monomial.
1. Find the value of (5a*y.
By §31, (_5a4)3=[(5)xa4]3
= ( 5)8 X («4)8(§ 94) =  125ai2(§93).
2. Find the value of ( 2 u^ny.
, We have, (  2 m^ny = (  2)* x (m^)* x w* = 16 wi^w*.
64 ALGEBRA
. 96. From §§93 and 94 and the examples of § 95, we have
the following rule for raising a rational and integral monomial
(§ 63) to any power whose exponent is a positive integer :
Raise the absolute value of the numerical coefficient to the
required power, and multiply the exponent of each letter by the
exponent of the required power.
Give to every power of a positive term, and to every even power
of a 7iegative term, the positive sign ; and to every odd power of
a 7iegative term the negative sign.
EXERCISE 25
Expand the following :
1. (a^y*z^y. 5. (Ja^b^^'f. 9. (a^b^cy.
2. (m^n'py\ 6. ( nV/)i«. 10. (a^^y^z^y\
3. (ab'cy. 7. (2m'xy. 11. (Sm^n^x^.
4. (^ 11 a^fy. 8. (4a^2/0' 12. (2amV)^
97. Square of a Binomial.
Let it be required to square a\b.
a + b
a^ + ab
ab\b'
Whence, (a \by == a^ + 2 ab + b\ (1)
That is, the square of the sum of two numbers equals the square
of the first, plus twice the product of the first by the second, plus
the square of the second,
1. Square 3 a + 2 &.
We have, (3 a + 2 6)2 = (3 aY + 2(3 a) (2 6) + (2 6)2
= 9a2 + i2a6 + 4 62.
Let it be required to square a — b.
SPECIAL METHODS 65
a — b
a — b
a^ —ab
 a6 + b\
Whence, {abf = a' 2 ab + b\ (2)
That is, the square of the difference of tivo numbers equals the
square of the first, minus twice the product of the first by the sec
ond, plus the square of the second.
In the remainder of the work we shall use the expression " the differ
ence of a and &" to denote the remainder obtained by subtracting h from a.
The result (2) may also be derived by substituting — 6 for 6, in equa
tion (1).
2. Square 4 a?^ — 5.
We have, (4 x^  5)2 = (4 ^2)2 _ 2(4 x"^) (5) + 52
= 16 x*  40 a;2 + 25.
If the first term of the binomial is negative, it should be en
closed, negative sign and all, in parentheses, before applying
the rules.
3. Square 2a^ + 9. ^ "^ ^^
We have, (_ 2 a^ f 9)2 = [( 2 a^) + gp
' = (2a3)2 + 2(2a3)(9)4.92
= 4 a6 _ 36 a3 + 81. .
EXERCISE 26
Expand the following :
1. {a^2)\ 7. (7a;f3a^l 13. (4 a^  11 .7^)2.
2. (a; 5)2. 8. (^n^ + llny. 14. {^ ax 12 byf.
3. {Qx^lyf. 9. (2a«7 62c)2. 15. {^n' + lOn^.
4. (3 + 8n2)2. 10. (_4m43n3)2. 16. (8a;^ + 9a;y.
5. (_m^ + 4p^y. 11. {Qofy + x'yy. 17. (7 a^m^  13 6 V) I
/6. (9a6l)l 12. (5a6 48 6c)2. 18. (6a^ll a;2;)2.
19. {bx^ + ^y^y. 20. (2a9a'')l
66 ALGEBRA
98. Product of the Sum and Difference of Two Numbers.
Let it be required to multiply a + 6 by a — 6.
o? + ah
abb'
Whence, (a {b)(ab) = a' b\ .
That is, the product of the sum and difference of two numbers
equals the difference of their squares.
1. Multiply 6a4563 by 6a56l
By the rule,
(6 a + 5 63) (6 a  5 63) = (6 ay  (5 63)2 = 36 a2  25 5»,
2. Multiply a^ + 4bya;2_4
( x2 + 4) ( x2  4) = [( x2) + 4] [( a:2)  4]
=:(_a;2)2_42 = aj416.
3. Expand (x + y + z)(x — y{z).
ix + y + z)ixy + z) = [(x + z) + y;\ [(x + z)y^
= (X + 0)2  1/2
=:X^ + 2XZ + Z^ 2/2.
4. Expand (a + 5 — c) (ct — & + c).
By § 52, (a + 6  c) (a  6 + c) = [a + (6  c)] [a  (6  c)]
= a2  (6  c)2, by the rule,
= a2 _ (52 _ 2 6c + c2)
= ^2  62 + 2 6c  c2.
EXERCISE 27
Expand the following :
1. (7a + 2b)(7a2b). V3. (3a;3 + 8 2/;22)(3a53_g2^^2)^
2. (9m2 + 4)(9m24). 4. (a« + 6)(a36).
SPECIAL METHODS 67
5. (llm'{5n')(llm'5n'). 7. (5 a' + 12 b^iBa' 12 b^.
■' 9. (10 m*?i+ 13 a^) ( 10 m%  13 a^).
10. (ab{c){abc). 13. (1 +a &)(1 a + &).
11. {x'{x + l){x' + xl). '14. (a2 + 3a+l)(a23a + l).
12. (a; + 2/ + ^)(«2/2;) 15. (x + y + ^){xy 'd).
16. (a;2 + a^ + 2/2)(a;2a;2/ + 2/^).
17. .(a' + 5a4)(a25a + 4).
18. (4ic2 + 3a;7)(4x23aj7).
V 19. (m^ + 5 m^n^ + 2 n^) (m^  5 mV  2 ti^).
99. Product of Two Binomials having the Same First Term.
Let it be required to multiply x\a'bj x\b.
x{a
x + b
3?\ ax
+ bx\ah
Whence, (x + a) (a; + 6) = x^ H (a + b)x + ab.
That is, the product of two binomials having the same first term
equals the square of the first ternV) plus the algebraic sum of the
second terms multiplied by the first term, plus the product of the
second terms.
1. Multiply a; — 5 by a; + 3.
The coeflB.cieut of x is the sum of — 5 and + 3, or — 2.
The last term is the product of — 5 and + 3, or — 15.
Whence, (x  5)(x + S) = x^  2x  15.
2. Multiply a; 5 by a; 3.
The coefficient of x is the sum of — 5 and — 3, or — 8.
The last term is the product of — 5 and — 3, or 15.
Whence, («  5)(x  3) = x^  8 x + 15.
68 ALGEBRA
3. Multiply ab — 4: by ab{7.
The coefficient of ab is the sum of — 4 and 7, or 3.
The last term is the product of — 4 and 7, or — 28.
Whence, (ab  4) (a& + 7) = ^252 ^Sab 28.
4. Multiply x^\6y^ hy x^\Sy\
The coefficient of x^ is the sum of 6 y^ and 8 y^, or 14 y^.
The last term is the product of 6 y^ and 8 y^, or 48 y^.
Whence, (x^ + 6 y^) {x^ \8y^) = x*^ 14 x^^ + 48 ^.
EXERCISE 28
Expand the following by inspection :
1. (x + S)(x + 4:). m. (xy + 7)(xye),
12. (x2)(x\5). 12. (a + Sx)(a + 9x).
3. (xll)ixl). 13. (x9y)(x5y).
^4. (a7)(a + 2). v 14. (m^fGyiXm^Tw).
V 5. (a2 + 15)(a' + 1). V 15. (a + 6 + 2)(a + & + 13).
6. (m33)(m3 + 8). 16. (x"{10y^^)(x'"' 9f).
J. (x2)(x6). 17. (a^9b%a' + Sb*).
8. (a'" + 10)(a'" + 2). 18. (m/i 14a;?/)(m7i4a;2/).
9. (mn7)(mn3). V19. (7/171 3)(m n + 11).
VlO. (ab + l){ab3). 20. (a^ft + 11 c3)(a26 12 c^).
ICX). Product of Two Binomials of the Form mx + n and px + g.
We find by multiplication :
mx{n
X
px\q
mpa? + npic
+ mqx\nq
mpx^ + (?ip + mq) x\nq.
SPECIAL METHODS 69
The first term of this result, mpx^, is the product of the first
terms of the binomial factors, and the last term, nq, the product
of the second terms.
The middle term, {np + mq)x, is the sum of the products of
the terms, in the binomial factors, connected by cross lines.
Ex. Multiply 3aj + 4by2aj — 5.
The first term is the product of 3 x and 2 ic, or 6 aj2.
The coefficient of x is the sum of 4 x 2 and 3 x (— 5) ; that is, 8 — 15,
or — 7.
The last term is the product of 4 and — 5, or — 20.
Whence, (35c + 4)(2a;  5) = 6x2  7ic  20.
EXERCISE 29
Expand the following by inspection :
1. (x + 6)(3x + 2). 9. (2aa;3)(5aa; + 6).
2. (2x^l)(lxl). 10. (3a?42n)(10a;n).
3. (2x5)(4aj + 3). 11. (4aj32/)(9a; + 2y).
4. (4a3)(5a3), 12. (7a 2m)(7a4m).
5. (4m + l)(4m + 3). 13. {^x"" \y){^x'' ^y).
6. (3 7i + 2)(57i2). 14. (6a2 + aj2)(8a25a^).
7. (2a2l)(lla24) 16. {^m^ 2n%10'm?7 n"),
8. (5aj^ + 6)(6a;4+l). 16. {^axZhy){^ax + bhy)'.
17. [6(m + w)5][(m + n)2].
18. [3(a&)4][4(a6)3].
101. We find by division,
— — = a6  = a46,
a\o a — o
That is.
If the difference of the squares of two numbers be divided by the
sum of the numbers, the quotient is the difference of the numbers.
70 ALGEBRA
If the difference of the squares oftivo numbers be divided by the
difference of the numbers, the quotient is the sum of the numbers.
1. Divide 25 yh' 9hj5yz' 3.
By § 96, 25 y^si^ is the square of 5 yz^ ; then, by the second rule,
^^y'''^ = 5yz^ + S.
byz^'6 ■
2. Divide x^ — {y—zf by x{{y — z).
By the first rule, ^^ " Cy  ^^ = x (y  z) = x  y + z.
x + iy z)
EXERCISE 30
Find, without actual division, the values of the following:
S6a'^121b*'^ n 225 a^' 100 b^^
VI. "^ — ^. vSi.
a24
a + 2
a^9
xS
25 n^ 
1
5n'
1
16 x^ 
■81
6aPllb^^ ' 15a« + 10
8n^Har' ' * 14mn^16a;«
v^*3 25n^l y l144a^^&^ ^^ 4a^(6cV
' " " ' ' '112 a"6^ * . * 2 a  (6  c) *
. 16a;^81 g 49.rV°1692^ ,« a^(m + 3n)^
' 4a^ + 9 * ^' ' 7a^2/^ + 132* ' * a + (m + 3w)*
V^ 13 (q + a;y(&y)'
(a + a;) + (&2/)
102. We find by division,
a+b a—b
That is,
ijf the sum of the cubes of two numbers be divided by the sum
of the numbers, the quotient is the square of the first number,
minus the product of the first by the second^ plus the square
of the second number.
SPECIAL METHODS 71
If the difference of the cubes of two numbers be divided by the
difference of the numbers, the quotient is the square of the first
number, plus the product of the first by the second, plus the square
of the second number.
1. Divide l + Sa^ by l42a.
By § 96, 8 a^ is the cube of 2 a ; then, by the first' rule,
l+2a l+2a ^ ^
2. Divide 27 x'64.y^ by 3x'4:f.
By the second rule,
27xe64y9 ^ (3.^)3(4yB)8 ^
3x24^/3 3x24?/3 ^ ^ ^^ ^^ ^ ''^^ ^ '
= 9 x* + 12 xV + 16 y«.
EXERCISE 31
Eind, without actual division, the values of the following :
• 1 ^ + 1 g £^!±A'. 11 27a;«125i/^
* ic + l* ' a=^ + 62* • Zx^hy
2 ^^' ^ 7 ^^±125. 12 343 m^r^^ + 8 1^3
* 1 — a * a + 5 * 7mn42p
• 3 ^'27 g 64 a?^"*! ^3 64a^6^ + 216c^
* n3* * 4a^J * * 4a26 + 6c3
^ 4 8 + m^ ,^ g a^6«216 ^14 a;^«  1000 y^g^^
24m^* * a66 * * a;«102/V
5 a;V^g9 ,Q 343m« + n« ^^5 729ftV4512 3/«
* xY^' ' lm? + n * * 9a3aj + 82/'
103. We find by actual division,
a'b'
a\b
a^a'b\ab'l^.
^ = a' + a'b^ab^\b^
72 • ALGEBRA
a+b
^'^' a^^a'b{a'b' + a¥\b'', etc.
a — b
In these results, we observe the following laws :
I. The exponent of a in the Jlrst term of the quotient is less by
1 than its exponent in the dividend, and decreases by 1 in each
succeeding term.
II. Tlie exponent of b in the sec&nd term of the quotient is 1,
and increases by 1 m each succeeding term.
III. If the divisor is a—b, all the terms of the quotient are
positive; if the divisor is a\b, the terms of the quotient are
alternately positive and negative.
A general proof of these laws will be found in § 466.
1. Divide a^ — b'' hj a — b.
By the above laws,
a — b
2. Divide 16 a;^ 81 by 2 a; + 3.
We have, 16 o^^  81 ^ ^2 .)^  3^
2a; + 3 2 a; + 3
= (2 x)3  (2 a;)2 . 3 + 2 X . 32  33
= 8 a;3 _ 12 a:2 + 18 X  27.
EXERCISE 32
Find, without actual division, the values of the following :
1 ^* ~ ^* 3 ^^~ ^ . 5 a^ — / ,
a—b ' x — 1 ' (xf + y^'
m + w' ^ 1 + a' ^^^ a«6c2*
SPECIAL METHODS
73
"■ ?5f
15.
m^729n^
J^ 7/1^" 81
©fi^fi*
i^
3a^2/
9 «^^'.
a — ic
j3 256a5^2/r
17.
a> + 128 6^*
a + 262
10. »' + !.
71 + 1
.. a' + 2^Sx'^
' a + 3a^
18.
64a;«729
2a;3
104. The following statements will be found to be true if n
is any positive integer :
I. a" — b"" is always 'divisible by a — b.
Thus, a^ — b^, a^ — b% a* — b% etc., are divisible hy a — b.
II. a^ — b"" is divisible by a\b ifn is even.
Thus, a — b^, a'* — b^, a^ — b^, etc., are divisible by a 4 6.
III. a** + 6" /s divisible by a\b ifn is odd.
Thus, a^ + b^, a^ + &'^, a' + 6'', etc., are divisible by a + 6.
IV. a"* + 6** IS divisible by neither a^b nor a — b if n is
even.
Thus, a^{b% a^ + 6^, a^ + b^, etc., are divisible by neither
a + b nor a — b.
Proofs of the above statements will be found in § 467.
74 ALGEBRA
VIII. FACTORING
105. To Factor an algebraic expression is to find two or more
expressions which, when multiplied together, shall produce the
given expression.
In the present chapter we consider only the separation of rational and
integral expressions (§ 63), with integral numerical coefficients, into fac
tors of the same form.
A Common Factor of two or more expressions is an expression
which will exactly divide each of them.
106. It is not always possible to factor an expression ; there
are, however, certain forms which can always be factored;
these will be considered in the present treatise.
107. Case I. When the terms of the expression have a com
mon factor.
1. Factor 14 a6^ 35 a«62.
Each term contains the monomial factor 7 ah\
Dividing the expression by 7 ah'^^ we have 2 62 _ 5 a\
Then, 14 a6*  35 a^&a _ 7 ^^i (2 &2 _ 5 «2).
2. Factor (2 m + 3)^2 + (2 m 4 3) 2/'.
The terms have the common binomial factor 2 m + 3.
Dividing the expression by 2 m + 3, we have x'^ + y\
Then, (2 m + 3) x2 + (2 m + 3) y3= (2 w + 3) {pi? + y^).
3. Factor (a — h) m \ (h — a) n.
By § 52, 5 _ a = _ (a  &).
Then, (a  6) m + (&  a) w = (a  6) wi  (a  6) »
= (a6)(TOw).
We may also solve Ex. 3 as follows :
(a  6) TO + (6  a) w = (6  a) n  (6  a) m = (J)'a){n' m).
FACTORING 75
4. Factor 5a(x — y)—Sa(x{'y).
= a(6x — 5y — Sx — Sy)
I =a(2a:8y)=2a(x42/).
EXERCISE 33
Factor the following :
^ 1. 63x^54 3;^ 5. (a2)6^ (a2yd^
w 2. a'5a'2a' + 3a\ 6. (3 a; + 5)m+ (3 a; + 5).
3. m^n^ + m^n"^ — mn®. i 7. (m — n) {x + y) — {n — m) z.
4. 24 a^2/'  40 a;y + 56 icY 8. a(a22) +3(2a2).
^'9. (x 4 2/) (m 4 ^) + (a? + 2/) W — *^)
^10. a(& + c)a(5c). 13. 5(2ici/J) 5(a; + 32/).
«*ll. 3aj2(a;l)i(la;). 14. (a + m)23(a + m).
12. 6(3a + 46) + 6(5a26). 15. x2(52/2;3) a^(22/ + ;2).
16. (mn)342m(mny.
17. 3a^+'7a»+'b + a^.
V 18. (a&)(m2 + a;2)(a6)(m2?/;2).
19. (mny~2m{m nf Jfrn^im nf.
108. The terms of a polynomial may sometimes be so
arranged as to show a common binomial factor; and the ex
pression can then be factored as in § 107.
1. Factor ah — ay + hx — xy.
By § 107, ah  ay \hx~xy = a{'by) + x{h  y).
The terms now have the common factor b — y.
Whence, ab — ay {bx — xy = (a + x)(b — y).
2. Factor a3 + 2a23a6.
If the third term is negative, it is convenient to enclose the last two
terms in parentheses preceded by a — sign.
76 ALGEBRA
Thus, a^\2a^3a6 = (a^ + 2a'^)(Sa + 6)
= a2(a + 2)  3(a + 2) = (a2  3)(a + 2).
EXERCISE 34
Factor the following :
^ 1. ac t ad \ be \ bd. V"3. mx \ my — nx —ny.
3. xy — Zx{2y — &. 4. a6 — a — 56 + 5.
5. 8 0^2/ + 12 ay + 10 6a; + 15 ab.
6. m* + 6 m^—Tm — 42.
7. 610a + 27a245a^
8. 20a628ad5 6c + 7cd'
9. m^ — m^7i + mn^ — n^.
10. a^6^a35Vd3a25Vd2__c6^6^
11. 63 + 36x^ + 560^ + 320^.
12. 48 a;?/ + 18 ria; — 88 mi/ — 33 m?i.
13. mx + my + nx + ny +px + py.
, 14. ax — ay + az — bx + by — bz.
15." 3 am — 6 an + Abm — S b7i + cm — 2 en.
V 16. ax +.ay —az— bx — by + bz + cx + cy — cz.
109. If an expression can be resolved into two equal fac
tors, it is said to be a perfect square, and one of the equal
factors is called its square root.
Thus, since 9 a'^b^ is equal to 3 a^6 X 3 a^b, it is a perfect
square, and 3a^6 is its square root.
9 05*62 is also equal to (— 3 a%) x (— 3 a^b); so that — 3 a26 is also its
square root ; in the examples of the present chapter, we shall consider
the positive square root only.
110. The following rule for extracting the positive square
root of a monomial perfect square is evident from § 109 :
FACTORING 77
j Extract the square root of the numerical coefficient, and divide
the exponent of each letter by 2.
Thus, the square root of 25 a'^6V is 5 a^b^c.
111. It follows from § 97 that a trinomial is a perfect square
when its first and last terms are perfect squares and positive,
and the second term plus or minus twice the product of their
square roots.
Thus, in the expression 4: x — 12 xy \ 9 y^, the square root of
the first term is 2 x, and of the last term 3 y ; and the second
term is equal to — 2 (2 x) (3 y).
Whence, 4: x^ — 12 xy \ 9 y^ is a, perfect square.
112. To find the square root of a trinomial perfect square,
we reverse the rule of § 97 :
j Extract the square roots (§ 110) of the first and third terms^
and connect the results by the sign of the second term.
1. Find the square root of 4: x^ { 12 xy ^ 9 y^.
By the rule, the result is2x\Sy.
(The expression may be written in the form
(_2x)2 + 2(2a:)(3y) + (3 2/)2,
which shows that (— 2 ic) + (— 3 ?/), or —2x — Sy, is also its square
root ; but the first form is simpler, and will be used in all the examples
of the present chapter.)
2. Find the square root of m^ — 2mn\ n^.
By the rule, the result is m — n.
(The expression may also be written w^ — 2 mn + m^ ; in which case, by
the rule, its square root is w — m.)
113. Case II. When the expression is a trinomial perfect
square. y
1. Factor 25 a^ + 40 ab' ^\ i' a H /> ^
By § 112, the square root of the expression is 6 a + 6*.
Then, 25 a^ + 40 ah'^ + &* = (5 a ^62)2.
2. Factor m^ — 4 m%^ + 4 n*.
78 ALGEBRA
By § 112, the square root of the expression is either w^ — 2n2, oi
2 n2  m2.
Then, m*  4 m'^n^ + 4 w^ = (m2  2 n^y, or (2 w2  m^y.
3. Factor a^ 2 %2;) + (?/ 2)2.
We have x^2 x(y e) + (y 0)2
= lxiyz)y=(ixy + zyi
or, =[(y  0)  a;]2 = (2/   ic)2,
4. Factor _ 9 a* 6 a^  1.
 9 a*  6 a2 _ 1 = _ ( 9 ^4 4. 6 a2 4. 1 ) = _ (3 ^2 + 1 ) 2.
EXERCISE 35
Factor the following:
1. q^ + Sx + 16. 5. a^2/2 414a^ + 49.
2. 96a + a'. 6. 36 0^132 ab^ 121b' .
3. m2 + 10mn425w3^ 7. 16 a' + 24: ax 9 a^.
4. 4a«4a36c2 + 6V. 8. 81 m^ + 180 m^ + 100 n^
^9.  25 a;io  60 x^fz'  36 ?/V.
^10. 64:a'a^2A0abxy + 225by.
11. 49 m^" + 168 m*^^ + 144 a^^.
12. 100 a2&2 I 180 a6c2 + 81 c^
13. 144a;V312a^"2/^^ + 169«*».
^14.  121 aV + 220 a'b'mn 100 b*n\
15. 169 a«62 + 364 a^6c2#4 196 cW.
^ 16. (a^ + yf + 22(0^ + 2/) + 121. ^;
17. a'Sa(mn) + 16(mnyj^
^18. 9a;26% + ^) + (2/ + 2)^
19. (m — ?i) — 2(m — n)w + w^.
'^20. 25(a + 2>)' + 40(a + 6)c + 16c2.
FACTORING 79
21. 36(aaj)284(aa;)2/ + 49/.
22. 49 m^ f 42 m(m + a;) + 9(m + a;)2.
v^3. (a + &)' + 4(a + &)(a&) + 4(a6)2.
24. 9(a; + 2/)'12(a;42/)(a^2/)+4(a;2/)^
114. Case III. When the expression is the difference of
two perfkct squares. ..,,
By §98, U\b^=(a + b)(ab).
Hence, to obtain the factors, we reverse the rule of § 98 :
Extract the square root of the first square, and of the second
square; add the results for one factor, and subtract the second
result from the first for the other,
1. Factor 36a'b^49c^
The square root of 36 a^b^ is 6 aft^, and of 49 c^ is 7 c^.
Then, 36 a^b^  49 c^ = (6 ab^ + 7 c^) (6 ab^  7 c^).
2. Factor (2xSyy{x y)\
By the rule, (2 x  3 y)2  (x  y)'^
= [(2x32/) + (x2/)][(2x32,)(x2^)]
= (2x3y + xy)(2x3yx + y)
= (3x4?/)(x2?/).
A polynomial of more than two terms may sometimes be
expressed as the difference of two perfect squares, and factored
by the rule of Case III.
3. Factor 2 mri + m^  1 + nl
The first, second, and last terms may be grouped together in the order
m2 + 2 TOW + n2 ; which expression, by § 112, is the square of m + w.
Thus, 2 TOn + to2  1 + w2 = (to2 + 2 mw + n2)  1
= (m + n)2l
= (to + W + 1) (to + 7i — 1).
4. Factor 12y + ic29/4.
80 ALGEBRA
122/ + x2  9^2 _ 4 = a:2  9?/2 + 12 ?/  4
= a;2(92/212 2/44)
=:x2(3?/2)2, by§112,
= (x + Sy2)(xSy + 2).
5. Factor a2c2 + 62 _c?2_ 2 ccZ 2 (i6.
a2  c2 + 62 _ ^2 _ 2 cd  2 a6
= a2  2 a6 + 62 _ c^^2cdd^
= a2  2 a6 + 62  (c2 + 2 cd + ^2) = (a  6)2 (c + d)2, by § 112
= [(«6) + (c + d)][(«^>)(c + «!)]
= (a — 6 + c + d)(a — 6 — c — d).
EXERCISE 36
Factor the following :
1. a^^m^\ 3. n«9. 5. l64mV.
2. 4a2l. 4. 1625a«. 6. 36x'121y\
7. 181 a«6V. . >/ 19. (ic  yf  (m + ^0'
8. 49a«144 6V^. 20. {2a + xy  {b {Syy.
9. 100mV2169nV«. 21. (a  6)^  (c  d^.
10. 225a.Y196 2^ 22. (2m + ny  (m\2ny.
11. 324 a^™6io« _ 625. 23. (6a\xy  (aSxf.
12. 36l£ci*2562/^c^«. 24. (9 a;  5 ?/)  (6 a;  7 ?/)'.
13. (abyc',~\ 25. 25(8 a3 6)29(4 a+5 6)2
14. (5x + yya^.l 26. (a+6c)=^<a6 + <5)2.
V15. c4'*(m + 7i)l } 27. (w+w+3)2(mw4)2.
16. a2(62c)^\ 28. a' ^ 2 ab \ b^  c\
Vl7. (x + 4.v)2922J 29. x'y^2yzz\
18. 36 m^  (2 m  3)2. 30. m^ 7i2 + 2wpi>2.
FACTORING 81
31, a'{b'l2ab, 34. 9 a' { 16b' 25 c' {24: ab.
32/ y' + 2xy4.\x\ 35. 9 a' i2ab b^
33/4:7)1^ — 4:mn\n'—p^. 36. 4:m' —p' — 9 n' — 6np.
37. 12yz + 16x'9z'4y\
y38. m^—2mn{n' — a:^ + 2xy — y\
39. a'{2ab + b'c'2cdd\
N/40. a2 + a:2_52_^2_j_2tta; + 26?/.
V/41. x'y'\7n^l2mx2y.
42. a24aa; + 4a^&2_f.662/92/2.
43. 16a2_8a6 + &'c'10cd25d2.
^44. 2SxyS6z' + 49y'' + 60z25 + 4x'.
115. Case IY. When the expression is in the form
x'^\axhf\y^.
Certain trinomials of the above form may be factored by
expressing them as the difference of two perfect squares, and
then employing § 114.
1. Factor a* + a262 + 6^
By § 111, a trinomial is a perfect square if its first and last terms are
perfect squares and positive, and its second term plus or minus twice the
product of their square roots.
The given expression can be made a perfect square by adding a%'^ to
its second term ; and this can be done provided we subtract a?h^ from the
result.
Thus, a* + a262 + ^4 = (^4 + 2 ^252 + 54) _ ^252
= (a2 + 62)2 _ a%2^ by § 112,
= (a2 _{. 52 ^ a6) (a2 + ftf ah), by § 114,
= («2 + a6 + 62) (^2 _ah + 62).
2. Factor 9 05*  37V + 4.
The expression will be a perfect square if its second term is — 12 x'^.
82 ALGEBRA
Thus, 9x*37jc2j4 = (9x* 12x2 + 4)25a;2
= (3 x2  2)2  (5 xy
= (3x2 + 5x2)(3a;2_5aj_2).
(The expression may also be factored as follows :
9 ic*  37 x2 + 4 =(9 x* + 12 ^2 4 4)  49 x'^
= (3x2 + 2)2_(7x)2=(3.r2 + 7a: + 2)(3x27a;+2)
Several expressions in Exercise 37 may be factored in two different ways.
The factoring of trinomials of the form x* + ax'^y^ + y*, when the factors
involve surds, will be considered in § 300.)
EXERCISE 37
Factor the following :
\ 1. .ic\+5a^ + 9. ^^5. 9a;^ + 6a;y + 49y*.
2. a^21c&2_^36 6^ 6. 16 a^  81 a^ } 16.
3. 4  33 ar^ + 4 a;\ ^^1. 6464m2 + 25m^
4. 25m*14mV + w^ 8. 49 a^  127 a V + 81 iB*.
Factor each of the following in two different ways (compare
Ex. 2, § 115) ;
9. a;^17a^ + 16. 11. 16 m*  104 mV + 25 a;*.
10. 9148a2 464a^ 12. 36a*97aW + 36m^
116. Case V. When the expression is in the form
x^ \ ax + b.
We saw, in § 99, that the product of two binomials of the
form X \ m and x \ n, was in the form a^ + aa; j & ; where
the coefficient of x was the algebraic sum of the second terms
of the binomials, and the third term the product of the second
terms of the binomials.
In certain cases, it is possible to reverse the process, and
resolve a trinomial of the form x^ \ ax{h into two binomial
factors of the form x \m and x \n.
\^ FACTORING 83
To obtain the second terms of the binomials, we simply re
verse the rule of § 99, and Jind two numbers whose algebraic sum
is the coefficient of x, and whose product is the last term of the tri
nomial.
The numbers may be found by inspection.
1. Factor a;2 4. 14 a; 4. 45.
We find two numbers whose sum is 14 and product 45.
By inspection, we determine that these numbers are 9 and 5.
Whence, a;2 + 14 x + 45 = (x + 9) (ic + 5).
2. Factor a^ — 5 ic + 4.
We find two numbers whose sum is — 5 and product 4.
Since the sum is negative, and the product positive, the numbers must
both be negative.
By inspection, we determine that the numbers are — 4 and — 1.
Whence, a;2 _ 5^; + 4 = (x  4)(x  1).
3. Factor a?^ + 6 a;^  16.
We find two numbers whose sum is 6 and product — 16.
Since the sum is positive, and the product negative, the numbers must
be of opposite sign ; and the positive number must have the greater abso
lute value.
By inspection, we determine that the numbers are + 8 and — 2.
Whence, x^ + Qx^ \Q = {y? + 8) {x^  2).
4. Factor x""  abx"  42 a'b^.
We find two numbers whose sum is — 1 and product — 42.
The numbers must be of opposite sign, and the negative number must
have tlie greater absolute value.
By inspection, we determine that the numbers are — 7 and + 6.
Whence, x^  abx^  42 ^252 _ (^.2 _ 7 ab)(c(^^ + 6 a?)).
5. Factor l^2a — 99a\
We find two numbers whose sum is + 2 and product — 99.
By inspection, we determine that the numbers are +11 and — 9.
Whence, 1 + 2 a  99 a^ = (1 + 11 o) (1  9 a).
If the x^ term is negative, the entire expression should be
enclosed in parentheses preceded by a — sign.
84 ALGEBRA
6. Factor 24^ + 5 xx^.
We have, 24 + 5 a:  x^ =  (a;2  5 x  24)
= (x8)(x + 3) = (8a;)(3 + x) J
changing the sign of each term of the first factor.
(In case the numbers are large, we may proceed as follows :
Required the numbers whose sum is — 26 and product — 192.
One of the numbers must be + , and the other — .
Taking in order, beginning with the factors + 1 x — 192, all possible
pairs of factors of — 192, of which one is + and the other — , we have :
+ 1 X  192.
' + 2 X  96.
+ 3x 64.
+ 4 X  48.
+ 6 X  32.
Since the sum of + 6 and — 32 is — 26, they are the numbers required.)
EXERCISE 38
Factor the following :
1. x'{4,x + S. 13. a^17a; + 52.
2. x'TxhlO. 14. a2 + 18a + 56.
. 3. a^ + TalS. '' 15. S4:{5xx\
4. m^ 14 m 15. 16. if\16y57._
5. 2/2162/ + 55. 17. x'lOxTB.
6. x' + lQx + Sd. 18. m2 + 19m + 90.
v7. 28 + 3cc2. 19. 9514 71^2.
8. 665nnK ^0. x^ 20 a; + 96.
9. a2  14 a + 48. 21. a^ + 21 a f 98.
10. x^\20xh51. 22. x'TxlS.
11. x^  12 X 4:5. 23. 105Smm'.
12. n2 + 14n32. 24. c^ 21 0^ + 104.
FACTORING 85
25. x'2Sx'{76. 43. l\5aUaK
26. a^' + a^llO. 44. m^  17 mn + 66 nl
27. n''16n'S0. 45. a^ + 12 ab + 27 b\
28. a^" + 18 a" + 65. 46. a;^  14 ma? + 40 m^.
29. cc2 + lla;"'12. 47. l9x36a;2.
30. c^19c2^ + 88. 48. m' + 3mn54:n\
v^31. a^/13a;?/3_3o; 49. ^2 _^ 12 a;^/ + 20 2/I
32. a'b^  23 ab' + 112. 50. aV  17 a6c + 60 c^.
 33. A2 + 25 na; 4154. 51. llSn6Sn\
' 34. 126 + 15?/'/. 62. a^ + 15 aa;  100 a^.
' 35. aV + 9aV162. 53. 1 + 17 mw + 70 mV.
• 36. m'^'*  23 77i3» + 120. '^ 54. a^^  17 a;^^^^ _^ 72 i/V.
• 37. (a + 6)2 + 14(a + 6)+24. 55. a^ + 6 a'"'^  91 fe^.
. 38. {xyy15{xy)lQ. 56. l3xy10^xY
39. (mn)2+21(mr6)130. 57. a^  32 a6c + 112 6V.
^0. (a + x)228(a + a;) + 192. 58. a^y + 29 a^y^  170 ^l
41. a2_^6aa; + 5a^. 59. y(2m + 3 n)a; + 6mn.
42. x'TxySy^ 60. x" {ab)xab.
117. Case VI. WJien the expression is in the form
ax^ +bx + c.
We saw, in § 100, that the product of two binomials of the
form mx + n and px + q, was in the form ax^ + bx + c', where
the first term was the product of the first terms of the bino
mial factors, and the last term the product of the second terms.
Also, the middle term was the sum of the products of the
terms, in the binomial factors, connected by cross lines.
In certain cases it is possible to resolve a trinomial of the
form ax^ + bx + c into two binomial factors of the form mx + ?i
and px + q.
86 ALGEBRA
1. Factor 3 a^ + 8 a; + 4.
The first terms of the binomial factors must be such that their product
is 3 a;2 . ^^gy can be only 3 x and x.
The second terms must be such that their product is 4.
The numbers whose product is 4 are 4 and 1,4 and —1,2 and 2, and
— 2 and — 2 ; the possible cases are represented below :
x + 4
x + 1
a;4
Sx + l
3a; + 4
3a;l
13 X
Ix
13 a;
x\
ic + 2
«2
3x4
3a; + 2
3x2
— Ix 8ic — 8a;
The corresponding middle term of the trinomial, obtained by cross
multiplication, as in § 100, is given in each case ; and only the factors
X + 2, 3 a; + 2 satisfy the condition that the middle term shall be 8 x.
Then, 3a;2 + Sx + 4 = (x + 2)(3a; + 2).
2. ractor6a;2_^_2.
The first terms of the factors must be 6 x and x, or 3 x and 2 x.
The second terms must be 2 and — 1, or — 2 and 1.
The possible cases are given below :
6x + 2
6xl
6x2
6x + l
x1
x + 2
x + 1
x2
4x
11 X
4x
llx
3x + 2
3xl
3x2
3x + l
2xl
2x + 2
2x+l
2x2
X 4x —X — 4x
Only the factors 3 x — 2 and 2 x + 1 satisfy the condition that the
middle term shall be — x.
Then, 6x2 x  2 = (3x  2)(2x + 1).
The following suggestions will be found of service :
(a) If the last term of the trinomial is positive, the last terms
of the factors will be both f, or both —, according as the middle
term of the trinomial is + or — .
FACTORING 87
Thus, in Ex. 1, we need not have tried the numbers — 1 and — 4, nor
— 2 and — 2 ; this would have left only three cases to consider.
(b) If the last term of the trinomial is negativey the last terms
of the factors will be one \~, the other — .
If the X term is negative, the entire expression should be enclosed in
parentheses preceded by a — sign.
If the coefficient of a:^ is a perfect square, and the coefficient
of X divisible by the square root of the coefficient of x^, the ex
pression may be readily factored by the method of § 116.
3. Factor 9 a;2 _ 18 a; + 5.
In this case, 18 is divisible by the square root of 9.
We have 9 a:2  18 x + 5 = (3 x)2  6(3 x) + 6.
We find two numbers whose sum is — 6, and product 5.
The numbers are — 5 and — 1.
Then, 9x2  ISx + 5 = (3a:  5)(3a;  1).
^EXERCISE 3# / > 3
Factor the following :
VI. 2,x' + 9x\9. 12. 10a^39a; + 14.
:>2. 3ar2lla;20. >1S. 12 a;^ + 11 a; + 2.
3. 4a:228a; + 45. 14. 20 a V  23 aa; f 6.
4. ex'hTxS, n5. 36a;212a;35.
5. 5a;236^' + 36. 16. 6aj15a^.
6. 16a;2 + 56a; + 33. 17. 5 + 9x18ar^.
7. 8n2 + 18n5. 18. 72H7a;49x^
8. 4.x^Sx7. >19. 24a;217wa; + 3n2.
9. 9a;2 + 12a;32. 20. 28a^a;2.
10. 6x' + 7ax^2a^. 21. 21 a^"» + 23 a; V + ^ ^^
11. 25x'25mx6m\' 22. 18 a^  27 afta;  35 r/6.
v23 24 tt^ + 26 tt^ 5.
88 ALGEBRA
118. It is not possible to factor every expression of the form
a^ + aa; H 6 by the method of § 116.
Thus, let it be required to factor a;^ + 18 a; + 35.
We must find two numbers whose sum is 18, and product 35.
The only pairs of positive integral factors of 35 are 7 and 5,
and 35 and 1 ; and in neither case is the sum 18.
It is also impossible to factor every expression of the form
axr \hx\c by the method of § 117.
Thus, it is impossible to find two binomial factors of the
expression 4 a;^ + 4 a; — 1 by the method of § 117.
In § 298 will be given a general method for the factoring of
any expression of the forms x^ \ax \h, or ax^ \hx\c.
119. If an expression can be "resolved into three equal fac
tors, it is said to be a perject cube, and one of the equal factors
is called its cube root.
Thus, since 27 o!^W is equal to 3 a6 x 3 ah x 3 o?h, it is a
perfect cube, and 3 a^h is its cube root.
120. The following rule for extracting the cube root of a
positive monomial perfect cube is evident from § 119 :
Extract the cube root of the numerical coefficient^ and divide
the exponent of each letter by 3.
Thus, the cube root of 125 a^6V is 5 a^bh.
121. Case YII. When the expression is the sum or difference
of two perfect cubes.
By § 102, the sum or difference of two perfect cubes is divis
ible by the sum or difference, respectively, of their cube roots ;
in either case the quotient may be obtained by the rules of § 102.
1. Factor a;^ 27 2/V.
By § 120, the cube root of x^ is x^, and of 27 yV is 3 yH.
Then one factor is x^ — 3 y^z.
Dividing a^ — 27 y'^z'^ by a;^ — 3 y^z., the quotient is
X* + 3 xhiH + 9 ?/622 (§ 102).
Then, x^  27 y^z^ = (x^ _ 3 yH) (cc* 4 3 x^yH + 9 fz'^).
^^V_^^^;
FACTORING 89
2. Factor a« 4 &^
One factor is a^ 4. ^2^
Dividing a^ + h^ by a^ + &'^, the quotient is a* — a^ft^ + 54,
Then, a^ + 56 ^ (^2 + 52) (a* _ a252 + ^4).
3. Factor {x + a)''' — {x — of.
(x + a)3(xa)3
= [(cc + a)  (xa)][(x + a)2+ (a;+a)(a;a) + (xa)2]
= (ic + a  ic + a) (x2 + 2 ax + a2 + x2  a2 + x2  2 ax + a2)
— " "< V" •*' T^ "* <
!•
EXERCISE ^
^0
a.  cxlr
Factor the following :
1. a? + b\
i^v
Sa^ + l.
9. 64m3 — n^
2. ^y\
.6.
l27nl
vlO. a«6«216cl
3. l + m«n^
7.
a« + l.
11. 8m3^ + 27n^.
4. a^ft^c^.
8.
ccy + z«.
12. 27iB«125/«.
13. 64 + 125a365.
18.
729a«6« + 8c^dl
14. 343a«64m^
vl9.
(^
4_2/)3 + (a;2/)3.
15. 125a^+216/;2^
20.
m^
— (m + w)^.
16. 27a«512 6».
21.
27
(a6)3 + 8 6^
17. 216aV343n«
1
22.
(2
a + a^)^— (a + 2a;)^
1/23. {hx2yf{?>x^yf.
122. Case VIII. TF7ie?i the expression is the sum or differ
ence of two equal odd powers of t2vo numbers.
By § 104, the sum or difference of two equal odd powers of
two numbers is divisible by the sum or difference, respectively,
of the numbers.
The quotient may be obtained by the laws of § 103.
Ex. Factor a^ + 6^
By § 104, one factor is a + &.
90 '^ ALGEBRA
(/■
Dividing a^ + ft^ by a + &, the quotient is
a*  a% + a2&2 _ ah^ + 6*. (§ 103)
Then, (a^ + 6^) ^ (« + 5) (^4 _ ^85 _f. ^252 _ ^h^ _j. 54).
EXERCISE 41
Factor the following :
vl. ar' + Z. 5. 1+a;^ 9. 32a«6«.
2. a^1. v6. a;^ + 7i9^ >10. 243a^ + 2/^.
3. 1mV. 7. a«l. v^ 11. m" + 128n^
4. a76^ 8. n«^ + 32. v 12. 32 a^^^^n _ 243 ciop.
123. By application of the rules already given, an expres
sion may often be resolved into more than two factors.
If the terms of the expression have a common factor, the
method of § 107 should always be applied first.
f
1. Factor 2 aa^if — 8 axy^.
By § 107, 2 ax^y^  8 axy^ = 2 axy^x"^  4y^)
= 2 axy\x + 2 ?/) (x  2 y), by § 114.
2. Factor .a«5^
By § 114, a6  &6 = (a^ + b^) (a^  63).
Whence, by § 121,
a6b^ = (a\ h){a^  ah + h'^){a  h){aP + a6 + 62).
3. Factor ajs/.
By § 114, a^  ^ = (X* + y4)(x4  y^)
= (a;* + t){'^^ + y^){^ + y){^  2/).
4. Factor 3 (m + w)^ _ 2 (m^  n^).
3(m + w)2  2(m2  rfi) = 3(m + n)2  2(m + w)(wi  w)
= (m4 w)[3(w + n)2(mn)]
= (m + w) (3 ?n 4 3 n  2 m + 2 w)
= (w + n)(»i + bn).
FACTORING 91
5. Factor a(al) 6 (61).
a(a  1)  6(6  1) = a2  a  62 + 5
= a2 _ 62 _ 05 + 5
= (a + 6)(a6)(a6)
= (a6)(a + 6l).
The following is a list of the typeforms in factoring, con
sidered in the present chapter :
ax \ ay — az.
(§ 107)
jr^ + fljr + b.
(§ 116)
fl + 2 a6 4 61
(§ 113)
ax^ ^ bx + c.
(§ 117)
fl^  2 a6 + b\
(§ 113)
a' + b\
(§ 121)
a'  b\
(§ 114)
a'  b\
(§ 121)
x' + axy +/.
(§ 115)
a"  b\
(§ 122)
a" + 6% n odd. (§ 122)
MISCELLANEOUS AND REVIEW EXAMPLES
EXERCISE 42
Factor the following :
1. 0^24.210^ + 98, 4. 49aV210a26«c + 225 6'\
2. a220a69. 5. 21a^ {Uo}24.alQ.
3. 12aj«18a5^ + 6a;^9a^. 76. a'  22 a^ \ 120.
>7. 49 a;2__ 49^.^12.
8. 7a;2(3a26)3a^(2a36).
9. x'16. 15, 45n« + 18n2207i8.
10. 24 a6 18 a?/ 20 6a; +15 0^2/. 16. 9620aa2.
^11. 27a3 + 1000. A7. 36 a;^  69 a;^ __ 25.
12. 646 m' 250 mril 18. 2a'xSa'a:^}2a^afiSaxl
13. (x'{x2y(x'x\3y. 19. 128a;y+288a;y+162a;/.
14. 64: n\ 20. 2 xi/  2 ic^^^ _ 254 a;^^^.
92 ALGEBRA
21. m'^l. 24. 121m« + 22m*l.
22. a^'l. 25. 36x^\24.a^21x\
23. a%' SO a^bc^\ 216 c\ 26. a'b^ + aY  ¥x^  x'f.
27. (a\ 2 by + S(a^ 2b){2 ab) +16(2 a by,
28. 4a;(a&c) + 02/(6 + ca).
29. (m + 7i)*2(m + ^)^4(m + w)^.
30. x^16xy^64:y^ 32. a;626^27.
31. Slm'256n^ 33. (a; + 2 2/)^ + (3 x  2/)^
34. (a + 2a;)2 + 10(a + 2x)144.
35. 27 x''75y^120yzASz\ 39. 49 a^^^ _f_ 12 a^^e _^ 4 a^^^^
36. (a' {4.ab + b')'(a' jb'y. 40. 14 a;^ _ 25 ^^ 4. 6.
37. (16m" + n2)264mV. 41. a''x'\
38. 49tt2 + 43662_28a. 42. x^'2x' + l.
43. 9 aV  16 a'd'  36 6V + 64 bH\
44. mV243mV. 46. a' + 12Sb\
45. _7aj226a; + 8. 47. 4S x^y  52 xY  ^^0 xy\
48. Kesolve a^ — 81 into two factors, one of which is a — 3.
49. Eesolve of — 64 into two factors, one of which is x\2.
50. Eesolve x^ — y^ into two factors, one of which is x — y.
51. Eesolve a^ + 1 into two factors, one of which is a + 1
52. Eesolve l\x^ into three factors by the method of
Case VII.
53. Eesolve a^ — 512 into three factors.
Factor the following :
54. a^ — m^ \ a \ m.
55. (x2f 4a?)237(ar^ + 4x) + 160.
56. iii*'1024. 57. m^ + m + x^ix.
FACTORING 93
58. a^(^  4. b'c^  S a'd^\ 32 b'd^
59. (m7i)(aj22/2) + (« + ^)(m2w2).
60. (xiy + 6(xiy + 9(xl).
61. (m + n) (771^ — a^) — (m + a?) (m^ — n^.
62. a24&2_a26. 63. (a;^ + 4 2/'  i^y  16 a;y .
64. {x^ 9xy{4.(x^9x) 140.
65. d'b'h27aySb^a^2Wa^f.
66. (m^ + m)2 + 2 (m^ + m)(m + 1) + (m + 1)'.
67. (2a;2_3)2ar^. 69. (4: a'  b'  9y  36 b\
68. 64aV + 8a3_8ar'l. 70. (x + 2yyx(x'4:y^).
71. 16aj2_.2/2_25;32_i^83.^_j_10^.
72. (a2 + 6a + 8)214(a2 + 6a + 8)15.
73. (l + ar^) + (l+a;)3. 75. (a^ + /) _ a;^/ (x + 2/).
74. a*9 + 2a(a2 + 3). ' 76. (a^ 8 m«)a(a2m)2.
77. 9a2(3a + 2)2 + 6a(3a + 2) + l.
78. m»m^ + 32m332. 80. m\m^p) +n^(np). ^
79. a(ac)6(6c). 81. a;^ + 8 a;« + a^" + 8.
82. (27m^a^) + (3m\x){9'm'6mx{x^.
83. (4a2 + 9)2_24a(4a2 + 9) + 144a2.
84. 16a2 + 962_25c24d224a620cd
85. m9 + m^64m«64.
86. (a;2^2/0'4a;y(aj2 42/2).
87. a'\a''b{a^b^{a^b^{ab^{b\
88. (87i3_27)+(27i3)(4n2 44716).
89. a^ + 2ar^ + 2a; + l.
(By altering the order of the terms, this may be written
x8+l + (2a;2 + 2x), or (x + l)(x'^  x + 1) \ 2x {x\ 1),
and X + 1 is a factor of the given expression.)
94 ALGEBRA
90. x^3x''\3xl.
92. S c^ + 36 x'y + 5Axy' } 27 if .
Additional methods in factoring will be found in §§ 298 to
300, and in Chapter XXXIV.
124. By § 54, (+ a) x (+ 6) = + ab, (+ a)x(b) =  ab,
{a)x(+b)=ab, { a)x{ b) = + ab.
Hence, in the indicated product of two factors, the signs of
both factors may be changed without altering the product; but if
the sign of either one be changed, the sign of the product will be
changed.
If either factor is a polynomial, care must be taken, on
changing its sign, to change the sign of each of its terms.
Thus, the result of Ex. 3, § 107, may be written in the forms
(b — a)(n — m), — (b — a)(m — n), or — (a — b)(n — m).
In like manner, in the indicated product of more than two
factors, the signs of any even number of them 7nay be changed
without altering the product; but if the signs of any odd number
of them be changed, the sign of the product will be changed (§ b5).
Thus, {a — b){c — d)(e — /) may be written in the forms
(abXdc)(fe),
(ba)(cd)(fe),
— (6 — a)(d — c)(/— e), etc.
SOLUTION OF EQUATIONS BY FACTORING
125. Let it be required to solve the equation
(a;3)(2a; + 5) = 0.
It is evident that the equation will be satisfied when x has
such a value that one of the factors of the first member is
equal to zero ; for if any factor of a product is equal to zero,
the product is equal to zero.
FACTORING 96
Hence, the equation will be satisfied when x has such a value
that either aj_3 = 0, (1)
or 2x + 5 = 0. (2)
5
Solving (1) and (2), we have a; = 3 or — •
It will be observed that the roots are obtained by placing the
factors of the first member separately equal to zero, and solving
the resulting equations.
126. Examples. "^
1 . Solve the equation a^ — 5 a; — 24 = 0.
Factoring the first member, (x  8) (x + 3) = 0. (§ 116)
Placing the factors separately equal to (§ 125), we have
X — 8 = 0, whence x = 8 ;
and X + 3 = 0, whence x = — 3.
2. Solve the equation 4 a;^ — 2 a; = 0.
Factoring the first member, 2 x(2 x — 1) = 0.
Placing the factors separately equal to 0, we have
2 X = 0, whence x = ;
and 2 x.— 1=0, whence x =  •
2
3. Solve the equation ar' + 4a^ — a; — 4 = 0.
Factoring the first member, we have by § 108,
(x + 4) (x2  1) = 0, or (X + 4) (X + 1) (x 1) = 0.
Then, x + 4 = 0, whence x = — 4 ;
X + 1 = 0, whence x = — 1 ;
and X — 1 = 0, whence x = 1.
4. Solve the equation a^ _ 27  (a^ + 9 a;  36) == 0.
Factoring the first member, we have by §§ 116 and 121,
(x  3)(x2 + 3x + 9)  (X  3)(x + 12) = 0.
Or, (x3)(x2 + 3x + 9x12) = 0.
Or, (x3)(x2 + 2x3)=0.
96 ALGEBRA
Or, (x3)(a; + 3)(xl) = 0.
Placing the factors separately equal to 0, x = 3, — 3, or 1.
The pupil should endeavor to put down the values of x without actu
ally placing the factors equal to 0, as shown in Ex. 4.
EXERCISE 43
Solve the following equations :
1. aj2_^7a; = 0. ' 11. a;^ + 18 a^ + 32 a;^ ^ 0^
2. 5a34a^ = 0. '^ ^ 12. x' 13x' + S6 = 0.
3. 3a^108ic = 0. 13. 8 a;2_i0a; + 3=:0.
4. (3ic2)(4a;^25) = 0. 14. 6 x' + 7 x + 2 = 0.
5 x^ 15 X + 54: = 0. 15. 3x'mx4:m^ = 0.
6. a^ + 23a; + 102 = 0. 16. lOx^f 7 a;12 = 0.
7. x2 + 4a;96 = 0. " 17. 15x^+x2 = 0.
8. aj^x110 = 0. ^ ' 18. 12a;329aj2 + 15a; = 0.
9. x'^ax2a^ = 0. 19. x" ax{bxab = 0.
10. (5a; + l)(a^6ic91) = 0. 20. x^ + mx \ nx { mn = 0.
21. a^2cx8aj + 16c = 0.
22. aj^ + 3 ma; — Sm^a;— 15m^ = 0.
23. 27a^ + 18x2_3^_2 = o.
24. (aj2)24(a^2)+3 = 0.
25. (4 a;2  49)(a;2 _ 3 ^ _ lo^)(3 ^,2 _^ ^4 ^^ _ 15) ^ q
26. (a:2)(5a;2 + 3a;4)(a^4) = 0.
27. (aj2l)(a^9) + 3(a;l)(a; + 3) = 0.
HIGHEST COMMON FACTOK 97
IX. HIGHEST COMMON FACTOR. LOWEST
COMMON MULTIPLE
(We consider in the present chapter the Highest Common Factor and
Lowest Common Multiple of Monomials^ or of polynomials which can
be readily factored by inspection.
The Highest Common Factor and Lowest Common Multiple of poly
nomials which cannot be readily factored by inspection, are considered
in §§ 439 to 443.)
HIGHEST COMMON FACTOR
127. The Highest Common Factor (H. C. F.) of two or more
expressions is their common factor of highest degree (§ 64).
If several common factors are of equally high degree, it is understood
that the highest common factor is the one having the numerical coefficient
of greatest absolute value in its term of highest degree.
For example, if the common factors v^ere 6 aj — 3 and 2 a: — 1, the
former would be the H. C. F.
128. Two expressions are said to be prime to each other when
unity is their highest common factor.
129. Case I. Highest Common Factor of Monomials.
Ex, Required the H. C. F. of 42 a?h\ 70 d'hc, and 98 a'hH\
By the rule of Arithmetic, the H. C. F. of 42, 70, and 98 is 14.
It is evident by inspection that the expression of highest degree which
will exactly divide a%'^, a%c, and a%hP is a%.
Then, the H. C. F. of the given expressions is 14 a^h.
It will be observed, in the above result, that the exponent of
each letter is the loivest exponent with which it occurs in any of the
given expressions.
EXERCISE 44
Find the H. C. F. of the following:
1. 14a^2/, 21xy\ 2. Ua'b\ 112 6V.
98 ALGEBRA
3. 60(xyy, S4:(xyy. 5. 72 a'b^, 27a%^, 99 a'b*.
4. 108 m^ny, 90 7nhipl 6. Ux'yz% SSxYz', 110 ar^T/'^
7. 32a'x\ 128 aWa^, 192 a^Y
8. 136 a%V, 516%)i«, 119 c^??^^.
9. 72xyz', 168a^?/V, 120ajV2jr
10. 26(a J6)2(a6/, 91{a + by{aby.
130. Case II. Highest Common Factor of Polynomials which
can be readily factored by Inspection.
1. Kequired the H. C. F. of
5 x^y  45 T'y and 10 a^/  40 xY  210 xy^
By §§ 107, 114, and 116, 5 x^y  45x^y = 6x^y(x^  9)
= 5x^y(ix + 3)(xS); (1)
and 10 x^y^  40 xV _ 210 xy^ = 10 xy'^ix'^  4 x  21)
= 10xy\x7)(xhS). (2)
The H. C. F. of the numerical coefficients 5 and 10 is 5.
It is evident by inspection that the H. C. F. of the literal portions of the
expressions (1) and (2) is xy(x + 3).
Then, the H. C. F. of the given expressions is 5xy{x + 3).
It is sometimes necessary to change the form of the factors in
finding the H. C. F. of expressions.
2. FindtheH.C.F. of a2 + 2a3 and 1al
By §116, a2 + 2a3 = (al)(a + 3).
By §121, la^ = (la){l + a + a^).
By § 124, the factors of the first expression can be put in the form
(la)(3 + a).
Hence, the H. C. F. is 1  a.
EXERCISE 45
Find the H. C. F. of the following :
1. SOicy + lOary, 15a;y30a;/.
HIGHEST COMMON FACTOR 99
2. a'ieb', a' + Sab\16b\
3. m"14m + 45, m'10mj25.
5. a3 + 64, a'7aA4..
6. 9a^, aj2aj6.
7. ac — 6c — ad + bd, d^ — cl
8. a;2 f 13 a; + 22, 2 aj^ + 9 a; + 10.
9. 3ac4ad66c + 86d, aH7a&1862.
10. x^ + y''z'^2xy, a? y'' z" + 2yz.
11. 3ar'16a'?/ + 5 2/^ a;^ + 10 a;?/  75 /.
12. m^^m\ m^ + 4m2 + 16.
13. 2a^7ajH6, 6aj2llx + 3.
14. 2ar'13a;i/ + 62/S xy^^o?.
15. llla + 18a2, 8a«l, 18a25a2.
16. Sa?2Qo?b\20ab\ 12 a^ 10 a^b 2^ db\
17. a;2 + 18a; + 77, a;^ + 22 a; + 121, a^ + a^HO.
18. 16m29n2, 16m224mn + 9 n^ 9 mn^  12 rri^n.
19. ar^27, a^6a; + 9, 2 aa;6 a 6a; + 3 6.
20. 27a3 + 863, 9a246^ 9aH12a6 + 462.
21. a23a18, 2a2a21, 3a2 + 4a15.
22. 2a^12a^ + 16aj, 3 a;*  3 a^  36 ;t^, 5 a:^ 4 5 a;^  100 a:^^
23. 125m*8m, 10 m^ \ m^  2 m, 25 m^  20 rrv' { ^ m.
24. a^ + 3a240, a*2o, a'^a'5a5.
25. 2a:3_aj2_g3,_j_3^ 6.T219a; + 8, 4a;2_^3^__5^
26. a'(b + cy, {bay(^, b'(acy.
27. Sx^y\xY, 64:xy{2xf, 24: a^y  SO aPy^  21 xy^. y
28. 2a'^ + 17a436, 4a24a99, 6a2 + 25a 9. '"'^'^
100 ALGEBRA
LOWEST COMMON MULTIPLE
131. A Common Multiple of two or more expressions is an
expression which is exactly divisible by each of them.
132. The Lowest Common Multiple (L. C. M.) of two or more
expressions is their common multiple of lowest degree.
If several common multiples are of equally low degree, it is understood
that the lowest common multiple is the one having the numerical coeffi
cient of least absolute value in its term of highest degree.
For example, if the common multiples were 4 a:— 2 and 6x — 3, the
former would be the L. C. M.
133. Case I. Lowest Common Multiple of Monomials.
Ex. Required the L. C. M. of 36 a% 60 aV, and 84 cv?.
By the rule of Arithmetic, the L. C. M. of 36, 60, and 84 is 1260.
It is evident by inspection that the expression of lowest degree which
is exactly divisible by a%, aPy'^, and cx^ is a^cx^y'^.
Then, the L. C. M. of the given expressions is 1260 aHx^y^.
It will be observed, in the above result, that the exponent of
each letter is the highest exponent with which it occurs in any of
the given expressions.
EXERCISE 46
Find the L. C. M. of the following :
1. 6 a^f, 6 xy. 5. 105 a% 70 b'c, 63 c'a.
2. lSa%4:5b'c. 6. 50 xy, 2i a^f, 4:0 a^y\
3. 2Sa^,36y\ 7. 21 ab*, 35 b'c% 91 a'(^.
4. 42 m%^ 98 ny. 8. 56 a'b^, 84 ba^, 48 xy.
9. 60 a'bc', 75 a'b% 90 aVcZ«.
10. 99 m^nx«, 66 m^nV, 165 wV/.
134. Case II. Lowest Common Multiple of Polynomials which
can be readily factored by Inspection.
1. Kequired the L. C. M. of
ar^ — 5a; + 6, ar^ — 4a;4, and a^ — 9 x.
LOWEST COMMON MULTIPLE 101
By § 116, a:2  5 x + 6 = (X  3) (a;  2).
By § 113, x24x + 4=(x 2)2.
By § 114, x89x = x(x + 3)(x3).
It is evident by inspection that the L. C. M. of these expressions is
x(x2)2(x + 3)(x3).
It is sometimes necessary to change the form of the factors.
2. Find the L. C. M. of acbcad^ bd and b  a^.
By § 108, achcad^hd = {a b) (c d).
By §114, 62a2= (& + a)(&a).
By § 124, the factors of the first expression can be written
\ba)idc).
Hence, the L. C. M. is (6 + a) (6  a) (dc), or (62 _ a^) (dc).
EXERCISE 47
Find the L. C. M. of the following :
2. o?b\2a^b\2a^b''{ab\
3. m26m + 9, m2llm + 24.
4. a*  49 a'bS a' + 12 a^b + 35 o?b\
5. 2a^ + 2a;284a;, 3x33a^90aj.
6. a^o^,a^a?x\ao?o?.
7. 1 + 27 0.3, l5a;24aj2.
8. ac — 3ad2bc + Q>bd,^ac + ad — Q>bc — 2bd.
10. a2_73a + 10, 105a + 2a2a^  l^ \^ ^ ^
11. a^ + 8, 4a;2(a^ + 4)2. vi "^^
12. 2a^ + 3a;35, 2ar^ + 19a: + 45.
13. 9n27n + 8, 3rj22n16. . , .
102 ALGEBRA
14. 16x'25y',12x^\15xy,Sxy10y\
15. x' 15x\50,x'{2xS5,af3x 70.
16. a'4:ab + 4:b',a^8 W, a% + 2 a^^^ 4. 4 ab\
17. m^ — 10 m^i + 21 n^, m^ — 5 mn — 24: 71^, m'^ — Sln*,
18. a;2 + 5a; + 6, a^2a;8, a^ + 2 a;2 + 5 a; + 10.
19. 9a6^4a^6, 8 ac + 2 ad126c3 6d.
20. a'lQa, a'Sa^4.a, a' { 5 a^ \ 4. a.
21. 27 n"" + 64 71, 18 n^  32 ^2, 9 n^ + 21 72* + 12 711
22. 9x2 + 300^ + 25, 6x''{7x5, 10a:29a; + 2.
23. 71^571 + 6, 9n^~n\ 10n2n\
24. aj37/3, aj2_2a;2/ + /, x'\xy + y\
25. 3ac+ac«6 6c2 6rZ, ac4ad2 6c + 8 6fZ, 3c2llcd4c?2.
26. 2x^x15, 2x^7x^3, 2aj2_9a,^9^
27. a'{4.b'9c^4tab,a^4:b^9(^+12bc,a'4:b'{9c^6ac.
28. 3 771^ + m^Ti — 2 mTi^, 6 771^71 + 11 mTi^ + 5 71^,
9 m^n + 5 771^71^ — 4 7W7i^.
29. 32a« + 4a^ 12 a4 + 12 a^ + 3 a^ 32a^ + 8a3 + 2a.
FRACTIONS 103
X. FRACTIONS
135. The quotient of a divided by & is written  (§ 6).
h
The expression  is called a Fraction ; the dividend a is called
h
the numerator, and the divisor h the denominator.
l^he numerator and denominator are called the terms of the
fraction.
136. It follows from § 69, (3), that
If the terms of a fraction he both multiplied, or both divided, by
the same expression, the value of the fraction is not changed.
137.
By
the Rule of
Signs
in Division
(§ 68),
+ a
+ &
— a
b
b
— a
That is, if the signs of both terms of a fraction be changed, the
sign before the fraction is not changed ; but if the sign of either
one be changed, the sigii before the fraction is changed.
If either term is a polynomial, care must be taken, on chang
ing its sign, to change the sign of each of its terms.
Thus, the fraction ^~ , by changing the signs of both
c—d b—a
numerator and denominator, can be written — (§ 51).
d — c
138. It follows from §§ 124 and 137 that if either term of
a fraction is the indicated product of tivo or more expressions,
the signs of any even number of them may be changed without
changing the sign before the fraction ; but if the signs of any odd
number of them be changed, the sign before the fraction is changed.
Thus, the fraction ^^ may be written
(cd)(ef)
a — b b — a b — a, .^
(dc)(fey (dc)(efy (dc)(/e)'
104 ALGEBRA
EXERCISE 48
Write each of the following in three other ways without
changing its value :
 a 2 Vi±^. 3 _^ 4 2a;— 7 ^ ^ 6x5
2 7 2x x^2 {x3)(y\4:)
6. Write (^ ^  1) («  ^) i^ four other ways without
(x+5)(y2)
changing its value.
REDUCTION OF FRACTIONS
139. Reduction of a Fraction to its Lowest Terms.
A fraction is said to be in its lowest terms when its numerator
and denominator are prime to each other (§ 128).
(We consider in the present chapter those cases only in which the
numerator and denominator can be readily factored by inspection.
The cases in which the numerator and denominator cannot be readily
factored by inspection are considered in § 444.)
140. By § 136, dividing both terms of a fraction by the
same expression, or cancelling common factors in the numera
tor and denominator, does not alter the value of the fraction.
We then have the following rule :
Resolve both numerator and denominator into their fciCtorSf and
cancel all that are common to both.
1. Reduce ^ ^ „ ^^o to its lowest terms.
40 a^b^cH^
We have.
40 aWcH^ 23 X 5 X a^h'^cH^ 5 cd^
by cancelling the common factor 2^ x d'^lP'C.
X? — 21
2. Reduce — to its lowest terms.
x?2x3
By §§ 121 and 116, x3  27 ^(x  3)(x^ 4 3^ + 9)^:.^ + 3x + 9,
^^^ 'x22x3 (x3)(x + l) « + l
 _, . ax — bx — ay^by^ ., , , ,
3. Reduce to ? to its lowest terms.
b^ — a^
By §§ 108 and 114,
FRACTIONS 105
axbx ay + by (a b)(x — y)
52 _ «2 (^) 4. d) (6  a)
By § 138, the signs of the terms of the factors of the numerator can be
changed without altering the value of the fraction ; and in this way the
first factor of the numerator becomes the same as the second factor of the
denominator.
axbx ay + by _ (b  a)(y  x) _ y x
men, 52 _ ^2 (& + «)(&_«)& + «*
If all the factors of the numerator are cancelled, 1 remains to form a
numerator ; if all the factors of the denominator are cancelled, it is a case
of exact division.
EXERCISE 49
Keduce each of the following to its lowest terms :
. 5 xYz' « 54 mri" ^ 126 a^6V ^ 90 aVn\
' Sxfz^' ' 99 mV* * 14 aV * * 36 amhi^'
o 12 a'b'  63 a^j/V g 26 m^ny g 88 x'y'z^
' 42 62c3' • 84 0.^2'' ' 130m*nV* ' 66a^yz''
g 120 a^5VQ ^Q 15x'y\10xY ^ a;^9a; + 18 .
■ 75a6V ' ■ 6 x'y^ + 4. a^y' ' ' x'' + x12'
12 a^ + lla6 + 28 6^ , 20 3a^4a^3a + 4
• a« + 14a'6 + 49a62* ' ' 9 a3 + 9 a^ie a16*
^3 64a^ + 72a;^?/48a;/ ^i 4m^ + 16myi + 15n^
64 x^y — Sly^ ' 6m^ — mn — 15 n^
14 ^^^ + ^^^ ~ 56 m^yt^ ,^ no 16 a;'^ + 4 a^ + 1
w? — 64 mn^ 8 a;^ — 1
15 a^ + &' 23 a^9/2^ + 6y2!
a2_2«^,_3ft2* • a;2_9^2^22_2a;2;'
16 ac + 3ad426c + 66cZ . ^^ (a2 5)^ (3 c d/
3acad+66c26d* ' (a + d)^ (2 & + 3 c)^*
. 8a^125 25 g^ + 28 ««&« + 27 6«
• 2a^ + a;'15a;* * a^ 4 9 a^ft^ ^ 81 6* *
18 a^ + a12 / og 25 a^
• 3a213a + 12' ' ar^lla; + 30'
19 (a^49)(a^16a; + 63) 07 9a;^49/
(«214a;+49)(a;22a;63)* ' 28 a;/  12 a^2/ '
106
28.
Sb'a^
a'^dabUb^
29
27a ^,
4a;2_9a,_9
ALGEBRA
30. tziSM^^,
21 a; 10x2
31.
15a;?/20«2l2/+28
141. Reduction of a Fraction to an Integral or Mixed Expression.
A Mixed Expression is a polynomial consisting of a rational
and integral expression (§ 63), with one or more fractions.
Thus, a + , and  \ ^ are mixed expressions.
c 6 x — y
142. A fraction may be reduced to an integral or mixed
expression by the operation of division, if the degree (§ 64)
of the numerator is equal to, or greater than, that of the
denominator.
1. Reduce '^^ to a mixed expression.
ox
By§72, 6x^415x2^6x_^ 15^_A^2x + 5A.
•^ "* ' 3x 3a: 3a; 3x 3x
2. Reduce — . ' "^ to a mixed expression.
4a;2 + 3
4x2 + 3)12x38x2 + 4a;5(3x2
12 x^ +9x
 8 x2  5 X
8x2 6
5x+l
Since the dividend is equal to the product of the divisor and quotient,
plus the remainder, we have
12x88x2 + 4x5 = (4x2 + 3)(3x2) + (5x + l).
Dividing both members by 4x2 + 3, we have
12x3  8x2 4 4x 5 _ g ^ _ 2 5x + l
4x2 + 3 4x2 + 3 *
Thus, a remainder of lower degree than the divisor may be written over
the divisor in the form of a fraction, and the result added to the quotient.
If the first term of the numerator is negative, as in Ex. 2, it is usual to
change the sign o/ each term oj the numerator^ changing the sign before
the fraction (§137).
FRACTIONS 107
Thus, 12a^8_8:,2 + 4a;5^3 _5xl,
' 4 x'^ + 3 4 x2 + 3
EXERCISE 50
Reduce each of the following to a mixed expression :
1 15m2 + 12m4 o 30a«5a^ + 15a + 7
1. ^. ^
3 9^+2 g 49a^ g 14a^ + 39a^ + 4a19
*3a; — 1* *7i» + 32/* * 2a + 5
'aj + 2/ ' a — b ' 5x\2
g a^ + 8&^ _ g m^nV ^^ 3a^ + 8a^4 ^^
a — 26 m + ri a^f2a — 3
12 150)^60.^20 a;^7 ^3 24 a;^ + 21 a; + 19
3a;24 ' * 4o^2aj + 5
14.
6 g^  17 a'b  21 a'b^ + 19 a6^ + 22 b*
2a^5ab6b^
143. Reduction of Fractions to their Lowest Common Denominator.
To reduce fractions to their Lowest Common Denominator
(L. C. D.) is to express them as equivalent fractions, each having
for a denominator the L. C. M. of the given denominators.
Let it be required to reduce — %, ;— ^^, and —^ to their
T , ^ • 4. 3 a^b^ 2 ab^ 4 a^b
lowest common denominator.
The L. C. M. of 3 ab^, 2 ab% and 4 a^b is 12 a^b^ (§ 133).
By § 136, if the terms of a fraction be both multiplied by the
same expression, the value of the fraction is not changed.
Multiplying both terms of ^^, by 4 a, both terms of ^,
by 6 a^b, and both terms of — ^ by 3 b^, we have
4a^6
16 ace? 18 a^bm , 15 b^n
12 a'b'' 12 a'b'' 12 a'b''
108 ALGEBRA
It will be seen that the terms of each fraction are multiplied
by an expression, which is obtained by dividing the L. C. D. by
the denominator of this fraction.
Whence the following rule.
Find the L. C. M. of the given denominators.
Multiply both terms of each fraction by the quotient obtained
by dividing the L. C. D. by the denominator of this fraction.
Before applying the rule, each fraction should be reduced to
its lowest terms.
144. Ex. Reduce ^ ^ and — — — to their lowest com
, . , a^ — 4 cr — oa{6
mon denominator.
We have, a^ 4. = (a{2){a2),
and a25a + 6 = (a2)(a3).
Then, the L. C. D. is (a + 2) (a 2) (a  3). (§ 134)
Dividing the L. C. D. by (a + 2) (a — 2), the quotient is a — 3 ; dividing
it by (a — 2) (a — 3), the quotient is a + 2.
