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ELEMENTS OF ALGEBRA
BY
JAMES M. TAYLOR, A.M., LL.D.
PROFESSOR OF MATHEMATICS, COLGATE UNIVERSITY
3>»ic
Boston
ALLYN AND BACON
1900
COPTKIGHT, 19 00,
BY JAMES M. TAYLOR,
NorfajootJ 5|«ss
J. S. Gushing & Co. - Berwick & Smith
Norwood Mass. U.S.A.
T3 '
PREFACE
The author has aimed to make this treatment of algebra
so simple that the pupil can begin the book to advantage
immediately upon completing an ordinary course in arith-
metic j and, at the same time, so scientific that he will
have nothing to unlearn as he advances in the study of
mathematics. Great care has been taken to develop the
subject logically, yet the immaturity of the pupil has been
constantly kept in mind, and every legitimate aid has been
given him. Simplicity has been attained not by using
inexact statements and mechanical methods, but by avoid-
ing many of the outgrown phrases of traditional algebra,
by giving demonstrations and explanations in full, and by
making fundamental concepts clear and tangible. An
introductory chapter explains the meaning and advantages
of the literal notation, and illustrates the use of the equa-
tion in solving arithmetic problems. In Chapter II real
numbers are first considered, and are defined as multiples
of the quality-units, + 1 and — 1, and the pupil is drilled
in the use of particular real numbers before he is required
to represent general real numbers by letters.
General principles are first illustrated by particular ex-
amples, the study of which prepares the pupil to grasp
the meaning of the formal statement of the principles, and
iv PREFACE
makes it less likely that lie will memorize without com-
prehending the demonstrations which follow. With this
arrangement the reproduction of the demonstrations may
be left for the review; but the pupil should become
familiar with each principle and definition before a new
one is considered. When the demonstrations are not re-
produced, it is recommended that the proofs be carefully
read and discussed in class, so that the pupil may be
fully convinced that the principles are true. He should
then be required to state the authorities for each step in
the proof when the steps are given.
The identity and the equation are sharply distinguished.
Two groups of principles are stated, the first for proving
the identity, the second for solving the equation.
The need of the principles of the equivalency of equa-
tions and systems is clearly shown. These principles are
fully illustrated and proved, and upon them are based the
methods of solving equations and systems of equations.
In the chapter on factoring, the formation of equations
with given roots serves as an introduction to the converse
problem of finding the roots of a given quadratic or higher
equation, and to the method of making factoring funda-
mental in the study and solution of quadratic and higher
equations and systems.
The graph is used to illustrate the meaning of equations
in two unknowns, of systems of equations and of equivalent
systems; it also serves to make clear some of the general
properties of equations in one unknown.
The theory of limits is given as briefly as is thought to
be consistent with clearness. It is used in proving the
PREFACE V
laws of incommensurable numbers and in evaluating ex-
pressions which assume the indeterminate form 0/0.
The treatment of imaginary numbers affords a good
illustration of the advantages derived from regarding alge-
braic numbers as arithmetic multiples of quality-units.
When a pupil understands that the quality-units V— 1
and — V— 1 include the idea of the arithmetic one and
that of oppositeness to each other, that (V— 1)^ = —1, and
(V— 1)*= +1, he has mastered all that is new in imagi-
naries, and can then state the general laws for products
and quotients of imaginary and real numbers (§§ 274, 276).
This concept makes for simplicity, for it enables us to
express general laws which are true for real, imaginary,
and complex numbers, and it clearly separates the problem
of finding the arithmetical value of a result from that of
finding its quality. Graphic representations are used to
illustrate the meaning and reality of imaginary and com-
plex numbers.
Special attention is invited to the brevity and complete-
ness of the demonstrations of the, principles of proportion,
the early introduction of the remainder theorem, the use
of type-forms in factoring, and the treatment of fractional
and irrational equations.
The methods of working examples have been chosen for
their simplicity and the scope of their application. The
problems are varied, interesting, well graded, and not so
difficult as to discourage the beginner. Many exercises
contain easy examples which, especially in the review,
should be used for oral work. Suggestions as to the
method of attack are freely given ; rules are stated only
Vi PREFACE
for the most difficult operations, but not until after these
have been illustrated by particular examples.
The author has sought to treat each subject with suffi-
cient fulness to meet the college entrance requirements,
and more subjects are given than are ordinarily considered
as a part of elementary algebra.
The author is indebted to many teachers for valuable
suggestions, but especially to his assistant, Mr. C. D. Kings-
ley, who has carefully read all the manuscript and most
of the proof sheets.
JAMES M. TAYLOR.
Colgate University,
June, 1900.
CONTENTS
CHAPTER I
Introduction
SECTIONS PAOB
1-8. Symbols of Number and of Operations .... 1
9-18. Equalities ; Identities and Equations .... 8
19. Problems solved by Equations 15
CHAPTER II
Positive and Negative Numbers
20-31. Real Numbers and their Symbols 20
32, 33. Principles of Identities 28
CHAPTER III
Addition, Subtraction, and Multiplication op Bbal
Numbers
34-38. Fundamental Laws of Addition ....
39-44. Definition and Fundamental Laws of Subtraction .
45-50. Definition and Fundamental Laws of Multiplication
51-53. Bases and Integral Positive Exponents .
54, 55. Change of Quality of Expressions ....
56, 57. Two Uses of + and — . Abbreviated Notation
58-60. Coefficients. The Distributive Law
30
33
36
40
42
43
44
CHAPTER IV
Addition and Subtraction of Integral Literal
Expressions
61-64. Addition of Similar Terms 49
65, 66. Addition of Polynomials 52
67. Subtraction of One Expression from Another ... 54
68-70. Removal and Insertion of Signs of Grouping ... 56
vii
viii CONTENTS
CHAPTER V
Multiplication of Integral Literal Expressions
SECTIONS PAGE
71-73. Degree of an Expression. Homogeneous Expressions 61
74-76. a- 0=0. 0» = 0. a'«-a" = a"»+" 62
77-81. Product of One Integral Expression by Another . . 63
82. Multiplication by Detached Coefficients ... 69
CHAPTER VI
Division of Integral Literal Expressions
83-
-89.
90,
91.
92-
-94.
95-
-97.
98,
99.
Definition and Laws of Division 71
Definition and Product of Fractions .... 74
«»"/«"= rt"*-". Distributive Law 75
Division of One Expression by Another ... 77
0/a=0. Division by Detached Coefficients . . 84
CHAPTER VII
Linear Equations in One Unknown
100-103. Definitions, Degree, and Root 87
104-110. Principles of Equivalence of Equations ... 88
111-114. Solution of Linear Equations 94
CHAPTER VIII
Problems solved by Linear Equations in One
Unknown
115, 116. Problems solved by Equations. Interest Formulas . 99
CHAPTER IX
Powers, Products, Quotients
117-119. Powers of Powers and of Products .... 109
120,121. Square of Any Expression Ill
122, 123. The Product (a -f &) (a - b) and (x -\-a) {x + b) . 113
124-128. Powers of a -f 6. Like Powers 115
129, 130. The Quotients (a" - 6«) ^ (a - 6) and (a'* ± &") -^
(a + b) 118
131-133. The Remainder and the Factor Theorem . . .122
CONTENTS
IX
CHAPTER X
Factors of Integral Literal Expressions
SECTIONS
134-136. Monomials and Common Factors .
137, 138. Perfect Squares
139, 140. Type-forms x"^ -\-px+ q and ax"^ + 6x + c
141. Type-form a" — ^", where n is Even
142, 143. Type-forms a*' T 6", where n is Odd
144, 145. Type-forms a* + ha'^b^ + &* and a" + 6», n Even
146, 147. Perfect Cube. Summary. Special Devices
148, 149. Formation and Solution of Equations .
PAGE
125
127
130
134
136
138
139
144
CHAPTER XI
Highest Common Factor and Lowest Common
Multiple
150-153. H. C. F. by Factoring
154-157. H. C. F. by Division
158, 159. L. C. M. by Factoring
160-163. L. C. M. by H. C. F.
150
152
158
160
CHAPTER XII
Fractions
164-169. Definition and Quality of Fractious .
170-178. Reduction of Fractions ....
179. Addition and Subtraction of Fractions
180-184. Multiplication and Division of Fractions
185, 186. Complex Fractions. Powers of Fractions
163
165
170
177
181
CHAPTER XIII
Fractional Equations
187-190. Equivalence of Fractional Equations. Problems
187
CONTENTS
CHAPTER XIV
Systems of Linear Equations
SECTIONS
PAGE
191-195.
Equivalent Equations in Two Unknowns .
. 197
196-199.
Systems and Equivalent Systems .
. 199
200-203.
Elimination by Substitution ....
. 201
204-207.
Elimination by Addition or Subtraction
. 205
208, 209.
Systems in Three or More Unknowns .
. 213
210,211.
Systems of Fractional Equations .
CHAPTER XV
Problems solved by Systems
. 216
212.
A Determinate Problem ....
. 220
CHAPTER XVI
Evolution. Irrational Numbers
213-219. An nth Root. A Rational Number. Number of Roots 230
220-222. Principal Root. Evolution 232
223-225. Root of a Power, of a Product, of a Quotient . . 232
226,227. </«'=(</«)'' \/^« = V« 234
228, 229. Square Root of Expressions 235
230-232. Cube and Higher Roots of Expressions . . .239
233-237. Square and Cube Roots of Decimal Numbers . . 244
238-245. Irrational Numbers. Their Approximate Values . 249
CHAPTER XVII
Surds
246-250. Definitions
251, 252. Reduction of Surds . .
253, 254. Addition and Subtraction of Surds
255-257. Product of Surds. Conjugate Surds .
258-260. Division of Surds. Root of Monomial Surd
261-263. Properties of Quadratic Surds
253
254
255
257
260
264
CONTENTS XI
CHAPTER XVIII
Imaginary and Complex Numbers
8KCTION8 PA6I
264-269. Imaginary Units. Imaginary Numbers . . . 266
270,271. Laws of Imaginary Numbers. Powersoft. . . 269
272, 273. V^^a = i • y/a. Addition and Subtraction . . 270
274. Product of Two or More Quality-numbers . . . 270
275, 276. Quotient of One Quality-number by Another . . 271
277-286. Complex Numbers 272
CHAPTER XIX
Quadratic Equations in One Unknown
287, 288. Type-form. Completing the Square . . .278
289. Solution by Factoring 279
290. Principle of Equivalency of Equations .... 280
291-293. Solution of General Quadratic Equation . . . 282
294-296. Properties of Roots. Factors of Expressions . . 285
CHAPTER XX
Problems
297. Solution of Problems by Equations . . . .291
CHAPTER XXI
Irrational Equations
298, 299. Definition. Principle of Equivalence . . . .300
300. Irrational Equations in Quadratic Form . . . 304
CHAPTER XXn
Higher Equations
301. Higher Equations in Quadratic Form .... 306
302. Binomial Equations 308
xu
CONTENTS
CHAPTER XXIII
Systems involving Quadratic and Higher Equations
SECTIONS PAGE
303-305. Solution by Factoring .
306. Solution by Substitution
307. Solution by Addition or Subtraction
308. Systems of Symmetrical Equations
309, 310. Solution by Division .
311
313
315
318
321
CHAPTER XXIV
Inequalities
311-314. Definitions and Principles .
315, 316. Proof and Solution of Inequalities
327
329
CHAPTER XXV
Ratio and Proportion
317-323. Definitions. Properties of Ratios
324-332. Definition and Properties of Proportions
333, 334. Continual Proportion. Mean Proportional
332
334
338
CHAPTER XXVI
Theory of Exponents
335-339. Meaning of Fractional and Negative Exponents
340-343. Properties of Exponents ....
344-352. Proofs of Laws of Exponents
341
343
345
CHAPTER XXVII
Indeterminate Equations and Systems
353, 354. Division 'hj Zero. Forms 0/0, a/0 .
355. Impossible Equations and Systems
356-358. Defective and Indeterminate Systems .
354
355
356
CONTENTS
XlU
CHAPTER XXVIII
Theory of Limits
SECTIONS PAGB
359-361. Variables. Limits of Variables 361
362-375. Principles of Limits 362
376-379. Infinitesimals. Infinites 367
380-382. Indeterminate Forms 369
383-388. Proofs of tlie Laws of Incommensurable Numbei-s . 371
389-398. Variation, Direct, Inverse, Joint 374
CHAPTER XXIX
The Progressions
399-405. Arithmetic Progression 380
406-414. Geometric Progression 386
415-419. Harmonic Progression 392
CHAPTER XXX
Permutations and Combinations
420-424. Permutations
426-428. Combinations
395
400
CHAPTER XXXI
Binomial Theorem
429-435. Exponent a Positive Integer 404
436. Exponent Fractional or Negative 408
CHAPTER XXXII
Logarithms
437,438. Definitions 412
439-445. Properties of Logarithms to Any Base .... 413
446. Base Greater than 1 416
447-455. Common Logarithms 416
456. Exponential Equations 422
457-461. Compound Interest and Annuities .... 424
XIV
CONTENTS
CHAPTER XXXIII
Graphic Solution of Equations and Systems
462. Location of a Point in a Plane
463. Graphic Solution of Equations in x and y
464. Graphic Solution of Systems of Equations
465-468. Graphic Solution of Equations in x
PAGB
430
432
437
441
CHAPTER XXXIV
Theory of Equations
469. Horner's Method of Synthetic Division . . . 444
470-474. / (x) = 0 has w Roots 446
475-477. Equal and Imaginary Roots • 448
478-480. Relation of Roots and Coefficients . . . .452
481, 482. Positive, Negative, and Commensurable Roots . . 454
483. Limits of Roots . . . . • . . . .457
484. The Roots of One Equation Multiples of those of
Another 458
485-487. Location of Real Roots 460
ELEMENTS OF ALGEBRA
CHAPTER I
INTRODUCTION
1. Arithmetic number. In Arithmetic we have seen that
by taking a group of ones we can obtain any tvhole number ;
and that by dividing one into equal parts, and taking a
group of these parts, we can obtain any fractional number.
Hence, the primary unit of arithmetic number is one, 1.
A whole number, or an integer, is one or an aggregate
of ones.
A fractional unit is one of the equal parts of one.
A fractional number is a fractional unit, or any aggregate
of fractional units which does not equal a whole number.
In writing fractions we often use the sign /; thus,
3/2 denotes f.
The numbers defined above answer the single question,
' How many ? ' and are called arithmetic, or absolute, numbers.
Arithmetic numbers are used to express how many times one quan-
tity contains another of the same kind. By diminishing indefinitely
the fractional unit we can obtain a series of numbers in which the
difference between successive numbers will be as small as we please.
2. A numeral is any symbol which is used to denote a
particular number, and which is never used to denote any
other number. The more common numerals are the Arabic
figures, 1, 2, 3, etc., and the Roman letters, I, V, X, etc.
1
2 ELEMENTS OF ALGEBRA
By the use of numerals, as we have seen in Arithmetic, we can
state on\y particular problems. To state and solve general problems,
and to investigate the general properties of numbers, mathematicians
have invented the literal notation.
3. Letters denoting numbers. — An important step in en-
larging the notation of number is the use of a letter, as
a, 6" X, or y, to denote any number whatever or an unknown
number.
E.g., just as heretofore we have spoken of 5 dollars, of 8| miles,
etc., so sometimes we shall speak of a dollars, meaning any number
whatever of dollars ; of x miles, meaning any number of miles or an
unknown number of miles, etc.
Just as when we say the number 4, or simply 4, we mean
the number denoted by the figure 4 ; so when for brevity we
say the number a, or simply a, we mean the number denoted
by the letter a.
The following simple examples will illustrate how letters
are used to denote any number whatever in the statement of
general arithmetic problems.
Ex. 1. If one merchant has 50 dollars and another has 25 dollars,
the two together have 50 + 25 dollars. If one merchant has m dollars
and another has n dollars, the two together have m + n dollars.
Here m or n denotes any whole or fractional number ; and m -\- n
denotes the sum of these numbers.
Ex. 2. If a drover buys 5 horses at 50 dollars each, he pays 50 x 5
dollars for the horses. If a drover buys y horses at x dollars apiece,
he pays x •>< y dollars for the horses.
Here y denotes any whole number, x any whole or fractional num-
ber, and XX y their product.
Ex. 3. If 60 dollars is divided equally among 5 boys, each boy re-
ceives 60-4-6 dollars. If x dollars is divided equally among n boys,
each boy receives x -^ n dollars.
Ex. 4. li m men earn n dollars in one day, each man earns « -5- m
dollars in one day, and therefore x men will earn w -f- w x as dollars in
one day.
INTRODUCTION 3
In these examples the reasoning is the same whether the
numbers are denoted by figures or by letters.
When letters are used in its statement, each problem is a
general problem, and includes an unlimited number of par-
ticular problems.
4. The following signs, or symbols, of operation, with
which, as has been assumed, the pupil is already familiar,
are common to all branches of mathematics.
The sign of addition, +, read 'jylus/ indicates that the
number after the sign is to be added to the number before it.
E.g., S -^ i means that 4 is to be added to 3. a + 6, read 'a
plus 6,' means that the number denoted by b is to be added to the num-
ber denoted by a ; or, more briefly, it means that b is to be added to a.
The sign of subtraction, — , read ^minus,^ indicates that
the number after the sign is to be subtracted from the num-
ber before it.
E.g.., a-\-b — c, read 'a plus b minus c,' means that 6 is to be
added to a, and then c subtracted from this sum.
The sign of multiplication, x , or a point above the line,
read ^multiplied by, or Unto/ indicates that the number
before it is to be multiplied by the number after it.
The sign of multiplication is usually omitted between
two letters or a figure and a letter.
E.g.^ 2ab, read '2a6,' means 2xaxb; 7a6c, read ' 7 a&c,'
means 7 • a • 6 • c. The sign of multiplication cannot be omitted
between two factors when both are denoted by figures ; for by the
notation of Arithmetic, 54 means 60 + 4, not 6x4.
The sign of division, -i-, read ^divided by,' or ' by,' indicates
that the number before it is to be divided by the number
after it.
E.g.,, a -i- b X c -i- d, read ' a by & into c by d,' denotes that a is to
be divided by 6, the result multiplied by c, and then this result
divided by cl.
4 ELEMENTS OF ALGEBRA
Observe that in a series of additions and subtractions, or
in a series of multiplications and divisions^ the operations are
to he performed from left to right.
Exercise 1.
1. If a boy has 5 marbles and wins 4 more, how many-
marbles has he ? If he has a marbles and wins h more,
how many marbles has he ?
2. One part of 25 is 7. What is the other part ? One
part of 25 is n. What is the other part ? One part of the
number m is n. What is the other part ?
3. The difference of two numbers is 6, and the smaller
is 12. What is the greater ? The difference of two num-
bers is n, and the smaller is x. What is the greater ?
4. How old will a man be in 6 years, if his present age
is 36 years ? How old will a man be in c years, if his
present age is x years ?
5. In 10 years a man will be 50 years old. What is his
present age ? In & years a man will be m years old. What
is his present age ?
6. The length of a room is x feet, and its width is h
feet less than its length. What is its width ?
7. One number is x, and a second number is y times as
great. What is the second number ?
8. One number, x, is y times as great as a second num-
ber. What is the second number ?
9. The number which contains 4 units and 5 tens is
10 X 5 + 4. Write the number which contains x units and
y tens.
10. Write a number containing a? units, y tens, and v
hundreds.
INTRODUCTION 6
11. Of three consecutive whole numbers 6 is the second;
what are the first and the third ? If the second is m, what
are the first and the third ?
12. Of three consecutive whole numbers 7 is the first;
what are the second and the third ? If the first whole
number is x, what are the second and the third ?
13. Of three consecutive even integers, 8 is the third;
what are the first and the second ? If the third integer is
m, what are the first and the second ?
14. If a goat costs x dollars, and a cow costs 4 times as
much as a goat, and a horse costs 3 times as much as a cow,
how much does a horse cost ?
15. In example 14, how much do a goat, a cow, and a
horse together cost ?
16. A is ic years old, B is 17 years older than A, and C's
age equals the sum of B's age and A's age. How old is C ?
17. If m sheep cost x dollars, and n cows cost y dollars,
what would c sheep and b cows cost ?
18. A travelled a hours at the rate of m miles an hour,
and B travelled b hours at the rate of y miles an hour. How
many miles did A and B together travel ?
19. A rides his bicycle n yards ; the circumference of
each wheel is m feet. How many revolutions does each
wheel make in going this distance ?
5. A mathematical expression is any symbol or combina-
tion of symbols which denotes a number.
If all the symbols of number in an expression are
numerals, the expression is called a numeral expression.
An expression which involves one or more letters is
called a literal expression.
The number denoted by a numeral expression is a par-
ticular, or a fixed, number. For sake of distinction, the
6 ELEMENTS OF ALGEBRA
number which is denoted by a literal expression is called a
general, or an arbitrary^ number.
By the value of an expression we mean the number de-
noted by it.
E.g., 4, 5 — 3, and 7x5 + 4x2 are numeral expressions, and each
denotes a particular, or fixed, number ; while a, a +4, and ax-\-b — c-i-y
are literal expressions, and each denotes a general number.
6. An axiom is a truth so obvious that it may be taken
for granted.
Two numbers are said to be equal when they bear the
same relation to the same unit.
E.g., 4x3 and 6 x 2 are equal numbers, since each is 12 times 1.
I and I are equal numbers, since each is 6 times \.
The axioms concerning equal numbers, which are most
frequently used in Algebra, as in Arithmetic, are the fol-
lowing :
1. Any number is equal to itself.
2. Any number is equal to the sum of all its j)cirts.
3. If each of two numbers is equal to the same number,
they are equal to each other.
4. If equal numbers are added to equal numbers, the sums
are equal.
5. If equal numbers are subtracted from equal numbers,
the remainders are equal.
6. If equal numbers are multiplied by equal 7iumbers, the
products are equal.
7. If equal numbers are divided by equal numbers, except
zero, the quotients are equal.
E.g., 12 = 8 + 4, and 12 -- 4 = (8 + 4) -- 4.
Again, 2x0 = 5x0; but we cannot divide by 0 and say that 2 = 5.
8. The value of a mathematical expression is not changed
when, for any number in it, an equal number is substituted.
INTRODUCTION 1
7. The following signs of relation are common to all
branches of mathematics:
The sign of equality, =, read 'is equal to,^ is placed be-
tween two expressions to indicate that they denote equal
numbers.
The sign of inequality, >, read Us greater than/ is placed
between two expressions to indicate that the first denotes a
greater number than the second. The sign < is read 'is
less thanJ
E.g., 4 4- 8 > 10 is read ' 4 plus 8 is greater than 10 ' ;
and 7 — 2 < 12 is read ' 7 minus 2 is less than 12.'
Observe that in each case the small end of the symbol is toward
the less number.
The sign ^, read 'is not equal to/ is used in stating
that two numbers are unequal, without indicating which is
the greater. Thus, a^h\?, read ' a is not equal to 6.'
8. The signs of grouping are the parentheses ( ), the
brackets [ ], the braces \ \, and the vinculum .
Each of these symbols indicates that the expression in-
cluded by it is to be treated as a whole.
E.g., the expression 12 — (3 + 5) denotes that the sum 3 + 5 is to
be subtracted from 12 ; that is,
12 _ (3 + 5) = 12 - 8 = 4.
The expression [32 - (4 + 6) -r- 6] -=- 3 denotes that one-fifth of the
sum 4 + G is to be subtracted from 32, and the remainder divided by
3 ; that is,
[32 _(4 + 6)- 6]- 3 =[32 - 2]- 3 = 10.
Wlien one sign of grouping is used within another, to avoid ambi-
guity different forms must be used as above.
9. Classification of expressions. A term is any expression
in which the symbols of number are not connected by the
sign -I- or — ; as 4 x 5 ^ 2 or 3 a6 -h c.
8 ELEMENTS OF ALGEBRA
Hence the signs x and -;- indicate operations within a
term, and the parts of an expression which are connected
by the sign + or — are its terms.
E.g.,, each of the expressions 5, a, and 5 aj -r- a is a term.
The expression 2 ax + 3 6 -^ c consists of two terms, 2 ax and
Sh-^c.
In this definition of a term an expression ivithin a sign of
grouping must be considered as a single symbol of number.
Hence a factor or a divisor in a term can itself consist of
two or more terms.
E.g., the expression (a + &)(c + (?) is a term in which each of
the factors, a + b and c -\- d, consists of two terms.
A monomial is an expression of one term ; as 4, 6 xy, or
A polynomial is an expression of two or more terms ; as
4 + T or a + 3 a^?/ + 7 6.
A polynomial of two terms is called a binomial.
A polynomial of three terms is called a trinomial.
Observe that all operations within each of two terms
must be performed before performing the operation between
them.
E.g., the binomial 10 - (4 + 2) (7 - 3) -^ (6 + 2) denotes that. 4 + 2
is to be multiplied by 7 — 3, this product divided by 6 + 2, and the
resulting quotient subtracted from 10.
Exercise 2.
Express in its simplest form the number denoted by each
of the following numeral expressions :
1. 14 + (7 -4). 5. 18 -(6 -2)3.
2. 18 -(12 -7). 6. (6 + 9)^5.
3. (6 + 2) -(7 -3). 7. 16 -(7 -1)^3.
4. (3 + 8)3. 8. 22-(18-6)-^4.
INTRODUCTION 9
9. 12 +[4 -(5 -3)]. 11. 19 -[(2 + 4) -(5 -3)].
10. 18 -[8 -(4 + 2)]. 12. 22 - [23 - (7 - 4)] - 5.
13. How many terms in each of the expressions found in
examples 1 to 12 inclusive ?
14. Find the sum of 5 a; and 7 x.
Just as 5 + 7 = 12, so bx+1 x = V2 x.
Reduce each of the following expressions to its simplest
form :
15. 2x-{-4:X. 17. ^x-\-^x — (dx. 19. \x-ir\x-\-^x.
16. bx-\-lx. 18. ^x — ^x-\-2x. 20. la + ^a — \a.
10. An equality is the statement that two expressions
denote the same number. The expression to the left of the
sign of equality is called the first member of the equality,
and the expression to the right of this sign is called the
second member.
E.g., (5 + 3)0 = 72 is an equality ; of which (5 + 3)9 is the first
meuiber and 72 is the second member.
11. Zero is the number obtained by subtracting any num-
ber from itself ; that is, zero is dehned by the equality
a - a = 0. (1)
12. To find the value of a given literal expression when
each of its letters has some particular value, we substitute
for each letter its particular value, and simplify the result-
ing expression.
Ex. Find the value of the expression (x + y)z -h(a — b), when
a; = 6, ?/ = 3, 0 = 4, a = 9, 6 = 2.
Substituting, 6 for x, 3 for y, 4 for z, 9 for a, and 2 for b, in the
given expression, we obtain
(X + ?/)0 -4-(a - 6) = (6 + 3) X 4 -(9 - 2) (1)
= 36-7. (2)
10 ELEMENTS OF ALGEBRA
In the work above we have three equalities ; by axiom 8, the first
expression is equal to the second and the second is equal to the third ;
hence, by axiom 3, the first is equal to the third.
13. In working examples the student should give heed to
the following suggestions :
1. Too much importance cannot be attached to neatness
of style and arrangement. Neatness is in itself conducive
to accuracy.
2. It should be clearly brought out how each result fol-
lows from the one before it ; for this purpose it will some-
times be advisable to add short verbal explanations.
3. Unless the members are very short the signs of equal-
ity in the steps of the work should be placed one under the
other.
Exercise 3.
Find the .value of each of the following expressions when
a = 5, 6 = 3, c = 4, x = ^'.
1. a-\-h. 6. (a -f h)x. 11. x-^ {a — c).
2. a—b. 7. {a—b)c. 12. {a-\-b)(G -\- x).
3. a-\-b — c. 8. {a-\-b)^x. 13. (a — b) (x — c).
4. abc. 9. (a — b) -^ x. 14. [x — (b -{- 1)] a.
5. ab ~- c. 10. x^(a-\~c). 15. [a; + (a — c)] -j- a.
16. [3 6-(aj-a)]--c. 18. (9-a)(2b-c)(2x-3b).
17. (3a-2b)-h{x-b). 19. (3x-4.c)(3b-2 c)^(x-c).
20. [2 a - (3 6 - 2 c)] -- [(3 c - 3 6) (2 a - 3 6)].
21. [3ir-2(a-6)]-f-[(2ic-3&)(a-c)].
22. (2x-3b)(Aa-3x)-^{3x-3c-b).
14. A proof is a course of reasoning by which the truth
of a statement is made clear, or is established.
INTRODUCTION 11
15. Identical expressions. Two numeral expressions which
denote the same number or any two expressions which de-
note equal numbers for all values of their letters are called
identical expressions.
E.g., the numeral expressions 3G -^ 4 and 13 — 4 are identical, for
each denotes the number 9.
Again, the literal expressions 3 a; + 7 x and 6 x + 4 x are identical,
for each denotes the general number 10 x.
To prove that two expressions are identical, we reduce one
to the form of the other, or we reduce both to the same form.
Ex. Prove that the expressions 7x + 3x + 2x and 14 x — 2 x are
identical.
7x + 3x + 2x denotes 12 x, and 14 x — 2 x denotes 12 x ; hence,
by definition, the two expressions are identical.
An equality whose members are identical expressions is
called an identity.
The sign of identity, =, read Us identical with,'' is often
used instead of the sign = in writing a literal identity, i.e.,
one whose members involve one or more letters.
E.g., 9 + 6 = 6x3, (1)
or 3 X + 7 X = 8 X + 2 X, (2)
is an identity, (1) being numeral and (2) being literal.
Any equality which involves only numerals is an identity.
The sign = points out the fact that equality (2) is an identity.
The pupil should now prove the identities in Exercise 4.
16. Letters denoting unknowns. Any problem involves
one or more numbers whose values are given, and one or
more numbers whose values are to he found. Numbers given
are called knowns, numbers to he found are called unknoivns.
An unknown is usually denoted by one of the last letters
of the alphabet ; as x, y, z.
The following simple problems illustrate the advantage of
denoting an unknown by a letter.
Prob. 1. The sum of two numbers is 80, and the greater is 3 times the
less. Find the numbers.
12 ELEMENTS OF ALGEBBA
Let X = the less number ;
then, since the greater is three times the less,
^x = the greater number.
Hence their sum = a; + 3x = 4a;.
Therefore, by the conditions of the problem, we have
4 a; = 80. (1)
Divide by 4, x = 20, the less number.
Multiply by 3, Zx = 60, the greater number.
Observe that the numbers 20 and 60 satisfy the conditions of the
problem ; that is, 20 + 60 = 80, and 60 = 20 x 3.
Prob. 2. A farmer bought a horse, a cow, and a goat ; the horse cost
3 times as much as the cow, and the cow 4 times as much as the goat,
and all three together cost 255 dollars. What was the cost of each ?
Let X = the number of dollars the goat cost ;
then 4 ic = the number of dollars the cow cost,
and 12 x = the number of dollars the horse cost.
Hence the number of dollars all three cost
= x -}- 4 X + 12 a; = 17 ic.
Therefore, by the conditions of the problem, we have
17 X = 255. (2)
Divide by 17, x — 15.
Multiply by 4, 4 a; = 60.
Multiply by 3, 12 a; = 180.
Hence the goat cost $ 15, the cow f 60, and the horse •$ 180.
17. Equations. Any equality which is not an identity is
called an equation, as (1) or (2) in § 16.
A value of x in an equation in x is any number which
when substituted for x makes the equation an identity.
An equation in one unknown as x restricts x to one value
or to a definite number of values.
INTRODUCTION 13
E.g., if in the equation
2 a; + 2 = a; + 8, (1)
we put 6 for x, we obtain the identity
2x6 + 2 = 6 + 8.
Hence 6 is a value of x in equation (1); and, as will be seen
later, 6 is the only value of a; in (1).
An equation, as (1) or (2) in § 16, expresses in symbols
the conditions of a problem ; and it restricts its unknown to
such values as will satisfy these conditions.
Thus th« equation 4 a; = 80 restricts x to the one value 20 ;
and the equation 3 « = 15 restricts x to the one value 5.
The two kinds of equalities, equations and identities, must
be clearly distinguished the one from the other.
An equation states a condition, and the values of the
unknown which satisfy this condition are to be found ; while
an identity states that one of two expressions can be reduced
to the other, and is to be proved.
Exercise 4.
Prove each of the following identities :
1. 7x3x2 = (10-3)x6. 4. 2X+ 7 a; = (27 --3). a;.
2.88-4 =(7+4)x2. 5. 9a + loa =( 6x4) • a.
3. 204-6 =(12+5)x2. 6. 106 + 8 6 =(36^2) • 6.
By inspection find a value of x in each of the following
equations, and verify it by substitution :
7. a;-4 = 0. 10. 4a; = 20. 13. 3x + l = 10.
8. 2a; = 14. 11. 2a; + l = 7. 14. a; - 1 = 6.
9. 3a;-15 = 0. 12. 2a;+4 = 8. 15. 2a;-4 = 4.
16. 2 a; + 1 = a; + 3. 17. 2 a; - 1 = a; + 2.
14 ELEMENTS OF ALGEBRA
18. The following principles, which are proved in Chap-
ter VII, are used in finding the values of the unknown in
an equation :
(i) If the same number is added to or subtracted from both
members of an equation, the unknown has the same values in
the derived equation as in the given one.
(ii) If both members of an equation are multiplied or divided
by the same known number (except zero), the unkjioivn has the
same values in the derived equation as in the given one.
Ex. 1. Find the value of x in the equation
2a: + 5 = ll. (1)
Subtracting 5 from each member, we remove all the known terms
from the first member, and obtain
2 X = 6. (2)
Dividing each member by 2, we obtain
x=3. (3)
By principle (i), x has the same value in (2) as in (1); and by (ii),
X has the same value in (3) as in (2).
Hence 3 is the one and only value of x in equation (1).
Ex. 2. Find the value of x in the equation
4x-2 = a; + 4. (1)
Add 2, 4x = x + 6. (2)
Subtract ic, 3 a; = 6. (3)
Divide by 3, x = 2. (4)
By principle (i), x has the same value in (2) as in (1), and the
same in (3) as in (2) ; by (ii), x has the same value in (4) as in (3) .
Hence 2 is the one and only value of a; in (1).
Check : Putting 2 for x in (1), we obtain the identity
4x2-2 = 2 + 4.
Hence 2 is a value of x in (1).
IN TB OD UCTION 1 5
Ex. 3. Find the value of x in the equation
fx-fx = |. (1)
To clear (1) of fractions, we multiply both its members by 8,
I.e., by the least common multiple of its denominators.
12 X - 10 a; = 7, or 2 x = 7. (2)
Divide by 2, x = |. (3)
By (ii), X has the same values in (2) as in (1), and the same in
(8) as in (2); hence | is the one and only value of x in (1).
The foregoing examples illustrate the method of finding
the value of the unknown in a simple equation.
Exercise 5.
Find the value of x in each of the following equations :
1. 3a;-7=2aj + 3. 11. 5.x- - 2 = 3a; + 4.
2. 3a; + 4 = a;-hl0. 12. Tx- 9 = 17 + 2a;.
3. 4a; + 4 = a; + 7. 13. |ic-4 = 5-Ja;.
4. 7a; + 5 = a; + 23.
5. 8a; = 5a;+42.
6. 6.T — 5 = 4a: + l.
7. 18a;- 7 = 43 -7a;.
14.
4a;-3 = 7-^a;.
15.
l^-\ = \-\^-
16.
i^-\ = i-\x.
17.
7a; + 21 = 45-5a;.
18.
^ + f = H-i.'^'-
19.
\x + l=^\x + \.
20.
A^ + it = A^ + 4.
9. 19a; -11 = 15 + 6a;.
10. 3 a; + 15 = a; + 25.
19. Problems solved by equations. Read the problem care-
fully to find out exactly what it means; then state in
algebraic symbols just what it says.
To do this, let x denote the unknown number; or, if
there are two or more unknown numbers, let x or some
multiple of x denote one of them, and then express each of
the others in terms of x.
16 ELEMENTS OF ALGEBRA
By an equation express the condition which the problem
imposes on x.
Then find the value of x in this equation.
Exercise 6.
1. A line 30 inches long is divided into two parts, one
of which is double the other. How long are the parts ?
Let X = the number of inches in the second part ;
then 2 aj = the number of inches in the first part.
Hence the number of inches in the two parts = 2 a; + a; = oic.
Therefore, by the conditions of the problem, we have
3x = 30.
Divide by 3, x = 10, number in second part.
Multiply by 2, 2 a; = 20, number in first part.
2. A, B, and C together have f 90. B has twice as
much as A, and C has as much as A and B together. How
much has each ?
Let X = the number of dollars A has ;
then 2x = the number of dollars B has ;
hence 3 a; = the number of dollars C has.
.-. ic + 2 .r + 3 X = 90.
3. The sum of the ages of A and B is 67 years, and A
is 17 years older than B. What is the age of each ?
Ans. 42 and 25 years.
4. Three men, A, B, and C, trade in company and gain
^ 600, of which A is to have 3 times as much as B, and C
as much as A and B together. What is the share of each ?
Let X = the number of dollars B is to have, etc.
5. A farmer bought 3 cows for $ 180, and the prices
paid were as the numbers 1, 2, and 3. What was the cost
of each ?
INTRODUCTION 17
Let X — the number of dollars paid for the first ;
then 2x = the number of dollars paid for the second,
and 3 X = the number of dollars paid for the third.
6. Divide 500 into two parts which are as the numbers
1 and 4.
7. What number is that whose double exceeds its half
by 27?
8. Divide $ 575 between A and B so that A may receive
$ 75 more than B.
Let X = the number of dollars B receives ;
then X 4- 75 = the number of dollars A receives ;
hence 2 x + 76 = 675. (1)
9. Divide 105 into two parts whose difference is 45.
10. What number is that to which if 40 is added the
sum will be 3 times the original number ?
11. Divide $84 among A, B, and C, so that B shall
have $ 13 more than A, and C $ 16 more than B.
12. Three men, A, B, and C, contribute to an enterprise
$ 2400. B put in twice as much as A, and C put in as much
as A and B together. How much did each contribute ?
13. Find two numbers whose difference is 10, and one of
which is 3 times the other.
14. If two men, 150 miles apart, travel toward each
other, one at the rate of 2 miles an hour, and the other at
the rate of 3 miles an hour, in how many hours will they
meet?
15. A horse, carriage, and harness together are worth
% 625. The horse is worth 8 times as much as the harness,
and the carriage is worth $ 125 more than the harness.
Find the value of each. Ans. $ 400, $ 175, and $ 50.
18 ELEMENTS OF ALGEBBA
16. A man bought a cow, a sheep, and a hog for $ 80 ;
the cow cost $ 32 more than the sheep, and the sheep $ 6
more than the hog. Find the price of each.
Ans. f 50, $ 18, $ 12.
17. The sum of $6000 was divided among A, B, C,
and D; B received twice as much as A, C as much as A
and B together, and D as much as A, B, and C together.
How much did each receive ?
Ans. $ 500, $ 1000, $ 1500, $ 3000.
18. A man has two sons and one daughter. He wishes
to divide $ 12,000 among them so that the younger son
shall have twice as much as the daughter, and the older
son as much as both the other children. How much must
he give to each ?
19. Divide 90 into five parts so that the second shall be
5 times the first, the third shall be J of the first and second,
the fourth shall be ^ of the first, second, and third, and the
fifth shall be 2 times the sum of the other four.
20. A, B, and C enter into partnership to do business.
A furnishes 5 times as much capital as B, and C furnishes
\ as much as A and B together. They all together furnish
$ 18,900. How much does each furnish ?
21. A gentleman, dying, bequeathed his property of
$21,840 as follows: to his son 2 times as much as to his
daughter, and to his widow 1\ times as much as to both his
son and daughter. What was the share of each ?
22. A farmer purchased 100 bushels of grain. He bought
2 times as many bushels of corn as of oats, and 2\ times as
many bushels of wheat as of oats and corn. How many
bushels of each kind did he buy ?
23. Three candidates for an office polled the following
votes respectively : B received 3 times as many votes as A,
and C 1^ times as many as A and B together. The whole
INTRODUCTION 19
number of votes was 11,000. How many votes did each
receive ?
24. A banker loaned to each, of 4 men equal sums of
money. One man had the money 2 years, another 2^ years,
another 3^ years, and another 4^ years. The entire interest
money received was $ 275. How much did each man pay ?
Let X = the number of dollars in the yearly interest on the sum
loaned to each man.
25. A library contains 9 times as many historical works,
and 5 times as many scientific books, as works of fiction.
The historical works exceed the works of fiction and science
by 10,500 volumes. How many volumes are there of each ?
26. A drover, being asked how many sheep he had,
replied that if he had 3 times as many as he then had and
6 more, he would have 150. How many had he ?
27. The expenses of a manufacturer for 5 years were
$ 17,500. If they increased $ 500 annually, what were his
expenses each of the five years ?
28. A farmer had 590 sheep distributed in three fields.
In the first field there were 25 more than in the second, and
in the third there were 15 more than in the first. How
many sheep were in each field ?
29. Of a herd of cows, 280 are Jerseys, and these are
35% of the entire herd. How many cows in the herd ?
Let X = the number of cows in the entire herd ; then ^^^ x — 280.
30. A town lost 7% of its inhabitants, and then had 6045
inhabitants. What was its population before the loss ?
31. What number increased by -J- of 25% of itself equals
315?
32. The annual rent of a house is $240, and this is 8%
of its value. What is its value ?
CHAPTER II
POSITIVE AND NEGATIVE NUMBERS
20. Algebra treats of tlie equation, its nature, the methods
of solving it, and some of its applications.
21. In each of the equations thus far considered, the un-
known is an arithmetic number. But in many equations
the unknown cannot be an arithmetic, or absolute, number.
E.g., take the equation
^x = 2x-n. (1)
Subtracting 2 x from each member of (1), we obtain
a; = 0 - 5, or - 5. (2)
Hence the value of x in equation (1) is denoted by the
expression — 5, which has no meaning in Arithmetic.
If, therefore, such an equation as 3 x = 2 ic — 5 is to be of
any use, we must so enlarge our concept of number as to
give a meaning to such an expression as — 5.
To gain this larger idea of number let us first consider
opposite concrete quantities.
22. Positive and negative, or opposite, quantities. Two
quantities are said to be opposites, if, when combined (or
united as parts into one whole), any amount of the one
destroys, or annuls, an equal amount of the other. /
Of two opposite quantities, we call one positive and the
other negative.
JS.g.f debts and credits are opposites ; for when they are com-
bined, any amount of debt annuls an equal amount of credit. If we
call credits positive, debts will be negative.
20
POSITIVE AND NEGATIVE NUMBERS 21
Two forces acting in opposite directions are opposites ; for when
they are combined, any amount of the one annuls an equal amount
of the other. If one of these forces is called positive, the other is
called negative.
Distances measured or travelled in opposite directions are oppo-
sites ; for when they are combined, any distance travelled in the one
direction annuls an equal distance travelled in the opposite direction.
If one distance is called positive, the other is negative.
The sign -f- or the sign — is often written before the
measure of a concrete quantity to denote its quality, as
positive or negative. When thus used, the signs + and —
are read ^positive' and ^negative/ respectively, and are
called signs of quality.
E.g., if we call credits positive, + $5 will denote $5 of credit, and
— $4 will denote .^4 of debt. If + 8 inches denotes 8 inches to the
right, — 9 inches will denote 9 inches to the left. If + S° denotes 3°
above the zero point, — 7° will denote 7° below that point.
If 4- 400 years denotes 400 years after Christ, — 300 years will
denote 300 years before Christ.
In this chapter and the next we shall use as signs of
quality the small signs ■•" and ~, which, by their size and
position, are clearly distinguished from the signs of opera-
tion, + and — .
Exercise 7.
I
1. If credits are regarded as positive, what is denoted
by +^ 8 ? By "$ 11 ? By +$ 125 ? By -^ 175 ?
If debts are regarded as positive, what does each of the
above expressions denote ?
2. If degrees above the zero point are regarded as posi-
tive, what is denoted by +1°? By +22°? By "5°? By
-20°?
3. If distances measured from the point 0 to the right
are regarded as positive, what is denoted by '1 inches?
By +14 inches ? By "13 inches ?
22 ELEMENTS OF ALGEBRA
4. If distances north of the equator are regarded as posi-
tive, what is denoted by +300 miles ? By ~700 miles ?
State in symbols each of the following in two ways :
5. $45 gain and $ 2^ loss is equal to $ 20 gain.
+$45 +-|25 =+.$20, gain being positive ;
<n\ -$45 ++$25 =-$20, loss being positive.
6. $25 gain and $ 30 loss is equal to $ 5 loss.
23. Positive one and negative one. Just as from the con-
crete unit $1 or 1° we gain the idea of the unit 1, so from
1 he concrete positive and negative units +$ 1 and ~$ 1, or
+1° and "1°, we gain the idea of positive one, +1, and nega-
tive one, ~1.
Positive one, +1, and negative one, ~1, include both the idea
of the arithmetic one and that of 02)positeness to each other.
The units +1 and ~1 being opposites, each annuls the
other when added to it; that is, +1 + "! = 0, and ~1++1 = 0.
The units +1 and ~1 are called quality-units.
Of quality-units, "^1 is taken as the primary unit.
24. Positive and negative numbers. Just as we say that
+4 denotes 4 times +1, or 4 positive units ; so, enlarging the
meaning of times, we shall say that +(1) denotes -| times +1,
or I a positive unit, and "(f) denotes f times ~1, or f nega-
tive units.
Any arithmetic number of times the unit +1 is called a
positive number, as +5. Any arithmetic number of times
the unit ~1 is called a negative number, as ~4 or ~(f).
Observe that the only new idea in a positive or a negative
number is that of the quality-unit +1 or ~1.
A positive number and a negative number are opposite
numbers. Thus +5 and ~4 are opposite numbers.
A positive or a negative number answers the two ques-
tions, *How many?' and ^Of what quality?' Its arith-
POSITIVE AND NEGATIVE NUMBERS 23
metic, or absolute, value answers tlie first question, and its
quality-unit the second.
E. g. , the arithmetic value of +5 is 5, and its quality-unit is +1 ; the
arithmetic value of —(f) is |, and its quality is negative.
A positive or a negative number is integral or fractional
according as its arithmetic value is integral or fractional.
E.g.-i +(|) and -(|) are fractional numbers.
25. Symbols for positive and negative numbers. A figure
(or figures) with the sign +, or ~, prefixed denotes a particu-
lar positive, or a particular negative number. The figure
denotes the arithmetic value, and the sign +, or -, denotes
the quality-unit +1, or ~1.
E.g.^ each of the expressions +3, -7, +8, -5 denotes a particular
positive, or a particular negative, number.
A letter with the small sign + or ~ prefixed denotes a
general 2)ositive or a general negative number. The letter
denotes a general arithmetic number, and the sign + or ~,
denotes the quality-unit +1, or ~1.
E.g., the expression +a denotes a general positive number, the
letter a denoting a general arithmetic number, and the small sign +
the quality-unit +1.
A letter not preceded by a small sign +, or -, denotes any
number, positive or negative, integral or fractional.
E.g., a denotes +2, -3, +7, -9, or any other number, positive or
negative ; so also does h, x, y, or z.
Hence, a letter in Algebra denotes an algebraic number
except when, by the presence of a small sign (+ or ~) before
it, it is restricted to an arithmetic value.
26. To add one number to another is to unite the one
with the other into one whole or aggregate.
24 ELEMENTS OF ALGEBRA
As in Arithmetic, the two given numbers are called sum-
mands, and the result is called the sum.
Ex. 1. Add +6 to +4.
Four times the unit +1 plus 6 times the same unit is equal to 4 + 6
times that unit ; that is,
+1 X 4 + +1 X 6 = +1(4 + 6);
or +4 + +6 = +10.
Ex. 2. Add -5 to -7.
Seven times the unit -1 plus 5 times the same unit equals 7 + 4
times that unit ; that is,
-7 4- -5 = -12.
These examples illustrate the following principle :
27. To add one number to another of the same quality, find
the sum of their arithmetic values and jwefix to it the sign of
their common quality. Or stated in symbols,
+a + +6 = +(a + b), -a-^-b = "(a + b).
Proof a times the unit +1, or ~1, plus h times the same
unit is equal to a + 6 times that unit.
Exercise 8.
1. What is the arithmetic (or absolute) value and the
quality-unit of +7? Of -.15? Of "111? Of "a? Of
+(a + 2) ? Of -{a + &) ?
2. Find the sum of +5 and +7. Of +3 and +11. Of "3
and -16. Of "7 and "9. Of "10 and "12. Of +7 and +14.
3. Find the sum of +(i.) and +(f). Of +(f) and +(ii).
Of -(f) and -(t%). Of -(i) and "(i©.
Find the value of +a + +6,
4. When a = 43, 6 = 63. 5. When a = 23, 6 = 72.
Find the value of -a + -&,
6. When a = 15, 6 = 121 7. When a = f , 6 = i
POSITIVE AND NEGATIVE NUMBERS 25
What is the value of m -H w,
8. When m = +24, 91 = +32?
9. When m = -36,n= "22 ?
10. When m = +(f), n = +(-|) ?
11. When m =-(!), n=-(f) ?
28. r/ie s?<m of two opposite numbers which are equal
arithmetically is zero. Or stated in symbols,
^a-\--a = 0. (1)
Proof. Since ~a and +a are opposite numbers equal in
arithmetic value, they annul each other when added (§ 22).
E.g.^ -2 + +2 = 0, +5 + -5 = 0, -7 + +7 = 0, +8 + -8 = 0.
Ex. 1. Add -5 to +8.
When -5 is added to +8, the 5 negative units in -5 annul 5 of the
8 positive units in +8. There remain 8 — 6 positive units ; that is,
+8 + -5 = +(8 -5) = +3.
Ex. 2. Add -9 to +4.
AVhen -9 is added to +4, 4 of the 9 negative units in -9 annul the
4 positive units in +4. There remain 9 — 4 negative units ; that is,
+4 4.-9 = -(9 -4) =-6.
These examples illustrate the following principle :
29. To add one number to another of an opposite quality,
find the difference of their arithmetic values and prefix to it the
quality-sign of the number which is arithmetically the greater.
Or, stated in symbols,
^a-\--b = ^{a- b), when a>b. (1)
+a + -b = -{b - a), when a < b. (2)
Proof. AVhen a>b and 'b is added to +«, the b nega-
tive units in "6 annul b of the a positive units in +a.
26 ELEMENTS OF ALGEBRA
There remain a — h positive units; hence, +(a — 6) is the
sum.
When a < &, a of the h negative units in ~h annul the a
positive units in +a. There remain h — a negative units j
hence, ~(h ~ a) is the sum.
Exercise 9,
1. To make the sum zero, what number must be added
to +3 ? To -7 ? To -31 ? To +14 ? To +a ? To 'b ?
2. Find the sum of +8 and "6. Of +5 and "7. Of "8
and +4. Of +11 and "15. Of "5 and +17.
3. Find the sum of "(i) and +(f). Of "(f) and +(^).
Of +(f) and -(A). Of +a) and -(,4).
What is the vahie of +a + -b,
4. When a = 43, & = 23 ? 6. When a = 23, 6 = 43 ?
5. When a = 63, & = 43 ? 7. When a = 43, 5 = 63?
8. Write six different sums each of which denotes zero.
What is the vahie of x-\- y,
9. When x= -J, y= +9?
10. When x = +14, y = -19?
11. When x = -(-J^), y = -^{i)?
12. When a. = +(i-i), 2/ = -(!)?
30. The sign of continuation is ... or ---, either of which
is read, ' and so on,' or ' and so on to.'
Thus, 1, 2, 3, 4, ..-, is read, '1, 2, 3, 4, and so on' indefi-
nitely ; 2, 4, 6, 8, ... 32, is read, ' 2, 4, 6, 8, and so on to 32.'
The sign .-. stands for hence or therefore.
The sign •.• stands for since or because.
POSITIVE AND NEGATIVE NUMBERS 27
31. The integers of arithmetic number make up the
series (1).
0 12 3 4 5 6 7 8 9-.. /j^a
1 1 1 i 1 1 1 1 1 \ ^ ^
Writing the positive and the negative integers in opposite
directions from zero, we obtain series (2).
... -4 -3 -2 -1 0 +1 +2 +3 +4 ... (2\
1 i 1 1 1 1 1 \ 1 ^ ^
If the divisions of the lines in (1) and (2) be taken as units of
length, then each number in (1) expresses simply its distance from the
zero point ; while each number in (2) expresses not only its distance,
but also its direction, from the zero point, distances to the right being
regarded as positive.
Note. Arithmetic numbers are not positive numbers. An arith-
metic number has no quality.
If to any number in series (2) we add +1, we obtain the
next right-hand number.
E.g., -4 + +l = -3, -2+n = -l,
and so on for the entire series.
Hence, if we say that a number is increased by adding to
it +1, the numbers in series (2) increase from left to right ;
that is,
..., -3<-2, -2<-l, -1<0, 0<+l, +K+2....
We have, therefore, the following properties of positive
and negative numbers :
(i) Any positive number is greater than zero; while any
negative number is less than zero.
(ii) Of two positive numbers the greater has the greater
arithmetic value; while of two negative numbers the greater
has the less arithmetic value.
E.g., +4> 0 by +4, -4 < 0 by +4, -7< 0 by +7,
+4 > +2 by +2, -4 < "2 by +2, -7 < -3 by +4.
28 ELEMENTS OF ALGEBRA
Note. If we agreed to say that a number was increased by adding
to it -1, tlien the numbers in series (2) would increase from right to
left ; positive numbers would be less than zero, and negative numbers
greater than zero. By common consent, however, it is agreed to say
as above that a number is increased by adding to it +1, the primary
unit.
Exercise 10.
Which is the greater, and how much the greater,
1. +3 or +7 ? 4. 0 or +1 ? 7. "7 or +3 ?
2. +2 or -8 ? 5.0 or "1 ? 8. +2 or "3 ?
3. -11 or +2? 6. -5 or "9? 9. '5 or "11?
10. When is the product of two arithmetic fractional
numbers greater than each number ? Less than each num-
ber ? Greater than one and less than the other ? Can the
product of two arithmetic integral numbers ever be less
than either number ?
11. When is the sum of tw^o algebraic numbers greater
than each number? Less than each number? Greater
than one and less than the other ? Is the sum of two arith-
metic numbers always greater than each number ?
12. Multiplying by an arithmetic fractional number in-
volves what two operations with arithmetic whole numbers ?
Addition of algebraic whole numbers involves the one or the
other of what two operations with arithmetic numbers ?
32. In proving and using identities, the following princi-
ples concerning identical expressions will be useful.
These principles clearly follow from the definition of
identical expressions in § 15 and the axioms in § 6.
(i) Any expression is identical with itself.
(ii) If eoAih of two expressions is identical with a third,
they are identical with each other.
POSITIVE AND NEGATIVE NUMBERS 29
(iii) If two identical expressions are added to or subtracted
from tivo other identical expressions, the residting expressions
are identical.
(iv) If two identical expressions are multiplied by two
other identical expressions, the products are identical.
(v) If two identical expressions are divided by tico other
identical expressions, not denoting zero, the quotients are
identical.
(vi) If, for any expression in an identity, an identical ex-
pression is substituted, the resulting equality is an identity.
33. The converse of an identity is obtained by interchang-
ing its members ; that is, the converse oi A = B is B= A.
li A = B, then, from definition, B = A.
Hence, the proof of an identity proves its converse.
^.<7., in proving +a ++& = + (a + &),
we prove +(a + &)=+a ++b.
CHAPTER III
ADDITION, SUBTRACTION, AND MULTIPLICATION OF
REAL NUMBERS
34. The positive and negative numbers defined in Chai^-
ter II are together often called real numbers.
In performing any operation with real numbers, we must
keep in mind that any such number is simply an arithmetic
multiple of the quality unit ^1 or ~1, and that arithmetic
numbers are added, subtracted, multiplied, or divided in
Algebra just the same as in Arithmetic.
35. Addition. Observe that, by § § 27 and 29, the addition
of one real number to another is reduced to the additiortr of
one arithmetic number to another, or to the subtraction
of one arithmetic number from another.
To find the sum of three or more numbers we add the
second to the first, to this sum we add the third, and so on.
Ex. 1. +8 + - 5 + +6 + -7 = +3 + +6 + -7
= +9 + -7 = +2.
Ex. 2. -7 + +5 + -3 + +9 = -2 + -3 + +9
= -5 + +9 = +4.
36. The two following laws of addition are constantly
used in Arithmetic and Algebra :
The commutative law. The sum of tivo or more numbers
is the same in whatever order they are added.
That is, a -{-b + c = b -\-c -\-a =c -}- b -\-a= "'. (A)
30
ADDITION OF REAL NUMBERS 31
Thus, we can commute siimmands {change their order) to
suit our convenience or purpose.
E.g.^ in Arithmetic we write
^ + 3 + ^ + 2 + ^ = | + i + i + 3 + 2. (3)
Here by a change of order we can add the fractions first.
Prove each of the two following particular cases of {A) :
Ex.1. +4+-5++0=+4++6+-5. §14
Ex. 2. +2 +-3 ++4 =+2 ++4 +-3 =+4 +-3 ++2.
Proof of law (A). This law holds true for arithmetic
numbers, as is learned in Arithmetic; hence the total
number of positive units in a, h, c, etc., will be the same
in whatever order these summands are written. For the
same reason the total number of negative units in a, 6, c,
etc., will be the same in whatever order their summands
are written.
Hence the sum will be the same, however we change the
order of the summands; for equal numbers of opposite
units always annul each other.
The associative law. The sum of three or more numbers
is the same in whatever way the successive numbers are
grouped.
That is, a + 6 + c = a+(6+c). {B)
Thus we can associate successive summands (group them)
to suit our convenience or purpose.
Prove each of the two following particular cases of ( JB) :
Ex. 1. +4 +-6 ++6 =+4 + (-5 ++6).
Ex.2. -5 ++4 +-7 =-5 + (+4 +-7).
Proof of law (B). a -{- b +c = b -\- c + a by(^)
= (6 + c) + a by notation
= a-h(b-{-c) by (A)
A similar proof would apply to any other case.
32 ELEMENTS OF ALGEBRA
The rules for addition in Arithmetic are based on the commutative
and associative laws just given.
E.g., to add 45 and 23, we have
45 + 23 = 40 + 5 + 20 + 3 by notation
= 40 + 20 + 5 + 3 ^J (A)
= (40 + 20) + (5 + 3) by (^)
= 60 + 8 = 68.
Writing one number under the other and then grouping the vertical
columns, as we do in Arithmetic, is but a convenient way of applying
laws (A) and (5).
37. Since, by the laws of addition in § 36, we can change
the order of summands and group them to suit our purpose,
we have the following rule for adding three or more num-
bers, some of which are positive and some negative :
Add all the numbers of one quality, then add all the numbers
of the opposite quality, then add the two resulting sums. •
Ex. -5 ++9 +-11 ++6 =-5 +-11 ++9 ++6 by (^)
= -16++15=-l by (J?)
' In practice, the rearrangement and regrouping of the summands
should be done mentally and simultaneously.
Exercise 11.
By § 37 find the value of each of the following sums:
1. +19 +-7 ++5. 5. +4 +-5 ++6 +-8 ++7.
2. -12 ++9 +-4. 6. -9 ++6 +-11 ++12 +-4.
3. -22 ++5 ++7. 7. +15 +-9 ++7 +-8 ++11.
4. +42 +-9 +-3. 8. -(|)++(2)+-(|)++(|).
Find the value of x -\- y -\- z -{- v.
9. When a; ="25, 2/ ="^32, 2 =-45, v =+28.
10. When a; =+94, 2/ ="75, ^=+82, v =-Qb.
SUBTRACTION OF REAL NUMBERS 33
38. From the definition of zero it follows that
a + 0 = a.
That is, any number plas zero equals the number itself.
E.g., 7+0 = 7, 8 + 0 = 8.
Also, 9+(2-2)=9, 6+(5-5) = 6.
39. Subtraction is the inverse of addition. Given a sum
and one of its two parts, subtraction is the o^^eration of
finding the other part.
As in Arithmetic, the given sum is called the minuend,
the given part the subtrahend, and the required part the
remainder.
Hence, to subtract any subtrahend from any minuend is
to find a third number, the remainder, which added to the
subtrahend gives the minuend.
Ex. +9 = +9 + (+5 + -5) §§ 28, 38
= (+9+ +5) + -6. §36
Hence,
+9 ++5 is the number which must be added to -5 to obtain +9 ;
that is, +9 --5 = +9 ++5.
Here the remainder +9 ++5 is obtamed by adding to the minuend
+9, the subtrahend -5 with its quality changed.
This example illustrates the following rule :
40. To subtract one real number from another, add to the
minuend the subtrahend with its quality changed from + to ~,
or from ~ to '^.
That is, Jtf - +a = 1/ -f "a, (1)
M--a = M+^a, (2)
when M is any real number.
34 ELEMENTS OF ALGEBEA
Proof. If to the second member of (1) we add the sub-
trahend, '^a, we obtain the minuend M; that is,
(M-^-a)-^+a = M+(-a-^+a) by (5)
= if §§ 28, 38
also, (M-^+a) +-a = M+ (+a +"a) = M.
Hence, by § 39, the second member of (1) or (2) is a
remainder.
Ex.1. -4-+7 =-4+-7 =-11.
Ex.2. -5 --8 =-5 ++8 =+3.
Thus, subtracting any real number gives the same result
as adding its arithmetically equal opposite number.
E.g., subtracting §200 credit from an estate is equivalent to adding
$200 debt; and subtracting $300 debt is equivalent to adding .^300
credit.
Subtracting $100 income is equivalent to adding $100 expenditure.
Exercise 12.
Perform each of the following indicated subtractions :
1. +19 -+7. 4. +6-+7. 7. -20 --25.
2. -23 -+12. 5. +12 -+20. 8. "68 --98.
3. -16 --30. 6. -214 -+25. 9. -118--120.
What is the value of a — b,
10. When a =+5, 6 =+4? 12. When a ="4, 5 ="7?
11. When a =+7, 6 =+9? 13. When a ="14, 5 =-11?
14. From +4 +-8 ++9 +-3 subtract +7 +"2 ++9 +-8.
15. From -10 +-7 ++15 +"3 subtract +7 +"11 +"17.
41. When a monomial or the first term of a polynomial
is preceded by the sign of operation + or — , zero is to be
understood before this sign of operation.
SUBTRACTION OF REAL NUMBERS 35
Thus, — +a = 0 — +a = ~a, —~a = 0 — ~a= '^a.
Again, _ +5 + +7 = 0 - +5 + ^7 = "5 + +7.
42. Successive subtractions or successive additions and sub-
tractions can be performed from left to right, one at a time
in succession.
Ex.1. +8--3-+2--6=+ll-+2--6
=+9 --6 =+15.
We can, however, express each term to be subtracted, as
a term to be added, and then apply the principle in § 37, for
finding the sum of three or more numbers.
Ex.2. +8 --3 -+2 --6 =+8 ++3 +-2 ++6 (1)
=+17 +-2 =+15.
43. Commutative law of subtraction. Since each term to
be subtracted can be expressed as a term to be added, the
commutative law holds for subtraction as well as for addi-
tion, provided the sign of operation -[- or — before each term
is transferred with the term itself.
E.g., +7_-8+-9-+4=--8++7-+4+-9
= _+4_-8+-9++7.
Exercise 13.
Find the value of each of the following expressions :
1. +6-f-2-+3. 2. -14-+9-f"4. 3. +32 +"5 -"16.
4. +6--24-+3. 5. +4--2-+3+-2-+5++3--6.
6. +25 -+14 +-10 ++14 --5 -+18 ++16 +"18.
7. -35 +-5 --32 ++24 --14 +-28 --8.
44. From the definition of zero it follows that
a-0 = a.
That is, any number minus zero equals the number itself
36 ELEMENTS OF ALGEBRA
45. Multiplication. As in Arithmetic, the number multi-
plied is called the multiplicand, the number which multiplies
is called the multiplier, and the result the product.
Ill Arithmetic the product 9 x 3 is obtained by taking the multipli-
cand 9 three times, and the multiplier 3 is obtained by taking the pri-
mary unit 1 three times.
The product 9 x |- is obtained by dividing the multiplicand 9 by
3, and multiplying the result by 2, and the multiplier | is obtained by
dividing the primary unit 1 by 3, and multiplying the result by 2.
In each case we obtain the product by doing to the multiplicand
just what is done to the primary unit to obtain the multiplier.
Hence, we define multiplication as follows :
To multiply one number by another is to do to the multi-
plicand just what is done to the primary unit to obtain the
multiplier.
The multiplicand and the multiplier together are called
the factors of the product.
46. Multiplier any arithmetic number. Let a and b be any
two arithmetic numbers ; then b times a units of any Mnd is
equal to ab units of that Mnd; that is,
^axb = +{ab), (1)
and -a xb = -(ab). (2)
E.g., +4x5 = +20, -7x4 = -28, -(f) x 8 = -12.
47. Multiplier any positive or any negative number.
To obtain ^b from the primary unit +1 we take that unit
b times ; hence, by definition, to multiply any number by +6
we take that number b times;
that is, +a X +6 = +a X 6 = +(ab), (X)
and -a X +6 = -a X 5 = ~(ab). (2)
Hence, to multiply any number by +1 is to take that
number once ; that is, "*"a x "^1 = +a ; ~a x ■*"! = "a.
MULTIPLICATION OF REAL NUMBERS 37
To obtain ~b from the primary unit +1, we change the
quality of that unit and multiply the result by 6 ; hence, to
multiply any number by ~b, we change the quality of that
number and multiply the result by b ;
that is, +a X ~6 = ~a X 6 = ~{ab)j (3)
and ~a X "6 = +a X 6 = ^(a6). (4)
Hence, to multiply any number by ~1 is to change the
quality of that number; that is, "^a x ~1 = ~a; ~a x ~1 = ^a.
From identities (1) and (4) we have the law,
Tlie product of two real numbers like in quality is positive.
From identities (2) and (3) we have the law,
The product of two numbers opposite in quality is negative.
These two laws together are called the law of quality of
products.
From identities (1), (2), (3), (4), it follows that
The arithmetic value of the product of two real numbers is
the product of their arithmetic values.
E.g., +5 X +7 = +35, -6 x "8 = +48.
+4 X -9 = -36, -7 X +8 = -56.
Exercise 14.
Find the value of each of the following numeral ex-
pressions :
1. +2 X +4. 3. +9 X -8. 5. -21 X +3. 7. +22 x "6.
2. -2 X -7. 4. -11 X -1. 6. +31 X -1. 8. -32 x "4.
9. +10x-3 + -8x+2. 11. -6x-5+-^8x-4-+12x-5.
10. +14x-2 + -6x-5. 12. -9x+2 + +16x-4--14x+3.
88 ELEMENTS OF ALGEBRA
When a = +2, 6 = "4, m = "3, n = +9, x = +6, find the
value of each of the following literal expressions :
13. ab -f- mx. 16. ax — nb. 19. (a + b) (n -\- m).
14. ax + bm. 17. (a — b) x. 20. (a — b)(n — m).
15. am— bx, 18. (m — n)b. 21. (6 —a?) (771— n).
48. Continued products. By § 47, we obtain
+a X "^6 X +c = +(a?>) x +c = -^(abc).
+a X'^b x~c = +{ab) x~c = -(abc).
-^a x~b x~c = ~(ab) x~c = +(abc).
~a x~b x~c = +(ab) x^c = ~(abc).
From these and similar identities we have the following
laws which are more general than those in § 47 :
Aj^roduct which coyitains an odd number of negative factors
is negative; any other product is positive.
The arithmetic value of a product is the product of the arith-
metic values of its factors.
Ex. Find the value of +3 x "2 x +4 x -6 x -5.
The product is negative, since there is an odd number, 3, of nega-
tive factors ; its arithmetic value is 3 x 2 x 4 x 6 x 5, or 720.
Hence, +3 x -2 x +4 x "6 x - 5 = "720.
Exercise 15.
When a = -2, & = +4, c=-6, x = -S, y = ~5, find the
value of each of the following literal expressions:
1. abc.
4.
(6 + a) ex.
7. axy — be.
2. abxy.
5.
(x — y) abc.
8. xy — abc.
3. abcxy.
6.
(b 4- c) axy.
9. x-\- obey.
MULTIPLICATION OF MEAL NUMBERS 39
10. Prove +lx-lx-lx-l = -l; -lx-lx-lx-l = +l.
11. Prove +a X "6 X ~c = (+1 x "1 X ~1) (ahc).
12. Prove "a x "^6 X ~c x ~x = {-l x +1 X ~1 X ~1) (obex).
Examples 11 and 12 illustrate that the product of two or more num-
bers is equal to the product of their quality-units multiplied by the
product of their arithmetic values.
49. The two following laws of multiplication are con-
stantly used in Arithmetic and Algebra :
The commutative law. Tlie product of two or more num-
bers is the same in vjhatever order the factors are multiplied.
That is, abc = acb = cba="'. (A')
Prove each of the two following particular cases of (A') :
Ex. 1. +2 X -3 X +4 X -5 = -5 X -3 X +2 X +4.
Ex. 2. -3 X +7 X -2 X -1 = -2 X +7 X -1 X -3.
Proof In Arithmetic we have learned that this law
holds true for arithmetic numbers. Hence, by § 48, the
arithmetic value of a product of real numbers is the same
in whatever order the factors are multiplied.
From the law of quality, in § 48, it follows that the quality
of a product of real numbers will be the same in whatever
order the factors are multiplied.
Hence, a change of order of factors affects neither the
arithmetic value nor the quality of their product.
The associative law. Tlie product of three or more num-
bers is the same in whatever way the successive factors are
grouped.
That is, abc = a(bc). (B)
Prove each of the following particular cases of {B') :
Ex. 1. -3 X +4 X -2 = -3 X (+4 x "2).
Ex. 2. +5 X -0 x -1 X +2 = +5 X ("6 x -1 X +2).
40 ELEMENTS OF ALGEBUA
Proof, abc = hca by (A')
=\bc)a by notation
= a(bc) by (A')
Exercise 16.
By using the commutative and associative laws, find in
the simplest way the value of each of the following ex-
pressions :
1. +33 X -21 X -4. 4. +144 x "3 x -16|.
2. -123 X -33^ X +3. 5. "371 x "7 x "4.
3. +142 X -121 X -8. 6. -333i X -5 X -7 X +3.
50. A x>ii'oduct of two or more factors is multiplied by a
number if any one of the factors is multiplied by that number.
Proof (ab) x c = (ac)b = a(bc). § 49
51. Powers. A product of two or more equal factors is
called a power. Any number also is often called the first
power of itself.
E.g.j the product aa is called the second power of a.
The product bbb is called the third power of b.
aa is written a^ ; aaa is written a^ ;
ooa ••• to n factors, written a% is read 'the nth. power of a.'
a^ is often read ^the square of a/ and a^ 'the cube of a.'
In a*", a is called a base. Thus in 3^*, 3 is a base ; in a^,
a; is a base ; in y'", ?/ is a base.
52. A positive integral exponent is a whole number which
(written to the right and a little above a base) indicates
how many times the base is used as a factor, as 3 in a^,
or n in a".
MULTIPLICATION OF REAL NUMBERS 41
To avoid ambiguity, a base which is not denoted by a
single symbol must be enclosed within parentheses :
E.g., (-3)2 = -3 X -3 = +9, while -3^ = -1 (3 x 3) = -9.
Again, (4 x 5)2 = (4 x 5) (4 x 6) = 20 x 20 = 400,
while 4 X 52 = 4 X (5 X 5) = 4 X 25 = 100.
The meaning of fractional and negative exponents will be deter-
mined in a later chapter.
A power is said to be odd or even according as its expo-
nent is odd or even.
53. Quality of a power. An odd power of a negative base
is the only power which involves an odd number of negative
factors ; hence, by the law of quality in § 48 it follows that
An odd power of a negative base is negative, and an even
power positive ; any power of a positive base is positive.
E.g., any power of +1 is +1 ; any even power of -1 is +1 ; any odd
power of -1 is '1.
Exercise 17.
1. What number is the base and what the exponent in
-3^? In {Sy? In 3xy"? In (Sxyy? In (a + 6)"?
In a + b^? In (x")* ?
Find the value of each of the following expressions :
2. -3\ 3. (-3)*. 4. 17-32. 5. (17-3)2.
Express each of the following products by a base and
exponent :
6. (xy) (xy) (xy) ••• to 8 factors.
7. (a + 6) (a + &)(« + &)••• to 12 factors.
Express in symbols :
8. The sum of the cubes of x and y.
9. The cube of the sum of x and y.
42 ELEMENTS OF ALGEBRA
10. The sum of the squares of a, b, and c.
11. The square of the sum of a, b, and c.
54. If, in any one of the identities in § 48, the quality of
one factor is changed, the quality of the product is changed,
but its arithmetic value remains the same.
This illustrates the following principle :
TJie quality of any product is changed by changing the qual-
ity of one, or of any odd 7iumber, of its factors.
Proof By changing the quality of an odd number of
factors, the number of negative factors in the product is
changed from odd to even, or from even to odd ; hence, by
§ 48, the quality of the product is changed.
Note. When for brevity we speak of the quality of an expression,
we mean, of course, the quality of the number which the expression
denotes.
55. The quality of an expression is changed by changing the
quality of each of its terms.
Proof Changing the quality of a term does not affect its
arithmetic value. Hence, changing the quality of each term
of an expression will simply change a positive sum into an
arithmetically equal negative sum, or vice versa.
This principle is illustrated by the fact that if in a business account
we change debts into credits, and credits into debts, the balance will
not be changed in amount, but it will be changed from credits to debts,
or from debts to credits.
Ex. 1. Change in four ways the quality of -4 x +3 x -2. Of
-3x-6x-7. Of +ax+bx-c. Of +ax-bx-c. Of +xx+yx-2.
Ex. 2. Change in two ways the quality of -4 x +3 - -2 x +7.
Of +a X -b + -c X +x. Of -a X -X — +6 X +c.
MULTIPLICATION OF BEAL NUMBERS 43
56. Two uses of the signs + and -. Hereafter the larger
signs + and — will be used, not only as signs of operation,
but also with numerals as signs of quality.
To avoid ambiguity, parentheses will be used when needed.
Thus, in the expression,
(+4)-(+7) + (-3)-(-4),
each sign within parentheses denotes quality, and each without
denotes an operation.
Again, (- 3)ax -(+ 4)6y + (- 5)c2; = -3 ax - +4 6y +-5c5r.
A letter with the small sign + or " will continue to be
used to denote a general positive or a general negative
number.
57. Abbreviated notation. The sign — is never omitted.
But, for the sake of brevity, the sign -f has been omitted,
and is to be understood in the two following cases :
(i) When no sign is written before a monomial or before
the first term of a polynomial, the sign -f is to be under-
stood.
(ii) When only one sign is written between two successive
terms of a polynomial, the sign + is to be understood either
as a sign of operation or as a sign of quality.
E.g.^ 2 denotes + 2, 3 a denotes + 3 a, and a denotes + 1 a.
Again, 6 — 5 denotes the difference (+6) — (+5) or the sum
(+ 6) + (— 5) ; in each case the sign + is understood between 6 and 5 ;
in the first case as a sign of quality^ and in the second case as a sign
of operation.
Since (+6) - (+5) = (+6) + (-5),
6 — 5 denotes the same number whether it is regarded as ex-
pressing the difference (+6) — (+5) or the sum (+6) + (—5).
Again,
7-5-f8 = (4-7)-(+5)4-(-l-8), or (+7) + (-5) + (+8),
44 ELEMENTS OF ALGEBRA
according as we regard the written signs in the first expres-
sion as signs of operation or as signs of quality.
Hence, in the abridged notation, the written signs in any-
polynomial can be regarded either as signs of operation or
as signs of quality.
When all the written signs are regarded as signs of qual-
ity any polynomial becomes a sum.
E.g., -_5 + 3-2=(-5) + (+3) + (-2).
or the sum of the terms — 5, +3, and — 2.
Again, 7ac-4x + 3y = -i-7ac+(-4)x + (+ 3)?/,
or the sum of the terms + 7 ac, — 4 x, and + 3 y.
In general formulas, such as {A), (B), etc., it is usuall}-
better to regard the written signs as signs of operation ; but
in most other cases it is preferable to regard the written
signs as signs of quality and, therefore, to regard every
polynomial as a sum.
58. Coefficients. If a term is resolved into two factors,
either factor is called the coefficient, or the co-factor, of the
other.
E.g.^ in 4 a&c, + 4 is the coefficient of a&c, + 4 a of &c, + 4 a& of c,
dbc of + 4, and 6a of + 4 c.
A numeral coefficient is a coefficient expressed entirely by
numerals, and a sign of quality written or understood.
A literal coefficient is a coefficient which involves one or
more letters.
E.g.^ in —4xy, —4 is the numeral coefficient of xy ; x is the
literal coefficient of — 4 ?/, y of — 4tx, and —4x of y.
When in a term no numeral factor is written, 1 is understood, e.g.,
a denotes -h 1 • a and — a denotes — 1 • a ; abc denotes + 1 • abc
and — abc denotes — 1 • abc.
MULTIPLICATION OF HEAL NUMBERS 45
Exercise 18.
Find the value of each of the following expressions :
1. 15-9. 5. (-Il)x7.
2. -9+7. . 6. (-7) -(-4).
3.-8-6. 7. 9-74-4-3 + 5.
4. (-3) (-4) 8. 18-(-3)x(-4)-8.
9. 35_j.(_7)x6-r-15x(-2).
Find the value of a-[-h — c-\-d and a—{—h + c — d).
10. When a = 2, 6 = — 4, cz=^—Q>, d= —7.
11. When a= —7, b = — S, c = 5, d = — 6.
Find the value of x(y — v-\- z).
12. When x = 6, y = — 7, v = — 9, 2 = 8.
13. When a; = — 5, y = l^j v = — 4, 2 = — 7.
Find the value of x-T-{y — v — z).
14. When x = -10, y = -Q, v = -9, z = S.
15. When a; = -16, 2/ = - 10, ^ = -12, z = 6.
16. What is the coefficient of a in a? In —a? In
-7ay?
17. In the expression —Sab(x — y\ what is the coeffi-
cient of a; - 2/ ? Oib(x-y)? Of 8a? Oi -S(x-y)?
18. If the sum (x — y) -\- (x — y) -{- (x — y) -\ to a sum-
raands is expressed as a product, what is the coefficient of
x-y?
59. Having given a product and one factor, division is the
operation of finding the other factor. That is, if n is one
factor of m, m -^ n denotes the other factor ; whence
(m-i-n) X n= m. (1)
46 ELEMENTS OF ALGEBRA
60. The distributive law. The product of a polynomial by
a monomial is equal to the sum of the products obtained by
multiplying ea,ch term of the polynomial by the monomial;
and conversely.
That is, {a-{-b-\-c-{-"')x = ax-{-bx + cx+"' (0)
The distributive law lies at the basis of multiplication in Arithmetic,
e.g.^ if we wish to multiply any number as 248 by 7, we separate 248
into the parts 200, 40, and 8, multiply each of these parts by 7 and
add the results.
Thus, 248 X 7 = (200 + 40 + 8) x 7 (1)
= 200 X 7 + 40 X 7 + 8 X 7 . (2)
= 1400 + 280 + 56 = 1736.
We pass from (1) to (2) by the distributive law (C).
Prove each of the following particular cases of (O):
Ex. 1. (4-3 + 5). (-2) = 4(-2) + (-3).(-2)4-5(-2).
Ex. 2. (- 4 + 2 - 6)(- 3) = (- 4).(- 3)+ 2 (- 3) + (- 6).(- 3).
Ex. 3. (« + 6 + c)-3 = 3a + 36 + 3c.
Proof Let the multiplicand be any binomial a-\-b.
The proof involves three cases : when the multiplier is
(i) a positive integer, (ii) a positive fractional number,
(iii) a negative number.
(i) Let m be any positive whole number ; then
(a -h 6) m = (a + 6) -h (a + 6) H — to m summands § 47
= (a + a H — to m summands)
+ (6 + 6 H — to m summands) § 36
= am-\-bm. (1)
(ii) Let m and n be any positive whole numbers other
than zero; then — will denote any positive fractional
number.
MULTIPLICATION OF REAL NUMBERS 47
(a + b) (m ^ n)7i = (a + 6)m §§ 49, 59
= am -\- hm. by (1)
= a{m--rn)n-\-h(m^n)n §§ 49, 59
= [a {m -T-n)-\-b (711 h- n)'] n. by (1)
Dividing the first and last expressions by n, by (v) of
§ 32 we obtain
(a + b) (m -J- n) = a(m-i- n) -\- b (m h- n). (2)
Let r be any positive number, whole or fractional ; then,
from (1) and (2) we have
(a -\-b)r = ar + br. (3)
(iii) If the quality of equal numbers is changed from -f
to — , or from — to +, the resulting numbers will be equal.
Henee, changing the quality of both members of (3) we
have
(a + 6) (- r) = a (- r) + 6 (- r), §§ 54, 55
where — r is any negative number, whole or fractional.
A similar proof would apply to any polynomial as well
as to a + 6 ; hence the law as stated in (C).
Ex. 1. Multiply 3 a2 - 5 a + 3 6 by 2 X.
(3 a2 - 5 a + 3 6)(2 a;) = (3 a2)(2 a;) + (- 5 a)(2 x) + (S 6)(2 x)
= 6a2x- 10ax + 6bx.
Observe that in applying ( C) we regard a polynomial as a sum.
Ex. 2. Multiply 2 x^ - 3 x2 - 2 x by -3 a.
(2x3-3x2-2x)(-3a)=(2x8)C-3a) + (-3x2)(-3a) + (-2x)(-3a)
= - 6 ax3 4- 9 ax2 + 6 ax.
48 ELEMENTS OF ALGEBRA
Exercise 19.
Multiply :
1. a; + 2 by 3. 6. 2 cy -Ax hj -a.
2. 6a-75by-2. 7. 2 a- 3 & -c by - 2 a;.
3. 2 a; - 6 by - 5. 8. - 3 a; + 2 ?/ - 5 2; by 3 a.
4. 2x-5by-3a. 9. a;^- 3 a; + 4 by -2 a.
5. ax-3b hj -2c. 10. ar'-2 ?/-3 2; by - 5 a.
11. -2x'-hSxy-4:i/-x-{-2y-7 hj -3a,
CHAPTER IV
ADDITION AND SUBTRACTION OF INTEGRAL
LITERAL EXPRESSIONS
61. An integral literal expression is an expression which
involves only additions, subtractions, multiplications, and
positive integral powers of its letters.
Any expression which contains a literal divisor is called
a fractional literal expression.
E.g., a'^ + f and 4 x^ — | 6< are integral literal expressions ; while
- and are fractional literal expressions.
y 4-6
A letter can, in general, denote any integral or fractional
number ; hence, any literal expression can have any integral
or fractional value.
E.g., wlien x = I and y = |, the integral literal expression
x + y = ^+^ = f, a fractional number.
Also, when a* = 2 and y = 3, the integral expression | xy = ^.
Again, when x = 10 and y = 2, the fractional expression ? = 5.
y
Tlie pupil must clearly distinguish between integral and fractional
expressions and integral and fractional numbers.
62. Like or similar terms are terms which do not differ,
or which differ only in their coefficients.
E.g., 4 ab and 4 ab are like terms ; so also are 4 ab and — 10 ab.
Again, 6 axy and — 4 bxy are similar terms, if we regard 6 a and
— 4 &, respectively, as the coefficients of xy in the two terms ; but if
6 and — 4 be taken as the coefficients, these terms are dissimilar.
49
60 ELEMENTS OF ALGEBRA
63. Sum of similar terms. The converse of the distribu-
tive law in § 60 is
ax-\-hx-\-cx-\- '•• = (a-\-'b -{- C + '^•)x. (O)
That is, the sum of two or more similar terms is equal to
the sum of their coefficients into their co7nmon factor.
1. Find the sum of 7 a, — 5 a, 4 a.
(+ 7)a + (- 5)rt + (+4)a = (7 -5 + 4)a^6a.
2. Find the sum of 3 ah"^, - 5 ah"-, - 8 al)^.
(+ 3) a6--2 + (- 5) a62 + (- 8) a62= (3 - 5 - 8) ah'^= - 10 db^.
3. Find the sum of 7 (a — 6) , — 5 (a - 6), 4 (a - &).
( + 7)(a-6) + (-5)(a-5) + (+4)(a-&)=(7-5+4)(a-6) = 6('?'-6).
64. By § 57 the sum of two or more terms is indicated by
writing them in succession, each term being preceded by
the sign of quality of its numeral coefficient.
The sum of unlike terms can only be indicated.
E.g., the sum of — 5 c, 7 a, and — 9 6 is
-5c + 7a-9&, or 7rt-5c-9?).
Again, the sum of — 3 ax, — 5 by, and 6 cz is
— ^ax — ^hy-^Qcz, or 6 C0 — 3 ax — 5 by.
Exercise 20.
Find the sum of :
1. 2 a, -3 a, 5 a. 6. 4 aft^, -1 ab\ 3ab\
2. —4:X,2x,—x. 7. —Sx% 5x% —4 a;".
3. ab, -2 ah, 3 ah. 8. 2 oc^, - 5 ac^, - 8 acl
4. 2 a*, -3a^ 1 a\ 9. -5aV, -3aV, 9 aV.
5. a;", —2 a;'*, 4 a;". 10. 4&"2/'"^ — Td^y™, 9 5"2/'"'
ADDITION AND SUBTRACTION 51
11. 7aa^, — 5aic^, 4aa^, —9aa^, — 14aaj^, 25 aa?^.
12 9aa^, — aar', 4aic^, —Ihx^, —14 car'.
13. -3^2/2!, f a^2;, -^xyz, 6xyz, -^xyz, -^^-xyz.
14. (x-af, -2(x-a)\ 4.{x-af, -5{x-a)\ 12{x-aY.
15. (a^+2/2), -5(a^+/), 9(a:^+yO, -3(a^4-2/^, -l{^+fr
16. (aT^-2/'), -4(ar^-.v«), _3(a:«-r), -7(a^-2/«), 8(a^-2/3),
Simplify each of the following expressions by combining
like terms :
17. a^_7a^+4a^-5a^. 19. o?y--^a?f+4.a?f-l^y\
18. a;"— Sx^+Saj"— 7 a;". 20. aar'— 7a.-2+6a:2_5^
21. _9ic2_^i7^_^3a.o_^^_^^^2_5^2
22. Sab' -7 aV + 8 a/>- - 4 ah'' -\- 7 ca^ - 11 ca^.
23. -12a^-^4:a^-9x^-\-7a^ + Sa^-9a'' + 7a\
24. 7 aftcc? — 11 abed + 41 a6cd + 7xy — 20xy.
25. _5aj2_2a:2 + |a^ + 8/--|/-f/.
26. 7ar' + 2a2-5.'c2-3a2.
7 a:2 + 2 a2 - 5x2 - 3 a2 = 7 a;2 _ 5x2 + 2 a2 - 3 a2 by (^)
= 2x2-a2.
27. 7a6 — 5a;2/ + 3a6 + 2a^ — 6a5 — a^.
28. -9ax'-\-5bf-\-7ax^-3bf+llaa^ + 4:bf.
29. — 7 c/ — 4 a6 + 9 a.-2; 4- 11 C2/2 + 10 a6 — 5 a;2; — a6.
30. 2(a^-l) + 3(a2 + l)-4(x2-l)-5(a2 + l).
31. 3(a' + b')-4.(x-\-y)-7(a'-\-b')-^5(x-\-y),
32. Review this exercise, solving each example mentally.
52 ELEMENTS OF ALGEBRA
65. Addition of polynomials.
Ex. 1. Add - 3 x^ + 7 X to 5 a;2 _ 4 a;.
(5 a:2 - 4 x) + (- 3 x2 + 7 x)= 5 x2 - 4 x - 3 x- + 7 x by converse of {B)
= 2x2 + 3x. by (^), (S)
Ex. 2. Find the sum of
4 x2 - 3 xy + y2, _ 2 x2 - 5 xy - 6 2/2, and 2 x?/ - x2 - 3 62.
In adding polynomials, it is convenient to write them under each
other, placing like terms in the same column.
Thus, (4 x2 - 3 xy + 2/2) + ( _ 2 x2 - 5 x?/ - 6 2/2) + (2 ccy - x2 - 3 h'^)
can be written 4 x2 — 3 xy + 2/^
- 2 x2 - 5 xi/ - 6 2/2
- x2 + 2x2/ -3 62
x2 - 6 X2/ - 5 2/2 - 3 62.
Here the rows of terras are the groups of terms as given, while the
columns of terms are the groups of similar terms obtained by rearrang-
ing and regrouping by laws {A) and {B).
Since there is no carrying as in Arithmetic, the addition can be
performed from left to right, or from right to left.
66. When in a polynomial the exponents of some one
letter increase or decrease, from term to term, the polyno-
mial is said to be arranged in ascending^ or in descending^
powers of that letter.
This letter is called the letter of arrangement.
E.g., the polynomial x^ + 2 x^y + 3 xy'^ + 4 2/^ is arranged
in descending powers of x, x being the letter of arrangement ;
or, in ascending powers of 2/, y being the letter of arrangement.
In arranging a polynomial in ascending or descending
powers of any letter, we must first combine all the terms
which contain the same power of that letter.
In adding polynomials, it is usually convenient to arrange
them in ascending, or descending, powers of some letter,
as below:
ADDITION AND SUBTRACTION 63
Ex. 1. Find the sum of 2 x^ - 3 x^ + y, _ 4,
7 X - 4 x2 4- 5 x3 + 5, and 7 x^ - 4 x^ + 2 x - 1.
Arranging each polynomial in descending powers of x, we have
-3x3 + 2x2+ x-4
6x3-4x2+ 7x + 5
-4x3 + 7x2+ 2x-l
- 2 x3 + 5 x2 + 10 X
Exercise 21.
Find the sum of :
1. a + 26-3c, -3a + 6 + 2c, 2a-3& + c.
2. -Zx-\-2y-\-z, x-^y-\-2z, 2x + y-Sz.
3. -15a-196-18c, 14a4-1564-8c, o + 56 + 9c.
4. 5 aa; — 7 6?/ + C2;, aa; + 2 61/ — 02;, — 3ax + 2&2/ + 3c2;.
5. 20 p + q-r, p-20q + r, 2^ -\-q- 20 r.
6. — 5 a& + 6 6c — 7 ac, 8 a6 — 4 6c + 3 ac, — 2 06 — 2 6c
4- 4 ac.
7. pq + qr — pTf — pq -\- qr -{- pr^ pq — qr -{- pr.
8. 2 a6 4- 3 ac 4- 6 a6c, — 5 a6 + 2 6c — 5 a6c, 3 a6 — 2 6c
— 3ac.
9. x^ + xy — y-, —z^-\-yz-\- /, xz -\- z^ — 0^.
10. 5a»-3c» + d3, fo3_2a3 4-3^3, 4c3 -2a'»- 3d».
11. a,^ + 2/2_2ic?/, 2z2_3/-42/«, 2 ar^ - 2 2:2 - 3 ii'2;.
12. a.'3 + 3ar^2/ + 3a^', - 3 ar^3/ - 6 a^/ - a^, 3 a;^^, + 4 a^.
13. x'-^x'^y-hx'f, 3a;^+2a.V-6an/*, Zs^f+Qxy'-f.
14. a«-4a26 + 6a6c, a-6 - 10 a6c + c^, 63 + 3a26 + a6c.
15. 3a2-1062 + 5c2-76c, - a2 + 462- IOC24- 3a6,
c2+116c+8ac-2a6, 4c2-46c + ac, -2a2+662-9ac-6c.
54 ELEMENTS OF ALGEBRA
16. 4ic* + 12ar^-a;-10, llx" -2x^ - x' -\-'d, ^x" -3x'
+ 4.X, 4ar^-x^-5, Q,x^ - si? -\-2x- -1.
17. i:^-\x + \, -^x' + lx-^, Ix^ + lx^^.
18. la' + ^ah-^h^ lo?-ah-\b'', - a^- f a6 + 2 6=^.
19. _2^_a;2/ + 2/^3x2-|a?y-i2/', - f a^ + 2x'?/- |/.
20. -f a^-|a;/ + 22/^, |aj22^-|_aj?/2_|_i.2^a |ar''-2a;2^-f ^Z'.
21. a^-3ax2_|_5^'j^_^3^2a^ + 4ax2-6a2a;, Gaa^-Sa^a?
+ a^, — 2ar^ + 4a^a; — 5al
22. 3a;2 4-/-3?/2-2^ 2 a;^/ - 3 2/^ + 3 2/^, -4.x^-2xy-{-
23. Given a; = 6 + 2c — 3a, y = c + 2a — Sh, and
2; = (x-f26 — 3c; show that x-\-y -\-z = ().
24. Given a = 5a; — 32/ — 2 2;, b = 5y — 3z — 2x, and
c = 5z — Sx--2yj show that a + 6 + c = 0.
67. To subtract one expression from another, change the
sign before each term of the subtrahend from + to ^ or from
— to +, and add the result to the minuend.
Proof. Changing the sign before each term of the sub-
trahend changes the quality of the subtrahend (§ 55) ; and
by § 40 the minuend plus the subtrahend with its quality
changed is equal to the remainder.
Ex. 1. From — 5 x'^y take 4 x'^y.
- 5 ic2?/ - ( + 4 a;2i/) = - 5 ic2y + ( _ 4 a^Zj/)
= - 9 x'^y.
Ex. 2. From 5 x- + a;?/ - m take 2 x^ + 8 xy - 7 y2.
Changing the sign before each term of the subtrahend from + to -
or from — to +, and adding the result to the minuend, we have
5 a;2 + xy — m
- 2 x2 _ 8 xy + 7 1/2
3 x2 — 7 xy — m + 7 ?/2^ Remainder.
ADDITION AND SUBTRACTION 55
Note. The signs of the subtrahend need not be actually changed ;
the operation of changing the signs ought usually to be performed
mentally, as in the following example.
Ex. 3. From 2x* -Sx^ + 7 x-S take x* -2x^-9x + ^,
2x*^ -Sx^-\- 7x- 8
x^-2x^ -9x+4
x* + 2 a:8 - 3 a;2 + 16 a; - 12
Exercise 22.
1. From 4 a — 3 6 + c subtract 2a — Sb — c.
2. From 15x -\-10y — 18z subtract 2x — Sy + z,
3. From — 10 be -\- ab — A cd take — 11 ab-\- 6 cd.
4. From ab -\- cd — ac — bd take ab -\- cd-^ ac + bd.
5. From m^ + Sn^ subtract —4:m^—6n^-{-71x.
6. 7xy-(-Sxy) = ?
7. -9x'y-(-\-6x'y)-(-20x'y) = ?
8. 32a^-(-122/^-(-hl4a^)-(+92r) + (-2/) = ?
9. 28 a'b' - (+ 17 a'6-) - (- 19 ar^y) - (+ 15 a^y)
-(-5a'b')=:?
From
10. -8a^f-\-15x^y + lSxf take 4:a^i/^-{-Txh/-Sxf.
11. a^bc + b^ca + c^a6 take 3 a^bc — 5 &-ca — 4 c^a6.
12. -7a264-8aZ>2 + cd take 5arb -7 ab' + 6cd.
13. 10a262-|.i5a?>2_^3^25 take - 10 a-^^ + 15 oft^ - 8 a^fe
14. hs^c^-2 abc take a^-{-b^-S abc.
15. 7abc-Sa^-\-5b^-(^ take a^ + 6^ + <?' - 3 «6c.
16. ix'-^xy-^ 2/2 take _ f a;- + ;^ - y-.
56 ELEMENTS OF ALGEBRA
17. f 3^ — faa; take \ — \x^ — ^ax.
18. ^o? — 2aQi? — \a^x take \ o?x -\- \ a^ — ^ ax^.
If ^ = a2_4^5_352^ J5=a6-462-3a^
find the expression for
19. A + B+C+D. 22. J._5-C-Z).
20. A + B-\-C-D. 23. -^-J3H-C+^.
21. A^B-C-D. 24. _^ + 5-6' + Z>.
In solving example 20, under the values of J., B, and (7 write that
of D with its quality changed, and then add the results.
25. From 5a^ + 3a; — 1 take the sum of 2x — 5-{-7 a^
and 3aj2-f4-2a^ + a;.
26. From the sum of 2 a^ - 3 a^ + a - 2 and 2 + 8 a^ - a^
subtract 3 a — 7 a^ + 5 al
27. From the sum of 4:a^ + 3x-7, 2af ~Sx + 2x^~l,
and — 5a^H-2a7 — ar^ + 9 take the sum of 2 aj^ — 11 ic and
9a^ + 5a^ + 3-2a;.
68. Removal of signs of grouping. The converse of the
associative law for addition in § 36 is
a-^(b-\-c) = a + b-\-c. (1)
That is, a sign of grouping preceded by the sign + can be
removed if each enclosed term is left unchanged.
Observe that the sign + is understood before b within the
parentheses.
Ex. 1. a + (4 a - 7 1/ + 5 «)= a + 4x-7y4-5«.
Ex.2. z+(^-Sx-{:2y -ia)=z-Sx-\-2y-ia.
ADDITION AND SUBTRACTION 57
By the rule for subtraction in § 67, we have
A sign of grouping preceded by the sign — can be removed,
if the sign before each enclosed term is changed from + <o — ,
or from — to ■}-.
Ex.1. 5a-(3&-2a + 4c)=5rt-36 + 2a-4c (1)
= 7a-35-4c.
The sign + is understood before 3 b within the parentheses.
Ex.2. -(5?rt-4n)-(-3wi+7n)=-5m+4n + 3m-7n (2)
= - 2 m - 3 71.
69. Sometimes one sign of grouping is enclosed within
another ; in this case the different signs of grouping must
be of different shapes to avoid confusion.
When there are several signs of grouping they can be
removed one at a time by the rules of § 68 ; and it is better
for beginners to remove at every stage the innermost sign of
grouping.
Ex. Removing the signs of grouping, simplify the expression
a-lx + {y-ib-c)]-zl
a- [x + {y-(&-c)}-«] = a- [x + {y-6 + c}-2]
= « - [x-\-y — b-{-c-z'\
= a — x — y + b ~ c + z. (1)
In the above process the parentheses ( ) were removed first, Uien
the braces { }, and then the brackets [ ].