Then, by the rule, the required fractions are
, 4«Ca3) ^^^ Sa(a + 2)
(a + 2)(a2)(a3) (a + 2)(a2)(a3)'
EXERCISE 51
Reduce the following to their lowest common denominator :
.. Tab Sbc 2ca ^ 4.a^ 2
6 ' 10' 15' ', 4a29' 6a^9a
5 4 6 a 1 8mn 2
2«2
mn
2 ^ ^ ^ 6
2m^n 5 7)i^n^' 7 mn^ m—7i 2{m—rif Zim — iif
o 3 3^442; ^x — hy m 3yi 5
* ~Wxf' ^^yz" ' ' 71^8' n24n + 4*
. 11 c'p 9a'm Sbhi g 2 3a
■ 12 a^ft' 14 6V 21 c^a" * a^ {3a' + 2a{6' a' + 27'
2 4 6
x\2' x2' x'd'
FRACTIONS 109
^Q a + 36 a — 3 b a\4:b
d'7ab + 12b'' a^abVlb''' a"9&2*
a^4.3a;10' 2a^ + 7a;15' 2a^7a; + 6
ADDITION AND SUBTRACTION OF FRACTIONS
145. By §72, ^ + £^ = ^±^1^.
'' a a a a
We then have the following rule :
To add or subtract fractions, reduce them, if necessary, to
equivalent fractions having the lowest common denominator.
Add or subtract the numerator of each resulting fraction,
according as the sign before the fraction is \ or —, and write
the result over the lowest common denominator.
The final result should be reduced to its lowest terms.
146. Examples.
. o 14^ 4a + 3 , 1662
1. Simplify ■ ;, + ^ , 3 *
4a^6 Qab^
The L. C. D. is 12 a%^ ; multiplying the terms of the first fraction by
3 62, and the terms of the second by 2 a, we have
4a + 3 16&2 ^ 12 a62 + 9 &^ 2 g  12 ah"^
4a26 6a&3 12 a^ft^ 12 a2&3
^ 12 a62 4. 9 52 _^ 2 g  12 gftg ^ 9 62 + 2 q
12g263 12g263
If a fraction whose numerator is a polynomial is preceded
by a — sign, it is convenient to enclose the numerator in
parentheses preceded by a — sign, as shown in the last term
of the numerator in equation (A), of Ex. 2.
If this is not done, care must be taken to change the sign of
each term of the numerator before combining it with the other
numerators.
2. Simplify 5^4.v _7x2y.
^ •' 6 14
110 ALGEBRA
The L. C. D. is 42 ; whence,
6x4y 7 x2y _ 36x2Sy 21x6y
6 14 42 42
_ S5 X  28 y  (21 X  6 y)
42
(A)
_ 36 X  2S y  21 X h Q y _ U X  22 y _ 7 X  11 y
42 42 ~ 21 ■
3. Simplify
oc^ {X a? — X
We have, x"^ ■\ x = x{x + 1), and x^^ — x = x(x — 1).
Then, the L. C. D. is x(x ■\l){x 1), or x{x'^ — 1).
Multiplying the terras of the first fraction by x — 1, and the terms of
the second by x + 1, we have
1 1 _ x1 x + 1
x^\x x^x x{x^  1) x(a:2  1)
__ xl(x + l) _ xlxl _ 2
x(x^l) x(a;2l) x(x2l)*
By changing the sign of the numerator, at the same time changing the
o
sign before the fraction (§ 137), we may write the answer
x(x2 — 1)
Or, by changing the sign of the numerator, and of the factor x^ — 1 of
2
the denominator (§ 138), we may write it
4. Simplify __l___A__+ 1
a^Sa + 2 a24a + 3 a^Sa + G
We have, a^  3 a + 2 =(a  l)(a  2), a^ 4a + 3 = (a l)(a3),
and a2 5a + 6=(a 2)(a 3).
Then, the L. C. D. is (a  1) (a  2) (a  3).
Whence, \
a2_3aj_f.2 a24a + 3 a^ba + Q
g3 2(a2) a1
(al)(a2)(a3) (a l)(a _2)(a  3) (« _ i)(a2)(a3)
a32(a2)+al _ a32a + 44al
(a  l)(a  2)(a  3) (a  l)(a  2)(a  3)
« 0.
(al)(a2)(a3)
FRACTIONS 111
EXERCISE 52
Simplify the following :
1 4a;47 6x — 5 a 2m{5n 3m + 4yi ^
10 15 * ' 8mV 6mn^
2 3 5 5 5a7b a+6b
' 2a'b^ Ta'b' ' 27a 36b
Q 4a — 9 3a — 8 ^ g ^ — 2/ ■ y — ^ ^ i g — 3a;
9 12 ' xy 2yz 3zx
„ 2(6n{5) 3(n + 6) 4(5n4)
11 22 44 ' ,
g 3a2 4a7 7a3
3a« 7a2 9a *
g 8a; + l 10 y 9 9^ + 8
7x Uy 21 z '
10 ^^' + 3 _ 3 g^ + 1 _ 3a^2
6 a' 12 a^ 36 a*
11 ^x3 6x{5 5x\2 3 a; 10
5 10 15 20 *
io 3m2 7m8 , 9m + 4 lOm + 7
^^' 4 6— + "8 9^'
13 2a;4?/ 5a; + 4y 8rg — 3?/lla; — 2y
' ~~8~~ 16~" 24^ 32
14. _^+_^. 18. ^^ L.
5m — 2 2m + 3 2x\y 2x — y
15. i 1. 19. ^ ^
3a;7 4a; + 5 3a9 5a15
16 m 2 20 5a; 4a;^ + 3a; — 1
m + 2 m2* * a;3 a;2_^^_L2
y, a + 3 g 3 2i a; + 3 j?/ a; — 3 ? /
a — 3 a + 3 x — 3y x\3y
it
112
ALGEBRA
25.
33.
22.
23.
24.
a
a
a^ + 4 a ■
60 ar
4a
12
X
X
2.^
6
2x'd
2x + 3
4.x'
9
a — n
3a4
^^l
a5
71
2a427i 3aH3n 6a + 6w
2x
2x
Ux'^x^l 16i»2l
26.
3a;f2 9a^ + 4
3x2 9ar^4'
27.
28.
29.
30.
31.
32.
a^
+
a^ — xy'
x\y (a; + 1/)^ (x + yf
a\h . — 6 2 ah
2a2h 2a + 2b a^'h''
2a
3a2
a^
2al (2 a 1)2 (2aiy
m — 1 m + l . m — 6
m2
X
+
m+l m ^
m2 m^— 4
a^ 2a^
x\y x^ — xy\y^ a^{y^
b . c
+
(a + 6)(6 + c) (6 + c)(c + a) (c + «)(« + 6)
+
2a
a3b a\3b (a^Sbf
1 . 1
2 + 4a6 + 462 a2462 a'2ab
5a;44 3a;2 2a^hl9a;8
a;2_3a;_4
36.
37.
A 38
2x2 + 5a; + 3 4a;f8a;H3
34.
1
27
a;f3 a^ + 27
a; — 4
2
+
a; + l
3
a\x a— X a\2 X
1 1
FRACTIONS 113
39 1 X :ii? Q gl g + l a^~l
■ xl iB2l"^a^l ' a + l"^al a^ + l'
41.
4nl . 371+1
6w217n412 lOw^QiiQ
42 a4 3al 5a^9a4ll
• 2a3 a + 2 2a2 + a6
43^ a? + 4 a?2 ^ a^43
44.
J^ 1 27yi
1 1 2 mn a /v*^
147. In certain cases, the principles of §§ 137 and 138
enable us to change the form of a fraction to one which is
more convenient for the purposes of addition or subtraction.
1 o Tj; 3 , 26 fa
1. Simplify + ^ ^ '
Changing the signs of the terms in the second denominator, at the
same time changing the sign before the fraction (§ 137), we have
3 26 + <z
ah a'^b^
The L. C. D. is now a"^  b^.
Then — ^ 2fe + a ^ 3(fl + 6)(2& + a)
' ab a^b^ a"^  b^
__ Sa + Sb2ba ^ 2a + b
a2  62 ^2 _ 52'
2. Simplify ^ ^
(xy)(xz) (yx)(yz) (zx)(zy)
By § 138, we change the sign of the factor y — x in the second denomi
nator, at the same time changing the sign before the fraction ; and we
change the signs of both factors of the third denominator.
The expression then becomes
1.1 1
(xy)(xz) (xy)(yz) {xz){yz)
114 ALGEBRA
The L. C. D. is now (x — y^ix — z) {y — z); then the result
_ (yg)+(a;g)(a;y) _ y z + xzx + y
(xy)(xz)(yz) (x  y){x  z){y  z)
2y2z _ 2(yz) 5
(xy)ixz)iyz) {x y)(x  z)(j)  z) (xy){x~z)
EXERCISE 53
Simplify the following :
4 1 g a a 2g?
3a3 22a
2.
3a; 2
a;2_16 4a;
3.
a\h a — b
a'Sab 3b^ab
4.
5 8m + 6
2ml l4m2
_ 6x'SxS2
6.
3 + a3a a2_9
4 3a; 2
a^ — x 1 — x X
7. _i__ 1 I ^6
71 + 4 1— ?2 71^4371 — 4
Q 3a 2a 8 a6
a + 26 26a a'4:b^
2 1
9a;^16a; 43a; x
10. , ^ 4
(xy)(xz) {yx){yz)
a3_63 a2 + a6462 &_«
12. „ 3 .+ „ ^ .+ ^
a;^— 5a;46 a;^ — a; — 2 4 — a;^
jQ 3m + l, m — 4 3m^ — 2m — 4
3ml 52m 67?i217m + 5
14. 1 . 1 , 1
(a — b)(a — c) (b — c)(b — a) (c — a)(c — b)
148. Reduction of a Mixed Expression to a Fraction.
4 a; — t]
a; 41
EJa;. Eeduce 2 a; — 3 — A to a fractional form.
FKACTIONS 115
We may regard 2x — 3 as a fraction having the denominator 1, and
use the rule of § 145 ; thus,
2x 3 4x5 ^ C2a;3)Ca;+l)(4a;5)
x + 1 X + 1
_ 2x'^a;34a; + 5 _ 2x2 5 a; + 2 ^
x+1 x+1
EXERCISE 54
Reduce each of the following to a fractional form :
1. ^ + 3 + ^. 10. 2a.^5a; + ME±iIl.
4a; 4x49
2. 2a5^^^. 11. 3a^f8^^(^^^).
7a ^ la2
3. 3n + 4 + ?_. 12. ^^^±f.(2x + y).L^
Sn + l 3a;h52/
4. ^^^ + 1. 13. ^4.5 1.
2n*
af2m a6a — 5
K . 5a; + y 14. m^m^n + mn^yiS
^2/ ^ I 97
6. 53n + ^^i^. ^a^4_3a; + 9
5 + 371
7 2 2ar6 16. (^2j£^ + l.
8. ^ + 2.y + ,f + ^. 17. M±^^+«^+2a+4.
X — 2y ^ — ^a+4:
9. a46^l±Mll 18..2a;+5y ^^ + ^5.^^ .
MULTIPLICATION OF FRACTIONS
149. Required the product of  and .
Let x = .. (1)
116 ALGEBRA
I
Multiplying both members by b x d (Ax. 7, § 9),
XXbxd = xxbxd, or
d
ix')%>^'^y>^^xd
for the factors of a product may be written in any order.
Now since the product of the quotient and the divisor gives
the dividend (§ 67), we have
 xb = a, and  x d = c.
b d
Whence, (a) x (c) = x xb x d.
Dividing both members by 6 x c? (Ax. 8, § 9),
b xd
From (1) and (2), ^ x ^ = ^. (Ax. 4, § 9)
Then, to multiply fractions, multiply the numerators together
for the numerator of the product, and the denominators for its
denominator.
150. Since c may be regarded as a fraction having the
denominator 1, we have, by § 149,
XCX
Dividing both numerator and denominator by c (§ 136),
a ,, a
X c =
b b^c
Then, to multiply a fraction by a rational and integral expres
sion, if 2^ossible, divide the denominator of the fraction by the
expression; otherwise, multiply the numerator by the expression.
151. Common factors in the numerators and denominators
should be cancelled before performing the multiplication.
Mixed expressions should be expressed in a fractional form
(§ 148) before applying the rules.
FRACTIONS 117
1. Multiply ^ by ^^.
10 a^y ^ 3 6^g3 _ 2 X 5 X 3 X a^b^x^y _ 5 6% _
9 6a;2 4 a^y^ 3^ x 22 x a^ftxV 6 y '
The factors cancelled are 2, 3, a^, &, x^, and y.
2. Multiply together ^j±2^, 2^^^, aud J^f.
_5l±l£_x /o a;  4 \ x2  9
x2 + 2 X ^ 2a;6a; + 4 ^ a;^  9
X2 + X6 x3
^ x(x + 2) ^ a;  2 ^ (X 4 3) (x  3) ^ x
(x + 3)(x2) x3 (x + 2)(x2) x2*
The factors cancelled are x + 2, x — 2, x + 3, and x — 3,
3. Multiply ^±^^ hj ab.
Dividing the denominator by a — 6, ^^^ — x (a — 6) = ^— i
a^ — 62 a + 6
4. Multiply —HL. by m + w.
m — n
Multiplying the numerator by m + n, ^ x (m + n) = «*!_+:^.
EXERCISE 55
Simplify the following :
*. 8cd« 35a'b'' ' 15a« 126^ 7c**
« 5 a^ 9 6^ ^ 7 c* c 28 m^ 15 n« 5 a^
3 6^ 10 c^ 6 a* 25 nV 14 mV 21 mV
118 ALGEBRA
7. ^^x(2a6). 9. ^i!fl?x ''
a2b ' 47i2 n' + 7i^2
8. _^Zl2 ^3^^ ^^j a2_2a35 4a3_9a
12 ^«^5^ x «' + 3« + ^
14.
16.
18.
a^27 4a220a + 25
4m^ + 8m + 3 6m^9m
2m25m + 3 4m2l
25 «^ \ mx 4 na; 4 mn a^ — w?
x^ — mx — nx{ mn ^ — v?
a''2ab + b^c^ a^bc
a'\2ab\b^c^ ab + c'
17 16^4^20^+5^a;2 + 2a; + l
5x{5 6x^6 16ar^l
a2lla430^ a23a^, a' 9
\ 2a+3 7^ 4a5 y
a^Sf x + 2y \ ^x'2xy + 4:yy
^21 9a;^ + 12aa; + 4a^ a;'^ + aa^ ^/ 2a^ + 2aa;a^
x'a' 3a; + 2a V 3x + 2a J
22 2n^nS n^44n + 4 n^yi2
n^_8n2 + 16 n^ + n 2n2_3n'
DIVISION OF FRACTIONS
152. Required the quotient of ^ divided by ^«
6 a
FKACTIONS 119
Let 1^1 = .. (1)
Then since the dividend is the product of the divisor and
quotient (§ 67), we have
a c
h d
Multiplying both members by  (A x. 7, § 9),
c
bed c
(2)
From (1) and (2), ^^l = ^x^
babe
(Ax. 4, § 9)
Then, to divide one fraction by another^ multiply the dividend
by the divisor inverted.
153. Since c may be regarded as a fraction having the
denominator 1, we have, by § 152,
a a 1 a
^c = x = — •
b b c be
Dividing both numerator and denominator by c (§ 136),
a aTc
Therefore, to divide a fraction by a rational and integral
expression :
If possible, divide the numerator of the fraction by the expres
sion ; otherwise y multiply the denominator by the expression.
154. Mixed expressions should be expressed in a fractional
form (§ 148) before applying the rules.
1. Divide 1^ by li^.
Wehave 6q2fe 9^253 ^ 6^25 ^^ lOa;^ ^ 4y8
* SxV • iOa;V 5a.3y4 9^253 352^;
120 ALGEBRA
2. Divide 2 — by 3 — —
_ 2x + 22a; + 3 . Sx^  3  3a;2 + 13
X + 1 ■ x2  1
r2
l_ 5(x + l)(xl) _ xl
x + 1 10 2x5x(x+l) 2
3. Divide ^?L^bymn.
Dividing the numerator by m  w, ^^ ~ ^^  (m  w) = ^'^ "^.^^ t ^'^ 
4. Divide ^^ bya + &.
a — b
/j2 4^2 <22 I 7)2
Multiplying the denominator by a + 6, — — ^ (a + &) = — ^— •
If the numerator and denominator of the divisor are exactly
contained in the numerator and denominator, respectively, of
the dividend, it follows from § 149 that the numerator of the
quotient may he obtained by dividing the numerator of the divi
dend by the numerator of the divisor ; and the denominator of
the quotient by dividing the denominator of the dividend by the
denominator of the divisor.
5. Divide ^^iy\y3x + 2y^
ar — y^ * x — y
We have, 9x2 4y2 ^ 3x + 2y _ 3x  2y .
x2 — 2/2 x — y x + y
EXERCISE 56
Simplify the following:
. 45 a: V . o , , ' 2 12 a%' . 9a^&«
* 4tfn ' ' ' 55<fd' ' 22 c'd'
FRACTIONS 121
' a3 ^ ^ a2 + 6a + 9 aH3
4. ^ — ^y^(x + 3y). 9. f^ + ^^f^^.
g n23n40 , yi2 + 4n5 ^^ a^3a;y . a;^10a:y+21y^
4:% ' 5n ' a^ — y^ ' x^{xy\y^
' a»9a62 ' a + 36* ' 4a;3 • <. ^+ >
7 9a^4?/^ . 9a^ + 6ir.y j^g 8yi« + l . 4n^2n + l
16a^25/ ' SxylOy^' ' 2n^^4.n ' n2 + 4n44 ■
13. 2
2 + 8a;3a;'
9
H'W)
14. (x'y2 + 2yzz')^ ^~y~^^ .
x\y + z
j5 m^ + 2m^ + m + 2 . m^ + 3m^H2
m^ — m^ — m + l * m"* — 2m^ + l
jg 2a^a636^ . 3a^ + a62&^
9a22562 • 9a230a6 + 2562'
COMPLEX FRACTIONS'
155. A Complex Fraction is a fraction having one or more
fractions in either or both of its terms.
It is simply a case in division of fractions; its numerator
being the dividend, and its denominator the divisor.
a
1. Simplify
= axo^(§152)= '^
h— ^^ — ^ bd — c^ ^ bd
It is often advantageous to simplify a complex fraction by
multiplying its numerator and denominator by the L. C. M. of
their denominators (§ 136).
122 ALGEBRA
a a__
2.' Simplify — —'
a—b a+b
The L. C. M. of a + & and ab is (a + b)(a  &).
Multiplying both terms by (a + 6)(a — 6), we have
a a
ah a + b _ a(a + 6) g (g  &) _ a^ \ ab  a^ h ab _ 2 ab
b j^ a ~h{a + b)+a{ab) ab + b'^ + a^  ah a^ + b'^'
a — b a{b
3. Simplify ^ —
1+ ^
1+i
X
a; + l 05 + 1
. 1 ■■ ■ X x + l\x 2x+l
1+1 ^+1
In examples like the above, it is best to begin by simplifying the lowest
complex fraction.
1 X
Thus, we first multiply both terms of by x, giving — —  ; and
111 X r 1
1 ^ x + 1
then multiply both terms of — by x + 1, giving •
i.X X+i+X
x + 1
EXERCISE 57
Simplify the following :
X y 9a m
x — y_x\y a\b _ 5
a;4y a; — y ' ?4?:
X y a b
FRACTIONS
123
6.
3a . fy . 4rb
a
21j
X
^4.1_^
x^y^
+
xy ^ 1xy
9.
10.
l\xy xy
^ 1_
1 — a; 1 \x
1a^ 1+a^
14.
15.
16.
17.
18.
27 g^ h
6^ a
h a
4a^ — 2/^
3 x4.y "
2xy
a; + 4
ic — 4
5ic— 1 5a; + 1
a; 44 ic — 4 '
5a;l
2yz
5a; + l
L^1
a:^ + y^
2xy
3a
(a + 2)^ a + 2
2 a^ + 2 a  1 c
(
a2
11. 1
1+a
13.
^
^a
4al
1
2a + 5
3a2
3ar^Hy^ a; + y
/
124 ALGEBRA
a — oi a? — Q?
2j a{x g^ + a^
a — x g^ — a^ '
a + a; a^ H a^
MISCELLANEOUS AND REVIEW EXAMPLES
EXERCISE 57 a
Reduce each of the following to a mixed expression :
J ^x'2a?20 2 m' + n\ ^ 12 g^3 g^22 g + 8
12 a^ * ' m — n ' 3a^ — 5
Simplify the following :
^•5^:x^^^^> ^^•5^1^.^67
^ ;:rTT4"^('^ + ^) a;*2a^ + 8a;16
g*4w*
2 ma; /n o „\ 13.
(a_6)2_(cd)^
7. ^ ""^" ^(2m3a;). (6 + d)2 _ (a + cV
_1_,1_ _! L
8. a^5a;84 . a;47 , 14. 2a;"^3y 3a; 2y
27a;»8 3a;2 4.x^9y^'^ 9 x'4.y^'
9 3a; I 1 ^^ + ^  15 g^6cZ  g6^c  acd^ 4 &c^^ .
2 a; 3 ' hhd + ahc"  abd^  d'cd
16 /« + 2 2 Y g 3\
■ (^ g '^g_3Aa2 a + s)
17.
fl^_2ar^42a;l
a;^ + ar^+l
jg _J_ ,_i § llg56 .
' a — h b — a g+ft W—a^
19 27 a;^ 41 . 15a;^a;2
■ 25a;24 * 25 a;^  20 a; + 4*
FRACTIONS 125
20 (2 ^ + 4a^2n . (x + 1 x^\
' [^ a^^2xSj ' [x2^x + 4.J
ft, 2c~d c\2d
ac^2ad^2bc + 4:bd 2 acad+4.bc 2bd
\^ a;H4 J \ x\4: J
23 0^ + 3 , a + 2 g + l
25. t:^ ?^=:^
x^~{yzy (xzfy^
og (a+26 + 3cy(2a3&cy
• (3a462c)2(2a + 66h2c/
27. 11^+ ^
X + 3 X — 3 x\4 x — 4:
(First combine the first two fractions, then the last two, and then add
these results.)
28 3 3 5n' 5 n'
271 + 1 2711 8n« + l Sn^1
30 &  c c — g a—b
' (ab)(ac) (bc){ba) {c  d)(c  b)
3. (2x^\5x2y25 ^
' (3a^4(»3)216*
32 ft I 1 3 a^ + 2a
• a + 3 a3 a29 a^ + O '
(First add the first two fractions, to the result add the third fraction,
and to this result add the last fraction.)
33 3q 3a C^a' , 12 a*
a+b ab a'^y' a* +6*
(§
126 ALGEBRA
'^ >34. A.J—^ +
a 2 2 a^ + 4
2(al) 2(a + l) a^ + l o>\
35 1 1 , 1
39.
_2 1 1 2
2 a; + 2/ 3^ + 2?/
3g 6 g^  g  2 8a^18a5 4a^9
4a216a+15 12a25a2 4a2 + 8a + 3'
(SiJ
a; + 1\' /a?  1
i)
.^_2 a;3 ic4 . (a;  2) (aj  3) (a;  4)
40 a^ + 2 a; + 3 a;44 3 a^ 10 a^ 21 a; + 66
FRACTIONAL AND LITERAL LINEAR EQUATIONS 127
XI. FRACTIONAL AND LITERAL LINEAR
EQUATIONS
SOLUTION OF FRACTIONAL LINEAR EQUATIONS
156. If a fraction whose numerator is a polynomial is )re
ceded by a — sign, it is convenient, on clearing of fractions,
to enclose the numerator in parentheses, as shown in Ex. 1.
If this is not done, care must be taken to change the sign of
each term of the numerator when the denominator is removed.
1. Solve the equation — = 4 H ^ —
4 5 10
The L. C. M. of 4, 5, and 10 is 20.
Multiplying each term by 20, we have
15a; 5 (16X20) =80 + 14x + 10.
Whence, 15 a;  5  16 x + 20 = 80 + 14 a: + 10.
Transposing, 15 a:  16 x  14 x = 80 + 10 + 5  20.
Uniting terms, » — 15 x = 75.
Dividing by — 15, x = — 5.
2 5 2
2. Solve the equation = 0.
^ x2 x\2 af4:
The L. C. M. of X  2, X + 2, and x^  4 is x2  4.
Multiplying each term by x^ — 4, we have
2 (X + 2)  5 (x  2)  2 = 0.
Or, 2 X + 4  5 X + 10  2 = 0.
Transposing, and uniting terms, —Sx= — 12, and x = 4.
If the denominators are partly monomial and partly poly
nomial, it is often advantageous to clear of fractions at first
partially ; multiplying each term of the equation by the L. C. M.
of the monomial denominators.
128 ALGEBRA
o a 1 ^v ^ 6x\l 2x4: 2a;l
3. Solve the equation — ^ — = — ;: — •
^ 15 7 a; 16 5
Multiplying each term by 15, the L. C. M. of 15 and 5,
a , , 30 5C60 ^„ o
^"'^ lx16^'^ "•
Transposing, and uniting terms, 4 = ' ^~..^ •
7 a: — 16
Clearing of fractions, 28 ic  64 = 30 x  60.
Then, — 2 a; = 4, and x =
2.
EXERCISE 58
In Exs. 5, 11, 22, and 32, of the following set, other letters
than X are used to represent unknown numbers.
This is done repeatedly in the later portions of the work.
Solve the following equations :
. 1__1 = ^ ]_ 2 i ? ?.=,L.
' 2 9x 9 6x ' 5x 10a; 15a; 12*
3 A_ J__ ^ 7 ^11
' 3 a; 12 a; 8 ac 24 x 8 *
4. 4^ + ^^±l = ^. 7. a;i^ + ^^±^ = 2.
5 2 38
5. ^ + 3_?rt^ = 2v. 8. 3a^ + 2 5a;6 ^5
5 15 * * 5 8 4
6 Z£_§_£ZL?_2 = — 9 ^^1^ 7a;+4 3a;8 __Q
■ 4 7 14* * 9 12 8 '
.Q 5(xl) 2(a; + 2) ^, 5a;15
6 3 4
„ lli) + 12 4p6 5p9 ^ o ,
18 9 "^ 4
12 8a;l ll'a;7 13 a; + 3 ^ 14 a; + 38
3 5 10 15 * .
FRACTIONAL AND LITERAL LINEAR EQUATIONS 129
13 5a; + 4 16 a;h5 ^ 10^^9 4(3a;2)
3 9 5 15 '
14 3(a; + 7) 7x + 10 ^ 4:X7 2(7xl)
7x Sx 6 21 '
.. (3a;4)(3a; + l) (8 a; 11) (a; + l) ^ (5 a; 1)(4 a; 3)
2 4 8 *
16.
A_ = o.. 18. 21^!±I^±11=.3.
4a;3 7a;3 7iB2_4a;9
17 6 a; + 1 ^ 2a; + 3 ^^ 2a; + 7_ 10a;3
22.
9x5 Zx2 a;2_4 5a;(a; + 2)
OQ 8a; + 57 ^ 2a;15 2a; + 16
12 aj + 8 3
o. 12 a; — 5 3a; + 4 ^ 4a; — 5
21 3(3a;fl) 7
3/11 5y?. + 4 _ o 2^ ?^JIl? = a;_l_JL^±iL.
n5 71 + 8 * '9 3(3a; + 4)*
23 6a;^ + 23 a;l ^^ 05 5a;4 2a;  7 ^ 4ar^ a;
' (2a;3)2 2a;3 * * 7 14 7a;2*
26. ? i 4 ^ = 0.
2a; + l 3a; + 2 6ar^ + 7a; + 2
07 1 ^ I 1 _Q
6a;24 6a; + 3 6a;4 *
28.
4a; ^ + 3a; + 2 ^ 6a;^5a;4
4a; + 3 6a;5
29 8a;^ + 4 _ hx 3a;
16a;2_255 + 4a."^54a;*
30 7a; 5a; ^ 12(a;^l)
a; + 3 1a; a;2_^2a;3*
31.
2ar^a; + 3 2a;^f3a;l ^ 20a;^6a;43
3a; + 2 3a;2 9ar^4
130 ALGEBRA
22 3 a: — 5 _ 4a; + 2 _ 15 a; — 1 7
2 3a; + 2~ 10 5*
34.
x + 2 x + 3~xi6 x + T
(Eirst add the fractions in the first member ; then the fractions in the
second member.)
! x\l x — 6 x — 7
35.
36.
x — 2 x — 1 x — 4: x — 5
4aj + 7 8« + 4 12£c + l 5a;l
5 15 45 9(5a;42)
37 a^' + S 1 _ 2«l
2(a;3_8) 6(032) 3(a;2 + 2a; + 4)
«Q 2a;l 1 , 2a; + 3 , 5«2_.30^
a;2 2 3a; + 10 2 (a;  2) (3 a; + 10)
39 2a; + l 5a;6 ^2 23a;^10
' 3a;5 2a; + 7 635^ + 11a;  35*
157. Solution of Special Forms of Fractional Equations.
1. Solve the equation ^^ + K^ = 2.
^ 2a;3 a;2_^4
We divide each numerator by its corresponding denominator ; then
14— ?— + 1^^+1 = 2 orA__^±i = o
2x3 a;2 + 4 ' 2x3 «2 + 4
Clearing of fractions, 2 a;2 + 8  (2 a:2 4. 5 a;  12) = 0.
Then, 2a:2 + 82a;25x + 12 = 0; whence, a; = 4.
We reject a solution which does not satisfy the given equation.
2. Solve the equation + ""
xS x2 0.2535 + 6
Multiplying both members by (x  3)(x — 2), or aj2 — 5 as + 6,
x2 + «3 = 3x7.
FRACTIONAL AND LITERAL LINEAR EQUATIONS 131
Transposing, and uniting terms, —x = — 2, or x = 2.
If we substitute 2 for x, the fraction becomes —
x2
Since division by is impossible, the solution x = 2 does not satisfy
the given equation, and we reject it ; the equation has no solution.
3 4 2^'
"3. Solve the equation — \ = 4
« + 10 » + 6 aj + 8 x + 9
Adding the fractions in each member, we have
7x + 58 _ 7a; + 58
{x + 10) (X + 6) ~ (X + 8) (X + 9)*
Clearing of fractious, and transposing all terms to the first member,
(7x + 58)(xH8)(a; + 9)(7x + 58)(« + 10)(x + 6) = 0. (1)
Factoring, (7 a; + 58) [ (x + 8) (x + 9)  (x + 10) (x + 6) ] = 0.
Expanding, (7 x + 58) (x^ + 17 x\12 x^ Wx 60) = 0.
Or, (7x + 58)(x412) = 0.
This equation may be solved by the method of § 125.
Placing 7 X 4 58 = 0, we have x =  ^••
Placing X + 12 = 0, we have x = — 12.
158. If we should solve equation (1), in Ex. 3 of § 157, by
dividing both members by 7 a; + 58, we should have
 (x{S)(x + 9)(x + 10){x{6) = 0.
Then, a^ + 17 a; + 72a^16 aj60 = 0, or cc = 12.
In this way, the solution ic = — ^ is lost.
r It follows from this that it is never allowable to divide both
I members of an equation by any expressioii which involves the
unknown numbers, unless the expression be placed equal to and
\the root preserved, for in this way solutions are lost,
EXERCISE 59
Solve the following equations :
J 2£f7 2^^^g 2 4a; 4 11
2ajHl x2 x'{x20 x + 5 a;~4
132 ALGEBRA
3 _8 3_^J^ 5_ g 2xjS 2a;3 36 ^^
a;+3 x7 x\9 x2 ' 2xd 2x^'^ ^:^9~
4 a; + 3 a; + 4 a^ + 2 ^g .^ 2a;+5 3a?^H24a;419 ^ .^^
a; + 2 a; + 3 a;H4 * * a;+7 a;2+8a;+7
r 3.2 1.4
, 4 e a;2_2a.4.5 a^_^3a._7
a;+9 a;+4 a;+3 aj+18 0.2—20;— 3 x2+3a;+l
SOLUTION OF LITERAL LINEAR EQUATIONS
159. A Literal Equation is one in which some or all of the
known numbers are represented by letters ; as,
2a; + a = 62_^10.
Ex. Solve the equation — ^^^ = ^^±1'.
x — a x\a 01? — (T
Multiplying each term by x^ — a^,
x{x + a)  (x + 2 6) (x  a) = a2 + 62,
or, x2 + ax  (x2 I 2 &x  ax  2 a&) = a^ + 62,
or, x2 4 ax  x2  2 6x + ax + 2 a6 = a2 + 6^,
or, 2 ax  2 6x = a2  2 a6 + 62.
Factoring both members, 2 x(a — 6) = (a — 6)2.
Dividing by 2(a  6), x = ^^=^ = ^•
In solving fractional literal equations, we must reject any solution
which does not satisfy the given equation. Compare Ex. 2, § 157.
EXERCISE 60
Solve the following equations :
1. (aiB6)(6a; + a) = 6(a«26).
2. (a;2a6)2=(a; + a + 26)2.
3 3a; x — 2n _fy a 4a;H3a 6ct + 56 _q
2x + n 2x ' ' 4a;— 3a 6a — 5b
FRACTIONAL AND LITERAL LINEAR EQUATIONS 133
. — + — + " = a4& + c.
ao 00 ca
6 a7(a + 4 6) — 6^ x—b _ xja
a^ — b^ a\b a — b
y m^x\n _ n^x + m _m — n
mx nx mnx
g X — a x — b X — c _ bG(x \ b) — ah^ — a^c \ abx
b c a abc
Q 5 2 3 m
2a) + 5m 3 a;  4 m ~ 6 a^ + 7 ma?  20 m^*
10 ^ + 6 I g — 26 _ (2a — 6)a; + 3a6
' a; a; + a oc^ — a^
11 ^^ g^ + ?>^ _ g^ a; (g — 6)
12.
g g2 6^ &
& g — 6
a; — g x — b X
13_ «(»«) + H^&) ^„ + 6.
X — b x — a
14. (g + 6)(a;g + 6)(a6)a; + g262 = 2g(a; + g
15. (x jp + g)(a; p + g) + ^2 ^ (x p)(x + g).
16 ^^ + ^^ 4:X — 5n _ 10 n^
x\2n Sn — x a^ — nx — 6n^
jiy 3a; 5ga; — 26 ^ a + 3 6a; ax{2a? ~^h ^
'2 4 a 8?) 1fia/> *
6).
16 a6
b a— b
jg _a o___ g—
a;46 x[a a;Hg + 6
iQ a; + g a; — g 2ga; — 19g2 _^
« — 2g a;3g x^\ax — Qa?
20. J^ i_=_I L_.
a;— 2a 6a; + g 3a; — 8g 2x — Sa
134 ALGEBRA
21. 4 14 1
x — 4:n op\n x\4:n x\3n
22 a^ — 2 aa? + ft^ , x^\ax — 2a^ _ q
x^ — 2ax — Sa^ x^ \ ax \ 2 a^
no x\a x + b x — a — h _ o
x — a x — b x{a\b
24. x''^(xay+(xby = 3x(xa)(xb).
SOLUTION OF EQUATIONS INVOLVING DECIMALS
160. Ex. Solve the equation
.2 a? + .001  .03 a; = .113 a^  .0161.
Transposing, .2x .OZx .113 x =  .0161  .001.
Uniting terms, .057 x = — .0171.
Dividing by .057, x =  .3.
EXERCISE 61
Solve the following equations :
1. 7.98 a? 3.75 = .23 a; + .125.
2. 3 a; + . 052  7.8 a? = .04 5.82 a;. 0696.
3. .05i;1.82.7'y = .008v.504.
(Here, v represents the unknown number.)
4. .73 x + 8.86 = .6(2.3 a; .4).
5. .07(8aj5.7) = .8(5a: + .86) + 1.321.
6. 3.2 X. 84 + ^^ ^^^^ = .9 a;.
.9
„ 6.15 a; + .67 .6aj.81 5
3a;
— (
.3 ~J
3.1
.6 .03
V3 .9a;2.84 .8a;6.52 ^^^
9. 20.1a.:ni^=lM§?^^:^_.135.
FRACTIONAL AND LITERAL LINEAR EQUATIONS 135
PROBLEMS INVOLVING LINEAR EQUATIONS
161. The following problems lead both to integral and frac
tional equations; the former being somewhat more difficult
than those of Exercise 24.
1 . A can do a piece of work in 8 days which B can perform
in 10 days. In how many days can it be done by both working
together ?
Let X = the number of days required.
Then, _ = the part both can do in one day.
X
Also,  = the part A can do in one day,
8
and — = the part B can do in one day.
10
By the conditions, \ = —
8 10 X
Clearing of fractions, 5 x + 4 oj = 40, or 9 ic = 40.
Whence, x = 4, the number of days required.
2. The second digit of a number exceeds the first by 2 ; and
if the number, increased by 6, be divided by the sum of its
digits, the quotient is 5. Find the number.
Let X = the first digit.
Then, x\2 = the second digit,
and 2 a; + 2 = the sum of the digits.
The number itself is equal to 10 times the first digit, plus the second.
Then, 10 a; + (x + 2), or 11 x + 2 = the number.
By the conditions, lla; + 2 + 6 ^ ^
2x + 2
Whence, 11 x + 8 = lOx + 10, and x = 2.
Then, 11 x + 2 = 24, the number required.
136 ALGEBRA
3. Divide 44 into two parts such that one divided by the
other shall give 2 as a quotient and 5 as a remainder.
Let 11 = the divisor.
Then, 44 — ?i = the dividend.
Now since the dividend is equal to the product of the divisor and
quotient^ plus the remainder, we have
44 — w = 2 w + 5, whence — 3 « = — 39.
Then, w = 13, the divisor,
and 44 — w = 31, the dividend.
4. Two persons, A and B, 63 miles apart, start at the same
time and travel towards each other. A travels at the rate of
4 miles an hour, and B at the rate of 3 miles an hour. How
far will each have travelled when they meet ?
Let 4 X = the number of miles that A travels.
Then, 3x = the number of miles that B travels.
By the conditions, 4 x + 3 x = 63.
Then, 7 x = 63, and x = 9.
Whence, 4 x = 36, the number of miles that A travels,
and 3x = 27, the number of miles that B travels.
It is often advantageous, as in Ex. 4, to represent the unknown
number by some multiple of x instead of by x itself.
5. At what time between 3 and 4 o'clock are the hands of a
watch opposite to each other ?
Let X = the number of minutespaces passed over by the minutehand
from 3 o'clock to the required time.
Then, since the hourhand is 15 minutespaces in advance of the minute
hand at 3 o'clock, X — 15 — 30, or x — 45, will represent the number of
minutespaces passed over by the hourhand.
But the minutehand moves 12 times as fast as the hourhand.
Whence, x = 12 (x  45), or x = 12 x  540.
Then,  11 x =  540, and x = 49,Jt,
Then the required time is 49^^ minutes after 3 o'clock.
FRACTIONAL AND LITERAL LINEAR EQUATIONS 137
EXERCISE 62
1. The denominator of a fraction exceeds twice the numera
tor by 4. If the numerator be increased by 14, and the denomi
nator decreased by 9, the value of the fraction is J. Find the
fraction.
2. Divide 197 into two parts such that the smaller shall be
contained in the greater 5 times, with a remainder 23.
3. A piece of work can be done by A in 2f hours, and by B
in 4i hours ; in how many hours can the work be done by both
working together ?
4. The second digit of a number of two figures exceeds the
first by 5 ; and if the number, increased by 6, be divided by
the sum of the digits, the quotient is 4. Find the number.
5. At what time between 12 and 1 o'clock are the hands of
a watch opposite to each other ?
6. At what time between 7 and 8 o'clock is the minutehand
of a watch 10 minutes in advance of the hourhand ?
7. A piece of work can be done by A and B working together
in 10 days. After working together 7 days, A leaves, and B
finishes the work in 9 days. How long will A alone take to
do the work ?
8. Divide 54 into two parts such that twice the smaller shall
be 3 times as much above 29 as 4 times the greater is below
143.
9. At what time between 8 and 9 o'clock are the hands of a
watch together ?
10. The numerator of a fraction exceeds the denominator by
5. If the numerator be decreased by 9, and the denominator
increased by 6, the sum of the resulting fraction and the given
fraction is 2. Find the fraction.
11. At what time between 2 and 3 o'clock is the minute
hand of a watch 5 minutes behind the hourhand ?
138 ALGEBRA
12. The second digit of a number of two figures is i the
first; and if the number be divided by the difference of its
digits, the quotient is 15, and the remainder 3. Find the
number.
13. A garrison of 700 men has provisions for 11 days.
After 3 days, a certain number of men leave, and the pro
visions last 10 days after this time. How many men leave ?
14. A woman buys a certain number of eggs for $ 1.05 ; she
finds that 7 eggs cost as much more than 18 cents as 8 eggs
cost less than 27 cents. How many eggs did she buy ?
15. The width of a field is  its length. If the width were
increased by 5 feet, and the length by 10 feet, the area would
be increased by 400 square feet. Find the dimensions.
16. After A has travelled 7 hours at the rate of 10 miles in
3 hours, B sets out to overtake him, travelling at the rate of 9
miles in 2 hours. How far will each have travelled when B
overtakes A ?
17. The first digit of a number of three figures is f the
second, and exceeds the third digit by 2. If the number be
divided by the sum of its digits, the quotient is 38. Find the
number.
18. A, B, and C divide coins in the following way : as often
as A takes 5, B takes 4, and as often as A takes 6, C takes 7.
After the coins have been divided, A has 29 fewer than B and
C together. How many coins were there ?
19. A can do a piece of work in 3^ hours, B in 3 hours,
and C in 3f hours. In how many hours can it be done by all
working together ?
20. A man walks 13 miles, and returns in an hour less time
by a carriage, whose rate is f as great as his rate of walking.
Find his rate of walking.
21. At what times between 4 and 5 o'clock are the hands of
a watch at right angles to each other?
FRACTIONAL AND LITERAL LINEAR EQUATIONS 139
22. A man borrows a certain sum, paying interest at the
rate of 5%. After repaying $180, his interest rate on the
balance is reduced to 4^%, and his annual interest is now less
by $ 10.80. Find the sum borrowed.
23. The digits of a certain number are three consecutive
numbers, of which the middle digit is the greatest, and the
first digit the least. If the number be divided by the sum of
its digits, the quotient is ^ Find the number.
24. A certain number of apples were divided between three
boys. The first received onehalf the entire number, with one
apple additional, the second received onethird the remainder,
with one apple additional, and the third received the remain
der, 7. How many apples were there ?
25. A freight train runs 6 miles an hour less than a pas
senger train. It runs 80 miles in the same time that the
passenger train runs 112 miles. Find the rate of each train.
26. A and B each fire 40 times at a target ; A's hits are one
half as numerous as B's misses, and A's misses exceed by 15
the number of B's hits. How many times does each hit the
target ? »
27. A freight train travels from A to B at the rate of 12
miles an hour. After it has been gone 3 hours, an express
train leaves A for B, travelling at the rate of 45 miles an hour,
and reaches B 1 hour and 5 minutes ahead of the freight.
Find the distance from A to B, and the time taken by the
express train.
28. A tank has three taps. By the first it can be filled in
3 hours 10 minutes, by the second it can be filled in 4 hours
45 minutes, and by the third it can be emptied in 3 hours
48 minutes. How many hours will it take to fill it if all the
taps are open ?
29. A man invested a certain sum at 3f %, and \^ this sum
at 4^% ; after paying an income tax of 5%, his net annual
income is $ 195.70. How much did he invest in each way ?
140 ALGEBRA
30. A train leaves A for B, 210 miles distant, travelling at
the rate of 28 miles an hour. After it has been gone 1 hour
and. 15 minutes, another train starts from B for A, travelling
at the rate of 22 miles an hour. How many miles from B will
they meet ?
31. A can do a piece of work in  as many days as B, and
B can do it in  as many days as C. . Together they can do
the work in 3^^ days. In how many days can each alone do
the work ?
32. A vessel runs at the rate of llf miles an hour. It takes
just as long to run 23 miles up stream as 47 miles down
stream. Find the rate of the stream.
33. A man starts from his home to catch a train at the rate
of one yard in a second, and arrives 2 minutes late. If he had
walked at the rate of 4 yards in 3 seconds, he would have been
3^ minutes too early. Find the distance to the station.
34. A crew has bread for a voyage of 50 days, at IJ lb. each
a day. After 20 days, 7 men are lost in a storm, and the
remainder of the crew have a daily allowance of li lb. for the
balance of the voyage. Find the original nilmber of the crew.
35. A man invests $ 230 at 4^ %. He then invests a certain
part of a like sum at 3^%, and the balance at 5J%, and
obtains the same income. How much does he invest at each
rate ?
36. At what times between 5 and 6 o'clock do the hands of
a watch make an angle of 45° ?
37. At a certain time between 12 noon and 12.30 p.m., the
distance between the hands is f as great as it is 10 minutes
later. Find the time.
38. A woman sells half an egg more than half her eggs.
She then sells half an egg more than half her remaining eggs.
A third time she does the same, and now has 3 eggs left.
How many had she at first ?
FRACTIONAL AND LITlJRAL LINEAR EQUATIONS 141
39. A merchant increases his capital annually by \ of itself.
He adds to his capital $ 300 at the end of the first year, and
$350 at the end of the second; and finds at the end of the
third year that his capital is f of his original capital. Find
his original capital.
40. A and B together can do a piece of work in 5^ days,
B and C together in 6f days, and C and A together in 5 days.
In how many days can it be done by each working alone ?
41. A fox is pursued by a hound, and has a start of 77 of
her own leaps. The fox makes 5 leaps while the hound makes
4 ; but the hound in 5 leaps goes as far as the fox in 9. How
many leaps does each make before the hound catches the fox ?
42. A man puts a certain sum into a savings bank paying
4 % interest. At the end of a year he deposits the interest,
receiving interest on the entire amount. At the end of a
second year and a third year he does the same, and now has
$ 2812.16 in the bank. What was his original deposit ?
PROBLEMS IN PHYSICS
1. The density of a substance is defined as the number of
grams in one cubic centimeter. Hence the total number of
grams, M, in any body is equal to its density, D, multiplied by
its volume, F; or, to state this relation algebraically,
M=DV,
V being given in cubic centimeters, and D in grams.
Two blocks, one of iron and one of copper, weigh the same
number of grams; the iron has a volume of 10 cubic centi
meters and a density of 7.4 ; the copper has a density of 8.9.
Find the volume of the copper block.
2. When 100 grams of alcohol, of density .8, is poured into
a cylindrical vessel, it is found to fill it to a depth of 10 centi
meters. Find the area of the base of the cylinder in square
centimeters.
3. A cylindrical iron bar, 2 centimeters in diameter, has a
mass of 3 kilograms. Find the length of the bar.
Let TT = 3f
142 ALGEBRA
4. When a body is weighed under water, it is found to be
buoyed up by a force equal to the weight of the water which it
displaces.
If a boy can exert a lifting force of 120 pounds, how heavy
a stone can he lift to the surface of a pond, if the density of
stone is 2.5 and that of water 1 ?
5. When a straight bar is sup g ^ 2 ^ f
ported at some point, o (Fig. 1),
and masses m^, m^, etc., are hung
from the bar as indicated in the mi LJ fl m*. m&
figure, it is found that when the " ^*
bar is in equilibrium, the follow ^iQ !•
ing relation always holds,
mi X ao + W2 X bo = m^ X GO \ m^ X do \ m^ X eo.
If a teeter board is 10 feet long, where must the support be
placed in order that a 70pound boy at one end may balance a
60pound boy on the other end plus a 40pound boy 3 feet from
the other end ?
6. A bar 40 inches long is in equilibrium when weights of
6 pounds and 9 pounds hang from its two ends. Find the posi
tion of the support.
7. If in Fig. 1, ao = 100, bo = 40, co = 30, do = 60, eo = 110,
and if mj = 40, m2 = 60, mg = 60, m^ = 15, and m^ = 5, where
must a mass of 100 be placed in order to produce equilibrium ?
8. A gas expands ^73 ^^ '^^^ volume at 0° centigrade for
each degree of rise in its temperature ; i.e., the volume, F«, at
any temperature, t, is connected with the volume, Vo, at the
temperature 0° centigrade by the equation
or F.= Fo(l + 2kO
To what volume will 100 cubic centimeters of air at 0° expand
when the temperature rises to 50° centigrade ?
9. To what volume will 100 cubic centimeters of air at 50°
centigrade contract when the temperature falls to 0° centigrade ?
FRACTIONAL AND LITERAL LINEAR EQUATIONS 143
10. To what volume will 100 cubic centimeters of air at 50°
expand when the temperature changes to 75° ?
11. When a body in motion collides with a body at rest, the
momentum of the first body (i.e., the product of its mass, m^, by
its original velocity, v^) is found to be in every case exactly
equal to the total momentum of the two bodies after collision
(i.e., to the product of the mass, mg, of the second body times
the velocity, Vg, which it acquires, plus the product of rrii by
the velocity, ^3, which it retains after the collision). The alge
braic statement of this relation is
A billiard ball, the mass of which is 50 grams, and which
was moving at a velocity of 1500 centimeters a second, collided
with another ball at rest which weighed 30 grams. In the
collision the first ball imparted to the second a velocity of 1600
centimeters per second. Find the velocity of the first ball after
the collision.
PROBLEMS INVOLVING LITERAL EQUATIONS
162. Prob. Divide a into two parts such that m times the
first shall exceed n times the second by b.
Let X = one part.
Then, a — x = the other part.
By the conditions, mx = n(a — x) + b.
mx = an — nx + b,
mx ■\ nx = an + b.
x(to + 7i) = an + 6.
Whence, x = «!L±_^ , the first part. (1)
m + n
A„j „ „ ^ an + b am + an — an — b
And, a — x= a — = ■ —
m + n m \ n
am — b
m \ n
, the other part. (2)
144 ALGEBRA
The results can be used as formuloe for solving any problem of the
above form.
Thus, let it be required to divide 25 into two parts such that 4 times the
first shall exceed 3 times the second by 37.
Here, a = 25, m = 4, n = 3, and b = 37.
Substituting these values in (1) and (2),
the first part
^25_
X 3 + 37 _
7
_ 75 + 37 _
7
112 _
7
= 16,
and the second part
25.
X437
7
_ 100  37 .
7
_63_
7 "
= 9.
EXERCISE 63
1. Divide a into two parts whose quotient shall be m.
2. If A can do a piece of work in m hours, and A and B
together in n hours, in how many hours can B alone do the
work ?
3. Divide a into two parts such that the sum of onemth the
first and oner<th the second shall equal b.
10
4. A courier who travels a miles a day is followed by another
who travels Similes a day. How maij^ days must the second
start after the first to overtake him after c days ?
5. Divide a into three parts such that the first shall be
onemth the second and onenth the third.
6. The length of a field is m times its width. If the length
were increased by a feet, and the width by b feet, the area
would be increased by c square feet. Find the dimensions of
the field.
7. A courier who travels a miles a day is followed after b
days by another. How many miles a day must the second
courier travel to overtake the first after c days ?
8. If A can do a piece of work in a hours, B in 5 hours, C
in c hours, and D in d hours, how many hours will it take to
do the work if all work together ?
FRACTIONAL AND LITERAL LINEAR EQUATIONS 145
9. A vessel can be filled by two taps in a and h minutes,
respectively, and emptied by a third in c minutes. How many
minutes will it take to fill the tank if all the taps are open ?
10. Divide a into two parts such that one shall be m times
as much above h as the other lacks of c.
11. A can do a piece of work in onemth as many days as B,
and B can do it in oneTith as many as C. If they can do the
work in p days, working together, in how many days can each
alone do the work ?
12. A was m times as old as B a years ago, and will be n
times as old as B in 6 years. Find their ages at present.
13. How many minutes after n hours after 12 o'clock will
the hands of a watch be together ?
14. A and B together can do a piece of work in a hours,
B and C together in h hours, and A, B, and C together in c
hours. In how many hours can each alone do the work ?
15. How many minutes after 2 o'clock will the minutehand
of a watch be n minutes in advance of the hourhand ?
16. A and B together can do a piece of work in m days,
B and C together in n days, and C and A together in p days.
How many days will it take to do the work if all work
together ?
17. A sum of money, amounting to m dollars, consists
entirely of quarters and dimes, there being n more dimes than
quarters. How many are there of each ?
146 ALGEBRA
XIL SIMULTANEOUS LINEAR EQUATIONS
CONTAINING TWO OR MORE UNKNOWN NUMBERS
163. An equation containing two or more unknown numbers
is satisfied by an indefinitely great number of sets of values of
these numbers.
Consider, for example, the equation x\y = 5.
Putting x = l, we have 1 f ?/ = 5, or ?/ = 4.
Putting x = 2, we have 2 {y= 5, or y = S; etc.
Thus the equation is satisfied by the sets of values
x = l,y = 4:,
and x = 2, y = 3; etc.
An equation which is satisfied by an indefinitely great num
ber of sets of values of the unknown numbers involved, is
called an Indeterminate Equation.
• 164. Consider the equations
f ^ + 2/ = 5, (1)
[2x + 2y=10. (2)
Equation (1) can be made to take the form of (2) by multi
plying both members by 2 ; then, every set of values of x and
y which satisfies one of the equations also satisfies the other.
Such equations are called equivalent.
Again, consider the equations
^42/ = 5, (3)
[x^y = 3. (4)
In this case, it is not true that every set of values of x and y
which satisfies one of the equations also satisfies the other;
thus, equation (3) is satisfied by the set of values x =3, y = 2,
which does not satisfy (4).
If two equations, containing two or more unknown numbers,
are not equivalent, they are called Independent.
SLMULTANEOUS LINEAR EQUATIONS 147
165. Consider the equations
x + y = 5y (1)
x\y = 6. (2)
It is evidently impossible to find a set of values of x and y
which shall satisfy both (1) and (2).
Such equations are called Inconsistent.
166. A system of equations is called Simultaneous when each
contains two or more unknown numbers, and every equation
of the system is satisfied by the same set, or sets, of values
of the unknown numbers ; thus, each equation of the system
(x + y = 5,
\x~y = S,
is satisfied by the set of values x — 4:,y = l.
A Solution of a system of simultaneous equations is a set of
values of the unknown numbers which satisfies every equation
of the system ; to solve a system of simultaneous equations is
to find its solutions.
167. Two independent simultaneous equations of the form
ax{by = c may be solved by combining them in such a way as
to form a single equation containing but one unknown number.
This operation is called Elimination.
ELIMINATION BY ADDITION OR SUBTRACTION
168. 1. Solve the equations
Multiplying (1) by 4,
Multiplying (2) by 3,
Adding (3) and (4),
Whence,
Substituting x = 2 in (1),
Whence,
The above is an example of elimination by addition.
'5x3y=::19.
(1)
,7x\4.y= 2.
(2)
20 a:  12 ?/ = 76.
(3)
21x + 12y= 6.
(4)
41 X = 82.
(5)
x= 2.
(6)
10  3 ?/ = 19.
(7)
— Sy = 9, or'y = 
3.
(8)
15x + Sy= 1.
(1)
10x7y = 24:,
(2)
SOx + iey= 2.
(3)
30 a; 21y=72.
(4)
148 ALGEBRA
We speak of adding a system of equations when we mean placing the
sum of the first members equal to the sum of the second members.
Abbreviations of this kind are frequent in Algebra ; thus we speak of
multiplying an equation when we mean multiplying each of its terms.
2. Solve the equations
Multiplying (1) by 2,
Multiplying (2) by 3,
Subtracting (4) from (3), S7 y = 74, and y = 2.
Substituting y = 2 in (1), 15 a; + 16 = 1.
Whence, 15 x = — 15, and x = — 1.
The above is au example of elimination by subtraction.
From the above examples, we have the following rule :
If necessary, multiply the given equations by such numbers as
will make the coefficients of one of the unknown numbers in the
resulting equations of equal absolute value.
Add or subtract the resulting equations according as the coeffi
cients of equal absolute value are of unlike or like sign.
If the coefficients which are to be made of equal absolute value are
prime to each other, each may be used as the multiplier for the other
equation ; but if they are not prime to each other, such multipliers should
be used as will produce their lowest common multiple.
Thus, in Ex. 1, to make the coefficients of y of equal absolute value,
we multiply (1) by 4 and (2) by 3 ; but in Ex. 2, to make the coefficients
of X of equal absolute value, since the L.C.M. of 10 and 15 is 30, we mul
tiply (1) by 2 and (2) by 3.
EXERCISE 64
In several examples in the following set other letters than a?*]
and y are used to represent unknown numbers.
Solve by the method of addition or subtraction :
• l4i»f 2/ = 14. ' l7a;f42/ = 23.
■ l3a;82/ = 35. " 1 51/+ 9a; = 23.
SIMULTANEOUS LINEAR EQUATIONS
149
ilx
[Sx
15a;
I 6x
32/= 10.
5y = 5.
11.
'■1
15a;f Sy = 3.
12y = 5.
10x\15y = 22.
7x\20y =  4.
{>•{
6x\lly = 31.
6y — 11 X = 74.
9u + 6v = 16.
lSu{7v = 22.
19.
13 12a^ll2/ =
{12yllx = 21
8.1 *— 82/ =  3. i4_.
1 11 a; + 5 2/ = — 15.
24p 7^ =
52.
.18i> + 13i = 
34.
1 9aj142/ = 30. f 19a; + 202/ =
* l21a; + 132/= 67. ' l21a; + 162/ =
35.
57.
•^ fl3?n7w = 15. _ ri2a; + ll2/ =
10. 1 16. 1
I 8m47i= 9. l28aj172/ =
172.
60.
ELIMINATION BY SUBSTITUTION
169. Ex. Solve the equations \ ^
l82/5ic = 17.
(1)
(2)
Transposing — 5 a; in (2), 8 y = 5 x — 17.
Whence, y = ^^^^.
8
(3)
Substituting in (1), . 7a;  9^^^ ~ ^^) = 15.
(4)
Clearing of fractions, 56ic9(5ic17) = 120.
Or, 66 a;  45 X + 153 = 120.
Uniting terms, 11 a; = — 33.
Whence, a; = — 3.
(5)
Substituting x =  3 in (3), y= ~ ^^ ~
iI = 4.
(6)
From the above example, we have the following rule :
From one of the given equations find the value of one of the
unknown numbers in terms of the other, and substitute this value
in place of that number in the other equation.
' V
EXERCISE 65
Solve by the method of substitution
r x + 2y=:ll:
• [3x}5y = 29.
9.
^8e3/=47.
.6e7/=21.
(2x\ y= 8.
' [7x4.y = 4.3.
10.
'4.xlly = 
^9x^ Sy =
71. '
4.
(5x6y = 9.
' l3x52/ = 4.
11. ^
'6x\12y =
.32/ 4a^ = 
41.
9.
^ r3.T + 72/ = 12.
* \9yQ>x= 1.
12.
'7a;+ 62/ = 
. 9 a; + 10 2/ = 
13.
11.
r57H 2r = — 4.
1 8 m — 15 ?; =
' I12m+ 6v =
18.
11.
{3xby= 38.
' l3?/5aj = 26.
9a; + 82/ = 57.
' l6a; + 7?/ = 48.
7.
25a;12 2/ = 19.
15. ^
'18aj102/ =
15 2/14a; =
29.
.10a;+ 42/ =  1.
24.
8.
.10a; + 92/ = 6.
16.
7^92/ = 
. llfl; + 4 2/ = 
22.
■89.
ELIMINATION
BY COM
PARISON
■ 2 a;
170. Ex. Solve the equations ■
5y = 16.
\Ty= 5.
(1)
(2)
Transposing — by in (1),
2x=5
ylQ
"Whence,
x=^
y16
2
(3)
Transposing 7 ?/ in (2),
3x = 5
72/.
Whence,
x.= ^
3
(4)
Equa
ting values of x^
2
7y,
3
(5)
SIMULTANEOUS LINEAR EQUATIONS
151
Clearing of fractions,
15 y _ 48 = 10  14 y.
Transposing,
29 y = 58.
Whence,
y = 2.
(6)
Substituting y = 2 in (3),
. = l«l«.3.
(7)
From the above example, we have the following rule :
From each of the given equations, find the value of the same
unknoivn number in terms of the other, a7id place these values
equal to each other.
EXERCISE 66
Solve by the method of comparison :
I
3.
6.
8.
2x\3y = U.
ic f 4 2/ = 17.
6x 2/= 27.
Sy3x = 36.
3x\2y = 31.
7x3y = 34:.
5iB8?/ = 46.
2x\3y =  6.
5 x — 2y= 4.
4:x + 7y = 29.
jr3s = lS.
4r — 5s = — 7.
3xSy = 13.
6» — 4?/ = — 5.
807 7y =  6.
6x + lly = 37.
9.
10.
11.
12.
13.
14.
9x{5y = — 1.
9y6x= 13.
S x\ 5y = o.
12iK + 102/ = 7.
10/1+ 6k = 15.
Uk15h = 58.
Sx+ 9?/= 8.
10a? + 12 2/ = ll.
lld + 16^ = 64.
7 d  12 ^ = 13.
Sx7y = 68.
16y5x = S9.
(9x 6y = 17.
' l7a7415?/ = 46.
16.
\ioy
8 a; + 5 2/ = 49.
13xh6y = S6.
171. If the given equations are not in the form ax^by = c,
they should first be reduced to this form, when they may be
solved by either method of elimination.
152 ALGEBRA
\— — =0. (1)
1. Solve the equations j a; I 3 y\4: ^^
[x{y2)y(x5) = 13, (2)
Multiplying each term of (1) by (x + 3)(2/ + 4),
7 2/ + 28  3 X  9 = 0, or 7 y  3 X =  19. (3)
From (2), xy 2xxij + 6y = lS, or by 2x = lS. (4)
Multiplying (3) by 2, UyQx =  38. (5)
Multiplying (4) by 3, 15 2, _ 6 x =  39. (6)
Subtracting (5) from (6), y =: _ 1.
Substituting in (4) , _ 5 _ 2 x =  13.
Whence, 2x = — 8, or x = 4.
In solving fractional simultaneous equations, we reject any
solution which does not satisfy the given equations.
{2x\Sy =13. (1)
2. Solve the equations ] 1 1 _ ^ ,^,
[x2 y^~ ' ^"^
Multiplying each term of (2) by (x — 2) (y — 3), we have
?/  3 4 X  2 = 0, or y =  X + 6. (3)
Substituting in (1), 2 x  3 x + 15 = 13, or x = 2.
Substituting in (3), ?/ =  2 + 5 = 3.
This solution satisfies the first given equation, but not the second ; then
it must be rejected.
y EXERCISE 67
Solve the following :
1. \
2«_5j^_l
5 6 2'
X ■ 5y_5
6 T2*
3.
r ila;3y ^ 3a;4y
2. 11 8 * 4.
[ Sx5y = l,
' 4:X — Sy X
14
9
1.
2x
+ 32/ = 
10
e2t8
2
7*
5e + 2t =
7.
SIMULTANEOUS LINEAR EQUATIONS
11. '
153
5.
\ 2 5
8.
9.
10.
11.
x{5 y\l
= 0.
6. \
X ^ = 5.
11
9 x + 5
2 3
3 2/.
y r (2 a;  1)(2/  4)  (oj  5)(2 2/ + 5) = 121.
l 4a;32/ = 29.
8
07 — 3
9
y5
5
^2ajl 32/44
' cc 4 11 ?/ — 6
= 0.
= 0.
4.
10
7
x1
2
c?2n ^_1.
3cZ + /i + 3 5*
c; + 3 7i ^7
(^+4 7i_7 11 '
.08a; + .9i/ = .048.
.3 a; .35 2/ =.478.
12.
13.
{
15.
' 2x—3y 'ix^Qy ^ _1
4 "*" 3 ~ 2
5x{2y 7y3x ^S9
2 "^ 5 10*
a^f2/ ^ 1 .
cc — 2/ 10
3a; + 8 ^ 6.Tl
2/4 ~22/ + 3*
5a;i(3a522/ + 5)=ll.
(a^4 2/)K«^2/) = 16.
'8
x
4
^3
,,^,
6
aj
2/7
5
5
8
16.
17.
18.
\6x±By_3x^^ ^
16 5 ^
5x — 4:y _
5x\4:y
13
2fl; + 5y4l 3x42/3 _ ^, o,, o
a; — 4 2/ 4 6
8 a; 2 2/ 18
154
19.
20.
3xyl
ALGEBRA
1
3x + 5 2y
= 0.
. i(3^l..)i()=^
5x7y + 2 3x — 4:yj7
2/44.
21.
22.
23.
7x — 3y\4. ex — 5y\7 _ r>
4 5
5^_2^ cc 3 ?/
12 ^_3 2__
7 23 ~
4 2
U2 2/44(2a;2/)=0.
'21» + 1 5v + 2 63.^130v
3
21
3a; + l 2x{y x + 2y ^^
7 2 8
4 a; — 2 5x{Ay _x — y
2a; + 3 3a; + 4
2 5
17
= 0.
24. \4:y\5, 62/ + 7 2(4 2/ + 5) (6 2/ + 7)
t (a^_l)(2/ + 2)(a; + 3)(2/44) = 12.
25.
[ :08^±35 __ :13x±^ ^ .32 ^ + .17 2, + .21.
.15 .6
.02 a; + .17 .08 y  .47
.9
0.
172. Solution of Literal Simultaneous Equations.
In solving literal simultaneous linear equations, the method
of elimination by addition or subtraction is usually to be
preferred.
SIMULTANEOUS LINEAR EQUATIONS
155
Ex. Solve the equations
Multiplying (1) by b\
Multiplying (2) by 6,
Subtracting,
Whence, ,
Multiplying (1) by a',
Multiplying (2) by a,
Subtracting (3) from (4),
Whence,
J ax "1 by = c.
l a'x + h^y = c'.
ah'x + Wy  h'c.
a'hx + hh'y = &c'.
(a6'  a'h)x = h'c  he'.
h'c he'
ah'  a'h
aa'x + a'hy = ca'.
aa'x + ah'y — c'a.
(ah' — a'h)y =
y =
ca — ca'.
c'a — ca'
ah'  a'h
(1)
(2)
(3)
(4)
In solving fractional literal simultaneous equations, any
solution which does not satisfy the given equations must be
rejected. (Compare Ex. 2, § 171.)
1.
Solve the following
5x—6y=Sa.
EXERCISE 68
4:X+9y=7a
^ 2 raflj+62/ = l.
\cx{dv = l.
■I
r mx—ny
I mx\n'
= mn.
\n'y=m'n'.
1/3
J 8.
+ dy
aiX\a2y=bi.
a^ — ajy=b2.
5.
' 2ax — by _ ,
a
x±by_^^
3a+2
mi m^ mg
£. + l. = i.
Wi ^2 %
7'. ^'
m
n
7i\y m — x
m n
9
n\x
ai.
m
2/
V'lO.
bx—ay=b'^.
I {a—b)x + by—a^.
ax \by = 2 a.
b^y = a' + b\
(a + l)x f (a — 2)?/ = 3 a.
, (a + 3)ic + (a — 4)2/ = 7 a.
l a^a; —
a6(a — b)x + a6(a + &)?/ = a^ \2ab — W.
ax\by = 2.
156
ALGEBRA
• 16.
12. \
r m(x + 2/) + n(x — y) = 2.
13.
14.
15.
I 7nr(x \y) — n\x — y) — m— n.
(a + b)x \ (a  h)y = 2{o} 4 h^).
h _ a
x — a — b y — a\h
(a + h)x + (a  6)2/ = 2 c^  2 h\
y X _ 4 a?)
a — 6 a + 6 a^ — b^
bx\ay = 2.
. a6(a + 6)a; — ab{a — b)y = a^ + b^, '
y — a + b _ y — a ^
x + a\b x\b
a — x b
17
■(
y — b a
ay — bx=zd^\. b^.
(a + b)x{{ab)y = 2a?2b\
(a + b)x + (a — b)y = 2 a.
173. Certain equations in which the unknown numbers
occur in the denominators of fractions may be readily solved
without previously clearing of fractions.
,Ex. Solve the equations
Multiplying (1) by 5,
Multiplying (2) by 3,
Adding,
10
_9_
:8.
X
y
«+
15=
=1
X
y
50
X
45 _
' y
40.
24 45 _
x'^ y
3.
— = 37, 74 = 37 X, and x = 2.
Substituting in (1), 5   = 8,   = 3, and y =S.
y 9
(1)
(2)
SIMULTANEOUS LINEAR EQUATIONS
157
EXERCISE 69
Solve the following
1.
8.
X y
X y
I X ^
3. \
2. \
+ 22/:
10
3'
4.
ah _ ci + h ^
bx ay ah
h_
ax
a
x — a
b
X
a
by
+
 + 
a y
b^
a^b^
y + 'b
= 1.
= 1.
a b
a; y
\a' b' ,
4 = c'.
[ X y
9 11__11
d'^ s~ 2'
6 21
id'^ s
5.
6.
7.
2
3a;
3
4?/
1
12*
5
4a;
4
13
72*
8_
3
5y
89
30
5
6x
_6_
2/
59
18
9. \
p — q _q_ p^ — 2pq — q^
X y p(piq)
p_p^
X
y
10.
2pqq^
p{p + q)
24
2x\y x — 4,y
7 16
2x\y x — 4,y
 = 2.
= 3.
SIMULTANEOUS LINEAR EQUATIONS CONTAINING MORE
THAN TWO UNKNOWN NUMBERS
174. If we have three independent simultaneous equations,
containing three unknown numbers, we may combine any two
of them by one of the methods of elimination explained in
§§ 169 to 171, so as to obtain a single equation containing only
two unknown numbers.
We may then combine the remaining equation with either of
the other two, and obtain another equation containing the sa7ne
two unknown numbers.
By solving the two equations containing two unknown num
bers, we may obtain their values ; and substituting them in
either of the given equations, the value of the remaining
unknown number may be found.
158 ALGEBRA
We proceed in a similar manner when the number of equa
tions and of unknown numbers is greater than three.
The method of elimination by addition or subtraction is
usually the most convenient.
In solving fractional simultaneous equations, any solution which does
not satisfy the given equations must be rejected, (Compare Ex. 2, § 171.)
Q,X 4?/ —
72 = 17.
(1)
1. Solve the equations j 9ic— ly —
16 2 = 29.
(2)
lOcc hy
32 = 23.
(3)
Multiplying (1) by 3, \Sx\2y2\z =
51.
Multiplying (2) by 2, 18 ic  14 ^  32 z =
58.
Subtracting, 2y + Uz=
 7.
(4)
Multiply ing ( 1 ) by 5, SQx20ySbz =
85.
Co)
Multiplying (3) by 3, SOxloy  9z =
69.
(6)
Subtracting (5) from (6) by{26z=
16.
(7)
Multiplying (4) by 5, lOy + 6bz =■
35.
Multiplying (7) by 2, I0y + 52z=
32.
Subtracting, 3z =■
 3, or ^ = — 1.
Substituting in (7), 2y 11=
 7, or ?/ = 2.
Substituting in (1), 6xS + 7 =
17, or X = 3.
In certain cases the solution may be abridged by aid of the
artifice which is employed in the following example :
2. Solve the equations
lu\x + y = 6. (1)
xhy\z = 7. (2)
y{z + u = 8. (3)
_z + u\x = 9. (4)
Adding, Su + Sx + Sy + Sz = SO.
Whence, 2t \ x + y + z = 10. (5)
Subtracting (2) from (5), ' u = S.
Subtracting (3) from (5), x= 2.
Subtracting (4) from (5), y = 1.
Subtracting (1) from (5), z = 4.
SIMULTANEOUS LINEAR EQUATIONS
159
Solve the following :
(4:x3y= 1.
1. i 4.y'Sz = ~15.
[4^30;= 10.
f 4:X — 5y — 6z =
2. \ X— 2/4 2! = 
EXERCISE 70
r~. \
4.
6.
8.
22.
6.
= 22.
'<
l^x +
i3x\ y— z= 14.1
x\3y— z= 16.
[ x\ y—Sz = — 10.
r g^ hk = 24:.
l4.g + Shk = 61.
[Qg5hk = ll.
i^Sx + 5y= 1.
\ 9x\5z = 7.
[9y{3z= 2.
(5x— y{4:Z = — 5.
I x{SySz=:27.
{2x5y = 26.
7x + 6z=SS.
3 ^4
2/_4 2 + 2'
2ir + 4?/— 2! = — 2.
lSxSy\4:Z = 25.
[10x\iy9z = 30.