Verify (1) when a = 8, x = 3, y=-2, & = -3, c = -4, z = 7.
Removing the outer sign of grouping first, we have
«- [« + {y-C& - c)} - z^=a - X - {y -{b - c)} + z
= a - X — y -^ (b - c) -{- z
= a-x — y-\-b-c-^z.
In review, the student should begin with the outer sign of grouping,
as he can thereby soon learn to remove, without error, two or more
signs of grouping at a time.
58 ELEMENTS OF ALGEBRA
Exercise 23.
Simplify each of the following expressions by removing
the signs of grouping and combining like terms :
1. a-(6 + c) + (&-c-a).
2. Sx-(y-2x) + (z-{-y-5x),
3. z-\?j-(z-x)\.
4. 3x-\2y + 5z-{Sx^y)l.
5 a—[a — \a—(2a— a)]'].
Verify the results of examples 1 to 5 inclusive,
6. When a = 7, b = -3, c = 4., x=10,y =-5, z = -2.
7. When a = — 5j b = 2, c = — 1, x=—3, y=^, z=—7.
8. a + b — [a — b-\-la-\-b — {a — b)\'].
9. x-(y-z)-\-l2z-3y-5xl.
10. 2a-{36 + (4c-36 + 2a)|.
11. a-2b-\3a-{b-c)-5c\.
12. a-l3b-^\3c-(d-b) + a}-2a'].
13. 2x-(5y-3z-]-7)-l4.-{-\x-(3y + 2z-\-5)l].
14. 3a - [26- J4c- 12a- (4 ft- 8c) J -(6 6 -12c)].
15. - 115 X -{Uy - (15z -^12y) - (10 X- 15 z)}'].
16. — [a — 5a + (ic— a) — (ic— a) — aj — 2 a].
17. 2x-(3y-4.z)-\2x-(3y-\-4.z)\- \3y-{4:Z+2x)\.
70. Insertion of signs of grouping. Law (B) in § 36 is
a + 6 4- c = a + (6 + c).
That is, a7iy number of terms of a polynomial can be enclosed
within a sign of grouping preceded by the sign -[-, if each
enclosed term is left unchanged.
ADDITION AND SUBTRACTION 69
Ex. 1. 5x-7?/ + 4c-7& = 5a; + (-7y + 4c-7 6).
Ex. 2. 4a + 3c-5x-3y = 4a + (+3c-5x-3y).
From the rule for subtraction it follows that,
Any number of terms of a polynomial can be enclosed within
a sign of grouping preceded by the sign —, if the sign before
each enclosed term is changed from -\- to — or from — to +.
Ex. 1. 7x + 6y-5a + 7c = 7a;-(-6?/ + 5a-7c).
Ex. 2. ax^ — 2 ex — cx^ -\- hx^ — x = ax^ — cx^ + fex^ — 2 ex — x
= (a - c)x3 + 6x2 _ (2 ex + x)
= (a - c)x3 + 6x2 - (2 c f i)a;.
Exercise 24.
In each of the following expressions enclose the last four
terms within a sign of grouping preceded by the sign — ,
without changing the value of the expression :
1. 3x — 2a + 56 — 2/ + 2;.
2. a-b-x + Zb-z-\-2y.
3. Sy-\-2x + lz + a-\-2b-\-c.
4. 2z-7a;-2a-36-5c -9?/.
Simplify each of the following expressions by combining
the terms having the same powers of x, so as to have the
sign + before each sign of grouping :
5. ax^ + by? + ^-\-2bx-b^-\-2x^-Zx.
Ans. {a^2)x' + (b-b)x' + (2b-^)x + 5.
6. 3 6ar^ — 7 — 2 ic + a5 + 5 aa;' + co; — 4 0^ — 6af.
7. 2-7ar' + 5ax2_2ca; + 9ax'+7aj-3a:2
8. 2 cy^ — ^ abx -\- ^dx — 3 bx"^ — aV + x\
60 ELEMENTS OF ALGEBRA
Simplify each of the following expressions by combining
like terms in x so as to have the sign — before each sign of
grouping :
9. aa^-\-5a^-a^x*-2b:x^-Sx'-bx\
10. 7 a^ — 3 c^x — dhx' + 5 aa; + 7 a^ — ahca^,
11. ay? -{-o?^ — h^ — hy? — cx^.
12. ?>}y^x^ — hx — ax^ — cx^ — h(?x — l x^.
Simplify the following expressions, and in each result
add the terms involving like powers of x\
13. a:(?—2cx—{bQi?— \GX—dx—il)x^-\-Z c^)\ — {c^—hx)\.
14. 5 aa^ - (7 6aj - 7 ca;-) - J6 6a^ - (3 aa;^ + 2 aa;) - 4 cx'X.
Express in descending powers of x the sum of,
15. a^ci?—5x, 2ax^—5aa^, 2x^ — bx'^ — ax.
16. aa?^bx — c, qx — r—py?, ar4-2a; + 3.
17. pa? — qx, qa? —px, q — x^, jyy? + qm?.
18. 2aa^-3ca?2+^a;, 3y?a;- ma^-2ca^, a;-2a^-3a?'.
19. bx — ay? — bo?, Sx^ — 4:nx — 2ma?, 2a?— po?.
20. coi?^2ax-\'mo?, 4:X^'-bx^, 4:7ix-\-2px\ 3a?—2rx^~x.
CHAPTER V
MULTIPLICATION OF INTEGRAL LITERAL
EXPRESSIONS
71. The degree of an integral term is the number of its
literal factors. But we usually speak of the degree of a
term in regard to one or more of its letters.
E.g.^ 5 ax is of the second degree, and 7 a^^a is of the fifth degree.
Again 4 abx-y^, which is of the seventh degree, is of the first degree in
a, of the second degree in x, of the third degree in y, and of the fifth
degree in x and y.
72. The degree of a polynomial is the degree of its term
of highest degree.
E.g., the trinomial ax"^ + bx -\- c is of the first degree in a, 6, or c,
and of the second degree in x. The binomial ax^y + by-, which is of
the fourth degree, is of the second degree in x or y, and of the third
degree in x and y. The trinomial ax^ + 2 bxy -f cp^ is of the second
degree in x, in y, and i?i x and y.
73. An expression is said to be homogeneous in one or
more letters when all its terms are of the same degree in
these letters.
E.g., 2a^ -\-Sab + 4b^ is homogeneous in a and b ;
5 x^ + 3 x:^y -{-Sxy^ + y^ is homogeneous in x and y ;
and 0x2 ^ 2 bxy + cy^ is homogeneous in x and y.
Exercise 25.
What is the degree of the term 3 arba^y*,
1. In a? 3. In a;? 5. In a and 6? 7. In a, a?, and ?/ ?
2. In 6? 4. In y? 6. In x and y? 8. In b, x, and y?
61
62 ELEMENTS OF ALGEBRA
What is the degree of the trinomial
a'x'-\-7a'b'i^y'-5abxy%
9. In a;? 10. In a? 11. In a; and?/? 12. In 6 and y?
Write two trinomials of the third degree and homogeneous,
13. In a and b. 14. In x and y. 15. In a and x.
74. A product is zero when one of its factors is zero.
That is, a . 0 = 0 and 0 . a = 0.
Proof a'0 = a(b-h), §§ 11, 32
= ab-ab = 0. §§60,11
Similarly, 0 • a = (6 — 6) a = 0.
Conversely, when a product is 0, one or more of its factors
isO.
That is, if a • 6 = 0, then a = 0, or 6 = 0, or a = 0 and
6 = 0.
75. Any positive integral power of 0 is 0 ; that is 0" = 0.
Proof 0" = 0 • 0 . 0 .. . to ?i factors = 0. § 74
76. Product of powers of same base.
Ex. 1. 23 X 22 = (2 X 2 X 2)(2 x2) = 2 x 2 x 2 x 2 x 2 = 2^.
Ex. 2. a^a^ = (aaa) (aa) = aaaaa = a^.
These examples illustrate the following law of exponents.
The product of the mth poiver and the nth power of the
same base is equal to the (m + n)th power of that base, and
conversely.
That is, a'" ' a" = a'"+".
Proof a'^a'' = (aaa • • • to m factors) (axxa • • • to w factors) § 52
= aaa ••• torn + 91 factors §49
= a"'+\ § 52
MUL TIP Lie A TION 63
Ex. Multiply 3 a-x^ by — 4 a^x^y.
(3 aV) ( - 4 a%2y) = s a'^^^ {- i) a^x^y § 49
= 3(-4).a2«4.a;3a;2.y §49
= - 12 a^xSy. §§ 47, 76
This example illustrates the method of finding the
77. Product of two or more monomials. Using the commu-
tative and associative laws, we have the following rule :
Multiply together their numeral factors, observing the law
of quality ; after this write the product of their literal factors,
observing the law of exponents,
Ex. Multiply together — 5 ay^, — 2 a^x^, and — 9 az^y.
(_ 5 ay2)(_ 2 a2x8)(_ 9 axh,) == (_ 5)(- 2) (- 9) aa^Q;x?xhpy
= - 90 a*x5y3.
Exercise 26.
Find the product of :
1. a^ and a*. 7. —Za% and 12a5«.
2. a^ and a*. 8. — abed and — 3 aC-b'^c.
3. rf, jf, and y^ 9. 7 x^y'^7^ and — 5 3?yh.
4. aa; and 3 ax. 10. — 3 a^6V and 8 a^b^c*d.
5. — 2a6a; and —Tab. 11. 2 a^, — 4 a^ft, and 5ab^
6. 6 ic^y and — 10 axy. 12. — 5 ax, — 7 a% and 2 aic^.
13. 8 xy^, — 3 ar^i/, and — 3 xy.
14. - 7 aft^, _ 3 a'b% and - a^ftl
15. a^b% 2 aft'^c, and — 5 a6c.
16. — 7 x^y^, ay^y*, and aa^.
17. — a^bx, aWx, and — aa^.
18. — a^ar, — b'^x, and — a5?/.
64 ELEMENTS OF ALGEBRA
78. Multiplication of a polynomial by a monomial. The dis-
tributive law of multiplication is
(a -f- 6 -f c -|- •")x = ax -\- bx -\- ex -\- •••.
That is, to multiply a polynomial by a monomial, multi-
ply each term of the polynomial by the monomial, and add
the several products.
Ex. Multiply 2 ^2 _ 4 2,3 ijy _ 3 y^,
= _ 6 ?/^3 _!_ 12 yiz.
Writing the multiplier under the multiplicand, ^
the work can be arranged as at the right of the ~ ^^
page. —Qyz^-\- 12 y^z
Exercise 27.
Multiply :
1. 4a2-5a + 3& by 2 a\
2. 2a2 + 3a6 + 262 by -Sd'b\
3. bc-{-ca — ab by abc.
4. 2a^-3x^-\-5x-4:hj -Bx".
5. -4:X^-\-3a^-Sx^ + 4L hj -6i^.
6. 9gh-12ga-3gb by Sgh.
7. — a^6c + b^ca — c^ab by — ab.
8. — 5 £C2/^2; + 3 xyz^ — 8 x?y%— 7 a;^/^ by — 2 a;^^;.
9. ci^ftV -~ abc — ax —by — cz by — 5 abcxy.
10. f aV — f aa^ + ^aa; by — |a^a;.
11. -|i»?/2^iaaJ2/-2a/ + |a2^ by -faa^.
12. i^a^y-fa^/H-iaa^-^a^/by -i^a;^2/.
13. (x + yy-2a(x-\-y)-\-5a^ hj 2(x + y).
14. (aj + l)«-4a(a; + l)'-2a6 by -5a6(aj + l)'.
15. (a'-hiy-Sx{a' + iy-4:xy by _ 3 aj^^/ (a^ -f 1)4.
16. (a^ + yy- a(a^ -^yf + 3 a'b^ by - 4 a=^6* (a^ + 2//.
MULTIPLICATION 66.
Remove the signs of grouping, and simplify each of the
following expressions :
17. (a + h)c-(a-h)c. 19. ^(6 - 2 c) + |(c- 2 6).
18. 2(a-6)+4(a + 6). 20. 1 aQ) - c) -2h{a- c).
21. a^h^c^-dF) + (^d?{a?-W) + hh\d'-aF).
22. 2\Zab-4.a{c-2b)\.
23. 7ac-252c(a-36)-3(5c-26)a[.
79. To multiply one polynomial by another,
Multiply each term of the multiplicand by each term of the
multiplier J and add the resulting products.
Proof Let x-\-y-^z be the multiplicand, and a + & the
multiplier ; then by successive applications of the distribu-
tive law, we have
{x-\-y + z)(a + b) = x(a-\-b) + y{a-^b)+z{a-^b)
= xa -\- ya -{- za -{- xb + yb -\- zb. § 36
Similarly when each factor has any number of terms.
Ex. 1. Multiply -2x + 3y by 4x-7y.
=(-2a;).4a;+3y.4a;+(-2a;)(-7y)4-3y(-7y) (1)
= - 8 x2 + 12 xy + 14 xy - 21 2/2 (2)
= - 8 x2 + 26 xy - 21 y2. (3)
Performing tlie steps in (1) and (2) mentally, we can arrange the
work as below :
-2a; + 3y
4x — 7y
- 8 a;2 + 26 ary - 21 2/2
Observe that the first and last terms in the product are the products
of terms in the vertical lines, while the second term is the sum of
the products of the terms in the diagonal lines.
In this way solve the first 15 examples in Exercise. 28.
66 ELEMENTS OF ALGEBRA
Ex. 2. Multiply a;^ - 2 jc + x^ + 1 by 2 - x -\- x^.
Arranging both multiplicand and multiplier in descending powers
of ic, we have
x^ + x^-2x +1 (1)
x^-x +2
afi + x^-2x^+ a;'2 (2)
_ a4 _ a;3 + 2 a;2 _ x (3)
2 x3 + 2 a;-2 - 4 X + 2 (4)
x5 -x^ + 5x'^-bx + 2
Expression (2) is the product of (1) multiplied by x"^ ;
Expression (3) is the product of (1) multiplied by — x ; and
Expression (4) is the product of (1) multiplied by 2.
The sum of these partial products is the required product, by § 79.
A vertical line, or bar, is often a convenient sign of group
ing. Its use is illustrated in the next example.
Ex. 3. Multiply x^-2x^y + ^ xy'^ - y^ by x2 - 3 xy + ?/2.
x3 - 2 x2y + 3
x^-Sxy +
XJ/2 - y3
y^
x5-2x*.v+ 3
-3 +6
+ 1
x^y^- 1
- 9
- 2
x2|/3
+ 3
+ 3
xy*
-y'
x5 - 5 x^y + 10
x^y^ - 12
X22/3+6
xy^-y^
The sum of the numbers before each bar is the coefficient of the
literal factor after it.
In this example the multiplicand and the multiplier are both homo-
geneous. Observe that the product is homogeneous also.
This illustrates the following principle.
80. The product of two or more homogeneous expressions is
homogeneous.
Proof If the homogeneous multiplicand is of the nth
degree, and the homogeneous multiplier is of the mth
degree, then each term in the product will be of the
(m + n)th degree ; that is, the product will be a homo-
M UL TIP Lie A riON 67
geneous expression of the (?>i + n)th degree ; and so on for
any number of factors.
When the multiplicand and the multiplier are homogeneous, that
fact should be noted in every case by the pupil ; and if the product
obtained is not homogeneous, it is at once known that there is an error.
Exercise 28.
Multiply :
1. x-{-2y by x — 2y. 10. —a; + 7 by a; — 7.
2. 2x-\-Syhy3x — 2y. 11. —x—16 by —x-\-16.
3. a — 3 & by a + 36. 12. — a; + 21 by a; — 21.
4. x + 7bya;-G. 13. 2a + |& by 3 a + J &.
5. 3a;-7 by 2a;-l. 14. ^a- ^b hj ^a-^b.
6. 2.0; — 4 by 2a; + 6. 15. ax — by by ax-\-by.
7. 22/ + 5^> by 3?/ — 46. 16. a.'2 + a; + 1 by a; — 1.
8. 2 7/i2 + 5n2 by 2 m- — ir. 17. cr + a6 + 6- by a — b.
9. 3m2-l by 3?n- + l. 18. a'--a6 + 6- by a + 6.
19. x^-xY-\-y^ by a^ + 2/^.
20. a^-ab-^b'' hy o^-{-db-\-b\
21. a^ _ 2 ax + 4 aj2 by a^ + 2 aa; + 4 x^.
22. 10 a- + 12 a6 + 9 6^ by 4 a - 3 b.
23. a-x — a^ -\- si? — a? by x-\-a.
24. x^^x-2 by x--\-x-Q,.
25. 2a:3_3aj2^2a; by 2a^ + 3a; + 2.
26. a^ + 2a?b+2ab^ by a^ - 2 ab -\- 2 b\
27. Qi? — 3xy — 'ifhy—^-\-xy-\-y'^.
28. ar - 2 a;?/ + 2/- by ar' + 2 a.-?/ +/.
29. 27a.-3-36aa:2^48a2a;-64a3 by 3a; + 4a.
30. ab + cd + ac + bd by ab + cd — dc — bd.
68 ELEMENTS OF ALGEBRA
31. x^^ — x^y^ -{- x^'y* — xY' -\- y^ by x^-\-y~.
32. -23?y + y^-^^x'y-^-x'^-2xi/ hy x'^-\-2xy + f.
33. a^ -\- h^ -\- c^ — he — ca — ab by a-\-h + c.
34. a;"+2 _|_ 2 a;"+i — 3 aj'* — 1 by a; + l.
35. — ax^ + 3 aajy^ — 9 ay^ by — aa; — 3 ay^.
36. —a^y + y^ -\- x^y^ + x* — xy^ by ic + 2/.
37. ia^ + ia + i by ia-|.
38. |a^-2aj + f by \x^\.
39. f a^ + ic?/ 4- f 2/^ by i a; - 1 2/-
40. Ja^_|a^-f by ^a;2+|a^-|.
41. I a^ — aaj — I a^ by I a.-^ — 1^ aa? + I a^
42. laaj + fa.'^ + ^a^ by fa^ + f a;2_|^^
43. 3 a?"* - 2 x"^-^ + 4 a^'^-^ by 2 aj'^ + 3 x"*"^ - 4 a;'"-^
44. 3 a;'*-^ + a;'*-2 - 2 x^^-i - 4 a;'' by 2 a^"-^ + 3 a;"-'*.
45. 4 a V* — a^a^" 4- 5 a;" by a^a^^^-^ + 6 a;"-^
46. 3 a^'-'^y? — a"- V + a'* by aV'^ - 2 a;""^ - 3 ax''^"^.
47. 4 a^"*+i — 3 a^*" — 2 aj'"+^ + J a^'""^
by 1 a^-+i - 2 ar'^'+i - a^'"-^
48. 3(a4-?>)'-2(a4-&)'(aJ-2/)-4(a+6)(aj-2/)'+7(a;-2/)^
by 2{a^lSf{x-y)-Q>{a^h){x-y)\
81. Removal of signs of grouping.
Ex. Remove the signs of grouping, and simplify,
42_5[-12a;-3{-15x + 3(8-7-3 (k)}].
The expression = 42 - 6 [- 12 x - 3 {- 15 x + 3 (3 ic + 1)}]
= 42 _ 5 [- 12 X - 3{- 6 a: + 3}]
= 42 - 6 [6 X - 9]
= 87-30 X.
MULTIPLICATION 69
Exercise 29.
Eemove the signs of grouping, and simplify :
1. 36-S5a-[6a + 2(10-6)]S.
2. a — {h — c) — la-h — c — 2\h-\-c\~\.
3. 8(6 + c)-[-Sa-6-3(c-6 + a)J].
4. 2(36-5a)-7[a-6J2-(5a-6)j].
5. Q>\a-2[b-^{c + d)']\-^a-^[b-4.iG-\-d)^]\.
6. 5[a-2[a-2(a + a.')]|-4Ja-2[a-2(a + a;)"|5.
7. _l0Ja-6[a-(6-c)]5 + 60J6-(c + a)i.
8. _3J-2[-4(-a)]K5J-2[-2(-a)];.
9. _2S-l[-(x-2/)]S + S-2[-(a;-2/)]J.
Multiply together the following expressions, and arrange
each product in descending powers of x :
10. ax^-\-bx-\-l and ex + 2,
11. ax^ — 2bx-\-3c and x — 1.
12. ic* -h aaj^ — bx — c and ar' — a.^•^ — 6u; + c.
13. aa^ — i»2 -f- 3 a; — 6 and oar* + a^ -f- 3 a; + i^.
14. x'^ — €131? — bx^ -\- ex -{■ d and x^ 4- aar^ — 6a^ — ca; -f d.
82. Multiplication by detached coefficients.
The labor of multiplication is lessened by using the method
of detached eoefficients in the two following cases :
(i) When two polynomial factors contain but one letter.
Ex. 1. Multiply 4x^-Sx^ + 2X-5 by 5 a;^ + 3 x - 1.
Writing coefficients only, we proceed as below :
4_ 3+ 2- 5
6+ 3- 4
20 - 15 + 10 - 25
+ 12- 9+ 6-15
- 16 + 12 - 8 4- 20
20 - 3 - 15 - 7-23 + 20
70 ELEMENTS OF ALGEBRA
Inserting the literal factors, whose law of formation is seen by
inspection, we have for the complete product,
20x^-Sx^-l6x^-7x'^- 23 x + 20.
(ii) When each of two polynomial factors is homogeneous
and contains only two letters.
Ex. 2. Multiply 5 a* + 4 a^^ _ 3 ^fts + 2 6* by a^-2 h\
5_l_4_}_0— 3 + 2 I^ th^ first expression, the term con-
1 4. 0 — 2 taining a^h^ is lacking ; that is, its co-
efficient is zero, which is written in
"^ "^ ~ "^ the line of coefficients. In the second
~ ~ ~ expression, the term containing ah is
5 + 4 — 10 — 11 + 2 + 6 — 4 missing; hence its coefficient is zero.
In the method of detached coeffi-
cients, the zero coefficients must evidently be written with the other
coefficients.
Inserting the literal factors, whose law of formation is seen by
inspection, we have for the complete product,
5 a6 + 4 a55 _ 10 a*62 _ n a%^ + 2 a'^h^ + 6 a&s _ 4 56.
Observe that the entire number of coefficients (zero coefficients being
included) in the product is one less than the number of coefficients in
both the multiplicand and multiplier together.
Exercise 30.
1. Multiply ^J^2x^-x'^^x-l by x^-2x-3.
2. Multiply 3a^ + 2a2_5a-|.4 by 2a^-^a-2.
3. Multiply a^ + ^x'y-A.xif + ^Tf by 2x^-3xhj + f,
4. Multiply 3 a^ - 2 a^6 - 4 a^h^ - ab' by a^-2 h\
5. Multiply ^x'' -Z:^y -^1 xy^ ^-2f by y?-\-'6f.
6. Rework by detached coefficients those examples in ex-
ercise 28, from 19 to 42, to which the method is applicable.
CHAPTER VI
DIVISION OF INTEGRAL LITERAL EXPRESSIONS
83. Division is the inverse of multiplication. Having given
a product and one factor, division is the operation of finding
the other factor.
That is, to divide one number by another is to find a third
number which multiplied by the second number gives the
first.
Thus, - 12 -- 3 = - 4 ; for - 4 X 3 = - 12,
and _12h-(-3)= 4 ; f or 4x(-3) = -12.
As in Arithmetic, the given product is called the dividend,
the given factor the divisor, and the required factor the
quotient.
84. Law of Quality. In each of the following identities
the third number multiplied by the second gives the first;
hence by definition the third number in each case is the
quotient of the first divided by the second.
+(a6)-^-a=-6; "(aft) --+« =-6. J ^^
From identities (1) it follows that,
The quotient is positive 2vhen the dividend and the divisor
are like in quality ; and negative when they are opposite in
quality.
Tlie arithmetic value of the quotient is equal to the quotient
of the arithmetic value of the dividend by that of the divisor.
71
72 ELEMENTS OF ALGEBRA
Any number divided by ^1 is equal to the number itself.
Any number divided by ~1 is equal to its arithmetically
equal opposite number.
Exercise 31.
Perform each of the following indicated operations :
1. -25 -5. 5. 75 -(-25). 9. 21 - (- 1).
2. 36 ^(-6). 6. -72 -(-6). 10. - 36 -f- 4.
3. _5i^(_3). 7. _ 105 -(-21). 11. -1^|.
4. _33^(_l). 8. -144-24. 12. l-^(-|).
Find the value of (a? + ?/) — z,
13. When x = — 15, y = — S, z = 6.
14. When x = — AS, y = 6, z = —
Find the value of (x — y)-r- (a 4- b),
15. When 05 = 22, y = -2, a = 5, & = 3.
16. When a? = - 21, y = G, a = -7, b = 6.
85. From the definition of division we have
quotient x divisor = dividend.
That is, since the quotient of N divided by a is A^ — a,
we have,
(N^a)xa = N. (1)
86. The reciprocal of a number is 1 divided by that
number.
Since their product is + 1, any number and its reciprocal
have the same quality.
E.g., the reciprocal of 4 is ^ ; the reciprocal of — 4 is 1 -4-(— 4)
or — ^ ; and the reciprocal of — | is 1 -4- (— |), or — f.
DIVISION 73
87. Dividing by any number except zero gives the sams
result as multiplying by the reciprocal of thai number.
That is, N-^a = Nx(l^a). (1)
Proof. The second member of (1) multiplied by a is,
by § ^5, equal to N\ hence it is the quotient of N divided
by a.
Ex. 1. 16 H- 4 = IG X ^ = 4.
Ex.2. 16-(-4)=10 x(-i) = -4.
88. The commutative law for division.
Ex.1. _40^(-2)-(-5) = -40x(-^)x(-^) = -4. (1)
Ex.2. ^-^(_|)^(-|)=^x(-3)x(-^)=f (2)
Since we can change the order of the factors in the second member
of either (1) or (2), we can also change the order of the divisors in
the first member of either identity ; this illustrates that,
The commutative law holds for division as well as for
multiplication, provided the sign of operation, -r- or x, before
each number is transferred with the number itself.
That is, N xb^c = N^cxb. (1)
Proof Nxb-^c = Nxbx(l-hc) § 87
= Nx{l^c)xb §49
= N-r- cxb. § 87
Ex. (-60)x(-22)-f-(- l5) = (~60)-(- 15)x(-22)
= 4 x(-22) = -88.
89. A product of two or more factors is divided by a num-
ber if any one of the factors is divided by that number.
Proof {ab)-^c = a^cxb={a^c)b, §88
or {ab) -i- c = b -T- c X a = (b ^ G)a. § 88
74 ELEMENTS OF ALGEBRA
90. Any indicated quotient is called a fraction.
A quotient is often indicated by placing the dividend
over the divisor with a line between them.
E.g.^ a -^ b, -, and a/b are but different ways of indicating that
b
a is to be divided by b.
Each of these expressions is a fraction, a being the dividend, and
b the divisor. The dividend and divisor of a fraction are often called
its numerator and denominator respectively.
When the dividend or divisor consists of more than one
term, the horizontal dividing line in a fraction serves as a
sign both of division and of grouping.
E.g., in the fraction ^ ~ the horizontal dividing line takes the
c + d
place of both the sign of division and the two parentheses in the form
(a-6)^(c + c?), or (a- 6)/(c + d).
In § 1 any fractional number as 5/6 was regarded as
(1/6) X 5 ; but it can also be regarded as 5 -r- 6 ; for
]Sr-r-a = Nx(l-^a) = (1/a) x JV. §§ 87, 49
91. The product of two or more fractions is equal to the
product of their dividends divided by the product of their
divisors; and conversely.
mi- J. • a b c abc ,^.
That IS, _._._ = (1)
X y z xyz ^ ^
^ . a b c a b c ..„
Proof. -'~'-'X-y-z = ~'X'-'y'-'Z §49
'^ X y z ^ X y ^ z
= abc. § 85
Dividing each member by xyz, we obtain (1).
Ex 4 ^ 3 ^-2_ -(4x3x2)^ 8
_ 5 _ 7 _ 3 -(5x7x3) 35
DIVISION 75
92. Quotient of powers of the same base.
Ex. 1. a^ ^ a- = a^-'^ = a^ ; for a^ x a'^ = w'.
Ex. 2. a? -^ a^ = o?-^ = a* ; for a'^ x a^ = a?.
These examples illustrate the following law :
If /n > /I, the quotient of the mth power of any base divided
by the nth power of the same base is equal to the (m — n)th
power of that base; and convex
That is, a'" -T- a" = a'"'".
Proof a"*-" X a" = a*"-''+'' = a"*. § 76, 83
Ex. 1. Divide 20a*6S by -bah^.
91
20q^ft5_ 20 a^ 6^
— 6 ab^~ — 5 a 6*
= - 4 a362. §§ 84, 92
91
Ex. 2. Divide - ba^H^ by lla262a;2.
-ba%^x^_-b a^ b^ 7?
11 a262a;-^ ~ 11 ' d^' b'i' ic2
=-/x • 1 . 62 .x = -^bH.
These examples illustrate the following section.
93. The quotient of one monomial by another. By the con-
verse of § 91 we have the following rule :
Divide the numeral factor of the dividend by that of the
divisor, observing the law of quality; after this write the quo-
tient of their literal factors, obset^ving the laiv of exponents.
Ex. 1. - 84 a^xs -f- 12 a*x=- 7 ax"^.
Ex. 2. 77 a^x^y^ ^ ( - 7 ax^y) =-11 axyK
Check. Multiplying the obtained quotient by the given divisor, we
obtain the dividend ; hence, the division is correct.
T6 ELEMENTS OF ALGEBRA
Exercise 32.
Divide :
1. - 72 a^ by - 9 a. 6. 84 affz' by - 7 icyV.
2. 84a3by-7al 7. 28 a^d^ ^y _ 4 ^3^^
3. - 35 a^ by 7 ic^. . 8. - 35 a%^ by 5 ab.
4. 4a^6V by —ab^c\ 9. — 16a^/ by — 4ic/.
5. - 12 a^ft'^c^ by -Sa^ftc^. 10. 36 m%^2 ^^y 9 ^6^9^
11. 96a*afz* by 12 aV;^^
12. - 256 xyz^' by - 8 a;y;2«.
13. SAaWc' by 14a&V. 16. - 144 aV by -24aV.
14. - 16 %a;2 by - 2 a;?/. 17. - 3 x'^+^ hj 5 x'^+\
15. 50 yV by —Ba^y. 18. _ 4 a7'«+«2/'»+" by 7 x*"?/"*.
19. 5 a;"+V'+^ by - 8 aj^^/"*-
20. - 7 a;^+'^"'+2 by - 2 a;"- V"^-
21. — 42 x«+3a"*-i by — 7 a^'-^a'^-l
22. — 50 a7"+«?/'"+* by 25 a;"-*^"*-".
94. Distributive law for division. The quotient of one ex-
pression divided by another is equal to the sum of the results
obtained by dividing the parts of the first expression by the
second; and conversely.
That is, «±At£_±^^«_l_i + £+.., (o
X XXX
Principle (C) lies at the basis of division in Arithmetic; e.g., to
divide 894 by 6 we separate 894 into the parts 600, 240, and 54, divide
each of these parts by 6, and add the results.
Thus 8M^0Og^24O^54^j00^,„^9^j^g_
6 6 6 6
DIVISION 77
Proof. ±±±±SL±^=(a-^b + c+:.)^ §87
X X
= a- + /)- + ci+... §60
:- + -+- + -. §87
95. To divide a polynomial by a monomial.
By the distributive laAv for division, in § 94, we have the
following rule :
Divide each term of the polynomial by the monomial, and
add the resulting quotients.
Ex. 1. Divide l2x^-6ax^-2 a^x by 3 x.
3a; ~ 3x 3a; 3x ^
= 4a;2- 5rta;-|a2.
Ex. 2. Divide 12 a" + 0 a^ _ 6 gS by - 3 a2.
12q« + 9a*-6a5_ 12ff« 9 a* -6a^ „q.
= - 4 a - 3 a2 + 2 a8.
Exercise 33.
Divide :
1. 5x^-7 ax-i-4:X by X. 6. -24 a^-32 a;^ by -Sir^.
2. a;'''-7ar^ + 4aj*by a.-^. 7. a^-a-ft-a-ft^ by al
3. lOx^—Safi-^Sx* by a;^. 8. a^ — a6-ac by —a.
4. 27 3^-363.*^ by 9ar\ 9. .r^-a^-oa; by -a;.
5. 15 ar' - 25 a;* by -5. T^. 10. 3a;«-9.Tyby -3aj.
11. 4a^6^~8a%^ + Ga6^ by -2a6.
12. — 3a^ + f a^ — (j^2; by — f a;.
13. _5a^ + |.^y + j^0a, by -fa;.
14. ia.V-3.r^y-5.Ty by -^sc^y^
78 ELEMENTS OF ALGEBRA
15. J a^x — Jg- abx — | acx by | ax.
16. - 2 a^or^ + 1 aV by | a^x.
17. 25(aj + 2/)'-3a(a; + ?/)2 + 106(a^ + ?/) by 5(aj + 2/).
18. -S(a-by -12 x(a -by- 16 y (a -by by 4(a-6)2.
19. Ga^'" — 4 a-'" by 2 a".
20. 10 2/"+V - 15 r^'^' by - 5 y^z.
Divide 12 a;2«+y - 16 x^'^+Y - 20 aj^^+y by :
21. 4 a;". 22. — «'»+y.
23. — Sa^^n^^. 24. ia;2n-3^2^
25. Divide 4 a;-'^+y — 16 a;-"+V +i by 4aj2«2/«.
26. Divide - 15 a^+Y+^ + 21 x''+Y+'^ by 3 a;*+y+2,
96. To divide one polynomial by another.
Let it be required to divide
2 x^y"^ — x^y + x^ — xy^ + y^ by y"^ — xy + x^.
First arrange dividend and divisor in descending powers of «, for
convenience placing the divisor to the right of the dividend as below :
a* — x^y + 2 x'^y^ — xy'^ + y*
x^ — x^y + x'^y'^
x^ — xy + 2/2 Divisor
x^ + y'^ Quotient
x^y'^ — xy^ + 2/*
x^y^ — xy^ + y*
From the law of exponents we know that a;*, the term of the highest
degree in x in the dividend, is the product of the terms of highest
degree in the divisor and the quotient ; hence, the first term of the
quotient is x^ -r- x^, or x"^. Multiply the divisor by x^ and subtract the
result from the dividend.
The remainder, xV - xy^ + y'^, is the product of the divisor by the
other terms of the quotient ; hence, x^ij'^, the first term of the remain-
der, is the product of the first term of the divisor and the second
term of the quotient. Therefore the second term of the quotient is
a;2y2 ^ a;2^ or y"^. Multiplying the divisor by y^ and subtracting the
result from x^y^ — xy^ + ?/*, we have no remainder.
Hence the required quotient is x^ + y\
DIVISION 79
Observe that by the above process the dividend was separated into
the two parts ar* — x^y + x^y'^ and x'^y^ — xy^ + y^ ; hence, by the distrib-
utive law for division, we have
x^ — x^y + 2 x-y'^ — xy^ + y* _x* — x^y + x-y'^ x^y"^ — xy^ + y^
x/^ — xy + «/2 x^ — xy + y^ x^ — xy -{■ y^
= x2 + y2.
If the dividend and divisor were arranged in ascending powers of x,
the quotient would be obtained in the form y^ + x^.
Hence, to divide one polynomial by another, we have the
following rule :
Arrange the dividend and divisor in descending powers of
some common letter.
Divide the first term of the dividend by the first term of the
divisor, and write the result as the first term of the quotient.
Multiply the divisor by this first term of the quotient, and
subtract the resulting pj-oduct from the dividend.
Divide the first term of the remainder by the first teiin of
the divisor, and write the result as the second term of the
quotient.
Multiply the divisor by this second term of the quotient, and
subtract the resulting product from the remainder previously
obtained.
Treat the second remainder, if any, as a new dividend and
go on repeating the process iintil the remainder is zero, or is
of a loiver degree in the letter of arrangement than the divisor.
Ex. 1. Divide 2 a - 4 a^ + 3 a^ - 1 by 1 - a.
Arranging dividend and divisor in descending powers of a, we have
3a8-4a2H-2a-
-1
-a + 1
3a'-3a2
- a2 + 2 a
- a2-|- a
- 3 a-2 + a -
a -
a -
- 1
-1
80
ELEMENTS OF ALGEBRA
Ex. 2. Divide x^y'^ + x^ + y^ by y"- — xy + x'^.
Arranging dividend and divisor in descending powers of x, we have
ic* + x-y'^
xl^ — x^y + xhf'
x^y
x^y — x-y'^ + xy^
+ 2/*
X?/ + y2
yp- + xy ^
xhf- — xy^ + y^
Ex. 3. Divide 16 a* - 1 by 2 a - 1.
16 a4
-1
16 a4.
8^3
8a3_
-4a2
4a^
4a2_
-2a
2a-l
2rt-l
2a
8 rt3 4 4 ^2 + 2 a + 1
Exercise 34.
Divide :
1^ a;2 + 3a^ + 2 by a; + 1. 4. 3a;2 + 10a; + 3 by a; + 3.
2. a2_ii^_^3o by a-h. 5. oa^ + lla; + 2 by x^2.
3. aj2_7^_^i2 by a;-3. 6. 5a^ + 16a;H-3 by ic + 3.
7. 2x2-t-llx+5 by 2a.'-}- 1.
8. 2a^H-17a;-f 21 by 2ic-h3.
9. 4a)2-f-23ic + 15 by 4aj + 3.
10. 6a^-7aj-3 by 2a;-3.
11. 12a2-7aa;-12a.'2 by 3a-4a;.
12. 15 a^ -h 17 aic — 4 a^ by 3 a -f- 4 a;.
13. 12 a2 _ 11 ac - 36 c^ by 4 a - 9 c.
14. 60a;2-4ic?/-45?/2 by lOic-Oi/.
DIVISION 81
15. -A.xy-loy-^-^ijx' hy 12x-6y.
16. 10()a:3-3.T-13a;2 ^^^ 3-|_25x.
17. 16-96ic4-216ic2-216a:3 + 8i«^ by 2-3a;.
18. x'-x'-'^x-nhy x'-\-^x-\-^.
19. 22/3_32^2_gy_l by 22/--52/-1.
20. Q>m^-m'^-l^m-\-S by 3m2H-4m-l.
21. 6a'-13a^ + 4a3 + 3a2 by ^a^-2o?-a.
22. a.'* + .T3 + 7iv2-6a; + 8 by x' + 2x-[-^.
23. a^-a3-6a- + 15a-9 by a-4-2a-3.
24. a^ + Oa^ H-13a2 + 12a4-4 by a2 4-3a + 2.
25. 2a;^-ar»4-4a^ + 4a;-3 by ar^-a; + 3.
26. a;5-5a;^ + 9ii-3-6a.-2-a; + 2 by a^-3a; + 2.
27. ar'-4a;*4-3it'3-|.3a.-2_3a' + 2 by a^-a;-2.
28. 30a-*4-lla.'3-82x2_i2a; + 48 by 2.T-4 + 3a^.
29. 69?/-18-71 2/3 + 28^-352/2 ^^ 4^y2_x^yj^Q^
30. 0^--15^•^ + 4^^ + 7fc2-7A: + 2 by 3A:3-A: + 1.
31. 2x^-^x + x^-\-12-lx' hy x' + 2-3x.
32. a^-2a;^-7x'3 + 19x2_i0^. 1,^ ^_r'^_j_5
33. Ux^ + ^^a?y + l^^f-{-A^xi/-{-Uy' by 20.-2 + 50^2/
+ 72/'.
34. x^ -{- x^y - a?f -\- Qi? -2 xy'^ -\- f by a^ + a^-/.
35. x^ -2y^* -7 afy* -7 xy^'- -\-Uxy by x-2y\
36. a^ + 63 _^ c^ _ 3 ((6c by a + 6 + c.
82
ELEMENTS OF ALGEBRA
a3 _ 3 a6c + &^ + c3 a + h + c
aS + a'^b + a^c
a'^ - ab ~ ac + b'^ - be -\- c^
- a% - a2c - 3 abc + b^ -\- c^
-a^b-ab^- abc
- a^c + a62 _ 2 abc + b^ + c^
— a^c — abc — ac^
ab^ - abc + ac^ + b» -\- c^
ab-^ + b^ + b^c
- abc + ac2 - 6% + c-
- abc - b'^c - bc^
ac2+ 6c2 + c3
ac2 + ?)C2 + c^
Here we arranged the dividend and divisor in descending powers of
a, and gave b precedence to c throughout.
37. a^ -\- if — z^ -\- 3 xyz hj x-\-y — z.
38. S x^ — y^ -{- z^ -{- 6 xyz by y — z~2x.
39. 27a^-Sb^-^(^-\-lSahchj 3a-2b-\-c
40. a^ + 3a26 + 3ai!>2 + 63 + c3 by a + & + c.
41. a' -\-b' + c' -2b'<^ -2 a'c" -2a'b' by a + ^ + c.
42. ^:^-^-^^^xf^ + j\fhjix-^ly.
43. |a3_ 9^2^_^_2_7^^_27ar^ by ia-3a;.
44. ^aS-^i^a^ + ^-Va-eV by ^a-i-
45. I^V H-^a^ by ia^ + iac.
46. ^9^a4_|a^-7a2 + .|a4.i_§ by f a^-f-a.
47. 36a52+i2/' + i-4a^-6aj + i2/ by Gic-iy-i
48. ^8_o^5_2 4|ci,a;4by |a-fa;.
49. 43 a^'"-^ + 6 «2-+i _ 29 aj^- - 20 aj^--^ by 2 a;'" - 5 x'^-\
6 aj^'w+i - 29 x-^ + 43 a;2'«-i - 20 x^""-^ | 2 a;'" - 5 a;"*-^
6 a:2"*+i - 15 x^"* 3 x'^+^ - 7 a;"* + 4 x""-^
- 14 x2« + 43 ic2'»-i
- 14 a;2»» + 35 a;2'»-i
8 a;2»»-i — 20 a;2»*-2
8 a.2m-i _ 20 x^""-^
DIVISION 83
50. ic'*" — a^"^/"' + x'^if"^ — y^"" by x^ — 2/"*.
51. 6a3'»-25a-« + 27a''-o by 2 a'* -5.
52. 6a'"'-lla^ + 13a2« + 23a»"H-2-3a" by 3a"4-2.
53. 12 a;"+i + 8 a?" - 45 a;"-^ + 25 x''-'- by 6 a - 5.
54. i^ia' - Li3.ab -{-9 ac + 2b' - be by Ja-36+fc.
55. (5 + c) a^ — bcx -\- x^ — be (b -\- c) by a^^ — 6c.
56. a.'^ + (a + 6 + c) ar 4- (a6 + ac + 6c)a; + a6c by x-\-b.
57. a?^ 4- (a H- 6 — c) a^ + (a6 — ac — 6c) a; — a6c by x— e.
97. When, as in each example given above, the division
is exact, the quotient is the same whether the dividend
and divisor are arranged in deseending or in aseending
powers of any common letter. But when the division is
not exact, the partial quotient obtained with one arrange-
ment is not the same as that obtained with the other.
E.g., ^l±l = x-l + -^; (1)
while 1+^^=1 -a; + 2x2 -2x8+^^. (2)
1 + x 1+x ^ ^
Here the partial quotients x — 1 and 1 — x + 2 x^ — 2 x^ are evi-
dently unequal. The entire quotients, or the second membei-s of (1)
and (2), are, of course, identical.
In (1) the remainder is of a lower degree than the divisor.
In (2) the division can be carried to any number of terms.
When arranged in ascending powers of some common
letter, an expression of a lower degree can be divided by
one of a higher degree in the letter of arrangement.
E.g., -l_ = l + a; + aj2 + aj3 + ...+aj"-i+--^^.
1 — aj 1 — X
84 ELEMENTS OF ALaEBRA
Exercise 35,
Divide :
1. a^ -^ y^ hj X -\- y. 5. x^ — a^ hj x -\- a.
2. x^ -{- if hj X — y. 6. 1 by 1 + ic to 4 terms.
3. x^ — y^ bj X -\- y. 7. 1 + a; by 1 + ic^ to 5 terms.
4. x^ + y*hj x-}-y. 8. 1 -f- 2 a? by 1 — 3 a; to 4 terms.
9. Divide 2 by 1 -}- a; and thus reduce the second member
of identity (1) in § 97 to the form of the second member
of (2).
98. Zero divided by any number, except zero, is equal to
zero.
That is, when a^O, 0^a = Q. §§ 85, 74
Conversely, if a quotient is zero, the dividend is zero.
Zero as divisor will be considered in Chapter XXVII.
Prior to that chapter it will be assumed that any expression
used as a divisor does not denote zero.
99. Division by detached coefficients. In § 82 we considered
two cases in which the work of multiplication could be shortened by
using the method of detached coefficients. In the same two cases the
labor of division can be lessened by using detached coefficients and an
arrangement of terms known as Horner's method of synthetic division.
This method is illustrated by the following examples :
Ex. 1. Divide 2x^-1 x^-\-2x^- x^-Qx-^ 20 by 2x3-3x2+4aj-5.
cS i 3
S " 5
Quotient 1-2-4J 0 0 0 Remainder
Inserting in the quotient the literal factors, whose law of formation
Is seen by inspection, we have for the complete quotient x'^ — 2x — A.
Explanation. The modified divisor, or the column of figures to the
left of the vertical line, consists of the coefficients of the divisor, the
2
-7
+ 2
- 1
- 6
+ 20
3
-4
5
-6
+ 8
-10
-12
+ 16
-20
DIVISION 85
quality of each coefficient after the first being changed; this change
of quality enables us to replace the operation of subtraction by that of
addition at each successive stage of the work.
Observe that the number of coefficients in the quotient will be one
more than the number of coefficients in the dividend minus the num-
ber of coefficients in the divisor, in this case 1 + 6 — 4, or 3 (§ 82).
Thus, the numbers to the left of the vertical bar are the coefficients of
the quotient^ and those to the right of this bar are the coefficients
of the remainder.
Dividing the first coefficient of the dividend by the first coefficient
of the divisor, we obtain the first coefficient, 1, of the quotient. Multi-
plying the modified coefficients of the divisor (3, — 4, 5) by this first
coefficient of the quotient, we obtain line (1).
Adding the coefficients in the second column to the right of the
divisor, and dividing the sum by the first coefficient, 2, of the divisor,
we obtain — 2, which is the second coefficient of the quotient. Multi-
plying the modified coefficients of the divisor by this second coefficient
of the quotient, we obtain line (2).
Adding the coefficients in the third column and dividing the sum
by the first coefficient of the divisor, we obtain the thirds or last,
coefficient of the quotient. Multiplying the modified coefficients of
the divisor by this third coefficient of the quotient, we obtain line (3).
Lines (1), (2), and (3) are evidently the coefficients of the three
partial products obtained by multiplying the divisor by each term of
the quotient, the first term of each product being omitted and the
quality of the others being changed.
Hence by adding each of the vertical columns after the third, we
obtain the coefficients of the remainder.
Here the coefficients are all zero, and the division is exact.
Ex. 2. Divide 2x^-1 7^y+\2 v^y"^ - 8 x^y^ ■\- a; V by 2 x^ - 3 x^y - y^.
2
2
-7 +12
-8
+ 1
+ 0
+ 0
(1)
3
+ 3 0
+ 1
(2)
0
- 6
0
-2
(3)
1
+ 9
0
4-3
(4)
+ 3
0
+ 1
(5)
1 _2 + 3 +1 I +2 +3 +1
Inserting the literal factors, we have for the quotient x^ — 2 x'^y
+ 3 xy"^ + ?/'^, and for the remainder 2 x'^^/* + 3 a;?/^ + y^.
Explanation. The terms in xy^ and y^ are missing in the dividend,
and the term in xy"^ in the divisor; hence their zero coefficients are
86 ELEMENTS OF ALGEBRA
written with the other coefficients. The sums of the vertical columns
after the fourth give the coefficients of the remainder.
To find the remainder after one term of the quotient, add lines (1)
and (2) after the first vertical column ; to find the remainder after
two terms of the quotient, add lines (1), (2), and (3) after the first
two vertical columns ; to find the remainder after three terms of the
quotient, add lines (1), (2), (3), and (4) after the first three vertical
columns.
Exercise 86.
Divide :
1. a-^ - 4 or' + 2 a;2 + 4 a; 4- 1 by ^ - 2 a; - 1.
3. x^^-{-x^-\-l by 0^ + .1^ + 1.
by 2x^-3xy-\-Ay^
5. Sf-22xy'-^20a^rf^xy-7x'y + 6x^ .
by 4:y^ — 3xy -^2x^0
6. a' -3 a'b^ + 8 ab^ - 5b^ hj a^ -4.ab -{-b' to four terms
in the quotient.
CHAPTER VII
INTEGRAL LINEAR EQUATIONS IN ONE UNKNOWN
100. An integral equation is an equation all of whose terms
are integral in the unknown. (Review §§ 10, 16, 17.)
E.g.^ 2 a;2 + 3 = 2 X and ^ -}- — = x + 2 are integral equations.
2 0
101. The degree of an integral equation in one unknown
is the degree of its term of highest degree in the unknown.
A linear equation is an equation of the^rs^ degree.
A quadratic equation is an equation of the second degree.
A higher equation is an equation of a higher degree than
the second.
E.g., 3 ic + 1 = 4 and ax + 6 = 0 are linear equations in x.
6 x2 — 7 X = 1 and ax^ + 6x + c = 0 are quadratic equations in x.
6 x^ — 4 x^ + 3 X 4- 4 = 0 is a higher equation in x.
102. A root, or solution, of an equation in one unknown is
any value of the wiknown ; that is, it is any number which
when substituted for the unknown renders the equation an
identity.
E.g., 12 is a root of the equation
2 X - 5 = X + 7.
For, putting 12 for x in the equation, we obtain the identity
24 - 5 = 12 + 7.
Any root of an equation, since it satisfies the condition
expressed by the equation, is said to satisfy the equation.
87
88 ELEMENTS OF ALGEBBA
103. To solve an equation in one unknown is to find all
its roots. In solving equations we use the principles of
EQUIVALENT EQUATIONS.
104. Two equations in one unknown are said to be equiv-
alent, when every root of the first is a root of the second,
and every root of the second is a root of the first.
E.g.^ the equations
4a;-8 = 2-x (1)
and bx= 10 (2)
have the same root, i.e., are equivalent ; for 2 is a root of each equa-
tion, and, as will be seen later, 2 is the only root of either.
In solving equations we need to know what operations on
the members of an equation will make the derived equation
have the same root, or roots, as the given one.
Of such operations the most elementary and important
are found in §§ 105, 106, 108, 109.
105. Identical expressions.
If in the equation
4(x-l)-(3x-2)=3, (1)
we substitute for the first member the identical expression a; — 2, we
obtain the equivalent equation
X - 2 = 3. (2)
For, as is easily shown, 5 is a root of either equation ; and, as will
be seen later, 5 is the only root of either.
This example illustrates the following principle :
If , for any expression in an equation, an identical expres-
sion is substituted, the derived equation will he equivalent to
the given one.
INTEGRAL LINEAR EQUATIONS 89
That is, ii A = B denotes any equation in one unknown,
as X, and A = A'\ then the equations
A = B (1)
and A' = B (2)
have the same root, or roots.
Proof. To prove that equations (1) and (2) have the same
root, or roots, we must prove that every root of (1) is a root
of (2) ; and conversely that every root of (2) is a root of (1).
Since A and A^ are identical expressions, any value of x
which when substituted for x will make either one identical
with By will make the othei? identical with B (§ 32).
Hence, any root of (1) is a root of (2), (§ 102) ; and con-
versely any root of (2) is a root of (1) ; that is, equations
(1) and (2) have the same roots, i.e.y are equivalent.
E.g., since, 3(a; - 1) - {3 x - (2 + x)} = x - 1 ;
the equations 3(x - 1) - {3 x - (2 + x)} = 5 (1)
and x-l = 5 (2)
have the same root ; that is, we neither lose nor introduce a root by-
substituting for 3(x — 1) — {3 X — (2 + x)} in equation (1) its identical
expression x — 1.
106. Addition or Subtraction.
If to both members of the equation
2x-8 = 7-8 (1)
we add 8 + x, we obtain the equivalent equation
3x = 15.
For, as is easily shown, 5 is a root of each equation ; and, as will
be seen later, 5 is the only root of either equation.
This example illustrates the following principle :
If identical expressiojis are added to, or subtracted from,
both members of an equation, the derived equation will be
equivalent to the given one.
90 ELEMENTS OF ALGEBRA
That is, if 31= M', the equations
A=B (1)
and A±M=B±M' (2)
have the same root or roots.
Proof. Any root of (1) makes A = B. § 102
But, when A = B, A±M=B±M'. § 32, (iii)
Hence, any root of (1) is a root of (2).
Conversely, any root of (2) makes A ± M= B ± M'.
B\xt,when A ±M=B±M', A = B. § 32, (iii)
Hence, any root of (2) is a root of (1).
Therefore, equations (1) and (2) are equivalent. § 104
If, to each member of the equation
ax — b = ex — d, (1)
we add — ex and + 6, we obtain the equivalent equation
ax — ex = b — d. (2)
Adding — ex to both members of equation (1) removes the term
+ ex from the second member, and transfers it, with its sign changed
from + to — , to the first member. Likewise, adding + 6 to both
members of (1) removes the term — b froili the first member, and
transfers it, with its sign changed from — to +, to the second mem-
ber. This example illustrates the following important application of
the principle proved above.
If any term is transposed from one member of an equation
to the other, its sign being changed from -\- to —, or from —
to +, the derived equation has the same root or roots as the
given one.
107. An expression is said to be unknown, or knoivn, ac-
cording as it does, or does not, contain an unknown number.
E.g., if X is an unknown number, a: — 2 is an unknown expression ;
if a is a known number, 9 + 5 a is a known expression.
INTEGRAL LINEAR EQUATIONS 91
108. Multiplication.
If both members of the equation
2 4 3 12 ^^
are multiplied by 12, we obtain the equivalent equation
6 X + 9 = 4 X + 13. (2)
For, as is easily shown, 2 is a root of each equation, and, as will
be seen later, 2 is the only root of either.
This example illustrates the following principle :
If both members of an equation are multiplied by the same
known expression, not denoting zero, the derived equation will
be equivalent to the given one.
That is, if C represents any known expression, not denot-
ing zero, the equations
A = B
and
CA=CB
have the same roots.
(1)
(2)
Proof Any root of (1) makes A = B. § 102
But, when A = B, CA= CB. § 32, (iv)
Hence, any root of (1) is a root of (2).
Conversely, any root of (2) makes CA = CB. § 102
But, when CA = CB, A = B, since C ^ 0. § 32, (v)
Hence, any root of (2) is a root of (1).
Therefore, equations (1) and (2) are equivalent.
Ex. 1. Solve the equation (5 x - 12) -4- 6 = (x - 3) ^ 3. (1)
Multiply by 6, 5 x - 12 = 2 x - 6. (2)
Transpose terms, 5 x — 2 x = 12 — 6. (3)
Unite terms, 3 x = 6. (4)
Multiply by 1/3, x = 2. (5)
92 ELEMENTS OF ALGEBRA
Proof of equivalency.
Equation (2) has the same roots as (1) by §108, 'identical ex-
pressions. '
Equation (3) has the same roots as (2) by § 106, ' addition. '
Equation (4) has the same roots as (3) by §105, 'identical ex-
pressions.'
Equation (5) has the same roots as (4) by § 108, 'multiplication.'
Hence the one and only root of each of these equations is 2.
Ex. 2. Solve the equation ^^^ ^ ^ = \.
(1)
Multiply by 12, 3 (x + 1) - 4 (x - 1) = 12.
(2)
Remove (), 3a; + 3-4x4-4 = 12.
(3)
Transpose terms, 3ic — 4a; = 12 — 3 — 4.
(4)
Unite terms, — x = 5.
(5)
Multiply by - 1, x = — 6.
(6)
Proof of equivalency. Equation (2) has the same roots as (1) by
the principle of ' multiplication ' (§ 108) ; (3) as (2) by ' identical ex-
pressions' (§105); (4) as (3) by 'addition' (§106); (5) as (4) by
'identical expressions' (§105); and (6) as (5) by 'multiplication'
(§ 108). Hence the one and only root of each of these equations is —5.
The two following applications of the foregoing principle
are very important :
(i) When, to clear an equation of fractional coefficients,
we multiply both members by the L.C.M. of their known
denominators, the derived equation has the same roots as
the given one.
(ii) When the sign before each term of an equation is
changed from + to — , or from — to -f- (that is, when each
member is multiplied by — 1) the derived equation has the
same roots as the given one.
109. Roots introduced or lost.
If we multiply both members of the equation
3x-7 = 2x + 2 (1)
by the known expression 0, we obtain the identity
(3ic-7) x0=(2x + 2) xO. (2)
INTEGRAL LINEAR EQUATIONS 93
Observe that (1) restricts x to the one value 9, while (2) does not
restrict the value of x at all.
Again, if we multiply both members of the equation
6x-l=4x + 3 (3)
by the unknown expression a; — 5, we obtain the equation
(6x-l)(x-5) = (4x + 3)(x-5). (4)
Equation (4) has the two roots 2 and 5, while (3) has only the one
root 2. Hence, the root 5 was introduced into (4) by multiplying
(3) by the unknoion expression x -^ 5.
The two examples above illustrate why the multiplier in § 108 was
limited to a known expression, not denoting zero.
If we divided identity (2) by 0, we would obtain equation (1).
If we divided equation (4) by x — 5, we would obtain equation (3),
and one root would be lost by the operation.
This illustrates why the divisor in the following article is limited to
a known expression not denoting zero.
110. Division. If both members of an equation are divided
by the same known expression, not denoting zero, the derived
equation will be equivalent to the given one.
That is, if 0 represents any known expression, not denot-
ing zero, the equations
A = B (1)
and A^C = B-^G (2)
have the same roots.
Proof Any root of (1) makes A = B. § 102
But, when A = B, A-^C=B-i-C. § 32, (v)
Hence, any root of (1) is a root of (2).
Conversely, any root of (2) makes A-t-C=B^C. § 102
But, when A^C=B-^C, A = B. § 32, (iv)
Hence, any root of (2) is a root of (1).
Therefore, equations (1) and (2) are equivalent.
94 ELEMENTS OF ALGEBBA
Ex. 1. Solve the equation (x - 1) (x - 2) + 5 = (x + 1)2.
(1)
Eemove (), x^-Sx + 2-{-6 = x^ + 2x + l.
(2)
Tra.nspose terms, — Sx — 2x — 1— 2 — 5.
(3)
Unite terms, — 6x = -Q.
(4)
Divide by - 5, x = ^.
(5)
Proof of equivalency. No root is lost or introduced by any one of
the operations performed on the members of the equations from (1)
to (5); hence, the one and only root of (1) is |.
Ex.2. Solve3(x-l)-{3x-(2-x)} = 5. (1)
Remove ( ), Sx -S - Sx + 2 - x = 5. (2)
Transpose, 3 x - 3 x - x = 5 + 3 - 2 (3)
Unite terms, —x = 6. (4)
Divide by - 1, x=-6. (5)
Proof of equivalency. No root is lost or introduced by any one
of the operations performed on the members of the equations from
(1) to (5); hence, the one and only root of (1) is — 6.
Observe the difference in the meanings of equal, identical,
and equivalent. Equal applies to numbers, identical to ex-
pressions, and equivalent to equations. Two numbers are
equal or unequal, two expressions are identical or not iden-
tical, and two equations are equivalent or not equivalent,
i.e., have or have not the same roots. We should not apply
the word equivalent to numbers or expressions.
111. Erom the examples given above, it will be seen that
the different steps in the process of solving a linear equa-
tion are the following :
(i) Clear the equation of fractions, if there are any.
(ii) Remove parentheses, if there are any.
(iii) Transpose the unknown terms to one member of the
equation, and the known terms to the other member.
INTEGRAL LINEAR EQUATIONS 95
(iv) Unite like terms j and divide both members by the co-
efficient of the unknown.
Note. In order to form the habit of clear and accurate thinking,
the pupil should at first state the operation by which each equation is
derived from the preceding one, and note whether by this operation
any root is lost or introduced.
But as he advances he should perform the simpler steps mentally,
and apply two or more principles at the same time.
112. A numeral, or numerical, equation is an equation in
which, all the known numbers are denoted by numerals.
Exercise 37.
Solve each of the following numeral equations :
1. 3a;-5 = 2a;4-l. 8. x- (4:-2 x) = 7 (x-1).
2. 3x-\-^ = x + S. 9. 5(4-3a;) = 7(3-4a;).
3. 4:X-4: = x-7. 10. 4(l-a;) + 3(2 + a;) = 13.
4. 7 a; -5 = a; -23. 11. S(x -2) = 2(x - S).
5. 8 x + 42 = 5 X. 12. 2 x - (5 x -\- 5) = 7.
6. 5 a; -12 = 6 a.- -8. 13. 3(.c + 1) =- 5(a; - 1).
7. 7a; + 19 = 5a; + 7. 14. 7 (a; - 18) = 3 (a; - 14).
15. 2(a;-2) + 3(a;-3) + 4(.r-4)-20 = 0.
16. . 2 (a; - 1) - 3 (a; - 2) + 4 (a; - 3) + 2 = 0.
17. 5a; + 6(.T-f l)-7(a; + 2)-8(a;4-3) = 0.
18. 2a;-[3-|4a; + (a;-l)|-5] = 8.
19. (x - 1) (x - 2) = (x + 3) (x - 4).
20. 3 a^ = (a; + 1)' + (x + 2f + (a; + 3)1
21. (x - 2) (x - 5) 4- (x - 3) (x - 4) = 2 (a; - 4) (a; - 5).
22. 5(x-\-iy-\-7{x + sy = i2(x-^2y.
23. (x - 1) {x - 4) = 2 a; H- (x - 2) (x - 3).
96 ELEMENTS OF ALGEBRA
24. x/5-x/4: = l. 27. 2 x/3 + 5 = 5 x/6 + A.
25. {x-l)/2-\-{x-2)/3 = 3. 28. x/2-\-2x/3 = 5x/6 + 7.
26. 2x/3 + 4. = 5-{-x/3. 29. 3 x/A ~ 5 = 7 x/S - 6.
30. |-(2-a;)- |-(5ir + 21) = a! + 3.
31. J(^ + l)-|-^(a; + 2)+i(a^ + 4) + 8=:0.
32. K^-5)-K^-4)-i(^-3)-(x'-2).
33. ^(a)H-i)_J(2a;-i) + li-0.
34. (3x + 5)/8 - (21 + x)/2 = 5x- 15.
35. (a^ - 2)/3 - (12 - x)/2 =(5x- 36)/4 - 1.
36. (a.- -f 8)/4 - (5 a.- + 2)/3 = (14 - a;)/2 - 2.
37. (x - 15)/4 - (7 - 2 a.')/21 - 3 a;/14 + 1/2.
38. 5[4-(3a?-l)]=6(a^-ll) + 49.
39. (x - 2)/4 + 1/3 -lx-(2x- l)/3]- 0.
40. 3ip8-(a^/8+24)] = 3i(2i + a./4).
41. 5(3x-5)-17 = 8(3x-5)-2 (3 x - 5).
Combine the terms involving Sx — 6.
42. 2 (i» + 1) - 3 (ic + 1) + 9 (x + 1) + 18 = 7 (a; + 1).
43. X (x -\- 2) -^ x (x -\- 1) = (2 X — 1) (^x -{- 3).
44. x'-xll-x-2(3-x)'] = x-^l.
45. 3 (x - 1)/16 - 5 (a^ - 4)/12 = 2 (i» - 6)/5 + 5/48.
46. 0.5 a; + 3.75 == 5.25 ^ — 1.
To clear of fractions multiply by 4.
47. 2.25 X - 0.125 = 3x-\- 3.75.
48. 0.25 X + 4 - 0.375 x = 0.2 ic - 9.
49. 0.375 X - 1.875 = 0.12 x + 1.185.
50. O.W a; + 1.2 - 0.875 a; + 0.375 = 0.0625 x.
INTEGRAL LINEAR EQUATIONS 97
113. A literal equation is an equation in which one or
more of the known numbers are denoted by letters.
E.g., ax-\-2x + i = 0 and ax -\- b = ex are literal equations.
Ex. Solve the equation (2 ~ 5x)/a = (ex + 7)/b. (1)
Multiply by ab, (2 — 5x)b = (ex -\- 7) a. (2)
Remove ( ), 2b - 6bx = aex + 7 a. (3)
Transpose, 6bx — acx = 7 a — 2b. (4)
Unite terms, — (5 b + ac) x = 7 a — 2 &. (5)
Divkle by - (5 6 + «c), x = ?A=lI_«. (6)
56 + ac
Proof of equivaleney. No root is lost or introduced by any one of
the operations performed on the members of the equations from (1)
to (6); hence, the one and only root of (1) is jj^iven in (6).
114. Any linear equation in one unknown has one, and only
one, root.
Proof. By transposing and combining terms, any linear
equation can be reduced to an equation of the form
ax = c, (1)
which, by §§ 105, 106, is equivalent to the given equation.
Divide by a, x = c/a. (2)
Equation (2) is equivalent to (1) by § 110 ; hence, c/a is
the one and only root of equation (1), or of the given linear
equation.
If c = 0 and a^O, the root c/a is zero (§ 98).
If c ^ 0, then the smaller a is, the larger arithmetically
is the root c/a.
Observe that, if b is an arithmetic number, the linear
equation x — b = 0 has the arithmetic root b, while the
equation x-\-b = 0 has no arithmetic root.
98 ELEMENTS OF ALGEBRA
Exercise 38.
Solve each of the following literal equations :
1. ax -i- b^ = bx -{- a\ 3. (a -\-b)x + (b — a)x = b^
2. x/a + x/b = c, 4. 2(x — a)-\-3(x—2a) = 2 a.
5. (a-\- b)x-\- (a — b)x = al
6. (a + bx) (b 4- ax) = ab (x^ — 1).
7. (a — x) (a -\- x) = 2 a^ -{- 2 ax — x^.
8. i(x-\-a-\-b)-{-i{x-{-a-b) = b.
9. i{a-\-x)-\-l(2a-\-x) -\-i(3a + x)=3a.
10. ica -=- 6 + a?6 ^ a = a^ + &^.
11. (a + &aj) (& + ax) = ab(x^ — 1).
12. (a^ + x) (b^ -\-x) = (ab + xf.
13. (a; 4- a + ?>)' + (a^ + a - ^')' = 2 a^.
14. {x -a)(x-b)-\- (a + &)' = {x -\- a) (x -\- b).
15. aic (cc + a) + bx (ic 4- 6) = (a + ^) (a? + a) (cc + b).
16. What kind of a number is the root of a numeral
equation? Of a literal equation?
See § 5.
CHAPTER VIII
PROBLEMS SOLVED BY LINEAR EQUATIONS IN ONE
UNKNOWN
115. Having learned some of the properties of linear equa-
tions in one unknown, we return to the subject of solving
problems by equations, which was introduced in the first
chapter. Reread § 19.
Prob. 1. A has $80, and B has § 15. How much must A give to B
in order that he may have just 4 times as much as B ?
Let X = the number of dollars that A must give to B ;
then 80 — x = the number of dollars that A will have left,
and 16 + X = the number of dollars that B will have.
But A will then have 4 times as much as B ; that is,
80-a; = 4(15 + ic). (1)
From (1) X = 4.
Hence A must give ^4 to B.
Prob. 2. A man has 10 coins, some of which are half-dollars, and
the rest dimes, and the coins altogether are worth $ 6. How many
has he of each kind ?
Let X — the number of half-dollars ;
then 16 — a; = the number of dimes.
The X half-dollars are worth ^ x dollars,
and the 16 — a; dimes are worth ^^ (16 — x) dollars.
Now the coins altogether are worth $ 6 ; hence
ix + T^(16-x)=6.
From (1) 05 = 11, the number of half-dollars.
.'. 16 — X = 6, the number of dimes.
100 ELEMENTS OF ALGEBRA
Note. It should be remembered that any letter as x always denotes
a number, and not a concrete quantity. Observe, also, that in any
problem all concrete quantities of the same kind must be expressed in
terms of the same unit ; for example, in each of the above examples
all sums of money were expressed in terms of the unit, one dollar.
Prob. 3. A father is 7 times as old as his son, and in 5 years he
will be 4 times as old as his son. How old is each ?
Let X years = the son's age ;
then 7 x years = the father's age.
Hence (x + 5) years = the son's age after 5 years,
and (7 ic + 5) years = the father's age after 5 years.
.-. 7aj + 5 = 4(x + 5). (1)
From (1) x = 6. .: 7x = 35.
Hence the son is 5 years old, and the father 35.
Prob, 4. Divide 60 into two parts, so that three times the greater
may exceed 100 by as much as 8 times the less falls short of 200.
Let X = the greater part ; then 60 — x = the less.
Three times the greater part is 3 x, and
3 a; — 100 = the excess of 3 a; over 100.
Eight times the less part = 8 (60 — x) and
200 - 8 (60 - a;) = the defect of 8 (60 - x) from 200.
But this excess and defect are equal ; that is,
3x- 100 = 200 -8(60- x). (1)
From (1) X = 36, the greater number.
.*. 60 — X = 24, the less number.
Prob. 5. A could do a piece of work in 14 hours which B could do
in 6 hours. A began the work, but after a time B took his place,
and the whole work was finished in 10 hours from the beginning.
How long did A work ?
Let X = the number of hours that A worked ;
theu 10 — X = the number of hours that B worked.
G^
A
PROBLEMS SOLVED BY LmEAE EQUATIONS 101
Since A could do the whole work in 14 hours, in 1 hour he would
do 1/14 of it; hence
^ x = the part of the work done by A in x hours.
Since B could do the whole work in 6 hours, in 1 hour he would
do 1 /6 of it ; hence
1(10 — x) = the part of the work done by B in 10 — x hours.
But A and B together did the whole work ; hence
t1jX + K10-^)=1. (1)
From (1) z = 7. .-. 10 - x = 3.
Hence A worked 7 hours, and B worked 3.
Prob. 6. Find the time between 5 and 6 o'clock at which the hands
of a watch are together.
Suppose that the hands are together at x minutes after 5 o'clock.
At 6 o'clock the hour-hand is 25 minute-spaces ahead of the minute-
hand ; hence, while the minute-hand moves through x minute-spaces,
the hour-hand will move through a; — 25 such spaces. But the minute-
hand moves 12 times as fast as the hour-hand ; that is, in any given
time the minute-hand passes over 12 times as many minute-spaces as
the hour-hand. Hence
a; = 12 (a; - 25). (1)
From (1) x = 27^p
Hence the hands are together at 27y\ minutes past 5 o'clock.
Exercise 39.
1. Find two numbers whose sum is 72, and whose dif-
ference is 8. Ans. 40 and 32.
2. Divide 25 into two parts whose difference is 5.
3. Divide 12 into two parts whose difference is 16.
Ans. 14 and — 2.
4. The difference between two numbers is 8; if 2 be
added to the greater, the result will be 3 times the smaller.
Find the numbers.
102 ELEMENTS OF ALGEBRA
5. A man walks 12 miles, then travels a certain dis-
tance by train, and then twice as far by coach as by train.
If the whole journey is 78 miles, how far does he travel by
train ?
6. Find two numbers whose difference is 12, and whose
sum is equal to ^ their difference.
7. Find a number such that the sum of its sixth and
ninth parts will be equal to 15.
8. Find the number of which the eighth, sixth, and
fourth parts together make up — 13. A71S. — 24.
9. Find a number such that ^ of it shall exceed ^ of it
by 2. Ans. — 35.
10. Two numbers differ by 28, and one is -| of the other.
Find them.
11. A, B, and C have a certain sum of money between
them. A has ^ of the whole, B has \ of the whole, and C
has $ 50. How much have A and B ?
12. A and B together have $ 75, and A has $ 5 more
than B. How much has each?
13. A has $ 5 more than B, B has $ 20 more than C, and
they have $ 360 between them. How much has each ?
14. A has $ 15 more than B, B has $ 5 less than C, and
they have ^ Q^ between them. Hoav much has each ?
15. A has $100, and B has |20. How much must A
give B in order that B may have half as much as A?
16. The sum of two numbers is 38, and one of them
exceeds twice the other by 2. What are the numbers?
17. Find a number which when multiplied by 8 exceeds
27 as much as 27 exceeds the original number.
18. Find two numbers of which the sum is 31, and which
are such that one of them is less by 2 than half the other.
PROBLEMS SOLVED BY LINEAR EQUATIONS 103
19. Divide 100 into two parts such that twice one part is
equal to 3 times the other.
20. Four times the difference between the fourth and
fifth parts of a certain number exceeds by 4 the difference
between the third and seventh parts. Find the number.
21. Fifty times the difference between the seventh and
eighth parts of a certain number exceeds half the number
by 44. Find the number.
22. A father is 4 times as old as his son; in 24 years he
will be only twice as old. Find their ages.
23. A is 25 years older than B, and A's age is as much
above 20 as B's is below 85. Find their ages.
24. A's age is 6 times B's, and 15 years hence A will be
3 times as old as B. Find their ages.
25. Find a number such that if 5, 75, and 35 are added
to it, the product of the first and third sums will be equal
to the second sum multiplied by the number.
26. The difference between the squares of two consecu-
tive whole numbers is 121. Find the numbers.
27. Divide $380 between A, B, and C, so that B will
have i$30 more than A, and C will have $20 more than B.
28. The sum of the ages of A and B is 30 years, and 5
years hence A will be 3 times as old as B. Find their pres-
ent ages.
29. The length of a room exceeds its breadth by 3 feet ;
if the length had been increased by 3 feet, and the breadth
diminished by 2 feet, the area would not have been altered.
Find the length and breadth of the room.
30. The length of a room exceeds its breadth by 8 feet ;
if each had been increased by 2 feet, the area would have
been increased by 60 square feet. Find the dimensions of
the room.
104 ELEMENTS OF ALGEBRA
31. The width, of a room is | of its length. If the
width had been 3 feet more, and the length 3 feet less,
the room would have been square. Find the dimensions
of the room.
32. A, B, and C have $ 1285 between them ; A's share is
greater than | of B's by ^25, and C's is j\ of B's. Find
the share of each.
33. If silk costs 6 times as much as linen, and I spend
$66 in buying 40 yards of silk and 24 yards of linen, find
the cost of each per yard.
34. If $600 be divided among 10 men, 20 women, and
40 children, so that each man receives $ 15 more than each
child, and each woman receives as much as two children,
find what each receives.
35. Divide $152 among 5 men, 7 women, and 30 chil-
dren, giving to each man $4 more than to each woman, and
to each woman 3 times as much as to each child.
36. A sum of money is divided between three persons,
A, B, and C, so that A and B have $60 between them,
A and C have $ 65, and B and C have $ 75. How much
has each?
37. A dealer bought four horses for $1150; the second
cost him $ 60 more than the first, the third $ 30 more than
the second, and the fourth $ 10 more than the third. How
much did each cost ?
38. Two coaches start at the same time from York and
London, a distance of 200 miles, travelling one at 9J miles
an hour, the other at 9J miles an hour. In how many hours
after starting did they meet, and how far from London ?
Ans. 10| hours ; 98J miles from London.
39. A man leaves ^ of his property to his wife, l to his
son, and the remainder, which is $2500, to his daughter.
How much did he leave to his wife and son each ?
Let X = the number of dollars which he left in all.
PBOBLEMS SOLVED BY LINEAR EQUATIONS 105
40. A man divided his property between his three chil-
dren so that the eldest received twice as much as the second,
and the second twice as much as the youngest. The eldest
received ^3750 more than the youngest. How much did
each receive?
41. A third of the length of a post is in the mud, a
fourth is in the water, and 5 feet is above the water. Find
the length of the post.
42. A flock of sheep and goats together number 84.
There are 3 goats to every 4 sheep. How many are there
of each ?
43. Find the time between 3 and 4 at which the hands
of a clock are together.
44. A can do a piece of work in 30 days which B can do
in 20 days. A begins the work, but after a time B takes
his place, and the whole work is finished in 25 days from
the beginning. How long did A work ?
45. A can do a piece of work in 20 days which B can do
in 30 days. A begins work, but after a time B takes his
place and finishes it. B worked 10 days longer than A.
How long did A work?
46. One number exceeds another by 3, while its square
exceeds the square of the second by 99. Find the numbers.
47. Of two consecutive numbers, ^ of the greater exceeds
I of the less by 3. Find the numbers.
48. A garrison of 1000 men having provisions for GO
days was reinforced after 10 days, and from that time the
provisions lasted only 20 days. Find the number in the
reinforcement.
49. In a mixture of spirits and water half of the whole
plus 25 gallons was spirits; and a third of the whole minus
5 gallons was water. How many gallons were ther% of
each ?
106 ELEMENTS OF ALGEBRA
50. At 3 o'clock, A starts upon a journey at the rate of 4
miles an hour, and after 15 minutes B starts from the same
place, and follows A at the rate of 4J miles an hour. When
does B overtake A ?
51. A fish was caught whose tail weighed 9 pounds; his
head weighed as much as his tail and half his body, and his
body weighed as much as his head and tail. What did the
fish weigh ?
52. Find a number such that if | of it be subtracted
from 20, and ^j of the remainder from i of the original
number, 12 times the second remainder shall be half the
original number.
53. A cistern can be filled in half an hour by a pipe A,
and emptied in 20 minutes by another pipe B ; after A has
been opened 20 minutes, B is also opened for 12 minutes,
then A is closed, and B remains open for 5 minutes more,
after which there are 13 gallons in the cistern. What was
the capacity of the cistern ?
54. A father was 24 years old when his eldest son was
born ; and if both live till the father is twice as old as he
now is, the son will then be 8 times as old as now. Find
the father's present age.
55. If 19 lbs. of gold weigh 18 lbs. in water, and 10 lbs.
of silver weigh 9 lbs. in water, find the quantities of gold
and silver in a mass of gold and silver weighing 106 lbs. in
air, and 99 lbs. in water.
56. The sum of $ 1650 is laid out in two investments, by
one of which 15 per cent is gained, and by the other 8 per
cent is lost; and the amount of the returns is $1725. Find
the amount of each investment.
57. How many children are there in a family, if each son
has as many brothers as sisters, and each daughter has twice
as many brothers as sisters ?
PROBLEMS SOLVED BY LINEAR EQUATIONS 107
116. Interost formulas. In problems of interest, the quan-
tities involved are the principal, interest, rate, time, and
amount.
Let p = the number of dollars in the principal ;
r= the rate, or the ratio of the interest per annum
to the principal ;
t = the number of years in the time ;
i = the number of dollars in the interest for t years
at rate r.
a = the number of dollars in the amount, or the sum
of the principal and the interest ;
then /=prt; (1)
and a = p-\- prf. (2)
Proof. The interest on $p for one year is $pr', hence i,
or the interest on $p for t years, is $j^?*i; whence (1).
But a=2J-\-i; whence (2).
If any three of the four numbers /, p, r, t, or a, p, r, t are
given, the fourth can be found by solving equation (1) or (2).
Ex. Find the principal that will amount to $ 1584 in 5 years 4
months at G per cent.
Here a = 1584, < = 5|, r = 0.06. Substituting in (2), we have
1584 = p 4- P (0.06) (5|) = 1.32 p.
.-. p = 1584 - 1.32 = 1200.
Hence the principal is $ 1200.
Exercise 40.
1. Solve i =prt forp, r, and t.
2. Solve a = p -{-jvt forp, r, and t.
3. Find the interest on $ 4760 for 4 years 6 months at
5 1 per cent.
108 ELEMENTS OF ALGEBRA
4. Find the amount of $ 3500 for 5 years 4 months at 6
per cent.
5. Find the interest on $ 7240 for 3 years 3 months at
8 per cent.
6. The interest on $ 1250 for 8 months is $ 50. Find
the rate per cent.
7. The amount of $ 1050 for 2 years 6 months is $ 1260.
Find the rate.
8. The interest on $3420 at 6 per cent is $049.80.
Find the time.
9. A sum of money doubles in 12 years 6 months.
Find the rate.
10. Find the principal that will yield f 262.50 per month
at 7 per cent.
11. Find the time in which $ 1350 will amount to $ 1809
at 6 per cent.
12. The interest in 4 years 3 months at 4 per cent is
$ 2099.50. Find the principal.
13. Find the time in which a sum of money will double
itself at 6 per cent.
14. The interest on $ 1270 for 8 months is $76.20. Find
the rate.
15. At 4 per cent how much money is required to yield
$ 2500 interest annually ?
CHAPTER IX
POWERS, PRODUCTS, QUOTIENTS
117. Certain products and quotients are so frequently
required in Algebra that the student should prove and
memorize the identities by which they can be written out.
In this chapter the most important of these identities are
proved, and used in obtaining products and quotients.
In the next chapter the converses of these identities are
used for factoring.
118. Power of a power. The nth power of the mth power
of any base is equal to the mnth power of that base; and
conversely.
That is, (a/")" = a""".
Ex. 1. (23)2 = 23 X 23 = 23+8 _ 26.
Ex. 2. (a2)* = a^a^a'^a^ = a2+2+2+2 = ^8.
Ex. 3. (a2)6 = a-ixo = qjIj . conversely, a^xc = {a^y.
Proof (a**)" = a'^a*" ••• to ?i factors by notation
— ^m-f-m-f •••ton sunimandg g i^Ct
= a"•^
119. Power of a product. Tlie nth power of the product
of two or more factors is equal to the product of the nth powers
of those factors ; and conversely.
That is, {aby = a"b", (abc)" = a"b"c".
Ex.1. {ahcY = {abc){ahc){ahc) by notation
= {aaa) (bbb) (ccc) by laws (A') , (B')
= a363c3, by notation
109
110 ELEMENTS OF ALGEBRA
Proof.
{aby'=(ah) (ab) ••• to n factors by notation
= (aa ••• to n factors) (bb -■' to n factors) by (A')^ (B')
= a^"". by notation
Similarly for any number of factors.
Ex. 2. (— «)^ = (— 1)^«^= — «^ ; conversely, (— \ya^ =(— a)^.
Ex. 3. (2 x'-yy = 2i(x-2) V = 16 x^y* ;
conversely, 2*(x'^yy'^= (2 x~yy.
Ex. 4. (- 5x'^y^y= (- 5)3(x2)3(?/3)3^ _125iK6^9
Exercise 41.
Express as a power or as a product of powers each of
the following powers of products :
1. (-a^y. 7. (aa^y. ' 13. {ab-a^yy.
2. {-xy. 8. {-Q^y. 14. (-xh/'zy.
3. (-/)l 9. (-aa;2)^ 15. (-2a5V)^
4. (-z'f. 10. {-bfy. 16. (-2a26c«)^.
5. [(-a)^^ 11. (-2aary. 17. {-^Ci'x^yy.
6- [(-2)^'. 12. (-a2ic3)^ 18. {-a?Wxyy.
19. (- a)^ (- ay, (- a)^ (- d)\ (- a)«, (- a)^
20. (- aby, (- a5)3, (- aby, ( - a6)^ (- aby.
21. (-2 a3z>4)2^ (_ 2 a354^^3^ (_ 2 a^by.
22. As a power of tlie base 3^, express 3^ 3«, S^^^ 3^0, S^-.
23. As a power of the base x^, express x'^, x^, x^, x^^, x^.
24. Asa power of the base a?, express a^, a}^, a}^, aP-, oF*.
Express as a power of a product :
25. 6^x4^ 27. {-ay{-by 29. {-xy%f{-zy
26. 4^x(— 3)^ 28. (—ayb^ 30. a\x-\-yy.
POWERS, PRODUCTS, QUOTIEXTS 111
120. Square of binomials. By multiplication, we obtain
(a + 6)- = fl2 + 2a6 + 6-. (1)
Tliat is, the square of the sum of tivo numbers is equal to
the sum of their squares plus twice their product.
Ex. 1. (3 .X 4- 5 vY = (3 xy + (5 y^ + 2 (3 x) (5 y) by (1)
= 9 x2 + 25 y^ + 30 xy. § 1 19
Ex.2. (2x-Syy=[{2x-)-{-(-Zy)Y
= (2x)2+(-3y)--^ + 2(2x)(-3y) (2)
= 4 a;2 + 9 2/2 _ 12 xy. (3)
Ex.3. (a_6)2^[«+(_5)-|2
= a2+(-^)2 + 2rt(-6)
= a2 - 2 ah + b^.
In the examples of this chapter there are two steps :
First step. The application of an identity.
Second step. The simplification of the result obtained by
the first step.
E.g., in example 2, the application of identity (1) gives the result
in (2), and the simplification of this result gives (3).
At first the pupil should write out these steps separately ; later
he should apply the identity mentally, and write only the final result.
Observe the advantage gained in this chapter by regarding a poly-
nomial as a sum.
121. Square of polynomials. If in the identity
{a + x) - = a''^x' + 2ax (1)
we put b -{-y for x, we obtain
(a -h 6 -f /)- = a' ^(b-{-yy + 2a(b+y)
= a- + b- +/ + 2 a6 4- 2 a/ 4- 2 6/. (2)
And so on for a polynomial of any number of terms.
112 ELEMENTS OF ALGEBBA
Hence we infer that
The square of any polynomial is equal to the sum of the
squares of its several terms, plus the sum of the products of
twice each term into each of the terms which follow it.
Ex. (x2 + 2?/-3c)2
^(x2)2 4.(2?/)-^+(-3c)2 + 2a;2(2|/) + 2x2(-3c)+2(22/)(-3c)
= X* + 4 2/2 + 9 c2 + 4 ic^?/ - 6 cx2 - 12 cy.
Exercise 42.
Write the square of each of the following expressions :
1. 2a + x. 18. a«-2?/^ + 3c«.
2. X' + y-. 19. ^x^ — Qx — Q.
3. 3a- 5 6. 'ZO. x + y + z + v.
4. ^d^-Wx. 21. x + y — z-v.
5. -2a2 + 56l 22. x-y-z-v,
6. -aa^^ + V- 23. o.-^ + a^^ - 2 cc - 2.
7. 3a6c-4a;V 24. a + 26-3c + 4:d
8. -2 2;2-a&iKl 25. 1 + a) - a^^ _|_ ^^
9. -J^a? + ?>cz\ 26. 3a2-a;2 + c--2/.
10. a + h + c. 27. 4x2_3(^_4c_32/2.
11. a + &-c. 28. 3ir^-2a2 + 4 6-2A'.
12. a-6 + c. 29. 8 o.-^/ - 4 a^^s.
13. a-h-c. 30. |-a^2/^ + f^y.
14. a + 26 + 4. 31. 2x-'»-7.
15. £c + 2i/ + 3;2. 32. ijx-y^-4.x^y\
16. 2 4-2a;-3a^. 33. 4 a'-^^ - 3 a'-'^^H
17. 2a^-3a;-2. 34. 2(a + 1) - 5(6 + c).
POWERS, PRODUCTS, QUOTIENTS 113
122. Product of sum and difference. By multiplication, we
have
(a+b)(a-b) = a'-b\ (1)
That is, tJie product of the sum and the difference of the
same two numbers is equal to the square of the first number in
the difference J minus the square of the second.
Ex. 1. (2x-^ + 5 6y3)(2x2-6 6y8) = (2a;2)2_(5 6?^)2 by (1)
= 4a:4_25&2y6 §§119,118
By properly grouping terms, the product of two poly-
nomials can frequently be written as the product of the sum
and the difference of the same two numbers.
Ex. 2. (a + 6 + c){a + 6 - c)=[{a + h)+e] [(a + 6)- c]
= (a + 6)2-c2 by(l)
= a2 + 2a& + Z>2 -c2.
Ex. 3. (a + 6-c)(a-6 + c) = [a + (6 - c)J [a -(6 - c)]
= a2 - (6 - c)2
= a2 - 62 4. 2 6c - c2.
Exercise 43.
Write each of the following indicated products :
1. (b + a){a-b). 5. (x' + 4.if){x'-^f).
2. (5 + a;)(a;-5). 6. (3 or^ + 5 r') (3 ar' - 5 /).
3. (l4-3ic)(l-3«). 7. {^by-\-2ax)(2ax-Sby).
4. (62 + a2)(a2_62). 8. (4ca^ + 5 6-V)(4car-5i2^).
9. (a 4- 5 + c) (a - ft - c).
10. (l + 6_c)(l-6 + c).
11. (a-6 4-c)(a-ft-c).
12. (a; + 32/-22;)(.^•-3?/-f-2^).
13. (or^ + a^?/ + /) (x" -xy + f).
14. (2r' + 2/ + 2)(/-?/ + 2).
114 ELEMENTS OF ALGEBRA
15. (3a + 6-3c)(3a-6 + 3c).
16. (a2 + 3a-l)(a2-3a-l).
17. (a^-2a'-\-l)(a'-{-2a'-l).
18. (a^ - 62 _ c2) (a2 -{- b^ -\- c^).
19. (_a^_/ + 7)(a^-2/^ + 7).
20. (ax — hy-\- cz) {ax + hy — cz).
21. (3a; + 9-42/)(3a;-9+42/).
22. (l4-4a; + 32/ + 22)(l+4ic— 32/-20).
23. (a) + 22/ + «-6)(a; + 22/-a + &).
123. By multiplication, we obtain
{x + a){x^b) = X' + (a + 6) jr + ab.
That is, ^/ie product of two binomials having the same first
term is equal to the square of the first term, plus the sum of the
second terms into the first term, plus the product of the second
terms.
Ex. 1. (x + 7) (x + 5) = aj'-^ + (7 + 5) X + 7 X 5
= a;2 + 12 X + 35.
Ex. 2. (a: - 7) (a; - 5) = [X + ( - 7)] [a: + ( - 5)]
= x2 + (-7-5)x+(-7)(-5)
= x2 - 12 X + 35.
Ex.3. (a; + 7)(x-5)=x2 + (7-5)a; + 7(-5)
= a;2 + 2 a: - 35.
Ex.4. (x-7)(a; + 5)=a;2_2x-35.
Exercise 44.
Write each of the following indicated products :
1. (x4-8)(a; + 5). 4. (a; - 4) (aj + 11).
2. (a; - 3) (a; + 10). 5. (a + 9) (a -5).
3. (a; + 7)(a;-9). 6. (a-8)(a + 4).
POWERS, PRODUCTS, QUOTIENTS 115
7. (a -6) (a + 13). 9. (a - 9 6) (a - 8 6).
8. (x-3a)(x + 2d). 10. (3x -2y){Sx -\- y).
{Sx-2y)iSx + y) = (Sxy-^(-2ij + y)Sx-^(i-2y)y
= 9x^-3xy-2y-2.
11. (a -5 6) (a + 10 6). 17. {xy - 7 ab) (xy - 2 ab).
12. (x'-6)(a^ + 4:). 18. (x-4:ab)(x-h5ab).
13. (a2 + 2a;)(a2-5ic). 19. (xz - 9 ab) {xz -\- 11 ab).
14. (a;?/ - 9) (a.-2/ + 6). 20. (a" + c) (a" - 6).
15. (xy-6ab)(^xy-\-2ab). 21. (a;''+i-3)(ic"+i-8).
16. (ab-5)(ab-\-7). 22. (ic2«-i-6-)(a.-2"-i + c-).
23. (iB2 + 42/ + 42)(a:2_5y_52)^
Regarding 4(y + 2) and - 5(?/ + z) as the second terms, we have
[x2 + 4(2/ + 5!)][x-i - 5(?/ + 2)] =x* -(y + ^)x2 - 20(?/ + z)'^.
24. (a; + 2/ + 3)(aj + ?/-5). 26. (x'-y - 9)(a;-y + 8).
25. (a + 6-7)(a + 6-8). 27. (a-4-2 6)(a+6-26).
124. Cubes of binomials. By multiplication we obtain
(fl + bf = fl^ + 3 a-6 + 3 a6- + 6^ (1)
That is, ^/ie CM6e of the sum of two numbers is the cube of
the first, jjIus three times the square of the first into the second^
plus three times the first into the square of the second, plus the
cube of the second.
Ex. 1.
(2x + 3?/)3 = (2x)3+3(2x)2(3y)+3(2x)(3y)2 + (3y)3 by (1)
= 8 x3 + 36 x^y + 54 xy'^ + 27 y\
Ex. 2.
(2 X - 3 a)3 = [(2 X) + ( - 3 a)]3
= (2 x)8 + 3(2 x)2(- 3 a) + 3(2 x)(- 3 a)2 + (- 3 a)*
= 8 x3 - 36 x2a + 64 xa^ - 27 a^.
Ex. 3.
(a - 6)8= a3 - 3 a26 + 3 ab'^ - b\
116 ELEMENTS OF ALGEBRA
Observe that when the second term has a negative numeral coeffi-
cient, each even term in the result contains an odd power of this
coefficient, and therefore has a negative numeral coefficient.
The same is true in § 120.
125. The operation of raising a number to any required
jjower is called involution.
Exercise 45.
Write out the cube of each of the following expressions :
1. aj-f-l. 6. a — 2 b. 11. 2aa^-\-7n^n.
2. 2x-j-a. 7. ax + hy. 12. i a V - | 6 Y
3. a + 3 6. 8. 2 a — 3 6c. 13. 2 a;" + 5 2/".
4. i» — 1. ^. x^-\-^a. 14. 3.'c'"?/" + al
5. 3ic — a. 10. xy — 4:ab. 15. x"b — 3ay''+\
126. Powers of sums. By multiplication we obtain
(fl-f-6y = a^ + 4a«6+ 6a'b'-{^ ^ab' + b\
{a + 6)^ = a^ + 5 a'b + 10 a'b' + 10 a'b' + 5 ab' -h b'.
{a + 6/ = a« + 6 a'b + 15 a'b' + 20 a'b' + 15 a'b' + 6 ab' + b\
The expression obtained by performing the indicated
operation in (a-\-by is called the expansion of (a + by.
Thus, the second member of each of the above identities
is the expansion of its first member.
By inspection we discover in each of these expansions
the following laws of exponents and coefficients:
(i) The exponent of a in the first term is equal to the
exponent of the binomial, and it decreases by 1 from term to
term.
(ii) The exponent of b in the second term is 1, and increa^ses
by 1 from term to term.
POWERS, PRODUCTS, QUOTIENTS 117
(iii) The coefficient of the first tenn is 1, and that of the
second term is the exponent of the binomial.
(iv) If in any term the coefficient is multiplied by a^s
exponent, and. this jwoduct is divided by 6's exponent plus 1,
the result is the coefficient of the next term.
E.g., in the expansion of (a + 6)^, from the second term 5 a*6, by
(iv), we obtain 5 x 4 -=- 2, or 10, which is the coefficient of the third
term. From the third term 10 a^'fe'^ by (iv), we obtain 10 x 3 -4- 3,
or 10, which is the coefficient of the fourth term ; and so on. It can
be proved that the above laws hold for any power of a binomial.
In the expansion of (a + 6)* there are 5 terms, each of the fourth
degree in a and 6, and the first two coefficients are repeated in inverse
order after the third term. In the expansion of (a + 6)^ there are 6
terms, each of the fifth degree in a and 6, and the first three coef-
ficients are repeated in inverse order after the third term.
Observe that in each of the above expansions :
The sum of the exponents of a and b in any terra is equal
to the exponent of tlie binomial.
The number of terms is equal to the exponent of the
binomial plus 1.
The coefficients are repeated in the inverse order after
passing the middle terra or half the number of terms, so
that the coefficients of the last half of the expansion can
be written out from the first half.
Each expansion is homogeneous in a and h.
Ex. 1. (2x+3 6)4
= (2 x)4+4(2 a;)8(3 6) +6(2 x)2(3 6)2+4(2 a;)(3 6)8+ (3 6)*
= 16 X* + 96 x36 + 216 a;262 ^ 216 x62 + 81 6*.
Ex. 2. {2x-ay
= (2 x)8 + 6(2 xy{- a) + 10(2 x)X- a^ + 10(2 a;)2(- a)3
+ 5(2x)(-a)* + (-a)6
= 32 x6 - 80 x<a + 80 xH^ - 40 x%8 + 10 xa* - a\
Ex. 3. (a - 6)6
= a6 - 6 a66 + 15 a*62 - 20 a^h^ + 15 a%^ - 6 a66 + 6«-
118 ELEMENTS OF ALGEBBA
Observe that when the second term in the binomial has a
negative numeral coefficient, each even term in the expansion
contains an odd power of this coefficient, and therefore has
a negative numeral coefficient. Thus the expansion of
(x — yy differs from that of (x + yY only in the signs
before the even terms.
Exercise 46.
Expand each of the following powers :
1. (2x + iy. 5. (2 a -3)1 9. {2m^-any.
2. (2-3 7/y. 6. {a'-\-hcy, 10. {cC'-2hy.
3. {x^ + xyy. 7. (2a; + 3)^ 11. {ceb-x'^yf.
4. {2x + af. 8. (3a-2&)«. 12. {^a-c + x'f.
Smce any polynomial can be written as a binomial, the laws in § 126
can be used to expand a power of a polynomial.
E.g.^ we can write,
(2 a - c + a:-)3= [(2 a - c) + x'^Y
= (2 a - c)3 + 3(2 a - c)2(x2) + 3(2 a - c) (a;2)2^ (.^2)3
= 8 a3 - 12 a2c + 6 ac^ - c^ + 12 a^x^ - 12 acx'^
+ 3 c2x2 + 6 rtx4 - 3 cx^ + x6.
13. (a^ + .T4-l)3. 15. (3.T2-5i» + l)'l
14. (x2_^._^2)^ 16. (2;c-3« -1-6)3.
127. Two powers are said to be like, when their exponents
are equal ; and unlike, when their exponents are unequal.
E.g.^ a"^, x^ are like powers; a^, a^, a* are unlike powers.
128. By 6 of § 6, like powers of equal numbers are equal.
Hence, like powers of identical expressions are identical.
129. By division we obtain,
(fl4 _ ^4^^ ^(a-b) = a' + d'b -j-ab' + b';
(flS _ ^5) ^(a-b) = a' + w'b + a'b' + ab^ + b\
POWERS, PRODUCTS, QUOTIENTS 119
From the above identities we infer the following theorem :
The difference of the like x>owers of any tivo numbers, as a
and b, is exactly divisible by the difference of the numbers,
taken in the same order ; the laws of exponents and coeffi-
cients in the quotients being as follows :
(i) The exponent of a in the first term is 1 less than the
exx>onent of a in the dividend^ and it decreases by 1 from term
to term.