3p_l_4g + 57. = 10.
4p5g3r = 25.
5^_3g_4r = 21.
4:X — lly — 5z— 9.
8a;+4i/— 2 = 11.
Wx+ 7 2/ 4 6 2; = 64.
8X + 42/+ 32 =52.
y^l2z = 52.
36.
j^ll r 6a; 2/ + 3 2= 42.
10.
11.
\5x— 2/412 2 =
l9a;472/62 =
12.
13.
14.
15.
16.
nOx 5y z= 2.
I 6x172/ + 4 2 = 46.
r2a; + 5 2/+32 = 7.
22/ 4 2 = 23 X.
i 5a;+9 2/ = 5 + 7 2.
r5_8
a; 2/
8_3
2/ 2
< '7
= 3.
= 1.
?5 + ^ =
2 3x
3aj 2/
42/ ^
152 a;
ax{by =
by \ cz =
cz \ ax —
_3_
10'
7_
30'
J^
12'
g'^ + b\
abc
W + c^
abc
(?\a ^
abc
160
ALGEBRA
17.
19.
20.
21.
22.
4:e12t20w = 9.
8e 6ti10iv = 5.
23.
1 12 e 18^ 5 IV = 13.
24.
18. i
( u — x\y= 15.
x — y\z = — 12.
y — z\u=: 13.
I 2; — w p a; = — 14.
ri^i 1
X y z
1^1 1
y z X
1^11
iz X y
(ax — by = a^ — ab^.
\ ay — hz = oj^h — If.
'3u]x =
 5.
4:X — y =
21.
5y + z =
19.
6z — u =
39.
'l^l
= 3.
^M
31
12'
I 3^4
21
2*
ixy y
z 7
3
4 3
yz z
{X
13
3 '
5
15
Z^X X
2
2/_43
5 10
25.
26.
27.
28.
29.
X y
2/ 2;
 +  = c.
12 a;
r& a
 +  = c.
a; 2/
a c
2; a;
 4  = a.
12/ 2;
0— c c—a
J+^=6+c42a.
c—a a— 6
^ +T^=cHa42 6.
a— 6 6— c
3 +^=2.
x+y x—z
_6 L_ = i
xhy y—z
4 +^=2.
a; — 2; y — z
Sx+ 9y{15z = 29.
17a;10 2/+13;2 = 12.
lla;15?/+ 7z= 15.
w + 3a; — 22/2 = — 3.
2%a;2/ + 3;2= 23.
u{x\3y2z = 12.
[3u2x\y + z= 22.
SIMULTANEOUS LINEAR EQUATIONS 161
(x{y \z = 0.
30. \ (b\c)x\(c^a)y{(a + b)z = 0.
[ hex 4 cay + ahz = 1.
id v.s_ Q ( 6x\5y y—4:Z_ 14
31.
^ + = 28. 32.
4 6 3
6^3 4
3 5 5
5x
7x\Sy _ 4
6 9 ' ~ 9*
3 "^ 4 ~ 2*
PROBLEMS INVOLVING SIMULTANEOUS LINEAR EQUA
TIONS WITH TWO OR MORE UNKNOWN NUMBERS
175. In solving problems where two or more letters are
used to represent unknown numbers, we must obtain from
the conditions of the problem as many independent equations
(§ 164) as there are unknoion numbers to he deterrairied.
1. Divide 81 into two parts such that threefifths the greater
shall exceed fiveninths the less by 7.
Let
X = the greater part,
and
y = the less.
By the conditions,
x + y = Sl,
(1)
and
Sx^6y^
6 9 ^
(2)
Solving (1) and (2),
x = 45, y = S6.
2. If 3 be added to both numerator and denominator of a
fraction, its value is f ; and if 2 be subtracted from both nu
merator and denominator, its value is ^ ; find the fraction.
Let
n = the numerator,
and
d = the denominator.
By the conditions,
n + S 2
d + S 3'
and
w2 1
d2 2
Solving these equations,
n = 7, d = 12; then, the fraction is —
162 • ALGEBRA
3. A sum of money was divided equally between a certain
number of persons. Had there been 3 more, each would have
received $ 1 less ; had there been 6 fewer, each would have re
ceived $5 more. How many persons were there, and how
much did each receive ?
Let X = the number of persons,
and y = the number of dollars received by each.
Then, xy = the nuijiber of dollars divided.
Since the sum of money could be divided between x + S persons, each
of whom would receive y — 1 dollars, and between x — 6 persons, each of
whom would receive y + 6 dollars, (a;+3)(y — 1) and (a; — 6)(?/ + 5)
also represent the number of dollars divided.
Then, (x^S)(yl)=xy,
and (x — 6) (y + 5) = xy.
Solving these equations, x = 12, y = 5.
4. The sum of the three digits of a number is 13. If the
number, decreased by 8, be divided by the sum of its second
and third digits, the quotient is 25 ; and if 99 be added to the
number, the digits will be inverted. Find the number.
Let X = the first digit,
y = the second,
and z = the third.
Then, 100 x } 10 y { z = the number,
and 100 z + lOy \ x = the number with its digits inverted.
By the conditions of the problem,
x + y\z = lS,
100 X + 10 y + g  8 _ ,)^
2/ + S ~
and 100 X + 10 y + ^; + 99 = 100 ;? + 10 ?/ 4 a;.
Solving these equations, x = 2, y = S, z = S ; and the number is 283.
5. A crew can row 10 miles in 50 minutes down stream, and
12 miles in li hours against the stream. Find the rate in
miles per hour of the current, and of the crew in still water.
SIMULTANEOUS LINEAR EQUATIONS 163
Let X = number of miles an hour of the crew in still water,
and y = number of miles an hour of the current.
Then, x 4 y = number of miles an hour of the crew down stream,
and x — y = number of miles an hour of the crew up stream.
The number of miles an hour rowed by the crew is equal to the dis
tance in miles divided by the time in hours.
Then, x + ?/ = 10   = 12, .
and xy = 12 = S.
2
Solving these equations, 6 = 10, y = 2.
6. A train running from A to B meets with an accident
which causes its speed to be reduced to onethird of what it
was before, and it is in consequence 5 hours late. If the acci
dent had happened 60 miles nearer B, the train would have
been only 1 hour late. Find the rate of the train before the
accident, and the distance to B from the point of detention.
Let 3 X = the number of miles an hour of the train before the accident.
Then, x = the number of miles an hour after the accident.
Let y = the number of miles to B from the point of detention.
The train would have done the last y miles of its journey in ^ hours ;
3 X
but owing to the accident, it does the distance in " hours.
X
Then, l = ^^&, m
X 3 X ^ ^
If the accident had occurred 60 miles nearer J5, the distance to B from
the point of detention would have been y — 60 miles.
Had there been no accident, the train would have done this in ^~
?/ — 60 ^ *
hours, and the accident would have made the time ^ hours.
X
Then, y.Il^ = y^Z^+i, (2)
X 3 X
Subtracting (2) from (1), ^ = ^ + 4, or — = 4 ; whence, x = 10.
X 3x X
Then, the rate of the train before the accident was 30 miles an hour.
164 ALGEBRA
Substituting in (1), io~io'^ ^' ^^ TK ^ ' ^^^nce, y = 75.
EXERCISE 71
1. Divide 79 into two parts such that threesevenths the
less shall be less by 56 than fourthirds the greater.
2. If the numerator of a fraction be increased by 4, the
value of the fraction is f ; while if the denominator is de
creased by 3, the value of the fraction is . Find the fraction.
3. The sum of the two digits of a number is 14 ; and if 36
be added to the number, the digits will be inverted. Find the
number.
4. A's age is f of B's, and 15 years ago his age was \^ of
B's. Find their ages.
5. If the two digits of a number be inverted, the quotient of
the number thus formed, increased by 101, by the original num
ber is 2 ; and the sum of the digits exceeds twice the excess of
the tens' digit over the units' digit by 5. Find the number.
6. If 3 be added to the numerator of a fraction, and 7 sub
tracted from the denominator, its value is ^ ; and if 1 be sub
tracted from the numerator, and 7 added to the denominator,
its value is f. Find the fraction.
7. A's age is twice the sum of the ages of B and C ; two
years ago, A was 4 times as old as B, and four years ago, A
was 6 times as old as C. Find their ages.
8. If the greater of two numbers be divided by the less, the
quotient is 1, and the remainder 6. And if the greater, in
creased by 14, be divided by the less, diminished by 4, the
quotient is 5, and the remainder 4. Find the numbers.
9. If 8 yards of silk and 12 yards of woolen cost $ 27, and
12 yards of silk and 8 yards of woolen cost $ 28, find the price
per yard of the silk and of the woolen.
10. Find two numbers such that one shall be n times as
much greater than a as the other is less than a; and the quo
tient of their sum by their difference equal to b.
SIMULTANEOUS LINEAR EQUATIONS 165
11. A certain number of two digits exceeds three times the
sum of its digits by 4. If the digits be inverted, the sum of
the resulting number and the given number exceeds three
times the given number by 2. Find the number.
12. The sum of the three digits of a number is 16 ; the digit
in the tens' place exceeds that in the hundreds' place by 4 ;
and if 297 be added to the number, the digits will be inverted.
Find the number.
13. A rectangular field has the same area as another which
is 6 rods longer and 2 rods narrower, and also the same area as
a third which is 3 rods shorter and 2 rods wider. Find its
dimensions.
14. Find three numbers such that the first with onehalf the
second and onethird the third shall equal 29 ; the second with
onethird the first and onefourth the third shall equal 28;
and the third with onehalf the first and onethird the second
shall equal 36.
15. The circumference of the large wheel of a carriage is 55
inches more than that of the small wheel. The former makes
as many revolutions in going 250 feet as the latter does in
going 140 feet. Find the number of inches in the circumfer
ence of each wheel.
16. If the digits of a number of three figures be inverted,
the sum of the number thus formed and the original number
is 1615 ; the sum of the digits is 20, and if 99 be added to the
number, the digits will be inverted. Find the number.
17. A train leaves A for B, 112 miles distant, at 9 a.m., and
one hour later a train leaves B for A; they met at 12 noon.
If the second train had started at 9 a.m., and the first at 9.50
A.M., they would also have met at noon. Find their rates.
18. A boy has $ 1.50 with which he wishes to buy two kinds
of notebooks. If he asks for 14 of the first kind, and 11 of the
second, he will require 6 cents more ; and if he asks for 11 of
the first kind, and 14 of the second, he will have 6 cents over.
How much does each kind cost ?
166 ALGEBRA
19. A man invests $10,000, part at 4^%, and the rest at
3^%. He finds that six years' interest on the first invest
ment exceeds five years' interest on the second by $658.
How much does he invest at each rate ?
20. A man buys apples, some at 2 for 3 cents, and others at
3 for 2 cents, spending in all 80 cents. If he had bought J
as many of the first kind, and f as many of the second, he
would have spent 99 cents. How many of each kind did he
buy?
21. An annual income of $800 is obtained in part from
money invested at 3^%, and in part from money invested
at 3%. If the amount invested at the first rate were invested
at 3%, and the amount invested at the second rate were in
vested at 3%, the annual income would be $825. How
much is invested at each rate ?
22. A tank containing 864 gallons can be filled by two
pipes, A and B. After the pipes have been open together for
9 minutes, the pipe A is closed, and B finishes the work of
filling in 15 1 minutes.' If 15 minutes had elapsed before the
pipe A was closed, B would have finished in 2 J minutes.
How many gallons does each pipe fill in one minute ?
23. The contents of one barrel is f wine, and of another 
wine. How many gallons must be taken from each to fill a
barrel whose capacity is 24 gallons, so that the mixture may
be I wine ?
24. A boy spends his money for oranges. Had he bought
m more, each would have cost a cents less ; if n fewer, each
would have cost b cents more. How many did he buy, and at
what price ?
25. A vessel contains a mixture of wine and water. If 50
gallons of wine are added, there is J as much wine as water ;
if 50 gallons of water are added, there is 4 times as much
water as wine. Find the number of gallons of wine and water
at first.
SIMULTANEOUS LINEAR EQUATIONS 167
26. A man buys 15 bottles of sherry, and 20 bottles of
claret, for ^38. If the sherry had cost j as much, and the
claret ^ as much, the wine would have cost $38.50. Find the
cost per bottle of the sherry, and of the claret.
27. If a field were made a feet longer, and b feet wider, its
area would be increased by m square feet ; but if its length
were made c feet less, and its width d feet less, its area would
be decreased by n square feet. Find its dimensions.
28. If the numerator of a fraction be increased by a, and the
TYh
denominator by b, the value of the fraction is — : and if the
numerator be decreased by c, and the denominator by d, the value
of the fraction is — Find the numerator and denominator,
m
29. A certain number equals 59 times the sum of its three
digits. The sum of the digits exceeds twice the tens' digit by
3 ; and the sum of the hundreds' and tens' digits exceeds twice
the units' digit by 6. Find the number.
30. A piece of work can be done by A and B in 4 hours,
by B and C in 2 hours, and by A and C in 3 hours. In how
many hours can each alone do the work ?
31 . The numerator of a fraction has the same two digits as
the denominator, but in reversed order ; the denominator ex
ceeds the numerator by 9, and if 1 be added to the numerator
the value of the fraction is . Find the fraction.
32. A man walks from one place to another in 5^ hours. If
he had walked ^ of a mile an hour faster, the walk would have
taken 36 fewer minutes. How many miles did he walk, and
at what rate ?
33. A man invests a certain sum of money at a certain rate
of interest. If the principal had been $ 120l) greater, and the
rate 1 % greater, his income would have been increased by
$ 118. If the principal had been $ 3200 greater, aiid the rate
2 % greater, his income would have been increased by $ 312.
What sum did he invest, and at what rate ?
168 ALGEBRA
34. A sum of money at simple interest amounted to $ 1868.40
in 7 years, and to $ 2174.40 in 12 years. Find the principal
and the rate.
35. A and B together can do a piece of work in 3 hours.
If A works I as fast, and B f as fast, they can do it in the
same time. In how many hours can each alone do the work ?
36. Two men together can do a piece of work in 30 hours ;
they can also do it if the first man works 25^ hours, and. the
second 32 hours. In how many hours can each alone do the
work ?
37. A crew row IG^ miles up stream and 18 miles down
stream in 9 hours. They then row 21 miles up stream and 19 J
miles down stream in 11 hours. Find the rate in miles an
hour of the stream, and of the crew in still water.
38. A train travels from A to B, 228 miles, and another
from B to A. If the trains start at the same time, they will
meet 3 hours after. If the first train starts 3 hours after the
second, they will meet 2 hours after the second train starts.
Find the rates of the trains.
39. A man has quarterdollars, dimes, and halfdimes to the
value of $ 1.40, and has in all 12 coins. If he replaces the
quarters by dimes, and the dimes by quarters, the value of
the coins would be $ 1.55. How many has he of each ?
40. The middle digit of a number of three figures is onehalf
the sum of the other two digits. If the number be divided by
the sum of its digits, the quotient is 20, and the remainder 9 ;
and if 594 be added to the number, the digits will be invertedo
Find the number.
41. A certain number of workmen receive the same wages,
and receive together a certain sum. If there had been 9 more
men, and each ha'd received 30 cents less, the total received
would have been increased by $ 12.30. Had there been 8 fewer
men, and each had received 40 cents more, the total received
would have been decreased by f 13.20. How many men were
there, and how much did each receive ?
SIMULTANEOUS LINEAR EQUATIONS 169
42. A merchant has three casks of wine, containing together
66 gallons. He pours from the first into the second and third
as much as each of them contains ; he then pours from the sec
ond into the first and third as much as each of them then con
tains. There is now 8 times as much in the third cask as in
the second, and twice as much in the first as in the second.
How many gallons did each have at first ?
43. In a meeting of 600 persons, a measure is defeated by a
certain majority. It is afterwards successful by double this
majority, and the number of persons voting for it is  as great
as the number voting against on the former occasion. How
many voted for, and how many against, the measure on the
former occasion?
44. I bought apples at 3 for 5 cents, and oranges at 2 for 5
cents, spending in all $ 1.70. I sold threefourths of the apples
and onehalf of the oranges for $ 1.10, and made a profit of 5
cents on the latter transaction. How many did I buy of each ?
45. A gives to B and C as much money as each of them has;
B then gives to A and C as much money as each of them then
has ; C then gives to A and B as much money as each of them
then has. Each has now $ 8. How much had each at first ?
46. A has onehalf as many dimes as dollars, and B eight
sevenths as many dimes as dollars. They have together 3
more dollars than dimes, and B's money is 60 cents less than
A's. How much money has each ?
47. A man buys a certain number of $ 100 railway shares,
when at a certain rate per cent discount, for ^ 1050 ; and when
at a rate per cent premium twice as great, sells onehalf of them
for $ 1200. How many shares did he buy, and at what cost ?
48. A and B can do a piece of work in 4 hours, A and C
in J/ hours, A and D in i hours, and B and C in ^ hours.
How many hours will it take each alone to do the work ?
49. A and B run a race of 280 feet. The first heat, A gives
B a start of 70 feet, and neither wins the race. The second
heat, A gives B a start of 35 feet, and beats him by 6 seconds.
How many feet can each run in a second ?
170 ALGEBRA
50. A, B, C, and D play at cards. After B has won one
half of A's money, C onethird of B's, D onefourth of C's, and
A onefifth of D's, they have each $ 10, except B, who has $ 16.
How much had each at first ?
51. The sum of the four digits of a number is 14. The
sum of the last three digits exceeds twice the first by 2.
Twice the sum of the second and third digits exceeds 3 times
the sum of the first and fourth by 3. If 2727 be added to
the number, the digits will be inverted. Find the number.
52. A and B run a race of 210 yards. The first heat, A
gives B a start of 8 seconds, and beats him by 20 yards. The
second heat, A gives B a start of 70 yards, and is beaten by
2 seconds. How many yards can each run in a second ?
53. A sum of money consists of halfdollars, dimes, and
halfdimes. Its value is as many dimes as there are pieces of
money; and its value is also as many halfdollars as there are
dimes less 1. The number of dimes is 5 more than the num
ber of halfdollars. Find the number of each coin.
54. The forewheel of a carriage makes a revolutions more
than the hindwheel in travelling b feet. If the circumference
of the fore wheel were m times as great, and the circumference
of the hindwheel n times as great, the fore wheel would make
c revolutions more than the hindwheel in travelling d feet.
Find the circumference of each wheel.
55. A train running from A to B meets with an accident
which delays it a hours. It then proceeds at a rate onenth of
its former rate, and arrives at J5 6 hours late. Had the acci
dent occurred c miles nearer B, the train would have been d
hours late. Find the rate of the train before the accident,
and the distance to B from the point of detention.
56. A man buys 60 shares of stock, each having the par
value $100, part paying dividends at the rate of 3%, and the
remainder at the rate of 4^%. If the first part had paid divi
dends at the rate of 4^%, and the other at the rate of 3%,
the total annual income would have been $12 less. How many
shares of each kind did he buy ?
SIMULTANEOUS LINEAR EQUATIONS 171
176. Interpretation of Solutions.
1. The length of a field is 10 rods, and its breadth 8 rods;
how many rods must be added to the breadth so that the area
may be 60 square rods ?
Let X = number of rods to be added.
By the conditions, 10 (8 + ic) = 60.
Then, 80 + 10a; = 60, or x= 2.
This signifies that 2 rods must be subtracted from the breadth in order
that the area may be 60 square rods. (Compare § 16.)
If we should modify the problem so as to read :
" The length of a field is 10 rods, and its breadth 8 rods ; how many
rods must be subtracted from the breadth so that the area may be 60
square rods ?"
and let x denote the number of rods to be subtracted, we should find x = 2.
A negative result sometimes indicates that the problem is
impossible.
2. If 11 times the number of persons in a certain house,
increased by 18, be divided by 4, the result equals twice the
number increased by 3 ; find the number.
Let X = the number.
By the conditions, '^'^^^^^ = 2 x + 3.
4
Whence, llcc + 18 = 8x + 12, and a; = 2.
The negative result shows that the problem is impossible.
A problem may also be impossible when the solution is
fractional.
3. A man has two kinds of money : dimes and cents. The
total number of coins is 23, and their value 37 cents. How
many has he of each ?
Let X = number of dimes.
Then, 23 — x = number of cents.
The X dimes are worth 10 x cents.
172 ALGEBRA
14
Then, by the conditions, 10 x + 23 — aj = 37 ; and x = —
The fractional result shows that the problem is impossible.
EXERCISE 72
Interpret the solutions of the following :
1. If the length of a field is 12 rods, and its width 9 rods,
how many rods must be subtracted from the width so that the
area may be 144 square rods ?
2. A is 44 years of age, and B 12 years; how many years
ago was A 3 times as old as B ?
3. The number of apple and pear trees in an orchard is 23 ;
and 7 times the number of apple trees plus twice the number
of pear trees ecjuals 82. How many are there of each kind ?
4. The number of silver coins in a purse exceeds the num
ber of gold coins by 3, and 5 times the number of silver coins
exceeds 3 times the number of gold coins by 3. How many
are there of each kind ?
5. A's assets are double those of B. When A has gained
$ 250, and B $ 170, A's assets are 5 times those of B. Find
the assets of each.
6. A cistern has two pipes. When both are open, it is filled
in 7^ hours ; and the first pipe alone can fill it in 3 hours.
How many hours does the second pipe take to fill it ?
7. The numerator of a fraction is 4 times the denominator ;
and if the numerator be diminished by 9, and the denominator
by 15, the value of the fraction is . Find the fraction.
8. A and B are travelling due east at the rates of 4i and 3
miles an houi^ respectively. At noon A is 5 miles due east of
B. How many miles to the east of A's position at noon will
he overtake B ?
'9. A has $ 720, and B $ 300. After A has gained a certain
sum, and B has gained twothirds this sum, A has 3 times as
much money as B. How much did each gain ?
GRAPHICAL REPRESENTATION
173
XIII. GRAPHICAL REPRESENTATION
177. Rectangular Coordinates of a Point.
Lm
^
M
m.
A.
i/i
X
Let XX' and YY' be straight lines intersecting at right
angles at 0; let Pi be any point in the plane of XX' and YY',
and draw line Pi^i perpendicular to XX'.
Then, OMi and M^Pi are called the rectangular coordinates,
or simply the coordinates, of Pi; OM^ is called the abscissa,
and MiPi the ordinate.
178. It is understood, in the definitions of § 177, that
abscissas measured to the right of are positive, and to the
left, negative; also, that ordinates measured upwards from XX'
are positive, and downwards, negative. ^
Thus, let P2 be to the left of YY, and above XX', and P3 and
P4 below XX', respectively to the left and right of YY, an^"
draw lines P^M^, P^M^, and P^M^ perpendicular to XX'.
Let 0Mi = 5, MM = 3, M,0 = 5, 0M, = 2,
JfiPi = 3, M,P2 = 5, PsMs=S, P,M, = 4..
Then, the abscissa of Pi is f 5, and its ordinate + 3
the abscissa of Po is — 3, and its ordinate f 5
the abscissa of P. is — 5, and its ordinate — 3
the abscissa of P. is 4 2, and its ordinate — 4.
174
ALGEBRA
179. The lines of reference, XX' and YT', are called the
axis of X, and axis of T, respectively ; and the origin.
We express the fact that the abscissa of a point is b, and its
ordinate a, by saying that, for the point in question, x = b and
y = a; or, more concisely, we speak of the point as the point
(b, a) ; where the first term in parentheses is understood to be
the abscissa, and the second term the ordinate.
If a point lies upon XX', its ordinate is zero ; and if it lies
upon YY', its abscissa is zero.
The coordinates of the origin are (0, 0).
180. Plotting Points.
To plot a point when its coordinates are given, lay off the
abscissa to the right or left of 0, according
as it is H or — , and then draw a perpen
dicular, equal in length to the ordinate,
above or below XX', according as the
ordinate is  or — .
Thus, to plot the point (—2, 3), lay off
2 units to the left of O upon XX', and
then erect a perpendicular 3 units in length above XX'.
^
2,3)
Y'
■^
EXERCISE 73
Plot the following points :
1. (1, 4).
2. (2, 2).
3. (3, 6).
4. (2, 4).
5. (3, 1).
6. (4, 3).
7. (1, 2).
8. (4, 6).
9. (r, 3).
10. (6, 1).
11. (5, 0).
12. (0, 4).
13. (2, 0).
14. (0, 3).
GRAPH OF A LINEAR EQUATION INVOLVING TWO
UNKNOWN NUMBERS
181. Consider the equation y = x + 2.
GRAPHICAL REPKESENTATION
175
If we give any numerical value to x, we may, by aid of the
relation y = 0:{ 2, calculate a corresponding value for y.
If a; = 0,
If a; = 1,
If a; = 2,
2/ 2.
(1)
2/ = 3. . (B)
2/ = 4. . (C)
If a; = 3, 2/ = 5. (D)
If a^=l, 2/ = l. (E)
If a; =  2, 2/ = 0. (F)
Ifaj = _3, y = l; etc. ((9)
Now let these be regarded as the coordinates of points ; and
let the points be plotted, as explained in § 180.
Thus, to plot the point A^ lay oif 2 units above upon YY'.
The points will be found to lie on a certain line, GD, which
is called the Graph of the given equation.
By assuming fractional values for x, we may obtain intermediate
points of the graph.
EXERCISE 74
Eind by the above method the graphs of the following
equations :
1. 2/ = 2aj + 3. 3. 4?/ + a^ = 6. 5. y=:5x,
2. 2/ = 3a;4. 4. 3y2x = 12. 6. 3x^2y = 0.
182. AVe shall always find (and it can be proved) that a
linear equation, involving two unknown numbers, has a straight
line for a graph.
Then, since a straight line is determined by any two of its
points, it is sufficient, when finding the graph of a linear equa
tion involving two unknown numbers, to find two of its points,
and draw a straight line through them.
The points most easily determined are those in which the
graph intersects the axes.
For all points on OX, y = 0; hence, to find where the graph
cuts OX, put ?/ = 0, and calculate the value of x.
To find where the graph cuts OY, put a; = 0, and calculate
the value of y.
176
ALGEBRA
Ex. Plot the graph of 2cc + 3?/ = — 7.
7
Put ?/ = ; then 2 x = — 7, and x
2
Then plot A on OX',  units to the left
ofO. • 2 X
Put X = ; then 3 y = — 7 and y =*.
Then plot 5 on OT',  units below 0.
3
Draw the straight line AB \ this is the re
quired graph.
The above method cannot, of course, be used for a straight line passing
through the origin, nor for the equations of § 183.
183. Consider the equation y = 5.
This means that every point in the graph
has its ordinate equal to 5.
Then the graph is the straight line AB,
parallel to XX', and 5 units above it.
In like manner, the graph of x = — 3 is
the straight line CD, parallel to YY', and
3 units to the left of it.
The graph of y — is the axis of X, and the graph of a; = is the
axis of Y.
EXERCISE 75
Plot the graphs of the following equations :
1. 3x^2y = 6. 3. x = 2, 5. 16x27y:=72.
2. a;42/ = 4. 4. y = 4:. 6. Sx{Wy = 6.
c
Y
^^Tf
■^
A
"
D
Y'
INTERSECTIONS OF GRAPHS
184. Consider the equations
p y = 5. (AB)
U + 3y= 3. (CD)
Let AB be the graph of x—y=—5, x
and CD the graph oi x{3y = 3.
GRAPHICAL REPRESENTATION
177
Let AB and CD intersect at E.
Since E lies on each graph, its coordinates must satisfy both
given equations ; hence, to find the coordinates of E, we solve
the given equations.
In this case the solution is a; = — 3, y = 2; and it may be
verified in the figure that these are the coordinates of E.
We then have the following important principle :
If the graphs of tivo linear equations, with two unknown num'
hers, intersect, the coordinates of the point of intersection form a
solution of the equations represented by the graphs.
EXERCISE 76
Verify the principle of § 184 in the following equations :
■1
4 aj + 5 2/ = 24.
Sx
2 (3a^ + 72/ =
[Sx + 3y =
2y =  5.
5.
18.
3. i
f 5 ic — 4 2/ =
0.
[Tx{6y = 29. ■
9x\14:y = 25.
y= 22.
r 9 05 + 14 2/ =
As additional examples, the pupil might verify graphically
the solutions of Exs. 3, 8, 11, and 12, Exercise 65, and of
Exs. 7, 8, 9, and 16, Exercise 66.
185. Graphs of Inconsistent Linear Equations with Two Un
known Numbers.
Consider the equations
(Sx2y= 5. (AB)
[6x4.y = 7. {CD)
The first equation can be put in the form
6x — 4:y=10, by multiplying both mem
bers by 2.
Then, the given equations are inconsistent
(§ 165), and it is impossible to find any values
of X and y which satisfy both equations.
178
ALGEBRA
We shall always find that two inconsistent equations, with
two unknown numbers, are represented by parallel graphs; for
if the graphs could intersect at any point, the coordinates of
this point would be a solution of the given equations (§ 184).
186. Graphs of Indeterminate Linear Equations with Two Un
known Numbers.
Consider the equations
(3x2y= 5.
[6x4:y = 10.
The first equation can be put in the form
of the second, by multiplying both members
by 2, and the graphs coincide.
The given equations are not independent
(§ 164) ; in any similar case, we shall find that the graphs are
coincident.
EXERCISE 77
Verify the principles of §§ 185 and 186 in the following
equations:
(3x\4.y= 12. ^ f2x 7y = U.
2.
3x{4.y = 12.
2x 5y= 0.
6x15y = 30.
'■I
4a;14?/ = 28.
5x\ 6 2/ = 15.
15x\lSy = 4.5.
187. Graphical Representation of Linear Expressions involving
One Unknown Number.
Consider the expression 3x + 5.
Put 2/ = 3 a? + 5 ; and let the graph of
this equation be found as in § 183, /B
Putting y = 0, x = — ^; then the graph
cuts XX' I units to the left of 0.
Putting x = 0, y = 5; then the graph
cuts YY' 5 units above 0.
The graph is the straight line AB.
GRAPHICAL REPRESENTATION" 179
188. Graphical Representation of Roots of Equations (§ 81).
In order to find the abscissa of the point A (§ 187), where
the graph of 3 a? + 5 intersects XX', we solve the equation
3x + o = 0(§ 182).
That is, the abscissa of ^ is a root of the equation 3 a;+5=0.
Hence, the abscissa of the point in which the graph of the first
member of any linear equation, with one unknown number ^ inter
sects XX', is the root of the equation.
EXERCISE 78
Plot the graphs of the first members of the following equa
tions, and in each case verify the principle of § 188 :
1. 2a; + 7 = 0. 2. 5aj4 = 0.
(
180 ALGEBRA
XIV. INEQUALITIES
189. The Signs of Inequality, > and < , are read " is greater
than " and " is less than," respectively.
Thus, a > & is read " a is greater than 6 " ; a < 6 is read " a
is less than 6."
190. One number is said to be greater than another when
the remainder obtained by subtracting the second from the
first is a positive number.
One number is said to be less than another when the remain
der obtained by subtracting the second from the first is a
negative number.
Thus, if a — 6 is a positive number, a > 6 ; and if a — & is a
negative number, a<b.
191. An Inequality is a statement that one of two expres
sions is greater or less than another.
The First Member of an inequality is the expression to the
left of the sign of inequality, and the Second Member is the
expression to the right of that sign.
Any term of either member of an inequality is calle4 a term
of the inequality.
Two or more inequalities are said to subsist in the same sense
when the first member is the greater or the less in both.
Thus, a > 6 and G>d subsist in the same sense.
PROPERTIES OF INEQUALITIES
192. An inequality will continue in the same sense after the
same number has been added to, or subtracted from, both
members.
For consider the inequality a'>b.
By § 190, a — 6 is a positive number.
INEQUALITIES 181
Hence, eaxih of the numbers
(a 4 c) — (5 + c), and (a — c) — (6 — c)
is positive, since each is equal to a — b.
Therefore, a\c>b\c, and a — c>b — c. (§ 190)
193. It follows from § 192 that a term may be transposed
from one member of an inequality to the other by changing its
sign.
If the same term appears in both members of an inequality, affected
with the same sign, it may be cancelled.
194. If the signs of all the terms of an inequality be changed
the sign of inequality must be reversed. ^
For consider the inequality a — b>c — d.
Transposing every term, d — c>b — a. (§ 193)
That is, b — a<d — c,
195. An inequality will continue in the same sense after both
members have been multiplied or divided by the same positive
number.
For consider the inequality a > b.
By § 190, a — 5 is a positive number.
Hence, if m is a positive number, each of the numbers
m(a — b) and ~ , or ma — mb and , is positive.
m mm
Therefore, ma > mb, and — > — •
m m
196. It follows from §§ 194 and 195 that if both members of
an inequality be multiplied or divided by the same negative num
ber, the sign of inequality must be reversed.
197. If any number of inequalities, subsisting in the same
sense, be added member to member, the residting inequality will
also subsist in the same sense.
182 ALGEBRA
For consider the inequalities a>h, a^ > b', a" > b", •••.
Each of the numbers, a — b, a' — b', a" — b", •••, is positive.
Then, their sum ab + a' b' ja" b" + •,
or, . a\a' {a" { (6 + 6' + 6"+ •••)>
is a positive number.
Whence, a^a' + a" + "•>b + b' + b" + ••••
If two inequalities, subsisting in the same sense, be subtracted member
from member, the resulting inequality does not necessarily subsist in the
same sense.
Thus, if a > & and a' > 6', the numbers a — b and a' — b' are positive.
But (a — b)(a'  b'), or its equal, (a  a')(b  b'), may be posi
tive, negative, or zero ; and hence a — a' may be greater than, less than,
or equal to 6 — b'.
198. If a > 6 and a' >b', and each of the numbers a, a', b,
6', is positive, then ^^, ^ ^^,^
Since a' > b', and a is positive,
aa'>ab' (§195). (1)
Again, since a>b, and b' is positive,
ab'>bb\ (2)
From (1) and (2), aa' > 66'.
199. If we have any number of inequalities subsisting in
the same sense, as a>6, a' >6', a"> b", •••, and each of the
numbers a, a', a", •, 6, 6', 6", •••, is positive, then
aa'a"...>66'6"....
For by §198, aa'>66'.
Also, a">6".
Then by §198, aaV >66'6".
Continuing the process with the remaining inequalities, we
obtain finally
INEQUALITIES 183
200. Examples.
1. Find the limit of x in the inequality
Multiplying both members by 3 (§ 195), we have
21 X  23 < 2 «+ 15.
Transposing (§ 193), and uniting terms,
19 X < 38.
Dividing both members by 19 (§ 195),
x<2.
[This means that, for any value ofic<2,7x^<^ + 5. J
2. Find the limits of x and y in the following :
r3a; + 22/>37. (1)
l2a; + 32/ = 33. (2)
Multiply (1) by 3, 9 x + 6 y > 111.
Multiply (2) by 2, 4x + 6y= m.
Subtracting (§192), 5x> 45, and x>9.
Multiply (1) by 2, 6 x f 4 ?/ > 74.
Multiply (2) by 3, 6 x + 9 y = 99.
Subtracting, — 5 ?/ > — 25.
Divide both members by — 5, ?/ < 5 (§ 196).
(This means that any values of x and y which satisfy (2), also satisfy
(1), provided x is > 9, and y < 5.)
3. Between what limiting values of a; is a^ — 4 a; < 21 ?
Transposing 21, we have
x2  4 X is < 21, if x2  4 X  21 is < 0.
That is, if (x + 3)(x  7) is negative.
Now (x + 3) (x  7) is negative if x is between  3 and 7 ; for if x is
<  3, both X + 3 and x  7 are negative, and their product positive ; and
if X is > 7, both x + 3 and x — 7 are positive.
Hence, x^ — 4 x is < 21, if x is > 3, and < 7.
184 ALGEBRA
EXERCISE 79
Find the limits of x in the following :
1. (4a;+5)24<(8a; + 5)(2a; + 3).
2. (3 a; + 2) (ic + 3)  4 aj > (3 a;  2)(a;  3) + 36.
3. (x\4:)(5x2)\(2x 3)2 > (3 a; + 4)^  78.
4. (x  S)(x + 4)(a;  5) < (a; + l)(x  2){x  3).
5. a'(xl)<2 b''(2 x1)  ab, if a2b is positive.
6. ^~^ + 2 > ^"^^ , if m and % are positive, and m<^n.
n m
Find the limits of x and y in the following :
7 a; — 4 2/ > 41.
l3a;4T2/ = 35.
y r5a; + 62/<45. g^
\Sx4.y = ll.
9. Find the limits of x when
3a;ll<24lla;, and 5 aj + 23<20 x + 3.
10. If 6 times a certain positive integer, plus 14, is greater
than 13 times the integer, minus 63, and 17 times the integer,
minus 23, is greater than 8 times the integer, plus 31, what is
the integer ?
11. If 7 times the number of houses in a certain village,
plus 33, is less than 12 times the number, minus 82, and 9
times the number, minus 43, is less than 5 times the number,
plus 61, how many houses are there ?
12. A farmer has a number of cows such that 10 times their
number, plus 3, is less than 4 times the number, plus 79 ; and
14 times their number, minus 97, is greater than 6 times the
number, minus 5. How many cows has he?
13. Between what limiting values of a; is ar^H3a;<4?
14. Between what limiting values of a; is x^<Sx — 15.
15. Between what limiting values of a; is 3 a;^ + 19 a;< —20 ?
201. If a and b are unequal numbers,
a^{b^>2ab.
INEQUALITIES 185
For (aby>0', or, cc" 2 ab + b^>0.
Transposing —2ab, a^ \ b^ > 2 ab.
1. Prove that, if a does not equal 3,
(a42)(a2)>6a13.
By the above principle, if a does not equal 3,
a2 f 9 > 6 a.
Subtracting 13 from both members,
(j2_4;>6a_13, or (a + 2)(a  2) >6 « 13.
2. Prove that, if a and b are unequal positive numbers.
a^\b^>a'b + b'a.
We have, a^ + b'^>2 ah, or ^2 _ ab \b^> ab.
Multiplying both members by the positive number a + 6,
a^+b^> a% + b'^a.
EXERCISE 80
1. Prove that for any value of x, except {,
3a;(3a;10)>25.
2. Prove that for any value of x, except ,
4x(x5)>8a;49.
3. Prove that for any values of a and 6, if 4 a does not equal
^^' (4 a + 3 6) (4 a  3 6) > 6 & (4 a  3 6).
4. Prove that for any values of x and y,ii 5x does not equal
^^' 5 x(5 x 6 y) >2 y(5 xS y).
Prove that, if a and 6 are unequal positive numbers,
5. a% + ab'>2a'b'. 6. ?H>2.
b a
7. a^^a'b\ab^{b^>2ab{a\b).
186 ALGEBRA
^^ XV. INVOLUTION
202. Involution is the process of raising an expression to any
power whose exponent is a positive integer.
We gave in § 96 a rule for raising a monomial to any power
whose exponent is a positive integer.
203. Any Power of a Fraction.
Wehave, gy=?x?x^ axaxa a
b b b bxbxb W'
and a similar result holds for any positive integral power of  •
Then, a fraction may be raised to any power whose exponent is
a positive integer by raising both numerator and denominator to
the required power.
^ / 2 x'Y (2 x'\\. ... (2 x'Y 32 a^ ,. qan
EXERCISE 81
Find the values of the following :
^ r6a%^y • o / 3aV^^
/ 2 m'xy
V nhf J '
7 MJ \ Wy
o / 9 mny . (_±^y 6 f ^'^' \
\Sp' J' ■ V 5y'zy' ' [Sb'cny
204. Square of a Polynomial.
We find by actual multiplication :
a \b + c
a {b \c
a^ + ab i ac
+ ab +6^+ be
4 ac { bc\c^
a^\2ab{2ac\b^ + 2bc + (f
INVOLUTION 187
The result, for convenience of enunciation, may be written :
(a + 6 + c)2 = a + &' + c' + 2a6 + 2ac + 2 6c.
In like manner we find :
(af6 + c + (^)2 = a2 + &2 + c2 + d2
{2ab{2ac + 2ad{2bc^2bd + 2cd',
and so on.
We then have the following rule :
The square of a polyyiomial is equal to the sum of the squares
of its terms, together with twice the product of each term by each
of the following terms.
Ex. Expand (2x''^x B)\
The squares of the terms are 4 x*, 9 sc^, and 25.
Twice the product of the first term by each of the following terms gives
the results  12 x^ and  20 x^.
Twice the product of the second term by the following term gives the
result 30 x.
Then, (2 «2  3 «  5)2 = 4 a^4 + g 3^2 + 25  12 jc^  20 ic^ _^ 30 a;
= 4x*12a:3llx2 + 30a; + 25.
EXERCISE 82
Square each of the following :
\. a—b + c. 10. x^ — 4:xy — 5y\
2. x\yz. 11. 6a^\abSb^
3. n^3nl. 12. 2a^P8a + 9.
4. 3x{y + 2z. 13. 6 a^^  4 a.^ + 5 2/*.
5. l\Sx4:a^. 14. a^bcd.
6. 2a'" 5a" 1. 15. a 6 + c + d
7. 443m3 + 2m«. 16. a^ + a^ + a3.
8. 77i3_n2 + 6. ' 17. 2a;34a^3£cfl.
9. 2x + 3y\5z. 18. 32a + 4a25a^
188 ALGEBRA
19.
m + 4
.1 20. ^f « + 
m 3 X c
205.
Cube of a Binomial.
We find by
actual multiplication :
a +b
a^ + 2a'b{ ab^
a'b\2ab'\b^
aa?
That is, the cube of the sum of two numbers is equal to the
cube of the first, plus three times the square of the first times the
second, plus three times the first times the square of the second,
plus the cube of the second.
Again, (a by = a^2ab + b^
a b
a^2a^b\ ab^
 a^b^2ab^b^
(aby = a'3a'b{3ab'b^
That is, the cube of the difference of two numbers is equal to
the cube of the first, minus three times the square of the first
times the second, plus three times the first times the square of the
second, minus the cube of the second.
1. Find the cube oia + 2b.
We have, (a + 2 6)3 = a^ + 3 a%2 6) f 3 a(2 6)2 + (2 6)8
= a8 + 6 a26 + 12 a62 + 8 6».
2. Find the cube oi2o^5y\
(2 x8  5 2/2)3 = (2 a:3)3 _ 3(2 a:3)2(5 y^) + 3(2 aj8)(5 y2)2_ (5 ^2)8
= 8 x9  60 xhj^ + 150 x^y^  125 y^
The cube of a trinomial may be found by the above method,
if two of its terms be enclosed in parentheses; and regarded
as a single term.
INVOLUTION 189
3. Find the cube of a^  2 a;  1.
(x22a;l)8 = [(x22a;) ip
= (x2  2 x)3  3(^2  2 x)2 + 3(a:2 _ 2 a;)  1
=,afi  6x^ + 12 X^Sx^ 3(x*  4 ic3 + 4 x2) + 3(x2 _ 2 x) 1
= a;6  6 x5 + 12 cc*  8 x3  3 x* + 12 x3  12 a:2 4 3 x2  6 X  1
= x66x5 + 9x4 + 4x39x26xl.
EXERCISE 83
Cube each of the following :
1. a^bab\ 7. 2a{6x. 14. ab\c.
2. a 4 3. 8. 5m^3n^. 15. 1 — a — al
3. 2x + y. 9. 3a6a2. 16. a + 264c.
4. a5 6. 10. 2a^6 + 7c^ 17. x'hxS.
5. 6a^+l. 11. 8a2"» + 3a. 18. 3n + 2n2.
6. m47i3. 12. 9x'4.y\ 19. 2a2 + 3a4.
1 Q 'ni'^ 2 71
*'^' O 2 '
2n m^
190 ALGEBRA
t XVI. EVOLUTION
206. If an expression when raised to the nth. power, n being
a positive integer, is equal to another expression, the first
expression is said to be the nth Root of the second.
Thus, if a" = &, <x is the nth root of b.
Evolution is the process of finding any required root of an
expression.
207. The Radical Sign, ^, when written before an expres
sion, indicates some root of the expression.
Thus, Va indicates the second, or square root of a;
Va indicates the third, or cube root of a ;
Va indicates the fourth root of a ; and so on.
The index of a root is the number written over the radical
sign to indicate what root of the expression is taken.
If no index is expressed, the index 2 is understood.
An even root is one whose index is an even number ; an odd
root is one whose index is an odd number.
EVOLUTION OF MONOMIALS
208. We will now show how to find any root of a monomial,
which is a perfect power of the same degree as the index of
the required root.
1. Kequired the cube root of a^6V.
We have, (abH^y = a^b^c^.
Then, by § 206, \/a^b^ = ab^c\
2. Required the fifth root of 32a^
We have, (2a)5 = 32a^
Whence, \/ 32 a^ =  2 <?.
EVOLUTION 191
3. Eequired the fourth root of a^.
We have either (+ «)* or ( ay equal to a*.
Whence, Vo^ = ±a.
The sign ± , called the double sign, is prefixed to an expres
sion when we wish to indicate that it is either + or — .
209. From § 208, we have the following rule :
Extract the required root of the absolute value of the numerical
coefficient, and divide the exponent of each letter by the index of
the required root.
Give to every even root of a positive term the sign ± , and to
every odd root of any term the sign of the term itself.
1. Find the square root of 9 a''6V®.
By the rule, V9¥¥c^ = ± 3 a^b^c^.
2. Find the cube root of —64 x^y^.
V— 64 x^y^"* = — 4 x^y^.
The root of a large number may sometimes be found by
resolving it into its prime factors.
3. Find the square root of 254016.
We have, V254016 = VWxWxT^ =±2^xS^x7 =± 504.
4. Find the value of V72 X 75 X 135.
y/12 X 76 X 136 = \^(28 x 32) x (3 x 52) x (33 x 5)
= V28x 36 X 53 = 2 X 32 X 5 = 90.
EXERCISE 84
Find the values of the following :
1. V36^. 4. a/81 n%^y . 7. v'64 a'7i'\
2. a/64 a''b'c\ 5. Vl21 a'^b^'c*. 8. V^^^^2i3W.
3. sZxy^z''. 6. ^216a^yz''. 9. VlQ9x^y'
192 ALGEBRA
10. A/'128 m%2^ 13. V2916. 16. VSlx 64x324.
11. V343a^+^2/^. 14. V30625. 17. V84x 54x126.
12. ^625 a^'^b*^ 15. V86436. 18. ^/5S32,
19. Vl5 xy X 33 2/2! X 55 zx.
20. ^21952. 23. </l04976.
21. V627264. 24. a/59049.
22. ■v/112 X 168 X 252. 25. ■v/135 x 375 x 625.
26. V(a'  5 a 4 6)(a^ + 2 a  8)(a2 + a  12).
210. Any Root of a Fraction.
It follows from § 203 that, to find any root of a fraction,
each of whose terms is a perfect power of the same degree as
the index of the required root, extract the required root of both
numerator and denominator.
Ex.
27 a^b' V27a^ 3 ab^
64 c^ ^64? 4c3
EXERCISE 85
Find the values of the following:
1.
/64^ « 5/ 32 g^ K V
m"
49 2* >* b'<^ \256n«
" "V 125 6«* * ^ 2/'' * ' ^'729 6«"*
211. We have Vipiy' = Va^ = a*" = (v/a«)"».
^a;. Eequired the value of V (32 a^y.
We have, v^(32 aio)4 = ( ^^^32^* = (2 a2)4 = 16 a^.
This method of finding the root is shorter than raising
32 a^^ to the fourth power, and then taking the fifth root of
the result.
EVOLUTION 193
EXERCISE 86
Find the values of the following :
1. </(<i4ar. ^ ^ (243aVcy . 6. ^'/f l^Y.
2. V(I^)'. ^^ ^^ 27 W
3. </JW^¥f ^' ^(^'^ "*""")'• 7. V(a^2a6 + 6»)l
SQUARE ROOT OF A POLYNOMIAL
212. In § 112, we showed how to find the square root of a
trinomial perfect square.
The square roots of certain polynomials of the form.
a" \b' h c" + 2 ab + 2 ac4'2 be
can be found by inspection.
Ex. Find the square root of
9x^\y^^4.z^\6xy — 12 xz — 4: yz.
"We can write the expression as follows :
j (3x)2 + ?/2+(_20)2 + 2(3x> + 2(3x)(2;?) +22/(20).
By § 204, this is the square of Sx {■ y +(— 2 z).
Then, the square root of the expression is Sx + y — 2 z.
(The result could also have been obtained in the form 2z — y — 3x.)
r^ EXERCISE 87
Find the square roots of the following:
1. a'{b'\c'2ab2ac + 2bc.
2. a^ + 4:y^\9 + 4:Xy\6x\12y.
3. l+25m2 + 36n210m + 12w60m7i.
4. a' + Slb'\16{lSabSa72b,
5. 9x^{y^i25z^6xyS0xz + 10yz.
6. 36 m^ + 64 n^ + cc^ + 96 7nn — 12 mx — 16 nx,
7. 16 a^ + 9 6^ + 81 c" + 24 a'b'' + 72 aV f 54 ft^c^.
8. 25a;« + 49y« + 36;2«7Ox3/H6Oa^3^84 2/^0^
194 ALGEBRA
213. Square Root of any Polynomial Perfect Square.
By § 204, (a + & + c)2 = a + 2a& + 62_ 2ac 4 2 6c 4 c^
= a2 + (2a + 6)6 + (2a + 26 + c)c. (1)
Then, if the square of a trinomial be arranged in order of
powers of some letter :
I. The square root of the first term gives the first term of
the root, a.
II. If from (1) we subtract a^, we have
(2a + b)b\(2a\2b\c)c. (2)
The first term of this, when expanded, is 2 ah, if this be
divided by twice the first term of the root, 2 a, we have the
next term of the root, b.
III. If from (2) we subtract (2a + b) b, we have
(2a + 2 6 + c)c. (3)
The first term of this, when expanded, is 2ac; if this be
divided by twice the first term of the root, 2 a, we have the last
term of the root, c.
IV. If from (3) we subtract (2 a \ 2 b { c) c, there is no
remainder.
Similar considerations hold with respect to the square of a
polynomial of any number of terms.
214. The^ principles of § 213 may be used to find the square
root of a polynomial perfect square of any number of terms.
Let it be required to find the square root of
4tx' + 12x^7i^24.x^16.
4:x'\12x^ 7a^24a; + 16 2ar^+3a;4
2a^b = 4.x^{Sx
3x
12a^ 7x224.15 + 16, 1st Eem.
12a^4 9.^2
2a + 26 + c = 4ic26a54
4
 16 or  24 a; + 16, 2d Rem.
16ar'24ic + 16
EVOLUTION 195
The first term of the root is the square root of 4 ic*, or 2 x"^.
Subtracting the square of 2 x^, 4 x*, from the given expression, the first
remainder is \2x^ — 1 x^ — 24 x + 16.
Dividing the first term of this by twice the first term of the root, 4 x^,
we have the next term of the root, 3 x (§ 213, II).
Adding this to 4 x^ gives 4 x^ + 3 x ; multiplying the result by 3 x, and
subtracting the product, 12 x^ + 9 x^, from the first remainder, gives the
second remainder, — 16 x^ — 24 x + 16.
Dividing the first term of this by twice the first term of the root, 4 x^,
we have the last term of the root,  4 (§ 213, III).
If from the second remainder we subtract (4x2 + 6x — 4)(— 4), or
— 16 x2  24 X + 16, there is no remainder ; then, 2 x^ + 3 x  4 is the
required root (§ 213, IV).
The expressions 4 x^ and 4 x^ + 6 x are called trialdivisors^ and
4 x2 + 3 X and 4 x^ + 6 x — 4 complete divisors.
We then have the following rule for extracting the square
root of a polynomial perfect square :
Arrange the expression according to the powers of some letter.
Extract the square root of the first terrti, write the result as the
first term of the root, and subtract its square from the given
expression, arranging the remainder in the same order of powers
as the given expression.
Divide the first term of the remainder by twice the first term of
the root, and add the quotient to the part of the root already found,
and also to the trialdivisor.
Multiply the complete divisor by the term of the root last obtained,
and subtract the product from the remainder.
If other terms remain, proceed as before, doubling the part of
the root already found for the next trialdivisor.
215. Examples.
1. Find the square root of 9 a;^ + 30 aV + 25 a\
9 x* + 30 a^x^ + 25 a^  3 x^ + 5 g^
• 9x4
6x2 45a3
30 a3x2
30 a3cc2 + 25 a^
It is usual, in practice, to omit those terms, after the first, in each
remainder, which are merely repetitions of the terms in the given expres
sion ; thus, in the first remainder of Ex. 1, we leave out the term 25 a®.
196
ALGEBRA
It is also usual to leave out of the written work the multiplier of the
complete divisor.
f^^ 2. Find the square root of
Arranging according to the descending powers of x, we have
9 x6  12 a;5 + 28 a;*  22 a:3 + 20 ic2  8 ic + 1  3x^ 2x2 + 4a; 1
9x6
6x32x2
12x5
 12 x5 + 4 X*
6 x3  4 x2 + 4 X
24 X*
24 X*
16 x3 + 16 a;2
6 x3  4 x2 + 8 X  1
 6x3+ 4xz
 6x3+ 4x2
8x + l
It will be observed that each trialdivisor is equal to the pre
ceding complete divisor ivith its last term doubled.
If, in Ex. 2, we had written the expression
1  8 X + 20 x2  22 x3 + 28 x*  12 x6 + 9 x8,
the square root would have been obtained in the form 1— 4x + 2x2 — Sx^,
which is the negative of 3 x3 — 2 x2 + 4 x — 1.
EXERCISE 88
Find the square roots of the following :
1. x^\6a^{llx'\6x\l.
2. I_4a + 2a^ + 4a3 + a^
3. 9n^ + 12n«20n^16n + 16.
4. 5Qx' + 4.60x^24:Xj25x\
5. x^ \ y^ \ 4:Z^ — 2 xy } 4:xz — 4:yz.
6. Sa^4.a16a' + l\16a'\4.a^
7. 25x^ + 10a^yhxy\30x^y^\6xY^9f.
8. 36a2_f.25 62 + l6c260a648ac + 406c.
9. 9x* + 6a^y4t7a^y'^16xfi64:y\
10. 127^42n3 + 4_l9^^2 + 49^4
11. 16a' + 4:Sa^b + 60a'b' \36ab' h9b\
EVOLUTION 197
12. 4 n''  16 nV f 36 n'^x^  40 n^x^ + 25 ar^.
13. 30 xy  24 a;y _ 31 ^y _}_ 25 «» + 16 /.
14. 4a;2 + 20a; + 29 + — + i.
X x^
15. a^_2a^a*46a'^3a24a + 4.
16. 5x^23x'{12x\8a^22o(^\16x^ + 4L,
17 .2 2a5 , 136^ 46« , 46^
17. a ^4^ 9^ + 9^*
18 ^' ^' 41n^ 5n 25
■ 4 3 36 6 16*
19. 9 a« + 6 a'x + 31 aV  14 a^x^ + 17 aV  40 aa^ 4 16 a^.
* 16 4.y'^20y^ 5f 25y''
21 25 15a 41a^ 3a' ^ a'
'4 b 4.b^ 2^^ 16b^'
22. 44 a^^^ + 4 6«  30 a^ft + 4 a^d^ + 25 a^  16 a6^  31 a*b\
SQUARE ROOT OF AN ARITHMETICAL NUMBER
216. The square root of 100 is 10 ; of 10000 is 100 ; etc.
Hence, the square root of a number between 1 and 100 is
between 1 and 10; the square root of a number between 100
and 10000 is between 10 and 100; etc.
That is, the integral part of the square root of an integer of
one or two digits, contains one digit ; of an integer of three or
four digits, contains two digits ; and so on.
Hence, if a point be placed over every second digit of an integer,
beginning at the units^ place, the number of points shows the number
of digits in the integral part of its square root.
217. Square Root of any Integral Perfect Square.
The square root of an integral perfect square may be found
in the same way as the square root of a polynomial.
Required the square root of 106929.
198
ALGEBRA
2 a + 6 = 600 + 20
20
106929
a2 = 90000
2a+26 + c = 600 + 40 + 7
7
16929
12400
300 + 20 + 7
= a{'b\c
4529
4529
Pointing the number in accordance with the rule of § 216, we
find that there are three digits in its square root.
Let a represent the hundreds' digit of the root, with two
ciphers annexed; b the tens' digit, with one cipher annexed;
and c the units' digit.
Then, a must be the greatest multiple of 100 whose square is
less than 106929 ; this we find to be 300.
Subtracting a^, or 90000, from the given number, the result
is 16929.
Dividing this remainder by 2 a, or 600, we have the quotient
28+ ; which suggests that b equals 20.
Adding this to 2 a, or 600, and multiplying the result by b, or
20, we have 12400 ; which, subtracted from 16929, leaves 4529.
Since this remainder equals (2a + 2 6 + c)c (§ 213, III), we
can get c approximately by dividing it by 2 a + 2 6, or 600 + 40.
Dividing 4529 by 640, we have the quotient 7+; which
suggests that c equals 7.
Adding this to 600 + 40, multiplying the result by 7, and
subtracting the product, 4529, there is no remainder.
Then, 300 + 20 + 7, or 327, is the required square root.
218. Omitting the ciphers for the sake of brevity, and con
densing the operation, we may arrange the work of the example
of § 217 as follows :
106929 [327
9
62
169
124
647
4529
4529
EVOLUTION 199
The numbers 600 and 640 are called trialdivisors^ and the numbers
620 and 647 are called complete divisors.
We then have the following rule for finding the square root
of an integral perfect square :
Separate the number into periods by pointing every second digits
beginning with the units' place.
Find the greatest square in the lefthand period, and write its
square root as the first digit of the root; subtract the square of the
first rootdigit from the lefthand period, and to the result annex
the next period.
Divide this remainder, omitting the last digit, by twice the part
of the root already found, and annex the quotient to the root, and
also to the trialdivisor.
Multiply the complete divisor by the rootdigit last obtained, and
subtract the product from the remainder.
If other periods remain, proceed as before, doubling the part of
the root already found for the next trialdivisor.
Note 1. It sometimes happens that, on multiplying a complete divisor
by the digit of the root last obtained, the product is greater than the
remainder.
In such a case, the digit of the root last obtained is too great, and one
less must be substituted for it.
Note 2. If any rootdigit is 0, annex to the trialdivisor, and annex
to the remainder the next period. (See the illustrative example of § 220.)
219. Ex. Find the square root of 4624.
4624 L68
36
128
1024
1024
The greatest square in the lefthand period is 36.
Then the first digit of the root is 6.
Subtracting 62, or 36, from the lefthand period, the result is 10 ; to
this we annex the next period, 24.
Dividing this remainder, omitting the last digit, or 102, by tv^^ice the
part of the root already found, or 12, the quotient is 8 ; this we annex to
the root, and also to the trialdivisor.
200 ALGEBRA
Multiplying the complete divisor, 128, by 8, and subtracting the product
from the remainder, there is no remainder.
Then, 68 is the required square root.
220. We will now show how to find the square root of a
number which is not integral.
Ex. Find the square root of 49.449024.
We have, ^49:149024 = J4M49024 ^ V49449024.
V 1000000 vioooooo
49449024  7032
49
1403
4490
4209
14062
28124
28124
Since 14 is not contained in 4, we write as the second rootdigit, in
the above example ; we then annex to the trialdivisor 14, and annex to
the remainder the next period, 90. (See Note 2, § 218.)
Then,
'49.449024 = """" = 7.032
1000
mged as follows :
49.449024  7.032
49
1403
4490
4209
14062
28124
28124
Then, if a point he placed over every second digit of miy
number, beginning with the units' place, and extending in either
direction, the rule of § 218 may be applied to the result and the
decimal point inserted in its proper position in the root.
^^ EXERCISE 89
Find the square roots of the following :
1. 5776. 2. 15376. 3. 67081.
EVOLUTION
201
4.
8427.24.
8.
7974.49.
12.
.30316036.
5.
.165649.
9.
.00459684.
13.
39.375625.
6.
.133225.
10.
22014864.
14.
.000064272289.
7.
54.4644.
11.
1488.4164.
15.
889060.41.
221. Approximate Square Roots.
If there is a final remainder, the number has no exact square
root ; but we may continue the operation by annexing periods
of ciphers, and obtain an approximate root, correct to any desired
number of decimal places.
Ex. Find the square root of 12 to four decimal places.
3.4641 +
12.
9
00000000
64
3 00
2 56
686
4400
4116
6924
28400
27696
69281 I 70400
222. The approximate square root of a fraction may be ob
tained by taking the square root of the numerator, and then
of the denominator, and dividing the first result by the second.
If the denominator is not a perfect square, it is better to
reduce the fraction to an equivalent fraction whose denominator
I is a perfect square.
Ex. Find the value of Vf to five decimal places.
;
Vie
EXERCISE 90
Find the first five figures of the square root of :
1. 2. 4. 17. 7. .3. 10. .008.
2. 5. 5. 59. 8. .067. 11. .00095.
3. 11. 6. 75.8. 9. .46. 12. 96.756.
202 ALGEBRA
Find the first four figures of the square root of :
13. i. 15. II 17. f. 19. ii. 21. If
14. . 16. 1 18. . 20. i. 22. if.
CUBE ROOT OF A POLYNOMIAL
223. The cube roots of certain polynomials of the form
a^ + Sa'b\3ab^ + b^
can be found by inspection.
Ex. Find the cube root of 8 a^  36 a^b^ + 54 a&*  27 b\
We can write the expression as follows :
(2 ay  3 (2 a)2(3 62) + 3 (2 a) (3 62)2 _ (3 52)8.
By § 205, this is the cube of 2 a  3 62.
Then, the cube root of the expression is 2 a — 3 62.
EXERCISE 91
Find the cube roots of the following :
1. a3 + 6a2 + 12a + 8.
2. l9m^27m'27m^
3. 647i348n2 + 12nl.
4. 125a^ + 75afy + 15xy^ + f.
5. a« + 18a*6^4108a26«4216 6».
6. 125m3+150m2w + 60mn2 + 87i3.
7. 27 a^b^lOSa'b'c + lUabc' 64. (^,
8. m«21mV + 147mV343a;^2
224. Cube Root of any Polynomial Perfect Cube.
By§205, (a + 6 + c)«=[(a + &) + c7
= (a + 6)3 + 3(a + 6)2c + 3(a + 6)c2Hc3
= a» + 3a26 + 3a62 + &' + 3(a4&)'c + 3(af&)c' + c'
= a^ + (3 a2 + 3 a6 + 6^)6 + [3(a + bf + 3(a + &)c + c^Jc (1)
EVOLUTION 203
Then, if the cube of a trinomial be arranged in order of
powers of some letter :
I. The cube root of the first term gives the first term of the
cube root, a.
II. If from (1) we subtract a^, we have
(3 a^ + 3 a& + b^)b [f [3(a f bf + 3(a + b)c f c'y. (2)
The first term of this, "^when expanded, is 3 a^b ; if this be
divided by three times the square of the first term of the root,
3 a^, we have the next term of the root, b.
III. If from (2) we subtract (3 a^ + 3 a5 + b^b, we have
[3(a + 6)2 + 3(a + b)c + c']c. (3)
The first term of this, when expanded, is 3a^c; if this be
divided by three times the square of the first term of the root,
3 a^, we have the last term of the root, c.
IV. If from (3) we subtract [3(a + 6)^ + 3(a + 6)c + c2]c,
there is no remainder.
Similar considerations hold with respect to the cube of poly
nomial of any number of terms.
225. The principles of § 224 may be used to find the cube
root of a polynomial perfect cube of any number of terms.
Let it be required to find the cube root of
x^+6 #+ 3 X*  28 x8  9 x2 + 54 x 27
3 a2+3 a6 + 62= 3 x*+6 x3+4 x^
' 2x
6a:5+ 3cc428x39ic2+64x27
6x5 + 12x4+ 8x3
3(a+ 6)2=3 x*+12 x3+12 x2
3(a + 6)c+c2=  9x218x+9
3x4+12x8+ 3x318x+9
3
9 x436 x89 x2+54 x27^
9 x436 x89 x2+54 x~27
The first term of the root is the cube root of x^, or x'^.
Subtracting the cube of x^, or x^, from the given expression, the first
jmainder is 6 x& + 3 x*  28 xs  9 x'^ + 64 x  27.
204 ALGEBKA
Dividing the first term of this by three times the square of the first
term of the root, 3 x'^, we have the next term of the root, 2 x (§ 224, II).
Now, 3 a6 + 62 equals 3 x x^ x 2 ic + (2 x)2, or 6 x^ + 4 x^.
Adding this to 3 x^, multiplying the result by 2 x, and subtracting the
product, 6 x^ + 12 X* + 8 x^, from the first remainder, gives the second
remainder,  9 x*  36 x^  9 x2 + 54 x  27 (§ 224, III).
Dividing the first term of this by three times the square of the first
term of the root, 3x2, we have the last term of the root, —3.
Now, 3(a+6)2 equals 3(x2 + 2x)2, or 3 x* + 12x3+ 12x2; 3(a + 6)c
equals 3(x2 + 2 x) (  3), or  9 x2  18 x ; and d^ = 9.
Adding these results, we have 3 x* + 12 x^ + 3 x2 — 18 x + 9.
Subtracting from the second remainder the product of this by — 3, or
— 9 X* — 36 x^ — 9 x'^ + 54 X — 27, there is no remainder ; then, x2 + 2 x — 3
is the required root (§ 224, IV).
The expressions 3 x* and 3 x* + 12 x' + 12 x2 are called trialdivisors,
and the expressions 3 x* + 6 x^ + 4 x2 and 3 x* + 12 x^ + 3 x2  18 x + 9
complete divisors.
We then have the following rule for finding the cube root of
a polynomial perfect cube :
Arrange the expression according to the powers of some letter.
Extract the cube root of the first term, write the result as the
first term of the root, and subtract its cube from the given
expression; arranging the remainder in the same order of powers
as the given expression.
Divide the first teryn of the remainder by three times the square
of the first term of the root, and write the result as the next term
of the root.
Add to the trialdivisor three times the product of the term of
the root last obtained by the part of the root previously found, and
the square of the term of the root last obtained.
Multiply the complete divisor by the term of the root last
obtained, and subtract the product from the remainder.
If other terms remain, proceed as before, taking three times
the square of the part of the root already found for the next trial
divisor.
226. Examples.
1. Find the cube root of 8 a;^  36 xhj + 54 a^y  27 f.
EVOLUTION
205
8 x6  36 x*^ + 54 a;V _ 27 y^ \ 2x^Sy
8x6
12 X*  18 x2?/ + 9 2/2
 36 x%
 36 x^y + 54 x^y"^  27 yS
It is usual, in practice, to omit those terms, after the first, in each
remainder, which are merely repetitions of the terms in the given expres
sion ; and also to leave out of the written work the multiplier of the com
plete divisor.
2. Find the cube root of 40 aj^  6 or'  64 + ««  96 a;.
Arranging according to the descending powers of x, we have
x66x
X6
> + 40 x3  96 X  64 1 x2  2 X 
5
' f 12 x*  8 x8
^
3 x4  6 x3 + 4 x2
6x
6x
3x4
 12 x3 + 12 x2
 12 x2 + 24 X + 16
 12 x* f 48 x8
 12 x* F 48 x3  96 X  64
3x4
12x3 l24a
,+ 16
EXERCISE 92
Find the cube roots of the following :
1. 27a;«f27a;^h9aj2fl. 2. S a'  60 a'b +150 a'b'' 125 b\
3. S36xy'^3^3a:^64:f5SSx'y.
4. x^ + 9a^\30x' + 4.5a^ + S0a:^\9x\l.
5. Sn^12n'30n'' + 357i^ + 4.5n^2Tn27.
6. 9a' + 54.a'l2Sa^3d'+27a' + 6a,
7.
8.
27 12
aJl b^
16 64*
n^  12 n'x + 57 nV  136 n^a^ + 171 nV108 nar'+27 a^.
9. 135 a%^ f 12 a'b  125 b^ + Sa^59 a^b^+75 ab'54. a%K
10. 152a^2763aj2 4.27a;6h63a5^108a;108a;^.
ill. ^
15 a"
153a^^() 153_15 1.
8 4 ■ 4 ' ' ' 2a a' W
12. 64mH144m^h204m*hl71m3h92m2}36mh8.
13. a;96a;«M5a;^26a;«+39a^42a;^}37a^30a^f12aj
206
ALGEBRA
CUBE ROOT OF AN ARITHMETICAL NUMBER
227. The cube root of 1000 is 10; of 1000000 is 100; etc.
Hence, the cube root of a number between 1 and 1000 is
between 1 and 10; the cube root of a number between 1000
and 1000000 is between 10 and 100 ; etc.
That is, the integral part of the cube root of an integer of
one, two, or three digits, contains one digit ; of an integer of
four, five, or six digits, contains two digits ; and so on.
Hence, if a point he placed over every third digit of an integer ^
beginning at the units'' place, the number of points shows the number
of digits in the integral part of its cube root.
228. Cube Root of any Integral Perfect Cube.
The cube root of an integral perfect cube may be found in
the same way as the cube root of a polynomial.
Eequired the cube root of 12487168.
200 + 30 + 2
= a + & + c
12487168
a^ = 8000000
3^2 = 120000
Sab= 18000
¥ = 900
4487168
138900
30
4167000
3 (a + 6)2= 158700
3(a + 6)c= 1380
c2= 4
320168
160084
2
320168
Pointing the number in accordance with the rule of § 227, we
find that there are three digits in the cube root.
Let a represent the hundreds' digit of the root, with two
ciphers annexed ; b the tens' digit, with one cipher annexed ;
and c the units' digit.
Then, a must be the greatest multiple of 100 whose cube is
less than 12487168 : this we find to be 200.
EVOLUTION 207
Subtracting a^, or 8000000, from the given number, the result
is 4487168.
Dividing this by 3 a^, or 120000, we have the quotient 37+ ;
which suggests that h equals 30.
Adding to the divisor 120000, 3 ah, or 18000, and h\ or 900,
we have 138900.
Multiplying this by 6, or 30, and subtracting the product
4167000 from 4487168, we have 320168.
Since this remainder equals [3(a4 6)^ + 3 (a f 6) c + c^]c
(§ 224, III), we can get c approximately by dividing it by
3 (a + 6)2, or 158700.
Dividing 320168 by 158700, the quotient is 2+ ; which sug
gests that c equals 2.
Adding to the divisor 158700, 3(a + 6)c, or 1380, and c^, or
4, we have 160084 ; multiplying this by 2, and subtracting the
product, 320168, there is no remainder.
Then, 200 + 30 + 2, or 232, is the required cube root.
229. Omitting the ciphers for the sake of brevity, and con
densing the process, the work of the example of § 228 will
stand as follows :
12487168 [232
8
1200
4487
180
9
1389
4167
15870
320168
1380
4
16008
4
320168
The numbers 120000 and 168700 are called trialdivisors, and the
numbers 138900 and 160084 are called complete divisors.
We then have the following rule for finding the cube root of
an integral perfect cube :
Separate the number inta periods by pointing every third digit,
beginning with the units' place.
208 ALGEBRA
Find the greatest cube in the lefthand 2)eriod, and write its cube
root as the first digit of the root; subtract the cube of the first root
digit from the lefthand period, and to the result annex the next
period.
Divide this remainder by three times the square of the part of
the root already found, ivith ttvo ciphers annexed, and write the
quotient as the next digit of the root.
Add to the trialdivisor three times the product of the last root
digit by the part of the root previously found, with one cipher
annexed, and the square of the last rootdigit.
Multiply the complete divisor by the digit of the root last
obtained, aiid subtract the product from the remainder.
If other periods remain, proceed as before, taking three times
the square of the part of the root already found, with two ciphers
annexed, for the next trialdivisor.
Note 1. Note 1, § 218, applies with equal force to the above rule.
Note 2. If any rootfigure is 0, annex two ciphers to the trial
divisor, and annex to the remainder the next period.
230. In the example of § 228, the first complete divisor is
The next trialdivisor is 3 (a 4 b^, or 3 a^ + 6 a& + 3 b^.
This may be obtained from (1) by adding to it its second
term, and double its third term.
That is, if the first number and the double of the second number
required to complete any trialdivisor be added to the complete
divisor, the result, with two ciphers annexed, will give the next
trialdivisor.
This rule saves much labor in forming the trialdivisors.
231. Ex. Find the cube root of 157464.
157464 [54
125
7500
32464
600
16
8116
32464
EVOLUTION
209
232. We will now show how to find the cube root of a
number which is not integral.
Ex. Find the cube root of 8144.865728.
We have, v^8144. 865728 = ^
8144865728 V8144865728
1000000
v^lOOOOOO
8144865728  2012
8
120000
600
1
120601
144865
120601
600
2
24264728
12120300
12060
4
12132364
24264728
Since 1200 is not contained in 144, we write as the second rootdigit,
in the above example ; we then annex two ciphers to the trialdivisor
1200, and annex to the remainder the next period, 865. (Note 1, § 229.)
The second trialdivisor is formed by the rule of § 230.
Adding to the complete divisor 120601 the first number, 600, and twice
the second number, 2, required to complete the trialdivisor 120000, we
have 121203 ; annexing two ciphers to this, the result is 12120300.
Then, ^8144.865728 = ?^ = 20.12.
The work may be arranged as follows :
18144. 865728  20.12
8
120000
144 865
600
1
120601
120 601
600
24 264728
2
12120300
12060
4
12132364
24 264728
210
ALGEBRA
It follows from the above that, if a point he placed over every
third digit of any number, beginning with the units'* place, and
extending in either direction, the rule of § 229 may be applied to
the result, and the decimal point inserted in its proper position in
the root.
EXERCISE 93
Find the cube roots of the following :
1. 54872.
6.
3176523.
11. 331373888.
2. 262144.
7.
130323.843.
12. 37.595375.
3. 103.823.
8.
.102503232.
13. 667627.624.
4. .884736.
9.
.000356400829.
14. .964430272.
5. .000493039
10.
22.665187.
15. 3422470.843.
Find the first four figures of the cube root of :
16. 4.
18.
8.2.
20. J.
22. A.
17. 9.
19.
.03.
21. M.
23. .
233. If the index of the required root is the product of two
or more numbers, we may obtain the result by successive ea>
tractions of the simpler roots.
For by §206, C\/^)'»" = a. «
Taking the nth root of both members, j
CVS)»=V^. (1)
Taking the mth root of both members of (1),
Hence, the mnth root of an expression is equal to the mth root
of the nth root of the expression.
Thus, to find the fourth root of an expression, we find the
square root of its square root ; to find the sixth root, we find
the cube root of the square root, etc.
EVOLUTION 211
EXERCISE 94
Find the fourth roots of the following :
1 . a' 16 a'W + 96 a^h'  256 a%^ + 256 h^.
2. 81 a«  108 a' + 162 a«  120 a' + 91 a^  40 a^+lS o?^ a
+ 1.
3. 16 + 32 a;  72 a^  136 aj3 + 145 a;^+204 ^162 a;«108 x'
+ 81a^.
4. .011156640625.
Find the sixth roots of the following :
6. 64a;i2_^192a;i« + 240a^ + 160a^ + 60a;^412a;2 + l
6. a«  18 a^ + 135 a^  540 a^ + 1215 a?  1458 a 4 729.
7. 34296.447249.
234. By §206, (•v/^)" = a&.
Also, {y/a X y/ly = (v/a)" X (V&)" = «&.
Then, (v/^)« = ( v^ x ■>/&)".
Whence, Voft = Va x ">/&.
Y
212 ALGEBRA
XVII. THEORY OF EXPONENTS
235. In the preceding portions of the work, an exponent has
been considered only as a positive integer.
Thus, if m is a positive integer,
a"* = a X a X a X ••• to m factors. (§ 11)
The following results have been proved to hold for any
positive integral values of m and n :
orxor = «"*+" (§ bOf). (1)
{w^y = or^ (§ 93). (2)
236. It is necessary to employ exponents which are not
positive integers; and we now proceed to define them, and
prove the rules for their use.
In determining what meanings to assign to the new forms, it
will be convenient to have them such that the above law for
multiplication shall hold with respect to them.
We shall therefore assume equation (1), § 235, to hold for
all values of m and n, and find what meanings must be attached
in consequence to fractional^ negative, and zero exponents.
237. Meaning of a Fractional Exponent.
Let it be required to find the meaning of a^.
If (1), § 235, is to hold for all values of m and n,
a^ X a^ X a^ = a^^^"^^ = a^.
Then, the third power of a^ equals a^.
Hence, a^ must be the cube root of a^, or (3 = V^.
We will now consider the general case.
p
Let it be required to find the meaning of a^, where p and q
are any positive integers.
THEORY OF EXPONENTS 213
If (1), § 235, is to hold for all values of m and w,
? P P ?+?+?+... to ?t.rm« P^q
a^ X a^ X a'^ X ••• to g factors = a* ' * = a' = a^.
I'
Then, the gth power of a* equals a^.
p p
Hence, a' must be the ^th root of a^, or a' —{/a^.
i. Hence, in a fractional exponent, the numerator denotes a
power, and the denommator a root.
For example, a^ = \/a^ ; 6^ = V6^ ; x^ = y/x ; etc.
EXERCISE 95
Express the following with radical signs :
1. aK 3. 7 mi 5. ah^. 7. 8aW. 9. x^yh^.
2. x^. 4. 5xk 6. a^V'^ 8. lOw^cc^^. 10. 2a'"62cT.
Express the following with fractional exponents :
11. ^6. 13. V^3. 15. S</F. 17. 9^m^;?.
12. </a. 14. ^n^ 16. 4^/p. 18. ^^"v//.
19. ^a>y6^«. 20. ^s/^Vy^V^^.
^^,  238. Meaning of a Zero Exponent.
i,(sA^< If (1), § 235, is to hold for all values of m and n, we have
a'^x a^ = a"^^^ = a*".
Whence, a» = — = 1. '
a"*
We must then define a^ as being equal to 1.
239. Meaning of a Negative Exponent.
et it be required to find the meaning of a~^.
If (1), § 235, is to hold for all values of m and n,
a^ X a^ = a^+^ = a« = 1 (§ 238).
214 ALGEBRA
1
Whence,
.3
y3
Whence, a~'=
a"
We will now consider the general case.
Let it be required to find the meaning of a~% where s repre
sents a positive integer or a positive fraction.
If (1), § 235, is to hold for all values of m and n,
a' xa' = a'+' = a« = 1 (§ 238).
1_^
a"'
We must then define a~' as being equal to 1 divided by a'.
For example, a"^ = — ; a »=__; 3331^"^ = — _j etc.
240. It follows from § 239 that
Any f(xctor of the numerator of a fraction may be transferred
to the denominator, or any factor of the denominator to the
numerator, if the sign of its exponent be changed.
Thus, a?l^J^^^^^^^ ,te.
EXERCISE 96
Express with positive exponents :
1.
ojy.
5.
a^m\
9.
m~^n^.
2.
ahK
6.
miV.
10.
8a^6^V.
3.
m *?l^
7.
^a^n\
11.
6 mT^nT^Q^.
4.
Sn^a;.
8.
5 a; V^2;^
12.
la^n^x^.
Transfer all literal factors from the denominators to the
numerators in the following:
13. i. 14. 4 15 ^ 16. A
THEORY OF EXPONENTS 215
17. ^. 18. ^^. 19. ^^. 20. ^^"'^"l
^ Transfer all literal factors from the mimerators to the
denominators in the following:
21 i^ 23 ^~^^ 25 ^ a~^6^ 07 9 m~^yi~^
22. ^. 24. ^. 26. ^. 28. A^^.
241. We obtained the definitions of fractional, zero, and
negative exponents by supposing equation (1), § 235, to hold
for such exponents.
Then, for any values of m and n^
oTxa^'^z or+\ (1)
The formal proof of this result for positive or negative, integral or
fractional, values of m and n will be found in § 445.
1. Find the value of a^ x a~^.
We have, a^ x a~^ = a^s — ^58,
2. Find the value of a x Va^.
By §237, axVa^ = axa^ = a}^^ = a^,
3. Multiply a + 2 a* 3 a^ by 2  4 a"^  6 a"^.
4 a ^6 a
2 a + 4 a^  Qa^
 4 a^  8 a^ + 12
6 a^  12 + 18 a"^
2 a 20a^ f 18 a
It must be carefully observed, in examples like the above, that the
zero power of any number equals 1 (§ 238). >
216 ALGEBRA
EXERCISE 97
Multiply the following :
1. a^ by a^. ^' 4. n^ by n^ 7. 2x~^ hj 7 x\
5. 3 a' by a'l 8. x^ by J/^.
6. m by 4 m ^". 9. v^ by a^.
13. 3a?"%~^ by 4:X~^y.
,, '(/u. a^^/^ by a«V^.
15. m~^n~^^ by
\f '•
X' by xl
4 3.
a^ by a~^.
Vi
\j 10.
m^ by — •
11.
8Va3 by
12. 6 a^z, by a^i'^. 3 nr'n^
16. a;^ — 2 ic32/3 + 4 2/^ by x^\2yK
17. 2n^56nHy 3n?4.
18. 4a'' + 10a2 + 25 by 2 a' 5.
19. a^  ah^ 4 &"^ by a^ + ah^ + 6^
20. x~^ — x~^y^ + 2 x~^y^ by a;"^^/^ — a?"^^/* + 2 2/.
21. a'^ft'^ + 4 a^ + 3 a~h^ by a~^  4 a'^^^  3 &«.
22. x^4.xi^ + Qx^ hj 2xi^xiSx'^.
23. a"^  2 a^ni + 3 a'^Ti^ by 2 a^ri^ + 4 or^^  6 n^.
24. 2 a^  a^a?  5 a^a;^ by 4 a"^aj~^ + 2 a^a;"* + 10 a"i
242. To prove — = a"*"" /or aZ/ values of m and n.
ft
By § 240, ^ = a X a« = a^% by (1), § 241.
The proof of this result in the case where m and n are positive integers,
and m > n, is given in § 70. .
1. Find the value of — • 
We have, «I = oi+2^of
\
THEORY OF EXPONENTS
217
2. Find the value of
Va'
<^' a^
3. Divide 18 xy^  23 + x~Hj + 6 x'Y
by 3 x^y~'^ \x'^ — 2 x~*y.
18 a;i/2  23 + X ^y + 6 x'^y^
18 xy2 + 6 x^yi  12
3 x'^y'^ + x^ 2x ^y
6 x^i/i 2x ^3x ^y
6 x^?/i  11 + X ^?/ + 6 x12/2
6xV
1 _
2 + 4x'
%

93x
93x"
% + 6 X
% + 6x
.ly2
It is important to arrange the dividend, divisor, and each remainder in
the same order of powers of some common letter.
EXERCISE 98
Divide the following :
1. ar' by x^. 4. m~^ by y/m.
5. a~^ by
'^a^
7. Va' by ■^/a^
8. Sy/mF' by 2 m^.
9. 9a~^b'hy3a'bi
2. a^ by ai . a,
3. 71 by n"i 6. rc^ by a;"rV
10. ic"' — 2 a;"' — 8 x^ by a?"^ — 4A
11. a'b'' by a^5l 12. a^1 by x^ + 1.
13.
7 + n^ by n^ + S + n^
^14. a^ + 4a^2a^12a^49 by a* + 2a^3
\/15. 8m^ + 12m^n^ + 6mM4^^ by 2m^ + ni
16. a^?/i2_;i^;^^4y6_^l ^y a.V' + 3a^7/^a^2/"'
17. a"* + 2 a^62 + 9 b'' by a^ + 2 a'h'' + 3 a"^&2.
218 ALGEBRA
18. 4 aV^  17 a^x^ + 16 a~^a^ by 2 a*  a^  4 a" V.
19. 9 m*n~3 — 10 m^n^ + m~*n by 3 m^?i^ — 4 mn^ + m%.
243. We will now show how to prove equation (2), § 235,
for any values of m and n.
We will consider three cases, in each of which m may have
any value, positive or negative, integral or fractional.
I. Let ?i be a positive integer.
The proof given in § 93 holds if n is a positive integer, what
ever the value of m.
II. Let n = — , where p and q are positive integers.
Then, by the definition of § 237,
(ay = V(^^ = ^/aFp (§ 243, I) = a"^.
III. Let n = — s, where s is a positive number.
Then, by the definition of § 239,
(a)* = i = —(% 243, I or II) = a^\
Therefore, the result holds for all values of m and n.
1. Find the value of (a^)^
We have, (a2)5 = a^xs = ^lo.
2. Find the value of (a^)"^.
(a3)^ = a"^^^ = a.
3. Find the value of (Va)l
EXERCISE 99
Find the values of the following :
1. (a^)*. 2. (xy. 3. (x^)i 4. (a'yi
THEORY OF EXPONENTS 219
5. (mfy. 8. (a^)^o. H. (%i)^ 14^
7. (a»)i "■ Vn^J' 13. (x'y^. 16. (a»^V=i;
244. The value of a numerical expression affected with a
fractional exponent may be found by first, if possible, extract
ing the root indicated by the denominator, and then raising the
result to the power indicated by the numerator.
Ex. Find the value of ( 8)i
By § 243, ( 8)^ = [( 8)^]2 = (^/38)2 = (_ 2)2 = 4.
EXERCISE 100
Find the values of the following :
1. 27l 5. 811 9. 2564. 13. 2431
2. lel 6. (32)1 10. (512)^. 14. (128)7.
3. 64l 7. 361 11. 9l ^^ 15. 729^
4. 64l 8. (216)1 12. (8)"^. 16. 5121
245. We will now show how to prove the result
(aby = a^'b'',
for any fractional or negative value of n.
The proof of this result in the case where n is any positive
integer, was given in § 94.
I. Let n = ^, where p and q are any positive integers.
By § 243, [(abyy = (aby = a^b^ (§ 94). (1)
By § 94, (Jbly = {a^yib^y = a^hK (2)
From (1) and (2), \_{aby ]' = (a« 6 * )*.
220 ALGEBRA
Taking the qth. root of both members, we have
p p p
(aby =a^h^.
II. Let n = — s, where s is any positive integer or positive
fraction.
Then, (a&) = ^ = 4^(§§ 94, or 245, I) = a6
EXERCISE 101
Find the values of the following :
1. {ahy. 3. (ic'V)^ 5. (nV^)".
2. (m^n^)\ 4. {a^x^y^ 6. (a/o^^^^I
^IVIISCELLANEOUS EXAMPLES
EXERCISE 102
Square the following by the rule of § 97 :
^ 1. 3a^ + 4 6"l 2. 5 m^w^  8 m^n*.
3. Square ft^^,^ _ 2 a'  a,i6^ by the rule of § 204.
4. Expand {^ x^y~^ \ 1 z"^ {4. x^y~^  1 z^) by the rule of
§98.
Find the value of :
^ 25a«49m^ ^ .,. , o . ...
5. ~, by the rule of § 101.
5 a3 _ 7 y^ii •
^ 8 i«2 4 27 ?r^
2 a;* + 3 2/~* i ?' ) i V / '^
 ^' ^ z^ ■ 8 —, 1, by the rule of § 103.
x^x 3 a* + 6"^
9. {3x^4.yfy. 10. (a263 + 2a«6y.
THEORY OF EXPONENTS 221
Find the square roots of the following :
11. 16 a^mK 12. 49a;%2~l 13. ^? •
4 b^n^
14. 9x^6x'i + 25Sx^ + 16x\
15. 4a^420a5 + 21a^10a^ + l.
16. a^63  6 a'62 + 5 ^^ + 12 a,"* + 4 a~^6. •
Find the cube roots of the following :
_17. 8a^r' 18. 64a'^6M. 19. ?I^?L!^.
x^y 2
20. 27aj^454a;^2/"* + 36a;^2/"^ + 8 2/^.
21. x^ 6 x'^ + 21 x^ Ux^ + 63 x^ 54:x^ + 27 x^.
Simplify the following, expressing all the results with posi
tive exponents :
^. [^(.VO^^(»V)]« 23. ^x^.
■y/lf</c yjo?
k24. [a"i X (a0"+^] x [(a*^)"' x (a"^)!].
25. (x'^ xx^^y. 31, JLt^ aMi&!.
a^ + b^ a6
(^m+nX 2m //i 2m\ m— n
^ l^j ■ 32. ai±6* + ?l±61*.
n+1
a^fti a"^6"^
27. (a'*^ ^ a"+^) 2" . §!?
1 1
33 ^:^i^»^^
mn
28. (a;'"'fic"')'^+«fa; ». x^^' + l a^" — 1
29. [^(0.^0? . 34. ^^xj—^ + l.
30 ^^ + y^ a? 4 y . 35 g^ + 2 6^ 7 a^6^ + 6b^
222 ALGEBRA
0. wa ^ XVIII. SURDS .. ^ . ; _ f>
246. A Surd is the indicated root of a number, or expression,
which is not a perfect power of the degree denoted by the index
of the radical sign ; as V2, VS, or ^x + y.
fK^Jix^*^ 247. A monomial is said to be rational when it is rational
^^Jt^U ^^^ integral (§ 63), or else a fraction whose terms are rational
, Jt3. and integral.
wvT ic t>v^.A polynomial is said to be rational when each of its terms is
rational.
An expression is said to be irrational when it involves surds ;
as 2 + V3, or Va 4 1 — Va.
248. A rational number is a positive or negative integer, or
a positive or negative fraction.
An irrational number is a numerical expression involving
surds ; as V 3, or 2 + V5.
249. If a surd is in the form bVa, b is called the coefficient
of the surd, and ri the index.
250. The degree of a surd is denoted by its index ; thus, V5
is a surd of the third degree.
A quadratic surd is a surd of the second degree.
REDUCTION OF A SURD TO ITS SIMPLEST FORM
/ 251. A surd is said to be in its simplest form when the
expression under the radical sign is rational and integral
(§ 63), is not a perfect power of the degree denoted by any
factor of the index of the surd, and has no factor which is a
perfect power of the same degree as the surd.
252. Case I. When the expression under the radical sign is
a perfect power of the degree denoted by a factor of the index.
Ex. Reduce V8 to its simplest form.
We have, \/8 = v^ = 2 "^ (§ 2.37) = 2^ = \/2.
SURDS 223
'Exercise 103
Eeduce the following to their simplest forms :
1. </2E. 5. a/49. 9. ^"243. 13. a/216 aV.
2. <m. 6. ^^. 10. '</U3. 14. a/64^^.
3. v/i2i. 7. a/64. 11. ■v^l44^. 15. ^^8^^^
9/T777^ o 10
125. 8. V81. 12. V27nV. 16. V625^.
253. Case II. TF/ie)2 the expression under the radical sign is
rational and integral, and has a factor which is a perfect power
of the same degree as the surd.
1. Reduce V54 to its simplest form.
We have, \/54 = v/27"x2 = v^ x ^ (§ 234) = 3^2,
2. Reduce V3 a^b  12 a^b^ + 12 ab^ to its simplest form.
V3 a^b  12 a262 + 12 aft^ = y (a2 _ 4 a& + 4 b'')S ah
= Va2  4 a6 + 4 62>/3a& = (a  2 6) VSoft.
We then have the following rule :
Resolve the expression under the radical sign into two factors,
the second of which contains no factor which is a perfect power of
the same degree as the surd.
Extract the required root of the first factor, and multiply the
result by the indicated root of the second.
If the expression under the radical sign has a numerical
factor which cannot be readily factored by inspection, it is
convenient to resolve it into its prime factors.
3. Reduce Vl944 to its simplest form.
4. Reduce Vl25 x 147 to its simplest form.
Vl25 X 147 = V68,x 3 X 72 = V52 x 72 x y/bxZ = 5 x 7 X Vl5 =; 35yi6.
224
ALGEBRA
^ EXERCISE 104
Reduce the following to their simplest forms :
1. V90. 5. ^56. a^l9^. 13. V242ar'/.
2. V72. ""6. 7VW. ^'^10. a/432. ^ 14. ^SOOa^ftV.
3. V96. 7. 9^/81. ' 11. ^256. 15. a/162 m^n^^
4. a/75. 8. a/48. 12. V500 a^6^ 16. ^160 ^fz\
17.
18.
23.
24.
25.
V75ar't/8100a;Y.
a/128 a^63+320 a%\
19. A/(3a; + 2?/)(9ar24 2/').
20. A/3a324a2+48a.
N
21. a/18 a^6 + 60 a^i^ + 50 a6^.
22. a/(2 «2 + a;  15)(2 a^  19 a; + 35).
V896.
a/2268.
26. a/98 X 196.
27. V432x504.
V5145.
28. V1372.
29. a/7875.
30. a/375 X 405.
^31. a/54 X 63x336.
32. V63 xf X 175 2/2' x 875 ^o^.
254. Case III. Wlien the expression under the radical sign
is a fraction.
In this case, we multiply both terms of the fraction by such an
expression as will make the denominator a perfect power of the
same degree as the surd, and then proceed as in § 253.
Ex. Reduce \/ —  to its simplest form.
^8a^
Multiplying both terms of the fraction by 2 a, we have
EXERCISE 105
Reduce the following to their simplest forms :
1. A/f 2. Vf. 3. a/^ 4. a/.
5. Vf.
SURDS 225
^7. ■^.;s'^^"12. v^Y
/ 23
* ^32a^*
25 6
3/T 13. V. ly ^/llaV 20. ^^^.
q 3/13 ' H , ,
^ '15 J^ \18 A^^ 21 ^C^^.
^22 a;^ / 3 a^  18 eg + 27
255. To Introduce the Coefficient of a Surd under the Radical
Sign.
The coefficient of a surd may be introduced under the
radical sign by raising it to the power denoted by the index.
Ex. Introduce the coefficient of 2V3 under the radical
^^^^' 2^3 = v^ X ^3 = VISITS (§ 234) = ^M.
A rational expression (§ 247) may be expressed in the form of a surd
of any degree by raising it to the power denoted by the index, and writ
ing the result under the corresponding radical sign.
i^ EXERCISE 106
Introduce the coefficients d^B^following under the radical
signs : ^jp,,
1. 3V7. 3. 4a/5. 5. 4v'5. 7. 2^/3.
2. QVQ. 4. 5^/7. 6. 2v'8. 8. 9xV2^.
9. lOa^ftVe^.
!*• (2»+i)Vj=E3
10. 6»/v'T?.
11. 5a»«\/2A
12. Za?V^/Zl^. '" « + l^a*4a + 3
11. 5 an?i/2 a'n. „ g1 / g^ + 3 a + 2
"• a + lVa^_4a
13. /„ .^./S+^. ifi g^/o (=^2)'
'• («^)\& !«• ^W^
(^1)'
226 ALGEBRA
ADDITION AND SUBTRACTION OF SURDS
256. Similar Surds are surds which do not differ at all, or
differ only in their coefficients ; as 2 V<xa.*^ and S^vax^
Dissimilar Surds are surds which are not similar.
257. To add or subtract similar surds (§ 256), add or sub
tract their coefficients, and multiply the result by their common
surd part.
1. Required the sum of V20 and ViS. *
Reducing each surd to its simplest form (§ 253),
V^ + V45 = V4^^ + V9ir5 = 2 VS + 3 V5 = 5 Vs.
2. Simplify VJ + VV.
=:iV2+lV6_§V^ = lV6lV2.
2 3 4 3 4
We then have the following rule :
Reduce each surd to its simplest form.
Add or subtract the similar surds, and indicate the addition or
subtraction of the dissimilar.
EXERCISE 107
Simplify the following :
^l. V8+V32. 3. V300V147. 5. a/T35^40.
2. V28V63. 4. a/2 j ^128. 6. ^80 + ^405.
7. V3 4V192V243. 8. V250  V90  VlTG.
9. V + V«^ ^ V^ + Vf. wll. ^1^^.
12. V99V275 4V396. 14. V^VV + VV.
Ms. a/56 4 v/189f a/162. 15. V+VV^.
16. V72^a;V98«3 4.a;V200«.
SURDS 227
17. a V80 a'b' + ab V270 a'b' + b' V640 a'b.
118. V27^M36»V + V48ic/ + 64^3.
21. ■v/l28 + ^250\/432^88.
^ 22. V50 a' + V72 6^ _ V50 a^ + 120 a6 + 72 d^.
^23. ^96 + ^486 a/6.
24. V294V216+V405V600.
^25. V52^2_5yjX7^_y'j26a^ + aV56a^.
26. VI + VIVSVS. 27. VIIVSVI+V?.
28. V50a3« + 40a;2_f_8^_y32aj3_48x2 + 18x.
y 29. V125 x^  150 a;?/ + 45 / + V5a;2 + 60 icy + ISO/.
^x — y ^x\y x^ — y^
TO REDUCE SURDS OF DIFFERENT DEGREES TO EQUIVA
LENT SURDS OF THE SAME DEGREE
258. Ex. Reduce V2, ^3, and ^5 to equivalent surds of
the same degree.
By § 237, V2 = 2^ = 2X2 = '^26 = ^/64.
We then have the following rule :
Express the surds with fractional exponents, reduce these to
their lowest common denominator, and express the resulting
expressions with radical sigyis.
^ / The relative magnitudes of surds may be determined by reducing them,
\yE^necessary, to equivalent surds of the same degree.
Thus, in the above example, \/l25 is greater than \/8l, and Vsl than
\ Then, y/b is greater than \/3, and Vz than V2.
228 ALGEBRA
EXERCISE 108
Reduce the following to equivalent surds of the same degree :
ca.
^1. V2 and V7. 6. </ah, Vbc, and V
v2. V3 and a/4. ^. V2^, ^3b, and v^Sc.
3. a/2 and ■^/3. 8. a/2^, "v/S/, and y/W?.
\ 4. v^S and a/5. 9. a/^ThI and ^^^^.
5. a/3 and \/6. 10. >Jx — y and \/aJ + 2/.
11. Which is the greater, VG or a/14?
^ 12. Which is the greater, a/2 or ^5 ?
13. Which is the greater, a/3 or a/7 ?
14. Arrange in order of magnitude a/2, a/13, and a/31.
15. Arrange in order of magnitude a/4, a/6, and a/15.
Arrange in order of magnitude a/2, a/3, and a/10,
MULTIPLICATION OF SURDS
259. 1. Multiply V6 by VlS.
By § 234, \/6 X \/l5 = VCxlS = \/2x3x3x6 = V32x2x5 =1 VlO. .
2. Multiply a/2^ by y/^K ^^0
Reducing to equivalent surds of the same degree (§ 258),
V2a X y/Tcfi = (2 a)^ x (4 a^)^ = (2 ay x (4 a^y = v^(2a)8 x v^(4 a'^Y
= v^28 a8 X 24 a* = v/26 a6 X 2 a = 2a\/Ui.
We then have the following rule:
Tb multiply together two or more surds, reduce them, if neces
sary, to syrds of the same degree.
^ultij^Jtog ether the expressions under the radical signs, and
write the result under the common radical sign.
The result should be reduced to its simplest form.
SURDS
229
3. Multiply V5 by y/B.
By § 237, V5 = 6^ = 6^ = y/bK
Then, V5xv'5 = v'Px^ = \^ = 5^ = 5^ = \^=v^.
4. Multiply2V3 + 3V2by3V3V2.
2V3 + 3V2
SVS V2
18 + 9\/6
2\/66
18 + 7V66 = 12 + 7V6.
To multiply a surd of the second degree by itself simply removes the
radical sign ; thus, \/3 x V3 = 3.
5. Multiply 3 VlT^  Wx by VI + a; + 2V^.
3vr+x4Vx
vT+^+2Vi
3(l + x)4VxT^
+ 6\/x + x28x
8(1 + x) + 2 Vx + x2 _8x = 36a; + 2 Vx + a;^.
EXERCISE 109
Multiply the following :
j 1. V5 by V20.
\?. ^/49¥2 by VTx.
3. Vl5 by V27.
4. Vi8 by V42.
10.
11.
12.
13.
v^5. VlOSby Vl92.ll^*^^^14.
\s. ^72 by ^81. ,>^ 3:^15.
16.
17.
18.
V?by Vf_
^/98 by a/343.
\/63by
7. \/55xy by V66 2/2.
8. VsEhyy/TE.
9. a/84 by a/ISO.
'' ' ^135.
V6by A^.
V3 »2/ by y/Tyz.
s/M by A/i2.
a/135 by ^45.
V20 by ^.
230 ALGEBRA
V 19. V5d' by ^125^. ^ 23. s/^, Vyi, and ^^.
20. v/Q by ^27. 24. V20,>25, and ^5.
21. VS by ^^,. 25. Vl, ^6, and J/6.
^ 22. V by ^S ^26. Vi5, ^^, and ^^.
x>27. 6 + 3V2 and 4 + 5V2.
28. 4Va3V6and7Va + 2V6.
^29. 2V58^/3and9V54V3.
30. 5V2 + 6V6andl0V27V6.
31. 2a/9 + 9^7 and 8^3^49.
32. 4V^V^ + 3\/^and4VicHVy3V^.
33. 3 Va + 2 H 4 Vci  1 and 6 Va + 2 + ^^a^^,
^34. V2+V5+V7and V2V5V7.
/35. 4V3VSand2V9V^.
36. 3V3 + 2V64V8and3V32V6 + 4V8.
37. 6V5  5V7VI0 and 6V5 + 5V7 4 ViO.
1^38. 8Vl2 + 7V204V24and5V33N/5 + 2V6.
,39. 6V + 8V + llVand3V4V5V.
DIVISION OF MONOMIAL SURDS
260. By §234, ■\/^ = VaxV5.
Whence, 2U^^^,
' n/—
We then have the following rule :
To divide tvjo monomial surds, reduce them, if necessary, to
surds of the same degree.
Divide the expression under the radical sign in the dividend by
the expression under the radical sign in the divisor, and write the
result under the common radical sign.
The result should be reduced to its simplest form.
SURDS
231
1. Divide 7405 by </5.
We have,
v^iOS
» Ft
^5 ^ 5
2. Divide v^ by V6.
Reducing to surds of the same degree (§ 258),
4^
^4
mi
VQ qI (2x3)^ </W^^^
3. Divide VlO by ^40.
/ 2^
'23x33
f=^=i^.
"We have.
Then,
VIo = 10^ = 10^ = v^Io3 = V(2>rEy.
^40 ^ 23 X 6
\
Divide the following :
1. V90 by V5. 3.
2. V24 by Vl8. 4.
7. </32 by \/2.
8. a/186 by •^.
V9. V7 by v'lQ.
10. ^42^ by ^56^.
a/686 by ^63.
V25^ by ^25^.
VS by Vf.
EXERCISE 110
VTO by V 63
^144 by </9.
11.
12.
13.
14.
15. </M by ^S
l^V by V5f
5. </3 by a/192.
6. a/48 by ^56.
16. V6ab' by ^96 ftc^,
\/3a^ by V2a.
\7
18.
19.
v^20.
21.
22.
23.
V27^ by ^36^.
V by 7.
<m by V.
\/12V5^ by a/4^^.
by ^.
_. by ^.
24. a/15^2 by ^/ioS^.
INVOLUTION OF SURDS
261. 1. Kaise ■V^12 to the third power.
(%^)« =(12^)8 = 12^ (§ 243) = 12^ =\/i2 = 2V3.
232 ALGEBRA
2. Raise V2 to the fourth power.
( ^2)* = (2^)4 = 2^ = v^ = v^.
Then, to raise a surd to any positive integral power,
I If possible, divide the index of the surd by the exponent of the
required power ; otherwise, raise the expression under the radical
sign to the required power.
The rules of § § 97 and 98 should be used to find the value of
any product which comes under them.
3. Expand ( V6  VS)^.
By §97, (V6\/3)2 = (\/6)22\/6x V3+(V3)2
= 6  2\/PV2 + 3 = 9 Q^/2.
4. Expand (4 + ^) (4^).
By § 98, (4+ v^)(4 Vb) =42 (^)2=16 ^, by the above rule.
%^_^\ EXERCISE III
Find the values of the following : *
1. (^/2)*. 6. (5^5«^)2. vll. (^a^^/2E^y,^^
/2. (5^6)2. 7. (•v/^'^)^ 12. (4^^729)3.
3. (^4a;f 3 2/)l ^8. (^72^^)^ 13. (7 + 2V2)2.
4. (^32)1 Z : 9. {VU^f. V14. (4V55)2.
.b. (V2^^y. 10. {y/ly. Vl5. (3V6 + 6V3)^
16. (9V74VIl)l ^18. (4V^^:^ + 3V^T6)'.
17. (V5a; + 2V3^)2. Vl9. (6 45V2)(65V2).
v^20. (4V^ + 3V^^^)(4Va3V^:^^).
'/ 21. (V2a; + 2/ + V2a;2/)(V2a; + 2/V2a;y).
22. (5V3ic44: + 4V5a;2)(5V3a; + 44V5a;2).
^23. ^4 + 2V3x^42V3. ^ 24. (^ + ^9)(v^4\/9).
> 25. V3V5 + 2V7 X V3V52V7.
26. Expand (2 V2 + V6  V3)2, by the rule of § 204,,
SURDS 233
EVOLUTION OF SURDS
262. 1. Extract the cube root of \/27a^.
\/( v/27x8) = ( \/(3^P = [(3 x)^]* = (3 cc)^ = \/3«.
2. Extract the fifth root of V6.
V{ \/6) = (6*)^ = 6^^ = v^.
Then, to extract any root of a surd,
\ If possible, extract the ^^equired root of the expression under the
radical sign ; otherwise, multiply the index of the surd by the index
of the required root.
If the surd has a coefl&cient which is not a perfect power of the degree
denoted by the index of the required root, it should be introduced under
the radical sign (§ 255) before applying the rule.
Thus, ' v/(4V2) = ^(V32) = \/2.
\^,_,^S_ EXERCISE 112
Find the values of the following : ^ a, 'iT .
1. V(>^y ^/5. V(^/9a2+12a + 4). 9. Vil^a's/^).
2. V(^i3). v6. a/(a/49). ^7 ^10. V(2xV^y
3. ^{VW^y i 7. a/(81^I6). 3 a 11. </(^3^).
v4. ^/(■</24S^), 8. ^/(2^/3a^). 12. 's/(2n'Vie^').
\
REDUCTION OF A FRACTION WHOSE DENOMINATOR IS
IRRATIONAL (§ 247) TO AN EQUIVALENT FRACTION
HAVING A RATIONAL DENOMINATOR
263. Case I. When the denominator is a monomial.
The reduction may be effected by multiplying both terms of
the fraction by a surd of the same degB*e as the denominator,
having under its radical sign such an expression as will make \
the denominator of the resulting fraction rational.
234 ALGEBRA
Ex. Keduce — == to an equivalent fraction having a
rational denominator.
Multiplying both terms by Vq a, we have
5 ^ 6</9a _ 5</9a ^ 5v^9g .
EXERCISE 113
Keduce each, of the following to an equivalent fraction hav
ing a rational denominator ;
1. A. 3. ^. 5. ^  ^
VE' </6a' V25 V27
2 2 . 4 4=. 6. 4^. 8. ^ .
264. Case II. Whe^i the denominator is a binomial contain
ing only surds of the second degree.
1. Eeduce ~ _ to an equivalent fraction having a
rational denominator.
Multiplying both terms by 5 — V2, we have
5_V2 ^ (5V2)2 ^ 2510V2 + 2 ..gg^ ggN^ ^TlOV^
5 + V2 (6+V2)(6V2) 252 ' ^ 23 '
2. Reduce ^^ V q— ^^ ^^^ equivalent fraction having
2Va — 3Va— 6
a rational denominator.
Multiplying both terms by 2\/a + 3 Va — b,
3Va  2Va  6 ^ (3v^  2 V^"^r6)(2v^ + SV^i^^)
2y/a  SVa  b (2Va  3V^"I^)(2Va + 3Va^^)
6 q + SVa Vo"^^  6(a  6) _ 6 & + SVa^  a6
4a9(a6) 9b 5a
SURDS 235
We then have the following rule :
Multiply both terms of the fraction by the denominator with the
sign between its terms reversed.
EXERCISE 114
Keduce each of the following to an equivalent fraction hav
ing a rational denominator :
^ 8 ^ " V^+ Vy . i^ 7^ 3V5V3
V6 + 2 * V^V^ * 4V5 + 5V3
2 7  5 V106V2 ^ g 3Va^
53V2 ■ * VIO + 2V2 * 4Va^ir3
Vti . ^g^ 2V7 + 3V3 , ,Q Va;Va; + y
m + Vn 2V7 — 3V3 VaJ + Vaj + y
jQ V9a^23a 13. ?_
V9a^2 + 3a
11. Va^ + / + y .
14.
VVll + 3VVll
3
Va^ + 2/2 + Var^
f
Vic2 + 2/2Va:2
f
Vafl42Va
1
2Va;2Va; + 2 * 4 Vo + l + 3 Va^=3
265. If the denominator is a trinomial, containing only surds
of the second degree, the fraction may be reduced to an equiva
lent fraction having a rational denominator by two applications
of the rule of § 264.
Ex. Reduce —^ — _~ to an equivalent fraction having
4 + V3V7
a rational denominator.
Multiplying both terms by 4 + \/3 + \/7, we have
4V3.^V7 ^ (4\/8\/7)(4 + V3 + V7) ^ 42(V3 + V7)2 .^qq.
4 + V3\/7 (4 + V3V7)(4 + \/3 4V7) (4 + V3)2 (V7)2
236 ALGEBRA
Then 4  VS  \/7 ^ 16  (10 + 2 \/2l) ^ 6  2\/21 ^ 3  V21
4 + V3V7 19 + 8\/37 12 + 8\/3 6 + 4V3
Multiplying both terms of the latter by 6 — 4\/3,
4 _ V3 _ V7 ^ (3  V2T) (6  4 V3)
44.V3V7 62(4V3)2
_ 18  6 V2I  12\/3 + 4 V63 ^  9 + 3V2l + 6\/3  6V7
 12 6 '
The example may also be solved by multiplying both terms of the given
fraction by 4  \/3 + V7, or by 4  >/3  VT^.
EXERCISE 115
Eeduce each of the following to an equivalent fraction having
a rational denominator :
1. I 3. 12
2+V2 + V3 V5_V3V2
2 6 4 V6+V33V2
3+V5V2* . * .V6V§ + 3V2'
The reduction of a fraction having an irrational denominator to an
equivalent fraction having a rational denominator, when the denominator
is the sum of a rational expression and a surd of the nth degree, or of two
surds of the nth degree, will be found in § 446.
266. The approximate value of a fraction whose denominator
is irrational may be conveniently found by reducing it to an
equivalent fraction with a rational denominator.
Ex. Find the approximate value of to three places
of decimals. 2 — V2
1 ^ 2+'\/2 ^ 2+V2 _ 2 + 1.414... ^ J ^^^ ^^^
2_V2 (2V2)(2+\/2) 42 2
EXERCISE 116
Find the values of the following to three places of decimals :
^ 1. A. . 2. ^ « 1
V6 3V3 5 + 2V7
a , , SURDS 237
4. _A_. ^6.^ V7V2 g 4V55V3
■^•v/49 ' V7 + V2 ' 4V5 + 5V3*
/5 23 ,7 2V6 + V3 g 4 V7 + 7 V3
_<^ ' V53V2' * 2V6V3' ' 3V75V3*
/^ PROPERTIES OF QUAliliATIC SURDS (§250)
267. A quadratic surd cannot equal the sum of a rational
expression and a quadratic s^ird.
For, if possible, let Va = 6 + Vc,
where 6 is a rational expression, and Va and Vc quadratic
surds.
Squaring both members, a — ¥ + 2h^c\c,
or, 2 & Vc =ia — h'^ — c.
Whence, Vc =
g — 0^ — c
26
That is, a quadratic surd "equal to a rational expression.
But this is impossible ; whence, Va cannot equal h + Vc.
268. i/" a + V6 = c + V<^, where a and c are rational ex
pressions, and V& and 's/d quadratic surds, then
a = c, and V& = Vc?.
If a does not equal c, let a=:c}x; then, x is rational.
Substituting this value in the given equation,
c\x + V6 = c + Vd, or ic + Vb = V^.
But this is impossible by § 267.
Then, a = c, and therefore V& = V^.
269. If V a + V6 = V^ + Vy, where a, b, x, and y are rational
expressions, then Va — V^ = Va; — V^.
Squaring both members of the given equation,
a + V6 = a; + 2 VaJ2/ + y.
238 ALGEBRA
Whence, by § 268, a:=x\'yf
and V6 = 2^/xy.
Subtracting, a — V6 = x — 2^'xy + y.
Extracting the square root of both members,
V a _ V6 = Vaj — V^.
270. Square Root of a Binomial Surd.
The preceding principles may be used to find the square
roots of certain expressions which are in the form of the sum
of a rational expression and a quadratic surd.
Ex. Eind the square root of 13 — V160.
Assume, ^13  VIM = y/xVy. (1)
Then, by § 269, ^\^+V)M = Vx + Vy. (2)
Multiply (1) by (2), V169  160 = x~y. (§ 98)
Or, xy = Z. • (3)
Squaring (1), 13  \/l60 = a:  2 Vxy + y.
Whence, by § 268, x + y = 12>. (4)
Adding (3) and (4), 2 x = 16, or x = 8.
Subtracting (3) from (4), 2 y = 10, or y = 6.
Substitute in (1) , ^IZy/lim =V% Vb = 2^/2 y/l.
271. Examples like that of § 270 may be solved by inspec
tion, by putting the given expression into the form of a tri
nomial perfect square (§ 111), as follows :
Reduce the surd term so that its coefficient may be 2.
Separate the rational term into two parts whose product shall he
the expression under the radical sign of the surd term.
Extract the square root of each part, and connect the results by
the sign of the surd term (§ 112).
1. Extract the square root of 8 + V48.
We have, • \/48 = 2\/l2.
SURDS 239
We then separate 8 into two parts whose product is 12.
The parts are 6 and 2 ; whence,
V8 + V48 = V6 + 2 Vl2 + 2 = V6 + V2.
2. Extract the square root of 22  3 V32.
We have, 3\/32 = V9 x 8 x 4 = 2V72.
We then separate 22 into two parts whose product is 72.
The parts are 18 and 4 ; whence,
V22Z3V32 = V 18  2 V72 + 4 = \/l8  V4 = 3 V2  2.
Vh ^^
%0 EXERCISE 117
Find the square roots of the following :
^1. 15 + 2V54. 7. 30V500. 13. 455V80.
2. 212V80. 8. 13 + VI68. 14. 34 412V8.
3. 532V52. 9. 24 + 2ViiO. 15. 61 + 28V3.
>4. 23 + 6VIO. 10. 444V72. 16. 53V600.
>5. 3810V13. 11. 5520V6. 17. 605VI08.
>6. 29 + 2V54. 12. 55 + 3V24. 18. 54h3Vi28.
19. 4a2V4a29. 20. 4.(2xy)\2Vl5ay^12xy.
Solution of Equations having the Unknown Numbers under
Radical Signs.
272. 1. Solve the equation Vo^ — 5 — a; = — 1.
Transposing — cc, Vx^ — 5 = x — 1.
Squaring both members, x^ — 6 = x^ — 2x + 1.
Transposing, 2x = 6; whence, x = S.
(Substituting 3 for x in the given first member, and taking the positive
value of the square root, the first member becomes
Vo^Ts _3 = 23 = l;
which shows that the solution oj = 3 is correct.)
240 ALGEBRA
We then have the following rule :
Transpose the terms of the equation so that a surd term may
stand alone in one member; then raise both members to a power
of the same degree as the surd.
If surd terms still remain, repeat the operation.
The equation should be simplified as much as possible before perform
ing the involution.
2. Solve the equation V2a; — lf V2~a+~6 = 7.
Transposing V2a; — 1, V2 cc + 6 = 7 — \/2 x  1.
Squaring, 2a; + 6 = 4914V'2a;l+2xl.
Transposing, 14 \/2 x — 1 = 42, or v 2 x — 1 = 3.
Squaring, 2 x — 1 = 9 ; whence, cc = 5.
1
3. Solve the equation VaJ — 2 — Va; =
Va?2
Clearing of fractions, x — 2— Vic^ — 2x = l.
Transposing, — Va;"^ — 2 a; = 3 — x.
Squaring, x^ — 2 cc = 9 — 6 x + a;2.
9
Transposing, • 4 x = 9, and x = .
9 4
(If we put X = , the given equation becomes
If we take the positive value of each square root, the above is not a true
equation.
But a square root may be taken as either positive or negative ; and if
we take the negative value of v  , and the positive value of v/  , the first
1 3
member of (1) becomes , or —2, and the second member becomes
1 ^ ^ 9
— or — 2 ; then the 3olution x =  is correct.)
4. Solve the equation V2 — 3a; + Vl44a; = V34a;.
SURDS 241
Squaring both members,
23x + 2V23xVl + 4x + l + 4a; = 3 + x.
Whence, 2 V2 SxVl + 4x = ;
or, * V23x\/l +4x = 0.
Squaring, (2  3x)(l + 4a;) = 0.
2
Solving as in § 126, 23x = 0, orx = ;
and H4x = 0, or x = .
EXERCISE 118
Solve the following equations :
^1. V4ic + l+5 = 0. V5. V« + Va; + 9 = 2.
2. \/7a;82 = 6. 6. V372V3T=l.
3. ■Vl6x'\l4.x = 3. 7. V^Tl3Va;5 = 3.
v4. VSa^{36x'3 = 2x. V 8. V5a;19 V5a;+14=3.
yo ^ ^^ =V32fl;.
Va; + 4 V32a;
^ 10. , =: = :' .11. ViC— 5 + Va; = =:•
■VSx\2V3x 4 ^ ^^
Vl2. V6a;V6xll =
V6a;ll
10
13. V2sV2s + 5 = 
V2S45
4/14. Vx^ Va? + 21 = 2 Vx. "^cu^^
3VTT2^ + 4 VH2a^ + 6
V 15.
6Vl42a;l 2VH2
X
3V^_+4^3V^+_5^ ^^^ ■y/2axh^2a + x _^
5V^2 5V^3 ' V2aa;V2aHa;
242 ALGEBRA
18. V2ri — ic — Vn — fl7 = 3Vw.
J119. Va^f7a;4 + Va^3a; + l = 5. V/.>v r.f. ^—
20. Va;2a4V^= ^
Va; — 2 a
\/21. Va; 4 a + V a; + 2 a = V4 a; + 5 a.
22. Vaa;+V6a; = V2a + 26.
23. Va; 4 V5  2 a; = V5 — a;.
24. V4a;3V3xl=V7a;4.
25. V4i) + 1  Vp  8 = V9i?  83.
26. Va; — 2 a — Va; — 6 a = 2 Va; — 5 a.
27. V(3 a + ^/Saxj x^=VxVSa.
28. V»+V4a4aj=V4644a?.
29. V2 aa; + 6 + V2 aa;  6 = 2 V2 ax  3 &.
'30. Va — a; f V6 — a; = Va + 6 — 2 a;.
31. V2a;45a + V3a; + 4& = V5aj45a + 4 6.
32. V2a;l+V3a: + 2=V3a;2V2a; + 3.
33. V2a;45 + V3a52=V(5a; + 3+V24a;2_^15).
IMAGINARY NUMBERS O/^t/ vmK /'^^^^
273. It is impossible to find an even root of a negative
number; for no number when raised to an even power can
produce a negative result (§ 96).
^/■A' An Imaginary Number is an indicated even root of a negative
number; as V— 2, or ^—3.
In contradistinction, rational and irrational numbers (§ 248)
are called reed numbers.
274. An imaginary number of the form V— a is called a
► piire imaginary number, and an expression of the form
a + 6V— 1 a complex number.
SURDS 243
275. Meaning of a Pure Imaginary Number.
If Va is real (§ 273), we define Va as an expression such
that, when raised to the second power, the result is a (§ 206).
To find what meaning to attach to a pure imaginary number,
we assume the above principle to hold when V« is imaginary.
Thus, V— 2 means an expression such that, when raised to
the second power, the result is — 2 ; that is, (V— 2)^ = — 2.
In like manner, (V— 1)^ = — 1 ; etc.
OPERATIONS WITH IMAGINARY NUMBERS
276. By §275, (V^=5. (1)
Also, (V5V^' = (V5)2(V^2^5(l)=5. (2)
From (1) and (2), ( V^' = (V5 V^'.
Whence, V— 5=V5V— 1.
Then, every imaginary square root can he expressed as the
product of a real number by V— 1.,
V— 1 is called the imaginary unit; it is usually represented by i.
277. Addition and Subtraction of Imaginary Numbers.
Pure imaginary numbers may be added and subtra<;ted in
the same manner as surds.
1. Add V  4 and V^^^.
By §276, yr^ + V^^^ = 2V^l + 6^
2. Subtract 3  V9 from 1 + V 16.
In adding or subtracting complex numbers, we assume that the rules
for adding or subtracting real numbers may be applied without change.
Then, I + V^^iq _ (^^  \/~^) = 1 + ^V^  3^+ ^y/^^
EXERCISE 119
Simplify the following :
1. V^ + V^^^25. i 2. V^r5+V45.
244
3. V^r27_V12.
ALGEBRA , ^.^ ""
i iJ> ll
6. V64 + V100 + V121.
7. 2V165V49 + 3V81. ,
XT'
4. V(a; + l)'Var^.
W i
10. Va'2alVa2 + 2al.
11. Add 5 + V^^ to 3 + V^^16. T r t'lft
12. Add 6V^=^ to 1V^49.
13. Subtract 2 + V^^ from 8V^25.
14. Subtract 4V81 from 7+V^^^.
278. Positive Integral Powers of V— 1.
By §275, {V^y^l,
Then,
(V3i)3=(V3T)2x V^ri =(_i)xV3T = _V3ij
(V=T)^=(V^)^x(V:rT)2=(_l)x(_i) = i;
(V3i)«=(V3I)4x V^Ti = 1 xV^=V^::i; etc.
Thu s, the first four positive integral powers of V— 1 are
V— 1, —1, — V— 1, and 1; and for higher powers these
terms recur in the same order.
279. Multiplication of Imaginary Numbers.
The product of two or more imaginary square roots can be
obtained by aid of the principles of §§ 276 and 278.
1. Multiply V"^ by V"^.
By §276, V^r2xvCr3 = V2V:rTxV3v^ZT
= V2 \/3( V'^y = V6(  1) (§ 278) =  V6.
SURDS
^
'% ''
245
2. Multiply together V9, V16, and V25.
= 60(\/^n:)3 = 60(V"^) (§278) =60 v/^^.
3. Multiply 2 + 5V^=^ by 43V^^.
In multiplying complex numbers, we assume that the rules for multi
plying real numbers may be applied without change.
2 4 5V5
4_ zv^rs
8 + 20V5
15(5)
8 + 14V 5 + 75 = 83 + 14 V^.
4. Expand (V^^ + 2V^^)' by the rule of § 97.
( V35 + 2\/^)2 = ( V^2 + 4\/5 V^Tx VSV^^ + 4(\/^2
=  6 + 4 vT5(\/^=T)2 + 4( 3) =  5  4 Vl5  12 =  17  4 Vl5.
EXERCISE 120
Multiply the following
^1. V^3 by V^=^.
3.
36 by V25.
5. V14 by ^56.
6. _V^=^147 by V^^^45.
V81a^ by V 121x2. 7^ 5_.4V1 by 23V"=^.
4. _V3i5 by V^^. 8. 6 + V^3 by 7 + 4V^3.
9. 3V^^2V^ by 9V^^ + 6V^.
10. SVTTV^^^ by V^^5V^^.
V^^^ V^^TF, and V^=^9?.
V— 6,
11.
12.
2, V462, and
V^^27, and
V54.
13. V27+V^^^ by V"3V^^.
Sl4. 2V^^V^6 by V^ni + 4V^.
15. V^^^^, V^=49, V^=^, and V^^^IOO.
/16. V2, V3, V^6, and V10.
246
ALGEBRA
Expand the following by the rules of §§ 97, 98:
•17. (5lV^l
^19. (4V^^ + 3V^)2.
^^23. (4V^=^ + 5
24. (8V^=^ + 3
20. (3V52V^^y.
21. (7+2V^(72V^.
22. (Va4?>)(Va6).
— i/)(4v — ic — 5v^— y).
/ 25. (3V^=34V3)2 + (3V=^V3)2.
Expand the following by the rules of § 205 :
^26. (1V^^ 27. (2 + V=^)».
28. Expand (3 V^^  V^  2 V^3)2 by the rule of § 204.
280. Division of Imaginary Numbers.
1. Divide V40 by V5.
\/^r40 Vio V^^ V40
By § 276,
5
V5 v^n
2. Divide Vl5 by V"^.
Vl5 ^ _vT5(l) ^ \/l5(V^
V5
D^
= VS = 2 \/2.
278) =  V5 ^^=3 =  \/^.
3. Reduce — to an equivalent fraction having a
real denominator.
We multiply both terms of the fraction by the denominator with the
sign between its terms reversed ; multiplying both terms by V3 — V— 2,
V3
^= C^^^)^ (§98)
V3 + Vr2 (V3)2_(V3r2)2
_ (V3)2_2V3 V^=^+(
3(2)
_ 32\/^2 _ l2V^r6
3 + 2 6
2}f
(§97)
SURDS
247
EXERCISE 121
Divide the following :
^1. v=n:5 by v^^.
^. V48 by V^.
^3. V=^ by V8.
7. V343V63 by
4. —V—Qxy by —^2yz.
^15. Vl80 by V^n^.
6. Vi32 by V^=^.
"^^. V288V300 by V6.
Eeduce each_of_ilie following to an equivalent fraction hav
ing a real denominator :
3V32
9.
V
10.
1V
3V
3+V^^
11.
12.
3V3+2V6
2V^^47V^3
4V^=^3V^^
281. The complex numbers a + ftV— 1 and a — 6 V— 1 are
called Conjugate.
We have (a{b V"^) + (a  6 V"^^) = 2 a.
Also, (a + b V"^) X (a  6 V^^)
= a^  &2<( V^^)2 = a2 _^ 62 (§ 275).
Hence, the sum and product of two conjugate complex numbers
are real.
248 ALGEBRA
XIX. QUADRATIC EQUATIONS
I. A Quadratic Equation is an equation of the second degree
(§ 83), with one or more unknown numbers.
A Pure Quadratic Equation is a quadratic equation involving
only the square of the unknown number ; as, 2 a^ = 5.
An Affected Quadratic Equation is a quadratic equation involv
ing both the square and the first power of the unknown number ;
as, 2a^3aj5 = 0.
In § 126, we showed how to solve quadratic equations of the forms
ax2 \bx = 0, ax^ + c = 0, x^\ax + b = 0, and ax^\bx + c = 0,
when the first members could be resolved into factors.
PURE QUADRATIC EQUATIONS
283. Let it be required to solve the equation
a^ = 4
Taking the square root of each member, we have
±«=±2;
for the square root of a number may be either f or — (§ 208).
But the equations —x = 2 and — x= — 2 are the same as
x = — 2 and x = 2, respectively, with all signs changed.
We then get all the values of x by equating the positive square
root of the lirst member to ± the square root of the second.
284. A pure quadratic equation may be solved by reducing
it, if necessary, to the form a^ = a, and then equating jc to ±
the square root of a (§ 283).
1. Solve the equation 3 a;^ _^ 7 ^ ^ _^ 35^
Clearing of fractions, 12 a;^ + 28 = 5 tc* 4. 140.
Transposing, and uniting terms, .7 x^ = 112, or x^ = 16.
Equating a; to ± the square root of 16, x = ± 4.
QUADRATIC EQUATIONS 249
2. Solve the equation 7 x^ — 5 = 5 oiy^ — IS.
Transposing, and uniting terms, 2 a:2 — _ 8, or x^ = — 4.
Equating x to ± the square root of — 4, x = ± V— 4
= ±2\/:ri(§276).
In this case, both values of x are imaginary (§ 274) ; it is impossible to
find a real value of x which will satisfy the given equation.
In solving fractional quadratic equations, any solution which does not
satisfy the given equation must be rejected.
Thus, let it be required to solve the equation
x27 1 1
x^ + x2 x + 2 x1
Multiplying both members by (x + 2) (a; — 1), or x^ + jc — 2,
x^7 = xlx2, or x2 = 4.
Extracting square roots, x = ±2.
The solution x = — 2 does not satisfy by the given equation ; the only
solution is X = 2.
EXERCISE 122
Solve the following equations :
1. 2a;2 + 27 = 7a^53. 2. A_15 = _?.
4.af 8a^ 3
3. 5(2x3)\2x(4.x\l)=12x7,
V 4. 2(3x5y + 3(x\ 10)2 ^ 434^
^ ' 3 4.x 9 4 a;* j^^
6. 6V5ar^9 = 12. V?^
9. (2 a; + 7) (5 a; 6) 24 a; = (4 a; 3) (7 a; + 5) 59.
10 4^2 + 3 8f2l ^ 1 jj 3a x + 5h ^^
7 2 14* * a;56 3a + 105
250 ALGEBRA
\J12 3a^ + 7 5a^ 43 ^ 40)^10
7 14 35 *
13. {x{a){x^b) + {xa){xb)=x' + a^{ly^.
14. 3V^Tl+V3a.' + 7 = l.
15.
lOa^3 5x^i6 6x^1
18 9 9x^2 '
16. (k + l)(k2)(kS)(kl)(k + 2)(k\3) = ^52.
^17. 2xV^^TS2xy^^+2 = l.
18 3^!zi^_if^!±3^2aj* + 12_^
0,^ + 5 a^_5 a;4_25
3a^l ^ a;^ + 3 ^q a
2aj^5a;2 + l 2x^5 ' ' x2 x + 3
19 a;^ + 3a^l ^ a;^ + 3 ^q x'x^2 x'j^x^ ^j^
21.
Va2 + ax + a^4Va2aa; + a;2 = a(l + V3). (^^
22 J^ ^ = _^!:zlL_. Z.^''
• x + 3 x5 x'2x^l5 /C^f ^.,
(^ ^"^^
AFFECTED QUADRATIC EQUATIONS > ^f / Z'
285. First Method of Completing the Square. V* / ^
By transposing the terms involving x to the first member, and
all other terms to the second, and then dividing both members
by the coefficient of x^, any affected quadratic equation can be
reduced to the form x^{px = q.
We then add to both members such an expression as will
make the first member a trinomial perfect square (§ 111) ; an
operation which is termed completing the square.
Ex. Solve the equation x^ + Sx = 4:.
A trinomial is a perfect square when its first and third terms
are perfect squares and positive, and its second term plus or
minus twice the product of their square roots (§ 111).
Then, the square root of the third term is equal to the second
term divided by twice the square root of the first.
QUADRATIC EQUATIONS 251
Hence, the square root of the expression which must be added
to x^ \^x to make it a perfect square is Sxi2x, or f.
Adding to both members the square of f , we have
Equating the square root of the first member to ± the square
root of the second (compare § 283), we have
Transposing f, « = — fffor— f —  = lor— 4.
We then have the following rule :
Reduce the equation to tJie form oi? + px — q.
Complete the sq^iare, by adding to both members the square of
onehalf the coefficient of x.
Equate the square root of the first member to ± the square root
of the second, and solve the linear equations thus formed.
286. 1. Solve the equation 3 cc^ — 8 aj = — 4.
Dividing by 3, a;2_^^_,
which is in the form x^ +px = q.
4
Adding to both members the square of , we have
o
¥(ir=if=t'
Equating the square root of the first member to ± the square root
Transposing , x = ± = 2or.
If the coefficient of oc^ is negative, the sign of each term must
be changed.
2. Solve the equation — 9 a^ — 21 a; = 10.
Dividing by  9, x2 + ^ = ^
252 ALGEBRA
7
Adding to both members the square of ,
6
^+'^+(l)'=f+*i='
3 \6 9 36 36
7 3
Extracting square roots, x +  = ±'
6 6
Then. . = _Z± = _„._.
EXERCISE 123
Solve the following equations :
1. x^\Sx = 9. ' 8. 3a^ + Sx = 4:,
2. x^4.x = 60. /9. 2a^ + 9ic = 4.
3. x'dx^U. . 10. 4a^+15a;25 = 0.
4. x'+llx = 24., 11. 9ar^ + 14 = 27a;.
5. a^a; = 20. 12. 18 = 4a?2 _^21x.
6. 3x210x = 3. 13. 127a;10a;2 = 0.
7. 5a^ + 3a; = 8. 14. 19^ = 12^5.
287. If the coefficient of a^ is a perfect square, it is con
venient to complete the square directly by the principle stated
in § 285 ; that is, by adding to both members the square of the
quotient obtained by dividing the coefficient of x by twice the square
root of the coefficient of x^.
1 . Solve the equation 9 a^ — 5 a? = 4.
Adding to both members the square of —  — , or ,
^ ^ 2x3' 6'
\Q) 36
169
36
C 1 O
Extracting square roots, 3 x —  = i — •
6 6
Then, 3cc=±— =3 or , and x=l or .
6 6 3 9
QUADRATIC EQUATIONS 253
If the coefficient of ^ is not a perfect square, it may be made
so by multiplication.
2. Solve the equation 8 ar^ — 15 a; = 2.
Multiplying each term by 2, 16 aj^ — 30 a; = 4.
Adding to both members the square of , or — ,
2x4 4
16.^30..(^y = 4.f = f,
15 17
Extracting square roots, 4 x —  — ±—.
4 4
Then, 4a; = — ± — = 8 or 1, and a; = 2 or i.
4 4 2 4
If the coefficient of x^ is negative, the sign of each term must be
changed.
EXERCISE 124
Solve the following equations :
1. 4a;27x = 3. 8. 36 a;^  36 or = 7.
2. 9a^ + 22a; = 8. 9. Vlx^^x=\.
3. 16a^8a; = 35. 10. 49 ^^ .f 49 /i + 10 = 0.
4. '^y?^\^x = X 11. 64ar^ + 15 = 64a;.
5. 3a;28a; = 3. /12. 12 = 23e5e2.
6. 18a^5a; = 2. 13. 28a;32x2_3^0.
7. 25a;2 + 15a; = 4. 14. 25 a; = 50 a;^ 2.
288. Second Method of Completing the Square.
Every affected quadratic equation can be reduced to the form
a'31? + 6a; 4 c = 0, or aa^ ( 6a; = — c.
Multiplying both members by 4 a, we have
4 aV H 4 abx = — 4 ac.
We complete the square by adding to both members the square
of ^^^^(§287), or 6.
254 ALGEBRA
Then, 4 o V + 4 ahx + 52 ^ &2 _ 4 ^^^
Extracting square roots, 2ax\'b = ± Vb^ — 4 ac.
Transposing, 2ax = —b± Vb^ — 4 ac.
„7., —b± V6 — 4 ac
Whence, x = =^^— — •
2a
We then have the following rule :
Reduce the equation to the form ax^ {■bx = — c.
Multiply both members by four times the coefficient of 7?, and
add to each the square of the coefficient of x in the given equation.
The advantage of this method over the preceding is in
avoiding fractions in conipleting the square, v
1. Solve the equation 23i^ — 7x = — 3,
Multiplying both members by 4 x 2, or 8,
16x^66x = 2i.
Adding to both members the square of 7,
16x2  56x + 7^ =  24 + 49 = 25.
Extracting square roots, 4 x — 7 = ± 5.
Then, 4 x = 7 ± 5 = 12 or 2, and x = 3 or i.
2'
If the coefficient of x in the given equation is even, fractions
may be avoided, and the rule modified, as follows :
Multiply both members by the coefficient of x^, and add to each
the square of half the coefficient of x in the given equatio7i.
2. Solve the equation 15 a^^ __ 28 a; = 32.
Multiplying both members by 15, and adding to each the square of 14,
152x2 + 15 (28 X) + 142 = 480 + 196 = 676.
Extracting square roots, 15x + 14 = ± 26.
Then, 15 x =  14 ± 26 = 12 or  40, and x =  or  §•
5 3
The method of completing the square exemplified in the present section
is called the Hindoo Method.
QUADRATIC EQUATIONS 255
EXERCISE 125
Solve the following (equations :
^ 1. x\7x=18. 9. 12x'llx = 2.
2. 3x'2x = 4:0. 10. 6x'13x = 6.
3. 4.x'Sx = 10. 11. 2r215r + 25 = 0.
4. 4.x^Sx = 4.5. 12. 15a:2 426aj + 7 = 0.
5. 8a;2 + 2a; = 3. 13. 5a;2+ 48 = 32a;.
^6. 9a^ + 18a; = 8. 14. 13a;== lOcc^S.
7. 9aj2 + 4a; = 5. 15. 3 = 6 a;^ + 17 a;.
^>i8. 7g2 + 20g = 12. 16. 27i»9 8^^2 = 0.
289. Solution of Affected Quadratic Equations by Formula.
It follows from § 288 that, if ax^ \bx\c = 0, S^^w^mO
then ^^ 6±V64a c. ^t*^.,^^ (1)
Z ft
This result may be used as a formula for the solution of any
affected quadratic equation in the form aaf \bx + c = 0.
1. Solve the equation 2x^\5x — lS = 0.
Here, .a = 2, 6 = 5, and c = — 18 ; substituting in (1),
_5j:V25 + 144 ^ 5 ±13 ^^ ^^ 9
4 4 2
2. Solve the equation — 5 o?^ + 14 a; + 3 = 0.
Here, a = — 5, 6 = 14, c = 3 ; substituting in (1),
 14 j V196 + 60 ^  14 zfc 16 ^ 1
10 ~ 10 5
3. Solve the equation 110 a^  21 a; =  1.
Here, a = 110, & =  21, c = 1 ; then,
21 i V441  440 21 ± 1 1 «^ 1
a; =: — £= = = — or — •
220 220 10 11 •
Particular attention must be paid to the signs of the coefficients in
making the substitution.
256 ALGEBRA
EXERCISE 126
Solve the following by formula :
1. aj2_i2a;_f32 = 0. : 8. 40 17 a;5a^ = 0.
2. a^ + 7a;30 = 0. ^j'^ 9. 36p'\S6p = 5.
3. 2a;23a;20 = 0. H^ .^ 10. SOic^ + l = 17a;.
4. 3x^x4. = 0. I / *'ll. 19a; = 8aj2_^6.
5. 4a!2_5^_2i = o^ 3^.^ 12. 15 + 22aj 48x2 = 0.
6. 20a^ + ajl = 0. ^^. . <^ 13. l^x' + 2Qx^S.
7. 9a.218a; + 8 = 0. v^ ^ 14. 37a; = 6a^ + 6.
3 ' J
EXERCISE 127
The following miscellaneous equations may be solved by either of the
preceding methods, preference being given to the one best adapted to the
example under consideration.
In solving any fractional equation, we reject any solution which does
not satisfy the given equation. (Compare last example, § 284.)
10 3 ^ 13 ^ 1
3* '6a; ^x^ 18*
5 . A + i^ = _15
12a;* * U 3 6*
5. (3a; + 2)(2a;3) = (4a;l)214.
6. a;(5a; + 22)+35 = (2a; + 5)l
7. (a; + 4)(2a;l) + (2a;l)(3a; + 2) = (3a; + 2)(4a;l)49.
8. ^1:^^ = 1. 10. _A_4._A_ = 3.
6~ X 8 — a;
6a; + 5 _ 4a; + 4
4a;3~ a;3 *
15 3a; 4 — ^x _ 5
14. (a; 4)3 (a; + 3)3 = —217. * 45a; 3a; ~~6*
1.
a;2
2
7a;
6
2.
7
6a;2
1
2
24 24 _ ^
a;2 X
10.
d + 3 (^ + 4 3
d2 d 2
11.
(a; + l)(a; + 3) = 12
> + (^
+ 7)V2.
V5a;23a;41 =
3a;
7. _
QUADRATIC EQUATIONS 257
16. V5a; + ll=V3x + l + 2. a;2 a; + 4 ^ 7
17. V7548 V5s4 = 2. * a; + 5 a;3~ 3
19. V8 a;3_35 a;2+55 a;  57 = 2 a;  3.
20. 3V^=1 4^ = 4. 22. 28(3. + 10)_ 25 ,
^x — 1 8 0^^ — 27 a;(2a; — 3)
21 2a; + l 3a;2 ^17 03 _J^ 1 ^ 14
• 3a;2 2a; + l 4' * x'Sx x'^Ax ISa;^*
24. ^+ ^^ 1^
a;2 24(a; + 2) a^4
25 5 7 ^ 8^^13i;64
* 2vf3 3?;4 6v' + v12 '
26 1^— i__ lU 3/^—1— i^
* 3V4a:l 2/ V3a; + 1 3j
27. 2V3^T4 + 3V3^+7= ^
28.
V3a; + 4
2a^4a;3 ^ a^4a; + 2
2x'2x\S x^3x[2
29. 1— _i_ = l+ ^
a^4 3(a; + 2) 2 a;
30. J^Ii+J^±5 = ^
31. V2^ + 2V2a; + 5 = 2V6a; + 4.
32. V8a; + 7 = V4a; + 3 + V2a; + 2.
33. V2x247a; + 7 = 6V2a^9icl.
34. V65aj + V27a; = Vl2 + 6a;.
35. ^±^+^±^^.^±3 = 3.
ic — 1 x — 2 x — 3
(Compare Ex. 1, § 167.)
Q/j X — 2 x\2 x—2 I
(B — 4 a;H3 a; — 6
258 ALGEBRA
37.
38.
"^ 1
x1
x'^x
x'x
1
x~l '
X
X
X
x' + 2x
x' + ^x
2
x^2
X + 3
+ 6*
07 + 4
X
,.5
x
on ^f^ _ i^ __ JO — iJ _ M ^
X X+^ X X—5
(First combine the first two fractions, and then the last two.)
290. Solution of Literal Affected Quadratic Equations.
For the solution of literal affected quadratic equations, the
methods of § 288 are usually most convenient.
1. Solve the equation x^{ax—bx — ab = 0.
We may write the equation x^ { (a — b)x = ab.
Multiplying both members by 4 times the coefficient of x^^
4a;2 + 4(a b)x = 4ab.
Adding to both members the square of a — b,
4 ic2 + 4(a  b)x + (a  6)2 = 4 a& + a2 _ 2 a6 + b'^
= a2 + 2 a& + 62.
Extracting square root, 2 x + (^a — b) = ±(a + b).
Or, 2x = {ab)±{a + b).
Then, * 2x = a + b + a\b = 2b,
or 2x = a + bab = 2a.'
Whence, a; = 6 or a.
If several terms contain the same power of x, the coefficient of that
power should be enclosed in parentheses, as shown in Ex. 1.
The above equation can be solved more easily by the method of § 126 ;
thus, by § 108, the equation may be written
(a; + a) (x  6) = 0. w
Then, x + a = 0, orx=— a;
and « — 6 = 0, or a; = 6.
QUADRATIC EQUATIONS 259
Several equations in Exercise 128 may be solved most easily by the
method of § 126.
2. Solve the equation (m — l)x^ — 2 m^x =: — 4:m^.
Multiplying both members by w — 1, and adding to both the square
of m2,
(m  1)%2 — 2 TO2(m  l)x + w* = — 4 m^(m  1) + m*
= m* — 4 m^ + 4 m^.
Extracting square root, (m — l)x — m^ = ± (m^ — 2m).
Then, (m — l)aj = m^ + m^ — 2 w or m^ — m^ f 2 m
= 2m(TO — 1) or 2 m.
Whence, x = 2 m or — ^.
TO — 1
In solving any fractional equation, we reject any solution which does
not satisfy the given equation.
EXERCISE 128
Solve the following equations :
X 1. a^ 4 2 mx = 1 — m^. / 4. x^ {• nx i x = — n.
2. x^ — 2ax = — 6a\9. ^5. ay^ — m^nx \ mn^x = m^n^.
^. x^\(a — b)x = ab. ,6. a^ — 4aa; — 10a; = — 40 a.
/L 6a^ + 4aaj15 6a; = 10a6. ^
y8. amx^ — anx — bmx { bn = 0. J$k ♦ y
. Va+a; — •v2«= — /lO. M^^ , =« ^*f
. Va+x 2a; + a ^i4a 3
yii. (a\xy\{bxy = (a + bf. . i^f, / 1
yi2. V(a + 26)x2a6 = a;46.
il3. (a2a2)a;2(5al)ic = 6..
y^4. a^ — (m — p) a; + (m — n) (ri — p) = 0.
^15. (a + 6)a;2+(3ate&)a; = 2a.
/16. (b + c)x^{a{'C)x = ha,
260 ALGEBRA
16 a
^j^7. VaJ + a + 2 V^ + 6a
^ ■\Jx 4 a
18. Va; — a, + v2a; + 3a = V5a.
19.
20.
X
a + & ^ 2(a^ + &') .
a 4 6 ic a^—lP'
1 _^1 1 1.
a—b—x a b x
21. ^^ + ? = 2.