(ii) The exponent ofb is 1 in the second term and increases
by 1 from term to term.
(iii) The numeral coefficient of each term is +1. The
number of terms is equal to the exponent of a in the dividend.
Or stated in symbols the theorem is
°"~f = a" ^+ a"--b + a"-^b'' -f ... + ab"-'-+ b"-\ (1)
a — b
where n is any positive integer.
Proof Multiplying the second member of (1) by the
divisor a — b, we obtain the dividend a" — 6".
Hence (1) is an identity (§ 83).
Ex 1 27 a'^-/>'^_(.^> «)«-?>«
3a -6 (3a)- 6
= (3a)2 -f (3a)6 + 62 = 9rt2 + 3a?) + &2.
Ex. 2.
•^?-^^^^l^^^^= (2 a)4 + (2 a)8(3 a-) + (2 a)2(3 a;)2+ 2 a(3 x)8 + (3 a-)4
(2 a) -(3 a;)
= 16 a< + 24 aH + 36 a%2 + 54 ax^ + 81 «*.
Observing that each term in the dividend is the fifth power of the
corresponding term in the divisor, we write the quotient by taking the
proper powere of 2 a and 3 x.
The quotient is homogeneous when the dividend is homogeneous.
120 ELEMENTS OF ALGEBRA
Exercise 47.
Write each of the following indicated quotients :
, ^x^-4.f ^ a^-16 ,, 8aV-3437>3
12.
3. -^ 8. ^- 13.
^^ ^^^ ^^^ g^ .^^ . ^^^
9a;^-
-4/
30^^-
-22/
81a*
-16 6«
9a2
-46«
xY-
-a«
xy-
■ a
64aV
-8a^
4a-
-2a;
l--7292/^
a*-
16
a —
■2
81 a^
'-1
3a
-1
a;^-
32
iC —
• 2
32 or'
'-1
2x
-1
243 <
:i^-32
h'
1-92/ 3a-26
130. By division, we obtain :
^^^ I (a* - 6*) -- (a + 5) = a^ - a-6 + ^^^ - h^.
{a^ 4- 53) ^ (f^ ^ 5) ^ ^2 _ «5 _^ 52.
2aV
-Ih
64a«63
-S^f
4a^6
-2x'y
ar^"+2 -
4
a;«+i -
2
^+3_
y,.
a,r.+ l_
2,..
^4n+4 _
if"
^n+1 _
r
{{a
+ 6^) -- (a -h 6) = a* - a^ft + a^ft^ _ ah^ + &*•
From the identities in (i) and (ii) we infer the two fol-
lowing theorems:
(i) Tlie difference of the like even powers of two numbers
is exactly divisible by the sum of the numbers.
(ii) The sum of the like odd powers of two numbers is
exactly divisible by the sum of the numbers.
In each quotientj the laws of exponents and the number of
terms are the same as in § 129.
The numeral coefficient of any odd term is -f- 1, and that
of any even term is — 1.
POWERS, PRODUCTS, QUOTIENTS 121
Or stated in symbols, when n is even, the theorem is
°"'^f = a"-' - a"-'b + a"-'b' + ab"-' - b"-\ (1)
a + 6
and, when n is odd, the theorem is
^"~^f = a"-^ - a"-'b + a"-'b' ab"' -h b"-\ (2)
Proof. Multiplying the second member of (1) or (2) by
the divisor a -\- b, we obtain the dividend a" — b" or a* + 6^
Hence (1) and (2) are identities (§ 83).
j,^ J 16x4-81?/^_(2a;)*-(3y)*
2x + Sy 2x + Sy
= (2 a:)8 - (2 xy(Z y) -\- (2 x) (3 y)2 - (3 y)»
= 8 x3 _ 12 x-2y -1- 18 a-y2 - 27 ys.
^^ 2. i«!^ijt21«je!^=(2a62)2_(2a&2)(6ca;3) + (6cx8)a
2 aft* + C ca:*
= 4 a'^b* - 12 aft2cx8 + 36 c'^x\
Exercise 48.
Write each of the following indicated quotients :
J 1 - a'b^ ^ x' + l jg a' + 32
' l-\-ab' ' x + l' ' a-\-2'
2aa; + 32^* * x-\-2' ' a6 + 3
Oa^^-'-lGy^ g 1±A^'. 15 729 + 8 6«
3aj3 + 42/ " * l+2a ' 9 + 2b^
^ a*b^-a^f' ^Q a^y> + 216y' ^^ a'%'^-\-S2a^
ab^-\-x^f' ' xy-\-6z ' ' a-b^'-^2x
a.y-16m« J J (W-hSc^. ^j 16xy-2o6a\
xif + 27n^' ' rt-V + 2c' * 2x^-\-4:d'
6 ?Vzil. 12 tl±l. 18 ^'^'' + ^y"-
19. Make a list of the identities proved in this chapter.
122 ELEMENTS OF ALGEBBA
' 131. The remainder theorem. The result obtained by sub-
stituting a for X in any integral expression in x is the same
as the remainder arising from dividing the expression by
X — a.
E.g., dividing 2 x^ — x^ — 6 by x — 2 until the remainder does not
contain aj, we obtain the remainder 6, and 6 is what 2 x^ — x^ — 6
equals when x — 2.
Again, dividing x^ + a^ by x — a, we obtain the remainder 2 a^,
and 2 d^ is what x^ + a^ equals when x = a.
Proof Let P denote any integral expression in x.
Divide Phj x — a until the remainder does not contain x.
Let Q denote the quotient and R the remainder ; then
P=q{x~a) + R. (1)
Let P]„ (read ^ P for x = a^) denote the value of P when
a is substituted for x.
Put a for X in (1); then, observing that Q']a{a—a) is
zero, and that R does not contain x, we have
Pla^R- (2)
132. The factor theorem. If any integral expression in x
becomes zero when a is substituted for x, the expression is
exactly divisible by x — a.
Proof From P]„ =Ri\i% 131, it follows that if P], = 0,
the remainder is zero, and the division is exact.
Ex. 1. The expression x^ — a^ becomes zero when a is put for x ;
hence x^ — a^ is exactly divisible by x — a.
Ex. 2. The expression x'^ + y^ becomes zero when — y is put for x ;
hence x'^ + y"^ is exactly divisible \)j x — {— y), or x + y.
Ex. 3. The expression a" — &« becomes zero when & is put for a ;
hence a" — &»* is exactly divisible by a — 5.
Ex. 4. When n is odd, a" + 6" becomes zero when — & is put for
a ; hence a" + 6" is exactly divisible by a — (— &), or a -\-h.
POWERS, PRODUCTS, QUOTIENTS 123
Exercise 49.
By § 132, prove that each of the following dividends is
exactly divisible by the corresponding divisor :
1. (a^_a._6)-f-(a;-3). 3. (a^ _ 14 a; - 8) -- (a; - 4).
2. (a^-2a;-15)-=-(a;-5). 4. (s^ - Sx" + 4:)^{x-{-l).
5. (2i>:*-3i>^-4:X-\-5)-h(x-l).
6. (x*-2a'a^-\-ia^x-3a*)-i-(x-a).
7. (x^-3a^x'-7a^x-6a'')-ir{x-\-a).
When n is odd, by §§ 131 and 132, prove:
8. a;" + a" is exactly divisible by x -{- a, but not by
X— a.
9. af — a" is exactly divisible by x — a, but not by
x + a.
When n is even, prove :
10. x"" — a" is exactly divisible by both x-\-a and x — a.
11. a;** -fa" is not exactly divisible by either a; + a or
X — a.
133. The following examples illustrate how the formulas
in § 129 or § 130 often aid in writing out the partial quo-
tient and the remainder, when a division is not exact.
Ex. 1. Divide a'^ -{- b^ by a - b.
Adding to the dividend zero in the form — b^ + b^, we have
a^ + fe^-a* - &2 + 2 62_^ ^ ^ ^ 2 62
a — b a — b
Ex. 2. Divide a^ + 1 by a - 1.
Adding to the dividend zero in the form — 1 + 1 , we have
a— la— 1 a— 1
Ex.3. ?i±i = 5izii±2 = ^_^. + ^_l+^_.
x-^\ x-\-\ x + l
Ex.4. «i+3^a»-8 + 11^^3 + 2« + 4+^J-
0-2
a^ + 4
x-2
a;2 + 5
x^l
0.-^ + 3
x-\
y? H- cc"
124 ELEMENTS OF ALGEBRA
Exercise 50.
Write the partial quotients and the remainders :
1. :._^. 5. t^. 9. t±L
x — 2 x-\-l
2. - ■ -- 6. ^^!^1±/. 10. •:?i_±_^.
ax — y^ X — a
x'^ -{■■ a'^ x^ — a^
X — a x-\- a
4. :^^. 8. ^^ + ^. 12. ^^±^.
X — a x-\- a x—2
Simplify each of the following expressions :
13. (l + a)2-(l + a)(l-a). 15. (a + h)\a-h)\
14. {\-xy+(l+x'){l-x'). 16. (22/-3a)2(2 2/+3ay
17. {x — a){x-^a){x^ + a^){x'^-\-a'^).
18. {x' - a-) (aj« + a«) (a;^ + a^) (a;^ + a^).
19. (a;2 - X + 1) (x" ^x + l){x'-x^ + 1).
20. (x-\-y--\- z) {x + z — y) {y -\- z — x) {x-{-y — z).
Write each of the following indicated quotients :
2j a%^'-a^y^ ^^ g^" - 64 a;^»+«
ab'^ -x?f' ' a"' - 2 a;"+^ '
22 64ft^^-729a;^» 512^ _ 729/"+^
2a2 + 3ar' ' ' b^-^-3y^'+'
23 XV + 128 2^ ^5m^l0n._32a;5n^«+5
it'?/ + 2 * a"*62«, _ 2 a;n^»i+l
2^ a^V"+&^^'" 2g a;^" + (a + &y'"+^,
CHAPTER X
FACTORS OF INTEGRAL LITERAL EXPRESSIONS
134. The problem of multiplication is ' given two or more
factors, to find their product.' The converse problem, ' given
a product, to find its factors,' is the problem of factoring.
Reread §§ 33, 117.
Certain forms of products which frequently occur are
called type-forms, as a- + 2 ab + 6? or a- — b\
135. Any monomial is readily resolved into its factors.
E.g., the factors of 5a; (a + y) are 6, x, and a -\- y.
The factors of xy are a; and y or — a; and — y ; but we usually use
the factors x and y because of their simpler form, unless there is some
special reason for using — x and — y.
Again, the factors of x^ are x and x or — x and — x ; that is, x^ is
the square of x or — a:.
136. The converse of the distributive law is
ax + bx -\- ex -\- " =(a -{- b -h c -h'--)x. (1)
Hence, anjj factor ichich is common to all the terms of a
polynomial is a factor of the j^olynomial.
Ex. 1. Factor 3 ax^ + 0 a^x - 9 a^x^.
Here .3 ax is seen tp be a factor of each term ; hence
Zax^-\-6a^x-9 a^x^ = a;(3 ax) + 2 a (3 ax) + ( - 3 a*a;*) (3 ax)
= (x-\-2a-Sa-^x^)Sax.
Hence the required factors are 3, a, x, and x + 2a — S a^7?.
In identity (1) the letters x, a, 6, c, ••• can stand for any binomial
or polynomial.
125
126 ELEMENTS OF ALGEBBA
Ex.2. ractora;(« -3 6)-2?/(a -35).
The binomial a — 3 & is a factor of each term ; hence
x{a-Zh)-2 y{a -^h) = {x-2y) {a - 3 6).
Ex.3. y -x-2a{y - x)^\{y -x)-2a{]} -X)
= {\-2a){y-x).
Ex. 4. Eactor {y - x){a^ ^h)-2(y- x)(a^ - b).
The expression = (a2 _]_ ft) (^ _ a;) _ 2 (a^ _ ft) (^ _ a;)
= [a2 + ft_2(«2_ft)](y_a;)
= (Sb-a'^){y-x).
Ex. 5. a2(n - a;) - 6-(.x - «) = a2(w -x)+ b'\n - x)
= («2+ ft'2)(^ _ x).
Exercise 51.
Factor each of the following expressions :
1. 3 a; + 3. 13. 2 a'*^/" + 6 «"+y +^
2. a^ + 5 a;. A71S. 2, a% ?/". 1 + 3 a^
3. ab-\- be. 14. aa;"*+y+^ -f bx'^+^y''+\
4. 4 a^ - 6 a^d. ^l^^s. 3^"*+^ 2/"^^ « + bxy.
5. 2ax-\-3x\ 15. 6 ?/'«+-- 3 2/^
6. 7a«-21a2?). I6. 8x2"-4a;».
7. a^-5a^22/ + 20ar^/. 17. 7a;-+i-14a^.
8. 5 aa^ - 10 a^x _ 5 a^a^l 18. x(a + 1) - y(a + 1).
9. 38a«6«-57a^6l 19. y (x - a) - x + a.
10. 3a3&-6a2&2 + 9a263. 20. y (a? - a) - (a - a^).
11. 15 a^^ - 10 a^c + 5 aU 21. 4 (a^ + 1)^ - 6 (a^ + 1).
12. H a'x - 4: a'y- 12 a'b'. 22. x(y -bf - c(b - y).
FACTORS OF EXPEESSIONS 127
137. Trinomials of the type-form a- + 2 a6 -f b\
The converse of the identity in § 120 is
a^ + 2 a6 + 6' = (a + by. (1)
That is, a trinomial, tivo of ivhose terms are the squares of
two numbers respectively, and the remaining term is twice the
product of these numbers, is equal to the, square of the sum of
these numbers.
Ex. 1. Factor 9 x2 + 24 x + 16.
9 x'^ is the square of 8 a;, 16 is the square of 4, and
24ic = 2.3x.4.
.-. 0a;2 + 24x + 16 = (3ic + 4)2. (1)
Or, 9 x2 is the square of — 3 x, 16 is the square of — 4, and
24a; = 2(-3x)(-4).
.-. 9x2-f 24x+16 = (-3x-4)2. (2)
The factors in either (1) or (2) are correct, but unless there is some
reason to the contrary we usually take the simpler factors given in (1).
Ex. 2. Factor 36 «» + 6* - 12 a^b^.
36 a* is the square of 6 a- or — 6 «-, and b* is the square of b^ or
To obtain the term — 12 a^b^ we must take either 6 a^ and — b^ or
- 6 a^ and 6- ; that is,
- 12 a'62 = 2 . 6 rt2(- 62)^ or 2(- 6 a2)ft2.
.-. 36 a* + 6* - 12 a262 = (6 rt2 _ ^2)2^ or ( - 6 a2 + 62)2.
Any polynomial which is to be factored should be first
examined for any factors common to all its terms.
Ex. 3. - 3 a^ 4- 3 a^b^ - lb a'^b^ = - 3 a3(a2 - 10 ax^ + 25 6«)
= - 3 «3(a - 5 63)2.
In identity (1), a and 6 can denote any binomial or polynomial.
Ex. 4. (x-2y)2 + 2(x-2y)(3?/-2x) + (3y-2x)2
= [(X - 2 2/) + (3 y - 2 x)]2 - (y _ x)2.
128 ELEMENTS OF ALGEBRA
Exercise 52.
Factor each of the following expressions :
1. a2 + 6a + 9. 13. 4.xY-x^-^y^.
2. ic2_|_i2ic + 36. 14. Sx--4.x'-4:.
3. a'- + 25 + 10a;. 15. xY -\- x\j -^ \ xf.
4. x^ -\- 121 - 22 X. 16. 4.a'x^-\-4:abxy-{-by.
5. a2 + 49-14a. 17. 9 a" + 25 b'^ - 30 ab.
6. a2_|.25-10a. 18. 25 a'^a^ - 30 a^b-x + 9 b\
7. l-8a; + 16ar^. 19. 25a^ + 25b' - 50a'b.
8. 4a2 + 962_12a6. 20. a^ + 25 ?>« - 10 a5^
9. 9 a* + 24 a^d^ + 16 6*. 21. ^a' + ^^b^ - ^^a^b.
10. x^-^^y^-^xy. 22. 4 a.y - 4 a;y + ar^.
11. 5a^-10a-6 + 56l 23.* (a -\- bf -\- 2 {a + b) + 1.
12. a^ - 6 a^^ + 9 afel 24. (2x-a)2-8(a-2a;)4-16.
(2 X - rt)2 - 8(a - 2 x) + 16 = (2 X - «)2 + 8(2 a; - a) + 16.
= (2x-a + 4)2.
25. (aj^4-2aj.'?/ + /)« + (a; + ?/)5l
(a;2 + 2 x?/ + ?/2)a + (x + j/) &2 = (x + ?/)2a + (x + ?/)&2.
= (ax + ay + &2)(a; + 2/).
26. x\x + 2)-\-2(x-{-2y + 2x{x + 2).
27. 7?i^ 4- 2 m?i + n^ —p {711 + n) . 29. x"" -\-2 x'^y"' + 2/^"*.
28. a{b-c)-Q)'-2bc+c'). 30. 36 a;'^+2 _ 43 ^«+i _l_ Ig ^n
138. A perfect square which contains only two different
powers of some one letter can often be reduced to the type^
form o? -{-2ab -\-W by first writing the polynomial in de-
scending powers of that letter.
FACTORS OF EXPRESSIONS 129
Ex. 1. Factor x^ -h y^ + z- -{- 2xy - 2xz - 2yz.
The expression contains only two different powers of x ; hence, we
arrange the expression in descending powers of x, as follows •
The expression = x^ + 2x(y-z)-\-(y^-\-z'^ — 2 yz)
= x^ + 2xiy-z) + (y-zy
= (X + y - zy.
We could have arranged this expression in descending powers of
y or z.
Ex. 2. Factor a* + 4 6* + 9 c* + 4 a-b^ - 0 a-c^ - 12 b^d^.
Arranging the expression in descending powers of a, we have
The expression = a* + 2a^(2 b'^ - 3 c2) + (4 6* -f 9 c* - 12 b^c^)
= a*-\-2 a2(2 b^ - 3 c^) + (2 62 - 3 c2)2
= (a2 + 2 62 _ 3 c2)2.
Ex. 3. Factor
a6 _ 2 a^ + 3 a* + 2 a\b - 1) + a^Cl - 2 6) + 2 a6 + 62.
The expression contains only two different powers of 6 ; hence, we
arrange it in descending powers of 6, as follows :
62 + 2 6(a3 _ a2 + a) + (a6 _ 2 a5 4. 3 a* - 2 a3 + a2).
This expression is a perfect square, if its last term is the square of
a' - a- + a. By § 121, we have
(a« - a2 + rt)2 = a« - 2 a^ + 3 a* - 2 rt'' -f d^.
Hence the given expression is identical with
62 + 2 6(a8 - a2 + o) + (a^ - a'^ + a)2,
or (6 + rt3 _ ct2 4. a)2.
Exercise 53.
Factor each of the following expressions :
1. c^-6c{a-\-b)-\-9(a + by.
2. a* + 6--|-4c2 + 2a6-f-'4«f + 46c.
3. 4a2 4-624-9c^ + 66c-12ac-4a6.
4. 4 a* + 6* + c-* - 2 6-(r - 4 aV + 4 a-ft^.
130 ELEMENTS OF ALGEBBA
5. a^ -\- 4: y^ + 9 z^ + 4. xy + 6 xz-^ 12 yz.
6. 25 a* + 9 &^ + 4 c^ - 12 b'c' + 20 c^a^ - 30 a'b'.
7. 6 aca^ 4- 4 5V + aV« + 9 c- - 12 ^ca^^ - 4 abx\
8 . - 6 ft^c^ 4- 9 c4 + 6^ _ 12 c^a^ -f 4 a^ + 4 a-6l
9. 6 aft^c _ 4 a^ftc + aW' + 4 a^c^ + 9 b\^ - 12 abc\
Note. The products in exercise 53 can be factored by using the
converse of § 121.
139. Trinomials of the type-form jr- 4- yojr + q.
The converse of the identity in § 123 is
X' + (a -\-b)x-^ab = (x + a) (x + b). (1)
Any trinomial in the form x' +^9a; 4- g can be written in
the form x^ -\- (a -{- b) x -{- ab and factored by (1), when we
know the two factors of q whose sum is p.
The two factors of q whose sum is p can often be found
by inspection as below :
Ex. 1. Factor x^ + 7 x -{- 12.
Here p = 7 and q = 12.
The two factors of + 12 are both +, or both — ; hence, as their
sum is + 7, both are +. The pairs of positive whole numbers whose
product is 12, are 12 and 1, 6 and 2, 4 and 3; since 4 + 3 z= 7, 3 and 4
are the two factors of 12 whose sum is 7.
.-. a;2 + 7 a; + 12 = x^ + (3 + 4)a; + 3 x 4
= (x+3)(x + 4). by (1)
Ex. 2. Factor x^-9x + 20.
The two factors of + 20 are both + or both — ; hence, as their
sum is — 9, both are — . The pairs of negative whole numbers whose
product is 20 are — 20 and — 1, — 10 and —2,-5 and — 4 ; since
(— 5)+(— 4) = — 9, — 5 and — 4 are the two factors of 20 whose sum
is -9.
,, a;2-9a; + 20= x2 +(- 5 - 4)a: + (- 5) -(-4)
= (a:-6)(x-4). by (1)
FACTORS OF EXPRFSSIONS 131
Ex. 3. Factor x^ + 6 x - 27.
The two factors of — 27 are opposite numbers ; hence, as their sum
is + 6, the positive factor is arithmetically the larger. The pairs of
whole numbers whose product is — 27, the larger arithmetically being
+ , are 27 and — 1,9 and — 3 ; since 9 + ( — 3) = 6, 9 and — 3 are the
required factors.
.-. x2 + 6 a; - 27 = x2 + (9 - 3)x + 9 . ( - 3)
= (x + 9)(x-3). by(l)
Ex. 4. Factor a^x^ - 6 ax - 84.
The two factors of — 84 are opposite numbers ; hence, as their sum
is — 5, the negative factor is arithmetically the larger. The pairs of
whole numbers whose product is — 84, the larger arithmetically being
-, are - 84 and +1, - 42 and +2, - 28 and + 3,-21 and + 4,
- 14 and + 6,-12 and 7 ; since - 12 + 7 = 5, - 12 and + 7 are the
required factors.
.-. (ax)2 - 5(ax) - 84 = (ax - 12) (ax + 7).
Ex. 5. 9 x* - 12 X - 77 = (3 x)2 - 4(3 x) - 77
= (3x-ll)(3x + 7).
Ex. 6. Factor x^ - 32 xy - 105 y^.
The two factors of — 105 y^ whose sum is — 32 y are 3 y and — 35 y.
.-. x2 - 32 xy - 105 y'^={x + 3 y)(x - 35 y).
Ex.7. 4a-a2 + 21=-(a2-4a-21)
= -(a-7)(a + 3)
= (7-a)(a + 3).
Exercise 54.
Factor each of the following expressions :
1. a^ 4- 4 a; + 3. 6. ar + 2a; — 3.
2. a^ — 4 a; -1-3. 7. a:^-fa; — 6.
3. a.-2-h9a;-|-20. 8. a^-|-4a;-5.
4. a.'2_lla;-f-18. 9. a^4-2a;-35.
5. x^-Sx + 15. 10. x'-Sx-lO.
132 ELEMENTS OF ALGEBRA
11. x-x'-i-Q. 35. 4a?2_l2ic-91.
12. a^ + 5a; + 14. 36. oc^-20xy-96y\
13. ar^ + 18i»4-72. 37. x" - 26 xy + 169 y^
14. a;-ic2-M32. 38. a^- 23 0^2/ + 132 2/'.
15. a^-5i»-84. 39. 4 a^ + 20 a??/ + 21 2/1
16. a^ + 5aj-150. 40. 9 of - 39 xy -{- 22 y\
17. a;2 _ 25 a; + 150. 41. ar' + 43 a;2/ + 390 2/^^.
18. a^2 + lla!-180. 42. a" -20 abx-{- 75 b^x\
19. a;-ar^-f-156. 43. a^ - 29 a& + 54 ftl
20. a;2_3ia; + 240. 44. 130 + 31 a;2/ + a^2/^.
21. a;2_34a;4.288. 45. a^ + 12 a6aj - 28 ?/-a.'l
22. a.-2-35a;-200. 46. .'c^ + 13 aV - 300 «^
23. ar*- 17 a; -200. 47. x^ - a'x^ - 462 a\
24. aV-21aa; + 108. 48. a;^ - a^a^ - 132 a*.
25. aV - 21 aa; + 80. 49. 143 - 24 a;a + a^a^.
26. a^^ + 21 aa; + 90. 50. 216 + 35 a? + a^.
27. a^x^ - 19 aa; + 78. 51. e5 + Sxy- xy.
28. aV + 30aa? + 225. 52. 110 - a? -a;^
29. a^x^ + 54 aa;+ 729. 53. 98-7a?-a^.
30. aV - 38 aa; + 361. 54. 380-aj-a^.
31. a^y^-5xy-24:. 55. 120 - 7 aa; - a^a^.
32. 4a:2^i2aj-55. 56. 105 + 16cy-cy.
33. 9a^ + 6a;-35. 57. (a; + 2/)^ + 6 (a; + 2/) + 8.
34. 16a^ + 8a;-15. 58. (a- 6)2+8 (a- 6) +15.
140. Trinomials of the type-form ax^ -\- bx -\-c.
Multiplying and dividing ax^ -{-bx + chj a, we obtain
aa^ + &a; + c = \_{axy + h (ax) + ac] -4- a. (1)
FACTORS OF EXPEESSIONS 133
By § 139, the trinomial in brackets can be factored by
finding the two factors of ac whose sum is b.
Ex. 1. 3 ic2 _ 16 X + 5 = [(3 a;)2 - 16(3 x) + 15] -- 3
= (3a;- 16)(3x- l)-3 §139
= (x-5)(3x-l).
Ex. 2. 6 x2 4- 32 a; - 21 = [(5 x)^ + 32(5 x) - 105] -- 5
= (5 a; + 35) (5 X - 3) ^ 5
= (a; + 7)(5a;-3).
Ex. 3. 3 x2 - 1*7 xy + 10y^ = [(3 x)^ - 17 y(3 x)+ 30 2/2] ^ 3
= {x-6y)(Sx-2y).
Exercise 55.
Factor each of the following expressions :
1. 2x^-\-3x-{-l. 13. 3a^ + 13a;-30.
2. 3a^ + 5a;-h2. 14. Qa^-{.7x-S.
3. 3a^ + 10a; + 3. 15. 3 a^x^ + 23 ax -\- U.
4. 3a^ + 8x + 4. 16. 3 a^o.-^ + 19 oa; - 14.
5. 2ar^ + 7a;H-6. 17. 6 a'x'' - 31 ax -^ 35.
6. 2a^ + llx + 5. 18. 3a^ + 41a; + 26.
7. 6ar^-{- lla; + 2. 19. 4a^ + 23a; + 15.
8. 2ic2_^3^_2. 20. 3ar_13a; + 14.
9. 4ic2 + lla;-3. 21. 2a^-5xy-3f.
10. 2xr-{-lox-S. 22. 3a^ - 17a;^ + lOyl
11. 3x^ + 7x-6. 23. 12ar'-23x?/ + 10/.
12. 2ar^ + a;-28. 24. 24 ar _ 29 a;?/ - 4 2/2.
Factor each of the following miscellaneous expressions :
25. 2a;(7i-l)-2(l-»). 28. 7 ar^ - 15 a?^ - 18 2/'.
26. c}j"' — ay'^+-+ny"'+\ 29. 5x(a—2y)—2(2y—a).
27. 9 «* + 16 6^ - 24 a2^2^ 30. y?/^ -\-f/^^xy/3.
134 ELEMENTS OF ALGEBRA
31. (a;-3)2 + 4(3-a:) + 4. 37. 12^2 + 50aj - 50. '
32. 1^2x^ + x — l. 38. aa^ + (a — 6)aj — &.
33. aaj^ + (a + 6) a? + 6. 39. x^^2xy—4:XZ—4.yz+A:z\
34. i»(a; — a)2 — 2/(a; — a). 40. S?/"'-^— 32/*"+^+42/'"+^.
35. x^-^-^xr+^-5x^^\ 41. (a + 6)2 + 5 (a + 6) -24.
36. 121 ar' + 81/ + 198 a;?/. 42. (a; -?/)'- 4 (a; - 1/) - 21.
141. Binomials of the type-form a" — 6", where /i is even.
The converse of the identity in § 122 is
a^ - 6^ = (a + b) (a - b). (1)
That is, the difference of the squares of any tivo numbers
is equal to the product of the sumi and the difference of the
numbers.
Ex. 1. 9 a%^ - 4 c2 = (3 a%^y - (2 c)2
= (3 a353 + 2 c) (3 a%^ -2 c) by (1)
The letters a and h in (1) stand for any expressions.
Ex. 2. a2-4ay + 4?/2-9c2=(a-2y)2_(3c)2
= (rt-2y + 3c)(a-2?/-3c) by (1)
Ex. 3. 9ic2+12a6-9a2-462 = (3a;)2-(3a-26)2
= (3x + 3a-2&)(3x-3a4-26).
In factoring a given expression, it may be necessary to use the same
principle two or more times in succession as below :
Ex. 4. (X2 - ?/ + ^2)2 _ 4 a;2^2
= (X2 - ?/2 + ^2 ^ 2 XZ) (X2 - ?/2 + ^2 _ 2 OJ^)
= [(x + 0)2_y2j[;(a;_^)2_y2j
= (x + ;s + y) (x + ^ - y) (a: - ^ + 2/) (x - 0 - ?/) .
Whenever n is even, a"" — 6" should be factored as the
difference of two squares.
Ex. 5. X* - «* = (a;2)2 _ (^2)2
= (X2 + a-2) (X2 - ^2)
= (x2 + a^) {X 4- a) (x - a).
FACTORS OF EXPRESSIONS 135
Exercise 56.
Factor each of the following expressions :
1. a'-9, 6. 9a2-1662. 11. Sab^-lSa\
2. 2oa'-b\ 7. 81/-9a^. 12. 108 ar^ - 3 a.-^.
3. 16-61 8. 36 ar- 49/. 13. 7a:^-28ar'.
4. a^- 9 2/2 9^ Aa-b''-9c\ 14. 32 a^- 8 arY
5. 64a.'2-49&l 10. 4a^-9a^. 15. 7xyz^-7a^f.
16. a^ + 2ab-{-b'--(^. 23. 49 ar' - 1 + 14 a;i/ + ?/2^
17. a2-2a64-&'-cl 24. a^ _ ig ^^.2 _|_ g ^^ _^ 9 ^2
18. a2_ j2_25c-c2. 25. ar^- 9 ?/'+ 10 ax + 25 a^.
19. a^-b^-{.2bc-c^. 26. 6" - a- - 4ar^ + 4oa;.
20. ar^ + 4a;i/-a=* + 4/. 27. 9 0^- 4 a^- 9 a^^. 12 aa;.
21. ar^-l + 10ca; + 25c2. 28. Actr - y- -9z^ -\-Qyz.
22. l-{-2ab-a'-b\ 29. c^ - 25a2_ 952 ^30a6.
30. a^ + b''-{-2ab-c^-(f-2cd.
31. a2 4-6--2a6-a^-/-2ajy.
32. m^-^ir-2mn-a^-b--{-2ab.
33. a2_,_,i2_2rt,i_^>2_^^2_2 5,,i.
34. 16a2 + 8aa; + ar'-26v-62-2/l
35. 9a'-\-12ab-\-4.b^-(c-{-x-2yy.
36. (a4-2' + c)2-ar-?/" + 2a;?/.
37. (x + Syy-4:f. 39. (5 a^ + 2 y)^ - (3 a; - ?/)2.
38. 9a2-(3a-56)2. 40. (2a;+a-3)2-(3-2a;)2.
41. ixa* — ^\xb^ 45. 5 — 80 a;*. 49. a^" — /".
42. |aV-fa/. 46. aV - 16 6y. 50. 9.T'*-a;"+2.
43. 16 X* — y\ 47. .^•^" — /". 51.4 ar"+3 — x'^+K
44. a* -81. 48. ar''+2_y2n-2 53. x^''+^y^-a^y\
53. a;2 4- / + «« + 2 a^ — 2 a.-2; + 2 2/2 — 16.
54. a* + 4 62 + 9c2_ 4a6 + 6ac- 126c -a:2_2a:2/_ 2,2.
136 ELEMENTS OF ALGEBRA
142. Binomials of the type-form a" — 6", where n is odd.
When n = 3, by § 129 we have
Ex. 1. 243-8 cfi = (7)3 - (2 aY
= (7 - 2 a) [72 + 7 (2 a) + (2 a)2]
= (7-2a)(49+14a + 4a2).
Ex. 2. 125 - 8 a^66 = (5)3 _ (2 a262)8
= (5 - 2 a262) [52 + 5(2 a262) + (2 a^lP')'^']
= (5 - 2 a252) (25 + 10 a262 + 4 a^M).
Ex.3. (1 -2a;)8-64x3 = (l-2a;)3-(4x)3
= (l-2x-45c)[(l-2a;)2
+ (l-2x)(4a:) + (4x)2]
= (l-6ic)(l + 12a;2).
When n = 5, by § 129 we have
a^-b'={a- b) {a' + a'b + a'b' + ab' + b').
Ex. 4. 2 a^ - 64 65 = 2 [a^ - (2 6)5]
^ 2 (a - 2 6) [a* + a3(2 b) + a2(2 5)2
+ a(2 6)8 + (2 6)4]
=2(a - 2 6) (a* + 2 a35 + 4 ^252 + 8 aft^ + 16 6*).
From identity (1) in § 129, we have
a** - 6" = (a - b) (a«-i + a^'-^b -\-a^-%^-\ h ab""-^ + &""0,
when n is any positive integer.
143. Binomials of the t3rpe-form a" -f 6", where n is odd.
When n == 3, by § 130 we have
^3 + 53= (aH-6)(a2-a6 + 62).
Ex.1. 8a:3+27?/3 = (2a;)3 + (3?/)3.
= (2 a; + 3 y)[(2 a:)2 -(2 x)(3 2/) + (3 ?/)2]
= (2 a; + 3 ?/) (4 a;2 - 6 a;?/ + 9 ?/2).
FACTORS OF EXPRESSIONS 137
When n = 5, by § 130 we have
a!^^b'=(a-\- b)(a* - a^b + a'b^ - ab^ -\- 6*).
From identity (2) in § 130 we have, when n is odd,
a" -f 6" = (a + b) (a^^^ - a'^-^b H a6"-2 + 6"-!).
Exercise 57.
Factor each of the following expressions :
1. a^-1. 9. 216-0)3. 17. 32a« + l.
2. 27 -a^. 10. 27 718 + 1. 18. a^&« + 243.
3. a'-Sb^ 11. 8a^-27a«. 19. 1024 ar"^ - 32 2/^.
4. 125-a»6\ 12. a^6*'-ay8. 20. x'-yl
5. x-^ + l. 13. 40a3-13o6^ 21. x^ -1.
6. y3 + 27. 14. 27n3-f 64c». 22. j;^ -f 128.
7. 8ac» + 64. 15. 2/^-1. 23. 1 - (x + yf.
8. 343- 8 al 16. ar^- 32. 24. x^ - f.
When n is even, x" — y» should ^rs^ be factored as the difference of
two squares (§ 141).
a^ - y« = (a;8 - y3)(a;3 + y3) § 141
= (a; - y)(a;2 + xy + y2)(x + y) (x* - xy + y^).
25. a;«-l. 28. x^f-a^b\ 31. 81aV-l.
26. a«-64. 29. a;*-16 6\ 32. a«-729 6«.
27. »«-64/. 30. 16a;*-81a^ 33. 64ic«-729/.
34. (3 + 2a)3-64. 38. aV - (6 - c)*.
35. a^-(x-{-yy. 39. af^f-(xy + iy.
36. a;^_(a_6)«. 40. 16 a^ - (?/ + 2 2)*.
37. a5«_^3._2 5y. 41. 27 aj^^ - (a + &)«.
138 ELEMENTS OF ALGEBRA
144. A trinomial of the type-form a^ + ha^b'^ + 6* can be
factored by writing it as the difference of two squares.
Note. The two factors of a* + ha'^lP' + 6* are real and equal when
A = 2 ; real and unequal when ^ < 2, and complex when ^, > 2.
In all the examples given the factors are real and unequal, but as
some of them involve surds this article and the next should be omitted
until Chapter XVII. has been studied.
Ex. 1. Factor m* + m'^n'^ + w*.
Adding mhi'^ — mhi'^, we obtain
wi* + n* + m^n^ = m'^ + w* + 2 m^n"^ - mH^
= (m2 + n^y - (mw)2
= (wi2 4- n^ + mn) {m^ + n^ — m?i).
Ex. 2. m* - 6 ni^n^ -\- n'^ = m* + # - 2 m^n^ -Sm^ri^
= (m2- w2)2 -(mny/S)^
= (m'^-n^ + 17171 ^S){m^-n^-mn V^) .
Or, m* - 5 mhi^ + w* = w* + w* + 2 m2n2 - 7 m'^n^
= (m2 + n2 + mw V7)(to2+ ti^ -mn^Jl).
Ex. 3. 4 x* + 9 a* - 21 a2ic2 = 4x* + 9 a* - 12 a2x2 - 9a2x2
= (2x2-3a2)2-(3ax)2
EE (2 a;2 - 3 a2 + 3 ax)(2 ic2-3 a2-3 ax).
Exercise 58.
"Factor each of the following expressions :
1. x^j^x' + l. 9. 25a3*-44i»2/ + 162/'.
2. .T*-3a^4-9. 10. 4x^-4a^/ + 92/^
3. x* + 9a^ + 25. 11. ^x''-12x'y'' + Uy\
4. a;* + 9fl^ + 25. 12. 16 x^ - a^/ + 2/'-
5. a;* - 11 aV + a^ 13. 25 a;* - 29 a^2/' + 4 .v'-
6. a;* + (4-c2)a^2/' + 42/'. 14. x^'-s^y^ + y^.
7. (a; + 2/y4-(a5 + 2/)' + l- 15. aj^ + a^?/' + 2/'-
8. 9aj^ + 3a^?/2 + 42/^ 16. a^+a;y + /.
FACTORS OF EXPBESSIONS 139
145. Binomials of the type-form a" -f 6", where n is even.
(i) a* + b' = a* + 2a-b^-{-b^-2d'b^
= (a' + by - (ab^2y
= (a^ + b'-\- ab-y/2) (a' + b'- aby/2).
This method can be employed whenever w is a multiple of
4, as when n is 8, 12, 16, etc.
(ii) a« + 6«=(ay + W
= {a? + b^)[a^-o?b''+b'']
= {o? + b^lia' + by - (a6 V3)T
= (a2 + b^ (a« + 6^ 4- a6 V^) («' + ?>' - a6 V3).
This method can be employed when n is even and one
of its two factors is odd, as when n is 10, 12, 14, etc.
Exercise 59.
Factor each of the following expressions :
1. x^ + 1. 5. 01? -a\ 9. 3?-\-a\
2. x^ H- cy. 6. a^ + 1. 10. a^ + 1.
3. nx^-\-a\ 7. a^ + 64. 11. a;»« + ai«.
4. a^-1. 8. af-\-cy. 12. a;" + a".
146. Perfect cubes. The converse of identity (1) in § 124 is
a^ + 3a'b-\-3 ab^ + b^ = (a-\- bf.
Hence, if the four terms of the cube of a binomial are arranged
according to the powers of some letter, their extreme terms are the
cubes of the terms of the binomial.
E.g., if 64 a» - 144 a'^b + 108 ab^ - 27 68 is a perfect cube, it is the
cube of 4 a — 3 6 ; for when its four terms are arranged in descending
powers of a, the extreme terms are the cubes of 4 a and — 3 6
respectively.
The expression is a perfect cube ; for
(4 a - 3 6)8 = 64 a^ - 144 a'^b + 108 ab^ - 27 b^
140 ELEMENTS OF ALGEBRA
If a perfect cube which contains only three different
powers of some letter is arranged according to the powers
of that letter, its factors will often become obvious.
E.g., if we arrange the expression,
a^+b^ + c^ + S a% + 3 oT-c + 3 a62 + 3 ac^ + 6 ahc + 3 62c + 3 ftc^,
according to the three different powers of a, we have
a3 + 3 a2 (6 + c) + 3 a (62 + c2 + 2 ah) + (63 + 3 62c + 3 6c2 + c^),
or «3_^3rt2(54.c) + 3a(6 + c)2+(6+c)3,
which is seen to be (a + 6 -f c)^.
Exercise 60.
Factor each of the following expressions :
1. a3 + 3a2-f-3a + l. 3. Sm^ - 12m2 -f 6m - 1.
2. a^ + 6a;2 + 12a; + 8. 4. aV - 3 a V^/^ + 3 aa;^^ - /.
5 . 64 a« + 108 ah^ - 144 a% - 27 h\
6. a;3- 24 0^2/ + 192 a?/ -512 2/3.
7. o? + 6 a^ft - 3 a^c + 12 aV - 12 a6c + 3 ac'-{- 8 6^-12 6^0
^_3^2 3^_^ 8a^-4'ar^7/2+ 2a.y4_|!.
842 J -r^ y 21
10. 24 62a.-3_36 6V + 18 6=^a;-3 62.
11. a2_|_2a6 + 4c2 + 4ac + 46c + 6».
12. 2 ax^ + 4 aa;22/2 - 4 aifz" -\- 2 az^ -\- 2 ai/ - 4: ax'z^
13. 3 6aj4 - 6 hxhj + 12 a^^, _ 12 a^^/ + 3 62/' -f 12 a6ar^.
14. a?x'-^d'x^y^-21a'xi/~21aY'
15. a^ + 3a^2/ + 3 V - 3 aa.-' J^f-^a^f-^ axy + 3 a^2/
-{-Sa'x-a^
FACTORS OF EXPRESSIONS 141
147. Summary. To factor any given expression by the
foregoing methods, the pupil should first note whether the
expression is in any one of the following forms :
(i) A sum of terms having a common factor. § 136
(ii) A perfect power. §§ 137, 138, 146
(iii) A difference of squares. § 141
(iv) The type-form
ax^-\-bx-\-c or x^ -\- j^x -\- q. §§ 139, 140
(v) The type-form a" - ?/' or a" + 6", ?i odd. §§ 142, 143
(vi) The type-form
a* -f /ia-62 + h* or a" + 6", n even. §§ 144, 145
When a factorable expression has no one of these forms,
our first aim is to reduce it to one of them. In this reduc-
tion much will in the end depend upon the ingenuity of
the student. No definite directions which are applicable to
all cases can be given. The two following devices will in
many cases prove useful :
(i) The factors of an expression will frequently become
obvious when the exjJi'ession is aii'cmged in ascending or
descending powers of one of its letters^ particularly when the
expression contains only one power of that letter.
Ex. 1. Factor ax + 6y + 6x + ay.
Arranging in powers of x, we have
ax -\- hy -\- hx + ay = {a -\- b) x ■{- {a + b) y
= (a + b)(ix + y).
Ex. 2. Factor ax^ — x — a -\- I.
Arranging in powers of a, we have
ax3 - X - a -f 1 = (x3 - 1) a - (X - 1)
= (x-l)[a(x2-Fx + l)-l]
= (x - l)(ax2 + ax -I- a - 1).
142 ELEMENTS OF ALGEBRA
Ex. 3. Factor «2 (^x-y)-\^ x^ (?/-«)+ y^ (a -x).
Arranging in powers of a, we have
the given expression =a'^(x — y)— a (x^ — y^) + xy (x — y)
= (.^-y) [«^ -(x + y)a-\- xyl
= (x-y){a-x)(a-y).
(ii) Another device consists in adding to the given, expres-
sion some form of zero ; as, y^ — y^, or —1+1.
Ex. 1. Factor x'^ - Sy^ - z"^ -2xy + ^ yz.
Arranging in descending powers of x and adding y^ _ y2^ y^^e obtain
the given expression =x^ — 2 xy + y^ — (4 y"^ -\- z"^ -\- i yz)
~{x-yy-{2y-zy
= (x - y + 2 y - z)(x - y - 2 y + z)
= (x + y - z)(x - S y i- z).
Ex. 2. Factor x^ - 3 ic + 2.
Adding — 1 + 1, we obtain
x^-Sx-\-2^(x^-l)-S(x-l)
= (x-l)(x^-]-x-]-l-S)
= ix-l)(x-l)(x + 2).
Ex. 3. Factor x^ -Sx^-{-4.
Adding x^ — x^, or putting —2x'^ — x^ for — 3 x"^, we obtain
a;3 _ 3 a;2 + 4 = a;3 - 2 x2 - x2 4- 4
= (X - 2) x2 - (x2 - 4)
= (x-2)(x2-x-2)
= (x-2)(x-2)(x + l).
Exercise 61.
Factor each of the following expressions :
1. a^ 4- a& -h oc -h be. 4. mx — my — nx -\- ny.
2. orc^ 4- acd + ahc -\-hd. 5. Sax — bx — S ay -\- by.
3. a^ H- 3 a + ac 4- 3 c. 6. 6oi^-{-Sxy — 2ax — ay.
FACTORS OF EXPRESSIONS 143
7. aaP — 3 bxy — axy + 3 by^.
8. 2 ao/*^ + 3 axy — 2 6x?/ — 3 by\
9. ama:^ + bma^ — anxy — bny\
10. ax — bx -{- by -\- cy — ex — ay.
11. a^x -\- abx -\- ac -\- dby -\- b^y -\- be.
12. iB3 + ar^_4a;-4. 23. 2a;3 - 3ar^- 2a; + 3.
13. 57^-x^-^x-\-l. 24. a:3 ^ 5^ _ ^2^, _ ^25^
14. aa?-\-b3^-\-a + b. 25. a^d^ _ a.2 - 52 4. 1.
15. aar^ + by^ + {a + b) xy. 26. 6a;^ + ay? + bx-\-a.
16. a262 4_rt2_j_ j2^ 1^ 27. x? — f-Yxz — yz.
17. a* 4- a'62 - 6 V - c^ 28. 1 + &^ - (a' + a6)ar'.
18. a'' — a — c^ -}- c. 29. aV + acd + abc + bd.
19. a^ — 52 _ (a — 6)2. 30. ac-{-bd — ad — be.
20. a2-62 + 5c-ca. 31. a<? + b(P - ad^ - be\
21. aa;^ + a.-^ + a + 1. 32. a^x — d^a; + a^?/ — b^y.
22. ar'-Sar^ + aj-S. 33. a^ar' - c'ar^ - ay 4- c-y .
34. a'xf - ay - b'j^ + by.
35. acar' — 6ca; -f ada; — bd.
36. C'tf - C^ _ ^2g3^ ^ ^2^
37. l-a6a;^ + (&-a0^-
38. a^-b^-\-(r-d^-2(ac-bd).
39. 4 a^ft^ _ (^2 _|. ^2 _ ^2)2^
40. (a'-b'^c'-cr'y-{2a^-2bdy.
41. a;^ + a:'// + a^z' + 2/2^. 43. x* — 14 a^y + y\
42. a;(a; + 2)-2/(2/ + 2)- 44. xry^ - x^z^ - fz^ + si^.
144 ELEMENTS OF ALGEBRA
45. l — 2ax — (c — a^) x^ + aca?.
46. ax{f-\-h^)-\-hy{h^-\-a^y).
47. 2ix? — 4: x^y — yrz -\-2xy^ -\-2 xyz — yH.
48. {^^^xY-2{y?^^x)-lh.
49. (a2-2a)2-2(a2-2a)-3.
50. (i«2 4-4a; + 8)2 + 3a?(a.'2-f 4a; + 8)4-2ic2.
51. a^-6aj2 + 16.
52. a)3_15aj2 4-250. 53. a^Y + 13 icV + 49 1/^
54. 36 a;Y + 3 a.y H- ?/^
55. Resolve c^ — 64 a^ — a*' 4- 64 into six factors.
56. Resolve ^ ■\-x^ — 16 ar^ — 16 into five factors.
57. Resolve 16 a;^ - 81 a;^ - 16 x"" + 81 into five factors.
58. Resolve x^ + aj^ + 64 x^ + 64 into four factors.
59. Resolve x^ + ^]f' — 8 oi^if — Sy^ into four factors.
60. Factor a~(b - c) -{- b^c - a) -{- c' (a-b).
148. Formation of equations with given roots.
The linear equation whose root is 4 is evidently a; — 4 = 0. (1)
The linear equation whose root is — 2 is evidently aj + 2 = 0. (2)
Multiplying together the corresponding members of (1) and (2), we
obtain the quadratic equation (x — 4) (x -\- 2) = 0. (3)
When x = 4:, (3) becomes the identity (4 - 4) (4 + 2) = 0.
When x = -2, (3) becomes the identity (2 - 4) (- 2 + 2) = 0.
No other value of x will render either factor in (3) equal to 0.
Hence 4 and — 2 are the two and only roots of (3).
The quadratic equation (3) therefore is equivalent to, i.e. has the
same roots as, the two linear equations (1) and (2) together.