(c + a — c a; + 6 — c
22 ^^ — ^^ _L '^^ + n ^ 10^
3a7Hn 2x— Sn 3
23. a='c2(l + a?)'  bH%l  xf = 0.
y24. ^!^ = _i^. 25. 2a^ + l ^ 2n4l ,
26. Vma; + V(m — n)x + mn = 2m.
V 27.
x — a x{ a a^ — 5a^
x\a x — a x^ — a^
^^^^y^^t>'a.^ CV
V 28 ^' + ^ 3 a; — 2 g _ 21 aa; — 4 g^ ^ I'iC y<
2x — 3a 3a; + g 6a;2 — 7 ga; — 3 g^ 1— 't'V
^f^9. (g6f 2c)a^(2g + 6 + c)a; = a26 + c.'^ '
30. J_ + lt^ + i = 0.
^ l^g a M\b b
a; g g + o
« PROBLEMS IN PHYSICS
owing equations occur in the study of physics.
Solve in the first six equations for the number which appears to the
second power.
1. S.^\gt\ 3. F='^. ^ •^=^
QUADRATIC EQUATIONS 261
7. Solve the following equation for a ; ^ = 7r\/
8. Solve the following equation for t; S = Vot {^ gt\
9. Solve the following equation for s ; V=^2gs.
10. In problem 1 solve for g ; in problem 7 solve for I ; in
problem 2 solve for m ; in problem 4 solve for M ; in problem
6 solve for Z.
PROBLEMS INVOLVING QUADRATIC EQUATIONS WITH
ONE UNKNOWN NUMBER
291. In solving problems which involve quadratic equations,
there will usually be two values of the unknown number; only
those values should be retained which satisfy the conditions of
the problem.
1. A man sold a watch for $21, and lost as many per cent
as the watch cost dollars. Find the cost of the watch.
Let X = number of dollars the watch cost.
Then, x = the per cent of loss,
r
and X X ^, or — = number of dollars lost.
100 100
By the conditions, ^^ = ic — 21.
100
Solving, X = 30 or 70. ^
Then, the cost of the watch was either $ 30 or $ 70 ; for either of these
answers satisfies the conditions of the problem.
2. A farmer bought some sheep for $ 72. "^f h^ had bought
6 more for the same money, they would have ddlrt him $ 1
apiece less. How many did he buy ?
Let n = number bought.
72
Then, — = number of dollars paid for one,
n
72
and = number of dollars paid for^Oine if there
^ "^ had been 6 more.
262 ALGEBRA
72 72
By the conditions, — = h 1.
n n + 6
Solving, w = 18 or — 24.
Only the positive value is admissible, for the negative value does not
satisfy the conditions of the problem.
Therefore, the number of sheep was 18.
If, in the enunciation of the problem, the words "6 more" had been
changed to "6 fewer," and "$1 apiece less" to "1 apiece more," we
should have found the answer 24.
3. If 3 times the square of the number of trees in an orchard
be increased by 14, the result equals 23 times the number ; find
the number.
Let X = number of trees.
By the conditions, S x"^ + U = 23 x.
Solving, a; = 7 or ^.
Only the first value of x is admissible, for the fractional value does not
satisfy the conditions of the problem.
Then, the number of trees is 7.
4. If the square of the number of dollars in a man's assets
equals 5 times the number increased by 150, find the number.
Let X = number of dollars in his assets.
By the conditions, cc^ = 5 x + 150.
Solving, X = 15 or — 10.
This means that he has assets of $ 15, or liabilities of $ 10.
EXERCISE 129
1. What number added to its reciprocal gives 2^?
2. Divide the number 24 into two parts such that twice the
square of the greater shall exceed o times the square of the
less by 45.
3. Find three consecutive numbers such that the sum of
their squares shall be 434.
QUADRATIC EQUATIONS 263
4. Find two numbers whose difference is 7, and the differ
ence of their cubes 721.
5. Find five consecutive numbers such that the quotient of
the first by the second, added to the quotient of the fifth by
the fourth, shall equal f .
6. Find four consecutive numbers such that if the sum of
the squares of the second and fourth be divided by the sum
of the squares of the first and third, the quotient shall be i§.
7. The area of a certain square field exceeds that of another
square field by 1008 square yards, and the perimeter of the
greater exceeds onehalf that of the smaller by 120 yards.
Find the side of each field.
8. A fast train runs 8 miles an hour faster than a slow
train, and takes 3 fewer hours to travel 288 miles. Find the
rates of the trains.
9. The perimeter of a rectangular field is 180 feet, and its
area 1800 square feet. Find its dimensions.
\/ 10. A merchant sold goods for $22.75, and lost as many
per cent as the goods cost dollars. What was the cost ?
u 11. A merchant sold two pieces of cloth of different quality
for $ 105, the poorer containing 28 yards. He received for the
finer as many dollars a yard as there were yards in the piece ;
and 7 yards of the poorer sold for as much as 2 yards of the
finer. Find the value of each piece.
l^ 12. A merchant sold goods for % 65.25, and gained as many
per cent as the goods cost dollars. What was the cost ?
13. A has fivefourths as much money as B. After giving A
% 6, B's money is equal to A's multiplied by a fraction whose
numerator is 15, and whose denominator is the number of dol
lars A had at first. How much had each at first ?
^' 14. A and B set out at the same time from places 247 miles
apart, and travel towards each other. A's rate is 9 miles an
hour; and B's rate in miles an hour is less by 3 than the
number of hours at the end of which they meet. Find B's rate.
264 ALGEBRA
15. A man buys a certain number of shares of stock, pay
ing for each as many dollars as he buys shares. After the
price has advanced onefifth as many dollars per share as he
has shares, he sells, and gains $ 980. How many shares did
he buy ?
16. The two digits of a number differ by 1; and if the
square of the number be added to the square of the given
number with its digits reversed, the sum is 585. Find the
number.
17. A gives $ 112, in equal amounts, to a certain number of
persons. B gives the same sum, in equal amounts, to 14 more
persons, and gives to each $4 less than A. How much does
A give to each person ?
18. The telegraph poles along a certain road are at equal
intervals. If the intervals between the poles were increased by
22 feet, there would be 8 fewer in a mile. How many are
there in a mile ?
19. A merchant bought a cask of wine for $ 48. Having lost
4 gallons by leakage, he sells the remainder at $ 2 a gallon
above cost, and makes a profit of 25% on his entire outlay.
How many gallons did the cask contain ?
20. The men in a regiment can be arranged in a column
twice as long as it is wide. If their number were less by 224,
they could be arranged in a hollow square 4 deep, having in
each outer side of the square as many men as there were in
the length of the column. Find the number of men.
21. The denominator of a fraction exceeds twice the
numerator by 2, and the d,ifference between the fraction and its
reciprocal is j. Find the fraction.
22. A man started to walk 3 miles, intending to arrive at a
certain time. After walking a mile, he was detained 10 min
utes, and was in consequence obliged to walk the rest of the
way a mile an hour faster. Find his original speed.
QUADRATIC EQUATIONS 265
23. A regimentj in solid square, has 24 fewer men in front
than v/hen in a hollow square 6 deep. How many men are
there in the regiment ?
24. A rectangular field is surrounded by a fence 160 feet
long. The cost of this fence, at 96 cents a foot, was onetenth
as many dollars as there are square feet in the area of the
field. Find the dimensions of the field.
25. A tank can be filled by one pipe in 4 hours less time
than by another ; and if the pipes are open together li hours,
the tank is filled. In how many hours can each pipe alone fill
it ? Interpret the negative answer.
26. A crew can row down stream 18 miles, and back again,
in 7 hours. Their rate up stream is 1^ miles an hour less
than the rate of the stream. Find the rate of the stream, and
of the crew in still water.
27. A man put $ 5000 into a savingsbank paying a certain
rate of interest. At the end of a year, he withdrew $ 375, leav
ing the remainder at interest. At the end of another year, the
amount due him was $ 4968. Find the rate of interest.
28. A square garden has a square plot of grass at the cen
tre, surrounded by a path 4 feet in width. The area of the
garden outside the path exceeds by 768 square feet the area of
the path ; and the side of the garden is less by 16 feet than
three times the side of the plot. Find the dimensions of the
garden.
29. A merchant has a cask full of wine. He draws out 6
gallons, and fills the cask with water. Again he draws out 6
gallons, and fills the cask with water. There are now 25 gal
lons of pure wine in the cask. How many gallons does the
cask hold ?
30. A and B sell a quantity of corn for $ 22, A selling 10
bushels more than B. If A had sold as many bushels as 13
did, he would have received $ 8 ; while if B had sold as many
bushels as A did, he would have received ^ 15. How many
bushels did each sell, and at what price ?
266 • ALGEBRA
31. Two men are employed to do a certain piece of work.
The first receives $ 48 ; and the second, who works 6 fewer
days, receives $ 27. If the second had worked all the time, and
the first 6 fewer days, they would have received equal amounts.
How many days did each work, and at what wages ?
32. A carriagewheel, 15 feet in circumference, revolves in
a certain number of seconds. If it revolved in a time longer
by one second, the carriage would travel 14400 fewer feet in
an hour. In how many seconds does it revolve ?
PROBLEMS IN PHYSICS
1. When a body falls from rest from any point above the
earth's surface, the distance, S, which it traverses in any num
ber of seconds, t, is found to be given by the equation
in which g represents the velocity which the body acquires
in one second. The value of g is 32.15 feet, or 980 centi
meters.
A stone fell from a balloon a mile high ; how much time
elapsed before it reached the earth ? •
2. If a body is thrown downward with an initial velocity, Vq,
then the space it passes over in t seconds is found to be given
by the equation
S = Vot{igf.
If the stone mentioned in Problem 1 had been thrown down
from the balloon with a velocity of 40 feet per second, how
many seconds would have elapsed before it reached the earth ?
3. In the equation ^ = 7r\/, t represents the time required
by a pendulum to make one vibration, I represents the length
of the pendulum, and g is the same as in Problem 1. Find the
length of a pendulum which beats seconds.
4. If a pendulum which beats seconds is found to be 99.3
centimeters long, find from the above equation the value of g.
QUADRATIC EQUATIONS 267
5. In the equation F= — ~, M and m represent the masses
(I
of any two attracting bodies, as, for instance, the earth and the
moon, d represents the distance between these bodies, and F the
force with which they attract each other.
If the moon had twice its present mass and were twice as far
from the earth as at present, how much greater or less would
the force of the earth's attraction be upon it than at present ?
6. In the equation E = \ mv^, E represents the energy of a
moving body, the mass of which is m and the velocity is v.
Compare the energies of two bodies, one of which has twice the
mass and twice the velocity of the other.
7. When a bullet is shot upward with a velocity, v, the height,
8, to which it rises is given by the equation ^
Find with what velocity a body must be thrown upward to
rise to the height of the Washington Monument (b6b feet).
(See Problem 1.)
268 ALGEBRA
XX. EQUATIONS SOLVED LIKE
QUADRATICS
292. Equations in the Quadratic Form.
An equation is said to be in the quadratic form when it is
expressed in three terms, two of which contain the unknown
number, and the expoyient of the unknown 7iumber in one of
these terms is twice its' exponent in the other; as,
a;«6i^=16; a^{x^72 = 0; etc.
293. Ec^uations in the quadratic form may be readily solved
by the rules for quadratics.
1. Solve the equation x^ — 6a:^ = 16.
Completing the square by the rule of § 285,
x6 _ 6 a;3 + 9 = 16 + 9 = 25.
Extracting square roots, x^ — 3 = ± 6.
Then, ' a;^ z= 3 ± 5 = 8 or  2.
Extracting cube roots, x = 2 or — \/2.
There are also four imaginary roots, which may be found by the
method of §301.
2. Solve the equation 2x\ SVx = 27.
Since Vx is the same as x^, this is in the quadratic form.
Multiplying by 8, and adding 3^ to both members (§ 288),
16 X + 24\/S + 9 = 216 + 9 = 225.
Extracting square roots, 4 Vx + 3 = ± 15.
Then, 4 Vx =  3 ± 15 = 12 or  18.
Whence, Vx = 3 or  , and x = 9 or — .
3. Solve the equation 16 x'^ — 22 x~^ = 3.
EQUATIONS SOLVED LIKE QUADRATICS 269
Multiplying by 16, and adding ll"^ to both members,
162 x"^ _ 16 X 22 x~* + 112 ^ 48 + 121 = 169.
_3
Extracting square roots, 16 x ^ — 11 == i 13.
Then, 16 x~^ = 11 ± 13 = 24 or  2, and x~^ =  ^^ ~ *
Extracting cube roots, a:"^ ~ ( 9 ) ^ °^ ~ i'
Raising to the fourth power, x^ " (9 )^ ^^ i«*
. l = (?)tori,a„d. = ()iorl6.
P
To solve an equation of the form aj* = a, first extract the root corre
sponding to the numerator of the fractional exponent, and afterwards
raise to the power corresponding to the denominator ; careful attention
must be given to algebraic signs ; see §§ 96 and 209.
EXERCISE 130
Solve the following equations :
1. a;4  29 aj2 =  100. 5. a^^  63 a;"^  64 = 0.
2. x'^ + 19x'=216. ^6. 3x'\Ux' = 5. '
3. aj^10a;^ + 9 = 0. 7. 5 a;"^ + 7 ic"^ =  2. ^,
4. a;^+33a^^ = 32. 8. 4v'^ + 6 = ll^^.
ulO. 9(aj^ + lf=(a;34)2 + llflj3_5. ^ )(^4 /ir)(
11. 6h2 = llVTi. ,17. 32V^_33 = ^.
12. x^' + 2Ux^ = 243, ^^
3n 3n / 18. 161 0^ + 5 =  32 OJ^^
13. Sx^Ax' =16.
14. 2.«35. + 48 = 0.  19 308 = 64a.t.
15. 27x3 + 46 = g. ^^ V^ + V^^^_^V6.
16. 16a;»33»*243 = 0. * Va Va? V6 Va
\ 
270 ALGEBRA
^ ♦
21. V6+V^jV4Va; =
12
V4Va;
/^22. V3Va; + lfVV^4 = V4Va;+5.
294. An equation may sometimes be solved with reference
to an expression, by regarding it as a single letter.
1 . Solve the equation {x  5)^  3(a;  5)^ = 40.
Multiplying by 4, and adding 3^ to both members,
4(0;  5)3  12(x  5) 2 + 32 = 160 + 9 = 169.
Extracting square roots, 2(x — 5)^ — 3 = ± 13.
Then,
2Ca;5)2 = 3dbl3 = 16 or
10.
Whence,
{x 6)^ = 8 or  5.
Extracting cube roots,
(x _ 5)^ = 2 or  VI,
Squaring,
X  5 = 4 or v'25.
Whence,
a; = 9 or 5 + v'25.
Certain equations of the fourth degree may be solved by, the.
rules for quadratics.
2. Solve the equation a;* + 12a^ + 34aj2 12 a; 35 = 0.
The equation may be written
(x* + 12 x3 + 36 x2)  2 x2  12 X = 35.
Or, (x2 + 6x)22(x2 + 6x)=35.
Completing the square, (x2 + 6 x)2  2(x2 + 6 x) + 1 = 36.
Extracting square roots, (x2 + 6 x)  1 = ± 6.
Then, x2 + 6 x = 7 or  6.
Completing the square, x^ + 6 x + 9 = 16 or 4.
Extracting square roots, x + 3=±4or ±2.
Then, x =  3 ± 4 or  3 ± 2 = 1,  7,  1, or  5.
In solving equations like the above, the first step is to complete the
square with reference to the x* and x^ terms ; by § 287, the third term of
the square is the square of the quotient obtained by dividing the x^ term
by twice the square root of the x* term.
EQUATIONS SOLVED LIKE QUADRATICS 271
3. Solve the equation af — 6x\ 5 Va;^ — 6 a; + 20 = 46.
Adding 20 to both members,
(x2  6 x + 20) + 5Va;2  6 x + 20 = 66.
Completing the square,
(a;2 _ 6 X + 20)+ 5 Vx2  6 X + 20 +^ = 6Q+^=z^.
4 4 4
17
Extracting square roots, Vx^ — 6x + 20 +  = ±
Then, Vx'^  6 x + 20 = 6 or  11.
Squaring, ^ x2  6 x + 20 = 36 or 121.
Completing the square, x^ — 6 x + 9 = 25 or 110.
Extracting square roots, x — S=±5 or ± VllO.
Then, x = 8,  2, or 3 ± vHO.
In solving equations of the above form, add such an expression to both
members that the expression without the radical sign in the first member
may be the same as that within, or some multiple of it.
4. Solve the equation — h — —  = •
ar — xx^ — S2
^2 o
Representing — by y, the equation becomes
x^ — X
y +  = l, or 2?/2 + 2 = 5y.
y 2
Solving this, w =  or 2 : that is, 5!jZ§ = 1 or 2.
^ ' ^ 2 ' ' x2  X 2
Taking first value, 2 x^ — 6 = x^ — x, or x^ + x = 6.
Solving, X = 2 or — 3.
Taking second value, x^ — ,3 = 2 x^ — 2 x, or —3^+2 x=3.
Solving, , X = 1 ± V^^.
EXERCISE 131
Solve the following equations :
1. (2a^3a?)=^8(2a^3a;) = 9.
272 ALGEBRA
'2. a^H 103^^4 23 0.2100^24 = 0.
3. x412ar^ + 14ar^+132aj135 = 0. %u
•^4. 5 a; + 12 + 5 V5 x + 12 =  4.
5 a^'3 2 a; ^ 17
* 2x x'S '~ 4*
6. V5^+I + 3a/5¥+I = 10.
7. 3a;2 + aj + 5V3^T^T6 = 30.
8. 8a.*2l + 6W8^^^1 = 8a^.
9. a;*2aaj317aV+18a3a; + 72a^ = 0.
/lO. (7a;6)^5(7a;6)* = 6,"
11 ^' + 2 2d5 ^35
' 2 d  5 d^ + 2 6 '
^2. £c2^7Va;24a) + ll = 4a;23.
^. Var^3a;3 = a;23aj23.
14. (2a;23a;l)37(2a^3i»l)^ = 8.
A5. Ss/x'12x7s/x'12x = 2.
^16, 7c'lS](^ + 109k^252k\lS0 = 0,
17. 2ar' + 4a; + Var^H2x3 = 9.
18. 7 («3 _ 28)t 4 8 (o^  28)^ =  1.
19. (3^^ + 15)* 5 (3 ^ + 15)^ = 24.
20. 9a;^.12a;3_35^2^26a; + 40 = 0.
21 a:^5a; + l x^2x^2 ^ 8
* ic2_2a; + 2 aj2_5^_^,^ 3*
22. 9(x + a)^22b\x^a)^\Sb* = 0,
23. ar^ + l+Va;28ajH37 = 8(a;412).
24. 25 (a; + 1)"^  15 (x + 1)~^ =  2.
25. J?+? JIZ
THEORY OF QUADRATIC EQUATIONS 273
XXI. THEORY OF QUADRATIC EQUATIONS
295. Number of Roots. JU„,<j^. 'T3utr Cl*1 (jtvX,!^ A^'rCv
A quadratic equation mrmot have more than two different roots.
Every quadratic equation can be reduced to the form
g,oif + bx\c = 0. ^^v ■» '•*«
'^'^Tti possible, let this have three different roots, r^, rg, and r^.
Then, by § 81, ar,' + br^ 4 c = 0, (1)
ar2' + br2{c = 0, (2)
and ars' + brs\c = 0, (3)
Subtracting (2) from (1), a (?v — ^2^) + ^ (n — **2) = 0.
Then, a (r^ + rg) (rj  rg) + 6 (rj  ^2) = 0,
or, (ri — rg) (an + a^a + 6) = 0.
Then, by §126, either ri — r2=0, or ari+ar2+6=0.
But ri — 7'2 cannot equal 0, for, by hypothesis, ri and rg are
•different.
Whence, ar^ + arg + 6 = 0. (4)
In like manner, by subtracting (3) from (1), we have
ar^ + arg + 6 = 0. (5)
Subtracting (5) from (4), ar2— a?3=0, or rg— r3=0.
But this is impossible, for, by hypothesis, ^v and r^ are
different; hence, a quadratic equation cannot have more than
two different roots.
296. Sum of Roots and Product of Roots.
Let ?'i and rg denote the roots of ax^ + 6a; + c = 0.
By §289, ^^^ fe4:Vi'4ac ^ ^^^ ^^^ _&_ V^ _4«c .
2 CL 2i a
274 ALGEBRA
Adding these values, rj + rg = ^ — =
2a a
Multiplying them together, ^
4a^ 4a^ a
Hence, if a quadratic equation is m the form ax^ {hx\c — 0,
the sum of the roots equals minus the coefficient of x divided "
by the coefficient of a^, and the product of the roots equals the
independent term divided by the coefficient of x^. t
1. Find by inspection the sum and product of the roots of
Jl **
The sum of the roots is , and their product ^^^ — , or — 5. 9i^^^ 3. ft.
2. One root of the equation 6 a^ 4 31 a; = — 35 is — ; find
the other.
The equation can be written 6 cc^ + 31 aj 4. 35 = 0.
31
Then, the sum of the roots is
6
31 / 7\ 31 7
Hence, the other root is ( — ), or 1,
6 \ 2/' 6 2'
or.
We may also find the other root by dividing the product of the roots,
^,byL
6 "^ 2
We may find the values of certain other expressions which
are symmetrical in the roots of the quadratic.
3. If ri and rg are the roots of aa^ + 6a; + c = 0, find the value
of ri^ + rjrg + ?'2^.
We have, ri^ + rir^ + r2^ = (n \ rzY  rir2.
b c *
But, n + ra = — , and nr^ = —
a a
Whence, rv' + r.r, ^r,^ = ^^^ = ^^^.
THEORY OF QUADRATIC EQUATIONS 275
EXERCISE 132
Find by inspection the sum and product of the roots of :
> 1. a;2 + 8aj47=0. ^.^ 5. 2xUx' = 7. .'^ 
>2. x' + x20 = 0.l:~ 6. 10 + 12aj15a^ = 0. I
3. a.'26a; + l = 0. ^ >7. Sx^2 = x. ^ ~ ^
4. 4a;2.r5 = 0. 4,  ^^' 8. 9 mV + 21 w,?2c)+ 5 n^ = 0.
/' 9. One root of a;^ + 7 a; = 98 is 7 ; find the other. S   
y\ 10. One root of 28 ar^ — a; — 15 = is — f ; find the other, s = i^
(4,1. One root of 5 a^  17 a; + 6 = is  ; find the other, ^^^"t
If Ti and 72 are the roots of aoc^ \ bx { c = 0, find the values
of: c^.^^ a^
12. ri+lL. 13.'i + i. 14. 1 + 1. 15. r,3 + r/.
297. Formation of Quadratic Equations.
By aid of the principles of § 296, a quadratic equation may
be formed which shall have any required roots.
For, let rj and r^ denote the roots of the equation
aaj2 + 6a; + c = 0, ora^^^^^c^O. (1)
a a
Then, by § 296,  =  rj  rg, and  = r^r^.
Substituting these values in (1), we have
sc^ — riX — r^ + TiTz = 0.
Or, by § 108, (a;  r,) (x  n) = 0.
Therefore, to form a quadratic equation which shall have
any required roots.
Subtract each of the roots from x, and place the product of the
resulting expressions equal to zero.
Ex, Form the quadratic whose roots shall be 4 and — J.
276 ALGEBRA
By the rule, {xi){x+] = 0.
Multiplying by 4, (x  4) (4 a: + 7) = ; or, 4 x2  9x  28 = 0.
EXERCISE 133
Form the quadratic equations whose roots shall be :
"1. 5,8. 3. 1, f. 5. i, . 7. I, f.
2. 4,3. 4. 6, L3. ,6. 1/, 0. 8. i,j\.
9. a{2b, a2b. .11. _45V3, 45V3.
VlO. 3mn,m + An. >12. V^jh^V^, V^^V^K
2 ' 2
V ^ . ^ .> ^ .^ / 'j^ 7 FACTORING
298. Factoring of Quadratic Expressions.
A quadratic expression is an expression of the form
ax^ \hx\c.
In § 117, we showed how to factor certain expressions of this
form by inspection ; we will now derive a rule for factoring any
quadratic expression ; we have,
V a a)
\_ a \2aJ 4:a^ a J
W 2aJ 4.a' J
V 2a^ 2a J\ ^ 2a 2a J'
by § 114.
But by § 289, the roots of ax^ + &a; + c = are
^ _i.vV4ac^„, b ^b^4.ac
2~a'^ 2a ^""^ ~21l ^2^^
THEORY OF QUADRATIC EQUATIONS 277
Hence, to factor a quadratic expression, place it equal to zero,
and solve the equation thus formed.
Then the required factors are the coefficient of x^ in the given
expression, x minus the first root, and x minus the second.
1. Factor6a^ + 7i»3.
Solving the equation 6 a;2 + 7 a;  3 = 0, by § 289,
7i:Vi9T72 ^ 7i:ll ^l ^^ 3_
12 12 3 2
Then, . 6 a;2 + 7 a;  3 = 6 f x  1 Vx + ^^
= 3(a:)x2^a;+U(3:rl)C2x + 3).
2. Factor 4 4 13 a? 12 a^.
Solving the equation 4 + 13 x  12 x^ = 0, by § 289,
 13 ± V169 + 192 ^  13 ± 19 ^ _ 1 ^j. 4
24 24 4 3*
Whence, 4 + 13a;  12a;2 = 12(x + Vx ^
=Ki)x()el)
= (l + 4«)(43a;).
3. Factor 2ar'3a;2/2/7a; + 42/ + 6.
We solve 2 a;2  a:(3y + 7) 2 ^2 + 4 ^ _!_ 6 _ 0.
By § 289,
^ 3y + 7±V(3y + 7)2 + 16y2_32y48
4
_ 3 y + 7 zb V25 y2 +10 y . 1 ^ 3 y + 7 ± (5 y + 1)
4 4
SJM^ or =l^±i = 2y + 2or ZliL±3.
4 4^2
Then, 2x23a;i/2i/2_7ic + 4?/ + 6
= 2[x(2y + 2)][a; ~^ + ^ "
2
:(a;2y2)(2a; + y3).
^^278
ALGEBRA
/
Tac
EXERCISE 134
itor the following :
1.
0^414^ + 33.
12.
la^lSaxjSa^.
2.
a^_13 0^440.
13.
21a^ + 5ic6.
3.
a^a42.
14.
12ic2_l6a;35.
4.
2x'^llx6.
15.
3 4.13a;_30a;2.
5.
4a^ + 19a; + 12.
16.
285^12^1
6.
8m214m + 5.
17.
16x^'\S0x\9,
7.
QSxaf,
18.
18a^31aj + 6.
8.
10a^ + 39x + 14.
19.
6a^23a;35.
9.
40 419a;3a^.
20.
2013 a;15aj2.
10.
SBxQx".
21.
20a^^SSmx\7m\
11.
5a^21nx\lSnK 22.
24:x'3Sxyz\15yh'.
23. x^xy6y^6x}13y\5.
24. i»2_3a;y_42/2 4_e^_4^4_3^
^ 25. x^6xy + 5y'2x2yS.
26. 2a2 + 5a6 + 262 + 7a.45& + 3.
27. 3x^\Txy6y^10xSy{S.
28. 272/7a; + 3^2 4.^2/4a;2.
29. 6a;22a;2/202/' + 5a;2 + 232/2;622.
299. If the coefficient of x^ is a perfect square, it is prefer
able to factor the expression by completing the square as in
§ 287, and then using § 114.
1. Factor 9a^9a;4.
By § 287, 9 a;2 — 9 a; will become a perfect square by adding to it the
9 ^
square of , or —
2x8' 2
THEORY OF QUADRATIC EQUATIONS 279
Then, q x^ 9 x ^ = 9 x"^ 9 x + (^Y 4:= ^8^:^
(^^M)(«^M) «^^*>
= (3x + l)(3x4).
If the oc^ term is negative, the entire expression should be
enclosed in parentheses preceded by a — sign.
2. Factor 3  12 a;  4 a^.
3 _ 12 X  4 a:2 = (4 a:2 + 12 a;  3)
=  (4 a:2 + 12 X + 9  9  3)
= [(2x + 3)2_12]
= (2 X + 3 f VT2) X ( 1)(2 X + 3  VT2)
= (2V3 + 3f2x)(2V332x).
EXERCISE 135
Factor the following :
1. ^x'Uxl. . 7. lf2a;a^.
2. 9a^21a; + 10. 8. 16 x^  16 x \ 1.
3. ar^ + a;12. 9. 6~5x25x'.
4. 16a^ + 40a; + 21. 10. 4.x^ + 9x9.
5. 9.^2 + 24052. 11. 36 a:^ __ 72 a^ I 29.
6. 4x2 + 20a; + 19. 12. 25a^10xll.
300. We will now take up the factoring of expressions of
the forms x'^ + aa^y^ f y^, or x*^ \ y*, when the factors involve
surds. (Compare § 115.)
1. Factor a^ + 2 a'b^ + 25 b\
a* + 2 a262 + 25 64 = (a* + 10 a262 + 25 &4)  8 a262
= (a2+5 62)2_(a6V8)2
= (a2 + 5 &2 + a&V8)(a2 + 5 ^2 _ ahVS)
= (a2 + 2 a&\/2 + 5 62) (^2 _ 2 ahV2 + 5 62). *
280 ALGEBRA
2. Factor x* + l.
= (X2+1)2(XV^)2
= (a;2 + X V2 + 1) (a;2 _ a, V2 + 1).
EXERCISE 136
In each of the following obtain two sets of factors, when
this can be done without bringing in imaginary numbers :
1. x^7a^h4:. 4. 4a4 46a2 + 9.
2. a'{b\ 5. 36«^92a^ + 49.
3. Qm^llm^ + l. 6. 25 m^ + 28 mV + 16 w^
301. Solution of Equations by Factoring.
In § 126, we showed how to solve equations whose first mem
bers could be resolved by inspection into first degree factors,
and whose second members were zero.
We will now take up equations whose first members can be
resolved into factors partly of the first and partly of the
second, or entirely of the second degree.
1. Solve the equation x^ + 1 = 0.
Factoring the first member, (x \ 1) (x^ — x j 1) = 0.
Then, ic+l=0, ora; = l;
and x2  x + 1 = ; whence, by § 289, x = 1 ± ^1  ^ = l±y/S ^
2: Solve the equation a;* + 1 = 0.
.By Ex. 2, § 300, x^ + 1 = (a;2 + x\/2 + 1) (x2  x\/2 + 1).
Solving x2 + x\/2 + 1 = 0, we have
^ ^ V^ J:V2"^^ _  V2 J: \/^r2 _
2 2
Solving x2  x\/2 + 1 = 0, we have
^ ^ V2 ±y/2^^ ^ \/2 J: V^
2 2
THEORY OF QUADRATIC EQUATIONS 281
The above examples illustrate the important principle that the degree
of an equation indicates the number of its roots ; thus, an equation of the
third degree has three roots ; of the fourth degree, four roots ; etc.
The roots are not necessarily unequal ; thus, the equation x^— 2 x+l =
may be written (x — l){x — \) = 0, and its two roots are 1 and 1.
EXERCISE 137
Solve the following equations :
1. 5a;33a;29a; = 0. 11. a^  6a.2 + 1 = 0.
/2. (ajh4)(2x2_^5a? + 25)=0. 12. x^ox' + l^O,
V3. (9x24)(llar^+8a;4)=0. 13. e>4.:^ 125 = 0.
4. a;4llaV12a* = 0. 14. x^ 10^2 + 9 = 0.
5. a;^81 = 0. 15. a;^ 20x^4 16 = 0.
6. a;^ + 2a^ + 2aj + 4 = 0. 16. 9a;* + 5a;2 + 4 = 0.
1/7. x«l = 0. 17. a;6729 = 0.
1^. x^ + 8aj = 0. 18. 0^8256 = 0.
9. 5a)34a.'^ + 60a.48 = 0. a,^42a^ + 4 ^4
10. 2l7? + ^a^ = 0. ' a^2a;4 x"
20. 9aj*a;2f 4 = 0.
21. ■Va;^ + l4V9a;4a^ = 2a;l.
302. Discussion of General Equation.
By § 289, the roots oi ax^ \ hx \ c = are
2a 2a
We will now discuss these results for all possible real values
of a, b, and c.
I. 6^ _ 4 (5^c positive.
In this case, Vi and rg are real and unequal.
II. 62_4ac = 0.
In this case, rj and rg are ?'ea/ and egwaZ.
III. 6^ — 4 ac negative.
In this case, rj and rg are imaginary (§ 273). £^^1. n— ^rw^JL
IV. 6 = 0.
In this case, the equation takes the form , —
ax^ + c = J whence, a; = ± y ' ' '
If a and c are of unlike sign, the roots are real, equal in abso
lute value, and unlike in sign.
If a and c are of like sign, both roots are imagi7iary.
V. c = 0.
In this case, the equation takes the form ^ * ~ ^^ U* V
aa^6a; = 0; whence, a; = or .^ '^
a
Hence, the roots are both real, one being zero. *^
VI. 6 = 0, and c = 0.
In this case, the equation takes the form ax^ = 0. '^ r ^ ^y'
Hence, both roots equal zero. ^^
The roots are both rational, or both irrational, according, as
6^ — 4 ac is, or is not, a perfect square.
Ex. Determine by inspection the nature of the roots of
2x2_5ir18 = 0.
Here a =2, & =  5, c =  18 ; and 62 _ 4 ^c = 25 + 144 = 169.
Since h^ — i^ac is positive, the roots are real and unequal.
Since h^ — 4iac is a perfect square, both roots are rational.
EXERCISE 138
Determine by inspection the nature of the roots of the
following: ,,JL,>^, _, ,v
1. &x' + llx\^ = 0. /61 6. aj2_i9a;4. 125 = 0. if
2. 6aj2 + a; = o:"..^.... ;.^.^ 7. ^x'4.x = 0. i^^'tS
0^ ~ ^(> AM>3r
4. 12«219« + 4 = 0.''^:^ ^^J^ 16a;2 4.24a;H9 = 0. ^
5. 25a;2_4=:0. >^. 11.^0. 30a^30 = llaj. J2 7 2.
THEORY OF QUADRATIC EQUATIONS
283
GRAPHICAL REPRESENTATION OF QUADRATIC EXPRES
SIONS WITH ONE UNKNOWN NUMBER
303. The graph of a quadratic expression, with one unknown
number, x, may be found by putting y equal to the expression,
and finding the graph of the resulting equation as in § 181.
1. Find the graph of x^ — 2 a; — 3.
Put y = ic2  2 a;  3.
If X = 0,
2/ = 3. {A)
If x=l,
y = 4. (B)
If X = 2,
y = s. (C)
lix = 3,
y = o. (D)
If a; = 4,
y = 5. (E)
If a: =  1,
y = 0. (F)
If a; = 2,
y = 5. (G)
The graph is the curve GBE.
By taking other values for a;, the curve may be traced beyond E and ;
it extends in eitlier direction to an indefinitely great distance from XX'.
[We may determine the lowest point of the curve hy the artifice of com
pleting the square, as follows :
We have, x^  2 x  3 := (a:^  2 a; + 1)  1  3 = (x  1)2  4.
The latter expression has its negative value of greatest absolute value
when X = 1, being then equal to — 4.
Then, the lowest point of the curve has the coordinates (1, — 4); and
is therefore the point B'].
2. Find the graph of 2x2 + a; 3.
Put y = 2x2 + x3.
If X = 0, y=S.'
If X = 1, y = 0..
If X = 2, y = 7.
If x = l, y=2.
If X =  2, y = 3.
The graph is the curve ABC.
[To find the lowest point of the curve, we have
2x2 + x3 = 2fx2 + ^V3 = 2fx2 + ^+i^A_3 = 2fx + y— .
V 2^ V 2 16; 16 V 4/ 8
284
ALGEBRA
The latter expression has its negative value of greatest absolute value
1 25
when X = — , being then equal to
4 8
Then, the lowest point has the coordinates
)•]
304. The principle of § 188 holds for the graph of the first
member of any quadratic equation, with one unknown number.
Thus, the graph of a^ — 2 x — 3 (§ 303) intersects the axis
XX' at points whose abscissas are 3 and — 1, and the equation
ic^ — 2a7 — 3 = has the roots 3 and — 1.
Again, the graph of 2 x^{x — 3 intersects XX' at the point
whose abscissa is 1, and between the points whose abscissas are
— 1 and — 2 ; and the equation 2 cc^ + a? — 3 = has one root
equal to 1, and one between — 1 and — 2.
EXERCISE 139
Find the graph of the first member of each of the following
equations, and verify the principle of § 188 in the results :
1. x^5x\A = 0. 5. 4.x'^Tx = 0.
2. x^\x6 = 0. 6. 2a^lla;6 = 0.
3. a;2f7a; + 10 = 0. 7. 6x2 + 5a;6 = 0.
4. Sx'4:X = 0. 8. Sx'Ux15 = 0.
305. Graphs of the First Members of Quadratic Equations hav
ing Equal or Imaginary Roots.
1. Consider the equation a?^ — 4 a? + 4 = 0.
We may write the equation (a; — 2)(x— 2) = 0.
Then, by § 126, the roots are 2 and 2.
To find the graph of the first member, put
y = (x 2)2.
If x = 0, y = 4. If x = 2, y = 0.
If x = l, y = l. If x = S, y = l; etc. X
The graph is the curve ABC, which extends
to an indefinitely great distance from XX'.
Since (x — 2)2 cannot be negative for any value of x, y cannot be nega
tive ; and the grapli is tamjent to XX'.
THEORY OF QUADRATIC EQUATIONS
285
It is evident from this that, if a quadratic equation, with one
unknown nuinber, has equal roots, the graph of its first member
is tangent to XX'.
2. Consider the equation ic^ + a; + 2=0.
Solving,
 1 j V 7
!> i
•^ o
r
To find the graph of the first member, put
1/ = x2 + X + 2.
If x = 0,2/ = 2. Ifa;=1, y = 2.
If X = 1, y = 4. If X =  2, y = 4 ; etc.
The graph is the curve ABC, which extends
to an indefinitely great distance from XX'.
We have, x2 + x + 2 = (x^ + x + ^^  ^ + 2 = (x + ^V + ^•
Since (x +  ) + cannot be zero or negative for any value of x, y
V , 2/ 4
cannot be zero or negative, and the graph does not intersect XX'.
It is evident from this that, if a quadratic equation, with one
unknown number, has imaginary roots, the graph of its first
member does not intersect XX'.
EXERCISE 140
Find the graphs of the first members of the following, and
in each case verify the above principles :
2. a^ + 3a;f4 = 0. 4. 2x^Ax + 5 = 0.
286 ALGEBRA
XXII. SIMULTANEOUS QUADRATIC
EQUATIONS
306. On the use of the double signs ± and T .
If two or more equations involve double signs, it will be
understood that the equations can be read in two ways ; tirstj
reading all the n2:>per signs together; second, reading all the
lower signs together.
Thus, the equations x = ±2, y = ±3, can be read either
x = ^2,y = + 3,OTx = 2,y = 3.
Also, the equations x = ±2, y=:f 3, can be read either
x = j2,y = 3,ovx = 2,y = {3.
307. Two equations of the second degree (§ 83) with two
unknown numbers will generally produce, by elimination, an
equation of the fourth degree with one unknown number.
Consider, for example, the equations
fx^^y=a. (1)
\x\f = b. (2)
From (1), y = a — x^', substituting in (2),
x\a^ — 2 a'y? \x^—h\
an equation of the fourth degree in x.
The methods already given are, therefore, not sufficient for
the solution of every system of simultaneous quadratic equa
tions, with two unknown numbers.
In certain cases, however, the solution may be effected.
308. Case I. When each equation is in the form ■.. ,
ax^ + by = c.
In this case, either cc or y^ can be eliminated by addition or
subtraction.
SIMULTANEOUS QUADRATIC EQUATIONS 287
1. Solve the equations i„ ^ ^^ ^
1 3 ?/2 — 11 03^ = 4.
(1)
(2)
Multiply (1) by 3, 9x^ + 12 y^ = 228.
Multiply (2) by 4, 12 y^ 44 x2 = 16.
Subtracting, 53 x^ = 212.
Then, x^ = 4, and a; = ± 2.
Substituting x = ± 2 in (1), 12 + iy^ = 76, or ^y'^ = 64.
Then, y^ = 16, and 2/ = ± 4.
The solution is x = 2, ?/ = ± 4 ; or, x = — 2, ?/ = ± 4.
In this case there are four possible sets of values of x and y which
satisfy the given equations :
1. x = 2, ?/ = 4. 3. x = 2, y = 4.
2. x = 2, y=4. 4. x = 2, ?/ = 4.
It would not be correct to leave the result in the form x = ± 2, y = ± 4,
for this represents only the first and fourth of the above sets of values.
The method of elimination by addition or subtraction may be
used in other examples.
.« ^ , , . r3a^42/ = 47. (1)
" 2. Solve the equations i „ ^ ^ ,/^
^ l7aj2 + 62/ = 33. (2)
Multiply (1) by 3, 9 x2  12 y = 141.
Multiply (2) by 2, Ux^+12y= 66.
Adding, 23x2 = 207.
Then, x2 = 9, and x = ± 3.
Substituting x = ± 3 in (1), 27 — 4 ?/ = 47, and ?/ = — 5.
It is possible to eliminate one unknown number, in the above examples,
by substitution (§ 169), or by comparison (§ 170).
EXERCISE 141 
/^ ISolve the following equations :
(Sx'\2y' = 66. (Sx5y' = 116.
l9a^ + 52/' = 189. " ' l7a; + 4/ = 121.
288 ALGEBRA
fll(c262/' = 84.
4. ^ ^ _ ^_, „ „ 7.
3 a;2  2 iK.V = 24.
3. 1 6.
4:X^ — 5xy = 46.
8/i227A:2=:6. I .^2 _^ a:?/ + 3 2/' = 27.
la^_32/2 = _95. • 1 a^2/ = 17.
4 a;2 _ ^2 ^ 3 ^2 , 10 a6 + 3 62.
5.
9. ,
42/'a;2 = 3a10a64362.
309. Case. II. When one equation is of the second degree,
and the other of the first.
Equations of this kind may be solved by finding one of the
unknown numbers in terras of the other from the first degree
equation, and substituting this value in the other equation.
2 cc^ — xy = 6y. (1)
x+2y = 7^ (2)
From (2), 2y = 7  x, or y = '^^^  (3)
Ex, Solve the equations
Substituting in (1), 2x^x(  —  \ = 6 [ 'L^^ ] •
Clearing of fractions, 4 a;2  7 a; + x^ = 42  6 ib, ot 6x^x=i2.
Solving, x = S or — — .
Substituting in (3), y = ^? or ——2. = 2 or — .
2 2 10
The solution isa; = 3, w = 2: or, a; = ,w = —
Certain examples where one equation is of the third degree and the
other of/the first may be solved by the method of Case II.
y ■ '
y EXERCISE 142
1»Solve the following equations :
/^ ^ ra^432/2 = 37. ^ p + 2/ = _4.
la;— 2?/ =9. * 1 ic?/ = — 45.
SIMULTANEOUS QUADRATIC EQUATIONS 289
3.
5.
X— y = 5.
x^xy\2y^ = 8.
3x\y = 10,
X \ y = 7.
x^ — xy\y^ = 124.
a; + 2/=8.
, 2e3^ = 5.
344.
y
x — y = a\2h.
[o^ + y'' = o? + 2ab\2h\
rar^ 2/3 = 34
U w =8.
10.
f2a; 3?/_o
12a; 3^
2.
a;?/ = a^ + a — 2.
3x + 42/ = 7a + 2.
a; 40
21
310. Case III. When the given equations are symmetrical
with respect to x and y ; that is, when x and y can be interchanged
without changhig the equation.
Equations of this kind may be solved by combining them in
such a way as to obtain the values oi x\y and x — y.
1. Solve the equations \
x\y=:2.
xy=^ — 15.
Squaring (1), x^
+ 2 5cy + ?/2 = 4.
Multiplying (2) by 4,
4 xy =  60.
Subtracting, x2
 2 xy + y2 = 64.
Extracting square roots,
x2/=±8.
Adding (1) and (3),
2 a; = 2 ± 8 = 10 or  6.
Whence,
X = 6 or  3.
Subtracting (3) from (1),
2 y = 2 =F 8 =  6 or 10.
Whence,
?/ =  3 or 5.
The solution is a; = 5, y =
 3 ; or, a; =  3, y = h.
(1)
(2)
(3)
290 ALGEBRA .
In subtracting ± 8 from 2, we have 2 T 8, in accordance with the nota
tion explained in § 306.
In operating with double signs, ± is changed to =F , and T to ± , when
ever 4 should be changed to — .
(The above equations may also be solved by the method of Case II ; but
the symmetrical method is shorter and neater.)
, . . (x' + i/ = 50. (1)
2. Solve the equations i
^ I xy = 7. (2)
Multiply (2) by 2, 2xy= 14. (3)
Add (1) and (3), x'^ + 2xy + y^ = S6, or x+ y =±6. (4)
Subtract (3) from (1), x^ 2xy + y^ = 64, or x  y =±8. (5)
Add (4) and (5), 2 x = 6 ± 8, or  6 ± 8.
Whence, a; = 7, — 1, 1, or — 7.
Subtract (5) from (4), 2 ?/ = 6 T 8, or  6 T 8.
Whence, y = —l, 7, —7, or 1.
The solution is a? = ± 7, 2/ = =F 1 ; or, a; = ± 1, y =^7.
Certain examples in which one equation is of the tJiird degree,
and the other of the first or second, may be solved by the
method of Case III.
(a^f =56. (1)
3. Solve the equations i . „
^ \^ + xy\f = 2S. (2)
Divide (1) by (2), xy = 2. (3)
Squaring (3) , x'^  2 xy \ y^ = 4. (4)
Subtract (4) from (2), Sxy = 24, or xy = 8. (5)
Add (2) and (5), x^ + 2 xy + y^ = S6, or x + y =± 6. (6)
Add (3) and (6), 2ic = 2±6 = 8 or 4.
Whence, a; = 4 or — 2.
Subtract (3) from (6), 2 y = ± 6  2 = 4 or  8.
Whence, y = 2 or — 4.
The solution is x = i, y = 2 ; or, x = — 2, y = — i. '
(If we interchange x and y in equation (1), it becomes
y'^ — x^ = 56, or x^ — y^ = 56,
which is not the same as (1).
SIMULTANEOUS QUADRATIC EQUATIONS 291
Thus, the equation (1) is not symmetrical with respect to x and y ; but
the method of Case 111 may often be used when either or both of the given
equations are symmetrical, except with respect to the signs of the terms.)
We may advantageously use the method of Case III to solve
certain equations which are not symmetrical with respect to x
and 2/; as, for example, the equations
ic  2 ?/ =  4.
iB2 + 42/' = 40.
EXERCISE 143
Solve the following equations by the symmetrical method
rx2 + 2/2 = 29.
2.
'I
5.
X +2/ =3.
05 — 2/ = 11.
a^ = 28.
g2+s2^130.
q s =8.
xy = 12.
x^y^ = 35.
,a?\xy^y'^ = l.
^ ar^2/3 = 26.
U  2/ = 2.
I xy =
8.
2n 1.
xy = n" — n — 2.
x^ \ xy \ y^ = 63.
x~y = S,
x"
10.
11.
•m:
16.
xy + f = a^{3b\
x\y = 2a.
a^ + 2/3^280.
[x^xy^y^ = 2S.
x^ + xy\y^ = l.
^ x^ — xy \ y^ = 19. ■
^3 + 71^ = 407.
+ 71=11.
x' + 9y' = 50.
x3y = 0.
xy = — 16.
I 2 a; + 2/ = 14.
36^2 + 64 2/' = 85.
ex + 8 2/ = 11.
x^Sy^ = lS9.
X 2y =9.
311. Case IV. When each equation is of the second degree,
and homogeneous; that is, ichen each term involving the unknown
numbers is of the second degree with respect to them (§ 65).
292 ALGEBRA
Certain equations of tliis form may be solved by the method of Case I
or Case III. (See Exs. 1, § 308, and 2, § 310.)
The method of Case IV should be used only when the example cannot
be solved by Cases I or III.
x'2xy =5. (1)
0^ + ^ = 29. (2)
Patting in the given equations y = vx,
Ex. Solve the equations \
we have x^ — 2vx^= 6: or, x^ = • • (3)
9Q
and «2 4. ^2x2 = 29 ; or, x"^
K oq
Equating values of x^, —  — = , or 5 v^ {■ 5S v = 2i.
1 — 2 ■« 1 + ^2
2
Solving this equation, ?j =  or — 12.
5
Substituting these values in (3), we have
^ or — ^ = 25or ; then, x=±5 or ± ^
1_4 1 + 24 5'  ^V6
5
Substituting the values of v and x in the equation y = vx,
2. , ^N _ ,0/ , 1 \_ , o..^i2_
V6
y = (±5) or 12/±— W±2orT
The solution isx = ±6, y =±2; or, a; = ± i^, y = T — •
In finding y from the equation y = vx, care must be taken to multiply
each value of x by the value of v which was used to obtain it.
The given equations may also be solved as follows :
Dividing (1) by (2), ^^~^^J^ = — , or 29 a;2  58 a;w = 5 x2 + 5 y2.
3.2 4. 2/2 29
Then, 5y2 + 58a;y  24a;2 = 0, or (5 2/  2x)(2/ + 12 x) = 0.
Placing 5y2x=:0, y = — ; substituting in (1),
5
a;2 _ 1^ = 5^ or x2 = 25.
5
Then, x = ± 5, and y = ?5= ±2.
5 '
SIMULTANEOUS QUADRATIC EQUATIONS 293
Placing 2/ + 12 tc = 0, y = V2x; substituting in (1),
ic2 4 24 x2 = 5, or x^ = •
5
Then, x = ±!, and ?/ = 12a: = T
V6 V5 ^
EXERCISE 144
Solve the following equations:
I aj2 + ^2 = 25, rp2^pg5g2 = 25.
fa;2 + 3a;?/ = 5. f i«2_ 2 a;?/ 4 2/^ =  41.
l2a^2/2/' = 24. ' [x'5xy {Sy' = 5S.
r 5x22/2 = 9. 2x2 + 7x2/ + 42/2 = 2.
lx?/3/ = 90. * l3x2 + 8x?/4/ = 72.
ra;2^^^^2/2 = 19. r4x22x2/2/' = 16.
I 2a^ + x2/ =2. * 1 5x'7xy = 36.
^' I Sxy + y' = 2S. ' iBa^Sxy 72y' = 3S.
312. Special Methods for the Solution of Simultaneous Equa
tions of Higher Degree.
No general rules can be given for examples which do not
come under the cases just considered; various artifices are
employed, familiarity with which can only be gained by
experience.
[ :x^f = 19. (1)
1. Solve the equations i „ <, ^ /on
[xy — xy^ = 6. {Z) ,
Multiply (2) by 3, Zx'^yS xy^ = 18. (3)
Subtract (3) from (1), x^Sx'^y + 3 xy'^ y^ = l.
Extracting cube roots, x — y = 1. (4)^
Dividing (2) by (4), xy = 6. (5)
Solving equations (4) and (5) by the method of § 310, we find x = S,
y = 2; or, X =  2, y =  3.
294 ALGEBRA
r* ^ 1 , . ( x^{y^ = 9xy.
2. Solve the equations i
[ x\y = 6.
Putting x = u + V and y = u — v,
(u + vy + (w  vy = 9(m + V) (u  v), or 2 m8 + 6 uv"^ = 9(^2  ^2) ; (1)
and ' (u \ v)\{u — v)=6,2u = 6, or u = 3.
Putting w = 3 in (1), 54 + 18 ^2 _ g^g _ ^2).
Whence, v"^ = 1, or y = ± 1.
Therefore, x = u + v = S±l = 4:or2;
and yz=u — v = ST'i=2 or 4.
The solution is x = i, y = 2 ; or, x = 2, y = i.
The artifice of substituting ii + v and u — v for x and y is advantageous
in any case where the given equations are symmetrical (§ 310) with
respect to x and y. See also Ex. 4.
3. Solve the equations I ' ^
I xy = i^. (2)
Multiplying (2) by 2, 2xy = \2. (3)
Add (1) and (3), a;2 + 2 a;?/ + ?/2 + 2 a; + 2 ?/ = 35.
Or, (a; + y)2 + 2(a: + y)=35.
Completing the square, (x + ?/)2 + 2(ic + y)+ 1 =36.
Then, (x + y) + 1 = ± 6 ; and a; + y = 5 or  7. (4)
Squaring (4) , x2 + 2 xy + ?/2 = 25 or 49.
Multiplying (2) by 4, 4a;y =24.
Subtracting, x^2xy + y'^ = \ or 25.
Whence, a;  y = ± 1 or ± 5. (5)
Adding (4) and (5), 2 x = 5 ± 1, or  7 i 5.
Whence, x = 3, 2, 1, or 6.
Subtracting (5) from (4), 2 ?/ = 5 T 1, or  7 T 5.
Whence, 2/ = 2, 3, 6, or 1.
The solution ls.x=3, y=2; x=2, ?/=3 ; x=l, 2/=6; or x=6,
?/ = !.
/I a 1 XT, .• fiC^ + 2/^ = 97.
4. Solve the equations \
Ix \y = — 1.
SIMULTANEOUS QUADRATIC EQUATIONS 295
Putting x=u + v and y = uv,
(u + vy \(u vy = 97, or 2 m* + 12 mV + 2 v* = 97,
and
Xu + v)\(u — v) = — l, 2w = — 1, or u =
(1)
Substituting value of w in (1),  + 3 ^2 _j. 2 ^;4  97.
Solving this,
^2 = 25 Qj, _31. ^^^ ^,::3i5> ^j. _j_
4 4 2
Then, a; = w + v = i±, or
2 2
and.
,^^_, = _1^, orlT
2
'^^31
^ = 2, 3, or
 3, 2, or
The solution is x = 2, ^
y =
31
1
31
3, y=.2; x =
_ _ 1 + v/^3l
V31
2
i±V
31
2
itV:
31
2
l+x/:
31
MISCELLANEOUS AND REVIEW EXAMPLES
EXERCISE 145
Solve the following equations :
r2a;2/ + ic = 36.
" [xy — Sy = —,5. ,
2.
"XL
3.
4.
xy = 6.
if + V = \\K
a^ — Sxy — 4:y^ = 0,
Sx — 5y =46.
rar^22/2 + 3a5 = 8.
1
^
= 74
1
1
= 12
. aj
V
2x_ _11
I. \y
9.
10
xy = l.
x = 
y
2 36
2/4 = —
= 17.
p^ + 2/' = 17
I a;y = 3.
11. P
4dhA;3dA; = 6.
5A; + 2(?A; = 10.
296
ALGEBRA
12.
13.
iT xy y^
X y
I ^ + \^y == 5.
/^
14.
15.
16.
llx^~xy — y^ = 45.
7x^ + 3xy2y^ = 20.
a^+4:Xy = lS,
[2xy + 9y^ = S7.
a^y^  24: xy^ 95 = 0,
Sx2y=lS.
(3x'5xy = 2a'\13ub~7b\
1 x + y = 3{ab).
18.
19.
20.
3xi2y 3x2y ^41
3x2y 3x + 2y 20*
Sy^ + 3x' = 29.
21
■I
22.
— = 6a2.
xy
.x\y = 5 axy.
i+J=^^^
ix^ y
' 6^ + 9^2 _^4e = 9.
e^42^ = 2.
i»3 42/' = 2a^.424a.
.a^^ + a;/ = 2a38a.
26.
27.
28.
23 I V2x^9 = 3y + 6.
1 V«^17/ = a;25.
' 3x^—xy— xz = 4:.
5x2y = l.
4a;H32 = — 5.
jc2/^ = 56.
+ 2/ = l.
30
31
■ {
■ 1
■ I
r
25.
«2 ^_19
^"^«~"6"*
1 + 1=1.
a; !/ 6
3a^ + 3/ = 10ci^.
1 + 1 = 1
X y 3
^y 4 2/^aj = 42.
1+1=1.
£c 2/ 6
5g2 + gs3s2 = 27.
4g2_4^g_^352^72.
y^\4:Xy — 3y = 4:2.
2y^ — xy + 5y = — 10.
16a^y^104:xy = 105.
xy=z2.
32^ {^^ + ^¥ + 2/^ = 481.
^ ' \ x"
I x
33.
aj2/ + / = 37.
9a^13ajy3a;=123.
a^2/+4/+22/=125.
* Divide the first equation by the second.
SIMULTANEOUS QUADRATIC EQUATIONS 297
34.
35.
36
37
38
7f xy Y
^=12.
xy f
. I x — 3y + z = 17.
[ x\y — 3z = 13.
1 xy = j\.
+ y) + xy =11.
yy+a^y^ = 61.
39.
40.
41,
42.
43.
xy{x — y) = — 20.
aj2 + 2/^ = a;2/ + l.
+ 2/^ = a^2/' + l.
r 05^ + 2/5 = 211.
1 a^ + 2/ = l.
0^2^ — a? = — 6.
ar^ = 72.
11 1 ^ a'+a^h''+h\
Q? xy y^ a^W
i_l_Ll=^
■a262+64
a^ fl??/ 2/^
a^^^
PROBLEMS IN PHYSICS
1. From the equations v = gt and S = ^gf, find v in terms
of S and gr.
^
find H in
2. From the equations C — — and ^(7
terms of C, i^, and t.
3. From the equations E = F'aS, F = ma, S = \ af, and 'y = at,
find ^ in terms of m and v.
313. Problems involving Simultaneous Equations of Higher
Degree.
In solving problems which involve simultaneous equations
of higher degree, only those solutions should be retained which
satisfy the conditions of the problem. (Compare § 176.)
EXERCISE 146 '^
^ 1. The difference of the squares of two numbers is 56, and
the difference of the numbers is ^ their sum. Find the numbers.
2. The sum of the squares of two numbers is 61, and the
product of their squares is 900. Find the numbers.
298 ALGEBRA
3. The product of the sum of two numbers by the smaller
is 21, and the product of their difference by the greater is 4.
Find the numbers.
} 4. The sum of the cubes of two numbers is 224 ; and if the
product of the numbers be subtracted from the sum of their
squares, the remainder is 28. Find the numbers.
5. Two numbers are expressed by the same two digits in
reverse order. The sum of the numbers equals the square of
the sum of the digits, and the difference of the numbers equals
5 times the square of the smaller digit. Find the numbers.
6. The square of the sum of two numbers exceeds their
product by 84 ; and the sum of the numbers, plus the square
root of their product, equals 14. Find the numbers.
7. The difference of the cubes of two numbers is 342 ; and
if the product of the numbers be multiplied by their difference,
the result is 42. Find the numbers.
8. A party at a hotel spent a certain sum. Had there been
5 more, and each had spent 50 cents less, the bill would have
been $24.75. Had there been 3 fewer, and each had spent
50 cents more, the bill would have been $ 9.75. How many
were there, and what did each spend ?
9. The simple interest of $700, for a certain number of
years, at a certain rate, is $ 182. If the time were 4 years less,
and the rate 1^% more, the interest would be $ 133. Find the
time and the rate.
10. If the digits of a number of two figures be inverted, the
quotient of this number by the given number is If, and their
product 1008. Find the number.
11. The square of the smaller of two numbers, added to twice
their product, gives 7 times the smaller number ; and the square
of the greater exceeds the product of the numbers by 6 times
the smaller number. Find the numbers.
12. A rectangular piece of cloth, when wet, shrinks onesixth
in its length, and onetwelfth in its width. If the area is
diminished by 12 square feet, and the length of the four sides
by 6^ feet, find the original dimensions.
SIMULTANEOUS QUADRATIC EQUATIONS 299
13. A and B travel from P to Q, 14 miles, at uniform rates,
B taking onethird of an hour longer than A. to perform the
journey. On the return, each travels one mile an hour faster,
and B now takes onefourth of an hour longer than A. Find
their rates of travelling.
14. A and B run a race of two miles, B winning by two
minutes. A now inct'eases his speed by two miles an hour,
and B diminishes his by the same amount, and A wins by two
minutes. Find their original rates.
15. A man ascends the last half of a mountain at a rate one
half mile an hour less than his rate during the first half, and
reaches the top in 3J hours. On the descent, his rate is one
mile an hour greater than during the first half of the ascent,
and he accomplishes it in 2i hours. Find the distance to the
top, and his rate during the first half of the ascent.
16. The square of the second digit of a number of three
digits exceeds twice the sum of the first and third by 3. The
sum of the first and second digits exceeds 4 times the third by
1 ; and if 495 be subtracted from the number, the digits will be
inverted. Find the number.
17. A ship has provisions for 36 days. If the crew were 16
greater, and the daily ration onehalf pound less, the provisions
would last 30 days ; if the crew were 2 fewer, and the daily
ration one pound greater, they would last 24 days. Find the
number of men, and the daily ration.
18. A man lends f 2100 in two amounts, at different rates of
interest, and the two sums produce equal returns. If the first
portion had been loaned at the second rate, it would have pro
duced $48; and if the second portion had been loaned at the
first rate, it would have produced $ 27. Find the rates.
19. A can do a piece of work in 2 hours less time than B ;
and together they can do the work in li hours less time than
A alone. How long does each alone take to do the work ?
300
ALGEBRA
GRAPHICAL REPRESENTATION OF SIMULTANEOUS QUAD
RATIC EQUATIONS WITH TWO UNKNOWN NUMBERS
314. 1. Consider the equation x^\y'^ = 25.
This means that, for any point on the graph, the square of the abscissa,
plus the square of the ordinate, equals 25.
But the square of the abscissa of any point,
plus the square of the ordinate, equals the
square of the distance of the point from the
origin ; for the distance is the hypotenuse of
a right triangle, whose other two sides are the
abscissa and ordinate.
Then, the square of the distance from of
any point on the graph is 25 ; or, the distance
from of any point on the graph is 5.
Thus, the graph is a circle of radius 5, having its centre at 0.
(The graph of any equation of the form x"^ {■ y'^ = a i^ di. circle.)
2. Consider the equation 2/^ = 4 a? + 4.
nx = 0, 2/2 = 4, or ?/=±2. {A,B)
ft _
If cc = 1, ?/2 = 8, or y = ± 2 V2. (C, D)
If x = \, y =0. (E)
etc.
The graph extends indefinitely to the right of
YT.
If X is negative, and < — 1, y^ is negative, and
therefore y imaginary ; then, no part of the graph lies to the left of E.
(The graph of Ex. 2 is 2^ parabola; as also is the graph of any equation
of the form y"^ = ax or y^ = ax { b.
The graphs of §§ 303 and 305 are parabolas.)
3. Consider the equation a;^ + 4 ?/^ = 4.
In this case, it is convenient to first
locate the points where the graph inter
sects the axes.
If i/ = 0.
x2 = 4,
or
X
±2.
{A, A')
If X = 0,
41/2 = 4,
or
y =
±1.
(B, B')
Putting X
= ±1, 4
[y2
= 3
2/2 =
J, or 2/:
or2/ = ±^. (C, A CD')
SIMULTANEOUS QUADRATIC EQUATIONS 301
If X has any value >2, or < — 2, y''^ is negative, and y imaginary ; then,
no part of the graph Ues to the right of A, or left of AK
If y has any value > 1, or <  1, x^ is negative, and x imaginary ; then,
no part of the graph lies above B^ or below B'.
(The graph of Ex. 3 is an ellipse ; as also is the graph of any equation
of the form ax^ ^ by^ = c.)
4. Consider the equation a^ 
Here, x"^  l = 2y^, oi y^ = ^^~^
2 7/^1
2
If x=±l, 2/2=0, or y=0. {A, A')
If X has any value between 1 and
— 1, 1/2 is negative, and y imaginary.
Then, no part of the graph lies be
tween A and A'.
If x=±2, 2/2 =, or y = ±\^
The graph has two branches, BAC and B'A'C, each of which extends
to an indefinitely great distance from O.
(The graph of Ex. 4 is a hyperbola ; as also is the graph of any equation
of the form ax^ — by^ = c, or xy — a.)
(B, C, B', C)
5. 4a^49/=36.
6. 4ic2_2^2^_4
EXERCISE 147
Plot the graphs of the following :
1. xy = Q. 3. a;2^2/' = 4.
2. x'^^y. 4. 2/2 = 5a;l.
315. Graphical Representation of Solutions of Simultaneous
Quadratic Equations.
r 2/^ = 4 ic.
1. Consider the equations i •
ydx — y = o.
The graph of ?/2 = 4 x is the parabola A OB.
The graph oiZx — y = b is the straight line AB^
intersecting the parabola at the points A and JB,
respectively.
To find the coordinates of A and J5, we proceed
as in § 184 ; that is, we solve the given equations.
The solution is x
(§ 309).
25
1, ^ = 2; or, x = —, y =
10
302
ALGEBRA
It may be verified in the figure that these are the coordinates of A and
B^ respectively.
Hence, if any two graphs intersect, the coordinates of any point
of intersection form a solution of the set of equations represented
by the graphs.
2. Consider the equations
(x^ + f = 17.
[ xy = 4:.
The graph of cc^ + ^/^ = 17 is the circle
AD, whose centre is at 0, and radius \/l7.
The graph of xy = 4: is a hyperbola,
having its branches in Ihe angles XOY
and X'OT, respectively, and intersecting
the circle at the points A and B in angle
XO F, and at the points C and D in angle X' T'.
The solution of the given equations is (§310),
X = 4:, y = l ; cc=l, ?/ = 4; ic = — 1, ?/= — 4; and 5C = — 4, y = —l.
It may be verified in the figure that these are the coordinates of A, B,
C, and Z>, respectively.
EXERCISE 148
Find the graphs of the following sets of equations, and in
each case verify the principle of § 315 :
9aj2f/=148.
xy=S.
2/2 = 29.
xy = 10.
raj2 44?/2 = 4.
I x — y = l.
J r x'4.y=7.
'' l2icf3^=4.
I 1'*
' 1 3x24/= 24.
g r 0.2.^2/^ = 13.
* [4:x9v = 6.
316. 1. Consider the equations
rx2f42/2 = 4. (1)
[2x + 3y = 5. (2)
The graph of a;^^ 4 2/2—4 jg h^q ellipse
AB.
The graph of 2x + 3?/ = — 5 is the
straight line CD.
SIMULTANEOUS QUADRATIC EQUATIONS 303
To solve the given equations, we have, from (2), x= ^ •
Substituting in (1), ^^' + '^^ + ^^ + 4y^ = ^.
Then, 25 ^/^ + 30 y + 9 = 0, or (5 ?/ + 3)(5 y + 3) = 0.
o g
This equation has equal roots; the only value of i/ is —  ; and x = •
5 o
The line has but one point in common with the ellipse, and is tangent
to it.
Then, if the equation obtained by eliminating one of the un
known numbers has equal roots, the graphs are tangent to each other.
2. Consider the equations
f9a;22/^ = 9. (1) \l
I x2y = ~2. \\y
The graph of 9 ic^ — ?/2 = — 9 is a hyperbola,
having its branches above and below XX', re
spectively.
The graph of x — 2y = — 2 is the straight line
AB.
To solve the given equations, we substitute
a; = 2?/2 in (1).
Then, 9(4 2/2_ 8 ?/ + 4)?/2 = 9,
or 35 ^2 _ 72 y + 45 = 0.
This equation has imaginanj roots, which shows that the line does not
intersect the hyperbola.
In general, if the equation obtained by eliminating one of the
unkiiown numbers has imaginary roots, the graphs do not intersect.
Exercise i49
Find the graphs of the following sets of equations, and in
each case verify the principles of § 316 :
\sxy = ^. ' I
x'y
5 if — 4 ?/ =
{fZx = S. I x'^ + f=\.
' I x + 2y^2. ' [2y'3x = 5.
304 ALGEBRA
XXIII. VARIABLES AND LIMITS
317. A variable number, or simply a variable, is a number
which may assume, under the conditions imposed upon it, an
indefinitely great number of different values.
A constant is a number which remains unchanged throughout
the same discussion.
318. A limit of a variable is a constant number, the differ
ence between which and the variable may be made less than
any assigned number, however small.
Suppose, for example, that a point moves from A towards
B under the condition that it shall move, during successive
equal intervals of time, first
from A to C, halfway between f ^ L___L__f
A and B', then to D, halfway
between C and B ; then to E, halfway between D and B ; and
so on indefinitely.
In this case, the distance between the moving point and B
can be made less than any assigned number, however small.
Hence, the distance from A to the moving point is a vari
able which approaches the constant value AB as a limit.
Again, the distance from the moving point to 5 is a variable
which approaches the limit 0.
319. Interpretation of •
Consider the series of fractions , — , — , ——, •••.
3' .3' .03' .003'
Here each denominator after the first is onetenth of the
preceding denominator.
It is evident that, by sufficiently continuing the series, the
denominator may be made less than any assigned number,
however small, and the value of the fraction greater than any
assigned number, however great.
VARIABLES AND LIMITS 305
In other words,
If the numerator of a fraction remains constant, while the
denominator approaches the limit 0, the value of the fraction
increases without limit.
It is customary to express this principle as follows :
a
The symbol oo is called Infinity ; it simply stands for that which is
greater than any number, however great, and has no fixed value.
320. Interpretation of — •
GO
Consider the series of fractions
3' 30' 300' 3000'
Here each denominator after the first is ten times the pre
ceding denominator.
It is evident that, by sufficiently continuing the series, the
denominator may be made greater than any assigned number,
however great, and the value of the fraction less than any
assigned number, however small.
In other words,
If the numerator of a fraction remains constant, while the
denominator increases without limit, the value of the fraction
approaches the limit 0.
It is customary to express this principle as follows :
^ = 0.
00
321. No literal meaning can be attached to such results as
a a p,
 = oo, or  = 0;
00
for there can be no such thing as division unless the divisor is
2ifi7iite number.
If such forms occur in mathematical investigations, they
must be interpreted as indicated in §§ 319 and 320. (Com
pare § 420.)
306 ALGEBRA
THE PROBLEM OF THE COURIERS
322. The following discussion will further illustrate the
form  , besides furnishing an interpretation of the form  •
The Problem of the Couriers.
Two couriers, A and B, are travelling along the same road in.
the same direction, RW, at the rates of m and n miles an hour,
respectively. If at any time, say 12 o'clock, A is at P, and B
is a miles beyond him at Q, after how many hours, and how
many miles beyond P, are they together ?
R P Q R
I J I I
Let A and B meet x hours after 12 o'clock, and y miles
beyond P.
They will then meet y — a miles beyond Q.
Since A travels mx miles, and B nx miles, in x hours, we
^^^® f y = mx,
y — a = nx.
Solving these equations, we obtain
m — 7i
We will now discuss these results under different hypotheses.
1. m>n.
In this case, the values of x and y are positive.
This means that the couriers meet at some time after 12, at
some point to the right of P.
This agrees with the hypothesis made ; for if m is greater
than n, A is travelling faster than B ; and he must overtake
him at some point beyond their positions at 12 o'clock.
2. m<n.
In this case, the values of x and y are negative.
This means that the couriers met at some time before 12, at
some point to the left of P. (Compare § 16.)
VARIABLES AND LIMITS 307
This agrees with the hypothesis made ; for if m is less than
n, A is travelling more slowly than B ; and they must have
been together before 12 o'clock, and before they could have
advanced as far as P.
3. a = 0, and m>n or m<n.
In this case, a? = and y = 0.
This means that the travellers are together at 12 o'clock, at
the point P.
This agrees with the hypothesis made ; for if a = 0, and m
and n are unequal, the couriers are together at 12 o'clock, and
are travelling at unequal rates ; and they could not have been
together before 12, and will not be together afterwards.
4. m = 71, and a not equal to 0.
In this case, the values of x and. 2/ take the forms  and — ,
respectively.
li m — n approaches the limit 0, the values of x and y increase
without limit (§ 319) ; hence, if m = n, no fixed values can be
assigned to x and y, and the problem is impossible.
In this case, the result in the form  indicates that the given
problem is impossible.
This agrees with the hypothesis made ; for if m = n, and a
is not zero, the couriers are a miles apart at 12 o'clock, and are
travelling at the same rate ; and they never could have been,
and never will be together.
5. m = n, and a = 0.
In this case, the values of x and y take the form —
^
If a = 0, and m = n, the couriers are together at 12 o'clock,
and travelling at the same rate.
Hence, they always have been, and always will be, together.
In this case, the number of solutions is indefinitely great ;
for any value of x whatever, together with the corresponding
value of y, will satisfy the given conditions.
In this case, the result in the form  indicates that the number
of solutions is indejinitely great.
808 , ALGEBRA
XXIV. INDETERMINATE EQUATIONS
323. It was shown, in § 163, that a single equation involving
two or more unknown numbers is satisfied by an indefinitely
great number of sets of values of these numbers.
If, however, the unknown numbers are required to satisfy
other conditions, the number of solutions may be finite.
We shall consider in the present chapter the solution of
indeterminate linear equations, in which the unknown numbers
are restricted to positive integral values.
324. Solution of Indeterminate Linear Equations in Positive
Integers.
1. Solve 7 ic + 5 2/ = 118 in positive integers.
Dividing by 6, the smaller of the two coefficients,
aj + 2^+2/ = 23 + ; or, ^J^=2Zxy.
Since, by the conditions of the problem, x and y must be positive inte
2 X 3
gers, must be an integer.
5
Let this integer be represented by p.
Then, ^^~^ = p, or2a;3 = 5p. . (1)
5
Dividing (1) by 2, x 1 i= 2i) H 1 ; or, xl2p = ^^.
Since x and p are integers, ic — 1— 2j5isan integer ; and therefore
r) 4 1
^^—  — must be an integer.
z
Let this integer be represented by q.
Then, 2_+i = gr, or p = 2q h
Substituting in (1), 2 ic  3 = 10 g'  5.
Whence, xhq\. (2)
INDETERMINATE EQUATIONS 309
Substituting this value in the given equation,
35 g  7 + 5 y = 118 ; or, y = 25  7 g. (3)
Equations (2) and (3) form the general solution in integers of the
given equation.
By giving to q the value zero, or any positive or negative integer, we
shall obtain sets of integral values of x and y which satisfy the given
equation.
If q is zero, or any negative integer, x will be negative.
If q is any positive integer >3, ?/ will be negative.
Hence, the only positive integral values of x and y which satisfy the
given equation are those obtained from the values 1, 2, 3 of g'.
That is, a; = 4, y = 18 ; x = 9, y = 11 ; and a: = 14, ?/ = 4.
2. Solve 5 a; — 7 2/ = 11 in least positive integers.
Dividing by 5, the coefficient of smaller absolute value,
5 5 6
2 w 4 1
Then, ^ ^ must be an integer.
5
Let ?^±^ =i? ; or, 2 ?/ + 1 = 5p.
5
Dividing by 2, i/+i = 2p+^; or, y2j) = ^=^.
Then, ^ "~ must be an integer.
Let ^ ~ = g ; or, jp = 2 g + 1.
Then, y = 5p1 ^ 10g + 5l ^^^^^_
Then, from the given equation, x = ^ "j" — = 7 g + 5.
5
The solution in least positive integers is when ^=0 ; that is, x=6,y=2.
3. In how many ways can the sum of f 15 be paid with
dollars, halfdollars, and dimes, the number of dimes being
equal to the number of dollars and halfdollars together ?
Let X = number of dollars,
y = number of halfdollars,
and z = number of dimes.
Adding,
llx + 6ij + z = 150 + z,
or,
nx + 6y = 150.
Dividing by 6,
x + ^+y = 25.
310 ALGEBRA
riOx+52/+2r=150,
By the conditions, 1
I a: + ?/ = 0. (1)
(2)
5 X
Then, — must be an integer ; or, x must be a multiple of 6.
6
Let X = 6p, where p is an integer.
Substitute in (2), 66p + 6y = 150, or ?/ = 25  lip.
Substitute in (1), s = 6p {26  Up = 25  bp.
The only positive integral solutions are when j? = 1 or 2.
Then, the number of ways is two ; either 6 dollars, 14 halfdollars, and
20 dimes ; or, 12 dollars, 3 halfdollars, and 15 dimes.
EXERCISE 150
Solve the following in positive integers :
1. Sx\5y = 29. 7. 2Sx\9y = Wl.
2. 7x\2y = S9. 8. 8 x + 71 2/ = 1933.
V 3. 6x\29y = 2T4:, ^ f 8ajll2/ + 2^ = 10.
4. 4a; + 3l2/ = 473. ' I 2x9 y{z = S.
5. 42i» + ll2/ = 664. ^^ r3a^32/H7;2 = 101.
6. 10a; + 72/ = 297. * [4.x^2y 3z = 5,
Solve the following in least positive integers :
11. 6x7y = 4:. 14. 8a;3l2/ = 10.
V^12. 5a;8?/ = 17. 15. SO x  IS y =. 115.
13. 14aj52/ = 64. 16. 15a!38 2/ = 47.
"^ 17. In how many different ways can f 1.65 be paid with
quarterdollars and dimes ?
18. In how many different ways can 41 shillings be paid
with halfcrowns, worth 2^ shillings each, and twoshilling
pieces ?
INDETERMINATE EQUATIONS 311
J
19. Find two fractions whose denominators are 5 and 7,
respectively, whose numerators are the smallest possible posi
tive integers, and whose difference is ^^.
20. In how many different ways can $7.15 be paid with
fiftycent, twentyfive cent, and twentycent pieces, so that
twice the number of fiftycent pieces, plus twice the number of
twentycent pieces, shall exceed Ihe number of twentyfive cent
pieces by 31 ?
21. A farmer purchased a certain number of pigs, sheep, and
calves for f 138. The pigs cost $ 4 each, the sheep f 7 each,
and the calves $9 each; and the whole number of animals
purchased was 23. How many of each did he buy ?
22. In how many ways can $ 10.00 be paid with twentyfive
cent, twentycent, and fivecent pieces, so that 3 times the num
ber of twentyfive cent pieces, plus 15 times the number of
twentycent pieces, shall exceed the number of fivecent pieces
by 33 ?
312 ALGEBRA
XXV. RATIO AND PROPORTION
RATIO
325. The Ratio of one nuin'ber a to another number & is the
quotient of a divided by h.
Thus, the ratio of a to 6 is  ; it is also expressed a : h,
b
In the ratio a:b, a is called the Jirst term, or antecedent^ and
h the second term, or consequent.
If a and b are positive numbers, and a>b,  is called a ratio
of greater inequality; if a is <b, it is called a ratio of less
inequality.
326. A ratio of greater inequality is decreased, and one of less
iyiequality is increased, by adding the same positive number to
each of its terms.
Let a and b be positive numbers, a being > b, and x a positive
number.
Since a>b, ax> bx. (§ 195)
Adding ab to both members (§ 192),
ab + ax >ab\ bx, or a(b + a?) > b{a f x).
Dividing both members by b(b + x), we have
^>P^' (§195)
b b\x
•»
In like manner, if a is < &,  < ^^+^.
' b b^x
PROPORTION
327. A Proportion is an equation whose members are equal
ratios.
RATIO AND PROPORTION 313
Thus, it a:b and c : d are equal ratios
is a proportion.
a:o = c:d, or  = .
b d'
328. In the proportion a:b = c:d, a is called the first term,
b the second, c the third, and d the fourth.
The first and third terms of a proportion are called the ante
cedents, and the second and fourth terms the consequeyits.
The first and fourth terms are called the extremes, and the
second and third terms the means.
329. If the means of a proportion are equal, either mean is
called the Mean Proportional between the first and last terms,
and the last term is called the Third Proportional to the first
and second terms.
Thus, in the proportion a:b = b:c,bis the mean proportional
between a and c, and c is the third proportional to a and b.
The Fourth Proportional to three numbers is the fourth term
of a proportion whose first three terms are the three numbers
taken in their order.
Thus, in the proportion a:b = c:d, d is the fourth propor
tional to a, b, and c.
330. A Continued Proportion is a series of equal ratios, in
which each consequent is the same as the next antecedent j as,
a:b = b:c = c:d=:d:e.
PROPERTIES OF PROPORTIONS
331. In any proportion, the product of the extremes is equal to
the product of the means.
Let the proportion be a:b = c:d.
a_c
b~d
Then by § 327,
Clearing of fractions, ad = be.
314 ALGEBRA
332. From the equation ad = be (§ 331), we obtain
be 7 ad ad, , ■, be
a = —y b = — , c = — , and a = — .
deb a
That is, 171 any proportion, either extx^rne_eguols^thejgrg^^
of the means divided by the otlier^xtreme^i^giuA^UJxBi^mEaji^^^
tisprodMe^of^the^sctrein^
333. (Converse of § 331.) If the product of tivo numbers be
equal to the product of two others, one pair may be made the
extremes, and the other pair the means, of a proportion.
Let
ad = bc.
Dividing by bd.
ad _hc . ci_c
hd^ii °^" b~d
Whence, by §[327j
a:b = c:d.
In like manner, we
may prove that
a:c = b:d,
c:d = a:b, etc.
334. In any proportion, the terms are in proportion by Alter
nation ; that is, the means can be interchanged.
Let the proportion be a : b = c : d.
Then by §331, ad = bc.
Whence, by § 333, a : c = 6 : d
In like manner, it may be proved that the extremes can be inter
changed. '^'\o^^'
335. In any proportion, the terms are in propoHion by Inver
sion ; that is, th,e second term is to the first as the fourth term is
to the third.
Let the proportion be a:b = c: d.
Then, by §331, ad=bc.
Whence, by § 333, b:a = d:c.
RATIO AND PROPORTION 315
It follows from § 335 that, in any proportion, the means can be written
as the extremes, and the extremes as the means.
336. The mean proportional between two numbers is equal to
the square root of their product.
Let the proportion be a:b — b:c.
Then by § 331, b" = ac, and b = Vac.
337. In any proportion, the terms are in proportion by Com
position; that is, the sum of the first two terms is to the first
term as the sum of the last two terms is to the third term.
Let the proportion be a:b = c:d.
Then, ad = be.
Adding each member of the equation to ac,
ac \ ad = ac { be, or a(c f c?) = c(a + b).
Then by §333, a\b : a = c]d: c.
We may also prove a + b : b = c + d : d.
338. In any proportion, the terms are in proportion by Division;
that is, the difference of the first two terms is to the first term as
the difference of the last tivo terms is to the third term.
Let the proportion be a : b = c : d.
Then, ad = be.
Subtracting each member of the equation from ac,
ac — ad = ac — be, or a(c — d) — c(a — b).
Then, a — b : a = c — d:c.
We may also prove a — h : b = c — d : d.
339. hi an?/ proportion, the terms are in proportion by Com
position and Division ; that is, the sum of the first two terms is
to their difference as the sum of the last tivo terms is to their
difference.
Let the proportion be a:b =c:d.
316 ALGEBRA
Then by § 337, ^^±^ = ^±^. (1)
a c ^ ^
And by § 338, ^^^ = ^—^^ (2)
a c ^
Dividing (1) by (2),
a \b c\ d
a—b c—d
Whence, a\b:a — b = c\d:c — d.
340. In any propoHion, if the first two terms be multiplied by
any number, as also the last two, the resulting numbers will be in
proportion.
Let the proportion be  =  ; then, ^ = ^.
b d mb nd
(Either m or n may be unity ; that is, the terms of either ratio may be
multiplied without multiplying the terms of the other.)
341. I7i any proportion, if the first and third terms be multi
plied by any number, as also the second and fourth terms, the
resulting numbers will be m proportion.
Let the proportion be  =  ; then, 'B^^'^.
b d nb nd
(Either to or n may be unity. )
342. In any number of proportions, the products of the cor
responding terms are in proportion.
Let the proportions be  = , and  = 2.
^ ^ b d' f h
Multiplying, ^x^ = ^xf, or^^ = ^.
b f d h bf dh
In like manner, the theorem may be proved for any number
of proportions.
343. In any proportion, like powers or like roots of the terms
are in proportion.
R ATIO AND PROPORTION 317
■ ■ » I I n ' I r I r ■ ■ ' ■ ■ '^^ • •■' r t
Let the proportion be  =  ; then, — = — .
T Tl 'Vet \/g
In like manner, = •
V6 ^/d
344. In a series of equal ratios, any antecedent is to its con
sequent as the sum of all the ayitecedents is to the sum of all the
consequents.
Let a:b = c:d = e:f
Then by §331, ad = bc,
and af=be.
Also, ab = ba.
Adding, a(b{d{f) = b(a + c\e).
Whence,* a:b = a\c\e:b\d+f (§333)
In like manner, the theorem may be proved for any number
of equal ratios.
345. If three numbers are in continued proportion, the first is
to the third as the square of the first is to the square of the second.
Let the proportion be a: b = b :c; oi  = 
b c
Then, a b_a a ,a^^
b c b b c ¥
346. If four numbers are in continued proportion, the first is
to the fourth as the cube of the first is to the cube of the second.
Let the proportion be a:b = b : c = c: d; ov  =  = .
bed
Then, ^x^^ = ^X^X^, or^ = ^.
b Q d b b b d b^
318 ALGEBRA
347. Examples.
1. li x:y = (xizy: (y { zy, prove z the mean proportional
between x and y.
From the given proportion, by § 331,
yix + 0)2 = x(y + zy.
Or x"^ + 2 xyz + yz^ = a;^^ _(_ 2 aj?/^ 4 x^!^.
Transposing, x^y — xy^ = xz'^ — xjz'^.
Dividing by aj — ?/, xy = z^.
Therefore, z is the mean proportional between x and y (§ 336).
The theorem of § 339 saves work in the solution of a certain
class of fractional equations.
± Sol^e the equation ?^± = ?^^:i^.
^ 2x3 26+a
Regarding this as a proportion, we have by composition and division,
4x 46 ^„2x 26 ^i,^„„^ „ 3 6
— = , or — = ; whence, a; =
6 2a' 3 a a
3. Prove that if  = t, then *
h d
(^b'':a'3ab = c'd':c'3cd.
Let  =  = X, whence, a = 6a; ; then,
6 cZ c2_^
^2 _ ^2 ;,2a;2 _ 52 y.2 _ 1 (Z^ C2  (^2
a2  3 «6 62x2  3 62x x'^  3 x «! _ 3c c^ 3cd
d^^ d
Then, a2 _ ^2 . ^2 _ 3 ^^5 = c2  cZ2 : c2  3 cd
EXERCISE 151
1. Find the mean proportional between 18 and 32.
2. Find the third term of a proportion whose first, second,
and fourth terms are 24, 32, and 20, respectively.
3. Find the third proportional to J^ and ^.
4. Find the mean proportional between 1^ and 24.
RATIO AND PROPORTION 319
5. Find the fourth proportional to 4, 5f, and 1f.
6. Find the third proportional to ci^ + 8 and a + 2.
7. Find the mean proportional between
— and ^
x — 5 x\3
Solve the following equations :
SxS ^ 2x5 ^Q ^^2x^^Sx±2^
' 3x + 4:'~2x\7' ' x^2x3 3x2
12.
9 4a; + 7 ^ 7a4l n. «; + V3a;1 ^ a^ V2a; + 1 ^
• 4a;7 5a3* ' a^V3a;l a^+V2a; + l
a;4y _ a — 6
x — y a + 6 ♦
0? — g'^ _ 6^ + y ^
a? + a^ h^ — y
13. Find two numbers in the ratio 4 to 3, such that the dif
ference of their squares shall be 112.
14. Find two numbers such that, if 9 be added to the first,
and 7 subtracted from the second, they will be in the ratio 9:2;
while if 9 be subtracted from the first, and 7 added to the second,
they will be in the ratio 9 : 11.
15. Find two numbers in the ratio a : &, such that, if each be
increased by c, they shall be in the ratio m : n.
16. Find three numbers in continued proportion whose sum
is \7, such that the quotient of the first by the second shall
be.
17. What number must be added to each of the numbers a,
6, and c, so that the resulting numbers shall be in continued
proportion ?
18. Find a number such that, if it be subtracted from each
term of the ratio 8 : 5, the result is \\ of what it would have
been if the same number had been added to each term.
320 ALGEBRA
19. The second of two numbers is the mean proportional be
tween the other two. The third number exceeds the sum of
the other two by 20 ; and the sum of the first and third exceeds
three times the second by 4. Find the numbers.
20. If 8a — 56:7 a — 46 = 86 — 5c: 7 6 — 4 c, prove c the
third proportional to a and 6.
21. If ma{nb: pa — qb = mc indipc — qd, prove a:b = c:d.
22. If x{y:y ]z = Vxr — y^ : V^/^ — z^, prove y the mean
proportional between x and z.
23. Given (2 0^+2 ab)x\(a^+2b^)y=(a'b')x\{2a^\b^)y,
find the ratio of x to y.
24. If 4 silver coins and 11 copper coins are worth as much
as 2 gold coins, and 5 silver coins and 19 copper coins as much
as 3 gold coins, find the ratio of the value of a gold coin, and
the value of a silver coin, to the value of a copper coin.
TO a c
6 a
25. 3a446:3a46 = 3c + 4d:3c4d
26. d'5ab:2ab\7b^ = c'5cd:2cd + 7d\
27. a' + 6ab':a'bSb'=c'{6cd':c'dSd^
28. Each of two vessels contains a mixture of wine and
water. A mixture consisting of equal measures from the two
vessels is composed of wine and water in the ratio 3:4; another
mixture consisting of 2 measures from the first and 3 measures
from the second, is composed of wine and water in the ratio
2 : 3. Find the ratio of wine to water in each vessel.
VARIATION 321
J^XXVI. VARIATION
348. One variable number (§ 317) is said to vary directly as
another when the ratio of any two values of the first equals
the ratio of the corresponding values of the second.
It is usual to omit tlie word "directly," and simply say that one
number varies as another.
Thus, if a workman receives a fixed number of dollars per
diem, the number of dollars received in m days will be to the
number received in n days as m is to n.
Then, the ratio of any two numbers of dollars received equals
the ratio of the corresponding numbers of days worked.
Hence, the number of dollars which the workman receives
varies as the number of days during which he works.
349. The symbol oc is read ^^ varies as^' ; thus, accb is read
" a varies as 6."
350. One variable number is said to vary inversely as another
when the first varies directly as the reciprocal of the second.
Thus, the number of hours in which a railway train will
traverse a fixed route varies inversely as the speed; if the
speed be doubled, the train will traverse its route in onehalf
the number of hours.
351. One variable number is said to vary as two others jointly
when it varies directly as their product.
Thus, the number of dollars received by a workman in a
certain number of days varies jointly as the number which he
receives in one day, and the number of days during which he
works.
352. One variable number is said to vary directly as a sec
ond and inversely as a third, when it varies jointly as the
second and the reciprocal of the third.
322 ALGEBRA
Thus, the attraction of a body varies directly as the amount
of matter, and inversely as the square of the distance.
353. If xtjzy, then x equals y multiplied by a constant number.
Let x^ and y' denote a fixed pair of corresponding values of
X and y, and x and y any other pair.
By the definition of § 348, = ,\ or, a; = 7 y.
J ' y y" ' y'^
x'
Denoting the constant ratio — , by m, we have
x = my.
354. It follows from §§ 350, 351, 352, and 353 that :
1. If X varies inversely as y^ x = —
2. Ifx varies jointly as y and z, x = myz.
my
3. Ifx varies directly as y and iiiversely as z, x = — •
355. If xccy, and y^z, then xocz.
By § 353, ii xccy, x = my. (1)
And if 2/ QC 2, y = nz.
Substituting in (1), x = mnz.
Whence, by § 353, xazz.
356. If xccy when z is constant, and xgcz when y is constant,
then xccyz when both y and z vary.
Let y' and z' be the values of y and z, respectively, when x
has the value x'.
Let y be changed from y' to y", z remaining constantly equal
to z', and let x be changed in consequence from x' to X,
Then by §348, J=^. (1)
Now let z be changed from z' to z", y remaining constantly
equal to ?/'', and let x be changed in consequence from X to x".
VARIATION 323
Then, =^^. (2)
Multiplying (1) by (2), £; = lg. (3)
Now if both changes are made, that is, y from ?/' to 2/" and
z from z' to z'\ x is changed from x' to ic", and yz is changed
from y'z' to 2/"2!".
Then by (3), the ratio of any two values of x equals the
ratio of the corresponding values of yz ; and, by § 348, x oc yz.
The following is an illustration of the above theorem :
It is known, by Geometry, that the area of a triangle varies as the base
when the altitude is constant, and as the altitude when the base is
constant ; hence, when both base and altitude vary, the area varies as
their product.
357. Problems.
Problems in variation are readily solved by converting the
variation into an equation by aid of § § 353 or 354.
1. If cc varies inversely as y, and equals 9 when 2/ = 8, find
the value of x when y = l^.
If X varies inversely as y, x = — (§ 354).
Putting cc = 9 and y = 8, 9 =  , or m = 72.
8
Then, a; = ^ ; and, if y = 18, a: = ^ = 4.
y 18
2. Given that the area of a triangle varies jointly as its base
and altitude, what will be the base of a triangle whose altitude
is 12, equivalent to the sum of two triangles whose bases are
10 and 6, and altitudes 3 and 9, respectively ?
Let B^ H, and A denote the base, altitude, and area, respectively, of
any triangle, and B' the base of the required triangle.
Since A varies jointly as B and H, A = mBH (§ 354).
Therefore, the area of the first triangle is m x 10 x 3, or 30 m, and the
area of the second is m x 6 x 9, or 54 m.
Then, the area of the required triangle is 30 w f 54 m, or 84 m.
324 ALGEBRA
But, the area of the required triangle is also m y. B' x 12.
Therefore, 12 mB' = S4m, or B' =7,
EXERCISE 152
1. li y ccx, and x equals 6 when y equals 54, what is the
value of y when x equals 8 ?
2. If X varies inversely as y, and equals f when y = %, what
is the value of y when x — ^?
3. li ycc z^, and equals 40 when z = 10, what is the value of
y in terms of z^ ?
4. If z varies jointly as x and y, and equals f when a; = 
and 2/ = > what is the value of z when x= ^ and y = ^?
5. If X varies directly as y and inversely as z, and equals ^^
when y — 21 and 2! = 64, what is the value of x when 2/ = 9 and
2 = 32?
6. If x^ oc 2/^, and a; = 4 when 2/ = 4, what is the value of y
when a; =  ?
7. If 5 a; + 8 oc 6 2/— 1? and a;=6 when y=—3, what is the
value of X when y = 7?
8. The surface of a cube varies as the square of its edge. If
the surface of a cube whose edge is  feet is ^^ square feet, what
will be the edge of a cube whose surface is ^^ square feet ?
9. If 5 men in 6 days earn $ 57, how many days will it take
4 men to earn $ 76 ; it being given that the amount earned
varies jointly as the number of men, and the number of days
durjng which they work.
10. The volume of a sphere varies jointly as its diameter
and surface. If the volume of a sphere whose diameter is a,
and surface b, is c, what is the diameter of a sphere whose sur
face is p and volume q ?
11. The distance fallen by a body from rest varies as the
square of the time during which it falls. If it falls 579 feet
in 6 seconds, how long will it take to fall 402 J^ feet ?
VARIATION 325
12. A circular plate of lead, 17 inches in diameter, is melted
and formed into three circular plates of the same thickness.
If the diameters of two of the plates are 8 and 9 inches respec
tively, find the diameter of the other; it being given that the
area of a circle varies as the square of its diameter.
13. If y equals the sum of two numbers which vary directly
as x^ and inversely as x, respectively, and 2/ equals —53 when
X equals — 3, and ^^ when x equals 2, what is the value of y
when X equals ^^ ?
14. If X equals the sum of two numbers, one of which varies
directly as 2/^ and the other inversely as z^, and a? = 45 when
y = l and z = l, and ic = 40 when y = 2 and 2; = 3, find the value
of y when ic = 37 and z—1.
15. If y equals the sum of three numbers, the first of which
is constant, and the second and third vary as a?^ and a?j respec
tively, and ?/ = — 50 when « = 2, 30 when x = — 2, and 110
when a; = — 3, find the expression for y in terms of x.
16. The volume of a circular coin varies jointly as its thick
ness and the square of the radius of its face. Two coins whose
thicknesses are 5 and 7 units, and radii of faces 60 and 30
units, respectively, are melted and formed into 100 coins, each
3 units thick. Find the radius of the face of the new coin.
17. The weight of a spherical shell, 2 inches thick, is ^f of
its weight if solid. Find its diameter, it being given that the
volume of a sphere varies as the cube of its diameter.
PROBLEMS IN PHYSICS
1. When the force which stretches a spring, a straight wire,
or any elastic body is varied, it is found that the displacement
produced in the body is always directly proportional to the
force which acts upon it; i.e., if dj and dg represent any two
displacements, and j^ and /g respectively the forces which pro
duce them, then the algebraic statement of the above law is
■=^ (1)
326 ALGEBRA
If a force of 2 pounds stretches a given wire .01 inch, how
much will a force of 20 pounds stretch the same w^ire ?
2. If the same force is applied to two wires of the same
length and material, but of different diameters, D^ and D2, then
the displacements di and d^ are found to be inversely propor
tional to the squares of the diameters, i.e..
If a weight of 100 kilograms stretches a wire .5 millimeter
in diameter through 1 millimeter, how much elongation will
the same weight produce in a wire 1.5 millimeters in diameter ?
3. If the same force is applied to two wires of the same
diameter and material, but of different lengths, l^ and I2, then it
is found that , 7
f = f (3)
Erom (1), (2), and (3) and § 356, it follows that when lengths,
diameters, and forces are all different,
^ = ^ X  X ^^' (4)
^2 J2 ^2 ^1
If a force of 1 pound will stretch an iron wire which is
200 centimeters long and .5 millimeter in diameter through
1 millimeter, what force is required to stretch an iron wire
150 centimeters long and 1.25 millimeters in diameter through
.5 millimeter ?
4. When the temperature of a gas is constant, its volume
is found to be inversely proportional to the pressure to which
the gas is subjected, i.e., algebraically stated,
i^ = S. (5)
At the bottom of a lake 30 meters deep, where the pressure
is 4000 grams per square centimeter, a bubble of air has a vol
ume of 1 cubic centimeter as it escapes from a diver's suit. To
what volume will it have expanded when it reaches the surface
where the atmospheric pressure is about 1000 grams per square
centimeter ?
VARIATION 327
5. The electrical resistance of a wire varies directly as its
length and inversely as its area. If a copper wire 1 centimeter
in diameter has a resistance of 1 unit per mile, how many units
of resistance will a copper wire have which is 500 feet long
and 3 millimeters in diameter?
6. The illumination from a source of light varies inversely
as the square of the distance from the source. A book which
is now 10 inches from the source is moved 15 inches farther
away. How much will the light received be reduced ?
7. The period of vibration of a pendulum is found to vary
directly as the square root of its length. If a pendulum 1 meter
long ticks seconds, what will be the period of vibration of a
pendulum 30 centimeters long ?
8. The force with which the earth pulls on any body out
side of its surface is found to vary inversely as the square of
the distance from its center. If the surface of the earth is
4000 miles from the center, what would a pound weight weigh
15,000 miles from the earth ?
9. The number of vibrations made per second by a guitar
string of given diameter and material is inversely proportional
to its length and directly proportional to the square root of
the force with which it is stretched. If a string 3 feet long,
stretched with a force of 20 pounds, vibrates 400 times per
second, find the number of vibrations made by a string 1 foot
long, stretched by a force of 40 pounds.
GRAPHS IN PHYSICS
1. Graphical representation of a direct proportion.
When a man is running at a constant speed, the distance
which he travels in a given time is directly proportional to
his speed. The algebraic expression of this relation is
^ = h^ ov d = ms. (See § 353.)
^2 ^2
328
ALGEBRA
Now, if we plot successive values of the distance, d, which
correspond to various speeds, s, in precisely the same manner
in which we plotted successive
values of x and y in % 181, we
obtain as the graphical picture of
the relation between s and d a
straight line passing through the
origin. (See Fig. 1.)
This is the graph of any direct
proportion.
Fig. 1.
2. Graphical representation of an inverse proportion.
The volume which a given body of gas occupies when the
pressure to which it is subjected varies has been found to be
inversely proportional to the pressure under which the gas
stands; we have seen that the algebraic statement of this
relation is — i = ~
If we plot successive values of V and P in the manner indi
cated in § 181, we obtain a graph of the form shown in Fig. 2.
This is the graphical representation
of any inverse j^roportion; the curve
is called an equilateral hyperbola.
3. The path traversed by a falling
body projected horizontally.
When a body is thrown horizontally
from the top of a tower, if it were not
for gravity, it would move on in a
horizontal direction indefinitely, trav
ersing ' exactly the same distance in
each succeeding second.
Hence, if V represents the velocity
of projection, the horizontal distance,
H, which it would traverse in any
number of seconds, t, would be given
by the equation H= Vt.
or V= — .
V=1,
V=2,
V= 3,
P='
3*
F= 4, F=
r=
F=
F=
P
5, P
6, P
T, P:
8, P
VARIATION
329
On account of gravity, however, the body is pulled down
ward, and traverses in this direction in any number of seconds
a distance which is given by the equation S = ^ gt^.
To find the actual path taken by the body, we have only to
plot successive values of H and S, in the manner in which we
plotted the successive values of x and y, in § 181.
Thus, at the end of 1 second the vertical distance S^ is
given hj Si = g xl^ = g; at the end of 2 seconds we have
g ; at the end of 3 seconds, S^ = g y.2? = g\
etc.
1 j /»
at the end of 4 seconds, /S'4 = ^x4^ = — ^
On the other hand, at the end
of 1 second we have ^i=F; at
the end of 2 seconds, ^2 = 2 F; at
the end of 3 seconds, ^3 = 3 F;
at the end of 4 seconds, ^4 = 4 F
If, now, we plot these successive
values of H and >S', we obtain the
graph shown in Fig. 3.
This is the path of the body; it
is a parabola. (§ 314, Ex. 2.)
4. Orapli of relation between the
temperature and pressure existing
within an airtight boiler containing only water and water vapor.
One use of graphs in physics is to express a relation which
is found by experiment to exist between two quantities, which
cannot be represented by any simple algebraic equation.
For example, when the temperature of an airtight boiler
which contains only water and water vapor is raised, the pres
sure within the boiler increases also ; thus we find by direct
experiment that when the temperature of the boiler is 0° centi
grade, the pressure which the vapor exerts will support a
column of mercury 4.6 millimeters high.
Fig. 3.
330
ALGEBRA
When the temperature is raised to 10°, the mercury column
rises to 9.1 millimeters; at 30° the column is 31.5 milli
meters long, etc.
To obtain a simple and compact picture of the relation
between temperature and pressure, we plot a succession of
temperatures, e.g. 0°, 10°, 20°, 30°, 40°, 50°, 60°, 70°, 80°, 90°,
100°, in the manner in which
we plotted successive values of
a; in § 181, and then plot the
corresponding values of pres
sure obtained by experiment in
the manner in which w^e plotted
the ^'s in § 181 ; we obtain the
graph shown in Fig. 4.
From this graph we can find
at once the pressure which will
exist within the boiler at any
temperature.
For example, if we wish to know the pressure at 75° cen
tigrade, we observe where the vertical line which passes
through 75° cuts the curve and then run a horizontal line
from this point to the point of intersection with the line OP.
This point is found to be at 288 ; hence the pressure within
the boiler at 75° centigrade is 288 millimeters.
70 80 90 100
PROGRESSIONS 331
^^ XXVII. PROGRESSIONS
ARITHMETIC PROGRESSION
358. An Arithmetic Progression is a series of terms in which
each term, after the first, is obtained by adding to the preced
ing term a constant number called the Common Difference.
Thus, 1, 3, 5, 7, 9, 11, ••• is an arithmetic progression in
which the common difference is 2^
Again, 12, 9, 6, 3, 0, —3, ••• is an arithmetic progression in
which the common difference is — 3.
An Arithmetic Progression is also called an Arithmetic Senes.
359. Given the first term, a, the common difference, d, and the
number of terms, n, to find the last term, I.
The progression is a, a h c?, a + 2 d, a + 3 cZ, •••.
We observe that the coefficient of d in any term is less by 1
than the number of the term.
Then, in the nth term the coefficient of d will be n — 1.
That is, l = a\{nl)d. (I)
360. Given the first term, a, the last term, I, and the number of
terms, n, to find the sum of the terms, S.
8 = a + {a + d) + {a+2d)\"'^{ld){L
Writing the terms in reverse order,
S = l + {ld) + il2d) + '.' + {a + d)+a.
Adding these equations term by term,
2^=(a + + (a + + (« + +  + (« + + (« + 0
Therefore, 2S^n(a + l), and S = '^(a + l), (II)
361. Substituting in (II) the value of I from (I), we have
5 = [2a + («l)d].
332 ALGEBRA
362. Ex. Ill the progression 8, 5, 2, —1, —4, •.., to 27
terms, find the last term and the sum.
Here, a = 8, (^ = 5  8 =  3, w = 27.
Substitute in (I), z = 8 + (27  1) ( ^ 3) = 8  78 =  70.
Substitute in (II), ,S' = — (8  70) = 27 ( 31) =  837.
The common difference may be found by subtracting the first term
from tlie second, or any term from the next following term.
EXERCISE 153
In each of the following, find the last term and the sum :
1. 4, 9, 14, .. to 14 terms. 2. 9, 2, 5, ••• to 16 terms.
3.  51,  45,  39, ... to 15 terms.
4. I, V, 3, ... to 13 terms.
5. hh h '" "to 18 terms. 6. , JJ, fi ... to 17 terms.
* 7. V, \%y 1, — to 27 terms.
8 TO? i Ii ••• to 52 terms.
9. 3a + 4&, 8a + 26, 13a, ...to 10 terms.
10. ^Zl^I, ^, ^^2/, ...to 9 terms.
3 ' 6' 3 '
363. The first term, common difference, number of terms, last
term, and sum of the terms, are called the elements of the
progression.
If any three of the five elements of an arithmetic progres
sion are given, the other two may be found by substituting the
known values in the fundamental formulae (I) and (II), and
solving the resulting equations.
1. Given a = — f , n = 20, /S = — f ; find d and I.
Substituting the given values in (II),
_§^10f^4?V orl = ^+Z; then, I = ^ 
3 V 3 / 6 3 ' ' 3 6
3
6 2
PROGRESSIONS 333
3 5
Substituting tlie values of a, w, and I in (I),  = — + 19 ci.
2 3
Whence, i9dJ = § + ^ = ^, and <? = 1.
' 2 3 6 6
2. Given d = 3, Z = 39, /S^ = 264; find a and n.
Substituting in (I),  39 = a + (n  1)( 3), or a = 3 w  42. (1)
Substituting tlie values of Z, /S", and a in (II) ,
 264 =  (3 n  42  39), or  528 = 3 n^  81 n, or ii^ _ 27 w + 176 = 0.
2i
„., 27 ± V729  704 27 ± 5 ,^^11
Whence, n = — — = — ='^— = 16 or 11.
' 2 2
Substituting in (1), a = 48  42 or 33  42 = 6 or  9.
The solution is a = 6, n = 16 ; or, a = — 9, n = 11.
The significance of the two answ^ers is as follows :
If a = 6 and n = 16, the progression is 6, 3, 0, — 3, — 6, — 9, — 12,
15, _18, 21, 24, 27, 30, 33, 36, 39.
If a = — 9 and w = 11, the progression is
9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39.
In each of these the sum is — 264.
3: Given a=^, d = — ^^y ^ = —  ; find I and n.
Substitutmg in (I), Z = 1 + (71  1) (  ^) = ^^ (1)
Substituting the values of a, S, and I in (II),
i(h^)' '' ' = "('12^)' '' ^936 = 0.
Whence, ^ ^ 9 j: VsTTU? ^ 9^15 ^ 1, ^, _3^
2 2
The value w = — 3 must be rejected, for the number of terms in a
progression must be a positive integer.
512 7
Substituting w = 12 in (1), 1 =
12 12
A negative or fractional value of n must be rejected, together with all
other values dependent on it.
334 ALGEBRA
EXERCISE 154
1. Given d = S, ^ = 115, n = 15; find a and S.
2. Given d = 6, 71 = 14, aS = 616; find a and L
3. Given a = 69, n = 16, Z = 36; find d and S.
4. Given a =8, ?i = 25, >S = 2500; find d and ?.
5. Given a = f, l = \\ /S' = 78; find d and 71.
6. Given Z = H^, n = 13, AS = 2^f^; find a and d
7. Given a = — f, d = — ^, /S' = — ^^; find n and I
8. Given a = — f, 1 = ^, <^ = ; find n and >S^.
9. Given d = — ^, n = 55, S — — 165; find a and Z.
10. Given ? = ^V, n = 24, iS = 241; find a and c?.
11. Given Z = 9/, d = , .^ = ^f^; find a and n.
12. Given a = ^, l = U, S = ^^; find c? and n.
13. Given a — — ^2, n = 21, /S" =  ; find c? and I.
14. Given ? = f, d = ^, 8 = ^^^; find a and n.
15. Given a = %S d = i aS = ^; fiii^ n and /.
364. From (I) and (II), general formulce for the solution of
examples like the above may be readily derived.
Ex. Given a, d, and ^iS ; derive the formula for n.
By § 361, 2 ^S' = ri[2 a + (n  l)c?], or dn"^ \ {2 a  d)n = 2 S.
This is a quadratic in n, and may be solved by the method of § 288 ;
multiplying by 4 d, and adding (2 a — dy to both members,
4 d2n2 + 4 d(2 a  (^)?z + (2 a  (?)2 = 8 d^ + (2 a  d)2.
Extracting square roots, 2dn + 2a — <?= ± V8 d/S" + (2 a — d)^.
Whence, n = d 2a ± VSdS ^ {2a  d)\.
2d
EXERCISE 155
1. Given a, I, and n ; derive the formula for d.
PROGRESSIONS 335
2. Given a, n, and S; derive the formulae for d and I.
3. Given d, n, and S ; derive the f ormulie for a and I. ,
4. Given a, dj and / ; derive the formulae for n and S.
5. Given d, I, and n ; derive the formulae for a and JS.
6. Given I, n, and ^ ; derive the formulae for a and d.
7. Given a, d, and /iS ; derive the formula for I.
8. Given a, Z, and ^S ; derive the formulae for d and n.
9. Given d, I, and aS; derive the formulae for a and n.
365. Arithmetic Means.
We define inserting m arithmetic means between two given
numbers, a and b, as finding an arithmetic progression of m + 2
terms, whose first and last terms are a and b.
Ex. Insert 5 arithmetic means between 3 and — 5.
We find an arithmetic progression of 7 terms, in which a = 3, and
l = ~5; substituting n = 7, a = 3, and Z = — 5 in (I),
5 = 3 + 6^, or d = ^
o
The progression is 3, , ^, 1, I ^, 5.
o o o o
366. Let X denote the arithmetic mean between a and b.
Then, x — a = b — x, or 2x = a\b.
Whence, x = ^^±5 .
' • 2
That is, the arithmetic mean between two numbers equals one
half their sum.
EXERCISE 156
1. Insert 7 arithmetic means between 4 and 10.
2. Insert 6 arithmetic means between —  and — Y
3. Insert 9 arithmetic means between — ^ and 6.
336 ALGEBRA
4. Insert 8 arithmetic means between — 3 and — ^.
5. Insert 5 arithmetic means between f and — ^.
6. How many arithmetic means are inserted between — 
and j^, when the sum of the second and last is f?
7. If m arithmetic means are inserted between a and b, find
the first two.
Find the arithmetic mean between :
8. V a^^  ¥• 9. (3 m + nf and (m  3 nf.
367. Problems.
1. The sixth term of an arithmetic progression is f , and the
fifteenth term is ^. Find the first term.
By § 359, the sixth term is a + 5 d, and the fifteenth term a\ lid.
o
\a + Ud = f
(1)
Then by the conditions,
(2)
Subtracting (1) from (2),
9d = ; whence, d = .
2' 2
Substituting in (1),
^"^i"i'^^^^°^'^"~i'
2. Find four numbers in arithmetic progression such that
the product of the first and fourth shall be 45, and the product
of the second and third 77.
Let the numbers be x — Sy, x — y, x + y, and x + Sy.
 9 ?/2 = 45.
Then by the conditions, , , „ „„
Solving these equations, x = 9, y =±2 ; or, a; = — 9, ?/ = ± 2 (§ 308).
..Then the numbers are 3, 7, 11, 15; or, —3, —7, —11, —15.
In problems like the above, it is convenient to represent the unknown
numbers by symmetrical expressions.
Thus, if five numbers had been required, we should have represented
them by ic — 2 y, x — y, x, x ^ y, and x \ 2y.
PROGRESSIONS 337
EXERCISE 157
1. The fifth term of an arithmetic progression is ^, and
the thirteenth term . Find the twentysecond term.
2. Find the sum of all the odd integers, beginning with 1
and ending with 999.
3. How many positive integers of three digits are multiples
of 7 ? What is their sum ?
4. The first term of an arithmetic progression is 1, and the
sum of the sixth and tenth terms is 37. Find the second and
third terms.
5. The first term of an arithmetic progression of 11 terms
is , and the seventh term — 3. Find the sum of the terms.
6. In an arithmetic progression, the sum of the first and
last terms is twoninths the sum of all the terms. Find the
number of terms.
7. The seventh term of an arithmetic progression is — 37,
and the sum of the first 17 terms — 799. Find the sum of the
first 13 terms.
8. Find five numbers in arithmetic progression such that
the sum of the first, fourth, and fifth is 14, and the quotient of
the second by the fourth — ^.
9. How many arithmetic means are inserted between — 
and I, when their sum is y ?
10. If the constant difference of an arithmetic progression
equals twice the first term, the quotient of the sum of the terms
by the first term equals the square of the number of terms.
11. The sum of the first 10 terms of an arithmetic progression
is to the sum of the first 5 terms as 13 to 4. Find the ratio of
the first term to the common difference.
12. Find four numbers in arithmetic progression such that
the sum of the first and second shall be —1, and the product
of the second and fourth 24.
338 ALGEBRA
13. The last term of an arithmetic progression of 10 terms
is 29. The sum of the oddnumbered terms is to the
sum of the evennumbered terms as 14 is to 17. Find the first
term and the common differ'ence.
14. The sum of five numbers in arithmetic progression is 25,
and the sum of their squares is 135. Find the numbers.
15. A man travels ^^ miles. He travels 10 miles the first
day, and increases his speed onehalf mile in each succeeding
day. How many days does the journey require ?
16. Find the sum of the terms of an arithmetic progression
of 9 terms, in which 17 is the middle term.
17. Find three numbers in arithmetic progression, such that
the square of the first added to the product of the other two
gives 16, and the square of the second added to the product of
the other two gives 14.
18. If a person saves $ 120 each year, and puts this sum at
simple interest at 3% at the end of each year, to how much
will his property amount at the end of 18 years ?
19. A traveller sets out from a certain place, and goes
7 miles the first hour, 7^ the second hour, 8 the third hour,
and so on. After he has been gone 5 hours, another sets out,
and travels 16J miles an hour. How many hours after the
first starts are the travellers together ?
20. There are 12 equidistant balls in a straight line. A
person starts from a position in line with the balls, and beyond
them, his distance from the first ball being the same as the
distance between the balls, and picks them up in succession,
returning with each to his original position. He finds that he
has walked 780 feet. Find the distance between the balls.
GEOMETRIC PROGRESSION
368. A Geometric Progression is a series of terms in which
each term, after the first, is obtained by multiplying the
preceding term by a constant number called the Ratio.
PROGRESSIONS 339
Thus, 2, 6, 18, 54, 162, ••• is a geometric progression in which
the ratio is 3.
9, 3, 1, i, J, ••• is a geometric progression in which the ratio
isi.
— 3, 6, — 12, 24, — 48, ••• is a geometric progression in which,
the ratio is — 2.
A Geometric Progression is also called a Geometric Series.
369. Given the first term, a, the ratio, r, and the number of
terms, n, to find the last term, I.
The progression is a, ar, ar^, ai^, • • • .
We observe that the exponent of r in any term is less by 1
than the number of the term.
Then, in the wth term the exponent of r will be n — 1.
That is, I = ar"\ (I)
370. Given the first term, a, the last term, I, and the ratio, r, to
find the sum of the terms, S.
S = a + ar + ar' \ h ar""^ + ar"^ + ar~^ (1)
Multiplying each term by r,
rS = ar{ ai^ + ar^ + • • • + ar""^ + ar""^ + ar"". (2)
Subtracting (1) from (2), r8S = ar^a, or 8 = ^^^~^ '
But by (I), § 369, rl=^ar\
Therefore, ^ = 'l:z^. (II)
r — 1
The first term, ratio, number of terms, last term, and sum of the terms,
are called the elements of the progression.
371. Examples.
1. In the progression 3, J, •!■,•••, to 7 terms, find the last
term and the sum.
Here, a = S, r = , n = 7.
340 . ALGEBRA
Substituting in (I), I = ^(^)'=^5 = ^
lxA_3 J__3 2186
o ^. .. .• • /TT^ c ^ 243 729 729 1093
Substituting in (II), S = — = ^ = ^ = ^ 
S~ ~3 ~3
The ratio may be found by dividing the second term by the first, or
any term by the next preceding term.
2. In the progression — 2, 6, — 18, •••, to 8 terms, find the
last term and the sum.
Here, a = — 2, r = = — 3, w = 8 ; therefore,
; = _ 2( 3)7 =  2 X ( 2187) = 4374.
,,  3 X 4374 (2)  13122 + 2 000^
^ = — _3i = ~~:^ — =^^^^ ■
EXERCISE 158
Eind the last term and the sum of the following :
1. 1, — 2, 4, ... to 10 terms. 6. — f, i, — h ••• to 7 terms.
2. 6, 9,^, ••• to 7 terms. 7. 4, 3, , ••. to5terms.
3. 3, — 15, 75, ... to 5 terms. 8. — , f, — VS ••• to 8 terms.
4. 5, 20, 80, ...to 6 terms. 9. 2, f , 2%, ... to 6 terms.
5. ^, i, 1, ... to 9 terms. 10. f, —J, i, ... to 8 terms.
372. If any three of the five elements of a geometric pro
gression are given, the other two may be found by substituting
the given values in the fundamental formulae (I) and (II), and
solving the resulting equations.
But in certain cases the operation involves the solution of an
equation of a degree higher than the second ; and in others the
unknown number appears as an exponent, the solution of which
form of equation can usually only be effected by the aid of
logarithms (§ 437).
In all such cases in the present chapter, the equations may
be solved by inspection.
PROGRESSIONS 341
1. Given a = — 2, w = 5, Z = — 32^ find r and S.
Substituting the given values in (I), we have
— 32 = — 2 r* ; whence, r* = 16, or r = ± 2.
Substituting in (II),
If r= 2, ^ = 2(^2)(2)^_g4_^2=:_62.
z — 1
If r = 2 ^^ (2)(32)(2) ^64 + 2^_22^
21 3
The solution is r = 2, /S^ =  62 ; or, r =  2, /S' =  22.
The interpretation of the two answers is as follows :
If r= 2, the progression is —2, —4, —8, —16, —32, whose sum is —62.
If r= —2, the progression is —2, 4, —8, 16, —32, whose sum is —22.
2. Given a = 3, r = i 8 = ^^^^, find yi and ?.
lz3
Substituting in (II), ^ = _^ = ^.
~3~
WhencP 7_L96560. 6560 q_ 1
Whence, z + 9 ^ , or, Z _ ^ 9_ ^^^.
Substituting the values of a, r, and Z in (I) ,
__L=3flv^or, fiv^=^.
729 V 3/ ' ' V 3; 2187
Whence, by inspection, w — 1 = 7, or n = S.
From (I) and (II) general formulae may be derived for the solution of
cases like the above.
If the given elements are n, I, and S, equations for a and r may be
found, but there are no definite formulae for their values.
The same is the case when the given elements are a, w, and S.
The general formulae for n involve logarithms ; these cases are discussed
in § 437.
EXERCISE 159
1. Given r = 3, n = %,l = 2187 ; find a and S.
2. Given r = — 4, n = 5, S = — 410 ; find a and I.
342 ALGEBRA
3. Givena = 6,n = 6, Z = f4; find rand >S.
4. Given a = 3, r=^, 1 = y^ ; find n and S.
5. Given r = 2, 71 = 10, /S = iV>^; find a and L
6. Given a = f , n = 7, ? = f f ; find ? and S.
7. Given a = i ? = 2/^,/S' = fff; find r and w.
8. Given a = , r = , ,S = mf ; find I and n.
9. Given I =  768, r = 4, .S =  ^Q^^ ; find a and n.
10. Given a = , Z = 1458, S= ^^ ; find r and w.
11. Given a, r, and /S; derive the formula for I.
12. Given a, Z, and aS ; derive the formula for r.
13. Given r, I, and 8; derive the formula for a.
14. Given r, n, and I ; derive the formulae for a and S.
15. Given r, n, and S\ derive the formulae for a and I.
16. Given a, n, and Z ; derive the formulae for r and 8.
373. Sum of a Geometric Progression to Infinity.
The limit (§ 318) to which the sum of the terms of a decreas
ing geometric progression approaches, when the number of
terms is indefinitely increased, is called the sum of the series
to infinity.
Formula (II), § 370, may be written
a — rl ^
8 =
1r
It is evident that, by sufficiently continuing a decreasing
geometric progression, the absolute value of the last term may
be made less than any assigned number, however small.
Hence, when the number of terms is indefinitely increased,
Z, and therefore rl, approaches the limit 0.
Then, the fraction ^~^ approaches the limit — —
1— r 1—r
PROGRESSIONS 343
Therefore, the sum of a decreasing geometric progression to
infinity is given by the formula
S = ^^' (III)
1 — r
Ex. Find the sum of the series 4, — , ^j "to infinity.
Here, a = 4, r = —
3
2
4 12
Substituting in (III), S ^  
1+1 6
EXERCISE 160
Find the sum to infinity of the following :
1 6,2,1,.... 5. hU,U,
2. 12,3,1,.... 6. J^,^%., ^^, ....
.^111... 7_3_15 75 ...
4 —25 25 _ 50 ... 8 5 _ 5 10 ...
374. To j^nd ^^e value of a repeating decimal.
This is a case of finding the sum of a decreasing geometric
series to infinity, and may be solved by formula (III).
Ex. Find the value of .85151 ....
We have, .85151 ... = .8 + .051 + .00051 + ....
The terms after the first constitute a decreasing geometric progression,
in which a = .051, and r = .01.
.051 .051 51 17
Substituting in (III), S
.01 .99 990 330
8 17 281
Then, the value of the given decimal is 1 , or
^ 10 330 330
EXERCISE 161
Find the values of the following :
1. .7272 .... 3. .91777 .... 5. .23135135 •
2. .629629 .... 4. .75959 .... 6. .587474 ....
344 ALGEBRA
375. Geometric Means.
We define inserting m geometric means between two numbers,
a and b, as finding a geometric progression of m + 2 terms,
whose first and last terms are a and b.
Ex. Insert 5 geometric means between 2 and ^f .
We find a geometric progression of 7 terms, in which a = 2, and
128 128 '
I = — ; substituting w = 7, a = 2, and I = — — in (I),
729 729 '
1 = 2^; whence r« = ^, and. = ±
Theresultis2, ±^, 8 16 32 64 128.
'3 9 27 81' 243' 729
376. Let X denote the geometric mean between a and b.
Then,  = , or ic^ = ab.
a X
Whence, x = Va&.
That is, the geometric mean between two numbers is equal to
the square root of their product.
EXERCISE 162
1. Insert 4 geometric means between f and 24.
2. Insert 5 geometric means between — 3 and — 2187.
3. Insert 4 geometric means between ^ and — 320.
4. Insert 6 geometric means between — f and ^^\.
5. Insert 7 geometric means between — 48 and — ^^.
6. Insert 3 geometric means between J^^ and ■^^.
7. If m geometric means are inserted between a and b, what
are the last two means ?
Find the geometric mean between :
8. U and Y 9. ^^ and ^^^.
^8 xyy^ xy
10. a24a + 4 and 4a2 + 4a + l.
PROGRESSIONS 345
377. Problem.
Find 3 numbers in geometric progression such that their
sum shall be 14, and the sum of their squares 84.
Let the numbers be represented by a, ar, and ar^.
r a + ar+ar2 = 14. ^ (1)
Then, by the conditions, <
' ^ ' I a2 + aV2 + aV = 84. (2)
Divide (2) by (1), a  ar + ar2 = 6. (3)
Subtract (3) from (1), 2 ar = 8, or r = • (4)
a
Substituting in (1), a + 4 + — = 14, or a2  10 a + 16=0.
a
Solving this equation, a = 8 or 2.
4 4
 or 
8 2 2
4 4 1
Substituting in (4), r =  or  =  or 2.
Then, the numbers are 2, 4, and 8.
EXERCISE 163
1. The fifth term of a geometric progression is , and the
eighth term — ^^. Find the third term.
2. The product of the first five terms of a geometric pro
gression is 243. Find the third term.
3. Find four numbers in geometric progression such that
the sum of the first and fourth shall be 27, and of the second
and third 18.
4. Find an arithmetic progression whose first term is 2, and
whose first, fourth, and tenth terms form a geometric pro
gression.
5. The third term of a geometric progression is f , and the
seventh term Y^^'t find the ninth term.
6. The sum of the terms of a geometric progression whose
first term is 1, ratio 3, and number of terms 4, equals the sum
of the terms of an arithmetic progression whose first term is
4, and common difference 4. Find the number of terms of the
arithmetic progression.
346 ALGEBRA
7. The sum of the first four terms of a decreasing geometric
progression is to the sum to infinity as 16 to 25. Find the
ratio.
8. A man who saved every year fourthirds as much as in
the preceding year, had in four years saved $ 3500. How
much did he save the first year ?
9. The difference between two numbers is 16, and their
arithmetic mean exceeds their geometric mean by 2. Find the
numbers.
10. Find six numbers in geometric progression such that the
sum of the first and fourth shall be 9, and of the third and
sixth 36.
11. The digits of a number of three figures are in geometric
progression, and their sum is 7. If 297 be added to the num
ber, the digits will be reversed. Find the number.
12. There are three numbers in geometric progression whose
sum is ^. If the first be multiplied by f, the second by ,
and the third by ^^, the resulting numbers will be in arithmetic
progression. What are the numbers ?
HARMONIC PROGRESSION
378. A Harmonic Progression is a series of terms whose
reciprocals form an arithmetic progression.
Thus, 1, "I", ^, I, ^, ••• is a harmonic progression, because the
reciprocals of the terms, 1, 3, 5, 7, 9, •••, form an arithmetic
progression.
A Harmonic Progression is also called a Harmonic Series.
3i9. Any problem in harmonic progression, which is suscep
tible of solution, may be solved by taking the reciprocals of the
terms, and applying the formulae of the arithmetic progression.
There is, however, no general method for finding the sum of
the terms of a harmonic series.
Ex. In the progression 2, , , ••• to 36 terms, find the last
term.
PROGRESSIONS 347
Taking the reciprocals of the terms, we have the arithmetic progression
13 5
2' 2' 2' ""*
Here, a = , d = l, w = 36.
Substituting in (I), § 359, Z = 1 + (36  1) x 1 = — •
2
Then, — is the last term of the given harmonic series.
380. Harmonic Means.
We define inserting m harmonic means between two numbers, a
and b, as finding a harmonic progression of m + 2 terms, whose
first and last terms are a and b.
Ex. Insert 5 harmonic means between 2 and — 3.
We have to insert 5 arithmetic means between  and
2 3
Substituting a = ^, Z = , 7i = 7, in (I), § 359,
l = l + 6d,  = 6c?, or d = A.
3 2 6 ' 36
Then the arithmetic series is , — , , — , , , — •
2 36 9 12 18' 36 3
Therefore, the required harmonic series is
2, §^, ^, 12, 18, ^, 3.
' 13' 2' ' ' 7'
381. Let X denote the harmonic mean between a and b.
Then,  is the arithmetic mean between  and —
X a b
Then, by § 366, 1 = ^ = ^, and x = ^^.
^ ' X 2 2ab' a\b
EXERCISE 164
Find the last terms of the following :
1. I, J/, ¥? ••• to 19 terms.
348
ALGEBRA
2.
h h 2%. to 46 terms.
3.
h h ^^ ••' to 33 terms.
4.
IT? w TTJ"to 11 terms.
5.
4, 38,, 2^ ...to 28 terms.
6. Insert 7 harmonic means between — 4 and i.
7. Insert 8 harmonic means between —  and — .
8. Insert 6 harmonic means between i and — ^.
Find the harmonic mean between :
9. land . . 10. ^^±^ and "^
d' + b'
11. Find the next to the last term of the harmonic progres
sion a, b, •••to 71 terms.
12. If m harmonic means are inserted between a and b, what
is the third mean ?
13. The sixth term of a harmonic progression is ^, and the
eleventh term — f . Find the fourteenth term.
14. The geometric mean between two numbers is 4, and the
harmonic mean ^. Find the numbers.
382. If any three consecutive terms of a harmonic series be
taken, the first is to the third as the first minus the second is to the
second minus the third.
Let the terms be a, b, and c; then, since , , and  are in
arithmetic progression,
a b c
1111 b
= , or
c b b a be ab
Multiplying both members by r^^ — , we have
b — c
a _ a — b
c b — c
PROGRESSIONS 349
383. Let A, G, and H denote the arithmetic, geometric, and
harmonic means, respectively, between a and b.
Then, by §§ 366, 376, and 381,
A = ^, (y = V^, and^=2^^
a\b
But, ^X^ = a6=(V^)^.
' 2 a\b ^
Whence, A x H= G\ or G = VA x H.
That is, the geometric mean between two numbers is also the
geometric mean betiveen their arithmetic and harmonic means.
350 ALGEBRA
XXVIII. THE BINOMIAL THEOREM
POSITIVE INTEGRAL EXPONENT
384. A Series is a succession of terms.
A Finite Series is one having a limited number of terms.
An Infinite Series is one having an unlimited number of terms.
385. In §§97 and 205 we gave rules for finding the square
or cube of any binomial.
The Binomial Theorem is a formula by means of which any
power of a binomial may be expanded into a series.
386. Proof of the Binomial Theorem for a Positive Integral
Exponent.
The following are obtained by actual multiplication :
(a\xy = a''^2ax^3^;
(a \ xy = a* \ 4: a^x + 6 aV + 4 cta^ f x* ; etc.
In these results, we observe the following laws :
1. The number of terms is greater by 1 than the exponent
of the binomial.
2. The exponent of a in the first term is t^e same as the
exponent of the binomial, and decreases by 1 in each succeed
ing term.
3. The exponent of x in the second term is 1, and increases
by 1 in each succeeding term.
4. The coefficient of the first term is 1, and the coefficient of
the second term is the exponent of the binomial.
^ 5. If the coefficient of any term be multiplied by the expo
nent of a in that term, and the result divided by the exponent
of X in the term increased by 1, the qdotient will be the
coefficient of the next following term.
THE BINOMIAL THEOREM 351
387. If the laws of § 386 be assumed to hold for the expan
sion of (a + a?)", where n is any positive integer, the exponent
of a in the first term i^ w, in the second term n — 1, in the
third term n — 2, in the fourth term n — 3, etc.
The exponent of x in the second term is 1, in the third term
2, in the fourth term 3, etc.
The coefficient of the first term is 1 ; of the second term n.
Multiplying the coefficient of the second term, n, by n — 1,
the exponent of a in that term, and dividing the result by
the exponent of x in the term increased by 1, or 2, we have
'^y^~ ) as the coefficient of the third term ; and so on.
1.2
(A point is often used for the sign x ; thus, 1 • 2 signifies 1x2.)
Then, (a + xf = a''^ ua^r.^aj + '^^^~^] a"" V
. + «(«l)(r^2) ^.V+.. (1)
Multiplying both members of (1) by a\x, we have
(a 4 xy+^
= a^ + na^x + 'l(^^ay + ^(^ l)»2) ^nv 4. ...
\ • Z
Collecting the terms which contain like powers of a and a;,
we have,
(a + a;)"+^ = a"+^ + (n + 1) oiTx + [" ^ ^^ ~ *) + ^^1 a"" ^
r yi(nl)(n2) ^^C^ l) "1^n2^ . ...
[_ 1.2.3 ^ 1.2 J ^
= a"+i + (ri 4 1) a"a; + ^i f^^^ + 1] «""^^
n(nl)nj^^1 ,^ ^.,^
^ 1.2 [ 3 J
352 ALGEBRA
a^''x'
Then, (a + i»)"+i = a«+i + (^ + 1) a"x + n T^^ti
, 71 (71 — 1) Fn + 1~1 ,,_o o ,
= ay+' + (» + !) <t"a; 4 (" + ^) "• a V
^ („ + l)n(nl) ^„_y^,.._ (2)
It will be observed that this result is in accordance with
the laws of § 386 ; which proves that, if the laws hold for any
power of ft + a; whose exponent is a positive integer, they also
hold for a power whose exponent is greater by 1.
But the laws have been shown to hold for (a + xy, and
hence they also hold for (a + xy ; and since they hold for
^ (a + xy, they also hold for (a f xy ; and so on.
Therefore, the laws hold wfien the exponent is any positive
integer, and equation (1) is proved for every positive integral
value of n.
Equation (1) is called the Binomial Theorem.
In place of the denominators 12, 1.2.3, etc., it is usual to write
[2, [3, etc.
The symbol \n^ read "factorialn," signifies the product of the natural
numbers from 1 to w, inclusive.
The method of proof exemplified in § 387 is known as Mathematical
Induction.
A more complete form of the proof of § 387, in which the fifth law of
§ 386 is proved for any two consecutive terms, will be found in § 447.
388. Putting a = 1 in equation (1), § 387, we have
389. In expanding expressions by the Binomial Theorem,
it is convenient to obtain the exponents and coefficients of the
terms by aid of the laws of § 386.
THE BINOMIAL THEOREM 353
1. Expand (a\xy.
The exponent of a in the first term is 5, and decreases by 1 in each
succeeding term.
The exponent of x in the second term is 1, and increases by 1 in each
succeeding term.
The coefficient of the first term is 1 ; of the second, 5.
Multiplying 5, the coefficient of the second term, by 4, the exponent of
a in that term, and dividing the result by the exponent of x increased by
1, or 2, we have 10 as the coefficient of the third term ; and so on.
Then, {a {■ x)^ = a^ \ ^ a% + lo oF'x'^ + 10. a^x^ + 5 ax* + x^.
It will be observed that the coefficients of terms equally distant from
the ends of the expansion are equal ; this law will be proved in § 391.
Thus the coefficients of the latter half of an expansion may be written
out from the first half.
If the second term of the binomial is negative, it should
be enclosed, negative sign and all, in parentheses before
applying the laws ; in reducing, care must be taken to apply
the principles of § 96.
2. Expand (1  x)\
(la:)6 = [H(x)]6
= 16^6.15. (_a;) + ,1^.14. (^a:)2 + 20.l3. {xY
+ 15.12. (x)* + CI. (x)5 + (x)6
= 1  6 X + 16 x2  20 x3 + 15 X*  6 x5 + a;6.
If the first term of the binomial is an arithmetical number, it is con
venient to write the exponents at first without reduction ; the result
should afterwards be reduced to its simplest form.
If either term of the binomial has a coefficient or exponent
other than unity, it should be enclosed in parentheses before
applying the laws.
3. Expand (3 m2v^^y.
(3 w2  ^ny = [ (8 7^2) + ( _ n^) ]4
= (3 m'^y + 4(3 m2)3( _ n^) + 6(3 m2)2(_ n^y
+ 4(3 m2) (  n^y + (  n^y
= 81 »n8  108 mHi^ + 54 m*>i*  12 m^n+ n^.
354 ALGEBRA
f!'
A trinomial may be raised to any power by the Binomial
Theorem, if two of its terms be enclosed in parentheses, and
regarded as a single term ; but for second powers, the method
of § 204 is shorter.
4. Expand (x2x2y.
(a;2 _ 2 X  2)* = [(x2  2 x) + ( 2)]*
= (a:2  2 x)* + 4(a;2  2 a;)3(  2) + 6(x2  2 a;)2(  2)2
+ 4(^2 _ 2 x) ( 2)3 + ( 2)*
= x^  S x^ \2i x^  S2x^ h 16 x^
8(x66x6 + 12x*8a;3)
+ 24(a;4  4 x^ + 4 a;2)  32(x2 2x)+16
= x88 xH 16 a;6 + 16x556 ic*32 ic3+64 x2 + 64 x+ie.
EXERCISE 165
Expand the following :
1. (a\xy. 13^ (x^^sy.
21.
2. (nhiy
14. (ix^y.
(f'J
3. ayy. ' _ ' ^22. (sx^ — ^y
4. (a a;)'. 15. (Va ^Y. ^ 2 kV
5. (:^y + ^y. ^ ^'^'^ 23. (Zai + Vaf.
6. (x + 2 2,)'. 16. (mS + 4«})^ 24. fsVcP^Y
7. (2ay. 17. r.*,i+iY. .^ X, :S.
8. (3a^6^)^ ^ ^^ 25. ^^Va;^J.
9. (a4a^'^)^ 18. M + ^fY. .2m^^ .M^
10. {2x^^fy. ^ ^^ ^^ ^^ + 3^J*
11. (« + .)« 19. (5.^ny. ^^ /^__1_V
12. (a2H</^)6. 20. (2a:^ + 2/"^)'. " V Zi/hV
28. (a6)^ 29. {a + iy\
30. (iB2^a,^iy^ 32_ (a^_3a._i)4^ 34. (i^x^o?)'.
31. (2a; + a^)*. 33. {^x' + x^y. 35. (a;' + .T3)^
THE BINOMIAL THEOREM 355
390. To find the rtli or general term in the expansion of(a\xY.
The following laws hold for any term in the expansion of
(a\xY, in equation (1), § 387 :
1. The exponent of x is less by 1 than the number of the
term.
2. The exponent of a is n minus the exponent of x.
3. The last factor of the numerator is greater by 1 than the
exponent of a.
4. The last factor of the denominator is the same as the
exponent of x.
Therefore in the rth term, the exponent of x will be r — l.
The exponent of a will be n — (? — 1), or n — r \l.
The last factor of the numerator will be n — r \ 2.
The last factor of the denominator will be r — 1.
Hence, the rth term
^ n(n  1) (n  2) . . . (n  r + 2) nr+ Wl QN
1.2.3... (r1) * ^^
In finding any term of an expansion, it is convenient to obtain
the coefficient and exponents of the terms by the above laws.
Ex. Find the 8th term of (3 a^  hy\
We have, (3 a^  h^y^ = [(3 a^) + ( 6i)]ii.
In this case, n = 11, r = 8.
The exponent of (— &~i) is 8 — 1, or 7.
The exponent of (3 a^) is 11 — 7, or 4.
The first factor of the numerator is 11, and the last factor 4 + 1, or 5.
The last factor of the denominator is 7.
Then, the 8th term = 12 • 10 • 9 • 8  7 • 6 . 5 ^3 ^^y^^ b^y
1.2.3.4.5.6.7^ ^ ^
= 330(81 a2) ( _ 57) = _ 26730 a^ftT.
If the second term of the binomial is negative, it should be enclosed,
sign and all, in parentheses before applying the laws.
If either term of the binomial lias a coefficient or exponent other than
unity, it should be enclosed in parentheses before applying the laws.
356
ALGEBRA
Find the :
1. 4th term of (a f xj.
2. 6th term of {n + iy\
3. 5th term of (a  hf.
4. 7th term of (1  a^.
5. 8th term of (aj* + fY^
6. 5th term of (a^ + 2 x~^y\
7. 9th term of {x^  x'^y^
8. 10th term of (^^' + Y'.
\b aj
14. 8th term of
15. Middle term
EXERCISE 166
9. 10th term of /"v^m^^y'.
10. 6th term of (x'  4 yfy'.
11. Tthtermof (^m~* + ^^^^
12.
4th term
of (m*
— 5 mny^.
13.
9th term of fa^+
1 Y^
4^
(^■W
of (^3 a'
+
6*Y'
2; *
391. Multiplying both terms of the coefficient, in (1), § 390,
by the product of the natural numbers from 1 to n — r + 1,
inclusive, the coefficient of the rth term becomes
n(nl) ...{nr\2)'(nrhl) '"2 '1 _
I r1 xl 2... (nr + 1)
\n
r+1
Since the number of terms in the expansion is n + 1, the rth
term from the end is the (n — r\ 2)th from the beginning.
Then, to find the coefficient of the ?*th term from the end, we
put in the above formula n — r + 2 for r.
Then, the coefficient of the rth term from the end is
I 71 \ n
' Qp ' .
I yt — r + 2 — 1 I 71 — (yi — r + 2)4l '  n — r + 1  r — 1
Hence, in the expansion of (a + a?)", the coefficients of terms
equidistant from the ends of the expansion are equal.
UNDETERMINED COEFFICIENTS
367
XXIX. UNDETERMINED COEFFICIENTS
392. Infinite Series (§ 384) may be developed by Division,
or by Evolution.
Let it be required, for example, to divide 1 by 1 — a;.
la;)l(l + x + a^ + 
1a;
X
Then,
1x
l + a; + ar^ + a^ +
(1)
Again, let it be required to find the square root of 1 + a;,
l + a;
^2 8^
2 +
X +
2{x
x"
X y?
Then, V^^:^ = l +  +
(2)
It should be observed that the series, in (1) and (2), do not give the
values of the first members for every value of x ; thus, if x is a very
large number, they evidently do not do so.
EXERCISE 167
Expand each of the following to four terms :
1.
2.
3 + 4a;
l + 2a;*
3.
4a;
2 + 6a;a^
5. VTT6^.
4 2 + 4a;5a^, g Vr32^.
3_6a^ + 7a^
358 ALGEBRA
7. VT+a. 9. Va^ + a!2/ + 2/' H va^^ + l.
8. VI 5a. 10. V9a2+6. 12. A/a«^3 63.
CONVERGENCY AND DIVERGENCY OF SERIES
393. An intinite series is said to be Convergent when the sum
of the first ?i terms approaches a fixed finite number as a limit
(§ 318), when n is indefinitely increased.
An infinite series is said to be Divergent when the sum of the
first n terms can be made numerically greater than any assigned
number, however great, by taking n sufticiently great.
394. Consider, for example, the infinite series
1 \ X \ oir^ { a^ ] .
I. .Suppose X = Xi, where osi is numerically < 1.
The sum of the first n terms is now
1 + 0^1 + o^i^ +  + xr' = \=^^ (§ 103).
1 — iCj
If n be indefinitely increased, x^" decreases indefinitely in
absolute value, and approaches the limit 0.
Then the fraction approaches the limit
1 — i»l 1 — £Cl
That is, the sum of the first n terms approaches a fixed finite
number as a limit, when n is indefinitely increased.
Hence, the series is convergent when x is numerically < 1.
II. Suppose x = l.
In this case, each term of the series is equal to 1, and the
sum of the first n terms is equal to n; and this sum can be
made to exceed any assigned number, however great, by taking
n sufficiently great.
Hence, the series is divergent when x = l.
III. Suppose £c = — 1.
In this case, the series takes the form 1 — 1 + 1 — l4.', and
the sum of the first n terms is either 1 or according as n is
odd or even.
UNDETERMINED COEFFICIENTS 359
Hence, the series is neither convergent nor divergent when
An infinite series which is neither convergent nor divergent
is called an Oscillating Series.
IV. Suppose X = Xi, where x^ is numerically > 1.
The sum of the first n terms is now
1 + 0^1 + a^i^ +  + X,' = ^^ (§ 103).
By taking n sufficiently great, — — can be made to numeri
cally exceed any assigned number, however great.
Hence, the series is divergent when x is numerically > 1.
395. Consider the infinite series
l + x{x^\x^\ ,
developed by the fraction (§ 392).
1 — X
Let x= .1, in which case the series is convergent (§ 394).
The series now takes the form 1 + .1 + .01 + .001 H , while
the value of the fraction is — , or — •
In this case, however great the number of terms taken, their
sum will never exactly equal Y
But the sum approaches this value as a limit ; for the series
is a decreasing geometric progression, whose first term is 1, and
ratio .1 ; and, by § 373, its sum to infinity is , or — •
JL — .J. «7
Thus, if an infinite series is convergent, the greater the num
ber of terms taken, the more nearly does their sum approach
to the value of the expression from which the series was
developed.
Again, let x = 10, in which case the series is divergent.
The series now takes the form 1 f 10 + 100 + 1000 + •••,
while the value of the fraction is — , or — •
110 9
360 ALGEBRA
In this case the greater the number of terms taken, the
more does their sum diverge from the value — ^.
Thus, if an infinite series is divergent, the greater the number
of terms taken, the more does their sum diverge from the value
of the expression from which the series was developed.
It follows from the above that an infinite series cannot be
used for the purposes of demonstration, if it is divergent.