This example illustrates the following principle :
FACTORS OF EXPRESSIONS 145
The linear equations
x — a = 0, x—b — 0, X — c = 0, "' (1)
are jointly equivalent to the equation
(x -a)(x-b){x-c)'"= 0. (2)
Proof. The root of any one of the equations in (1)
renders one of the factors in (2) zero; hence by § 74 it
satisfies (2).
Conversely each root of (2) must render one factor of its
first member zero, and hence be a root of one of the equa-
tions (1).
Moreover, equations (1) have the same number of roots
as equation (2).
Hence the linear equations (1) are jointly equivalent
to equation (2).
Ex. Form an equation whose roots are 1 , — 3, and 4.
The linear equations whose roots are 1,-3, and 4, respectively are
X - 1 = 0, a; + 3 = 0, a; - 4 = 0. (1)
By § 148 the equation which is equivalent to equations (1) is
(«-l)(a; + 3)(x-4) = 0,
or a;»-2x2-llx + 12 = 0.
Observe, (i) that the second member of each of the equa-
tions (1) and (2) is 0, (ii) that equation (2) is formed from
equations (1) by multiplying together their corresponding
members, and (iii) that equations (1) are formed from (2)
by putting each factor of its first member equal to 0.
Exercise 62.
Form the equation whose roots are
1. -f-4, +3. 5. -2, 3.
2. -4, +3. 6. -2,-3.
3. 2, a 7. -7, 4.
4. 2, - 3. 8. 1, 2, 3.
9.
1,
-2,
—
3.
10.
—
1, -
.2
3.
11.
3,
-4,
5.
12.
1,
-2,
3,
4.
146 ELEMENTS OF ALGEBRA
149. To solve a quadratic or higher equation we must find
its equivalent linear equations.
For use in solving equations the principle proved in § 148
can be stated as follows :
If one member of an equation is zero and the other member
is the product of two or more integral factors, the equations
formed by putting each of these factors equal to zero are
together equivalent to the given equation.
E.g., the equation (x — 2) (cc + 3) (x — 4) = 0
is equivalent to the three linear equations,
cc-2 = 0, x + 3 = 0, a;-4 = 0.
Ex. 1. Solve the equation x^ = 4 aj + 12. (1)
Transpose, x^ - 4 a; - 12 = 0.
Factor the first member, (x + 2) (x - 6) = 0. (2)
Equation (2) is equivalent to the two linear equations,
x + 2 = 0, x-6 = 0.
Hence, the roots of (2), or (1), are - 2 and 6.
Ex. 2. Solve the higher equation x^ + x^ = 6 x. (1)
Transpose, x^ + x^ — 6 x = 0.
Factor, x(x - 2) (x + 3) = 0. (2)
Equation (2) is equivalent to the three linear equations,
x = 0, x-2 = 0, x + 3 = 0.
Hence, the roots of (2), or (1), are 0, 2, and — 3.
Ex. 3. Solve the equation 9 x^ = 4 x. (1)
Transpose, 9 x* — 4 x = 0.
Factor, x(3 x + 2) (3 x - 2) = 0. (2)
Equation (2) is equivalent to the three linear equations,
x = 0, 3x + 2=0, 3x-2 = 0.
Hence, the roots of (2), or (1), are 0, - f, and f.
FACTORS OF EXPRESSIONS 147
Ex. 4. Solve the equation 4 a.-* + 9 = 37 x^. (1)
Transpose, 4 ac* — 37 x^ _|_ 9 - o.
Factor, (2 x - 6)(2 a; + 6)(2 x - 1)(2 x + 1)= 0. (2)
Equation (2) is equivalent to the four linear equations,
2x-G = 0, 2x + 6 = 0, 2x-l=:0, 2x+l=0.
Hence the roots of (2) or (1) are 3, — 3, I, and - ^.
These examples illustrate the following rule for solving a
quadratic or higher equation in one unknown :
Transpose all the terms to one member.
Resolve this member into its linear factors in the unknown.
Solve the equations formed by equating to zero each of these
linear factors.
The problem of solving an equation is the converse to that
of fanning an equation with given roots.
If we multiply together the corresponding members of equations
X - 3 = 1 and X + 3 = 16, (1)
we obtain x"2 - 9 = 16, or x^ - 25 = 0. (2)
The roots of equations (1) are 4 and 13, and the roots of (2) are 6
and — 6.
Hence both roots of equations (1) are lost by multiplying together
their corresponding members.
Putting equations (1) in the form
X - 4 = 0 and x - 13 = 0,
and then multiplying them together, we obtain an equation equivalent
to equations (1).
This illustrates the importance of the form of the equations in § 148
Exercise 63.
Solve each of the following equations :
1. ar'-7if = 0. 4. x^-\-12x = -35.
2. x- + 9x = 0. 5. x^ = 6x + 91.
3. ic2 = 4a;-hl2. 6. ar2 + 12 = 7a;.
148 ELEMENTS OF ALGEBRA
7. a.'2 + 20 = 12a;. 33. 2x^-Sa^ = 5ax.
8. x' + 20 = 9x. 34. 12x^ + 3a' = lSax.
9. a.'2 + 28 = 11 ic. 35. 132.^2 + 0^ = 1.
10. x^-\-150 = 25x. 36. x^ + G00a^ = -^9ax.
11. 3a;2 = 10a.'-3. 37. or"^ - 3 a;^ = 10 a;.
12. 3aj2_^lla^ = 20. 38. 16 .^'3 -f- 3 a; = 16 «l
13. 4a;2 + 21a^ = 18. 39. 110 aj^ + a; = 21 icl
14. 3ar'-2a; = 96. 40. 5a;3 = 8a^ + 21 a;.
15. 15a^ + 4a' = 3. 41. 32 a; - 3 a;» = 10 a;*.
16. 6ar^-7a; = 3. 42. x^ -\- 2 a^x = 3 aa^.
17. 19a; = 4 -5a;2. 43. a."^-a^H-9 = 9a;.
18. 5x^-4.x = S3. 44. .T3 + 2a^-16a; = 32.
19. a;2 _^aa; = 42 a^. 45. x^-26x^-\-25 = 0.
20. a;2 _ 20 aa; = 96 a^ 46. aj* + 36 = 13a^.
21. 8a;2 + a.' = 30. 47. 36«* + 1 = 13a;2.
22. a; + 22 = 6a;l 48. a;<- aV + 4aV = 4cV.
23. 21-fa; = 2ar^. 49. a^ + 2a2 = 3aa;.
24. 3x^ + 3o = 22x. 50. 3562 = 9a^ + 66a;.
25. 6 aj2^ 55 a; = 50. 51. a^-2aa; + 4a6 = 26a!.
26. 6a;2_^6 = 13a;. 52. 3a^ - 2aa; - &a; = 0.
27. 19af'-39a; = -2. 53. a^-2aa; + 8a; = 16a.
28. 16x^-2ax = a\ 54. 36 a;^ - 35 6^ = 12 6a;.
29. 17a;2 + 8 = 70a;. 55. x" -{-2(b -e)x-^c' = 2bc.
30. 21a:2^-|^()^^_;^ ^Q x^-2(a-b)x-\-b^ = 2ab.
31. Gx'^llkx-^Tl^. 57. (a-xy+(x-by=^(a-by,
32. a.-^ - 23 aa; = - 132 a^. 58. a^ -]- x^ = 4: x -\- 4:.
FACTORS OF EXPRESSIONS 149
69. 5x'^ — x^ = ox — l. 65. bx^ + ax^ = bx -^ a.
60. x^ — x = (^ — c. 66. x^—Sx^z=4:X — 12.
61. x'-b'-^cx-bc. 67. x* -\- 36 = 13 x".
62. 2ic3_3^2^2a;-3. 68. 4x^ + 9=133^.
63. x^i-bx^ = a^x + a'b. 69. a;^ _^ o a:^ ^ ;l^g ^ ^ 32,
64. x'-\-5 = 5x^ + x. 70. Oa:^ + 27.^^ = .t- + 3.
71. Find two numbers one of which is three times the
other and whose product is 243.
72. Find two numbers whose sum is 18 and whose product
is 77.
73. A certain number is subtracted from 36, and the same
number is also subtracted from 30 ; and the product of the
remainders is 891. Find the number.
74. A rectangular court is 10 rods longer than it is
broad; its area is 375 square rods. Find its length and
breadth.
75. How many children are there in a family, when
eleven times the number is greater by five than twice the
square of the number ?
76. Eleven times the number of yards in the length of a
rod is greater by five than twice the square of the number
of yards. How long is the rod ?
77. The square of the number of dollars a man possesses
is greater by 1000 than thirty times the number. How
much is the man worth ?
Ans. The man may have $ 50 or he may owe f 20.
78. Find two numbers the sum of whose squares is 74,
and whose sum is 12.
CHAPTER XI
HIGHEST COMMON FACTORS AND LOWEST COMMON
MULTIPLES
150. A common factor of two or more expressions is an
expression which will exactly divide each of them.
E.g., a — X is a common factor of 6 (a — x) and a^ — x^.
151. Two or more expressions are said to be prime to one
another, when they have no common integral factor except 1.
E.g., xy and vz, 3 aP'h and 7 c^, or x^ + y'^ and x^ — y"^, are prime to
each other.
152. The highest common factor (H. C. F.) of two or more
integral literal expressions is the expression of highest
degree which will exactly divide each of them.
The numeral factor of the H. C. F. is the greatest common
measure (G. C. M.) of the numeral factors of the given
expressions.
E.g., y?y'^z^ is the H. C. F. of x^^V and x'^y^z^.
Again, 10 x^yz is the H. C. F. of 20 x'^yz and 30 x^y'^z^.
153. H. C. F. by factoring.
Ex. 1. Find the H. C. F. of 6 aP'hH^d^, 4 aHH, and 8 a%&d^.
The H. C. F. of these expressions cannot contain a higher power
of a than a?-, a higher power of c than <fi, and a higher power of d
than d ; and the G. C. M. of the numeral factors is 2.
Hence the H. C. F. of these expressions is 2 aP^c^d.
Observe that the power of each base in the H. C. F. is
the lowest power to which it occurs in any of the given
expressions.
150
HIGHEST COMMON FACTORS 151
Ex. 2. Find the H. C. F. of a*b-^ - a^b* and a*b^ + a^b*.
a^b^ - a%^ = a%^{a + b)(a-b);
a^b^-\-a^b* = a%^(a-\- b).
.'. H.C. F. = a-'62(rt + 6).
Ex.3. Find the U. C.F.olS a* + \oa^b -12 a^b-2, 6 a^ -SO a'^b -\-S6 ab^
and 8 65 _ 16 a*b - 24 a^'^.
3 a* + 15 a^b - 72 aHi^ = S - a\a 4- 8 6) (a - 3 6) ;
6 a3 - 3 a25 + 36a&2 = 6 • a(« - 2 6)(a - 3 ft);
8 a^ - 16 a45 _ 24 a^P- = 8 • a3(a + &) (a - 3 6).
.-. H.C. F.^rt(a-36).
Hence, to obtain the H. C. F. of two or more expressions,
ivejind the product of their common factors^ each to the lowest
power to which it occurs in any of them.
Exercise 64.
Find the H. C. F. of each of the following expressions :
1. al)^, a%. 10. a;"?/"'-\ a;"-'?/"-^^ a;"+'?/"'.
2. a*6^ a% ab\ 11. x" -{- jf\ x" - f.
3. a%x^, ab'^x^, a^b^x. 12. a^ - 27, 9 - al
4. 3 a*, 2a^ 4 a', a\ 13. a* - ?/S (a- -f /)'•
5. lOa;^ 15ar*, 5. 14. a'' — b'^, ax — bx.
6. lOd^y^, 20a;/e^ 30ary. 15. x'-l,x^-l.
7. 3 3^2/2^, l5 icyV, 10 a^/. 16. a*^ + 8, a' - a - 6.
8. 35 ay, 20 ay, 15 a^/a. 17. a- -{- ab, a" -{- b\
9. a^y-, a^^'»-\ a?y"+\ 18. a'-+3a;+2, a^+6a;-|-8.
19. a."^ + 1, a^ + aar + aa; + 1.
20. a;* + 7 3^2 + 12, a;* + 6 .r^ + 8.
21. 2&3 -h 3xhj + 2 xy% x' + 6 ar^ + 8 3^y\
152 ELEMENTS OF ALGEBRA
22. 3 a^ - 4 ab -^b^Aa'-B a^b + a'b''.
23. a^ — a^x, o? — ay?, a* — aa^.
24. a?-l,x^-}-l,x^-2x- 3.
25. 2 a^ - 7 X' 4- 3, 3 a;2 - 7 a.- - 6, 4 a^ - 17 » + 15.
26. 12a;2 + a;-l, lo x'' -\- S x -^ 1, Ga^^ + llaj + S.
27. 2a;2 + 9a; + 4, 2a;2 + lla?4-5, 2a^-3a;-2.
28. a'x - a?bx - 6 ab\ a'bx^ - 4 a6V + 3 6'V.
29. a^ — a^i/^ ^ + ^V + ^2/ + 2/^? ^^ — 2/^-
154. The polynomial factor of the H. C. F. of two expres-
sions can always be found by a process analogous to that
employed in arithmetic to find the G. C. M. of two numbers.
This process depends upon the two following principles :
(i) If one integral expression is exactly divisible by another,
the second is the H. C. F. of the two expressions.
E.g., (x^ — y^) -^{x^ + xy + y'^) = x — y ; hence by definition x^ +
xy -f y'^ is the H. C. F. of x^ — y^ and x^ -+- xy + y^.
(ii) If one integral expression is divided by another (of
the same or lower degree in the letter of arrangement), and
if there is a remainder, the H. C. F. of this remainder and the
divisor is the H. C.F. of the first tivo expressions.
E.g., the remainder obtained by dividing the expression
a;3_2a;2_5x + 6, or (x - l){x + 2){x -S), (1)
by. x'^-Sx-h 2, or (x - l)(x - 2), (2)
is . _ 4 a: + 4, or -4(a:-l). (3)
The H. C. F. of the remainder (3) and the divisor (2) is evidently
the same as the H. C. F. of the two expressions (1) and (2).
Proof of (ii). Let A and B denote any two integral literal
expressions arranged in descending powers of some common
letter, the degree of B not being higher than that of A.
HIGHEST COMMON FACTORS 153
Let Q be the quotient and R the remainder obtained by
dividing Ahy B\
then A = BQ+R. (1)
From (1) R = A- BQ. (2)
Every factor common to B and i2 is by § 136 a factor of
BQ -{- R, OT A] hence every factor common to B and R is
common to A and B.
Again, every factor common to A and B is by § 136 a
factor of A — BQy or R; hence every factor common to A
and B is common to B and R.
Hence, the H. C. F. of B and R is the H. C. F. of A and B.
The following example will illustrate the use of principles (i) and
(ii) in finding the H. C. F. of two expressions :
Ex. 1. Find the H. C. F. of x^ + a;' - 2 and 2i? + 2x^-Z.
Dividing x^ + 2 a;^ — 3 by x^ + x^ — 2 we obtain the remainder
a;2 - 1.
Hence, by (ii), the H. C. F of the remainder x^ — \ and the divisor
x* + x2 — 2 is the H. C. F. of the two given expressions.
Dividing x^ + x^ — 2 by x^ — 1 we obtain the second remainder
X - 1. Hence, by (ii), the H. C. F. of the second remainder x — 1, and
the second divisor x^ — 1 is the H. C. F. of x* + x'* — 2 and x* — 1,
and therefore the H. C. F. of the two given expressions.
But x* — 1 is exactly divisible by x — 1 ; hence, by (i), x — 1 is
the H. C. F. of x^ — 1 and x — 1, and therefore, by (ii), of the two
given expressions.
The work can be arranged as below :
a;8 + x2 - 2) x8 + 2 xa - 3 (1
X3 + X2 - 2
• X2 - 1) X8 + X2 - 2 (X + 1
X* -X
x2 + X - 2
X2 -1
X _ 1) x2 - 1 (X + 1
154 ELEMENTS OF ALGEBRA
Before employing the method given above, all monomial
factors should be removed from the given polynomials, and
the H. C. F. of these monomial factors found by factoring.
Ex. 2. Find the H. C. F. of
3 0%* + 3 d^x^ - 6 c^x and 6 cx^ + 12 ex* - 18 cx^.
3 c%4 + 3 c2x3 - 6 c% = 3 c% (x3 + x2 - 2),
and 6 c:*:^ _,. 12 ex* - 18 cx^ = 6 cx^ (x^ + 2 x2 - 3).
The H. C. F. of the monomial factors is 3 ex ; and by example 1,
the H. C. F. of the trinomial factors is x — 1.
Hence the H. C. F. of the given expressions is 3 cx(x — 1).
155. The H. C. F. of two expressions imll not he changed if
either expression is multiplied or divided by a factor which is
not a factor of the other exjjression.
Proof The factor introduced, by multiplication, into one
expression is not a factor of the other expression, and there-
fore will not be a factor of their H. C. F.
In like manner, the factor removed, by division, from one
expression is not a factor of the other expression, and there-
fore would not be a factor of their H. C. F.
The following examples illustrate how this principle frequently
simplifies the work of finding the H. C. F. of two expressions.
Ex. 1. Find the H. C. F. of
2 ax3 + 8 ax2 - 16 ax + 48 a and 4 ^2x4 _ 4 a^x^ -f 32 a2x - 32 a^.
4 aH^ - 4 a^x^ + 32 a^x - 32 a2 = 4 ^2 (^^a __ ^rj _|_ 8 x - 8),
and 2 ax3 + 8 ax2 - 16 ax + 48 a = 2 a (x3 + 4x2 - 8 X + 24).
The H. C. F. of the monomial factors is 2 a.
To find the H. C. F. of the polynomial factors we arrange each
expression in descending powers of x and proceed as in § 154.
x3 + 4 x2 - 8 X + 24) X* - x3 + 8 X - 8 (x - 5
xi + 4x3- 8x24-24x
-5x3+ 8x2 -16 X- 8
- 5 x3 - 20 x2 + 40 X - 120
28 x2 - 56 X + 112 = 28 (x2 - 2 X + 4)
HIGHEST COMMON FACTORS 155
By § 155 we reject the monomial factor 28, and continue the process
with x2 — 2 X + 4 as tlie second divisor.
x2 - 2 X + 4) x3 + 4 X--2 - 8 X + 24 (x + 6
x8 - 2 x2 + 4 X
6x2-12x + 24
6 x2 - 12 X + 24
By (i), x2 - 2 X 4- 4 is the H. C. F. of the first divisor and the first
remainder, and hence, by (ii), of the polynomial factors.
Therefore the H. C. F. sought is 2 a (x^ - 2 x + 4).
Ex, 2. Find the H. C. F. of
2 x2 - 5 X + 2 and x^ + 4 x^ - 4 x - 16.
Since 2 is not a factor of 2 x^ — 5 x + 2, we can by § 155 multiply
x** + 4 x2 — 4 X — 16 by 2, and thus avoid the inconvenience of
fractions.
The work may be written as below :
2x2-5x + 2)x8 + 4x2- 4x-16
Multiply by 2, 2
2 x8 + 8 x2 _ 8 X - 32 (x
2x8-5x2+ 2x
13x2-10x-32
Multiply by 2, 26 x2 - 20 x - 64 ( 13
26x2-65x + 26
Divide by 46, 45)45x-90
X - 2) 2 x2 - 5 X + 2 (2 X - 1
Hence the H. C. F. is x — 2.
Ex. 3. Find the H. C. F. of
2x8 + x2 - X - 2 and 3x3 - 2x2 -j- x - 2.
Multiply the last expression by 2.
2x8 + x2 - X - 2) 6x8 - 4x2 + 2x - 4 (3
6x8 + 3x2-3x-6
-7x2 + 5x + 2
156 ELEMENTS OF ALGEBBA
Multiply the first divisor by 7.
-7x2 + 5x + 2)14a;3 + 7^2 - 7a; - 14 (- 2x
17x2_3x-14
Multiply this remainder by 7,
119x2 - 21 X- 98 (-17
119x2 -85x- 34
Divide by 64, 04) 64 x - 64
X- l)-7x2+5x + 2(-7x-2
- 7 x2 + 7 X
-2x4-2
-2X-I-2
Hence the H. C. F. sought is x — 1.
In the above process of finding the H. C. F. of two inte-
gral expressions, each remainder is evidently of a lower
degree in the letter of arrangement than the preceding one.
Hence unless at some stage of the process the remainder
is zero, we must come at last to a remainder which does not
contain the letter of arrangement. In this case the given
expressions have no common polynomial factor containing
that letter; for by § 154 this last remainder contains all
the polynomial factors common to the given expressions.
156. By the foregoing principles we have the following
rule for finding the H. C. F. of two expressions :
Remove from the given expressions all monomial factors,
and set aside their II. C. F. as a factor of the required H. C. F.
Divide the expression of the higher degree arranged in
descending powers of the common letter of arrangement by the
other expression ; if both expressions are of the same degree
either can be taken as the first divisor.
Divide the first divisor by the first remainder; the second
divisor by the second remainder ; and so on, until the last
HIGHEST COMMON FACTORS 167
remainder is zero or does not contain the letter of arrange-
ment.
If the last remainder is zerOj the last divisor is the H. C. F.
sought; if the last remainder is not zero, the two expressions
have no common factor in the letter of arrangement.
Any dividend can he multiplied by any number which is not
a factor of the corresponding divisor; or any divisor can be
divided by any number which is not a factor of the correspond-
ing dividend.
157. Any factor common to three or more expressions
must be a factor of the H. C. F. of any two of them.
Hence, to find the H. C. F. of three expressions, we can
first find the H. C. F. of any two of them, and then find the
H. C. F. of this result and the third.
Ex. Find the H. C. F. of
x» + «« - a; - 1, x» + Sa;* - X - 3, and x« + iC'' - 2.
The H. C. F. of the first two expressions is x^ — 1 .
The H. C. F. of x^ - 1 and a:" + x- - 2 is a: - 1.
Hence the H. C. F. sought is x — 1.
Whenever the given expressions can be factored by inspection, their
II. C. F. should always be obtained by factoring.
Exercise 65.
Find the H. C. F. of the following expressions :
2. x^ — 5xy -^4:y% x^ — 5x^y-\-4: xif.
3. 2x'-6x + 2, 4.x^ + 12x'-x-^.
4. a^-5.T2-99.T + 40, .r» - 6 a^ - 86 a; -h 35.
5. ^^2x'-Sx-lQ, x^ + ^o?-^x-24..
6. ar'-a^-Saj-S, a^-4iB2_iia;-6.
7. cBH3ar^-8a;-24, a'» + 3a^-3a;-9.
158 ELEMENTS OF ALGEBRA
8. a^-{-3x'y-10afy% x'-3x''y + 2y^.
9. 2a2-5a + 2, 2a3-3a2-8a+.12.
10. 262_5^,^_2, 12b^-Sb^-Sb + 2.
11. a^-Ba^x-^Tax^-Sx^, a^ - 3 ax" -\- 2 a^,
12. x'^-2x^-4.x-7, x'^ + x^-3x^-x + 2.
13. r'-3a2i»-2a3^ .^-aa^--4a3.
14. 2x' + 4:a^-7x-U, 6a^-10x^-21x + 35.
15. 2a;^-2a.'3 + aj2 + 3a;-6, 4a;4-2a^+3a?-9.
16. 3aj3 + a;2+x-2, 2a^-a^-a;-3.
17. 3a^-3ax' + 2a^x-2a^, 3 a^ -i- 12 aa^ -^ 2 a^x -\- S a\
18. 3 a.*^ — 3 ay^y + a;?/^ — 2/^ 4 a^^/ — 5 .ti/^ + 2/^.
19. 12x^-15xy-{-3y'', 6a^-6a^y-^2xy^-2f.
20. 10a^ + 25aa^-5a^ 4a^4-9aa^-2a2aj-a3.
21. 6a3+13a2a;-9aa.'2-10a;«, 9a3H-12a2.r-ll aa^-lOa^.
22. 2a;* 4-9x3+140^ + 3, 2 + 9a; + 14ar^4-3a;^
23. 3a;* + 5a:3_7^_^2a; + 2, 2a;4 + 3a^- 2a;2 + 12a; + 5.
24. 2a^-lla^ + lla; + 4, 2aj*-3a^ + 7ar^ - 12a?-4.
25. 2a;* + 4a^ + 3a;2_2a._2, 3a;* + 6ar^ + 7a;^ + 2aj + 2.
26. x^ - 9 a; - 10, a^ - 7 X - 30, a^ - 11 a; + 10.
27. x2_|_^_6^ ^_2a^-a? + 2, x« + 3a^-6x-8.
28. ar^+7x-+5a;-l, x2+3x-3x^-l, 3a^+5x-+a;-l.
LOWEST COMMON MULTIPLE.
158. A common multiple of two or more integral expres-
sions is any integral expression which is exactly divisible
by each of them.
The lowest common multiple (L. C. M.) of two or more
integral literal expressions is the integral expression of
loivest degree, which is exactly divisible by each of them.
LOWEST COMMON MULTIPLE 159
The numeral factor of the L. C. M. is the least common
multiple (L. C. M.) of the numeral factors of the given
expressions.
E.g., a^b^ is the L. C. M. of a^b, ab^, and a^b^.
Again, the L. C. M. of 12 axy^z^ and 15 b^ is 60 ab^xy*^.
169. L. C. M. by factoring.
Ex. 1. Find the L. C. M. of aft^, a^bc% and ab^c^.
The L. C. M. of these expressions cannot contain a lower power of
a than a^, a lower power of b than b^, and a lower power of c than c^.
Hence, the required L. CM. is a^bV'.
Observe that the power of each base in the L. C. M. is
the highest power to which it occurs in any of the given
expressions.
When the expressions involve numeral factors, the L. C. M.
of these factors should be obtained as in Arithmetic.
Ex. 2. Find the L. CM. of x^ + 7 x + 12, x^ + 6 x + 8, and
5 x2 + 20 X + 20.
x2+7x + 12 = (x + 3)(x + 4);
x2 + 6 X + 8 = (X + 2) (x + 4) ;
6x2 + 20x4-20 = 5(x + 2)2.
.-. L. C M. = 6(x + 2)2(x + 3)(x + 4).
These examples illustrate the following rule :
To obtain the L. C. M. of two or more integral expressions,
multiply the L. C. M. of their mimerol factors by the product of
all their prime literal factors, each to the highest power to ivhich
it occurs in any one of them.
Proof. The L. C. M. by definition contains each factor
the greatest number of times that it occurs in any one of
the given expressions.
160 ELEMENTS OF ALGEBRA
Exercise 66.
Find the L. C. M. of the following expressions :
1. 4a^2/> 10a^2/^. 4. a^, x- — 3x.
2. 24a^6V, eOa^dV. 5. 21 o^, 7ar> + l).
3. 9 a'b^xy, S x^y^ 6. 6a^- 2ii-, Ga;^- 3 a;.
7. a^ + 2a;, a;2^3^_^2.
8. aj2-5a; + 4, a;2_6a; + 8.
9. a.-^ 4- 4 a; + 4, a;2 + 5 a; 4- 6.
10. a^-a;-6, a^ + i»-2, a;2_4^^3^
11. a^ + a;-20, ar^-10a;4-24, a;2-aj-30.
12. a;2_^a;-42, ar^-lla; + 30, a^ + 2aj-35.
13. 2a^ + 3.x' + l, 2a^+5a;4-2, a^ + 3a; + 2.
14. 5a;2+lla; + 2, 5a^ + 16a;4-3, a^ + 5aj + 6.
15. x^ — 7 xy -\- 12 y^, x^ — 6 xy -{- S y^, x^ — 5xy -\-6 y^.
16. 2a;2 + 3a;-2, 2a^ + 15a;-8, a^ + 10a;4-16.
17. 8a^-38x?/4-35/, 4 ar^ - a;^/ - 5 /, 2x'-5xy-7f\
160. L.C. M. by H. C. F. The L. C. M. of two expres-
sions can always be obtained by first finding their H. C. F.
Ex. 1. Find the L. C. M. ot x^ + x^ - 2 and x^ + 2 a;2 _ 3.
The H. C. F. of these expressions is found to be x — 1.
By division we find that
JC3 + ic2 - 2 = (x - 1) (x2 + 2 X + 2),
and x3 + 2x2-3 = (x-l)(x2 + 3x + 3).
Since x — 1 is the H. C. F. of the given expressions, their second
factors x2 + 2 X + 2 and x2 + 3 x + 2 have no common factor.
Hence, the required L. C. M. is
(x-l)(x2 + 2x + 2)(x2 + 3x + 3). (1)
LOWEST COMMON MULTIPLE 161
16L To find the L. C. M. of three expressions Af B, C,
we find 3f, the L. C. M. of A and B; then the L. C. M. of
M and C is the L. C. M. required.
Exercise 67.
Find the H. C. F. and L. C. M. of :
1. 2a^ + 3a;-20, 6ic«-25a^+21a;-|-10.
2. a^-15aj-f36, a^-3a^-2ar-|-6.
3. c)x^-x-2, 3a^-103^-7x-4:.
4. ix^ + x'-Ax-i, a^-{-6a^-\-llx + (i.
5. o^-x^-jx-^lo, x^ + x^-3x-^9.
6. a^-x^-\-x-\-3, x' + a^-Sx^-x-{-2.
7. x*-x^-\-Sx-S, x' + 4:X^-Sx-{-2A.
8. 6x^-{-x^-ox-2, 6x^-\-5a^-3x-2.
9. 4:a^-10x^-{-4:X + 2, 3a^ -2a^-3x-i-2.
10. x^-9xP-{-26x-24, x'-Ua^ + ATx-eO.
11. x^ — ax^ — a^x -\- a% x^+ax^~ a^x — a\
Find the L. C. M. of :
12. x^-(ya^-\-llx-6, ic^ - 9 o^ + 26 - 24,
ar^_8a^ + 19a;-12.
13. a^-5x^+9x-9, x^- x^-dx-\-9,
x*-Ax'-^12x-9.
162. The L. C. M. of two integral expressions is the product
of either expression into the quotient of the other divided by
the H. C. F. of the two expressions.
Proof Let A and B denote any two integral expressions,
H their H. C. F., and L their L. C. M.
162 ELEMENTS OF ALGEBRA
Then L by definition contains all the factors of A, and in
addition all the factors of B which are not in A ; that is,
the factors oi B-i- H.
Hence L = Ax(B-^H). (1)
163. From (1) in § 162 we obtain
AxB=LxH. (2)
That is, the product of two integral expressions is equal to
the product of their L. C. M. and their H. C. F.
CHAPTER XII
FRACTIONS
164. A fraction being an indicated quotient, the fraction
a/6 denotes that number which multiplied by the divisor
b is equal to the dividend a. Reread § 90.
E.g., - 8/4 = - 2, for - 2 x 4 = - 8.
165. Algebraic fractions. The fractions in arithmetic in-
volve only arithmetic numbers, and are called arithmetic
fractions.
In Chapter III. we used arithmetic fractions to denote
the arithmetic values of positive and negative numbers, the
quality being indicated by the sign ■•" or ~.
An algebraic fraction is one whose numerator and denomi-
nator are quality-numbers. The sign before an algebraic
fraction denotes the quality of its numeral coefficient.
Thus, — ^^^ denotes the product of — 1 and the fraction
(-4)/(+3).
166. By the law of quality in division it follows that —
Changing the quality of both the nuynerator and denominator
does not change the quality of the fraction.
Changing the quality of either the numerator or denominator
changes the quality of the fraction.
^* -8 8' h -b
But ^^ and -^^ are opposite in quality.
6 — b
163
-
-8
9
8
~9'
«_
— a
b
or --
a
-b
abc_
xyz~
,(z
a)(-6)(-
xyz
ic).
— abc
xyz
= -
(-
- abc
-x)yz
164 ELEMENTS OF ALGEBRA
167. By § 166 and the law of quality in § 48 it follows
that —
Changing the sign before a fraction and changing the
quality of either its nuynerator or denominator does not
change the quality of the term.
E.g.,
Again,
168. A fractional literal expression is an expression which
has one or more fractional literal terms.
E.g., and ax + by -\ are fractional expressions.
x-y a+b
An integral literal expression, as we have seen, denotes
any integral or fractional number; likewise a fractional
literal expression denotes any integral or fractional number.
E.g., a denotes 2, 5, 3/2, — 2/3, or any other number.
Again, when a = 6 and 6 = 2, a/b = 3 ;
when a = 12 and 6 = 3, a/b = 4 ;
when a = 3 and 6 = 5, a/b = 3/5 ; and so on.
169. A proper literal fraction is a fraction whose numer-
ator is of a lower degree than its denominator in a common
letter of arrangement.
An improper literal fraction is a fraction whose numerator
is of the same or of a higher degree than its denominator in
a common letter of arrangement.
1 J* 4- 1
E.g., — = — and "-^-^ are wroper literal fractions.
While — - — and — ^— t are improper literal fractions.
x-2 + 1 x2 + 3x-4
The value of a proper literal fraction may be either less or greater
than 1.
FRACTIONS 166
REDUCTION OF FRACTIONS.
170. To reduce an expression is to find an identical expres-
sion of some required form.
171. To reduce an improper fraction to an expression con-
taining no improper fractions,
Perform the indicated operation of division.
Sometimes the quotient can be obtained by inspection.
For examples, see §§ 129 and 133.
Exercise 68.
Reduce each of the following improper fractions to frac-
tional expressions containing no improper fractions :
, 5ar^-20a;-15 ^ -^ , ^ o
1. • 0. • y.
ox
x + 7
x + 2
x + 3*
5x-\-7
x' + a'
x-\-a
ar^ + a'^
x — a
^-a'
X + a
a^ + 16
10.
11.
12.
x* +
a'
X —
a
2x2
-7x
-1
x-3
a^-
■3x
X-
-2
Sor
+ 2x
4-1
x-3 x + 2 x-\-4:
4 g^ + 6 a6 + 9 6^ ^^ 60x^-17x^-4.-^ + 1
2a-3h ' ' 5x2 + 9x-2
172. The value of a fraction is not changed by multiplying
its numerator and denominator by the same number.
That is, a/b = am/(bm).
Proof - = - X -, m/m being a form of 1
b b m
= am/(bm). § 91
■^■' 4 4x6 20' x+y (x + y)(x + y) (x-Vy)'^'
166 ELEMENTS OF ALGEBRA
173. The value of a fraction is 7iot changed by dividing its
numerator and denominator by the same number.
That is, a/b = (a^ m)/(b ^ m).
Proof ^^'^^ = (^^^^0^^^ = ± § 172
E.g.,
b -i-m (b -i- m) m b
c -\- ex _{c + ex) -4- c _ 1 ■\-x
G + cy~ (c -\- cy) -^ c~ 1 -^ y
174. A fraction is said to be in its lowest terms Avhen its
numetator and denominator have no common factor.
175. To reduce a fraction to its lowest terms,
Divide its numerator and denominator by all their common
factors, or by their H. C. F. (§ 173).
Ex. 1. Reduce ^^ ^ to its lowest terms.
8 a'^xy^
The H. C. F. of the numerator and denominator is 4 axy^ ; and
4 ax^j^ _ 4 ax^y^ -^ 4 axy^ _ x^
8 a%y* ~ 8 a^xy^ -^ 4 axy^ ~ 2 ay^'
Ex. 2. Reduce £JZ^ to its lowest terms.
Factoring numerator and denominator, we obtain
a'^ — ax_ a(a — x) _ a
cfi — x'^~ {a ->t x){a — x)~ a -\- X
173
§173
Ex. 3.
X* - 1 _ x^-\
x^ — x^ — x"^ -{- X ic2(a:4 — 1) — x(x* — 1)
_ 1 _ 1
~ x^ — x~x{x — 1)
Exercise 69.
Reduce to its lowest terms each of the following fractions :
■ -ajy ■ 4aWc
FRACTIONS
167
xY^"^
3- r7-2-
— 5 ab^xifz
5 a^b*<^x}f
g Sa^bc*x^fz
4 a^b^ca^y^z^
10. «;-°^
a''* + a6
12. ^-^y".
_. a^ + 2a;
,. 4.T-16
15 2ar^-4a;\
ar* — 4 a;''
17.
20.
a-3
9-a-
18. ^'-^'.
0^—1
a;'* — a*
x^ — a^
16.
x-2
4.-x^'
15 g^ — 5 aa;
g^ - 2 oa; + a^
ar^-g''
g^ + 2g^6^ + 6^
61 -g*
l-5g + 6g»
• l_7g + 12g2
jr -\-6x — 55
l-9f-^20y*
• 1 + 62/^-552^
a:^ + :r^-2
2g a;'^ + 2a;>' + l
(a^+f)(x'+xy+y')
- ar^ — ga.*^ + 6^a; — ab^
a^ — aji^ — b^x + g6-
168 ELEMENTS OF ALGEBRA
176. When the factors of the numerator and denominator
of a fraction cannot be found by inspection, their H. C. F.
can be found by the method given in Chapter XI.
Ex. 1. Reduce ^^^ ~ '^^^^^ 23 a; -21 ^^ ^^ lowest terms.
15a;3- 38x^ -2x + 21
The H. C. r. is found to be 3 a; — 7 ; and by division we find
3x3 - 13 x2 + 23x - 21 = (x2 - 2 X + S)(Sx - 7)
15x3 - 38x2 - 2x + 21 = (5x2 - X - 3)(3x - 7).
3 x3 - 13 x2 + 23 X - 21 _ x2 - 2 X + 3
" 15x3 -38x2 -2x + 21" 5x2 -X- 3'
Exercise 70.
Reduce to the simplest form the following fractions:
a3_3a-f-2 ^ 2 0? ^ cux" -\- 4. a^x - 1 a^
!• - — :; :: — 7. r' "■
• a4 _ ^2 _ 12
o. — : r :• o*
2a^-3a2
+ 1
a« + 3a2-
-20
a>-a^-
12
4 x^ + 3 aa;2 + a
3
x''-^ax'-\-
a^x^
a*
4a^-10a;
2 + 4a
• + 2
^x'^-2y?
-3aj
+ 2
6a;^ + a^-
-5a;-
-2
a^_7aa^4_8a2a'_2a3
2a^ + 3a^ + 4a;-3
6a^ + a^_l
a;4 _ a^ _ a; _|_ 1
a,4_2a^-a^-2aj4-l
4a;* + lla^ + 25
4a;*-9a^-f30a;-25
i»4-20a.-2-15x + 4
^* 6a^-i-5a;2-3x-2* " x'*-f9a^+19ar'-9a;-20
177. Two or more fractions which have the same denomi-
nator are said to have a common denominator.
The lowest common denominator (L. C. D.) of two or more
fractions is the L. C. M. of their denominators.
E.g., the L. C. D. of the fractions — -^ and — ^-^— - is
(« _ 5)2^(j ^ ft)^ or the L. C. M. of the denominators ol^ — IP- and
(a - 6)2.
FRACTIONS 169
178. To reduce two or more fractions to identical fractions
having the L. C. D.,
Multiply both the numerator and the denominator of each
fraction by the quotient obtained by dividing their L. C. D. by
the denominator of thai fraction.
Proof The derived fractions have the L. C. D. ; and by
§ 172, each is identical with its corresponding given fraction.
Ex. 1. Reduce , — — -^ , and - — -^ to iden-
a^h{x + a) ah-\x-a) ah{x^ - d^)
tical fractions having the L. C. D.
The L. C. M. of the denominators is a'^b^{x^ - a*).
Dividing this L. C. D. by the denominator of each fraction, and
multiplying both the numerator and denominator by the quotient,
we obtain
X _xxh(x — a)_ bx(x — a)
a26(x + a) ~ a-^62(a;-2 _ a^) " a^ft^Cx^ - a^y
y -y X a(x + a)_ ay(x + a)
ab\x - a) ~ a^b'\x^ - a«) a^b^x:^ - a'^) '
and
_ z X ab _ abz
abix:'' - a?') a'^b\x^ - ««) a^b^x^ - «^)
Ex. 2. Reduce — — ;? -, — — ^ -, — — to iden-
«« - 5 a; + 6 x^-ix-\-S a;^ - 3 x + 2
tical fractions having the L. C. D.
The denominators equal
(x_3)(x-2), (x-3)(x-l), (x-2)(:«-l),
respectively. Hence their L.C. M. is (x — 3)(x — 2)(x — 1).
1 _ x-1
{X-
-3)(x-
-2) (X -
-3)(x-
-2)(x-
-1)
1
X -
-2
c^-
-3)(x-
-l)-(x-
-3)(x-
-2)(x-
-1)
1
X -
-3
(x-2)(x-l) (x-3)(x-2)(x- 1)
170 ELEMENTS OF ALGEBRA
Exercise 71.
Eeduce to identical fractions having the L. C. D. :
, 3 4 5 ^234
Ax 6a^' 12a^ * a - b' a + b' a' + b'
5a^ Sbx 1 cy — m ay ax^ xy^
Q>x^y Sfx 10 xz' ' r^' (1-a;/ (1-xf
a X O? a '^ '"'^ '*"^^
^' J • > — -»' *>• —y
X — ax — ax- — a^
ab m — n
7.
am — bm -\- an — bn 2o? — 2 ab
3 5 2
a^ + 3u; + 2' a;2 + 2a;-3' :f?^hx-^^
^ 2 a b 3a^ 5 6^
10.
a-b' 2b-2(K 4.(a'-b'')' 6(b'-a')
1 1 1
(x — a){x — by (b — x) (c — ic)' {x — c)(x — a)
ADDITION AND SUBTRACTION OF FRACTIONS.
179. The converse of the distributive law for division is
a b c _a-\-b — c
x x x~ x
Hence to add or subtract fractions,
Reduce the fractions, if they have not a C. D., to identical
fractions having the L. C. D. ; then add or subtract each nu-
merator as the sign before the fractiori directs, and write the
result over the L. C. D.
h c__hy cx_by + cx
and
X y xy xy xy
a b _ax bc_ax — be
c x" ex cx~ ex
FRACTIONS 171
Note. The student should remember that when either the numer-
ator or the denominator is a polynomial, the horizontal line in a frac-
tion is a sign of grouping as well as a sign of division.
Ex. 1. Combine and simplify 1-
The L. C. M. of the denominators is {x — y)(x-\- y); and
x-y x-\-y (x-y)(a: + y) (x-y){x + y)
_.a; + y-Ka;-y)_ 2x
1 1
Ex. 2. Combine and simplify
a;2-5x+ 6 = (a;-2)(x-3),
x^~lx^ 12 = (x-3)(x-4);
hence, the L. C. M. of the denominators is (x — 2)(x — 3)(a; — 4), and
the expression = ^-^^^ ^ ~
(x_2)(x-3)(x-4) (x-2)(x-3)(x-4)
_ (x-4)-(x-2) -2
(X - 2)(x - 3)(x - 4) - (X - 2)(x - 3)(x - 4)'
Ex. 3. Combine and simplify g^ZL^ _ «£_i^^ _ «fe - c^
he ca ah
The L. C. M. of the denominators is ahc, hence
the expression =«(«1:^M _ Hac~b') _ c(ab - c^)
dbc abc abc
_,a(a^ - be)- b(ac - b^)- c(ab - c^)
~ abc
_fl^ + b^ + c^-Sabc
~ abc
Exercise 72.
Combine and simplify :
, o, — 5b a — 3 b „ a — 36,3a — 6
1. ^' 1 z —
172 ELEMENTS OF ALGEBRA
6 a — 5 b 4 a — 76 2^~'^ q^ + ?/
3 2 ' '3 4 *
_ X x — 4:,x — 5 ^a — b „a — 6
5- T 3 1 — 8. 5 7
4 3b ex
2.T-3 a;4-2 5a; + 8
9 6 "^ 12 '
10 g — 26 a-5 6« + 7&
2 a 4 a 8 a
11 b -\- c . c-{-a a — b
2a Ab 3c
a — a;a4-ic a^ ~ a^
14. 1-
13.
X a 2 ax
2 gg - 62 62 _ g2 ^2 _ ^2
a'-^ 6^ c2
^^ 2a;-3y 3a;-2g 5
xy XZ X
15 a;-3a;^-9 8-a^
5a; 10a;2 15a^'^
16. l-^Mn^ + ^^^hl!.
a;?/ xy^ x^y^
Keduce to an improper fraction :
.1. a^ + 6^
17. a + 6 -^— •
a — 6
q + 6 a2_^ft2_^2_ 52_(^2^52)
1 a — b ~ a — b
~a — 6~6 — a
18. a-l4--4-T- 20. a;4-2?/4--^^-
a 4- 1 X — 2y
19. a + a;+-^. 21. x^ _ 3x - ^^^?^=^.
a — a; a; — 2
FRACTIONS 173
Gar*
22. a^ — 2 ax + 4:0^ —
23. x-a + y + '^'-^'y-^y"
a-^2x
HA
x-\- a
24. l + aj + ic' + a^' + r-^-
1 — a;
25. ^;+2a^ + l_ _
26. l + 2a; + 4a52 + -^-+i.
2ic — 1
ar^ + 13a;-5
27. cc2_2a; + 3-
28. Combine and simplify — ^ + ^^
a — ic a^ — a*
Beginners should always see to it that the denominators of the
fractions to be added or subtracted are all arranged in descending
powers, or all in ascending powers, of some particular letter of
arrangement.
Arranging the denominators in this example in descending powers
of a, we have
a ax _ a , — ««
a — X x^ - a^ a — X a^ — x^
_a(a + x) — ax _ a^
- rt2 _ a;'2 " a2 - x^'
Combine and simplify :
29. -^ + -*-. 33. ^- + -1-.
a — b b — a 3 + x a^—d
30. .-^_ + -«-. 34. l+?-ll^.
x — aa — x 1 — ojl-fa;
31. '^ + « . 36. ^- + — ?
32. -^ 2_. 3 a + 2b_a-2b_
1-x 1-x' a-2b a + 26
174 ELEMENTS OF ALGEBRA
x-\-y x — y ' 4y^
x — y x^y xr — y^
38. 2a ^ 26 ,C' ^h\
a-j-b a—b a? — b^
39. -^ 1- 1
a — 1 a (a — 1)
4.o?-b^ 2a-[-b
41. -^+ 1 1
a; — 1 ic — 2 ic4-2 x-\-l
The character of the denominators in this example suggests that it
is simpler first to combine the first and fourth fractions, next the
second and third, and then to combine these results, as below :
1 1 _x + l - (r<;-l)_ 2
x-\ x-\-\ a;2 _ 1 a;2 - r
^ 1 _a; + 2-(a;-2)_ 4
X - 2 a; + 2 " a:2 _ 4 "a;'^ _ 4'
and ^ . 4 ^ 2(a;2-4)+4(x2-l) ^ 6x^-12
x2-l x2-4~ (a;2_ i)(a;2_4) "a:*- 6x2 + 4'
42. ^ + ^?_+ 2 ^. 4
1— ic l + a; 1 + a^ l + ic*
Here it is simpler first to combine the first and second fractions,
next to combine this result and the third fraction, then this last result
and the fourth fraction, as below :
2 '
1-X 1+X 1-X2 1-X
2 2 _2(1 +x2)+2(l -x2)_ 4
1-X2 1 + X2~ l-X* ~1-X*
4 4 _4(1 + x*)+4(l -x4)^ 8
1 _ X* 1 + X4~ 1 - X8 1 - XS"
43.
FRACTIONS 175
2 1 ic + 6
x-2 x-\-2 ar^ + 4
44. -^ + _iL_+ 2a^
a — ic a + a; a^ + a^
45 3 -a; 3 + a; l-16a?
l-3a; l+3ic 9x2_i'
1 1 a.' + 3
46.
x-1 2(a; + l) 2(ar' + l)
.^ a , a , 2a2 , 4 a*
47. 1 r-^— — n-T—r-, — i*
a — X a-\-x a- -f ar a* -f a;*
48. ^-^+ 3 1
a; — 3 a; — 1 a; + 1 a;-|-3
49. 1 -_±,+6__4_^ 1
a; — 2 x — 1 X x-{-l x-\-2
50. , f ,+ 2 1
aj2_3a; + 2 ar^-a;-2 ar'-l
The expression
51.
52.
53.
54.
(a;_2)(a;-l) (a;-2)(x+l) (x-l){x + \)
2(x + l)+2(a;-l)-(a;-2)
(x-2)(x-l)(a;+l)
3 X + 2
(x-2)(x-l)(x+l)*
1.1
x^-dx-[-20 xr-llx-\-30
1 1
a^-
7x
4-12
a;2-
- 5a;
+ 6
1
1
2ar^
— X
-1
2x2^3..
-3
1
3
2aj2
— X
-1
6ar^
— X-
-2
4
3
55.
4-7a-2a2 3-a-lOa"
176 ELEMENTS OF ALGEBRA
56. ? ?