THE THEOREM OF UNDETERMINED COEFFICIENTS
396. An important method for expanding expressions into
series is based on the following theorem :
If the series A f Bx \ Cx^ + Da? \ is always equal to the
series A' + B'x + C'x^ + D'x^ + • • •, when x has any value which
makes both series convergent, the coefficients of like powers of x in
the series will be equal; that is, A=A', B = B', C= C, etc.
For the equation
AiBx+Cx'{Da? + '"=A' + B'x+C'a?\D'a?+'" (1)
is satisfied when x has any value which makes both members
convergent.
But both members are convergent when x = 0] for the sum
of all the terms of the infinite series a ^ bx { ca? ^ dx^ i
is equal to a when x = 0.
Then, the equation (1) is satisfied when x = 0.
Putting a; = 0, we have A = A'.
Subtracting A from the first member of the equation, and its
equal A' from the second member, we obtain
Bx + Cx^^Da?^'"=B'x\C'x'\D'a?+"'. (2)
Dividing each term by x,
B^Cx + Dx^h''. =B' h C'xjD'x'+'". (3)
This equation also is satisfied when x has any value which
makes both members convergent ; and putting a; = 0, we have
B = B'.
UNDETERMINED COEFFICIENTS 361
In like manner, we may prove (7= C", D = D', etc.
The proof of § 396 is open to objection iu one respect.
We know that (2) has the same roots as (1), including the root 0; but
when we divide by z, all that we know about the resulting equation is
that it has the same roots as (2) , except the root 0.
Thus, we do not know that is a root of (3) , though we assume it in
proving that B = B'.
A more rigorous proof of the Theorem of Undetermined Coefficients
will be found in § 450.
397. The theorem of § 396 holds when either or both of the
given series are finite.
EXPANSION OF FRACTIONS
2 _ 3aj2 _ gj3
398. 1. Expand — — in ascending powers of x.
1 — 2a; + 3ar
Assume ^ ~^^^ ~^\ = A] Bx + Cx'^ + Dx^ + Ex^ + ■«., (1)
where A^ J5, O, Z>, E^ •••, are numbers independent of x.
Clearing of fractions, and collecting the terms in the second member
involving like powers of x, we have
2_3x2a;3 = ^+ B\x+ C
2^1 2B
+ 3^
20
+ 35
x^^ E
2D
+ 30
a^ + .... (2)
A vertical line, called a 6ar, is often used in place of parentheses.
Thus, + B\x \& equivalent to (J5 — 2 A)x.
2a\
The second member of (1) must express the value of the fraction for
every value of x which makes the series convergent (§395); and there
fore equation (2) is satisfied when x has any value which makes the
second member convergent.
Then, by § 397, the coefficients of like powers of x in (2) must be
equal ; that is,
A= 2.
B2A= 0; or, 5 = 2^ =4.
(725 + 3.A = 3; or, O = 2 J5  3^  3 = 1.
I>20+35=l; or, i> = 2035l=15.
^21> + 30= 0;or, J? = 2D30 =27; etc.
assume
362 ALGEBRA
Substituting these values in (1), we have
23a;2x8 ^ 2 + 4 x  a:2  15x8  27a^ .
l2x + 3x2
The result may be verified by division.
The series expresses the value of the fraction only for such values of x
as make it convergent (§ 395) .
If the numerator and denominator contain only even powers
of X, the operation may be abridged by assuming a series con
taining only the even powers of a;.
Thus, if the fraction were — — —  — — — , we should
1 — 3 or \ 5 x'^
it equ al to ^ + 5a^ + Cx' \Dx^ + Ea^+"'.
In like manner, if the numerator contains only odd powers
of X, and the denominator only even powers, we should assutne
a series containing only the odd powers of x.
If every term of the numerator contains x, we may assume a
series commencing with the lowest power of x in the numerator.
If every term of the denominator contains x, we determine
by actual division what power of x will occur in the first term
of the expansion, and then assume the fraction equal to a series
commencing with this power of x, the , exponents of x in the
succeeding terms increasing by unity as before.
2. Expand — — — — in ascending powers of x.
O Xi — Xj
Dividing 1 by 3 x^, the quotient is —  ; we then assume,
o
^ .4x2 + 5xi + 0+i)x + ^x2+.. (3)
3 X2  X8
Clearing of fractions,
l = 3^ + 35x + 3Cx2 + 3i)x3 + 3^1x4+ .
 a\  b\  c\  b\
Equating coefficients of like powers of x,
3^=1, 35^ = 0, 305 = 0, 32) e = 0, 3^D = 0; etc.
Whence, A=\ B=.\ C = — , D = — , iE' = — , etc.
3 9 27 81' 243'
UNDETERMINED COEFFICIENTS
363
Substituting in (3) ,
3x2
T "9" 27 81 243
In Ex. 1, ^=22) — 3C; that is, the coeflBcient of x* equals twice the
coeificient of the preceding term, minus three times the coefficient of the
next but one preceding.
It is evident that this law holds for the succeeding terms ; thus, the
coefiBcient of x^ is 2 x ( 27) 3 x ( 15), or  9.
After the law of coefficients has been found in any expansion, the terms
may be found more easily than by long division ; and for this reason the
method of § 398 is to be preferred when a large number of terms is
required.
The law for Ex. 2 is that each coefficient is onethird the preceding.
EXERCISE 168
Expand each of the following to five terms in ascending
powers of x\
1.
2.
3.
5.
3 + 2a;
\x '
l6x
4 + a;^
lSx"'
2x
'SBx^'
l\4:X — xr
l_a;f 3a^*
g 2x{3x^ jj 3 + 4a^
9.
10.
1+20^2
l4a.2
l^4.xx'
X — 5x^ — 7 x*
\2x4.x'
x^3^
^^x2^
4 + a;5a^
12.
13.
2aj + 3a^
15.
2f3a^a^
3
2a^ + a;^*
2Jf.^xSx\
a^ — 5 ic^ + ic^
^^ l6a^ + 4a^
X'\x^2^ '
l2a; + 3a^
2a^ + 4:X^\a^'
EXPANSION OF SURDS
399. Ex. Expand v 1 — a; in ascending powers of x.
Assume y/l  x = A + Bx \ Cx^ + Dx^ + Ex^ + ....
(1)
Squaring both members, we have by § 204,
1  X = ^2
\2AB
x+ B^
+ 2AG
+ 2 AD
+ 2BC
X3+ (72
+ 2AE
+ 2BD
x* +
364 ALGEBRA
Equating coefficients of like powers of cc,
A^= 1 ; or, ^ = 1.
2^5 =  1; or, ^=_J_=_1.
2A 2
B^\2AC= 0; or, (7 =  — = 1.
2 A o
2AD + 2BC= 0; or, D=^=l.
A 16
C^ + 2AE + 2BD= 0: or, ^ =  .^1+^^^ = _ A etc.
2 A 128'
Substituting these values in (1), we have
Vnr^^i_?_^_^_6^
2 8 16 128
The result may be verified by Evolution.
The series expresses the value of Vl — a; only for such values of x as
make it convergent.
EXERCISE 169
Expand each of the following to five terms in ascending
powers of x:
1. Vl + 2a;. 3. Vl4iBhar^. 5. s/l + dx.
2. VI 3a;. 4. Vl^xx", 6. </lx2x'.
PARTIAL FRACTIONS
400. If the denominator of a fraction can be resolved into
factors, each of the first degree in x, and the numerator is of a
lower degree than the denominator, the Theorem of Undeter
mined Coefficients enables us to express the given fraction as
the sum of two or more partial fractions^ whose denominators
are factors of the given denominator, and whose numerators
are independent of x.
401. Case I. No factors of the denominator equal.
1. Separate — ^J^t into partial fractions.
^ (3a;l)(5a; + 2) ^
UNDETERMINED COEFFICIENTS 365
^^"°"' (3xl)(6» + 2) =3^3T + 5^Ti' ^^^
where ^ and £ are numbers independent of x.
Clearing of fractions, 19 x + 1 = ^(5 a; + 2) + .5(3 xV).
Or, 19 X + 1 = (5 ^ + 3 LB)a; + 2 ^  ^. (2)
The second member of (1) must express the value of the given fraction
for every value of x.
Hence, equation (2) is satisfied by every value of x ; and by § 397, the
coefBcients of like powers of x in the two members are equal.
That is, 5 ^ + 3 5 = 19,
and 2A B = \.
Solving these equations, we obtain J. = 2 and ^ = 3.
Substituting in (1), ,„ ^^^^.t ^ ^^ = ^^ + r^Z'
^ ^ (3xl)(5x + 2) 3xl 5x + 2
The result may be verified by finding the sum of the partial
fractions.
2. Separate — ^i into partial fractions.
The factors of 2 x  x^  x^ are x, 1  x, and 2 + x (§ 116).
Assume then ^ "^ +/ =^+_A + ^.
2xx2x8 X 1x 2 + x
Clearing of fractions, we have
X + 4 = ^(1  X) (2 + X) + 5x(2 + x) + Ox (1  X).
This equation, being satisfied by every value of x, is satisfied when x = 0.
Putting X = 0, we have 4 = 2^, or Jl = 2.
Again, the equation is satisfied when x = 1.
Putting X = 1, we have 5 = 3 5, or J5 = —
o
The equation is also satisfied when x = — 2.
Putting X =  2, we have 2 =  6 C, or O =  •
3
5 _1
Then, _^+^ = ? + J_ + ^^2^ '
2xx2x8 x"^lx"^2 + x x3(lx) 3 (2+x)
366 ALGEBRA
To find the value of A, in Ex. 2, we give to x such a value as will make
the coefficients of B and C equal to zero ; and we proceed in a similar
manner to find the values of B and C.
This method of finding J., B, and G is usually shorter than that used
in Ex. 1.
EXERCISE 170
Separate the following into partial fractions :
1.  — :; r* «*• ^ t^ — * 0
9iB2_4 x316a; ^ ^ 3 ax"  4. o?x
23a:425 ^ 6a;ll ^ 109a;
6a;2 + 5a; 6a^413aj + 6 5a^14a; + 8
12 + 17(c2a;2 g 4 + 14a;2a;2
(1 + 3 a;).(9 h 6 a;  8 a:2>^ (a^5 a;)(a^4)
402. Case II. All the factors of the denominator equal.
Let it be required to separate — p^ — into partial
fractions. \^~ )
Substituting 2/ + 3 for «, the fraction becomes
(y + 3yll(y + 3) + 26 ^ y^5y + 2 ^1 5 ^ 2
f f y y^ f'
Keplacing 2/ by a; — 3, the result takes the form
_J____5__ _2_
x3 (xZy^ {x^Y
This shows that the given fraction can be expressed as the
sum of three partial fractions, whose numerators are indepen
dent of X, and whose denominators are the powers of ic — 3
beginning with the first and ending with the third.
Similar considerations hold with respect to any example
under Case II; the number of partial fractions in any case
being the same as the number of equal factors in the denomi
nator of the given fraction.
Ex. Separate — — ^^—  into partial fractions.
^ (3a; + 5)2 ^
UNDETERMINED COEFFICIENTS 367
In accordance with the above principle, we assume the given fraction
equal to the sum of two partial fractions, whose denominators are the
powers of 3 x + 5 beginning with the first and ending with the second.
Thatis, 6^ + 5 ^_A_+ B
(3 X + 5)'^ 3 a; + 5 (3 x + 5)2
Clearing of fractions, 6x + 5 = J.(3a:4 5) + 5.
= ^Ax + bA^ B.
Equating coefficients of like powers of x,
3^ = 6,
and bA\B = b.
Solving these equations, A = 2 and 5 = — 5.
Whence, 6x + 5 _ 2 5
(3x+5)2 3x + 5 (3x + 5)2
EXERCISE 171
Separate the following into partial fractions :
J 24 a; + 2 3 12^f 7a;l . 16a^19
^x' + TZx^^ ' (l + 3xf ' ' {4:x3y'
2 2fl^lla; + 3 ^ 6x' + 12x10 g a^2af7x
(x4:y ' ' (sj2xy * * (x{iy
y 2a^18x'{24:X15 « 18 a; + 54 cc^ + 27 af
(a;2)^ ' ' (2 + 3 «)*
403. Case III. Some of the factors of the denominator equal.
Ex. Separate — ^^ into partial fractions.
X \X j— J. )
The method in Case III is a combination of the methods of Cases I and
II ; we assume,
a;2 _ 4 a; + 3 ^ ^ B G
X{x+iy X X+1 (X+l)2'
Clearing of fractions,
x2 _ 4 X + 3 = ^(x 4 1)2 f Bx{x + 1) + Cx
= (^ + S)x2 +{2A + B\Cyc\A.
368 ALGEBRA
Equating coefficients of like powers of x,
A\B = l,
2A + B\C = 4,
and ^ = 3.
Solving these equations, A = S, B = — 2^ and (7 = — 8.
Whence, ^^4x + 3_3 2 8
x{x +1)2 X a: + 1 (x + I)*
The following general rule for Case III will be found convenient :
A fraction of the form should be assumed
equal to (x + a)(x + 6) ... (x + m)'•...
x + a x\h x + m {x + my (x + my
Single factors like x \ a and x + & having single partial fractions cor
responding, arranged as in Case I ; and repeated factors like (x + my
having r partial fractions corresponding, arranged as in Case II.
EXERCISE 172
Separate the following into partial fractions :
x(x+3f ' ' a?(x\lf
o 3a^ + 7a^ + 24aj16 5 4a;3^ 29 a^36 a;9
a?(x^) x{xl){x^)
3 14aj253a;4 ^ llSx4.x'
(3 a; + 2) (2 a;  3)2 (8 aj2  2 a;  3) (2 x + 1)
404. If the degree of the numerator is equal to, or greater
than, that of the denominator, the preceding methods are
inapplicable.
In such a case, we divide the numerator by the denominator
until a remainder is obtained which is of a lower degree than
the denominator.
Ex. Separate — — into an integral expression and
partial fractions. ~ ^
UNDETERMINED COEFFICIENTS 369
Dividing a^ — 3 x^ — 1 by x^ — x, the quotient is x — 2, and the re
mainder — 2 X — 1 ; we then have
^ »''3^''=»2 + ^^^ (1)
X2 — X X'^  X
We can now separate ~ ^ ~ into partial fractions by the method
X''^ — X
1 3
of Case I : the result is
Substituting in (1), — — '— ^ = x  2 + —
X'^  X X X — 1
Another way to solve the above example is to combine the methods of
398 and 401, and assume the given fraction equal to
Ax + B + ^+ ^
X X — 1
EXERCISE 173
Separate each of the following into an integral expression
and two or more partial fractions :
1.
(a;2)(3a; + l) * :>^(xl)
(a; 43)^ ' * x'ix + lf
5 2a^8a;^ + 2a;^5a^fl2a;^fl; + 4 '
iC^(£C— 4)
405. If the denominator of a fraction can be resolved into
factors partly of the first and partly of the second, or all of the
second degree, in x, and the nurnerator is of a lower degree
than the denominator, the Theorem of Undetermined Coeffi
cients enables us to express the given fraction as the sum of
two or more partial fractions, whose denominators are factors
of the given denominator, and whose numerators are inde
pendent of X in the case of fractions corresponding to factors
of the first degree, and of the form Ax + B in the case of
fractions corresponding to factors of the second degree.
370 ALGEBRA
The only exceptions occur when the factors of the denominator are of
the second degree and all equal.
Ex. Separate — —  into partial fractions.
3/ p JL
The factors of the denominator are cc + 1 and x"^ — x + 1.
Assume then Ji— = A_ + Bx+C ,j.
x^\lx + lx^x\l ^ ^
Clearing of fractions, 1 = A(x'^  x 4 1) + {Bx + C) (a; + 1).
Or, lz={A + B)x'^ + {A + B + C)x\A\C.
Equating coefficients of like powers of cc,
A\ B = 0,
J. + 5 + C = 0,
and A\C=l.
1 1 2
Solving these equations, ^ =  , B = — , and G = —
3 3 3
Substituting in (1), —1 = ^^^
EXERCISE 174
Separate the following into partial fractions :
2705^ + 8
x'l
6 38a;4a^
REVERSION OF SERIES
406. To revert a given series y = a\ bx"^ + ca;" + '• • • is to ex
press ic as a series proceeding in ascending powers of y.
Ex. Eevert the series y = 2x~3x^{4:X^ — 5x^{ •••.
Assume x = Ay + By^ ^Ci/ + Dy^ + . (1)
3a^_4a;h4
a^1
1043aj
llaj2
{Sx2)(a^
xi2)
x'hSx
5
UNDETERMINED COEFFICIENTS
371
Substituting in this the given value of y,
x = A(2xSx'^\ix^bx^+ •••)
+ ^(4 a;2 4 9 a;4  12 x3 + 16 a^ + ...)
4 0(8x3 36aj4+...) + l>(16x* +...)+ —
That is,
2AXSA
445
x2+ 4A
x^ 6 A
12J5
+ 25 5
+ SC
36(7
+ 16i>
X4 +
Equating coeflBcients of like powers of
X,
2A =
= 1;
3^+45=
:0:
4A
125 + 80 =
:0;
5A+2bB
360+16D =
:0;
etc.
Solving, A =
, 5, 0
. 5
'16'
D =
35
"128
Substituting in (1), x =
2^^8^ ^16
^ ^128
y* +
, etc.
If the even powers of x are wanting in the given series, the
operation may be abridged by assuming x equal to a series con
taining only the odd powers of y.
EXERCISE 175
Revert each of the following to four terms :
1. y = aj+3ar'+5aj3f7a;*+.... ^ . x^ . a^
2. y = x2x^\3a^4.x'^"'.
4. y = 2x + 5x^{Sa^ + llx^ +
3. ,=.+_+_+_+....
^ 2 4 6 8
6. 3,=£+^+^V+
* [2 [3^[4^[5^
7. 2/ = 2a;4a^ + 6a^8a!' +
" 2468
372 ALGEBKA
XXX. THE BINOMIAL THEOREM
FRACTIONAL AND NEGATIVE EXPONENTS
407. It was proved in § 387 that, if ti is a positive integer,
(a + xy = a" + na'x + ^ ^1" ~ ^) a"^a^
1 • 2
n(yil) (712) ^^.3^
123
If n is a negative integer, or a positive or negative fraction,
the series in tlie second member is infinite ; for no one of the
expressions n — 1, n — 2, etc., can equal zero ; in this case, the
series gives the value of (a f xy, provided it is convergent.
As a rigorous proof of the Binomial Theorem for Fractional and Nega
tive Exponents is too difficult for pupils in preparatory schools, the author
has thought best to omit it ; any one desiring a rigorous algebraic proof of
the theorem, will find it in the author's Advanced Course in Algebra, § 675.
408. Examples.
In expanding expressions by the Binomial Theorem when
the exponent is fractional or negative, the exponents and
coefficients of the terms may be found by the laws of § 386,
which hold for all values of the exponent.
1. Expand (a + xy to five terms.
2
The exponent of a in the first term is , and decreases by 1 in each
succeeding term.
The exponent of x in the second term is 1, and increases by 1 in each
succeeding term.
The coefficient of the first term is 1 ; of the second term, —
2 i^
Multiplying , the coefficient of the second term, by — , the exponent
o o
of a in that term, and dividing the product by the exponent of x increased
by 1, or 2, we have — as the coefficient of the third term ; and so on.
y
Then, (a + x)^ = a^ + a~^ x  a'^x"^ + — a~^ x^  ^ a'^ x* 4 —.
' ^ ^ 3 9 81 ' 243
THE BINOMIAL THEOREM 373
2. Expand (1 + 2 a;:^)^ to five terms.
Enclosing 2 x~^ in parentheses, we have
(1+2 x~^)^ = [1 + (2 x~^)]2
= 12 _ 2 . 13 . (2 x~^) + 3 . 1* . (2 x'^y
 4 . 16 . (2 «"^)3 + 5 . 16 . (2 x~^y
= 1  4 a;"^ + 12 x^  32 x~^ + 80 x2 + ....
By writing the exponents of 1, in expanding [1 + (2x~^)]2, we can
make use of the fifth law of § 386.
3. Expand — to four terms.
Enclosing a"^ and — 3x^ in parentheses, we have
^ ^ [(ai)+(3a:br^
'^ai3x* (ai3xbi
= («^)'*  1 (a0"^ (  3 xb + 1 (a^r^ (  3 x^^
f(a^)'^(3xb«+...
= a^ + a^x^ + 2 aW + ^ a'^a; + ....
o
EXERCISE 176
Expand each of the following to five terms :
1. (. + .)t. 6. _^. ''' ^C(a65^cy].
^1. ,,^ 1
2. (l + a.)«. 7. («f+2 6)l (a.^22/V
3. (1  x)K 8. (a«  4 xY^. 13. (^e! + ^'^^.
4. ^^^76. 9. ^_^^^J 14. (^I_3n^)^.
_ _ . m^ + — 
(a + a^y V 4
5._A_. 10. rm + ^V^ 15.' ^ ^
374 ALGEBRA
409. The formula for the rth term of (a j xy (§ 390) holds
for fractional or negative values of n, since it was derived from
an expansion which holds for all values of the exponent.
Ex. Find the 7th term of (a  3 x'^^yi
Enclosing — 3 x~^ in parentheses, we have
(a  3 x~^)~3 = [a ^. (_ 3 x"2)]~3.
The exponent of ( 3 x~^) is 7  1, or 6.
1 19
The exponent of a is 6, or
3 3
1 IQ
The first factor of the numerator is — , and the last factor — — + 1,
or.
The last factor of the denominator is 6.
Hence, the 7th term
_ 1 . _ § . _ Z . _ 10 . _ 13 . _ 16
3 ' 3 ' 3 ' 3 * 3 ' 3 _i_9 _3
= 1.2.3.4.5.6 « ^ C3^ ^)'
38 ^ ^9
EXERCISE 177
Find the :
1. 6th term of (a 4 a^)l ^ 9th term of (a  a;)l
2. 5th term of (a — o)~i
3. 7th term of (1 + x)l
^ ^ , „ , 1 6. 11th term of V(m + nY.
2. 5th term of (a  o)"^. /
7. 7th term of (a'^  2 b^)\
8. 8th term of
4. 8th term of (1  x)\ (a^ f yiy
9. 10th term of (x^ + y^yK
10. 6th term of (a^  2 byK
. 11. 5th term of (m + 3 n^)i
12. 9th term of
THE BINOMIAL THEOREM 375
13. 11th term of ( aW  — V^
14. 10th term of {x~^  4 /)l
410. Extraction of Roots.
The Binomial Theorem may sometimes be used to find the
approximate root of a number which is not a perfect power of
the same degree as the index of the root.
Ex. Find ■\/25 approximately to five places of decimals.
The nearest perfect cube to 25 is 27.
We have V25 = V27  2 = [(S^) + ( 2)]
= (33)i + 1 (33)f (  2)  1 (33)t (  2)2
+ ^(3«)"^(2)»...
= 3 ^ 4 40
3 . 32 9 . 35 81 • 38
Expressing each fraction approximately to the nearest fifth decimal
place, we have
y/2b = 3  .07407  .00183  .00008 = 2.92402.
We then have the following rule :
Separate the given number into two parts, the first of which is
the nearest perfect power of the same degree as the required root,
and expand the result by the Binomial Theorem.
If the ratio of the second term of the binomial to the first is a small
proper fraction, the terms of the expansion diminish rapidly ; but if this
ratio is but little less than 1, it requires a great many terms to insure any
degree of accuracy.
EXERCISE 178
Find the approximate values of the following to five places
of decimals :
1. Vl7. 2. V51. 3. ^/60. 4. ^14. 5. <M. 6. ^35.
376 ALGEBRA
XXXI. LOGARITHMS
411. The Common System.
Every positive arithmetical number may be expressed, exactly
or approximately, as a power of 10.
Thus, 100 = 102; 13 = 10i"3»; etc.
When thus expressed, the corresponding exponent is called
its Logarithm to the Base 10.
Thus, 2 is the logarithm of 100 to the base 10; a relation
which is written logio 1^^ = 2, or simply log 100 = 2.
Logarithms of numbers to the base 10 are called Common
Logarithms, and, collectively, form the Common System.
They are the only ones used for numerical computations.
412. Any positive number, except unity, may be taken as
the base of a system of logarithms ; thus, if a' = m, where a
and m are positive numbers, then x = log^ m.
A negative number is not considered as having a logarithm.
413. By §§ 238 and 239,
10« = 1, 10^ = 1 =.1,
10^ = 10, 102 = ^^=.01,
102 = 100, 103 = r^ = .001, etc.
Whence, by the definition of § 411,
log 1 = 0, log .1 =  1 = 9  10,
log 10 = 1, log .01 =  2 = 8  10,
log 100 = 2, log .001 =  3 = 7  10, etc.
The second form for log .1, log .01, etc., is preferable in practice.
If no base is expressed, the base 10 is understood.
LOGARITHMS 377
414. It is evident from § 413 that the common logarithm of
a number greater than 1 is positive, and the logarithm of a
number between and 1 negative.
415. If a number is not an exact power of 10, its common
logarithm can only be expressed approximately ; the integral
part of the logarithm is called the characteristic, and the decimal
part the mantissa.
For example, log 13 = 1.1139.
Here, the characteristic is 1, and the mantissa .1139.
A negative logarithm is always expressed with a positive
mantissa, which is done by adding and subtracting 10.
Thus, the negative logarithm —2.5863 is written 7.4137 — 10.
In this case, 7 — 10 is the characteristic.
The negative logarithm 7.4137 — 10 is sometimes written 3.4137 ; the
negative sign over the characteristic showing that it alone is negative, the
mantissa being always positive.
For reasons which will appear, only the mantissa of the
logarithm is given in a table of logarithms of numbers; the
characteristic must be found by aid of the rules of §§ 416
and 417.
416. It is evident from § 413 that the logarithm of a
number between
1 and 10 is equal to + a decimal ;
10 and 100 is equal to 1 f a decimal ;
100 and 1000 is equal to 2 + a decimal ; etc.
Therefore, the characteristic of the logarithm of a number
with one place to the left of the decimal point is ; with two
places to the left of the decimal point is 1 ; with three places
to the left of the decimal point is 2 ; etc.
Hence, the characteristic of the logarithm of a number greater
than 1 is 1 less than the number of places to the left of the decimal
point.
For example, the characteristic of log 906328.51 is 5.
378 ALGEBRA
417. In like manner, the logarithm of a number between
1 and .1 is equal to 9 f a decimal — 10 ;
.1 and .01 is equal to 8 f a decimal — 10 ;
.01 and .001 is equal to 7 f a decimal — 10 ; etc.
Therefore, the characteristic of the logarithm of a decimal
with no ciphers between its decimal point and first significant
figure is 9, with — 10 after the mantissa ; of a decimal with
one cipher between its point and first significant figure is 8,
with —10 after the mantissa; of a decimal with two ciphers
between its point and first significant figure is 7, with —10
after the mantissa ; etc.
Hence, to find the characteristic of the logarithm of a number
less than 1, subtract the number of ciphers between the decimal
point and first significant figure from 9, writing — 10 after the
mantissa.
For example, the characteristic of log .007023 is 7, with —10
written after the mantissa.
PROPERTIES OF LOGARITHMS
418. In any system, the logarithm of 1 is 0.
For by § 238, a*^ = 1 ; whence, by § 412, log« 1 = 0.
419. In any system, the logarithm of the base is 1.
For, a^ = a ; whence, log^ a = 1.
420. In any system ijuhose base is greater than 1, the loganthm
of is —oo.
For if a is greater than 1, a"* = — = 1 = (§ 320).
a oo
Whence, by § 412, logaO =  oo.
No literal meaning can be attached to such a result as loga = — oo ;
it must be interpreted as follows :
If, in any system whose base is greater than unity, a number approaches
the limit 0, its logarithm is negative, and increases indefinitely in absolute
value. (Compare § 321.)
LOGARITHMS 379
421. In any system^ the logarithm of a product is equal to
the sum of the logarithms of its factors.
Assume the equations
a' = m
a^ = n
whence, by § 412, [ ^ " J°^« '^^
2/ = logaW.
Multiplying the assumed equations,
a" y^a^ = mn, or a''+^ = mn.
Whence, log^ mn = x^y= log„ m + loga n.
In like manner, the theorem may be proved for the product
of three or more factors.
By aid of § 421, the logarithm of a composite number may
be found when the logarithms of its factors are known.
Ex. Given log 2 = .3010, and log 3 = .4771 ; find log 72.
log 72 = log (2 X 2 X 2 X 3 X 3)
= log2 + log2 + log2 + log3 + log3
= 3 X log2 + 2 X log3 = .9030 + .9542 = 1.8672.
EXERCISE 179
Given
log 2 = .3010, log 3 = .4771, log 5 = .6990, log 7= .8451, find:
1. log 15. 4. log 125. 7. log 567. 10. log 187'5.
2. log 98. 5. log 315. 8. log 1225. 11. log 2646.
3. log 84. 6. log 392. 9. log 1372. 12. log 24696.
422. In any system, the logarithm of a fraction is equal to
the logarithm of the numerator minus the logarithm ' of the
denominator.
Assume the. equations
a'' = m\ ■ f£c = log„m,
I ; whence,
ay = n J
a; = log„m,
l2/ = log„n.
380 ALGEBRA
Dividing the assumed equations,
— = — , or a*^= •
a^ n n
Whence, log„ —z=x — y = log^m — log„n.
Ex. Given log 2 = .3010 ; find log 5.
log 6 = log— = log 10  log 2 = 1  .3010 = .6990.
2
EXERCISE 180
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find :
1. logY. 4. log 245. 7. log If. 10. log ^o^.
2. log 2^. 5. log85f. 8. log 375. 11. log 46f
^3. logllj. 6. log 175. 9. log If. 12. log 2Jf
423. In any system, the logarithm of any power of a number
is equal to the logarithm of the number multiplied by the exponent
of the power.
Assume the equation a* = m ; whence, x = log^ m.
Raising both members of the assumed equation to the^th
P ' a^ — m^ ; whence, log„ mP=px=p log„ m.
424. In any system, the logarithm of any root of a number
is equal to the logarithm of the number divided by the index of
the root.
— i 1
For, log, Vm = log„(m') = log„m (§ 423).
425. Examples.
1. Given log 2 = .3010 ; find log 2l
log 2^ = I X log 2 = ^ X .3010 = .5017.
3 3
To multiply a logarithm by a fraction, multiply first by the numerator,
and divide the result by the denominator.
LOGARITHMS 381
2. Given log 3 = .4771 ; find log ^'3.
Iog^^l2g3 = illl = .0596.
8 8
3. Given log 2 = .3010, log 3 = .4771, find log (2^ x 3^).
By § 421, log (23 x 3^) = log 2^ + log 3?
= i log 2 +  log 3 = .1003 + .5964 = .0907.
EXERCISE 181
Given log 2 = .3010, log 3 = .4771, log7 = .8451, find :
1. log 2«.
5. log42«. 9.
log 50'.
13. log a/8.
2. log5^
6. log 45^. 10.
log ^3.
14. log a/54.
3. log3i
7. log63l 11.
log a/5.
15. log a/225.
4. log 7l
8. log98i 12.
log ^/7.
16. log a/162.
17. log n 21. log ^^.
18. log()i ^2
19. Iog(3^xl00i). 22^ ^^^2±
20. log (5a/3). 5'
23.
24.
log^f
78
log ^' .
^^75
426. To prove the relation
log,m = J°^«
log.
Assume the equations
: whence,
'ic = log„w
. y = logjm
^j
From the assumed equations, a" =
X
Taking the
yth root of both members, oy = h
••
Therefore,
log„6 = , or 2/ =
y
X
log„6
That is.
log.m = ^^'
log
382 ALGEBRA
By aid of this relation, if the logarithm of a number m to
a certain base a is known, its logarithm to any other base h
may be found by dividing by the logarithm of h to the base a.
427. To prove the relation
log6axlog„6 = l.
Putting m = a in the result of § 426,
log.a = M^ = l(§419).
log, 6 log, 6
Whence, logj a x log,6 = 1.
428. In the common system, the mantissce of the logarithms of
numbers having the same sequence of figures are equal.
Suppose, for example, that log 3.053 = .4847.
Then, log 305.3 = log(100 x 3.053) = log 100 + log 3.053
= 2 4 .4847 = 2.4847 ;
log .03053 = log (.01 X 3.053) = log .01 + log 3.053
= 8 10 + . 4847 = 8.4847 10; etc.
It is evident from the above that, if a number be multiplied
or divided by any integral power of 10, producing another
number with the same sequence of figures, the mantissas of
their logarithms will be equal.
For this reason, only mantissse are given, in a table of Com
mon Logarithms ; for to find the logarithm of any number, we
have only to find the mantissa corresponding to its sequence of
figures, and then prefix the characteristic in accordance with
the rules of §§ 416 and 417.
This property of logarithms only holds for the common
system, and constitutes its superiority over other systems for
numerical computation.
429. Ex. Given log2=.3010, log3 = .4771; find log .00432.
We have log 432 = log (2* x 3^) = 4 log 2 + 3 log 3 = 2.0323.
LOGARITHMS 383
Then, by § 428, the mantissa of the result is .6353.
Whence, by § 417, log .00432 = 7.6353  10.
EXERCISE 182
Given log 2 = .3010, log 3 = .4771, log 7 = .8451, find :
1. log 2.7. 6. log .00000686. 11. log 337.5.
2. log 14.7. 7. log .00125. 12. log 3.888.
3. log .56. 8. log 5^70. 13. log (4.5)«.
4. log .0162. 9. log .0000588. . 14. log s/SA.
5. log 22.5. 10. log .000864. 15. log (24.3)1
USE OF THE TABLE
430. The table (pages 384 and 385) gives the mantissae of
the logarithms of all integers from 100 to 1000, calculated to
four places of decimals.
431. To find the logarithm of a number of three figures.
Look in the column headed " Ko." for the first two signifi
cant figures of the given number.
Then the required mantissa will be found in the corre
sponding horizontal line, in the vertical column headed by
the third figure of the number.
Finally, prefix the characteristic in accordance with the
rules of §§ 416 and 417.
For example, log 168 = 2.2253 ;
log .344 = 9.5366  10 ; etc.
For a number consisting of one or two significant figures, the
column headed may be used.
Thus, let it be required to find log 83 and log 9.
By § 428, log 83 has the same mantissa as log 830, and log 9
the same mantissa as log 900.
Hence, log 83 = 1.9191, and log 9 = 0.9542.
384
ALGEBRA
No.
1
2
3
4
6
6
7
8
9
lO
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
II
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0792
0828
0864
0899
©934^
09^
1004
1038
1072
1106
13
"39
"73
T206
1239
1271
1303
1335
1367
1399
1430
14
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
15
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
23
3617
3636
3655
3674
3692
37"
3729
3747
3766
3784
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
4200
4216
4232
4249
4265
4281
4298
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4942
4955
4969
4983
4997
501 1
5024
5038
32
5051
5065
5079
5092
5105
5"9
5132
5145
5159
5172
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
35
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
38
5798
5809
5821
5832
5843
5855
5877
5899
39
59"
5922
5933
5944
5955
5966
5977
5988
5999
6010
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
49
6902
691 1
6920
6928
6937
6946
6955
6964
6972
6981
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7235
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
Uo.
1
2
3
4
5
6
7
8
9
LOGARITHMS
385
No.
1
2
3
4
6
6
7
8
9
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
56
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
6x
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
68
8325
8331
8338
8344
8351
8357
8363
S370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
f932
8938
8943
8949
8954
8960
8965
8971
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9143
9149
9154
9159
9165
9170
9175
9180
9186
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
^^.P
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
9
No.
1
2
3
4
5
6
7
~8~
386 ALGEBRA
432. To find the logarithm of a number of more than three
figures.
1. Eequired the logarithm of 327.6.
We find from the table, log 327 = 2.5145,
log328 = 2.5159.
That is, an increase of one unit in the number produces an increase of
.0014 in the logarithm.
Then an increase of .6 of a unit in the nuflftber will increase the
logarithm by .6 x .0014, or .0008 to the neareaflourth decimal place.
Whence, log 327.6 = 2.5145 + .0008 = 2.5153.
In finding the logarithm of a number, the difference between the next
less and next greater mantissas is called the tabular dij0%Yence ; thus, in
Ex. 1, the tabular difference is .0014.
The subtraction may be performed mentally.
The following rule is derived from the above : J
Find from the table the ma7itissa of the first three significant
figures, and the tabular difference.
Midtiply the latter by the remaining figures of the number, with
a decimal point before them.
Add the result to the maritissa of the first three figures, and
prefix the proper characteristic.
In finding the correction to the nearest units' figure, the decimal por
tion should be omitted, provided that if it is .6, or greater than .5, the
units' figure is increased by 1 ; thus, 13.26 would be taken as 13, 30.5 as
31, and 22.803 as 23,
2. Find the logarithm of .021508. ^
Mantissa 215 = .3324 Tab. diff. = 21
2 _M^
.3326 Correction = 1.68 = 2, nearly.
The result is 8.3326  10.
EXERCISE 183
Find the logarithms of the following:
1. 64. 2. 3.7 3. 982. 4. .798.
LOGARITHMS 387
5. 1079. 9. .00005023. 13. 7.3165.
6. .6757. 10. .0002625. 14. .019608.
7. .09496.  11. 31.393. 15. 810.39.
8. 4.288. 12. 48387. 16. .0025446.
433. To find the number corresponding to a logarithm.
1. Required the number whose logarithm is 1.6571.
Find in the table the mantissa 6571.
In the corresponding line, in the column headed " No.," we find 45, the
first two figures of the required number, and at the head of the column we /
find 4, the third figure. \^ *
Since the characteristic is 1, there must be two places to the left ^^ *^f^ ^^
decimal point (§ 416).
of the ^'V
Hence, the number corresponding to 1.6571 is 45.4.
2. Required the number who^ Ipgarithm is 2.3934.
We find in the table the mantis&'3927 and 39^5.' ■" a
The numbers corresponding to the logarithms 2.3927 and 2.3945 are a'TK
247 and 248, respectively. "^ \ y
That is, an increase of .0018 in the mantissa produces an increase of "^^/^
one unit in the number corresponding. r^O ^
Then, an increase of .0007 in the mantissa will increase the number by \ ^
— of a unit, or .4, nearly. dPy^
Hence, the number corresponding is 247 + .4, or 247.4. « <^,
The following rule is derived from the above : A
Fi7id from the table the next less mantissa, the three figures j/
corresponding, and the tabular difference. ^^
Subtract the next less from the given mantissa, and divide the i
remainder by the tabular difference. ''i,^
Ayinex the quotient to the first three figures of the number, and
point off the result.
The rules for pointing off are the reverse of those of §§ 416 and 417 :
I. ijf— 10 is not written after the mantissa^ add 1 to the characteristic,
giving the number of places to the left of the decimal point.
II. If — \Q is written after the mantissa, subtract the ptositive part of
the characteristic from 9, giving the number of ciphers to be placed between
the decimal point and first significant figure.
388 ALGEBRA
3. Find tKe number whose logarithm is 8.5265 — 10.
5265
Next less mant. = 5263 . ; figures corresponding, 336.
Tab. diff. 13)2.00(.15 = .2, nearly.
13
70
By the above rule, there will be one cipher to be placed between the
decimal point and first significant figure ; the result is .03362.
The correction can usually be depended upon to only one 4ecimal
place ; the division should be carried to two places to determine the last
figure accurately.
EXERCISE 184
Find the numbers corresponding to the following :
1. 0.8189. 6. 8.795410. 11. 1.3019.
2. 7.606410. 7. 6.599310. 12. 4.252710.
3. 1.8767. 8. 9.943710. 13. 2.0159.
4. 2.6760. 9. 0.7781. 14. 3.726410.
5. 3.9826. 10. 5.457110. 15. 4.4929.
APPLICATIONS
434. The approximate value of a number in which the
operations indicated involve only multiplication, division, invo
lution, or evolution may be conveniently found by logarithms.
The utility of the process consists in the fact that addition
takes the place of multiplication, subtraction of division,
multiplication of involution, and division of evolution.
1. Find the value of .0631 x 7.208 x .51272.
By § 421, log (.0631 x 7.208 x .51272)
= log .0631 + log 7.208 + log .51272.
log .0631= 8.800010
'^ log 7.208= 0.8578
log .51272= 9.709910
Adding, log of result = 19.3677  20 = 9.3677  10 (See Note 1.)
Number corresponding to 9.3677  10 = .2332.
LOGARITHMS 389
Note 1. If the sum is a negative logarithm, it should be written in
such a form that the negative portion of the characteristic may be — 10.
Thus, 19.3677  20 is written 9.3677  10.
(In computations with fourplace logarithms, the result cannot usually
be depended upon to more than four significant figures.)
" 336.8
2. Find the value of
7984
By .§ 422, log ^ = log 336.8  log 7984.
log 336.8 = 12.5273 10
log 7984 = 3.9022
Subtracting, log of result = 8.6251  10 (See Note 2.)
Number corresponding = .04218.
Note 2. To subtract a greater logarithm from a less, or a negative
logarithm from a positive, increase the characteristic of the minuend by
10, writing — 10 after the mantissa to compensate.
Thus, to subtract 3.9022 from 2.5273, write the minuend in the form
12.5273  10 ; subtracting 3.9022 from this, the result is 8.6251  10.
3. Find the value of (.07396)^
By § 423, log (.07396)5 = 5 x log .07396.
log .07396 = 8.8690  10
44.345050
= 4.3450  10 = log .000002213.
4. Find the value of V.035063.
By § 424, log v^.035063 = i log .035063.
o
log .035063 = 8.5449 10
3 )28.544930 (See Note 3.)
9.5150 10 = log .3274.
Note 3. To divide a negative logarithm, write it in such a form that
the negative portion of the characteristic may be exactly divisible by the
divisor, with — 10 as the quotient.
Thus, to divide 8.5449 — 10 by 3, we write the logarithm in the form
28.5449  30 ; dividing this by 3, the quotient is 9.5150  10.
390
ALGEBRA
EXERCISE 185
A negative number has no common logarithm (§ 412); if such numbers
occur in computation, they may be treated as if they were positive, and
the sign of the result determined irrespective of the logarithmic work.
Thus, in Ex. 3 of the following set, to find the value of ( 95.86) x 3.3918
we find the value of 95.86 x 3.3918, and put a — sign before the result.
Find by logarithms the values of the following :
1. 4.253x7.104. 4. 54.029 x ( .0081487).
2. 6823.2 X .1634. 5. .040764 x .12896.
3. ( 95.86) X 3.3918. 6. (285.46) x ( .00070682).
8.
9.
10.
11,
5978
9.762'
21.658
45057 *
.06405
.002037'
38.19
.10792 *
670.43
^5382.3*
12
.000007913
.00082375
13. (88.08)1
14. (.09437)^
15. (3.625)^
16. (.4623)^
17. loot
18. (.09)1
19. (85.7)^.
20. (.000216)1
21. V7.
22. </3.
23. V"^.
24. VVd.
25. v/:2005.
26. V.08367.
27. V.00015027.
28. \/'^^^:0040628.
435. Arithmetical Complement.
The Arithmetical Complement of the logarithm of a number,
or, briefly, the Cologarithm of the number, is the logarithm of
the reciprocal of that number.
Thus, colog 409 = log ^^ = log 1  log 409.
log 1 = 10.  10 (See Ex. 2, § 434.)
log 409 = 2.6117
.. colog 409 = 7.388310.
1
Again, colog .067 = log
.067
log 1  log .067.
LOGARITHMS 891
log 1=10. 10
log .067 = 8.826110
. •. colog .067 = 1.1739.
It follows from the above that the cologarithm of a number
may be found by subtracting its logarithm from 10 — 10.
The cologarithm may be found by subtracting the last significant figure
of the logarithm from 10 and each of the others from 9,10 being
written after the result in the case of a positive logarithm.
.51384
Ex.
log
a. oiio V .
^^ 8.708 X .0946
.51384
8.708 X .0946
.log (.51384 x^^^^^x_^y
= log.51384 + log^}^^ + log^^l^^
= log .51384 + colog 8.708 + colog .0946.
log.
51.384
= 9.710910
colog
8.708
= 9.0601  10
colog
.0946
= 1.0241
9.7951  10 = log .6239.
It is evident from the above example that, to find the loga
rithm of a fraction whose terms are the products of factors, we
add together the logarithms of the factors of the numerator, and
the cologarithms of the factors of the denominator.
The value of the above fraction may be found without using cologa
rithms, by the following formula :
log i^l^§^ = log .51384  log (8.709 x .0946)
^ 8.709 X. 0946 ^ ^^ ^
= log .51384  (log 8.709 + log .0946).
The advantage in the use of cologarithms is that the written work of
computation is exhibited in a more compact form.
MISCELLANEOUS EXAMPLES
436. 1. Find the value of ^^^
3^
392 ALGEBRA
log ^ = log 2 + log v^5 + colog 3^ (§ 435)
3^ 1 ^
= log2 + Mog5 + ^colog3.
6 b
log 2= .3010
log 5= .6990; r 3 = .2330
colog 3 ^ 9.5229  10 ; x  = 9.6024  10
.1364 =logl.
2. Find the value of \l ~'^^!^^^ '
V 7.962
>'7.9
03_296 ^ 1 i,g :03296 ^ 1 ^^^^^ _ ^
962 3 ^ 7.962 3 ^ ^ & J
log .03296 = 8.5180  10
log 7.962 = 0.9010
3 )27.617030
9.2057 10 = log. 1606.
The result is  .1606.
EXERCISE 186
Find by logarithms the values of the following :
J 2078.5 X .05834 . « (.076917) x 26.3
.3583x346 * * .5478 x (3120.7)*
2.
(6.08) X. 1304 . .8102 x( 6.225)
4.046 X .0031095* ' ( .0721) x ( 17.976)'
5. 6^x5i .f. f 5510\^ 14 ^4/7 8/3
^ ^' ,3gy 15. V6X ^10x^2.
^ n/68 ^^' VsoTg* 16 f ^^^^ 'V ,
^' SIsS' ' \ 8.7 X. 06037
g _^^ ..12. ^^^=f^. jy VTMMe
(.1)1* ^' * ^/.0003867
9 (100)» , j3 (.03)^ 18. (^4582)^ .
^ _ .004 s/  1000  (.72346)^
LOGARITHMS 893
19. ( 143.59)" x(.00532)^ ^^ (.0462)^
20. ^ 40.954 X .0002098 . ' ^^8.27 x V:2296'
i" ^ OA ^ 7.92 X (.1 807)^
21. (3075.6)* X (.016432)^. 24. ^^^^^^^ Z_.
22 v^28l8 X v^MO , 25 27.931
^61021 \/M6 X (.03023)7
26. \/ .067268 x ^.4175 x a/.00263.
27 .0005616 x a/424:6 28 485.7 x (.7301)^ x S^TOOO
(6.73)^ X (.03194)^ (9.127)« x (.7095)^
EXPONENTIAL EQUATIONS
437. An Exponential Equation is an equation in which the
unknown number occurs as an exponent.
To solve an equation of this form, take the logarithms of both
members ; the result will be an equation which can be solved by
ordinary algebraic methods.
1. Given 31^ = 23 ; find the value of x.
Taking the logarithms of both members,
log (31^) = log 23 ; or a: log 31 = log 23 (§ 423).
Then, ^^I2g23^0617^9,3
log 31 1.4914
2. Solve the equation .2^ = 3.
Taking the logarithms of both members, x log .2 = log 3.
Then, x = i^ = — iII^^i^ = .0285 + .
log. 2 9.301010 .699
An equation of the form w" = h may be solved by inspection
if b can be expressed as an exact power of ^.
3. Solve the equation 16* = 128.
We may write the equation (2*)=« = 2^, or 2^" = 2^.
7
Then, by inspection, Ax = l \ and x = —
4
394 ALGEBRA
(If the equation were 16* = ^, we could write it (2*)* = — = 2"'^ ;
y^ 128 2'
then 4 x would equal — 7, and x
I)
EXERCISE 187
Solve the following equations :
1. 13^ = 8. 5. 34^1 = 42^+1 9. 32=gV
2. .06^ = .9. 6. 7^+2^.8^ 10. (tV)' = 8.
3. 9.347=^ = . 0625. 7. .2^+^ = .5^^ 11. (i)^=2V
4. .005038^ = 816.3. 8. 16^ = 32.
12. Given a, r, and l] derive the formula for n (§ 372).
13. Given a, r, and S ; derive the formula for n.
14. Given a, I, and S ; derive the formula for n.
15. Given ?*, I, and S, derive the formula for n.
438. 1. Find the logarithm of .3 to the base 7.
By S 426, log, .3 = Ip^ = «:«Ii^ = ^ =  ■^^^^
logio7 .8451 .84ol
Examples of this kind may be solved by inspection, if the
number can be expressed as an exact power of the base.
2. Find the logarithm of 128 to the base 16.
Let logic 128 = x ; then, by § 412, 16^ = 128.
Then, as in Ex. 3, § 437, x = ; that is, logie 128 = I
4 4
EXERCISE 188
Find the values of the following :
1. log7 59. 3. log.482. 5. log68 2.92.
2. log6.7. 4. log.9 .00453. 6. logsi .0604.
Find by inspection the values of the following :
7. log«125. 8. log«(). 9. log^;^(3). 10. log^,^_(rh)
MISCELLANEOUS TOPICS 395
XXXII. MISCELLANEOUS TOPICS
HIGHEST COMMON FACTOR AND LOWEST COMMON
MULTIPLE BY DIVISION
439. We will now show how to find the H. C. F. of two
polynomials which cannot be readily factored by inspection.
The rule in Arithmetic for the H. C. F. of two numbers is :
Divide the greater number by the less.
If there be a remainder, divide the divisor by it; and con
tinue thus to make the remainder the divisor, and the preceding
divisor the dividend, until there is no remainder.
The last divisor is the H. C. F. required.
Thus, let it be required to find the H. C. F. of 169 and 546.
169)546(3
507
39)169(4
156
13)39(3
39
Then, 13 is the H. C. F. required.
440. We will now prove that a rule similar to that of § 439
holds for the H. C. F. of two algebraic expressions.
Let A and B be two polynomials, arranged according to the
descending powers of some common letter.
Let the exponent of this letter in the first term of A be
equal to, or greater than, its exponent in the first term of B.
Suppose that B is contained \n A p times, with a remainder
(7; that C is contained in B q times, with a remainder D\ and
that D is contained in O r times, with no remainder.
To prove that D is the H. C. F. of A and B.
The operation of division is shown as follows.
396 ALGEBRA
B)A{p
pB
~~C)B(q
D)C{r
rD
We will first prove that Z> is a common factor of A and B.
Since the minuend is equal to the subtrahend plus the
remainder (§ 34),
A=pB + G, (1)
B=qC + D, (2)
and C = rD.
Substituting the value of C in (2), we obtain
B=qrD^D = D{qr + l). (3)
Substituting the values of B and C in (1), we have
A=pD{qr\l) + rD = D{pqr+p + r). (4)
From (3) and (4), i) is a common factor of A and B.
We will next prove that every common factor of A and B is
a factor of D.
Let F be any common factor of A and B ; and let
A = mF, and B = nF.
From the operation of division, we have
C=ApB, (5)
and D = BqC. (6)
Substituting the values of A and B in (5), we have
(7= mF — priF.
Substituting the values of B and C in (6), we have
D = nF — q{mF — pnF) = F(n — qm + 2^9'^)
MISCELLANEOUS TOPICS 397
Whence, i^is a factor oi D. >
Then, since every common factor of A and i9 is a factor of
D, and since D is itself a common factor of A and B, it follows
that D is the highest common factor of A and B.
We then have the following rule for the H. C. F. of two
polynomials, A and B, arranged according to the descending
powers of some common letter, the exponent of that letter in
the first term of A being equal to, or greater than, its exponent
in the first term of B :
Divide A by B.
If there be a remainder, divide the divisor by it; and continue
thus to make the remainder the divisor, and the preceding divisor
the dividend, until there is no remainder.
The last divisor is the H. C. F. required.
It is important to keep the work throughout in descending powers of
some common letter ; and each division sliould be continued until the
exponent of this letter in the first term of the remainder is less than its
exponent in the first term of the divisor.
Note 1. If the terms of one of the expressions have a common factor
which is not a common factor of the terms of the other, it may be re
moved ; for it can evidently form no part of the highest common factor.
In like manner, we may divide any remainder by a factor which is not
a factor of the preceding divisor.
1. Find the H. C. F. of
6 a^ 25 a; + 14 and 6 x^  7 x"  25 x + IS.
6ic225x+14)6a;3 7 x'^ 26x + lS(x + S
6 a;3  25 x2 + 14 a;
18 x2  39 X
18 x2  75 a; + 42
36 X  24
In accordance with Note 1, we divide this remainder by 12, giving
*~ * 3x2)6x225x + 14(2ic7
6x2 ix
21x
 21 X + 14
Then, 3 x  2 is the H. C. F. required.
398 ALGEBRA
Note 2. If the first term of the dividend, or of any remainder, is not
divisible by the first term of the divisor, it may be made so by multiply
ing the dividend or remainder by any term which is not a factor of the
divisor.
2. Find the H. C. F. of
3a^\d'b2 ab' and Aa^b + 2 w'b^  ab^ + b\
We remove the factor a from the first expression and the factor b from
the second (Note 1), and find the H. C. F. of
Sa^ + ab2b^ and 4a^\2 a^b  ab'^ + b^
Since 4 a^ is not divisible by 3 a'^, we multiply the second expression by
3 (Note 2).
4 a3 + 2 a^b  ab^ + b'^
3 a2 + a6  2 b)\2 d^ + %a%2> ab' + 3 Z>3(4 a
12 gs 4, 4 a2j) _ g a&2
2 a^b + 5 a62 _^ 3 ^3
Since 2 a% is not divisible by 3 a^, we multiply this remainder by
3 (Note 2).
2«26+ 5a&2^ 3&3
3
3 a2 + a6 _ 2 62)6 (^25 + 15 ^^2 + 9 63(2 ft
Qa^b+ 2ab^ 4 63
13 a62 + 13 68
We divide this remainder by 13 62 (Note 1), giving a + 6.
a + 6)3 a2 + a6  2 62(3 a  2 6
Sa^ + Sab
2ab
 2 a6  2 62
Then, a + 6 is the H. C. F. required.
Note 3. If the first term of any remainder is negative, the sign of
each term of the remainder may be changed.
Note 4. If the given expressions have a common factor which can
be seen by inspection, remove it, and find the H. C. F. of the resulting
expressions ; the result, multiplied by the common factor, will be the
H. C. F. of the given expressions.
MISCELLANEOUS TOPICS 399
3. Find the H. C. F. of
2x'h3x'6x{2x and 6 a;H 5 x^  2 o^  a;.
Removing the common factor x (Note 4), we find the H. C. F. of
2 x3 + 3 x2  6 a; + 2 and 6x^+ 5x^ 2xl.
2a;3 + 3x26x + 2)6x3 + 5ic2 2xl(3
6 x3 + 9 ic2  18 X + 6
 4 x2 + 16 X  7
The first term of this remainder being negative, we change the sign of
each of its terms (Note 3).
2x3+3x26x42
2
4x2 16x + 7)4x3+ 6x2 12 x + 4(x
4x316x2+ 7x
22x2 19 x+ 4
2
44x2 38x+ 8(11
44x2176x + 77
69 )138 X 69
2x 1
2x 1)4x2 16 x+ 7(2x7
4 x2 — 2 X
 14x
 14x + 7
The last divisor is 2 x — 1 ; multiplying this by x, the H. C. F. of the
given expressions is x(2 x — 1).
(In the above solution, we multiply 2 x3 + 3 x2 — 6 x + 2 by 2 in order
to make its first term divisible by 4 x2 ; and we multiply the remainder
22 x2 — 19 X + 4 by 2 to make its first term divisible by 4 x2.)
EXERCISE 189
Find the H. C. F. of the following :
1. 2a^ + a6, Aa^Sa\S.
2. Qx'lTx^lO, 9a^UxS.
3. x'6x27, a^2x'Sx + 21.
400 ALGEBRA
4. Gx^x2, Sx'Ux'x^e.
5. 2'ia'22ab7b', 32a'12ab5b\
6. 16a^hSx'y\13xy^ + 3f, 24.x^Ux'y\13xy'15 y^
7. 4:a^ + 4:X^3x, 6x^ \llx^x^6x.
8. AxFyWxy^jdy^ S x' IS x^'y + 25 x'y' 12 xf.
9. 6 a^ {5 a' 6 a' 3 a^ + 2 a", 9 a' ilS a'^5 a' 8 a 4.
10. 3 a^  13 a'b + 3 a'b' + 4 a&^ 9 a^b + 12 a^^s _ g at^ _ 5 b\
11. 4a^+ 9aj 9, 4x^ + 10ar''7a;2 + 9.
12. 6a*7a25a2 + 5a3, S a'Qa' 5 a' 9.
13. 3 ^3 + 8 w^a;  9 nx + 2 ar^,
6n' ^23 n^x\2 n^x" 13 nx^ + 2 x\
14. a3 + 9a2 413a15, a^ + 9 (1^ + 22 fr'^ + 9 a29 a.
15. m«27m^ m«44m^25m^ + 12 ml
16. 9a^ + 30a^&21a2&2f12a&^
16 a^b + 60 a^ft^ _ 20 ab^  16 &^
17. 4:x'llxy20y% 2 x^  4. x'y  17 x^ + xy^ \ 12 y".
18. 4a^ + 8a415a^ + 2a24a, Aa^12a'{9a^3a\2.
19. 3a;38a^ + 16a;8, 3x*5a:3 + 5r^lla; + 6.
20. 3a^^2_2a;y_7^^4_^7^2^5^3^^6^
3 a^2/' + 7 a;y + 5 ar^/ 5 aff2xy^
21. 2a;* + 5a^ + 4a^ + 7a; + 6, 2 a;*  5 aj^ 4. n a;2 _ 9 3, __ 9^
22. 6a;*4a^ + 3a^6aj4, 12 a;4 + 8 ar^3 aj^lO a;4.
23. 3a^8aj25a; + 6, a^ 5x^ + 5 a^ + «2_^7 a;3.
441. The H. C. F. of three expressions, which cannot be
readily factored by inspection, may be found as follows :
Let A, B, and C be the expressions.
Let G be the H. C. F. of A and B ; then, every common factor
of G and C is a common factor of A, B, and C.
MISCELLANEOUS TOPICS 401
But since every common factor of two expressions exactly
divides their H. C. F., every common factor of A, B, and C is
also a common factor of G and C.
Whence, the H. C. F. of G and C is the H. C. F. of A, B,
and C.
Hence, to find the H. C.F. of three expressions, find the H. C.F.
of two of them,, and then of this result and the third expression.
We proceed in a similar manner to find the H. C. F. of any
number of expressions.
Ex. Find the H. C. F. of
a;3_7a; + 6, ar^ + 3 a;^  16 ic + 12, and »» 5 a;2 + 7 a;3.
The H. C. F. of x3  7 X + 6 and x3 + 3 x2  16 X + 12 is x^  3 x + 2.
The H. C. F. of x2  3 X + 2 and x3  5 x2 + 7 X  3 is X  1.
EXERCISE 190
Find the H. C. F. of the following :
1. 6a;25a;25, 9x^ + 27 a; + 20, 12 ic^ + 11 a;  15.
2. 20a2H23a57 52, 28^^43 a6+9 6^ 24.a'\U ab5b'.
3. 5a233a14, 5a^13a^Ua{S, 5aM27a2+20a+4.
4. Sx'GxySBy^ 10 x" 27 xhj xy' ^15 f,
6x^13x^yrS xy + 20 y\
5. a53^4a;2llic + 30, ar*^ + 2 a;25 a;6, ar^  a;^  17 a;  15.
6. a3_8a2 + 20a16, a^ + 3a''4.a12,
a3_6a2 + lla6.
7. 3a3 + 17a26 + 18a62_85^ 6a^ 4a'6 19 a6 + 6 6^
8. 3a^a;2_38^_24^ 3^:3^5 ^_53 ^_40^
3a:3 + 26a;2_^61a.' + 30.
442. We will now show how to find the L.C.M. of two
expressions which cannot be readily factored by inspection.
402 ALGEBRA
Let A and B be any two expressions.
Let F be their H. C. F., and M their L. C. M.
Suppose that A = aF, and B = bF.
Then, AxB = abF\ (1)
Since F is the H. C. F. of A and jB, a and h have no common
factors 5 whence, the L. C. M. of aF and hF is a6i^.
That is, M=abF.
Multiplying each of these equals by F, we have
FxM=ahF\ (2)
From (1) and (2), AxB = FxM. (Ax. 4, § 9)
That is, the product of two expressions is equal to the product
of their H. C. F. and L. C. M.
Therefore, to find the L. C. M. of two expressions.
Divide their product by their highest common factor ; or,
Divide one of the expressio7is by their highest common factor ^
and multiply the quotient by the other expression.
Ex. Find the L. C. M. of
6a:217x + 12 and 12x24i»2L
6 a;2 _ 17 x+ 12)12x2 4x21(2
12 x2  34 a; + 24
15)30 X  45
2x 3)r)x217x + 12(3a;
6x2 9x
4
 8x
 8X + 12
Then, the H. C. F. of the expressions is 2 x — 3.
Dividing 6 x^ — 17 x + 1 2 by 2 x — 3, the quotient is 3 x — 4.
Then, the L. C. M. is (3 x 4) (12x2 4 x 21).
EXERCISE 191
Find the L. C. M. of the following :
1. Sx'{Ux24, 3i»2__23aj + 30.
MISCELLANEOUS TOPICS 403
2. 6x'31x7j + lSy% 9 a^ hl5xyUf.
3. 4aj2 + 13a; + 3, 4a^23a;6.
4. Sx' + 6x9, ex^ + Tx'Txe.
5. 3 a^  8 a'b + 4 a^^^ a^6  11 a'b^ f 22 aft^  8 b\
6. 6w3+n^i«llw^'6a^, 6 7i35 7i2iz;8na^2_p3a^.
7. 2a;^ + 7a^H7a^2__2a;^ 2 a;^ + a;'  10 a;^  8 a;.
8. 6a^ + aj217ic410, 3aj^ + 5ar^5a;2_5a; + 2.
9. 4.x'llx3, Sx^ + 6a^llx'23x5.
10. 2a;^ar'^7/4a?y43a;2/'j S x^y  10 x'y^ { 12 xf  10 y\
11. 6 m^— 17 mhi — 1 mn^\^ n^, 12 m^ — 13 m^nH21 myi^—Qn^.
12. 2ar'45a;*2ar^ + 3a^, 3 a;« + 8a5^ 2a;* + a^ GojI
13. a^2a^2a? + laQ, d" 4.a^ ^d" ^1 a2.
It follows from § 442 that, if two expressions are prime to
each other (§ 128), their product is their L. C. M.
443. The L. C. M. of three expressions may be found as
follows :
Let A, B, and C be the expressions.
Let M be the L. C. M. of A and B ; then every common mul
tiple of 3f and (7 is a common multiple of A, B, and C.
But since every common multiple of two expressions is ex
actly divisible by their L. G. M., every common multiple of A,
B, and C is also a common multiple of M and C.
Then, the L. C. M. of M and C is the L. C. M. of A, B, and C.
Hence, to find the L. C. M. of three expressions, find the L. C. M.
of two of them, and then of this result and the third expression.
We proceed in a similar manner to find the L. C. M. of any
number of expressions.
EXERCISE 192
Find the L. C. M. of the following :
1. 3a^4a;4, 3a;27x + 2, 3 x^  10 a?  8.
404 ALGEBRA
2. 2a«H3a49a^ 4a^ + 13a^ + 3a% 6 a^ + 13 a^lS a.
3. 37i2ll7i4, 47i222w + 24, 6?i2 + lln + 3.
4. 4a3^_4a243a + 20, 4 a3 + 20a2 + 13 a12,
4a« + 12a231a60.
5. 2a^5x + 3, 4aj34a^ + 3a;9, 4a^13aj + 6.
444. We will now show how to reduce a fraction to its low
est terms, when the numerator and denominator cannot be
readily factored by inspection.
By § 127, the H. C. F. of two expressions is their common
factor of highest degree, having the numerical coefficient of
greatest absolute value in its term of highest degree.
We then have the following rule :
Divide both numerator and denominator by their H. C. F.
Ex. Reduce ^ a^  H a2 + 7 a 6 ^^ .^^ lowest terms.
2 a^  a  3
By the rule of § 440, we find the H. C. F. of 6 a^ 11 a"^ + 7 a 6 and
2 a2  a  3 to be 2 a  3.
Dividing 6 a^ _ n ^^2 _,_ 7 ^ _ 6 \)y 2 a  3, the quotient is Sa^a +2.
Dividing 2 a^ — a — 3 by 2 a — 3, the quotient is a + 1.
6a3_iia2 + 7a_6 3a2a + 2
Then,
2a2a3 a + 1
EXERCISE 193
Reduce each of the followins: to its lowest terms
1.
6a2_5a4
2.
6 m2 h 7 m  20
4m2 + 16m + 15
3.
5a^lSab6b'
4a215a& + 9&2
4.
4a^4a;3
3a^M0a;f8
6.
5a^ + lla^^22a\~4:
5a3 + 8a2 + 16a8
7 4yi«6n^ + Syt6
n^Sn^2n + 4: '
8.
4a3 4.l3a26_4a&6ft3
4 0^3 _ 4 ^2 _ 5 ^ _ 1 8 a^ 4. 14 a5  15 ?/
MISCELLANEOUS TOPICS 405
9 ea^o^llx + Q ^Q 2a^9x'y2xy'15if
' 9a^lSx'^nx2' ' 2x^7 aryUxy'+5f'
PROOF OF (1), § 235, FOR ALL VALUES OF m AND n
445. I. Let m = ^ and n = , where p, q, r, and s are posi
re integers. p ^r ps I __
We have, a^ x a* = a«' x a'' = Va^' x Vo^'" (§ 237)
= Va^" X a^' (§ 234) = Va^^+''' (§ 56) =a '' (§ 237) = a' ».
We have now proved that (1), § 235, holds when m and w
are any positive integers or positive fractions.
II. Let m be a positive integer or fraction ; and let n = — q,
where g is a positive integer or fraction less than m.
By §§ 56, or 445, I, a*"* x a' = a"^^^^ = a"*.
Whence, a'"«= — = a'" x a"' (§ 240).
a'
That is, a'" x a~' = a'""''.
III. Let m be a positive integer or fraction ; and let n = q,
where g' is a positive integer or fraction greater than m.
By § 240, a'" x a^= ^— = ^— (§ 445, II) = a*""'.
IV. Let m = —p and n = — q, where p and g are positive inte
gers or fractions.
Then, a"^ x a"'? == — = — (§ § 56, or 445, I) = a"^"'.
Then, a"* x a" = a"*"^"* for all positive or negative, integral or
fractional, values of m and n. .
446. We will now show how to reduce a fraction whose
denominator is irrational to an equivalent fraction having a
rational denominator, when the denominator is the sum of a
rational expression and a surd of the nth degree, or of two
surds of the ?ith degree.
406 ALGEBRA
1. Reduce ^ to an equivalent fraction having a
2 + ^3 , ^
rational denominator.
1 1
We have,
2 + v^ 8^ + 3^
Now, (a + &) (a2  ab + b^) = a^ \ b^ (§ 102).
Then, if we multiply both terms by 8^  83 . S^ + 3^, the denomiDator
will become rational ; thus,
1 _ 8^  8^ ■ 3^ + 3^ _ (8^)2  8^ . 33 + 3^
8^ + 3^ (8^ + 3^) (8^ 8^. 33 + 3^) 8 + 3
42^/3 + v'9
2. Reduce — to an equivalent fraction haviner a
</T</E
rational denominator.
1 1
We have,
^_^5 7^5^
Now, (a  6) (a3 + a% + ab^ + b^) = a*  6* (§ 103).
Then, if we multiply both terms by
the denominator will become rational ; thus,
1 7^ + 7^ . 5* + 7^ . 5^ + 6*
7i _ 5? (7? _ 6?) (7? + 7^ . 6^ + 7* • 5^ + 6^)
_ \/7« + V^72 . 5 + VTT52 + \/p ^ v/343 + v^245 + Vm + C/125
75 ~ 2
The .method of § 446 can be applied to cases where the
denominator is in the form VaV&, or Va — \/b.
3. Reduce to an equivalent fraction having a
rational denominator.
The lowest common multiple of the indices 3 and 2 is 6.
MISCELLANEOUS TOPICS 407
We have, —z = — z — rzz = ; 7'
y/2 + VE ^22 + v^53 (22)i + (53)i
Now, (a + b) (a5 _ a*^ + QjS^a _ ^253 ^ 0^54 _ 55) = a^_ ^6.
Then, if we multiply both terms by
(22) f _ (22)t(53)^ + (22)1(53)1  (22)i(53)t + (22)i(53)t (53)t,
the denominator will become rational ; thus,
1 _ 2t  2t ■ 5t 4 2 • 5  2t • 5t + 2i • 5^  5t
(22)i+(53)5 (22)t + (53)t
_ 2 ♦ 2!  2 > 2t . 5t + 10  2^ . 5 « 5t + 2^ ♦ 52  52 . 5I
22 + 53
_ 2v^2v^^2T58 + l05v^^*^ + 25\/225\/5
~ 4 + 125
_ 10 + 2v^i2v^5005\/2000 + 25v^225\/5
129
EXERCISE 194
Keduce each of the following to an equivalent fraction having
a rational denominator :
1
1. J 3 ___L__. 5
2 ^_. 4. ^ 6. V3V2.
2V/4 a/3 4 a/4 V3 + V2
THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL
EXPONENTS
447. In the proof of § 387, we only considered the first four
terms of the expansion of (a + a;)*'+^, in equation (2).
To make the proof complete, we must show that the fifth law
of § 386 holds for any two consecutive terms, in equation (2).
Let P, Q, and R denote the coefficients of the terms involv
ing a'»''a^", a'*~''~V+\ and a""''~V+2, respectively, in the second
member of (1), § 386.
408 ALGEBRA
Thus, (a + «)" = a" + na^''^x + • • •
+ Pa^'af + Qa''' V+i + i2a"'V+2 + .... (3)
Multiplying both members by a + x, we have
(a + xy+^ = a'^+i + na^'x \ \ Qa"'"x'+i + Ra^'^xr^^ + ...
+ a'^ic H h Pa''iC+i + Qa"* V+2 + . . .
= a'*+i + (7i4l)a"a;+..
+ (P4 Q)a'*'aj'+i+(Q+i2)a"'V+2+ .... (4)
Since the fifth law of § 386 is assumed to hold with respect
to the second member of (3), we have
r + 1 r + 2
Therefore,
Q{nT1) Qin + V)
Q + E r + 2 r^'l nr
P+Q Q(rhl) I Q Q(n + 1) r + 2
n—r n—r
Whence, Q^]i=(p^Qy
n — r
n — r
r + 2
But n — r is the exponent of a in that term of (4) whose coeffi
cient is P^Q, and r + 2 is the exponent of x increased by 1.
Therefore, the fifth law holds with respect to any twp con
secutive terms in equation (2), § 387.
wp
/I
THE THEOREM OF UNDETERMINED COEFFICIENTS
448. Before giving the more rigorous proof of the Theorem
of Undetermined Coefficients, we will prove two theorems in
regard to infinite series.
First, if the infinite series
a \hx + cx^ { da^ \ "*
is convergent for some finite value of x, it is Jinite for this value
of X (§ 393), and therefore finite when x=0.
Hence, the series is convergent when a; = 0.
MISCELLANEOUS TOPICS 409
449. Second, if the infinite series
is convergent for some finite value of x, it equals when a; = 0.
For, ax + 6a^ + ca;^  • • • is finite for this value of x, and
hence a + 6aj + ca^ + ••• is finite for this value of x.
Then, a \ hx { cx^ \ • • • is finite when aj = ; and therefore
x{a\hx + C7? + •••), or ax + hy?\CQi^'\ •••, equals when a;=0.
450. Proof of the Theorem of Undetermined Coefficients (§ 396).
The equation
^ + ija; + (7x2 + Z>a^ + ... = ^' 4. B'x + Ox^ + Z)'a^ + ... (1)
is satisfied when x has any value which makes both members
convergent ; and since both members are convergent when x =
(§ 448), the equation is satisfied when x = 0.
Putting X = 0, we have by § 449,
5x + (7x2 + i)x3 + ... ^ 0, and B'x + C'x^ + D'x^ + ... = 0.
Whence, A = A'.
Subtracting A from the first member of (1), and its equal A'
from the second member, we have
Bx \ Cx" + Da^ + ... = B'x + C'x^ + D'^ + •.
Dividing each term by x,
B {. Cx h Dx' + •" = B' { C'x + D'x' + .... (2)
The members of this equation are finite for the same values
of X as the given series (§ 449).
Then, they are convergent, and therefore equal, for the same
values of x as the given series.
Then the equation (2) is satisfied when x = 0.
Putting x = 0, we have B=B'.
Proceeding in this way, we may prove C = C, etc.
410 ALGEBRA
XXXIII. THE FUNDAMENTAL LAWS FOR
ADDITION AND MULTIPLICATION
451. The Commutative Law for Addition.
If a man gains ^ 8, then loses $ 3, then gains $ 6, and finally
loses $ 2, the effect on his property will be the same in what
ever order the transactions occur.
Then, with the notation of § 16, the result of adding + $ 8,
— $ 3, + $ 6, and — $2, will be the same in whatever order
the transactions occur.
Then, omitting reference to the unit, the result of adding
+ 8, — 3, +6, and — 2 will be the same in whatever order the
numbers are taken.
This is the Commutative Law for Addition, which is :
The sum of any set of numbers will he the same in whatever
order they may he added.
452. The Associative Law for Addition.
The result of adding 6 + c to a is expressed a + (6 f c), which
equals (&+c)a by the Commutative Law for Addition (§ 451).
But (& + c) f a equals 6 + c + a, by the definition of § 3 ; and
h\c{a equals a+h\c, by the Commutative Law for Addition.
Whence, . a \ (h + c) = a \h \ c.
Then, to add the sum of a set of numbers, we add the num
bers separately.
This is the Associative Law for Addition.
453. The Commutative Law for Multiplication.
The product of a set of numbers will be the same in whatever
order they may be multiplied.
By § 55, the sigji of the product of any number of terms is
independent of their order ; hence, it is sufficient to prove the
commutative law for arithmetical numbers.
LAWS FOR ADDITION AND MULTIPLICATION 411
Let there be, in the figure, a stars in each row, a m 2^ row
and h rows. * # # # ..
We may find the entire number of stars by * * * ^ •
multiplying the number in each row, a, by the * * * ^ ••
number of rows, h. ' ' '
Thus, the entire number of stars is a x 6.
We may also find the entire number of stars by multiply
ing the number in each vertical column, b, by the number of
columns, a.
Thus, the entire number of stars is b x a.
Therefore, a xb = b x a.
This proves the law for the product of two positive integers.
Again, let c, d, e, and/ be any positive integers.
C S C X s
Then,  x — = ; for, to multiply two fractions, we
multiply the numerators together for the numerator of the
product, and the denominators together for its denominator.
Then,  x — = ; since the commutative law for multi
' d f fxd'
plication holds for the product of two positive integers.
C S 6 C
Hence,  x — = — x ; which proves the commutative law
(^ f f d
for the product of two positive fractions.
454. Thei Associative Law for Multiplication.
To midtiply by the product of a set of riumbers, we multiply by
the numbers of the set separately.
This law was assumed to hold in §§ 56 and 57.
The result of multiplying a by be is expressed a x (be), which
equals (be) X a, by the Commutative Law for Multiplication.
But by the definition of § 5, (be) x a equals bca, which equals
abc by the Commutative Law for Multiplication.
Whence, a x (be) = abc.
This proves the law for the product of three numbers.
412 ALGEBRA
The Commutative and Associative Laws for Multiplication may be
proved for the product of any number of arithmetical numbers.
(See the author's Advanced Course in Algebra, §§ 18 and 19.)
455. The Distributive Law for Multiplication.
The law is expressed (a + h)c = ac\bc (§ 40).
We will now prove this result for all values of a, b, and c.
I. Let a and b have any values, and let c be a positive
integer.
Then, (a + b)c = {a + b) { (a { b) \ ••• to c terms
= (a{a\ '•• to c terms) + (6 f & + • • • to c terms)
(by the Commutative and Associative Laws for Addition),
= ac f be.
II. Let a and b have any values, and let c = — , where e and
/ are positive integers. ^
Since the product of the quotient and divisor equals the
dividend, ^
Then, (a + b) x jX f=(a\b) xe = ae^be, by I.
Whence, (a + b) x ^X f=a x , x f+b x ^ X f.
J J J
Dividing each term by / (Ax. 8, § 9), we have
(a + 6)x^ = ax^ + 6x.
Thus, the result is proved when c is a positive integer or a
positive fraction.
III. Let a and b have any values, and let c = — g, where g is
a positive integer or fraction.
By § 54, (a + 5)( (7) =  (a + b)g^  (ag + bg), by I and II,
= agbg = a(g)^b(g). .
Thus, the distributive law is proved for all positive or nega
tive, integral or fractional, values of a, b, and c.
ADDITIONAL METHODS IN FACTORING 413
XXXIV. ADDITIONAL METHODS IN
FACTORING
456. The Remainder Theorem.
Let it be required to divide px^ \qx\ rhj x
py? \ qx\r
psi? — apx
px + {ap + q)
(ap + q)x J
(ap + q)x — pa^ — qa
pa^ \ qa{r, Remainder.
We observe that the final remainder,
pa^ j qa\ r,
is the same as the dividend with a substituted in place of x ;
this exemplifies the following law :
If any polynomial, involving x, be divided by x — a, the
remainder of the division equals the result obtained by substi
tuting a for X in the given polynomial.
This is called The Remainder Theorem.
To prove the theorem, let
px"" + 5af~^ H \rx{s
be any polynomial involving x.
Let the division of the polynomial hj x — a be carried on
until a remainder is obtained which does not contain x.
Let Q denote the quotient, and R the remainder.
Since the dividend equals the product of the quotient and
divisor, plus the remainder, we have
Q(x — a) \ R — px"" 4 qx""'^ H \rx\s.
Putting X equal to a, in the above equation, we have,
R = pa"" + qa""^ \ \ra\ s.
414 ALGEBRA
457. The Factor Theorem.
If any polynomial, involving x, becomes zero when x is put
eqital to a, the polynomial has x — a as a factor.
For, by §456, if the polynomial is divided by x — a, the
remainder is zero.
458. Examples. f
1. Find whether cc — 2 is a factor of a^ — 5 a^ + 8.
Substituting 2 for x, the expression x^ — bx^ \ S becomes
23 5 22 + 8, or 4.
Then, by § 456, if x^ — 5x^ + 8 be divided by x — 2, the remainder is
— 4 ; and x — 2 is not a factor.
2. Find whether m + n is a factor of
m
4 m^n + 2 m^n^ + 5 mn^ — 2 ?i^. (1)
Putting m = — w, the expression becomes
w* + 4 w* + 2 w*  5 w*  2 w*, or 0.
Then, by § 456, if the expression (1) be divided by m + n, the re
mainder is ; and w + w is a factor.
3. Prove that a is a factor of
(a\b\c) (ab + bc\ ca)  (a + 6) (6 + c) (c f a).
Putting a = 0, the expression becomes
(6 + 6)1)0 h{h\ c)c, or 0.
Then, by § 456, a — 0, or a, is a factor of the expression.
4. Factor a^3a;214a;8.
The positive and negative integral factors of 8 are 1, 2, 4, 8, — 1, — 2,
— 4, and — 8.
It is best to try the numbers in their order of absolute magnitude.
If X = 1, the expression becomes 1—3—14 — 8.
If X = — 1, the expression becomes —1 — 3 + 14 — 8.
If X = 2, the expression becomes 8 — 12 — 28 — 8.
If X = — 2, the expression becomes — 8 — 12 + 28 — 8, or 0.
This shows that x + 2 is a factor.
Dividing the expression by x + 2, the quotient is x^ — 5 x — 4.
Then, x^  3x2  14 x  8 =(x + 2)(x2  5x  4).
<^ 1
ADDITIONAL METHODS IN FACTORING 416
EXERCISE 195
Factor the following :
1. a^ + 1. 2. 0^^81. 3. a;«64.
4. a;^ + 4a^ + 7a;12. 8. a^18aj + 8.
5. x''Q(?\Qfx' + l^x\Q>. 9. a;35cc2_8ic + 48.
6. a?3x2llx10. 10. a^4_^8ar^ + 13a^'13a;4.
7. ar^9a^ + 15£c + 9. 11. x^ ^Qix" x30.
Find, without actual division,
.^ 12. Whether a;  3 is a factor of ar"  6 x" + 13 a;  12. :/ >i
13. Whether a? 4 2 is a factor of ar' + 7 a^ — 6.
14. Whether a; + 1 is a factor of a;'' — 4a^ + 2a52 — 2a5 — 9.
^15. Whether a; is a factor of x{y + zf +y{z + xf + z{x + yf.
16. Whether a is a factor of a\bcf[h\caf+c\ahf.
17. Whether a; — ?/ is a factor of (a; — yf + (2/,— =2)^ + (2 — xf.
18. Whether m H n is a factor of m{m ^ 2 Tif — n{2 m \ nf.
459. We will now give formal proofs of the statements of
§ 104.
Proof of 1.
If h be substituted for a in a'' — 6", the result is 6" — 6", or 0.
Then, by § 457, a" — 6" has a — Z> as a factor.
Proof of II.
If —h be substituted for a in a^'—b'', the result is (— ^)"— 6";
or, since w is even, b^'—b'*, or 0.
Then, by § 457, a" — 6" has a + 6 as a factor.
Proo/ of III.
If 6 be substituted for a in a'^+ft^the result is (6)" +6";
or, since n is odd, — &" I 6", or 0.
Then, a" 4 6" has a + 6 as a factor.
416 ALGEBRA
Proof of ly.
If — 6 or f 6 be substituted for a in a" + 6", the results are
(—&)" + ^" 01* &" + ^"^ respectively.
Since n is even, neither of these is zero.
Then, neither a + 6 nor a — 6 is a factor of a" + 6".
SYMMETRY
460. An expression containing two or more letters is said to
be symmetrical with respect to them, when any two of thepi
can be interchanged without altering the value of the expres
sion.
Thus, ah{hc\ ca is symmetrical with respect to the letters
a, 6, and c ; for if a and b be interchanged, the expression be
comes ba + ac i cb, which is equal to ab \bc\ ca.
And, in like manner, the expression is not altered in value if
we interchange b and c, or c and a.
461. Cyclosymmetry.
An expression containing n letters a, b, c, ">, m, n, is said to
be cyclosymmetrical with respect to them when, if a be replaced
by b, bhj c, •••, m by n, and n by a, the value of the expression
is not changed.
The above is called a cyclical interchange of letters.
Thus, the expression a^b + b^c + c^a is cyclosymmetrical with
respect to the letters a, b, and c ; for if a be replaced by b, b by c,
and c by a, the expression becomes b^c + c'^a + a^6, which is equal
to a^b + b^c { c^a.
The above expression is not symmetrical with respect to a, 6, and c ;
for if a and h be interchanged, the expression becomes b^a \ a^c + d^b,
which is not equal to a^h + b'^c + c^a.
462. It follows from §§ 460 and 461 that, if two expressions
are symmetrical or cyclosymmetrical, the results obtained by
adding, subtracting, multiplying, or dividing them are, respec
tively, symmetrical or cyclosymmetrical.
ADDITIONAL METHODS IN FACTORING 417
463. Applications.
The principle of symmetry is often useful in abridging alge
braic operations.
1. Expand (a + b^cy.
We have, (a \ b + cy = {a \ b + c)(a + b + c)(a + b { c).
This expression is symmetrical with respect to a, 6, and c (§ 460) , and
of the third degree.
There are three possible types of terms of the third degree in a, 6,
c ; terms like a^, terms like a%, and terms like abc.
It is evident that a^ has the coefficient 1 ; and so, by symmetry, b^ and
c^ have the coefficient 1.
The a^b terms may be obtained by multiplying the a's in any two fac
tors by the b in the remaining factor.
Then, it is evident that ab has the coefficient 3 ; and so, by symmetry,
have b'^a, b^c, c^b, c^a, and a^c.
Let m denote the coefficient of abc.
Then, (« + 6 + c)3
= a^ i b^ + c^ { S(a~b \ b'^a + b'^c + c% + d^a + a'^c) + mabc.
To determine m, we observe that the above equation holds for all values
of a, b, and c.
We may therefore let a = ?> = c = 1.
Then, 27 == 3 + 18 + m ; and m = Q.
Whence, (a + b + cy
= a^ + &3 4. c3 + 3(a26 + &% + b^^ + c2& + c^a + «%) + 6 abc.
2. Expand (x — y — zy \ (y — z — xy \ (z — x — yy.
This expression is symmetrical with respect to x, y, and z, and of the
second degree.
The possible types of terms of the second degree in x, y, and z are
terms like x^, and terms like xy.
It is evident, by the rule of § 204, that x^ has the coefficient 3 ; and so,
by symmetry, have y^ and z^.
Let m denote the coefficient of xy.
Then, (x y  z)'^ \ (y  z  xy +{z  x  yy
= S(x'^ + y^ {z^)+ m(xy^yz{zx).
To determine m, put x = y = z = l.
Then, 3 = 9 + Sm, or m = 2.
Whence, (x — y — zy \(y — z — xy ^(z — x — yy
= 3(x2 + 2/2 + s2)  2(xy + yz + zx).
418 ALGEBRA
3. Expand
(a + 6 + c)3 + (a + &  c)3 + (5 + c  a)3 + (c + a  by.
The expression is symmetrical with respect to a, 6, and c, and of the
third degree.
The possible types of terms are terms like a^, terms like a^b, and terms
like abc.
It is evident, by proceeding as in Ex. 1, that a^ has the coeflacient
1_.1_141, or2; and so, by symmetry, have b^ and c^.
Again, proceeding as in Ex. 1, it is evident that a^b has the coefficient 3
in the first term, 3 in the second, 3 in the third, and — 3 in the fourth.
Then, a'^b has the coefficient 3+3 + 33, or 6; and so by symmetry
have ba, b% c^b, d^a, and a^c.
Let m denote the coefficient of abc. —
Then, (a + 6 + c)3 + (a + 6  c)3 + (& + c  a)^ + (c + a  6)8
= 2{a^ + b'^\ c3) + 6(a26 + &% + bc + c^b + c'a + a^c) + mabc.
To determine m, let a = & = c = 1.
Then, 27 + 1 + 1 + 1 = 6 + 36 + m, or m =  12.
Then, (a + 6 + c)^ + (a + 6  c)3 + (6 + c  ay + (c + a  by
= 2(a3 + 63 + c3) + 6(^25 + b^a + b'^c + 0^6 + c^a + a'^c)  12 abc.
EXERCISE 196
1. In the expansion of an expression whicli is symmetrical
with respect to a, b, and c, what are the possible types of terras
of the fourth degree ? of the fifth degree ?
2. If one term of an expression which is symmetncal with
respect to a, b, and c is (2 a — b — c)(2 b — c — a), what are
the others?
3. Is the expression a(6 — c)^4 6(c — a)^ + c(a — &)^ sym
metrical with respect to a, b, and c?
4. Is the expression (sc^ — y'^y + (y^ — z^y + (z^ — x^^ sym
metrical with respect to x, y, and z ?
Expand the following by the symmetrical method :
5. (a + b + cy. 6. (a + 6 + c + d)l
ADDITIONAL METHODS IN FACTORING 419
7. (x + yzy^{y + zxy^{z + xy)\
8. (2 a  3 6  4 c)2 + (2 &  3 c  4 a)2 + (2 c  3 a  4 hf.
9. (a 4 6 4 c)3 f (a  6  c)'^ + (6  c  a)^ + (c  a  6)^.
10. {a\h^cdf\{h + c{day + (G + d^ahY
\{d\a + h — cf.
11. (aH6 + c + (?)^
12. (x + 2/ 2;)(i/H2;a;)(2 + i»?/).
13. (a + & + c)(a + h— c)(b \c a)(c \a — b).
14. (a^^y^ + z^\2xy + 2yz + 2zxy.
464. Factoring of Symmetrical Expressions.
The method of § 457 is advantageous in factoring symmet
rical expressions (§§ 460, 461).
1. Factor
a(b + cf + 5(c + a)2 + c(a + bf  a\b + c)  bXc + a)  c^a + &)•
The expression is symmetrical with respect to a, 6, and c.
Being of the third degree, the only literal factors which it can have are
three of the type a ; three of the type « + 6; ora + 6 + c, and a factor
of the second degree.
Patting a = 0, the expression becomes
6c2 + c62  h^c  c%, or 0.
Then, by § 457, a is a factor ; and, by symmetry, h and c are factors.
The expression, being of the third degree, can have no other literal fac
tor; but it may have a numerical factor.
Let the given expression = mahc.
To determine m, let a = 6 = c = 1.
Then, 4 + 4 + 4222 = wi, or w = 6.
Whence, the given expression = 6 dhc.
2. Factor x^ \ if \ z^ — ^ xyz.
The expression is symmetrical with respect to cc, ?/, and z.
The only literal factors which it can have are three of the type x ;
three of the type x + y\ or x \ y \ z, and a factor of the second degree.
;
420 ALGEBRA
It is evident that neither x^ y, nor is a factor.
Putting X equal to — y, the expression becomes
y^ + y^ + z^ + 'S x%
which is not 0.
Then, x {• y is not a factor (§ 457) ; and, by symmetry, neither y + z
nor z + X is a factor.
Putting X equal to — y — z, the expression becomes
{y  zY + y^ + z^ ^{y  z)yz
= y^  Sy^zSyz^  z^ + y^h z^ + 3 y'^z + 3 yz^ = 0.
Therefore, x + y \ z is a factor.
The other factor may be obtained by division, or by the following
process :
It is of the second degree ; and as it is symmetrical with respect to x,
y, and z, it must be of the form
m(x2 + 2/2 + z^) + n{xy + yz + zx).
It is evident that w = 1, as this is the only value which will give the
terms x^, y^, and z^ in the given expression.
Then,
a;3 + 2/3 4. 23 _ 3 xyz=(x\y + z) [x2 + y2 ^ z"^ ^ n (xy + yz + zx)^.
To determine 7i, let x = I, y = I, z = 0.
Then, 2 = 2(2 + w), or 1=2 + w, or n =  1.
Whence,
x^ \ y^ \ z^ — Z xyz = (x\ y + z) (x^ \ y"^ \ z"^  xy  yz  zx).
3. Factor ab(a — b) \ be {h — c) \ ca(G — a).
The expression is cyclosymmetrical (§ 461) with respect to a, &, and c.
It is evident that neither a, 6, nor c is a factor.
The expression becomes when a is replaced by b.
Then, a — 6 is a factor ; and, by symmetry, b — c and c — a are
factors.
The expression can have no other literal factor, but may have a numeri
cal one.
Let the given expression = m(a — b)(b — c)(c — a).
To determine m, let a = 2, & = 1, and c = 0.
Then, 2 = — 2 m, and m = — 1.
Then, the given expression = — (a — b)(b — c) (c — a).
ADDITIONAL METHODS IN FACTORING 421
EXERCISE 197
Factor the following :
1. m» + 2m2n42mri2 + n3.
2. (ab\bc^ca)(a+b+c)a\b+c)b%c\a)c\a + b).
3 x'(y^z)+y\z + x)\z\x^y){2xyz.
4. a(b + cy + b(c 4 a.)2 + c(a + 6)^4 a6c.
5. a\bc)^b\ca)\c\ab).
6. (a; + 2/ + 2)(a;.v + 2/2: + z^) (x + y)(y + ^)(^ + a?)
7. a6(a + 6) + 6c(6 + c) f ca{c + a) + 2 a6c.
8. (a; + 2/ + 2)3_a^_2/3_^
9. (x + y{z)(xy{yz{zx)xyz.
10. (a;y)3 4(2/;2)3 + (2aj)3.
11. a\b  c) + &'(c  a) + c3(a  6).
422 ALGEBRA
XXXV. MATHEMATICAL INDUCTION
465. In § 387 we gave an example of Mathematical Induc
tion, in proving the Binomial Theorem for a Positive Integral
Exponent ; in the present chapter, we will give other illustra
tions of the method.
466. AVe will now prove that the laws of § 103 hold univer
sally.
We will first prove, by Mathematical Induction, that they
hold for , where n is any positive integer.
Assurme the laws to hold for , where n is any positive
. , a — b
integer.
Then, ^^^ = a»i f a^% H c^^h^ + • •  + h^\ (1)
a — h
a — h a— h
^ a\a — h) h hjal' — 6^)
a~h
= a"" + 6(a"^ L a'^26 i a''^^ \ \ 6"^), by (1),
= a" _l_ a"i5 ^ cjt^ 2^2 ^ ... j^ ^n
This result is in accordance with the laws of § 103.
Hence, if the laws hold for the quotient of the difference of
two like powers of a and b divided by a — 6, they also hold for
the quotient of the difference of the next higher powers of a
and b divided hj a — b.
•^ fj,5 7^5
But we know that they hold for , and therefore they
a — b
A6 7j6 ,y6 7j6
hold for ; and since they hold for , they hold for
ab' ^ ab' ^
: and so on.
a — b
MATHEMATICAL INDUCTION 423
Qn Jyn.
Hence, the laws hold for , where n is any positive
. , a — h
integer.
Putting — h for h in (1), we have
a(6) \ Jr r\ }
If n is even, ( 6)" = 6^ and ( 6)"^ =  6"\
Whence, ^" ~ ^" = a^i  a^''^h + a^^d^ 6i. (2)
If n is odd, ( by =  6", and ( 6)"^ = + h''\
Whence, ^^i^±^ = a^  a^'b + a^''b' + 6"i. (3)
a\b
Equations (2) and (3) are in accordance with the laws of
§103.
467. We will now prove that the law of § 204 holds for the
square of a polynomial of any number of terms.
Assume the law to hold for the square of a polynomial of m
terms, where m is any positive integer ; that is,
(a\b + c] \l\my
= a^^b^{ ■"\m^\2a(b + c] hm)
426(c++m) + .= 42Zm. (1)
Then, (a + 6 + c+ • + m + n)^
= (a + b\c++my
42(ai6 + cH \m)n^n^, by § 97,
= a^ + b + c^\ \m^ + 7i^
\2a{b + c\ hm + w)
+ 2b(c\ \m\n)\ \2mn, by (1).
This result is in accordance with the law of § 204.
424 ALGEBRA
Hence, if the law holds for the square of a polynomial of m
terms, where m is any positive integer, it also holds for the
square of a polynomial of m + 1 terms.
But we know that the law holds for the square of a polyno
mial of three terms, and therefore it holds for the square of a
polynomial of four terms ; and since it holds for the square of
a polynomial of four terms, it also holds for the square of a
polynomial of five terms ; and so on.
Hence, the law holds for the square of any polynomial.
468. As another illustration of the method, we will prove
that the sum of the first n terms of the arithmetic progression,
a, a + d, a 4 2c?, •••,
is given by the formula na __ ^v^^— ) ^^ (Compare § 361.)
The sum of the first two terms is 2a\d, which can be
written in the form 2af ~ ^ d
z
Then, the formula holds for the sum of the first two terms.
Assume that the formula holds for the sum of the first ?i
terms.
That is, the sum of the first n terms = na f ^ ~" ^ d
2i
Now the {n + l)th term of the progression is a + nd.
Whence, the sum of the first {n + 1) terms equals
na + ^^^^^^^^(^ + a + nf« = (n + l)a + ^(n  1 + 2)
= («4l)« + ^^^d.
This result is in accordance with the formula.
Hence, if the formula holds for the sum of the first n terms,
it also holds for the sum of the first n\\ terms.
But we know that the formula holds for the sum of the first
two terms, and hence it holds for the sum of the first three
terms ; and since it holds for the sum of the first three terms,
it also holds for the sum of the first four terms ; and so on.
MATHEMATICAL INDUCTION 425
Hence, the formula holds for the sum of the first n terms,
where n is any positive integer.
EXERCISE 198
1. Prove that the sum of the first n terms of the series 1, 3,
5, ••• is n^.
2. Prove that the sum of the first n terms of the series 3, 6,
9, ...isM!L+D.
3. Prove that the sum of the first n terms of the series
111 . n
IS
1.2' 23' 34' n + 1
4. Prove, by mathematical induction, that the sum of the
first n terms of the geometric progression,
a, ar, ar^^ •••,
is given by the formula S = ^^^"^ ~ ^^ (§ 370).
r — 1
5. Prove that the sum of the first n terms of the series 2*^
42 62 ... is 2n(n + l)(2n + l) .
' ' 3
6. Prove that the sum of the first n terms of the series 1^
426 ALGEBRA
XXXVI. EQUIVALENT EQUATIONS
469. Two equations, each involving one or more unknown
numbers, are said to be Equivalent when every solution of the
first is a solution of the second, and every solution of the second
a solution of the first.
470. To solve an equation involving one unknown number, Xy
we transform it into a series of equations, which lead finally to
the value of x.
We have assumed, in passing from any equation to any other,
in this series, that every solution of the first was a solution of
the second, and every solution of the second a solution of the
first ; so that it was legitimate to use the second in place of the
first to find the value of the unknown number.
That is, we have assumed that the two equations were equiva
lent (§ 469).
We will now prove some theorems in regard to equivalent
equations.
471. If the same expression he added to both members of an
equation, the resulting equation will he equivalent to the first.
Let A=^B (1)
be an equation involving one or more unknown numbers.
To prove the equation A + C=B\C, (2)
where C is any expression, equivalent to (1).
Any solution of (1), when substituted for the unknown num
bers, makes A identically equal to B (§ 79).
It then makes A{C identically equal to jB + C (§ 84, 1).
Then it is a solution of (2).
Again, any solution of (2), when substituted for the unknown
numbers, makes A\ C identically equal to B \ C.
EQUIVALENT EQUATIONS 427
It then makes A identically equal to B (§ 84, 2).
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The principle of § 84, 1, is a special case of the above.
472. The demonstration of § 471 also proves that
If the same expression he subtracted from both members of an
equation, the resulting equation will be equivalent to the first.
The principle of § 84, 2, is a special case of this.
473. If the members of an equation be multiplied by the same
expi'ession, ichich is not zero, and does not involve the unknown
numbers, the resaltirig equation will be equivalent to the first.
Let A = B (1)
be an equation involving one or more unknown numbers.
To prove the equation AxG=Bx C, (2)
where C is not zero, and does not involve the unknown num
bers, equivalent to (1).
Any solution of (1), when substituted for the unknown num
bers, makes A identically equal to B.
It then makes Ax C identically equal to .B x (§ 84, 3).
Then it is a solution of (2).
Again, any solution of (2), when substituted for the unknown
numbers, makes Ax C identically equal to B x C.
It then makes A identically equal to B (§ 84, 4).
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The reason why the above does not hold for the multiplier zero is, that
the principle of § 84, 4, does not hold when the divisor is zero.
The principle of § 84, 3, is a special case of the above.
474. If the members of an equation be multiplied by an ex
pression which involves the unknown numbers, the resulting
equation is, in general, not equivalent to the first.
Consider, for example, the equation a; + 2 = 3 a? — 4. (1)
428 ALGEBRA
Now the equation
(x{.2)(xl) = (Sx^)(ixl), (2)
which is obtained from (1) by multiplying both members by
ic — 1, is satisfied by the value x = l, which does not satisfy (1).
Then (1) and (2) are not equivalent.
Thus it is never allowable to multiply both members of an inte
gral equation by an expression which involves the unknown
numbers ; for in this way additional solutions are introduced.
475. If the merribers of an equation he divided by the same ex
pression, which is not zero, and does not involve the unknown
numbers, the resulting equation will be equivalent to the first.
Let A = B (1)
be an equation involving one or more unknown numbers.
A B
To prove the equation — — —^ (2)
where C is not zero, and does not involve the unknown num
bers, equivalent to (1).
Any solution of (1), when substituted for the unknown num
bers, makes A identically equal to B.
A B
It then makes — identically equal to — (§ 84, 4).
G C
Then it is a solution of (2).
Again, any solution of (2), when substituted for the unknown
A ■ • B
numbers, makes — identically equal to — •
G G
It then makes A identically equal to B.
Then it is a solution of (1).
Therefore, (1) and (2) are equivalent.
The principle of § 84, 4, is a special case of the above.
476. If the members of an equation be divided by an ex
pression which involves the unknown numbers, the resulting
equation is, in general, not equivalent to the first.
EQUIVALENT EQUATIONS 429
Consider, for example, the equation
(a; + 2)(a;l) = (3a;4)(a;l). (1)
Also the equation x + 2 = Sx — 4., (2)
which is obtained from (1) by dividing both members by x — 1.
Now equation (1) is satisfied by the value x = l, which does
not satisfy (2).
Then (1) and (2) are not equivalent.
It follows from this that it is never allowable to divide both
members of an integral equation by an expression which in
volves the unknown numbers ; for in this way solutions are lost.
(Compare § 158.)
477. If both members of a fractional equation be multiplied by
the L.C.M. of the given denoyninators, the resulting equation is,
in general, equivalent to the first.
Let all the terms be transposed to the first member, and let
them be added, using for a common denominator the L. C. M.
of the given denominators.
The equation will then be in the form
1 = 0. (1)
We will now prove the equation
A = 0, (2)
which is obtained by multiplying (1) by the L. C. M. of the
given denominators, equivalent to (1), if A and B have no com
mon factor.
Any solution of (1), when substituted for the unknown num
bers, makes — identically equal to 0.
Then, it must make A identically equal to 0.
Then, it is a solution of (2).
Again, any solution of (2), when substituted for the unknown
numbers, makes A identically equal to 0.
430 ALGEBRA
Since A and B have no common factor, B cannot be when
this solution is substituted for the unknown numbers.
Then, any solution of (2), when substituted for the unknown
numbers, makes — identically equal to 0, and is a solution of (1).
B
Therefore, (1) and (2) are equivalent, if A and B have no
common factor.
If A and B have a common factor, (1) and (2) are not equivalent;
consider, for example, the equations
^^^^ = 0, and X  1 = 0.
The second equation is satisfied by the value x = 1, which does not
satisfy the first equation ; then, the equations are not equivalent.
478. A fractional equation may be cleared of fractions by
multiplying both members by any common multiple of the
denominators; but in this way additional solutions are intro
duced, and the resulting equation is not equivalent to the first.
Consider, for example, the equation
1
If we solve by multiplying both members by x^—1, the
L. C. M. of cc^ — 1 and a^ — 1, we find x = — 2.
If, however, we multiply both members by (x^ — l){x — 1),
we have
a^oc^{a^x=:2x^2x^2x^2, or a^{x2 = 0.
The latter equation may be solved as in § 126.
The factors oi xr \ x ~2 sue x\2 and x — 1.
Solving the equation .'^^ + 2 = 0, x= — 2.
Solving the equation .t — 1 = 0, a? = 1.
This gives the additional value x=l; and it is evident that
this does not satisfy the given equation.
479. If both members of an equation be raised to the same
positive integral power, the resulting equation ivill have all the
solutions of the given equation, and, in general, additional ones.
EQUIVALENT EQUATIONS 431
Consider, for example, the equation x = 3.
Squaring both members, we have
aj2 = 9, or a;9 = 0, or (x{3){xS) = 0.
The latter equation has the root 3, and, in addition, the
root — 3.
We will now consider the general case.
Let A = B (1)
be an equation involving one or more unknown numbers.
Raising both members to the nth power, n being a positive
integer, we have
A^' = B, or A  B'' = 0. (2)
Factoring the first number (§ 121),
(A  B) (A""' + A^'B + . . . + 5«"i) = 0. (3)
Now, equation (3) is satisfied when A=B.
Whence, equation (2) has all the solutions of (1).
But (3) is also satisfied when
so that (2) has also the solutions of this last equation, which,
in general, do not satisfy (1).
EQUIVALENT SYSTEMS OF EQUATIONS
480. Two systems of equations, involving two or more
unknown numbers, are said to be equivalent when every solu
tion of the first system is a solution of the second, and every
solution of the second a solution of the first.
481. //
are equations involving tioo or more unknown numbers, the
system of equations
A=0,
mA + nB = 0,
where m and n are any numbers, and n not equal to zero, is
equivalent to the Jirst system.
432 ALGEBRA
For any solution of the first system, when substituted for
the unknown numbers, makes A = and ^ = 0.
It then makes ^ = and mA + nB = 0.
Then, it is a solution of the second system.
Again, any solution of the second system, when substituted
for the unknown numbers, makes ^ = and mA + nB = 0.
It therefore makes nB = 0, oi B = 0.
Since it makes A = and 5 = 0, it is a solution of the first
system.
Hence, the systems are equivalent.
A similar result holds for a system of any number of equations.
Either m or n may be negative.
482. If either equation, in a system of tvjo, be solved for one
of the unknown numbers, and the value found be substituted for
this unknown number in the other equation, the resulting system
will be equivalent to the first.
Let H = ^^ W
la=A (2)
be equations involving two unknown numbers, x and y.
Let E be the value of x obtained by solving (1).
Let F=G be the equation obtained by substituting E for
X in (2).
To prove the system of equations
'x=E, ' (3)
\f=G, (4)
equivalent to the first system.
Any solution of the first system satisfies (3), for (3) is only
a form of (1).
Also, the values of x and y which form the solution make x
and E equal ; and hence satisfy the equation obtained by put
ting E for x in (2).
Then, any solution of the first system satisfies (4).
Again, any solution of the second system satisfies (1), for
(1) is only a form of (3).
EQUIVALENT EQUATIONS 433
Also, the values of x and y which form the solution make x
and E equal ; and hence satisfy the equation obtained by put
ting X for E in (4).
Then, any solution of the second system satisfies (2).
Hence, the systems are equivalent.
A similar result holds for a system of any number of equations,
involving any number of unknown numbers.
483. We will now apply the principles of § § 481 and 482
to show that the solutions of Ex. 1, § 168, and the examples of
§§ 169 and 170 are equivalent to the given equations. '
Ex. 1, § 168.
By § 481, the given system is equivalent to the system (1)
and (5), or to the system (1) and (6).
By § 482, the system (1) and (6) is equivalent to the system
(6) and (7), which is equivalent to the system (6) and (8).
Then, the given system is equivalent to the system (6) and (8).
Ex., § 169.
By § 482, the given system is equivalent to the system (3)
and (4), or to the system (3) and (5).
By § 482, the system (3) and (5) is equivalent to (5) and (6).
Ex., § 170.
The given system is equivalent to (3) and (4).
Now any values of x and y which satisfy (3) and (4) also
satisfy (3) and (5).
Then, the given system is equivalent to the system (3) and
(5), or to (3) and (6).
By*§ 482, the system (3) and (6) is equivalent to (6) and (7).
484. The principles of §§ 471, 472, 473, 475, 477, 479, 480,
and 481 hold for equations of any degree.
434 ALGEBRA
XXXVII. GRAPHICAL REPRESENTATION OF
IMAGINARY NUMBERS
485. Let be any point in the straight line XX'.
We may suppose any positive real number, f a, to be
represented by the distance from
O to ^, a units to the right of „> . ^ . ^ . ^
in OX, ^ A' a .a A '
Then, with the notation of § 16, any negative real number,
— a, may be represented by the distance from to A^ a units
to the left of in OX.
486. Since — a is the same as (+ a) x (— 1), it follows from
§ 485 that the product of + a by — 1 is represented by turning
the line OA which represents the number f a, through two
right angles, in a direction opposite to the motion of the hands
of a clock.
Then, in the product of any real number by —1, we may
regard — 1 as an operator which turns the line which repre
sents the first factor through two right angles, in a direction
opposite to the motion of the hands of a clock.
487. Graphical Representation of the Imaginary Unit i (§ 276).
By the definition of § 275, — 1 = i x i.
Then, since one multiplication by z, „
followed by another multiplication by *,
turns the line which represents the first +z
factor through two right angles, in a direc
tion opposite to the hands of a clock, we i
may regard multiplication by i as turning ''
the line through one right angle, in the
same direction. ^
Thus, let XX and YY' be straight lines intersecting at
right angles at 0.
^ai
^
0V(^ A.
ai
REPRESENTATION OF IMAGINARY NUMBERS 435
Then, if + a be represented by the line OA, where A is a
units to the right of in OX, +ai may be represented by
OB, and — ai by OB', where ^ is a units above, and B' a
units below, 0, in YY'.
Also, H i may be represented by OC, and — i by 0(7', where
C is one unit above, and C one unit below, 0, in YY'.
488. Graphical Representation of Complex Numbers.
We will now show how to represent the complex number
a + bi.
Let XX' and YY' be straight lines inter
secting at right angles at 0.
Let a be represented by OA, to the right ^_
of 0, if a is positive, to the left if a is
negative.
Let bi be represented by OB, above if 6 is positive, below
if b is negative.
Draw line AC equal and parallel to OB, on the same side of
XX' as OB, and line OC.
Then, OC is considered as representing the result of adding
bito a; that is, OC represents the complex number a + bi.
The figure represents the case in which both a and b are positive.
As another illustration, we will show how to represent the
complex number — 5 — 4 i
Lay off OA 5 units to the left of in
OX', and OB 4 units below in YY'.
Draw line AC below XX', equal and
parallel to OB, and line OC.
Then, 00 represents — 5 — 4 i.
The complex number a + bi, if « is positive and b negative, will be
represented by a line between OX and Y' ; and if a is negative and b
positive, by a line between Y and OX'.
EXERCISE 199
Represent the following graphically :
1. 3i. 2. 6i. 3. 4 + i. 4. l + 2i.
\''J ~^
X
y
1
c
J
<s>.
B
y'
436
ALGEBRA
5. 25i.
6. 53i.
7. 7 + 4i.
489. Graphical Representation of Addition.
We will now show how to represent the result of adding b to
a, where a and b are any two real, pure imaginary, or complex
numbers.
Let the line a be represented by OA, and
the line b by OB.
Draw the line AC equal and parallel to
OB, on the same side of OA as OB, and the
line 00.
Then, OG is considered as representing
the result of adding b to a; that is, 00 represents a\b.
The method of § 488 is a special case of the above.
If a and 6 are both real, B will fall in OA, or in AO produced
through 0.
The same will be true if a and 6 are both pure imaginary.
If one of the numbers, a and &, is real, and the other pure imaginary,
the lines OA and OB will be perpendicular.
As another illustration, we will show how to represent
graphically the sum of the complex numbers 2 — 5 1 and
4 + 3i.
The complex number 2 — 5i is repre
sented by the line OA, between OX and
OY'.
The complex number — 4 f 3 if is repre
sented by the line OB, between OT and
OX'.
Draw the line BC equal and parallel to
OA, on the same side of OB as OA, and the line OC.
Then, the line 00 represents the result of adding
to 2  5 1.
413/
490. Graphical Representation of Subtraction.
Let a and b be any two real, pure imaginary, or complex
numbers.
REPRESENTATION OF IMAGINARY NUMBERS 437
Let a be represented by OA, and b
by OB; and complete the parallelogram
OBAC.
By § 489, OA represents the result of
adding the number represented by OB to
the number represented by 00.
That is, if b be added to the number
represented by 0(7, the sum is equal to
a; hence, a — 6 is represented by the line OC.
EXERCISE 200
Represent the following graphically :
1. The sum of 4 i and 3 — 5i.
2. The sum of — 5 ^ and — 1 + 6 i.
3. The sum of 2 + 4 i and 5  3 i.
4. The sum of 6 + 2i and 47z.
5. Represent graphically the result of subtracting the sec
ond expression from the first, in each of the above examples.
438 ALGEBRA
XXXVIII. INDETERMINATE FORMS
491. In § 322, we found that the form  indicated an ex
pression which could have any value whatever ; but this is not
always the case.
^ rjf^ rt^ '
Consider, for example, the fraction — .
or — ax
If x = a, the fraction takes the form .
JSTow x^a^ ^ (x\a)(xa) ^ x{a ^
x^— ax x(x — a) x ^
which last expression is equal to the given fraction, provided
X does not equal a.
The fraction ^_ZL^ approaches the limit ^L+_^ or 2, when
X a
X approaches the limit a.
This limit we call the value of the given fraction ivhen x = a.
Then, the value of the given fraction when x = a h 2.
In any similar case, we cancel the factor which equals for the given
value of X, and find the limit approached by the result when x approaches
the given value as a limit.
EXERCISE 201
Find the values of the following :
1. ^«^^^' when^ = 2a. 3. /^li_ when a; =  4.
x'4:a' x'\2xS
'• 4^T3^ when . = 0. 4. g^,_^,^^,, when . = i.
g_ ^dh6^±12^±8^hen.: = 2.
x*8 X + 16
6. ^l^M±l£^whenx = 2.
x'Tx + G
INDETERMINATE FORMS 439
492. Other Indeterminate Forms.
Expressions taking the forms ^,
values of the letters involved, are also indeterminate.
Expressions taking the forms ^, x oo, or oo — oo, for certain
1. Find the value of (o(^ \ 8)(l + ^^ when x = 2.
This expression takes the form x oo, when x = — 2 (§ 319).
Now, (x3 + 8) (l + —^) = x3 + 8 • ^^ + ^
\ X + 2/
'^x + 2
= x3 + 8 4 ic2  2 X + 4 = x3 + x2  2 X + 12.
The latter expression approaches the limit —8 + 4 + 4 + 12, or 12,
when X approaches the limit — 2.
This limit we call the value of the expression when x = — 2 ; then, the
value of the expression when x = — 2, is 12.
In any similar case, we simplify as much as possible before finding the
limit.
2. Find the value of ^ ?^, when x==l.
1 — X 1 — or
The expression takes the form oo — oo, when x = 1 (§ 319).
2x 1+X2X 1x 1
Now,
1x 1X2 1X2 1X2 1+X
The latter expression approaches the limit  when x approaches the
limit 1. ^
Then, the value of the expression when x = 1, is  •
493. Another example in which the result is indeterminate
is the following :
Ex. Find the limit approached by the fraction "^ ^ when
X is indefinitely increased. ~ ^
Both numerator and denominator increase indefinitely in absolute value
when X is indefinitely increased.
Dividing each term of the fraction by x, ^ —  =
^ "^'25x2^
— o
X
440 ALGEBRA
The latter expression approaches the limit ^ (§ 320), or — , when
X is indefinitely increased. ~ '^
In any similar case, we divide both numerator and denominator of the
fraction by the highest power of x.
EXERCISE 202
Find the limits approached by the following when x is in
definitely increased:
1 4 + 5a;3^ 2 ^^ + 1 3 ^2x^
7  a; 4 4 0)2 ' ' 3 aj^  2* ' x^\bx^3
Find the values of the following :
1 12
a;_2 aj38
4. r when x = 2.
5. (2a^5a;3)/'2f^') whena; = 3.
HORNER'S SYNTHETIC DIVISION
441
XXXIX. HORNER'S SYNTHETIC DIVISION
494. Division by Detached Coefficients.
In finding the quotient of two expressions which are arranged
according to the same order of powers of some common letter,
the operation may be abridged by writing only the jiumerical
coefficients of the terms.
If the term involving any power is wanting, it may be sup
plied with the coefficient 0.
Ex. Divide 6 x^\2 a^9 a;^+5 a^ + 18 i»30 by 3 a^ + a^6.
3 + l40_6
2+03+5
6 + 2
9+ + 5 + 18
fO12
9 + 12
_9 3 + + 18
30
6 + 2
15 + 5
15j_5f 0
30
Then the quotient is 2 a^ — 3 a; + 5.
495. Horner's Synthetic Division.
Let it be required to divide 6 x^ — ic^ — 3 a^ + 10 a? — 12 by
2a^ + xS.
ex* a^3a^ + 10a;
6x*\3a^9x^
12
4ar^ + 6a;2
4a^2a^ +
ex
Sx' +
8a.'2 +
Ax
Ax
12
2a^ + a?3
3a^_2a; + 4
The dividend equals (2 a;^ + x — 3) times the quotient.
Then, we can find the quotient by subtracting from the
dividend + x times the quotient, and — 3 times the quotient,
and dividing the result by 2 a;^.
442 ALGEBRA
Or, we can find it by adding to the dividend —x times the
quotient, and + 3 times the quotient, and dividing the result
by2a^.
We may arrange the work as follows :
2a^
60)4 a?Zx'\10x12
— X
3a^ + 2aj2 4^
+ 3
+ 9a^ Qx + 12
^x'2x 44, quo.
We write the divisor in the lefthand column, with the sign
of each term after the first changed.
We get the first term of the quotient, 3 a^, by dividing the first
term of the dividend, 6 a?'^, by the first term of the divisor, 2 x^.
Since we have to add to the dividend — x times the quotient,
and +3 times it, we can put down the terms — 3a^ in the
second column, and 4 9 x" in the third, these being the prod
ucts of — a; and + 3 by the first term of the quotient.
Now we get the second term of the quotient by subtracting
3 ic^ from — a^, and dividing the result by 2 x^.
Then, in the synthetic method, we add the terms in the
second column, and divide the sum by 2 x^, giving — 2 a?.
We now write the term + 2 a^ in the second column, and
— 6a; in the third; these being the products of —a; and 3
by the second term of the quotient.
We get the third term of the quotient by subtracting — 9 a^
and — 2 a;^ from — 3 x^, and dividing the result by 2 x?.
Then, in the synthetic method, we add the terms in the
third column, and divide the sum by 2 oi?, giving 4.
We now write the term — 4 a; in the fourth column, and + 12
in the fifth ; these being the products of — a; and + 3 by the
third term of the quotient.
We have now added to the dividend — x times the quotient,
and + 3 times the quotient, and divided the result by 2 a^ ; so
that we have obtained the quotient.
The sum of the terms in the fourth and fifth columns is ; if this had
not been the case, there would have been a remainder.
HORNER'S SYNTHETIC DIVISION 443
496. We will now give some additional examples of the
method :
1. Divide 12 ar^ 11 aj^f 20 a; 9 by 3 ar2a; + 4.
3x2
4
12 a:3  11 a;2 + 20 X  9
+ 8x2 2 X
16x + 4
4 X — 1, quo. 2 X — 5, Rem.
"We write the divisor in the lefthand column, with the sign of each
term after the first changed.
Dividing 12 x^ by 3 x^ gives 4 x for the first term of the quotient.
We multiply + 2 x by 4 x and put the product, 8 x2, in the second
column ; and multiply — 4 by 4 x, and put the product, — 16 x, in the
third column.
We add the terms in the second column, giving — 3 x2, and divide the
result by 3 x2, giving — 1 as the second term of the quotient.
We multiply + 2 x by — 1 and put the product, — 2 x, in the third
column ; and multiply — 4 by — 1, and put the product, + 4, in the
fourth column.
Adding the terms in the third and fourth columns, the sum is 2 x — 5.
Then, the quotient is 4 x — 1, and the remainder 2 x — 5.
It is advantageous to use detached coefficients (§ 494) in the synthetic
method ; the work of Ex. 1 would then stand as follows :
. 3
+ 2
4 ^
41, +25
2. Divide
a«+2a^614a362__l5a6*5 6^ by a''^ah\h\
12
—
11 4. 20  9
+
8 2
16 + 4
+ 3a6
&2
^5 4. 2 a*&  14 aW + 15 ah^  5 65
+ 3a46 + 15a362  15 a&*
 amba%^ +5 66
a3 + 5a2&+ 5 63,
In the above solution, the sum of the terms in the third column is 0, so
that the third term of the quotient is 0.
We then add the terms in the fourth column and divide by a^^ giving
— 5 6^ as the fourth term of the quotient.
It is important, in an example like the above, to keep similar terms in
the same vertical column.
444
ALGEBEA
The work of Ex. 2 will appear as follows with detached coefficients
1
+ 3
1
1 + 214 + + 155
+ 3_.15 _15
 15 +5
1 + 5+05,
EXERCISE 203
Divide the following by synthetic division :
1. 12x'7a^23xS hj 4.x'5x3.
2. 4a*9a2+30a25 by 2a2+3a5.
3. 2a'a^b\Sab^5b^ by 2a^3ab + 5b\
4. 4 m^n* + n^ + 16 m'* by 2 mn^ + 4 m^ f 7i*.
'5. 6a^13x'20a^ + 55a^l4:x19 by 2a^7a; + 6.
6. Sx^4:x'ySxY18xy^{21y'
by 4 a^ — 2 0^2/ + 6 a;^^ — 7 2/"'.
7. 37a2 + 50+a^70a by 2a2+5 + a36a.
8. 2a'ab2ac6b'+llbc4.c' by 2a + 364c.
PERMUTATIONS AND COMBINATIONS 445
XL. PERMUTATIONS AND COMBINATIONS
497. The different orders in which things can be arranged
are called their Permutations.
Thus, the permutations of the letters a, 6, c, taken two at a
time, are ahy ac, ha, he, ca, ch ; and their permutations, taken
three at a time, are abc, ach, hac, hca, cah, cha.
498. The Combinations of things are the different collections
which can be formed from them without regard to the order
in which they are placed.
Thus, the combinations of the letters a, h, c, taken two at a
time, are ap, he, ca; for though ah and ha are different permu
tations, they form the same combination.
499. To find the number of permutations of n different things
taken two at a time.
Consider the n letters a, h, c, •••.
In making any particular permutation of two letters, the
first letter may be any one of the n; that is, the first place
can be filled in n different ways.
After the first place has been filled, the second place can be
filled with any one of the remaining n — 1 letters.
Then, the whole number of permutations of the letters taken
two at a time is n(n — 1).
We will now consider the general case.
500. To find the numher of permutations of n different things
takeri r at a time.
Consider the n letters a, h, c, •••.
In making any particular permutation of r letters, the first
letter may be any one of the n.
After the first place has been filled, the second place can be
filled with any one of the remaining n — 1 letters.
446 ALGEBRA
After the second place has been filled, the third place can be
filled in M — 2 different ways.
Continuing in this way, the ?*th place can be filled in
n — (r — l), or n — r\l different ways.
Then, the whole, number of permutations of the letters taken
r Sit a time is given by the formula
,P, = n(7i  l)(n  2) ... (n  r + 1). (1)
The number of permutations of n different things taken r at a time is
usually denoted by the symbol nPr
501. If all the letters are taken, r = n, and (1) becomes
.P„ = n(nl)(n2)...3.2.1=[n. (2)
Hence, the number of permutations of n different things taken
n at a time equals the product of the natural numbers from 1 to n
inclusive. (See note, page 352.)
502. To find the number of combinations of n different things
taken r at a time.
The number of permutations of n different things taken r at
a time is ^ ^^ ._ i) (^ _ 2) ... (^ _ r + 1) (§ 500).
But, by § 501, each combination of r different things may
have \r permutation?.
Hence, the number of combinations of n different things taken
r at a time equals the number of permutations divided by [r.
That is, „fi = "('^l)("2)(«r + l) . ^g^
[r
The number of combinations of n different things taken r at a time is
usually denoted by the symbol nCv
503. Multiplying both terms of the fraction (3) by the prod
uct of the natural numbers from 1 to n — r inclusive, we have
^ ^ n(nl)...(nr + l)(^r)...2.1 ^ 1^ .
[rxl.2...(nr) \r \nr
which is another form of the result.
PERMUTATIONS AND COMBINATIONS 447
504. The number of combinations of n different things taken r
at a time equals the number of combinations taken n — r at a time.
For, for every selection of r things out of n, we leave a selec
tion oi n — r things.
The theorem may also be proved by substituting n — r for r, in the
result of § 503.
505. Examples.
1. How many changes can be rung with 10 bells, taking 7 at
a time ?
Puttmg n = 10, r = 7, in (1), § 500,
10P7 = 10. 9. 8. 7. 6. 5. 4 = 604800.
2. How many different combinations can be formed with 16
letters, taking 12 at a time ?
By § 504, the number of combinations of 16 different things, taken 12
at a time, equals the number of combinations of 16 different things, taken
4 at a time.
Putting n = 16, r = 4, in (3), § 502,
r _ 16151413 _.,Qo^
''^' 1.2.3.4 ^^^^*
3. How many different words, each consisting of 4 consonants
and 2 vowels, can be formed from 8 consonants and 4 vowels ?
The number of combinations of the 8 consonants, taken 4 at a time, is
«ll^il^,or70.
1.2.3.4
The number of combinations of the 4 vowels, taken 2 at a time, is
, or 6.
1.2'
Any one of the 70 sets of consonants may be associated with any one
of the 6 sets of vowels ; hence, there are in all 70 x 6, or 420 sets, each
containing 4 consonants and 2 vowels.
But each set of 6 letters may have [6, or 720 different permutations
(§ 501).
Therefore, the whole number of different words is
420 X 720, or 302400.
448 ALGEBRA
EXERCISE 204
1. How many different permutations can be formed with
14 letters, taken 6 at a time ?
2. In how many different orders can the letters in the word
triangle be written, taken all together ?
3. How many combinations can be formed with 15 things,
taken 5 at a time ?
4. A certain play has 5 parts, to be taken by a company of
12 persons. In how many different ways can they be assigned ?
5. How many combinations can be formed with 17 things,
taken 11 at a time ?
6. How many different numbers, of 6 different figures each,
can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, if each
number begins with 1, and ends with 9 ?
7. How many even numbers, of 5 different figures each, can
be formed from the digits 4, 5, 6, 7, 8 ?
8. How many different words, of 8 different letters each,
can be formed from the letters in the word ploughed, if the
third letter is o, the fourth u, and the seventh e ?
9. How many different committees, of 8 persons each, can
be formed from a corporation of 14 persons ? In how many
will any particular individual be found ?
10. There are 11 points in a plane, no 3 in the same straight
line. How many different quadrilaterals can be formed, having
4 of the points for vertices ?
11. Erom a pack of 52 cards, how many different hands of
6 cards each can be dealt ?
12. A and B are in a company of 48 men. If the company
is divided into equal squads of 6, in how many of them will A
and B be in the same squad ?
13. How many different words, each having 5 consonants
and 1 vowel, can be formed from 13 consonants and 4 vowels ?
PERMUTATIONS AND COMBINATIONS 449
14. Out of 10 soldiers and 15 sailors, how many different
parties can be formed, each consisting of 3 soldiers and 3
sailors ?
15. A man has 22 friends, of whom 14 are males. In how
many ways can he invite 16 guests from them, so that 10 may
be males ?
16. From 3 sergeants, 8 corporals, and 16 privates, how many
different parties can be formed, each consisting of 1 sergeant,
2 corporals, and 5 privates ?
17. Out of 3 capitals, 6 consonants, and 4 vowels, how many
different words of 6 letters each can be formed, each beginning
with a capital, and having 3 consonants and 2 vowels ?
18. How many different words of 8 letters each can be
formed from 8 letters, if 4 of the letters cannot be separated ?
How many if these 4 can only be in one order ?
19. How many different numbers, of 7 figures each, can be
formed from the digits 1, 2,"~^ 4, 5, 6, 7, 8, 9, if the first,
fourth, and last digits are odd numbers ?
506. To find the number of permutations ofn things which are
not all different, taken all together.
Let there be n letters, of which p are a's, q are 6's, and r are
c's, the rest being all different.
Let N denote the number of permutations of these letters
taken all together.
S appose that, in any particular permutation of the n letters,
the p a's were replaced by p new letters, differing from each
other and also from the remaining letters.
Then, by simply altering the order of these p letters among
themselves, without changing the positions of any of the other
letters, we could from the original permutation form [p differ
ent permutations (§ 501).
If this were done in the case of each of the N original per
mutations, the whole number of permutations would be iVx [p.
450 ALGEBRA
Again, if in any one of the latter the q 6's were replaced by
q new letters, differing from each other and from the remain
ing letters, then by altering the order of these q letters among
themselves, we could from the original permutation form [g
different permutations ; and if this were done in the case of
each of the Nx\p_ permutations, the whole number of permu
tations would be Nx\^X\q
In like manner, if in each of the latter the r c's were replaced
by r new letters, differing from each other and from the remain
ing letters, and these r letters were permuted among them
selves, the whole number of permutations would be
We now have the original n letters replaced by n different
letters.
But the number of permutations of n different things taken
n at a time is [n (§ 501).
In
Therefore, Nx\2JX\qx\r = \n: or, N= , , . •
\p\q\L
Any other case can be treated in a similar manner.
Ex. How many permutations can be formed from the let
ters in the word Tennessee, taken all together ?
Here there are 4 e's, 2 n's, 2 s's, and 1 t.
Putting in the above formula w = 9, jp = 4, g = 2, r = 2, we have
[9 5.6.7.8.9
[4[2[2 22
= 3780.
EXERCISE 205
1. In how many different orders can the letters of the word
denomination be written ?
2. There are 4 white billiard balls exactly alike, and 3 red balls,
also alike ; in how many different orders can they be arranged ?
3. In how many different orders can the letters of the word
independence be written ?
PERMUTATIONS AND COMBINATIONS 451
4. How many different signals can be made with 7 flags, of
which 2 are blue, 3 red, and 2 white, if all are hoisted for each
signal ?
5. How many different numbers of 8 digits can be formed
from the digits 4, 4, 3, 3, 3, 2, 2, 1 ?
6. In how many different ways can 2 dimes, 3 quarters,
4 halves, and 5 dollars be distributed among 14 persons, so
that each may receive a coin?
507. To find for what value of r the number of combinations
ofn different things taken r at a time is greatest.
By § 502, the number of combinations of n different things^
taken r at a time, is
"^^ 1.2.3... (rl)r ^^
Also, the number of combinations of n different things, taken
r — 1 at a time, is
y,(,,_l)...[^_(^l) + l] n(nl) ...(71^ + 2) ,2>>
1.2.3... (r1) ' 1.2.3... (r1) ^^
The expression (1) is obtained by multiplying the expres
sion (2) by — , or — ' 1.
r r
The latter expression decreases as r increases.
If, then, we find the values of (1) corresponding to the val
ues 1, 2, 3, • • ., of /•, the results will continually increase so
long as ■ — IS > 1.
r
I. Suppose n even ; and let w = 2 m, where m is a positive
integer.
Then, "'• + ^ becomes ?JItszI+l.
r r
If r = m, ^~^"^ becomes ^^ , and is >1.
r 771
If r = m + 1, — — ^tL_ becomes — — — , and is < 1.
r m + 1
452 ALGEBRA
Then, ^C^ will have its greatest value when r=m — '
II. Suppose n odd ; and let n — 2m\l, where m is a posi
tive integer.
Then, ^1=1+1 becomes 2«^^r±2.
r r
If r = m, ^~ ^ "*" becomes ^"*" , and is >1.
r m
If r = m + 1, ^ ^^ ~ ^ + ^ becomes ^ "^ , and equals 1.
r m + 1
If r = m + 2, ^"^^^'^^ becomes ^^, and is < 1.
r m + 2
Then, „C^ will have its greatest value when r equals m
or m + 1 ; that is, ^'^—^ or ^^^ + 1.
1
Then, „(7,. will have its greatest value when r equals ^ ~
n4l ^
or I' ; the results being the same in these two cases.
EXPONENTIAL AND LOGARITHMIC SERIES 453
XLI. EXPONENTIAL AND LOGARITHMIC
SERIES
508. The Theorem of Limits.
If two expressions, containing the same variable (§ 317), are
equal for every value of the variable, and each approaches a
limit (§ 318), the limits are equal.
Let A and B be two expressions containing the same variable.
Let A and B be equal for every value of the variable, and
approach the limits A' and B\ respectively.
To prove A'=B\
Let ^'— ^ = m, and 5'— JB = w.
Then, m and n are variables which can be made less than
any assigned fixed number, however small (§ 318).
Then, either m — n is a variable which can be made less than
any assigned fixed number, however small, or else m — n = 0.
But m  w = A A  (B' B)
= A'AB'{B = A'B';
for, by hypothesis, A and B are equal for every value of the
variable.
But A'—B' is not a variable; and hence m — n is not a
variable.
Then, m — n is 0; and hence its equal, A'—B', is 0, or
A'=B'.
THE EXPONENTIAL SERIES
509. We have for all values of w and x.
1+^
n
=".!)■
454 ALGEBRA
Expanding both members by the Binomial Theorem,
[_ n \2 n^ [3 n^ j
^ , 1 , nx(nx — l) 1
n [2 n^
^ 3 n«^ * ^ ^
We may write equation (1) in the form
a;( a; ) x( x Ycc
= ^ +  + V^+ ^ f " + S (2)
which holds however great n may be.
Now let n be indefinitely increased.
1 2
Then, the limit of each of the terms , , etc., is (§ 320).
n n
Hence, the limiting value of the first member of (2) is
['+'+tt}
and the limiting value of the second member is
By the Theorem of Limits (§ 508), these limits are equal ;
that is.
Denoting the series in brackets by e, we obtain
EXPONENTIAL AND LOGARITHMIC SERIES 455
510. Putting mx for x, in (3), § 509, we have
e = Hmx + ^ + ^+.... (4)
Let m = logg a.
Then, by § 412, e"" = a, and e'"== = a*.
Substituting in (4), we "obtain
a^ = 1 + (log, a)x^ (log, ay^ + (log, af^ + • • • . (5)
This result is called the Exponential Series.
511. The system of logarithms which has e for its base
is called the Napierian System, from Napier, the inventor of
logarithms.
Napierian logarithms are also called Natural Logarithms.
The approximate value of e may be readily calculated from
the series of § 509.
and will be found to equal 2.7182818 ••..
THE LOGARITHMIC SERIES
512. To expand log, (1 + x) in ascendijig powers of x.
Substituting in (5), § 510, 1 + a; for a, and y for x,
(1 H a;)^ = 1 + [loge (1 + ic)] ?/ H terms in /, if, etc.
Expanding the first member by the Binomial Theorem,
= 1 + [log, (1 + a;)] y + terms in y\ y^, etc. (6)
This equation holds for every value of y which makes both
members convergent; and, by the Theorem of Undetermined"
Coefficients (§ 396), the coefficients of y in the two series are
equal. ^
456 ALGEBRA
That is, x^x^{^a^^^x'\ ... = \og,(l^x),
Z [^ [4
Or, log,(l + aj) = a. + J + .... (7)
This result is called the Logarithmic Series.
CALCULATION OF LOGARITHMS
513. The equation (7), § 512, can be used to calculate
Napierian Logarithms, if x is so taken that the second mem
ber is convergent; but unless x is small, it requires the sum
of a great many terms to insure any degree of accuracy.
We will now derive a more convenient series for the calcula
tion of Napierian Logarithms.
Bev ; 2 3 4 5 ^^
514. Putting — X for x, in (7), § 512, we have
log,(la;) = a;
Subtracting (8) from (7), we obtain
log.(l + a^)logXla;) = 2a^ + ?! + ?!+....
o o
Or (§422), log.l± = 2(x +  + J + ...). (9)
^ m — n
Let x = : then ~ — = ! — =  = —
m + 7i 1 — x m — n 2n n
m\n
Substituting these values in (9), we obtain
m — n 1 / m — nV 1 / m — n Y 1
m + n 3\m\nJ 5\m\nj J
m
, But by § 422, log,— = log, m — log, n ; whence,
n
log,m = log,» + 2r'5^ + lf2^
_m + 71 3\m\
n Y I 1 / m — nY ^
n) 5lm +
EXPONENTIAL AND LOGARITHMIC SERIES 457
515. Let it be required, for example, to calculate the
Napierian logarithm of 2 to six places of decimals.
Putting m = 2 and n = l in the result of § 514, we have
log,2 = log,l + 2
1 . iriy . 1/r
3~^3VSj '^BKSJ "^
••}
Or since log,l = (§418),
loge 2 = 2(.3333333 + .0123457 + .0008230 + .0000653
+ .0000056 + .0000005 + • • •)
= 2 X .3465734 = .6931468 = .693147,
correct to six places of decimals.
Having found log^2, we may calculate loge3 by putting m=3
and n = 2 in the result of § 514.
Proceeding in this way, we shall find log^ 10 = 2.302585 •••.
516. To calculate the common logarithm of a number, having
given its Napieriayi logarithm.
Putting 6 = 10 and a = e in the result of § 426,
Thus, logio 2 = .4342945 x .693147 = .301030.
The multiplier by .which logarithms of any system are
derived from Napierian logarithms is called the modulus of
that system.
Thus, .4342945 is the modulus of the common system.
Conversely, to find the Napierian logarithm of a number
when its common logarithm is given, we may either divide the
common logarithm by the modulus .4342945, or multiply it by
2.302585, the reciprocal of .4342945.
EXERCISE 206
Using the table of common logarithms, find the Napierian
logarithm of each of the following to four significant figures :
1. 10000. 2. .001. 3. 9.93.
458 * ALGEBRA
4. 243.6. 5. .04568. 6. .56734.
7. What is the characteristic of logs "^8 ?
8. What is the characteristic of logj500?
9. If log3 = .4771, how many digits are there in 3^^ ?
10. If log 8 = .9031, how many digits are there in 8^^?
INDEX
Addition, of fractions, 109.
of imaginary numbers, 243.
of monomials, 18.
of polynomials, 21.
of positive and negative num
bers, 12.
of similar terms, 19.
of surds, 226.
Affected quadratic equations, 250.
Any power, of a monomial, 63.
of a fraction, 186,
Any root, of a fraction, 192.
of a monomial, 190.
Approximate square root of an
arithmetical number, 201.
Associative Law, for addition, 410.
for multiplication, 411,
Calculation of Logarithms, 456.
Clearing of fractions, 53.
Commutative Law, for addition, 410.
for multiplication, 410.
Completing square, first method, 250.
second method, 253.
Cube, of a binomial, 188.
Cube root, of an arithmetical num
ber, 206.
of a polynomial, 202.
Definitions:
Abscissa, 173.
Absolute Value, 12.
Affected Quadratic Equation, 248.
Algebraic Expression, 9.
Arithmetic Means, 335.
Arithmetic Progression, 331.
Arithmetical Complement, 390.
Axiom, 2.
Binomial, 21.
Characteristic, 377.
Definitions — Continued
Coefficient, 17.
Combinations, 445.
Common Factor, 74.
Common Logarithm, 376.
Common Multiple, 100.
Complex Fraction, 121.
Complex Number, 242.
Convergent Series, 358.
Cyclosymmetric ExpreaBion, 416.
Degree of Equation, 52.
Degree of Expression, 41.
Divergent Series, 358.
Division, 42.
Equation, 2.
Equation in Quadratic Form, 268.
Equation of Condition, 51. ""^
Equivalent Equations, 146.
Equivalent Systems of Equations,
431.
Exponent, 7.
Exponential Equation, 393.
Exponential Series, 455.
Factor, 17.
Fraction, 103.
Fractional Exponent, 213.
Geometric Means, 344.
Geometric Progression, 338.
Graph, 175.
Harmonic Means, 347.
Harmonic Progression, 346.
Homogeneous Terms, 41.
Identical Equation, 51.
Imaginary Number, 242.
Inconsistent Equations, 147.
Independent Equations, 146.
Indeterminate Equations, 146.
Index of Root, 190.
459
460
INDEX
Definitions — Continued
Inequality, 180.
Infinite series, 350.
Infinity, 305.
Integral Equation, 51.
Irrational Number, 222.
Limit, 304.
Linear Equation, 52.
Literal Equation, 132.
Mantissa, 377.
Monomial, 17.
Napierian Logarithm, 455.
Negative Exponent, 214.
Negative Number, 12.
Negative Term, 17.
Numerical Equation, 51.
Ordinate, 173.
Perfect Cube, 88.
Perfect Square, 76.
Permutations, 445.
Polynomial, 21.
Positive Number, 12.
Positive Term, 17.
Proportion, 312.
Pure Quadratic Equation, 248.
Quadratic Equation, 248.
Quadratic Expression, 276.
Quadratic Surd, 222.
Rational and Integral, 40.
Rational Number, 222.
Rectangular Coordinates, 173.
Root of Equation, 52.
Series, 350.
Similar Surds, 226.
Similar Terms, 18.
Simultaneous Equations, 147."
Subtraction, 18.
Surd, 222.
Symmetrical Expression, 416.
Trinomial, 21.
Variable, 304.
Zero Exponent, 213.
Discussion of general quadratic equa
tion, 281.
Distributive Law, for multiplica
tion, 412.
Division, by detached coefficients,
441.
of fractions, 118.
of imaginary numbers, 246.
of monomials, 43.
of polynomials by monomials, 45.
of polynomials by polynomials, 46.
of surds, 230.
Elimination, by addition or subtrac
tion, 147.
by comparison, 150.
by substitution, 149.
Evolution of surds, 233.
Expansion, of fractions into series,
361.
of surds into series, 363.
Exponential equations, 393.
Extraction of roots by the Bino
mial Theorem, 375.
Factor Theorem, 414.
Factoring, of expressions whose
terms have a common factor,
75, 76.
of quadratic expressions, 276.
of symmetrical expressions, 419.
of the difference of two perfect
squares, 79.
of the difference of any two equal
odd powers, 89.
of the sum or difference of two
perfect cubes, 88,
of the type x* + ax'^y'^ + yS 81, 279.
of the type x'^ \ ax \ 6, 82, 276.
of the type ax'^ + 6x h c, 85, 276.
of trinomial perfect squares, 77.
Formation of quadratic equations,
275.
General term of binomial expan
sion, 355.
Graphical representation, of addi
tion of complex numbers, 436.
of complex numbers, 435.
of imaginary unit, 434.
of roots of equations, 179, 284.
of solutions of simultaneous linear
equations, 176.
INDEX
461
of solutions of simultaneous quad
ratic equations, 301.
Graph, of first member of a quad
ratic equation having equal or
imaginary roots, 284.
of inconsistent linear equations
with two unknown numbers,
177.
of indeterminate linear equations
with two unknown numbers,
178.
of a linear equation with two un
known numbers, 174.
of a linear expression involving
one unknown number, 178.
of a quadratic equation involving
two unknown numbers, 300.
of a quadratic expression involv
ing one unknown number, 283.
Graphs in Physics, 327.
Highest Common Factor, of expres
sions which can be readily fac
tored by inspection, 98.
by long division, 395.
Indeterminate form  , 438.
Indeterminate forms, x oo ,
CO  00 , 439.
Interpretation, of solutions, 171.
of the form ^, 304.
of the form  , 305.
GO
Introduction of the coefficient of a
surd under the radical sign, 225.
Involution of surds, 231.
Logarithm of a number to any base,
394.
Lowest Common Multiple, of ex
pressions which can be readily
factored by inspection, 100.
by long division, 401.
Meaning of a pure imaginary num
ber, 243.
Multiplication, of fractions, 115.
of imaginary numbers, 244.
of monomials, 33.
of polynomials by monomials,
34.
of polynomials by polynomials, 35.
of positive and negative numbers,
14.
of surds, 228.
Parentheses, insertion of, 30.
removal of, 28.
Partial fractions, 364.
Permutations of things not all dif
ferent taken all together, 449.
Physical Problems, 141, 260, 266,
297, 325.
Problem of the Couriers, 306.
Product, of the sum and difference
of two numbers, 6Q.
of two binomials having same first
term, 67.
Proof, of a"* X a" = «"*+»', for all
values of m and 7i, 405.
of Binomial Theorem, for a posi
tive integral exponent, 350.
Quadratic surds, 237.
Reduction, of fractions to integral
or mixed expressions, 106.
of fractions to their lowest com
mon denominator, 107.
of fractions to their lowest terms
when the numerator and de
nominator can be readily fac
tored by inspection, 104.
of fractions to their lowest terms
• when the numerator and de
nominator cannot be readily
factored by inspection, 404,
of fractions with irrational de
nominators to equivalent frac
tions with rational denomina
tors, 233.
of fractions with irrational de
nominators to equivalent frac
tions with rational denomina
tors, when the denominators
are in the forms a ± V&,
Va ± ^^b, or 7a ± V&, 405.
462
INDEX
Reduction, of mixed expressions to
fractions, 114.
of surds of different degrees to
equivalent surds of the same
degree, 227.
of surds to their simplest forms,
222.
Remainder Theorem, 413.
Repeating decimals, 343.
Reversion of series, 370.
Solution, of equations by factoring,
94, 280.
of equations having the unknown
numbers under radical signs,
239.
of equations involving decimals,
134.
of fractional linear equations,
127.
of integral linear equations,
54.
of literal affected quadratic equa
tions, 258.
of literal linear equations, 132.
of quadratic equations by form
ula, 255.
Square, of a binomial, 64.
of a polynomial, 186.
Square root, of an arithmetical
number, 197.
of a binomial surd, 238.
of a polynomial, 193.
Subtraction, of fractions, 109.
of imaginary numbers, 243.
of monomials, 24.
of polynomials, 26.
of surds, 226.
Sum of a geometric progression to
infinity, 342.
Sum and product of roots of quad
ratic equations, 273.
Theorem of Limits, 453.
Theorem of Undetermined Coeffi
cients (Rigorous), 409.
Transposing terms, 53.
Use of Table of Logarithms, 383.
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