5 + x-lS^ 2-h5x + 2x'
g^ 5x 15(a;-l) 9(x-\-S)
2(x-\-l)(x-3) 16(x-3)(x-2) 16 (a; +1) (a; -2)'
58. ^^^+ 1^ 12
x^-{-5x + 6 ar^ + 9ic + 14 a^H-10a;4-21
69. _^ + -A_+ 40.4-2
a^-1 2a; + l 2a^ + 3ic + l
24 a; 34-2a; 3-2a;
• 9_i2a;4-4a;2 3-2a; 3 + 2a;'
61. 1 2^1
a;^ 4- 5 aa; + 6 a^ ar^ -f 4 aa; + 3 a^ a;^ 4- 3 ax -\-2a^
62. —^ "'-" + 2 ■
(a; — 2 a)^ a^ — 5 aa; 4- 6 a^ a; — 3 a
63. 1_ 4 ^ 6 _ 4 ^ 1
a a4-l a + 2 a -^3 a 4-4
1 2 1
64. — :--, : — :4-
a^_5a; + 6 x'-4,x-\-3 '.x^-3x + 2
65. ,A^-;,-V+ ^
8_8x 8 4-8a; 4 4-4a.'2 24-2a;*
1 1 1 , 18
66. tt:-:; -tt: - o . .. +
6a-18 6a4-18 a^ 4-9 a^ 4-81
6T. , 1 . + ,. ,^ ,+ 1
(a — b)(a — c) (b — c) (6 — a) (c — a) (c — 5)
In examples of this kind it is best for beginners to arrange all the
factors in the denominators of the fractions so that a precedes b or c,
and b precedes c.
We therefore change b — a into — (« — 6), c — a into — (a — c),
and c — b into —(b — c). The expression then becomes
1 1 + 1
(a - &) (« — c) (a - &) (6 - c) (a - c) (6 - c)
FRACTIONS 177
The L. C. M, of the denominators is (a — b)(a — c) (6 — c) ;
••• the expressions (a - 6)(a - c)(6 - c) -^-
68. c . a . b
91
(6-c)(6-a) (c-d)(c-b) (a-b)(a-c)
69. ? + - + ^
(y-x)(z-x) (y-z){y-x) (z-x){z-y)
70. y + z I z + x ^ x-\-y
{y-x)(z-x) (y-z)(y-x) (z-x){z-y)
MULTIPLICATION AND DIVISION OF FRACTIONS.
180. Product of fractions. See § 91.
Ex 1 a; + 2 ^ X + 3 ^^ X + 4 _ (a: + 2) (x + 3) (x + 4) _ ^
■x + 3 x + 4 x + 2 (x + 3)(x + 4)(x + 2)
Ex. 2. Simplify ^^-^^ x -M±J/L. x -^M^
The factors common to numerator and denominator can be can-
celled before the multiplication is performed, as below :
The expression =^(2-ri) x ^(^^J^ ^ W
_ 2x
~3(x + 2j/)"
181. To multiply a fraction by any number, ,
Multiply the numerator^ or divide the denominator, by that
number.
Proof. ^xm^^X?^^ §91
0 0 1 b
= a/(b-^m). §173
8 62" 8 62 " 8 62^4'
Ex. ^x4=^«iAior ^«'
182. The reciprocal of a fraction is equal to the fraction
inverted.
That is, 1 -J- {a/b) = b/a.
178 ELEMENTS OF ALGEBRA
Proof, b/a multiplied by the divisor a/b is equal to the
dividend 1 ; hence b/a is the quotient.
183. To divide by a fraction,
Multiply by the reciprocal of the fraction.
Proof. Dividing by a number gives the same result as
multiplying by its reciprocal (§ 87).
Ex 1 3« ■ ^x_Sa ^^7 y_2lay
* 56 ■ ly~bh 2x~106x'
Ex 2 ^ — « . x^ — a^ _ X — a x + a
x^ + a^ x + a x^ -^ a^ x^ — a^
\
(x2 - ax + a^) (ofi -\-ax-^ a^)
1
a;4 + a'^x'^ + a*
184. To divide a fraction by any number,
Divide the numerator, or multiply the denominator, by that
number.
Proof «^m = -x- = — §91
b b m bm
= (a-^m)/b. §173
Exercise 73.
Simplify each of the following expressions :
1. l^x'p- 6. !^X^-^'-
3 c 4a a c a
^ 2a .6 c 5x ^ G? .^ o.
"' 3b^ ha^2y ^' b^^ f ' /
^ 2a 5c x'b ^ Sa^ ^ 2c Sa
o. X — X * o. — — — X
5b X y 4.b 7ax 7bx^
^ 2 a' . Sabx ^ 5f^^^21c, . 35 (fy
be c'y ' 7 a^ 4: ax 7 a^x
3axy ^ 6aV -^ 2 b , x 9a^
' 56^ ' lOba^' ' 3a' V 4 6^*
17.
x-2''
x — 2 X-
X — S ' X -
-4
-3'
18.
a^-a'
x + 2a^
ar-4a
^ x — a
19.
a' + a^
Aa-x)\
9ft
Ux"-
7x 2x-
1
FRACTIONS 179
x^ -\-xy xy — y^
_- a^ + 2a;,,g^-3a;
XT — ^y^ X -\- y
14 _a±h_ ab--^
a^-a'b ab + a^ " 12a^ + 24:x' ' x" + 2x
^g ar> + 3a;^ . x + S ^^ 16a^-9a^ . 4a;-3a
x + 4 ' x?-\-4:X ' «2-4 ' a;-2
,. a + 46 . ah-^AW ._ a2/>2-|.3a& aft + 3
(ji-^hab a3 + 5a-6 4a2-l 2a+l
ar'-14a;-15 . a^-12a;-45
• a:2_4a._i5 ' ar^_6a;-27*
24 ar^-6a.-^ + 36a; . a^ + 216a;
iB2 _ 49 ' Q^-x-^2
y?-x-2^ , x-\-\ .a^H-2a;-8
ar'-25 ' or^-h 5a; ' ar^ - a;- 2 *
ar'-lSx + SO . a;^-15a; + 56 a;4-5
• ar*_5a;_50 * ^-^x-1 x-l
27 af^-8a;-9 a;^ - 25 . a;^ + 4a; - 5
^-Vlx^l2 x^-l ' a^-9x + S
28 a;^-8a; ^ar^ + 2a; + l .a.-' + 2a; + 4^
a;^ — 4a; — 5 ar' — a;- — 2a;* x — 5
29 (g + sy . g-^-ft^ . (a-\-by
' (a - 6)» * (g2 - 62)2 • ^2 _^ ^2
. (a - cy -b'- ' b^-(c- ay
180 ELEMENTS OF ALGEBRA
31. ^^ + / X ^ ~ ^ • ^^ ~ ^''^^ "^ ^*-
' QC^ — y^ x-\-y ' X* -\- x^y^ + y^
First reduce each of the mixed expressions to fractions.
33. (a + ^!^\(b «''
a— bj\ a + 6
34. of_±xy^r_x_ y\
\x — y x-\-y)
^ -\- y^ \^ — y ^ + Vy
35 fci + ba — b'\^/'a-{-b a — b\
\a — b a -j- bj ' \a — b a -\- bj
"■(-¥)x(3^-')*J
2 a (a — by \ ctj \ «/
38 4a^ + a;-14 ^ 4fl;^ ^ a; - 2 _^ 2a^ + 4fl;
6 a;?/ — 14 2/ a^ — 4 4fl; — 7*3ar^ — a; — 14
x^ + a;-2 a^2 + 5a; + 4 . /a;^ + 3a; + 2 .^-{-3
• aj2_^._2o'^ or^-o; * l^a.-' - 2 a^ - 15 a^ J
._ 4a^-16a; + 15 a;^-6a;-7 ^, 4ar'-l
)•
2aj2^3a; + l 2a;2_l7a; + 21 4a^-20a; + 25
a^ 4- a6 — ac (a + c)^ — 6^ ab — b^ — be
a2-62_c2_26c 62_2?,c + c2-a**
43 a;^-64 ar'-H2a;-64 . a?^-16a; + 64
ic2 4-24a;4-128 aj3_64 • a^^4,x-^16'
FRACTIONS 181
185. A complex fraction is a fraction whose numerator and
denominator, either or both, are fractional expressions.
g + 6
E.g.^ , or /(- + -), is a complex fraction.
^ _L £ c — d/\x yl ^
X y
Observe that a heavy line is drawn between the numerator and
denominator of the complex fraction.
ay
bx
Ex.1. ^/^=^xy=
b/ y b z
Ex. 2.
+ x
X
x+j,^_x_
y y + x y
Sometimes the easiest way to simplify a complex fraction
is to multiply its numerator and denominator by the L. C. M.
of the denominators of their fractional terms.
a-\-x a — x a-\-x a — x\, .. ,
; — — ; — (a - x) (a +
a — x a -{- X \a—x a + x/
._.. „^.. .„_.. .^... ^)
Ex. 3. _ ,
a + xa-x (a-\-xa-x\. .. , ,
h — ; — h — ; — ) (a - x) (a -{- x)
a — x a + x \a-x a-\-xJ^ ^^ ^
- (a + x)2 + (a-a:)2
_ 2 ax
~ a2 -1- x2*
Here (a-x)(a + x) is the L. C. M. of the denominators of the
fractional terms in the numerator and denominator of the complex
fraction.
^^4 X _\ X J _x* + a-x^ _ x2
Ex. 5. Simplify
x + 2
x + 2
X4- 1
182 ELEMENTS OF ALGEBRA
In a fraction of this kind, called a continued fraction^ we first sim-
plify the lowest complex fraction as below :
X X
X4-2 =^ {x + 'l)x §83
^ _, 2 ^ + 1 (a;+2)x-(a:+l)
X
x^ + 2x
x^-\-x-l
xCx"^ -h X- I)
X (x'^ + X - 1) - (ic2 + 2 x)
Exercise 74.
Simplify each of the following fractional expressions:
3a + — - + -
8c ^ b d
4. ^r-r' 7.
Q ,76
3c+^
8a
5. ^.
1.
71 m
a 6
m n
2.
x-y
d
3.
46
"+3
10.
12 3
^—x
X
11.
2x'-x-6
S-'
X 8.
m
X
_1
a;
l+i
a;
ar X
a;^ X OCT
a -\-b a — b
a
12. -
^ _ a^ + 6^
13.
(a + 6)^
x^'^a'
a^ ax x^
FRACTIONS
183
14. 1 -f
l-\-x
2x2
15.
16.
a —
a +
17.
18.
19.
1-
1 +a;
1
X
X
x-2-
X —
x-1
x-2
1
x-\-
m
2/4-
ic +
x +
20.
21.
22.
23.
24.
25.
x — y —
x-y-\-
xy
x-y
x + y
x-\-y
xy
x + y
l-x , l + 2aj2l
1 l-fo: 1
a;-2
1
a;-2
a;-4
X
01? — y^
^ ~ 9
x--y^
x-\-l
2 0^ + 1*
4
ic — 4
x-2
X — b
X—4: —
X— 4:
a" + h^
2 ah
^\ aW . 4a6(a-f ?>) '
|i + _^+_-^Ui i_l
V ict + by]
1 +
a + 6
^^* U W"^S A2/"^W ^/
184
27.
28.
29.
30.
6«
ELEMENTS OF ALGEBRA
a-b a'- a^b' + b*
a^-W
a'
b' ' a' + a'b' + b'
x + 2 4aH-5\ /2 a; + 3 3 a; + 4
2aj + 3 5x-{-Q>] V3i» + 4 4.x-{-6
l-\-x l^a?\ fl-\-x^ 1-^x
1 + a? l + xV Kl + x" l-\-x'
a - by
a + b
a + b
a — bj a — b J
186. Power of a fraction. The nth power of a fraction is
equal to the nth power of its numerator divided by the nth
power of its denominator ; and conversely.
That is,
Proof
{a/by = ay b".
f«Y^^l.«.«...tonfactors
\bj b b b
Ex. 1.
Ex. 2.
(-
_ aaa • - • to n factors
~ bbb ••' to n factors
= a'yb\
27 a^^c^
by notation
§91
by notation
§§ 119, 186
§§ 118, 119
(x'^-7x+ 12)2
{X - 3)-^
^IM"^^^-^^-^'- 5
x-S
Exercise 75
186
Write each of the following powers as a quotient of
products :
2a\2 ^ r_2_ax\^ ^ /2a^Y
SbfJ' ' \3bYj'
1.
'zay
3 by'
f 3 ax'y
3.
V 2 62/7 ' V 2 6cVy *
FRACTIONS 186
'■(-?)■ -MS)' "-(-sr
Simplify each of the following expressions :
(»-a/ * (a; + 2/)" ' (a' -by
20. (^^-^)* . 22. ^^-+iYx'^^^.
(a^ + a + iy V2/ + V ^' + 1
\2ab' mhjy2ab' mh)
Expand each of the following powers :
-g.!)-. »-($-^- ...e.^jj.
-i^'W -■($-?)■ -(^^o■
V6 .yy \h a) \x a yj
Factor each of the following expressions :
34. i-^+i- ' 36. %-"^+i-
y y- y by b^
35. i!i4.1^'4.25. 37. -i^-^4-4.
y y 4/ y
186 ELEMENTS OF ALGEBRA
38. 51^ + 52£ + 4.
39. •^-2+-^.
25 9^
41. -^ + 1^ + 40^ + 8.
27 3
7/^ yiQ 9 2/^ x^ 2ax o?
42. 8ar^-4;.y + |V-A. 48. _--4.___.
* 642/3 8/^ ■ ' 25a2"^5a2'^a2 c^ *
Eeduce to its lowest terms each of the following fractions :
44.
45.
4aV 9?/'
6^ c^
46.
47.
40)2 4_^4aj_
a"
51.
a;2_ 30^2/ -28^2
64.
a.'« +
x'-
x'-l
a.-«-
x« +
x'-l
55.
a^-
-a^6
- ab* + b'
a'-
a3/>-
- a'b' +
a63
52 (^' - a^) (g 4- a;) gg (x-^y ^zf-jx-y -zf
(a^-^x'){a-x) ' 3x(y''-\-2yz-\-z^)
gg (x'-f)(x-y) g^ a'-16
{^-f){x^-y^) ' a*-4a3+8a2_l6a+16
CHAPTER XIII
FRACTIONAL EQUATIONS
187. A fractional equation is an equation one or both of
whose members are fractional with respect to an unknown.
E.g., = 4 is a fractional equation in x,
2a; - 1 X
while ^ -I- ^ = ^sd: — is an integral equation in x.
We cannot speak of the degree of a fractional equation.
The term degree as defined in § 101 applies only to an
integral equation.
188. If both members of an integral equation are mxdti plied
by the same unknown integral expression M, the derived equa-
tion has all the roots of the given equation, and, in addition,
those of M=0.
E.g.., if we multiply both members of the equation
2x+l=a; + 3 (1)
by ic — 5 ; the root 5 is introduced in the derived equation.
Proof If A and B denote integral expressions in the
unknown, and we multiply both members of
A = B (1)
by any unknown integral expression M, we obtain
AM= BM, or {A - B) M= 0. (2)
By § 149, (2) is equivalent to the two equations
^-^ = 0and3/=0.
187
188 ELEMENTS OF ALGEBRA
That is, the roots of M=0 are introduced in the derived
equation (2) by multiplying both members of (1) by M.
189. If both members of a fractional equation in one un-
known are multiplied by any integral expression which is
necessary to clear the equation of fractions, the derived integral
equation will be equivalent to the given fractional equation.
3
Ex. 1. Solve the equation = 5 — x. (1)
X —\
Multiplying by ic — 1 to clear (1) of fractions, we obtain
3 = 6 X - x2 - 5.
Transpose, x2 - 6 x + 8 = 0.
Factor, (x - 2) (x - 4) = 0. (2)
No root could be lost, nor could either root of (2) be introduced by
multiplying (1) by x — 1 ; hence (2) is equivalent to (1).
Therefore, the roots of (1) are 2 and 4.
Ex. 2. Solve -^ + -^ = 5. (1)
X — 5 X — 3
Multiplying by (x — 5) (x — 3) to clear (1) of fractions we obtain
3 (X - 3) + 2 X (x - 5) = 5 (X - 5) (x - 3).
. •. x2 - 11 X + 28 =^ 0.
... (x_4)(x-7)=0. (2)
No root could be lost nor could either root of (2) be introduced by
multiplying by x — 5 or x — 3 ; hence (2) is equivalent to (1).
Therefore, the roots of (1) are 4 and 7.
Proof By transposing to the first member all the terms
of any fractional equation, adding them, and reducing the
resulting fraction to its lowest terms, we derive an equation
of the form
A/B = 0, (1)
where J. and jB have no common factors.
FRACTIONAL EQUATIONS 189
By the preceding principles of equivalent equations, the
derived equation (1) is equivalent to the given fractional
equation.
We are to prove that (1) is equivalent to the equation
^ = 0. (2)
Any root of (1) reduces A/B to 0. But when A/B is
zero, A is zero ; hence any root of (1) is a root of (2).
To prove the converse we must first prove that when
A = 0, B=^0.
If A and B could become 0 for the same value of x, as a ;
then by § 132 they would have the factor x — a in common.
But by hypothesis A and B have no common factor ; hence
when A = 0, B^O.
Hence any root of (2) reduces ^ to 0 but not B to 0.
Therefore any root of (2) reduces A/B to 0 and is a root
of (1).
Hence equations (1) and (2) are equivalent.
Ex. 3. Solve 1 - -^ = — 6. (1)
Transposing and adding the fractions, we have
r'2 — 1
1 - + 0 = 0.
X — 1
. •. 1 - (x + 1) + G = 0, or a; = 6. (2)
By §§ 105 and 100, equation (2) is equivalent to (1) ; hence 0 is
the one and only root of (!)•
But if, as would be more natural for the beginner, we should
clear equation (1) of fractions by multiplying by x — 1, we would
obtain
x-l-x2 = -l-Gx + e.
Transpose, x^ - 7 x + 0 = 0.
...(x-i)(x-e)=0, (3)
of which the roots are 1 and 0.
100 ELEMENTS OF ALGEBRA
As was shown above, to clear equation (1) of fractions it was not
necessary to multiply by x — \ ; hence multiplying (1) by x — 1 is
the same as multiplying the equivalent integral equation (2) by ic — 1.
In clearing of fractions an equation in one unknown, to
avoid introducing roots, the following suggestions should
be heeded :
(i) Fractions having a common denominator should be
combined.
(ii) Factors common to the numerator and denominator
of any fraction should be cancelled.
(iii) When multiplying by a multiple of the denomi-
nators, we should always use the L. C. M.
Ex. 4. Solve -^— + ^^ = ^i-i + ^^=-^. (1)
x — 'l X — 1 x—\ X — iS
Transpose so that each member is a difference,
x X 4- 1 _x — 8 a; — 9
05 — 2 X — \ X— 6 X — 7
Combine, ^ = ? : (2)
(x _ 2) (X - 1) (x - 7) (X - 6) ' ^ ^
Clear of fractions, x2 - 1.3 x + 42 = x2 - 3 x + 2.
. •. 10 X = 40, or X = 4. (3)
Since the root 4 could not be introduced in clearing (2) of fractions,
4 is the root of (1).
Ex. 5. Solve x-^_^x±^^x±\ ^x + 3, ^^
x+lx+7x+3x+5
Transpose, x-^ _x±l^x±^ _x±b^ ^^
x+lx+3x+5x+7
Combine, ^- = ^ (3)
(X + 1) (X + 3) (X + 5) (X + 7)
Clear of fractions,.x2 + 12x +35= x^ -f 4 x + 3.
.•.8x = -32, or x = -4.
FRACTIONAL EQUATIONS 191
Or reducing the improper fractions in (2) to mixed expressions, we
have,
x+ I x+3 x+5 x+7
1 . I ^ + ^
x-\-l x + 3 x + 5 x + 7
Combining these fractions, we obtain equation (3) above.
Since the root — 4 could not be introduced in clearing (3) of
fractions, — 4 is the root of (I).
Exercise 76.
Solve each of the following fractional equations :
, 3x-16 5
X 3
5x-5
11.
x-\-l
a;-1^2
x + 1 3
x-2 ^
2x-5
= 3. 12.
X-
-5
2x-
-2
4a;-
-5
13.
14.
5. 2£:z3 = .^_,. 15.
3a;-4 6a;-7
6. -^— = -^-2. 16.
x-\-l x -\-2
7. ^^-1-1 = 1. 17.
X -\-l X
8. 10^4^ = ._A__7. 18. ^
a — 1 » + 1
1 . 1
4a;+6 6a;+4 2£c+3
19.
12 2
10. — 1 = — = -^ — 20.
3a;-f-9 5a;H-l x+3 Sx-5 4a; + 8
^ 1
5 7
2a;+3 4a?-f6 6a;+8
4.^
X _3
a; 4-1
a;-2-^'
6x
^ -5
x-7
x-6 ^•
2x
4a; ^
x + 3
x + 7 -
1 +
a; + 4
2 3
a;-f-6 a;-h5
3
2 1
x + 1
a; + 2 a;H-3
5 1
2x + 4:
_3 1 , 1
3 1
2xH-2
=5 1 +1.
4a; + 3 4a;
2x-5
2a;-7
3a;-7
3a;-5
6a;-2
_3a; + 7
192 ELEMENTS OF ALGEBRA
21. ^±l_^Z:i? = §. 25. 3'^-^-\-2^±l = 5.
ic — lie + 3a; ic + 1 x — 1
22. ^+2_£^ = _§_. 26. 5^!^^-2^^ = 3.
x-2 x + 2 x + 1 x + 2 x + S
x-\-3 ic — 4 ic x^ — 1 aj + l 1 — ic
24. -^ = 3^^ ^. 28. -^+ 1 ^
a;-h2 i«-2 ic + l x--9 x + 3 3
29.
30.
35.
36.
37.
38.
39.
3 g; -f 5 5 ^ 8 4- 3 a;
3a;-l l--9a;*-^~l + 3;»*
1111
a; + 5 x + 6 x + 6 aj + 8
31. 1 +_!=! + 1
x + 2 x + 10 a; + 4 a + S
32. -J_ + _1_ = ^_ + _L..
if — 5 37 + 2 .T — 4 x + 1
33 ^' .a; — 9_a; + l , x — S
x—2 x—1 x—1 X— 6
g^ a;4-3 a; — 6_a; + 4 . a; — 5
a; + l a; — 4 a; + 2 x — 3
x — 3 a; — 4_a^ — 6 x — 7
a; — 4 x — b x — 1 ^ — 8
X a; + l_a; — 8 a; — 9
x — 2 a;— l~a; — 6 x — 1
x-[-iS _ x — Q _ x — 4: _ a; — 15
a; + 4 x — 1 x — 5 x — 16
x—1 _ x — 9 _ x — 13 _ x — 15_
X — 9 a; — ll~x— 15 x — ll'
x-\-3 _ x + 6 ^ x-\-2 _ a;4-5_
x4-6 x + d~x-{-5 x-i-S
FJi ACTIONAL EQUATIONS 193
a? + 2 x — 7 x-^3 __x — 0
X x — 5 x-\-l if — 4
4a;_17 10a;_13 8aj-30,5a;
a;_4 2a;-3 2a;-7 x-l
^„ 5a;-8 , 6x-44 10a;-8 x-S
x-2 x-7. x-1 x-6
^^ S0±6x_^60±8x^^^_^ 48
cc-f-l x-i-3 ic-fl
44 25-^a; 16a^ + 4i^g 23
a; + l 3a; + 2 aj + l
45. 3_4-...^' =;^+ '
46.
4-2a; 8(l-x) 2 - a; 2-2a;
60 _ 10^ 8_^
X
4 5aj-30 3a;-12
47. -i 2- = ^ ?i-.
a! + 3 K + l 2x + 6 2x + 2
48. (2x-l)(3x + 8) j^Q_
6a;(a; + 4)
In the five following examples first reduce improper fractions to
mixed expressions.
4^ 5ar-64 2 ic - 11 4a;-55 aj~6
50.
51.
52.
a; -13
X-
-6
a;-
-14
a;
-7
x — H ,
.T-4
a;
-^-
a:-7
a;-9
a;- 10 '
a;-G
x-\-b _^x
• + c
2.
a;— c
a; — 6 x-\- c x
vnx nx
53. ; 1 = m 4- w.
m 4- a; n -\- x
194 ELEMENTS OF ALGEBRA
54.
55.
x — c_ x — b 2(b — c)
x—b X — c X — b — c
m -f- r n-\- r _m -\- n -\- 2 r
x-\-2 n x-\-2m x-\-m-\-n
190. Problems which lead to fractional equations.
Prob. 1. The quotient of a certain number increased by 7 divided
by the same number diminished by 5 is 4. Find the number.
Let X = the required number.
Then by the conditions of the problem, we have
^ + 7^1
x — b
Whence a; = 9, the required number.
Prob. 2. The value of a fraction is 1/4. If its numerator is dimin-
ished by 2 and its denominator is increased by 2, the resulting fraction
will be equal to 1/9. Find the fraction.
Let X =■ the numerator of the fraction ;
then 4 X = the denominator of the fraction ;
and, by the conditions of the problem, we have
x-2 ^1
4 a; + 2 9'
Whence a; = 4, and the required fraction is 4/16.
Exercise 77.
1. The value of a fraction is 1/7. If its numerator is
increased by 5 and its denominator by 15, the resulting
fraction will be equal to 1/5. Find the fraction.
2. The sum of two numbers is 20, and the quotient of
the less divided by the greater is 1/3. Find the numbers.
3. What number added to the numerator and denominator
of the fraction 3/7 will give a fraction equal to 2/3 ?
4. What number must be added to the numerator and
subtracted from the denominator of the fraction 5/11, to
give its reciprocal?
FRACTIONAL EQUATIONS 195
5. The reciprocal of a number is equal to 7 times the
reciprocal of the sum of the number and 5. Find the number.
6. A train ran 240 miles in a certain time. If it had
run 6 miles an hour faster, it would have run 48 miles
farther in the same time. Find the rate of the train.
7. A number has three digits which increase by 2 from
right to left. The quotient of the number divided by the
sum of the digits is 48. Find the number.
8. A steamer can run 18 miles an hour in still water.
If it can run 96 miles with the current in the same time
that it can run 48 miles against the current, what is the
rate of the current ?
9. A number of men have $ 80 to divide. If $ 150 were
divided among 2 more men, each one would receive $ 5 more.
Find the number of men.
10. The circumference of the hind wheel of a wagon
exceeds the circumference of the front wheel by 4 feet. In
running 200 yards the front wheel makes 10 more revolu-
tions than the hind wheel. What is the circumference of
each wheel ?
11. A number has two digits which increase by 4 from
right to left. If the digits are interchanged and the result-
ing number is divided by the first number the quotient will
be 4/7. Find the number.
12. A train runs 10 miles farther in an hour than a man
rides on a bicycle in the same time. If it takes the man 5
hours longer to ride 352 miles than it takes the train to run
the same distance, what is the rate of the train ?
13. A tank can be filled with one pipe in 30 minutes, by
a second pipe in 40 minutes, by a third in 50 minutes. How
long will it take to nil it with them all running together ?
14. A can do a piece of work in 3^ days, B in 2i days,
C in 3| days. If A, B, and C work together, how long will
it take to do the work ?
196 ELEMENTS OF ALGEBRA
15. A cistern can be filled in 15 minutes by two pipes, A
and B, running together ; after A has been running by itself
for 5 minutes, B is also turned on, and the cistern is filled
in 13 minutes more. In what time would it be filled by
each pipe separately ?
16. A man, woman, and child could reap a field in 30
hours, the man doing half as much again as the woman, and
the woman two-thirds as much again as the child. How
many, hours would they each take to do it separately ?
17. A and B ride 100 miles from P to Q. They ride
together at a uniform rate until they are within 30 miles of
Q, when A increases his rate by 1/5 of his previous rate.
When B is within 20 miles of Q he increases his rate by
1/2 of his previous rate, and arrives at Q 10 minutes earlier
than A. At what rate did A and B first ride ?
18. A and B can reap a field together in 12 hours, A and
C in 16 hours, and A by himself in 20 hours. In what
time could B and C together reap it ? In what time could
A, B, and C together reap it ?
19. The sum of two numbers is n, and the quotient of the
less divided by the greater is a/b. Find the numbers.
20. The reciprocal of a number is n times the reciprocal
of the sum of the number and a. Find the number.
21. A train ran a miles in a certain time. If it had run
b miles an hour faster, it would have run c miles further in
the same time. Find the rate of the train.
22. A steamer can run a miles an hour in still water. If
it can run b miles with the current in the same time that it
can run c miles against the current, what is the rate of the
current ?
23. The value of a fraction is 1/a. If its numerator is
increased by m and its denominator by n, the resulting
fraction will be equal to 1/6. Find the fraction.
CHAPTER XIV
SYSTEMS OF LINEAR EQUATIONS
191. Equations in two or more unknowns. In the equation
y = 3x + 2, (1)
where x and y are both unknowns, y has one and only one
value for each value of x.
E.g., when x = l, y = 5] when x = 2, y = S -, when x = 3,
?/ = 11 ; when a; = 4, y = 14, etc.
That is, equation (1) restricts x and y to sets of values.
In like manner, any equation in two or more unknowns
restricts its unknowns to sets of values.
192. A solution of an equation in two or more unknowns
is any set of values of the unknowns which renders the
equation an identity.
E.g., if in the equation
2/ = 3a; + 2 (1)
we put 3 for x aud 11 for y, we obtain the identity
11 = 3 X 3 + 2.
Hence 3 and 11, as a set of values of x and y, is one solution of (1) ;
2 and 8 is another solution ; and so on.
Note. The word solution denotes either the process of solving or
the result obtained by solving. The word is here used in the latter
sense.
A root of an equation in one unknown is often called a solution.
197
198 ELEMENTS OF ALGEBRA
193. The degree of an integral equation in two or more
unknowns is the degree of that term which is of the highest
degree in the unknowns.
E.g., ax -^ by = 7 is R linear equation in x and y ; while ax^ -\-hy = c
or cxy + 3 ic = 2 is a quadratic equation in x and y.
194. Two equations are said to be equivalent when every
solution of each equation is a solution of the other.
195. The following principles concerning the equivalence of
equations, which have been proved for equations in one
unknown, hold true for all equations :
(i) If for any exj^ression in an equation an identical expres-
sion is substituted, the derived equation will he equivalent to
the given one (^ 105).
(ii) If identical expressioiis are added to or subtracted from
both members of an equation, the derived equation will be
equivalent to the given one (§ 106).
(iii) If both members of an equation are midtiplied or
divided by the same known expression, not denoting zero, the
derived equation ivill be equivalent to the given one
(§§ 108, 110).
(iv) If one member of an equation is zero, and the other
member is the jyroduct of two or more integral factors, the
equations formed by putting each of these factors equal to zero
are together equivalent to the given equation (§ 149).
E.g., the equation
(ic + 2?/-4)(2a;-32/+ 1) = 0
is equivalent to the two equations
x + 2i/-4 = 0 and 2a;-32^ + l=0.
(v) If both members of an integral equation are multiplied
by the same unknown integral exjyressioyi M, the derived equa-
tion has all the solutions of the given equation, and in addition
those of M=0 (§188).
SYSTEMS OF LINEAR EQUATIONS 199
Proof. If in the proofs of these principles for equations in one
unknown we substitute the word "solution" for the word "root,"
the proofs will apply to equations in any number of unknowns.
Exercise 78.
Of the following equations state which are equivalent to
the equation 2 a; + ?/ = 3, and give the reason :
1. (4 a- -f 2 2/)/2 = 3. 4. 6 a; + 3^ = 9.
2. 'Sx-\-y = x + ^. 5. (2a;4-?/)/3 = l.
3. a; + y = 3 — x. Q. i:X + oy = (S-\-y.
State to what two linear equations each of the following
quadratic equations is equivalent, and give the reason :
7. (a;-2/)(a-+2y-hl)=0. 8. {■y-x)x-^2y{x-y)=0.
Obtain ten solutions of each of the following equations :
9. 2a;-hy = 3. 10. 2a; -f 3?/ = 6. 11. 2 a; — 3y = 4.
12. How many solutions has a single equation in two
unknowns ?
13. By (iii) in § 195, show that the two equations
ax -\-by = c and a'x -\- b'y = c'
are equivalent when a' /a = b'/b = c'/c.
196. Independent equations.
Prob. If the sum of two numbers is 10 and their difference is 4,
what are the two numbers ?
Let X = the less number
and y = the greater number.
Then by the Jirst condition we have the equation
y + x = \0; (1)
and by the second condition wc have the equation
y - X = 4. (2)
200 ELEMENTS OF ALGEBRA
In (1), when x = l, y ^9; when x = 2, y = S ; when x = 3,
/ = 7, etc.
In (2), when x = 1, y = 5 ; when x = 2, ?/ = 6;^when jr = 3,
/ = 7, etc.
Hence, 3 and 7 is a set of values of x and y which will satisfy each
of the two different conditions expressed by equations (1) and (2),
and are therefore the required numbers.
Equations, like (1) and (2), which express different condi-
tions are called independent equations.
Observe that independent equations express different rela-
tions between their unknowns, while equivalent equations
express the same relation.
Any solution as x = 3, ?/ = 7, can be written briefly 3, 7,
it being understood that the value of x is written first.
197. Systems of equations. Two or more equations are
said to be simultaneous, when the unknowns are restricted
to the set or sets of values which satisfy all the equations.
A group of two or more simultaneous equations is called
a system of equations.
E.g.^ equations (1) and (2) in § 196 are simultaneous^ and form a
system of equations.
198. A solution of a system of equations is a set of values
of its unknowns which satisfies all its equations.
E.g., 3, 7 is a solution of the system of equations, (1)
and (2), in § 196.
To solve a system of equations is to find all its solutions.
199. Equivalent systems. Two systems of equations are
said to be equivalent when every solution of each system is
a solution of the other system.
E.g.^ the systems (a) and (6)
x^2y = b, (1)1
4.x- y = 2, (2)}*^'^' i(5-2y)-y = 2, (4) j
x-^2y = b, (1)1 , x = 5-2y, (3)1^^^
[ («) ... ^ . ^ ... r (ft)
SYSTEMS OF LINEAR EQUATIONS 201
are equivalent ; for each system has the solution 1, 2 ; and, as will be
proved later, neither system has any other solution.
Observe that (3) is obtained by solving (1) for x, and (4) by put-
ting in (2) the value of x given in (3) ; x therefore does not appear in
(4), and is said to have been eliminated.
200. Elimination is the process of deriving from two or
more equations a new equation involving one less unknown
than the equations from which it is derived.
The unknown which does not appear in the derived equa-
tion is said to have been eliminated; as a; in § 199.
There are in common use two methods of elimination :
I. Elimination by substitution or comparison.
II. Elimination by addition or subtraction.
201. In this chapter we shall use three principles con-
cerning the equivalence of systems of equations. For con-
venience of reference we shall number them, (i), (ii), (iii).
(i) Equivalent equations. If any equcUion of a system is
replaced by an equivalent equation, the derived system will be
equivalent to the given system.
E.g., since equation (3) is equivalent to (1), and (4) to (2), system
(6) is equivalent to system (a).
3a; + 22/ = 8, (1)1 dx
4x-Sy = 5, (2)/^^^ 8x
+ 6y =
-Gy =
= 24,
= 10.
>>
The only solution of either system is 2, 1.
Ex. Solve the system 3 + 4 a; = 15,
Sl<-'
2 + 3y = 8.
From (1), a; = 3.
From (2), y = 2.
S)<'>
Since equation (3) is equivalent to (1), and (4) to (2), system (b)
is equivalent to system (a) ; hence, the one and only solution of system
(a) is 3, 2.
202 ELEMENTS OF ALGEBBA
Proof of (i). Let (1) and (2) be a system of equations in
two unknowns,
^ = ^' ^^A(a^ "^' = ^'' ^^^(h^
C=D, (2)r^ C' = Z>' (4)]^^
and let (3) be equivalent to (1), and (4) to (2) ; we are to
prove that system (h) is equivalent to system {a).
Since (1) and (3) have the same solutions, and (2) and (4)
also have the same solutions ; it follows that any solution
common to (1) and (2) will be common to (3) and (4) also ;
and conversely. Hence, systems (a) and (6) are equivalent.
202. The method of elimination by substitution depends
upon the following principle of equivalence of systems:
(ii) Substitution. If one equation of a system is solved for
one of its unknowns, and the value thus obtained is substituted
for this unknown in the other equation (or equations) of the
system, the derived system will be equivalent to the given one.
Ex. 1. Solve the system 2 a:; = 10, (1)
y = l2-lSx. (2) j
■(«)
From (1), X = 5. 1
Substituting this value of x in (2), we obtain I (?))
ij = 12 -16 = -3. J
By (ii), system (6) is equivalent to system (a).
Hence, the one and only solution of system (6) or (a) is 5, — 3.
Ex. 2. Solve the system Sx + by = 19, (1) 1
bx-4y = 1. (2) J
rrom(l), x = il9-by)/S. (3)1
Substituting this value for x in (2), we have [ (6)
(5/3)(19-5i/)-42/ = 7. (4) J
SYSTEMS OF LINEAR EQUATIONS 203
By (ii), system (6) is equivalent to system (a).
From (4), y = 2. (5)
Substituting 2 for y in (3), we obtain
x=(19-10)/3 = 3. (6)
(c)
By (ii), system (c) is equivalent to system (6) ; hence, the one and
only solution of system (c) or its equivalent system (a) is 3, 2.
Ex. 3. Solve the system 2x- 5y=l, (1)
7x + 3i/ = 24. (2)'
}(«:
From(l), x=(5y+l)/2. (3)1
Substituting this value of x in (2), we have > (/>)
(7/2)(5y+l)+3y = 24. (4) J
By (ii), system (6) is equivalent to system (a).
From (4), y = 1. (5)
Substituting this value of y in (3), we obtain (c)
x = (5 + l)/2 = 3. (6).
By (ii), system (c) is equivalent to (6), and therefore to (a).
Hence, the one and only solution of system (a) is 3, 1.
The foregoing examples illustrate the following rule for
eliminating by substitution.
From one of the equations find the value of the unknown to
he eliminated, in terms of the others; then substitute this value
for that unknown in the other equation or equations.
Proof of (ii). Let (1) and (2) be a system in two
unknowns, and let (3) be the equation obtained by solving
^ = A (1)K. =^ = P, (3)1
C = D, {2)]^' C' = D', (4)J
(b)
(1) for X, and (4) the equation obtained by substituting for x
in (2) its value F as given in (3); we are to prove that
system (6) is equivalent to system (a).
204
ELEMENTS OF ALGEBRA
Since x = Fis equivalent to equation (1), system (c) is by
(i) equivalent to system (a),
X=:F, (5)1
C = D.
(6) J
(«)
Any solution of system (c) renders x = F and C=D',
hence, any such solution must satisfy (6) after F has been
substituted for x (vi, § 32) ; therefore, any solution of
system (c) will be a solution of system (b).
Conversely, any solution of system (b) renders x = F and
C = D' ; hence, any such solution must satisfy (4) after x
has been substituted for F (vi, § 32) ; hence, any solution of
system (b) is a solution of (c).
Hence, system (b) is equivalent to system (c) or (a).
In like manner the theorem could be proved if the systems
(a) and (6) contained three or more equations.
Exercise 70.
Solve each of the following systems of equations by the
method of substitution :
1. 3 a; = 27,
2 a; 4- 3 2/ = 24.
2. 3 a; + 4 2/ = 58,'
2y = U,
3. 3 a; + 4 2/ = 10,1
4 a; + 2/ = 9. J
4. x-\-2y = lS,
Sx +
5.
2/ = 13,|
2/ = 14. J
4x + 72/ = 29,l
a; + 3 2/ = 11. J
6. 5 a; + 6 2/ = 17,1
6 a; + 5 2/ = 16. J
7. 8 a; -2/ = 34,1
ic + 8 2/ = 53. 1
8. 6^ -5 a; = 18,1
12 a; - 9 2/ = 0. J
9. 7aj + 42/ = l, I
9a; + 4^ = 3. J
10. x-lly = l,
111 2/ - 9 a; = 99.
11. 3 a; + 5 2/ = 19,1
5x — 4:y = 7. J
12. 8 x- 21 2/ = 5, 1
6 a; + 14 2/ = - 26. J
SYSTEMS OF LINEAR EQUATIONS 205
13. 3 a; -112/ = 0, I 16. 1 0^ + 3 2/ + 14 = 0,1
19 a;- 19 2/ = 8. J |a; + 52/ + 4 = 0. 1
14. ^^-^^y = ^'_ ] 17. i(^+x)=i(9+y), I
25 0.-17 2/ = 139. J i(ll+^+2/)=i(9+2/)
15. ^_^_1
3 6~2'
5 10 2 J
18. J(a. + 1)=4(2/ + 2),|
i(^ + 2/)=i(2/ + 2).J
19. (a; + l)(2/ + 5) = (.T + 5)(2/ + l),|
]
20. an/-(a;-l)(2/-l)=6(2/-l)
a; - 2/ = 1.
203. The following example illustrates a special form of
the method of elimination by substitution, which is called
elimination by comparison.
Ex. Solve the system
2x-3y = \,
6x-^2y = l2Q.
;:;)'•'
Solve (1) for a;,
Solve (2) for a;,
a; = (3y + l)/2.
x=026-2y)/5.
'>
Substituting in (4) the value of x given in (3), or, what amounts to
the same thing, putting these two values of x equal to each other, we
obtain
3y4-1^126-y ^r y = 13.
2 6
Substituting in (3), x = 40/2 = 20.
(c)
By principles (i) and (ii), systems (a) and (c) are equivalent ; or,
in other words, no solution has been either lost or introduced in pass-
ing from system (a) to system (c) ; hence, the one and only solution
of system (a) is 20, 13.
206
ELEMENTS OF ALGEBRA
Exercise 80.
Solve each of the followmg systems by the method of
comparison :
1. ^+2/ = 16,
4
2. X — y
6. ^ + ^ = 5,
5 2 •
X — y =z4.
7. l-^-TV?y = 3,|
4:X-y = 20. J
-? = 2.
8. ^ + -^^
0,
10,
3. ^ + 2/
1 i» + 2/ = 50
4. a? = 3 2/,
1 oj + 2/ = 34
5 it*
3 4
Sx-7y = 37. }
X -\-l _3 y — 5
10 ~~^
a; + 1 X — w
5.
2/
5 2/
10.
10 8
a^ + 3 8 - V I
3 (x + ?/) ^ a? + 3
8 ~ 5 '
204. The method of elimination by addition or subtraction
depends upon the following principle :
(iii) Addition. If an equation obtained by adding, or sub-
tracting, the corresponding members of two or more equation"}
of a system is put \n the place of any one of these equations,
the derived system will be equivalent to the given system.
Ex. 1. Solve the system 3 a; + 7 y = 27, (1) )
bx + 2y = \Q. (2) J
To eliminate x, we obtain from (1) and (2) equivalent equations in
which the coefficients of x are equal.
Multiply (1) by 5, 15 x + 35 «/ = 135, (3)
Multiply (2) by 3,
Ibx + Qy
48.
(4)
(&)
SYSTEMS OF LINEAR EQUATIONS 207
Subtract (4) from (3), 29 </ = 87, (5)
From (4), 15 x + 6 y = 48. (6)
From (5), y = 3,
Substitute in (6), x = 2.
lid)
Proof of equivalency. By (i), systems (a) and (&) are equivalent;
by (iii), system (c) is equivalent to (b) ; and by (ii) and (i), system
(d) is equivalent to (c).
Hence the one and only solution of (a) is 2, 3.
Ex.2. Solve the system 7 x + 2 y = 47, ^_, ,
ox — 4 y = 1.
(2)1
To eliminate y we obtain from (1) and (2) equivalent equations in
which the coefficients of y are arithmetically equal.
Multiply (1) by 2, 14 x + 4 y = 94. (3) )
From (2), 6x-4?/= 1. (4) J
Add (3) and (4), 19 x = 95, or x = 5. (5) |
Substitute in (4), y = 0. j
Proof of equivalency. By (i), system (h) is equivalent to (a) ; by
(iii), system (4) and (5) is equivalent to (6) ; and by (ii) and (i),
system (c) is equivalent to system, (4) and (5).
Ex. 3. Solve the system
(X - l)(y - 2)-(x - 2)(y - 1) = - 2, (1)
(X + 2)(y + 2)-(x - 2)(y - 2)= 32. . (2)
We first reduce (1) and (2) to the form (ix+ by = c.
|(«)
From(l), x-y = 2, ^^U (6)
From (2), x + y = 8. (4) J
Add (3) and (4), 2 x = 10, or x = 5. (5)
Subtract (3) from (4), 2 y = 6, or y = 3. (6)
By (i), system (&) is equivalent to (a) ; by (!) and (iii), either (5)
an^l (4) or (6) and (4) form a system equivalent to (6) ; hence the
solution of (a) is given in (5) and (6).
208 ELEMENTS OF ALGEBRA
Ex. 4. Solve the system
Sx-l^ = ^-^^, (1)
/ id
We first reduce (1) and (2) to the form ax + by = c.
(a)
From (1),
From (2),
Ux-2y = -Sl,
10x + 6y = S7.
:,h
Multiply (3) by 3,
i2x-6y = -dS.
(5)
Add (4) and (5),
Substitute in (3),
62 X = - 56, or ic =
y = 207/26.
- 14/13. )
}(C)
Since (3) and (5) are equivalent and (3) is the simpler equation,
to find y we substitute in (3) rather than in (5).
By (i), (ii), and (iii), no solution has been lost or introduced in
passing from system (a) to system (c).
Proof of (iii). Let (1) and (2) be a system in two
unknowns,
^ = ^' ^^Hia) ^ = ^' ^^Hcb)
C=D, (2)}^"^ A+C=B + D, (4)1^"^
and let (4) be obtained by adding the corresponding mem-
bers of (1) and (2) ; we are to prove that system (b) is equiv-
alent to system (a).
Any solution of (a) renders (1) and (2) identities. But,
if (1) and (2) are identities; by § 32, (3) and (4) are iden-
tities ; hence any solution of system (a) is a solution of
system (b).
Conversely, any solution of (6) renders (3) and (4) iden-
tities. But, if (3) and (4) are identities ; by § 32, (1) and (2)
are identities ; hence any solution of system (b) is a solution
of (a).
Hence (a) and (6) are equivalent systems.
In like manner the theorem could be proved if the sys-
tems (a) and (b) contained three or more equations.
SYSTEMS OF LINEAR EQUATIONS
209
The foregoing examples illustrate the following rule for
elimination by addition or subtraction :
Reduce the equations to the form ax -\-by = c.
Find the L. C. M. of the coefficients of the unknown to he
eliminated. Multiply both members of each equation by the
quotient of this L. C. M. divided by the coefficients of that un-
known in the equation.
Add or subtract the corresponding members of the equations
thus denved, according as the coefficients of the unknown to be
eliminated are opposite or like in quality.
Exercise 81.
Solve each of the following systems by the
addition or subtraction :
1. 3 a; + 4 2/ = 10,
4:X + y = 9.
2. x-\-2y = lS,
3x-{-y = li.
3. 2x — y = 9j
Sx-7y = 19.
4. 4 a; + 7 2/ = 29,
x-{-Sy = ll.
6. 2x + y = 10, 1
7 a; + 8 2/ = 53. J
6. 5 a; + 6 2/ = 17,
6 a; 4- 5 y = 16
7. 8x-2/ = 34, 1
x + Sy = 5S.}
;;1
8. 15 a; +7 2^ = 29,
9 aj + 15 2/ = 39,
:1
9. 14:X-Sy =
6 a; + 17 2/ =
10. 28 a: -23 2/
63 a; - 25 2^
11. S5x-{-17y
56x-lSy
12. 5 a; — 7 2/ =
7x + 5y =
13. 15 a; + 77 2/
55x-S3y
14. 5x = 7y —
21x-9y =
15. 6y — 5x =
12x-9y =
16. 21 a; -50 2/
2Sx-27y
method of
= 39,1
= 36.1
= 33, 1
= 101. J
= 86,1
= 17.1
u]
0,
74
= 92,
f
21,
= 7{
18,1
= 0.1
= 60, I
= 199. J
= 92,1
= 22.1
..)
210
ELEMENTS OF ALGEBBA
205. The following example illustrates a special form
of the method of elimination by addition, which is called
elimination by undetermined multipliers.
(3)
(4)
5/3.
Ex. Solve the system 3 x — 5 </ = 2,
Multiplying (1) by an arbitrary multiplier /c, we obtain
3 kx - 5 ky = 2k.
Adding (3) and (2), we obtain
(3 k + [>')x - {5 k -\- 2)y = 2k-\- 10.
Putting the coefficient of x in (4) equal to 0, we obtain k = -
Substituting — 5/3 for k in (4), we obtain y = 2. (5)
Putting the coefficient of y in (4) equal to 0, we obtain A: = — 2/5.
Substituting — 2/5 for k in (4), we obtain x = 4. (6)
The two equations, y ^2 and x = 4, which result from the two
different values of k in (4) form a system which by (i) and (iii) is
equivalent to (a).
No one method of elimination is preferable for all cases.
The learner should aim to select that method which is best
suited to the system to be solved.
Exercise 82.
Solve each of the following systems by that method which
is best suited to it :
1. 57 a; + 25^ = 3772,
25 0^4-57 2/ = 1148.
2. 93a;H-15i/ = 123,
15x4- 93 2/ = 201.
3. Wx-hl9y = lS,]
5.
+ 52/ = -4,
5
SYSTEMS OF LINEAR EQUATIONS
211
6. ''-±l-^iy = 2,
y + 11 x + l^
11 2
1.
2x — 5y x-^7 _ ^
8.
3
a:-2
4
2a;-5 ll-2y
= 0.J
9. ^-3^ = 0,
2x
7 13 -y
3 16
0.
10. ^-f = 4,
11. 2a;-f2/=:0,
i2/-3a; = 8.J
12. \x-\-\y = n,
13. 3. r — 7 2^ = 0,
f a; + |2/=7.
14. lx-\y = 0,
3
x + ^y = ll,]
15. aa; -h fty = (a + 6)2
ax — by= a^ — Ir.
16. «x + &y=a24-6V
hx-{-ay = 2db.
17. aa; + 6y = a--62,
6a;
+ 6y = a2-62 ,
+ a?/ = a2_62 J
18. x + y = a-^h,
ax — by = b^ — a'.
19. b-x-a'y = 0,
bx
4-a?/=:aH-6. J
20. x — y = a~b,
ax-by = 2a^-2b-. J
21. ax-by = a' + b^,)
x+ y = 2a. j
22. 6a; — a?/ = ft'', I
ax — by = al J
23. aa;H-6?/ = l, |
6a; + ay = l. J
24. (a + b)x-(a-b)y = 3ab,]
(a + 6)2/ - (a - 6)a; = ab. J
a^a; + 6^y = c*. J
26. ^ +
a 6 a6'
a? _ y^ 1
a' 6' a'6'"
212
ELEMENTS OF ALGEBRA
27.
Sx_^2y
a b
9x_6y_^
a b
28.
qx — rb =p{a — y), "
^+,.=,(1+1).
29.
-^-1 =1.
m' m
30.
a b
3a 66 3 J
31.
a 0
a' V J
32.
b a b
206. Two conditions are said to be consistent or inconsistent
according as they can or cannot be satisfied at the same
time.
Equations which express consistent conditions and there-
fore have one or more solutions in common are called con-
sistent equations. Thus the equations in any of the above
systems are consistent equations.
Equations which express inconsistent conditions and there-
fore have no solution in common are called inconsistent
equations.
E.g.., the equations
3 5c + 3y = 15,
(1)
(2)
express inconsistent conditions and have no solution in common. For
if a; -I- 2/ is 4, 3(x + ?/) is 12 and cannot therefore be 15.
207, Each of the foregoing systems of linear equations
illustrates the following theorem :
If the two equations of a system in two unknowns are linear,
independent, and consistent, the system has one, and only one,
solution.
SYSTEMS OF LINEAR EQUATIONS 213
Proof. By the principles of equivalent equations and (i)
in § 201, any system of two linear equations can be reduced
to an equivalent system of the form
)(«)
ax-\-hy = c, (1)
a'x + h'y = c'. (2)
Multiply (1) by h', ab'x + bb'y = b'c. (3)
Multiply (2) by b, a'bx + bb'y = be', (4)
Subtract (4) from (3), (ab' - a'b) x = b'c - be'. (5) ^
From (1), ax-\-by = c. (1) J
By (i) and (iii), system (6) is equivalent to system (a).
When ab' — a'b = 0 and b'c — be' = 0, (5) is an identity,
and system (b) or (a) has all the solutions of equation (1) ;
hence equations (1) and (2) are equivalent.
When ab' — a'b = 0 and b'c — be' ^ 0, no value of x will
satisfy (5) ; hence system (6) or (a) has no solution, and
equations (1) and (2) are inconsistent.
Hence (1) and (2) are not independent and consistent
unless
ab' - a'b ^ 0.
When ah' — a'b ^ 0, x has one, and only one, value in
(5), and this value of x will give one, and only one, value
for y in (1) ; hence system (6) or (a) has one, and only one,
solution.
When ab' — a'b ^ 0, from (5) we obtain
X = (b'c - be') / (ab' - a'b). (6)
Similarly, y = (ae' — a'e) / (ab' — a'b). (7)
208. Systems of three linear equations.
Ex. 1. Solve 6x + 2y-5z = lS, (1) 1
3a; + 3y-20=13, (2) i (a)
1x + 5y-Sz = 26. (3) I
(ft)
(0
214 ELEMENTS OF ALGEBRA
To eliminate y, we can proceed as follows :
Multiply (1) by 3, \^x + Qy -\^z = 39.
Multiply (2) by 2, Qx-^Qy - 4:Z = 2Q.
Subtract, 12 x - 11 2; = 13. (4)
Multiply (1) by 5, 30 a: + 10 «/ - 25 ^ = 65.
Multiply (3) by 2, Ux-{-10y- 6z = 52.
Subtract, 16 x - 19 0 = 13. (5)
Solving system (6), i.e., (4) and (5), we obtain
z=l, (6) I
x = 2. (7) J
From (6), (7), and (1), y = 3. (8)
The systems (6) and (c) are equivalent.
But (b) with (1) forms a system equivalent to (a); hence (c) with
(1), or (c) with (8), forms a system equivalent to (a).
Hence the solution of system (a) is 2, 3, 1.
Ex. 2. Solve 3 X + 2 y + 4 2: = 19, (1)1
2x+ 5^4-30 = 21, (2) K«)
Sx-y-\-z = i. (3) J
From (3), y = Sx + z-i. (4)
Substituting in (1) and (2) the value of y in (4), we obtain
3 X + 2(3 X + 0 - 4) + 4 2: = 19,
and 2x4-5(3x + ;3 -4)+32 = 21;
or, 9 X -i- 6 2; = 27, |
and 17x + 8^ = 41. J
Solving system (6), we obtain
(5)1
X = 1. (6) J ' '
From (4), (5), (6), y = 2. (7)
By (ii), system (6) with (4) forms a system equivalent to (a);
hence (c) with (4), or (c) with (7), forms a system equivalent to (a).
Hence the solution of (a) is 1, 2, 3.
SYSTEMS OF LINEAR EQUATIONS
215
The foregoing examples illustrate the following method
of solving a system of three linear equations :
From any two of the three equations derive an eq^iation,
eliminating an unknown; next from the third equation and
one of the other two derive a second equation eliminating the
same unknown.
Solve for these two unknowns the two equations thus derived,
and substitute the values of these two unknowns in the simplest
equation which contains the third unknown.
209. From a system of four linear equations we can elimi-
nate one of the four unknowns, and thiis obtain a new system
with three unknowns. Solving this new system, we can
substitute the values thus obtained in the simplest equation
which contains the fourth unknown.
Solve
Exercise 83.
1. .T-f .3?/ -1-42 = 14,]
or + 2?/ -1-2 = 7,
2x-\-y-\-2z=2.
2. a; -1-22/ + 22 = 11,
2x + y-\-z = l,
3a; + 42/-h2 = 14.
3. ^x-2y-\-z = 2,]
2aj-j-32/-2 = 5,
.T + 2/ -h 2 = 6.
4. x-\-y^z = l,
2a;-|-3 2/ + 2 = 4,
4 a; H- 9 2/ + 2 = 16.
5. 5x + 3y-\-7z = 2y
2a; -42/ + 92 = 7,
3x + 22/ + 62 = 3.
6. x-{-2y-3z = 6,
2.T + 42/-72 = 9,
3a; — ^ — 52 = 8.
7. x-2y + 3z = 2,
2a; — 3 2/ + 2 = 1,
3x-y-\-2z = 9.
8. 3x-\-2y-z = 20,
2x-\-3y-\-6z = 70
a; — 2/ + 62 = 41. J
216
ELEMENTS OF ALGEBRA
9. 2a; + 3?/ + 42 = 20,
3a; + 42/ + 5^ = 26,
Sx-{-5y-{-6z = 31.
10. Sx-4:y = 6z-16,
4:X-y = z-\-5,
a; = 32/ + 2(2-1). J
11. ax-\-by = l,
by -\- cz = 1,
cz + ax = 1.
12. cy -\-hz — he,
az -\- ex = ca,
hx -\- ay = ah.
13. a;-| = 6,
y--
^ 7
= 8. !-
= 10.
14. ^(a;+2!-5)=2/-2,
i(a;+;2_5)=2a;-ll,
2a;-ll = 9-(a;+22). .
15.
16.
17.
18.
19.
20.
a; + 20 = 12/ + 10,
a; + 20 = 2 2 + 5,
22 + 5 = 110 -(2/ + 2).
ax + &2/ = Ij 1
62/ + C2 = 1, I
C2 + aa; = 1. J
cy -i-bz = be,
az -^ ex = ca,
bx -\- ey = ab.
X — ay -\- a-z = a^,
X —by -\- bh = b^,
X — cy + c^2 = c^. J
•'c + 2/ + ^ "~ "^ = H>
a; + 2/ - 2 + w = 17,
a; — 2/ + 2 + w= 9,
■ X -\- y -\- z -\- u = 12.
x-\-y -{-z = %
x-\-y-{-u = l,
a; + 2 + 1< = 8,
2/ + 2 + ?^ = 9. J
SYSTEMS OF FRACTIONAL EQUATIONS.
210. In clearing of fractions the equations of a system,
no solution will be lost, but new solutions may be introduced
even when we clear of fractions in the simplest manner.
Ex. 1. Solve the system 4 a; — 2 ?/ =2,
5x + l
3y-l 8
Clearing (2) of fractions and transposing, we have
40 a: -33?/ =-19.
0)
(2)
(3)
(a)
SYSTEMS OF LINEAR EQUATIONS 217
The solution of system, (1) and (3), is 2, 3.
In clearing (2) of fractions we multiplied by the unknown factor
3 y — 1 ; hence any solution which was introduced will be a solution
of the equation 3y — 1=0, or3y + 0x — 1 = 0.
Since 2, 3 is not a solution of this equation, it was not introduced
in clearing (2) of fractions.
Hence 2, 3 is the one and only solution of system (a).
Ex. 2. Solve the system 5 « — y = 2, (1)
^ 4-^^ = 0. (2)
(a)
x-1 y-S
Clear (2) of fractions, z-\-y = 4. (3)
The solution of system, (1) and (3), is 1, 3.
To clear (2) of fractions we multiplied by the unknown factor
(a;— l)(y— 3), and 1, 3 is a solution of the equation (x— l)(y— 3) = 0.
Hence the solution 1, 3 may have been introduced by clearing (2)
of fractions.
By trial we find that 1,3 is not a solution of (2) ; hence the solu-
tion 1, 3 was introduced, and system (a) has no solution ; that is, its
equations are inconsistent.
211. A system of fractional equations which are linear in
the reciprocals of their unknowns is readily solved without
clearing of fractions, by treating these reciprocals as the
unknowns.
(«)
Ex. 1. Solve the system a/x + c/y =
:m,
(1)
b/x + d/y =
: n.
(2)
Multiply (1) by 6, ab(l/x) + cbil/y) =
: bm.
(3)
Multiply (2) by a, ab(\/x) + ad(\/y) =
-- an.
(4)
Subtract (4) from (3) , (6c - ad) (1 /y) =
: 6m-
- an.
.'. y =
6c-
bm -
-ad
-an
(5)
Multiply (1) by d, ad(\/x) + cd(\/y) =
- dm.
(6)
Multiply (2) by c, 6c(l/x) + cd(l/?/) =
= C7l.
(7)
Subtract (7) from (6), {ad - 6c)(l/x) =
- dm -
- en.
.'. x =
ad-
-be
(S)
dm
218
ELEMENTS OF ALGEBRA
Multiplying (1) and (2) by xy to clear them of fractions would give
us a system of quadratic equations and introduce the new solution 0, 0.
Ex. 2. Solve the system
36,
1 + 1 + 1
X tj z
1 + 5-1 = 28,
X y z
^y 2z
20.
Subtract (1) from (2), 2 (\/y)-2 (l/z) = -S,
Subtract (3) from (1), ^ (l/y)+ l(l/z')= W.
Solving system (6), y = 1/12,
0 = 1/16.
From (1), (6), and (7), x = 1/8.
(1)
(^)
(3)
4) J
5) J
(4)
(
(6)
(7)
(8)
(a)
(&)
Solve the system
Exercise 84.
1. «-» = !,
X y
5.
^ + 1« = 79,
X y
15+6^7.
X y
15-1 = 44.
a; ?/
2. 5_4=2,
X y
6.
6 7 „
18 + 18 = 10.
X y
2 + 11 = 3.
^ y
3.. 5_5 = 9j
a; y
7.
^ + 2 -7 1
1-2 = 5.
X y
7 1 -3.
6* 10?^ J
4. 5 + ^ = 3,
a; y
8.
14-2-3 1
2X + 32,-''
5-^ = 1
4a; oy
9.
SYSTEMS OF LINEAR EQUATIONS
219
9. - +- = a,
X y
X y
10.
m n
mr
11.
X y n
7i_m_ nr
X y m
mx ny
mx ny
12. - + f = 2,
ax by
— -- = 7.
ax by
13. A_2 ^4,
ax by
a; 2^
14. 1 = m + 71,
?ia; ?n.y
X y
1 2
15. i_-_j_4 = 0,
a; ^
1-1 + 1=0,
2/ 2
5 + 3
2! a;
14.
16. 1 + 1 + 1
X y z
36,
1 + ^-1 = 28,
X y z
a; 3y 2z
17. -^ + -^
J_ J J^^l
2x~^4.y 3z 4'
1^ 1^
X Sy
i-^ + i = 2A.
X by z
18. 3a; + 4^ = 11,
3y + l 7*
19.
2x-\-A
= 1,
= 2.
3y-l
lx-2
.V + 3
20.
x±2y±J^^
Sx-\-y-l '
3x+x±l = i
4a;-2/-2
21.
5a; + 2y 9a; + 4y
2a; ^2
7a; + 2/ 5*
22. 5x — Sy = S,
1 +^- = 0.
a; — 3 y — 4:
CHAPTER XV
PROBLEMS SOLVED BY SYSTEMS
212. A determinate jrroblem is one which has a finite num-
ber of solutions. Every determinate problem must contain
as many independent consistent conditions, expressed or
implied, as unknown numbers. If in any such problem we
denote each unknown by a letter, and express each condition
by an equation, we shall obtain as many independent con-
sistent equations as there are unknowns.
The solutions of the system of equations thus obtained
will give the solutions of the problem.
Prob. 1. Find two numbers such that twice the greater exceeds three
times the less by 6, and that twice the less exceeds the greater by 2.
Let X = the greater number, and y = the less.
Then, by the first condition, we have
2x-Sy = Q, (1).
and by the second condition we have \ (a)
2y-x = 2. (2) J
From system (a), x = 18, the greater number ;
and y = 10, the less number*
Prob. 2. A number expressed by two digits is equal to six times the
sum of its digits, and the digit in the tens' place is greater by one than
the digit in the units' place. Find the number.
Let X = the digit in tens' place,
and y = the digit in units' place.
220
PROBLEMS 221
Then, from the first condition, we have
10x + i/ = C(x + ?/), (1)-
and from the second condition we have ■ (a)
x-y = l. (2) .
From system (a), a; = 5, the digit in tens' place ;
and 2/ = 4, the digit in units' place.
That is, the required number is 54.
Prob. 3. If the numerator of a fraction is increased by 2 and the
denominator by 1, it becomes equal to 5/8, and if the numerator and
denominator are each diminished by 1, it becomes equal to 1/2. Find
the fraction.
Let X = the numerator, and y = the denominator ; then,
from the first condition, ^-±-? = -, (1) ]
and from the second, - — - = -• (2) J
The solution of sy.stem (a) is 8, 15 ; hence the fraction is 8/15.
Prob. 4. A man and a boy can do in 15 days a piece of work which
would be done in 2 days by 7 men and 9 boys. How long would it
take one boy or one man to do it.
Let X = the number of days it would take one man to do the whole
work, and y = the number of days it would take one boy.
Let the whole work be represented by 1.
Then in one day a man would do 1/x of the work, and a boy l/y
of it.
Hence, by the first condition, we have
15/x + Vo/y = 1, (1) ^
and by the second condition we have [ (a)
14/x + 18/?/ = 1. (2) J
The solution of system (a) is 20, 60.
Hence one man would do the work in 20 days, and one boy in 60
days.
222 ELEMENTS OF ALGEBRA
Exercise 85.
1. Six horses and 7 cows can be bought for $1250, and
13 cows and 11 horses can be bought for $2305. Find the
value of each animal.
2. Four times B's age exceeds A's age by 20 years, and
\ of A's age is less than B's age by 2 years. Find their
ages.
3. Find a fraction such that if 1 be added to its denomi-
nator it reduces to |-, and if 2 be added to its numerator it
reduces to f .
4. A man being asked his age, replied : " If you take
2 years from my present age the result will be double my
wife's age, and 3 years ago her age was \ of what mine will
be in 12 years." Find their ages.
5. One-eleventh of A's age is greater by 2 years than \
of B's, and twice B's age is equal to what A's age was 13
years ago. Find their ages.
6. In 8 hours A walks 12 miles more than B does in
7 hours ; and in 13 hours B walks 7 miles more than A does
in 9 hours. How many miles does each walk per hour ?
7. At an election the majority was 1G2, which was ^\ of
the whole number of voters. What was the number of the
votes on each side ?
8. A and B have $ 250 between them ; but if A were to
lose half his money^ and B | of his, they would then have
only $ 100. How much has each ?
9. A man bought 8 cows and 50 sheep for $ 1125. He
sold the cows at a profit of 20%, and the sheep at a profit
of 10%, and received in all $1287.50. What was the cost
of each cow and of each sheep ?
10. Twenty-eight tons of goods are to be carried in carts
and wagons, and it is found that this will require 15 carts
and 12 wagons, or else 24 carts and 8 wagons. How much
can each cart and each wagon carry ?
PROBLEMS 223
11. A and B can perform a certain task in 30 days, work-
ing together. After 12 days, however, B was called off, and
A finished it by himself 24 days after. How long would
each take to do the work alone ?
12. Find the fraction such that if you quadruple the
numerator and add 3 to the denominator the fraction will
be doubled, but if you add 2 to the numerator and quadruple
the denominator, the fraction will be halved.
13. The first edition of a book had 600 pages, and was
divided into two parts. In the second edition J of the
second part was omitted and 30 pages were added to the
first part. The change made the two parts of the same
length. How many pages were in each part in the first
edition ?
14. A marketman bought eggs, some at 3 for 5 cents, and
some at 4 for 5 cents, and paid for all $ 5.60; he afterwards
sold them at 24 cents a dozen, clearing $ 1.80. How many
eggs did he buy at each price ?
15. In a bag containing black and white balls, half the
number of white is equal to a third of the number of black ;
and twice the whole number of balls exceeds 3 times the
number of black balls by 4. How many balls does the bag
contain ?
16. A crew that can row 10 miles an hour down a river,
finds that it takes twice as long to row up the river as to
row down. Find the rate of the current.
17. A certain number between 10 and 100 is 8 times the
sum of its digits, and if 45 be subtracted from it the digits
will be reversed. Find the number.
18. If A were to receive $ 50 from B, he would then have
twice as much as B would have left; but if B were to receive
$50 from A, B woukl have 3 times as much as A would
have left. How -much has each?
224 ELEMENTS OF ALGEBRA
19. A farmer sold 30 bushels of wheat and 50 bushels of
barley for $93.75. He also sold at the same prices 50
bushels of wheat and 30 bushels of barley for $ 96.25.
What was the price of the wheat per bushel ? ,
20. One rectangle is of the same area as another which
is 6 yards longer and 4 yards narrower; it is also of the
same area as a third, which is 8 yards longer and 5 yards
narrower. What is the area of each ?
21. A boy rows 8 miles with the current in 1 hour 4 min-
utes, and returns against the current in 2|- hours. At what
rate would he row in still water ? What is the rate of the
current ?
22. A, B, C, D have $1450 among them; A has twice
as much as C, and B has 3 times as much as D ; also C and
D together have $ 250 less than A. Find how much each
has.
23. A, B, C, D have $1350 among them ; A has 3 times
as much as C, and B 5 times as much as D ; also A and B
together have $ 250 less than 8 times what 0 has. Find
how much each has.
24. A number consists of 2 digits followed by zero. If
the digits be interchanged, the number will be diminished
by 180 ; if the left-hand digit be halved, and the other digit
be interchanged with zero, the number will be diminished
by 454. Find the number.
25. A train travelled a certain distance at a uniform rate ;
had the speed been 6 miles an hour more, the journey would
have occupied 4 hours less ; and had the speed been 6 miles
an hour less, the journey would have occupied 6 hours more.
Find the distance.
Let X = the number of miles the train runs per hour,
and y = the number of hours the journey takes.
Then xi/ =(x + 6)(?/ - 4), ]
and xy=ix-Q)(y + 6).\
PROBLEMS 225
26. A traveller walks a certain distance ; had he gone ^
mile an hour faster, he would have walked it in ^ of the
time ; had he gone ^ mile an hour slower, he would have
been 2i hours longer on the road. Find the distance.
27. A man walks 35 miles, partly at the rate of 4 miles
an hour, and partly at 5 ; if he had walked at 5 miles an
hour when he walked at 4, and vice versa, he would have
covered 2 miles more in the same time. Find the time he
was walking.
28. A fishing-rod consists of two parts ; the length of the
upper part is f- that of the lower part ; and 9 times the upper
part together with 13 times the lower part exceeds 11 times
the whole rod by 36 inches. Find the lengths of the two parts.
29. A man put $12,000 at interest in three sums, the
first at 5 per cent, the second at 4 per cent, and the third
at 3 per cent, receiving for the whole $ 490 a year. The
sum at 5 per cent is half as much as the other two sums.
Find each of the three sums.
30. A, B, and C can together do a piece of work in 30
days ; A and B can together do it in 32 days ; B and C can
together do it in 120 days. Find the time in which each
alone could do the work.
31. A certain company in a hotel found, when they
came to pay their bills, that if there had been 3 more per-
sons to pay the same bill, they would have paid f 1 each
less than they did; and if there had been 2 fewer persons,
they would have paid $ 1 each more than they did. Find
the number of persons, and the number of dollars each paid.
32. A railway train, after travelling 1 hour, is detained
30 minutes, after which it proceeds at f of its former rate,
and arrives 20 minutes late. If the detention had occurred
10 miles farther on, the train would have arrived 5 minutes
later than it did. Find the first rate of the train, and the
distance travelled.
226 ELEMENTS OF ALGEBRA
Let X = the number of miles the train at first ran per hour ;
and y = the number of miles in the whole distance travelled.
Then y — x = the number of miles to be travelled after the de-
tention,
= the number of hours required to travel y — x miles
at the rate before the detention,
and ^^^ — -^^ = the number of hours required to travel y — x miles
at the rate after the detention.
y
— X
X
4(2/-
-X)
-X 4(y -x) ,
c 5x
_io
go'
Hy-x-
10)
5
5a;
60
10_
X
40.
5x
_ 5
60"
.-. x = 24
t, y-
= 44.
Hence ^Lr^ _ li^^^_2 = i!£. (i)
X 5x CO ■ ^ ^
Similarly, y - x - 10 _ 4(y -^x - 10) ^ |_^ ^2^
Subtract (2) from (1),
Hence the first rate was 24 miles an hour, and the distance travelled
was 44 miles.
33. A railway train, after travelling 1 hour, meets with
an accident which delays it 1 hour, after which it proceeds
at I of its former rate, and arrives at the terminus 3 hours
behind time ; had the accident occurred 50 miles farther on,
the train would have arrived 1 hour 20 minutes sooner.
Find the length of the line, and the original rate of the
train. Ans. 100 miles, 25 miles per hour.
34. A jockey has 2 horses and 2 saddles. The saddles
are worth $ 15 and $ 10 respectively. The value of the
better horse and better saddle is | that of the other horse
and saddle ; and the value of the better saddle and poorer
horse is |f that of the other horse and saddle. Find the
worth of each horse.
35. Five thousand dollars is divided among A, B, C,
and D. B gets half as much as A ; the excess of C's share
over D's share is equal to ^ of A's share, and if B's share
PROBLEMS 227
■were increased by $ 500 he would have as much as C and D
have between them. Find how much each gets.
36. A i^arty was composed of a certain number of men
and women, and, when 4 of the women were gone, it was
observed that there were left just half as many men again
as women ; they came back, however, with their husbands,
and now there were only a third as many men again as
women. What was the original number of each ?
37. Two vessels contain mixtures of wine and water; in
one there is 3 times as much wine as water, in the other
5 times as much water as wine. Find how much must be
drawn off from each to fill a third vessel which holds 7 gal-
lons, in order that its contents may be half wine and half
water.
38. There is a number of 3 digits, the last of which is
double the first ; when the number is divided by the sum of
the digits, the quotient is 22 ; and when by the product of
the last two, 11. Find the number.
39. Some smugglers found a cave which would exactly
hold the cargo of their boat ; viz. 13 bales of silk and 33
casks of rum. While unloading, a revenue cutter came
in sight, and they were obliged to sail away, having landed
only 9 casks and 5 bales, and filled J of the cave. How
many bales separately, or how many casks, would it contain ?
40. There are 2 alloys of silver and copper, of which one
contains twice as much copper as silver, and the other 3
times as much silver as copper. How much must be taken
from each to weigh a kilogram, of which the silver and the
copper shall be equal in weight ?
41. A person rows a distance of 20 miles, and back again,
in 10 hours, the stream flowing uniformly in the same
direction all the time ; and he finds that he can row 2 miles
against the stream in the same time that he rows 3 miles
with it. Find the time of his going and returning.
228 ELEMENTS OF ALGEBRA
42. A and B can do a piece of work in m days, A and C
can do the same piece in n days, and B and C can do it in
p days. Find in how many days each can do the work.
43. For $26.25 we can buy either 32 pounds of tea and
15 pounds of coffee, or 36 pounds of tea and 9 pounds of
coffee. Find the price of a pound of each.
44. A pound of tea and 3 pounds of sugar cost $1.50;
but if sugar were to rise 50 per cent, and tea 10 per cent,
they would cost $ 1.75. Find the price of tea and sugar.
45. A person possesses a certain capital which is invested
at a certain rate per cent. A second person has $5000
more capital than the first person, and invests it at 1 per cent
more; thus his income exceeds that of the first person by
$400. A third person has $7500 more capital than the
first, and invests it at 2 per cent more; thus his income
exceeds that of the first person by $ 750. Find the capital
of each person and the rate at which it is invested.
46. Two plugs are opened in the bottom of a cistern con-
taining 192 gallons of water ; after 3 hours one of the plugs
becomes stopped, and the cistern is emptied by the other in
11 more hours ; had 6 hours occurred before the stoppage,
it would have required only 6 hours more to empty the
cistern. How many gallons will each plug-hole discharge
in an hour, supposing the discharge uniform ?
47. A certain number of persons were divided into 3
classes, such that the majority of the first and second classes
together over the third was 10 less than 4 times the majority
of the second and third together over the first ; but if the
first class had 30 more, and the second and third together
29 less, the first would have outnumbered the last 2 classes
by 1. Find the number in each class when the whole num-
ber was 34 more than 8 times the majority of the third class
over the second.
PROBLEMS 229
48. Two persons, A and B, could finish a work in 7n
days ; they worked together n days, when A was called off,
and B finished it in p days. In what time could each do it ?
49. The fore-wheel of a carriage makes 6 revolutions
more than the hind-wheel in going 120 yards ; if the circum-
ference of the fore-wheel be increased by \ of its present
size, and the circumference of the hind-wheel by \ of its
present size, the 6 will be changed to 4. Required the cir-
cumference of each wheel.
CHAPTER XVI
EVOLUTION. IRRATIONAL NUMBERS
213. An nth root of a given number is a number whose
nth power is equal to the given number.
U.g., one second root of 4 is 2, since 2^ = 4.
Another second root of 4 is — 2, since (— 2)2 = 4.
A third root of — 8 is — 2, since (— 2)^ = — 8.
A second root of a number is usually called a square root ;
and a f^irc? root a c«6e root.
214. The radical sign, ^, written before a number, denotes
a root of that number.
The radicand is the number Avhose root is required.
The index is the number which, written before and a little
above the radical sign, indicates ivhat root is required.
When no index is written, 2 is understood.
E.g., -y/lG or ^16 denotes a second, or square, root of 16 ;
16 is the radicand, and 2 is the index.
The expression ^u denotes an nth root of u ; u is the
radicand, and n the index.
215. Since by definition {-^uY = u, it follows that ^u
is one of the n equal factors of u.
216. A rational, or commensurable, number is any whole or
fractional number.
A rational expression is one which ca7i be written without
using an indicated root. All the expressions in the previous
chapters are rational expressions.
230
EVOLUTION 231
217. A perfect nth. power is a number or expression whose
»ith root is a rational number or expression.
E.g., since ^'25 = 5, 25 is 2i perfect square.
Since V— Sa;^^ = — 2 icy^, _ 8 ic^?/^ is a perfect cube.
Prior to § 238 each radieand will be a perfect power of a
degree equal to the index of the root.
218. Two roots are said to be like or unlike according as
their indices are equal or unequal.
An even root is one whose index is even; as, -^a?.
An odd root is one whose index is odd; as, -y/27.
219. Number of roots.
(i) An arithmetic niuiiber has one, and only one, nth root.
Any odd power of a positive or negative base has the
same quality as the base itself; hence,
(ii) A positive or a negative number has one odd root of the
same quality as the number itself.
E.g., one value of \/+ 27 is + S, since (+ 3)3 = + 27.
Again, one value of v^— 32 is — 2, since (— 2)^ = — 32.
If two numbers, opposite in quality, are arithmetically
equal, their like even powers are the same positive number ;
hence,
(iii) A jyositive number has two even roots, which are arith-
metically equal, and opjwsite in quality.
E.g., two values of V+81 are + 9 and - 9, since (+ 9)2 or (- 9)'^
is + 81.
Again, two values of \/+ 81 are +3 and —3, since (+3)* or
(_3)Ms +81.
Any even power of a positive or a negative number is
positive; hence an even root of a negative number cannot be
a positive or a negative number.
Even roots of negative numbers give rise to new quality-
numbers,, which will be considered in Chapter XVIII.
232 ELEMENTS OF ALGEBRA
220. The principal root of a positive number is its positive
root.
The principal odd root of a yiegative number is its negative
root. E.g., + 4 is the principal square root of 16, and — 3
is the principal cube root of — 27.
Unless the contrary is stated, the radical sign will hereafter
be understood as denoting only the principal root.
221. The like principal roots of equal numbers are equal ;
hence,
The like principal roots of identical expressions are identical
expressions.
222. Evolution is the operation of finding any required
root of a number or expression.
In the statement of the following principles of joots, by
" the root " is meant " the principal root."
223. The exponent of any base in the root is equal to the
exponent of that base in the radicand divided by the index of
the root; and coyiversely.
That is, ^a"'" = a'". (1)
Proof By § 118, {aTf = a"*".
Hence, by ? 221, a"* = -^a"*", and conversely (1).
E.g., ^a^ = a^^^ = a^ ; ^x^^ = x^.
224. The nth root of a product is equal to the product of the
nth roots of its factors ; and conversely.
That is, V~^b=ya'^b. (1)
Proof By § 119, {^a • ^by =(^ay(-^by=ab.
Hence, by § 221, -^a • -y/b = Vab, and conversely (1).
Ex. 1. v/- 32 aio = V- 32 • ^a^^ = -2a^.
Ex. 2. V- aPh^ = </^-\ ■ ^«9 . ^66 = _ a%\
EVOLUTION 233
Observe that, from this principle, it follows that the ?ith
root of any real number is the nth root of its quality-unit
into the nth root of its arithmetic value. Thus,
^/■+27=^Ti-^27 = + 3; </^^^ = V^^ - ^S2 = -2.
225. The nth root of the quotient of two numbers is equal to
the quotient of their nth roots; and conversely.
That is, Va/b = ^a/^b. (1)
Proof By §186, f^X ^^^^
Hence, by § 221, ^a/^h = Va^, and conversely (1).
Ex.1. <p^^^=J^E3ggg §225
^ 216 ai2 ^216^
= (-5x2)7(6 a*). §224
Ex. 2. ^/TiJy^j^Eiggg §§ 167, 225
= -2xyV(a62;58). §224
Exercise 86.
Reduce to a rational form the following expressions :
1. V4^*. 8. ^-64a^/. 14 ^ /400cW^
2. V9a^/. 9. ^343 a^'b\ ,1125 a'b^
3. V25W«. 10. .^Sr^^ ''* V216^
'■f
12. ^/32^«. 1^- \_aW'
6. Sy-8a^/.
7. V-aW^.
... vf • "■ ^"
128
234 ELEMENTS OF ALGEBRA
19. J?-l 21. J- + ^- 23. J5-1.
^3 9 >i5 25 Ve 18
-• Vf-i- -• \fl- -• \
226. 77ie s^7i roo^ of the rth power of a number is equal to
the rth power of its sfh root ; and conversely.
That is,
^'a^^{-^ay.
(1)
Proof Let
</a = B',
(2)
then,
a = B\
§128
.'. a^={By = B-,
§§ 128, 118
Hence, by 221,
</a^=B\
(3)
From (2),
{^ay = B\
(4)
From (3) and (4), by § 32, we obtain (1).
Ex. 1. \/(64/125)-« = (V64/125)2
= (4/5)2 = 16/25.
Ex.2. V(81x2«c*)3=(>/8r^«c*)3
= (9x«c2)3 = 729x3«c6.
227. The sth root of the qth root of a nmnber is equal to
the qsth root of the number ; that is,
Proof If a number is resolved into q equal factors, and
then each one of these q equal factors is resolved into s
equal factors, the number will be resolved into qs equal
factors ; that is, , / y^ _ gy^
Ex. 1. ^ v(2^ ^^y^'^) = \/(2® ^y^^) = 2 xy^- § 227
Ex. 2. ^ V(25-^ X 93) = </(56 X 36) = 5 X 3 = 15.
EVOLUTION 235
Exercise 87.
Reduce each of the following expressions to a rational
form :
1. ^^{64. aV). ^ sW'\
2. V^'(27'x64^. \a«"ft««
3. V^(aWa-). '' ^V(729a«a-).
4. Vae/W- '^- ^(27/64)^.
6. <WW^'- ''• ^(^^^"^^)-
\.064a«' 14. ^(8 aV?/»)*.
228. Square root by inspection. When a perfect square
can be factored by inspection, its square root is found by
inspection.
Ex. 1. 30 rt» + ft* - 12 a2fta = (6 a^ - b^y-, or (ft2 - 6 a2)2.
.-. V(36 a» + ft* - 12 a2ft2) = 6 a^ - ft^, or ft2 - 6 a^.
Ex.- 2. Find the square root of the first eight expressions in each
of the exercises 52 and 53.
229. To show how to find the square root of any perfect
square, we must show how to reverse the process of squanng
any expression.
J5r.gr., squaring expression (1) we obtain expression (2).
x8 + rx2 + sx (1)
xfi + 2.ric6 + (r2 + 2 s)a;* + 2 rsx^ + sH^ (2)
Hence if (2) is taken as a radicand, (1) is its square root.
Now, the square root of the Jirst term of the radicand (2) is the
first term of the root (1).
If we subtract from (2) the square of the ^rs« term of (1), the^rsi
term of the remainder is 2 ra^. Dividing 2 rx^ by twice the first term
of the root, 2 x^, we obtain rx2, the second term of the root.
236 ELEMENTS OF ALGEBRA
If we subtract from the radicand the square of the sum of the Jirst
two terms of the root, (x^ + rx^)^, the first term of the remainder
is 2 sx^. Dividing 2 sx* by twice the first term of the root, we obtain
sx, the third term of the root.
This example illustrates the following principles (i) and (ii).
If the terms of a perfect square and its square root are
arranged in descending (or ascending) powers of some letter,
(i) TJie square root of the first term of the radicand is the
first term of the square root. ,
(ii) If the square of the first term in the root, or the square
of the sum of its first two or more terms is subtracted from the
radicand, and the first term of the remainder is divided by
twice the first term of the root, the quotient will be the next
term of the root.
Proof Let ^ stand for any number of terms of the
square root of any perfect square, and B for the rest ; the
terms of A and B being arranged in descending (or ascend-
ing) powers of the same letter, and every term of A being
of a higher (or lower) degree than any term of B.
By § 120, we have the identity
A' + 2AB-\-B' = (A + By. . (1)
Let A denote only the fij^st term of the root ; then, since
-^A^ =A, we have (i).
Let A denote the first one or more terms of the root;
then, if we subtract A^ from the radicand, the remainder is
2 AB 4- J5^. Let a denote the first term of A, and 5 the first
term of B ; then, supposing the remainder 2 AB -f- B^ to be
arranged in descending (or ascending) powers of the letter
of arrangement, 2ab will be its first term. Hence, as
2 ab -i- 2 a = b, we have (ii).
E.g., by (i), the first term of the square root of
16 x^ - 24 yx^ + 25 ?/%2 _ 12 y^x + 4:y^ (3)
is \/l6 X*, or 4 jr2 ; and by (ii) the second term is - 24 yx^ -7-2(4 x^),
or — Zyx.
EVOLUTION 237
The radicand (3) less (ix^-S yxY is 16 y'^x'^ - 12 t/^x + 4 y2.
Hence, by (ii), the next term of the root is 16 y'^j^^ -^ 2(4 x^), or 2/2.
The radicand (3) less (4 x2 - 3 ?/x + 2 y2)2 jg zero.
Hence, the square root of (3) is 4 x2 — 3 yx + 2 y2.
Instead of finding each square independently, some labor
can be saved by using the relation
A' + (2A + h)h = {A + h)\ (2)
and thus making use of the previous square. Thus the
work in the example above is usually written as below :
16 X* - 24 yx^ + 25 ^2^2 -\2yH-\-^ y*(4 x2 - 3 i/x + 2 2/2.
16 X*
8 x2 - 3 yx) - 24 yx^
- 24 yx^ + 9 y2x2
8x2-6yx +22/2 ) 162/2x2
16 ygx2 - 12 y^x + 4 y^
Subtracting from the radicand the square of the first term of the
root, (4 x2)2, the first term of the remainder is - 24 yx'.
By (ii), the second term of the root is - 24 yx' -^ 2(4 x2), or - 3 yx.
Write 2(4x2)-3yx to the left of the first remainder, multiply it
by - 3 yx, and subtract the product from the first remainder.
Then, by (2), we have subtracted in all
(4 x2)2 + (2 . 4 x2 - 3 yx) ( - 3 yx), or (4 x2 - 3 yx)2.
By (ii), the next terra of the root is 16 y2x2 -4-2(4 x2), or 2 y2.
Write 2(4x2 -3yx)+2y2 to the left of the second remainder,
multiply it by 2 y2, and subtract the product from the second remain-
der. Then, by (2), we have subtracted in all
(4 x2 - 3 yx)2 + (8 x2 - 6 ?yx + 2 y2)2 y2, or (4 x* - 3 yx -f 2 y2)2.
As there is no remainder, the required root is 4 x2 — 3 yx + 2 y2.
Observe that we could just as well write radicand (3) in ascending
powers of x, or what is the same thing, begin with its last term.
16x*-24yx8-f 25y2x2-12y8x + 4y*. (3)
Thus, by (i), the last terra of the square root of (3) is \/4y2, or 2 y2 ;
and, by (ii), the term before the last is - 12 y'x -r- 2 (2 y2), or - 3 yx,
which agrees with the result above.
238 ELEMENTS OF ALGEBRA
Ex. Find the square root of 4 x* — 8 r^ + 4 x + 1. (2)
The Jirst term of the root is 2 x'^, and the second term is — 8 x^-f-4 x^,
or — 2x.
The last term is ] or — 1. If the last term is — 1, the term before
the last is 4 a; H-(— 2), or — 2 x, which is the second term as found
above.
But (2 ic2 — 2 a; — 1)^ = the given expression ;
hence 2 x^ _ 2 x — 1 is the required root.
If we took — 2 ic^ as the first term of the root, the second term
would be 2 x, and the last term 1.
Note. In the following exercise the pupil should write out the
root at once by (i) and (ii), as in the example above ; but he should
be drilled also in arranging the work of finding and subtracting the
successive squares as on page 237.
Exercise 88.
Find the square root of the following expressions :
1. .T* + 2it'3 + 3a^ + 2a;-f 1.
2. 4:X* — Sa:^-\-4:X-\-l.
3. 9x'-S6x^-^72x + 3(J.
'4. 4x^ + 4:X^ — ^x-\-^Q.
5. x*-\-2x^y-{-3xY-^2xf + y*.
6. x*-2x^-{-^x'^--lx-\-j\.
7. 16-9Gx + 216x'-216x^-\-Slx\
8. l-\-4.x-{-10x'-^12x^-\-9x\
,. 4.^4_4^_,_3^2_^_^
10. l-xy-~\^-xPf + 2a^y^-\-4.xY.
11. x^-4:X^-{-6x*-Sx^ + 9x^-4:X-\-4:.
12. 9x^-12o(^ + 22x*-\-x^-{-12x-^4.
13. ct^-22a.'4 4-34a^4-121a^-374a; + 289.
14. a^ — ax-^^x^-{-Sa — 4:X + 16.
EVOLUTION 239
16. a^ + 2a;^-f a^-4ar'-12a;*-8a^4-4a^ + 16a;-fl6.
16. (l-^2x'y-4.x(l-x)(l + 2x).
17. x' -\-2a^(y + z)-{-x'(y' -hz'' -^^yz) -{-2xyz(y + z) + fz\
18. a^-2x + ^-{-\^x''-6a^.
19. -Sa^ + ^ + a*-5a-\-{ia\
20. ia;* + 4:x2 + ^aar^ + i«'-2a^-Jax.
21. 24 + 16|!!_8.^^ 32,.
ar y y X
In the polynomial
x3 + X-^ + X+l+- + -\ + ^ (1)
each term after the first is obtained by dividing the preceding term
by x; hence we regard all the terms in expression (1) as arranged
according to the powers of x.
Arranging the given expression according to the powers of x, we
have
x^ 8x o^ S2y my\
y^ y X x2
4 a; ar*
-'-i
9a2 6a 101
a^ 5 a; 25
4.x 4ar'
15a 9a*
23
' x" bx Zb 1
24. 4a.-* + 32ar^H-96+^ + i^.
«* ar
230. Cube root by inspection. When a perfect cube can be
factored by inspection, its cube root is found by inspection.
Ex. 1. 27 a^ - 54 a% + 36 a^b'^ - 8 a%^ = (3 ^2 _ 2 ahy.
.-. ^(27 a« - 54 a^h + 36 a»62 _ 8 a%^) = 3 a^ - 2 aft.
Ex. 2. Find the cube root of the first nine expressions in exer-
cise 60.
231. Cube root of any perfect cube. Let A stand for any
number of terms in the cube root of any perfect cube and
240 ELEMENTS OF ALGEBRA
B for the rest; the terms of A and B being arranged in
descending (or ascending) powers of the same letter, and
every term in A being of a higher (or a lower) degree than
any term in B.
By § 124 we have the identity
^3 + 3 A^B 4- 3 AB^ + l^ = {A-\- Bf. (1)
(i) Let A denote the first term of the root ; then from (1)
it follows that the cube root of the first tenn of the radicand
is the first term of the root.
(ii) Let A denote the first one or 7nore terms of the root ;
then if we subtract A^ from the radicand the remainder is
3A^B + SAB^ + B\ Let a denote the first term of A,
and b the first term of B; then supposing the remainder
3 A^B + 3 AB"^ + B^ to be arranged in descending (or ascend-
ing) powers of the letter of arrangement, 3a^b will be its
first term. But 3 a^6 -;- 3 a- = 6 ; hence.
If the cube of the first term of the root, or the cube of the
sum of its first two or more terms, is subtracted from the
radicand, and the first term of the remainder is divided by
three times the square of the first term of the root, the quotient
will be the next term of the root.
E.g., by (i), the^rs^ term of the cube root of
8 a;6 - 36 x5 + 66 a;4 - 63 ccS + 33 x2 - 9 X + 1
is v^8x6, or 2 ^2 ; and by (ii) the second term is - 36 x^ ^ 3 (2 x2)2,
or — 3 X.
The radicand less (2x'^-Sxy is 12 x* - S6x^ + Sox'^-9x + 1.
Hence, by (ii), the next term of the root is 12 x* -^ 3 (2 x^)^, or 1.
The radicand less (2 x^ — 3 x + 1)* is zero.
Hence the cube root is 2 x^ — 3 a; + 1.
Instead of finding each cube independently, some labor
can be saved by using the relation
A' -f (3 A' + 3^16 + b')b = (A + by,
EVOLUTION 241
and thus making use of the previous cubes. Thus the work
in the example above is usually written as below :
I 2a;2-3a;+l
Sofi-S6x^+6Qx^-6Sx^+3Sx^-9x-\-l
08= Sofi
3a2=12x*
3a6 + 62= -18x3+ 9x2
3.42^12x4-30x3+27x2
-36x5
-36x5+54x*-27x»
12x*-36x3
12x4-36x8-33aj2-9a; + l
3^& + &2= 6x2-9x+l
By (i), the first term of the root is y/8afi, i.e., a = 2x^.
Subtract (2x2)8; then, by (ii), the second term of the root is
-36x5-3(2x2)2, i.e., 6 = -3x.
Hence 3a2 + 3a6 + 52 = 12 x* - ISx^ + 9x2.
Multiply this sum by — 3 x, and subtract the product from the first
remainder. Then in all we have subtracted a* + (3 a2 + 3 aft + b^)b,
or (a + 6)3 ; that is, we have subtracted (2 x2 — 3 x)*, since a = 2 x^
and 6 = — 3 X.
Let A = the terms of the root already found = 2 x2 — 3x,
and b = the next term of the root = 12 x* -=- 3(2 x2)2 = 1 ;
then 3^2 4. 3^5 4. 52 = 12 X* - 36x« + 33x2 - 9 x + 1.
Multiply this sum by 1, and subtract the product from the second
remainder. Then in all we have subtracted
(A + 6)8, or (2 x2 - 3 X + I)'.
As there is no remainder, the required root is 2 x2 — 3 x + 1.
We could just as well write the radicand in ascending powers of x,
or, what is the same thing, begin with the last term.
8x«-36x5 + 66x*-63x8 + 33x2-9x + l. (1)
Thus, by (i), the last term of the cube root of (1) is -^1, or 1 ; and,
by (ii), the term before the last is — 9x -^ 3 • I2, or - 3x, which agrees
with the result above,
Ex. Find the cube root of
27 + 108X + 90x2 -80x3- 60x* + 48x5 - 8x«. (1)
The first term is 3, and the second is 108 x -f- 3 • 3*, or 4 x.
The last term is — 2 x2, and the term before the last is
48x5 -^3(- 2x2)2, or 4x.
Since (3 + 4 x — 2 x2)8 = the given expression ;
3 + 4x — 2x2 = the required root.
242 ELEMENTS OF ALGEBRA
Exercise 89.
Find the cube root of the following expressions :
1. l-{.3x-^6x'-^7x'-{-Qx' + 3x^-{-x^
2. f -3f -\-6y' - 7 f ^ 6f -8y + 1.
3. l-6x + 21i^-Uiif-{-()3x'-o4:X^ + 27x^.
4. 8 a« - 36 a^ + (^(^ a' - 63 a^ + 33 a^ - 9 a + 1.
5. Sx^-{-12x^-30x*-3ox^-^4.ox^-{-27x-27.
6. 27x^-27x^-lSx' + 17x^ + 6s^-3x-l.
7. 24 xY + 96 xY - 6 :t'V H" a^' - 9^ a.y + 64 / - 56 a^/.
8. 27 a;« _ 54 a^^a + 117 a;V - 116 x'a^ + 117 a^-a* - 54 xa'
+ 27 a«.
9. 216 + 342 x- + 171 X* + 27 x^-27^- 109 a.-^ - 108 a;.
10. a:3_9^^27_27
X x^
11. --6a;^ + 12ar2/'-8/.
12. ^V — + — -4-^ + — --•
l^ y' y X x? x^
13.
14.
15. — i , ^ + -^ + J7 + — .
232. Higher roots. The fourth, fifth, or any other root of
a perfect power can be obtained by a method based on one
of the following identities:
A' -{■4:A'B + 6^2j52 + 4.AE'-\'B*={A + By, (1)
A''{-5A'B-^10A'B'+10A'B'-\-5AB'+B'^iA+By. (2)
x'
27
|42.-
-7 +
18_
X
27 27
a^^'^ar'*
a
/^« 7 + ^'
6 6^
3 a'
b'
352 ?>3
GOa;^
2/^
80^3
2/^
90 ar^
2/^
+ 8a
f 108 a;
y
EVOLUTION 243
If the terms of a perfect fourth power are arranged in
descending (or ascending) powers of some letter, from (1)
it follows that the first term of the root is the fourth root
of the first term of the radicand ; that the second term of
the root is the second term of the radicand divided by four
times the cube of the fii'st term of the root ; and that the
last term of the root is the fourth root of the last term
of the radicand. Similarly for any other higher root,
E.g.^ the first term of the fourth root of
81a;*+ 108x3+54x2 + 12a; + l (1)
is v^81 X*, or 3ic ; the second term is 108 x^ ^ 4(3x)8, or 1, which we
know to be the last term of the root.
Since (3 a; +1)*= the radicand (1); 3ic+l is the fourth root
of (1).
Again the first term of the fifth root of
32x6- 80x* + 80x8 -40x2+ lOx-1 ^ (2)
is \/32x^, or 2 X ; the second term is - 80x* h- 5(2 x)*, or — 1, which
we know to be the last term of the root.
Since (2x - 1)5 = the radicand (2); 2x - 1 is the fifth root of (2).
The fourth root can also be obtained by finding the square
root of the square root; and the sixth root, by finding the
cube root of the square root. Similarly for any other root
whose index is not a prime number.
Exercise 00.
By inspection find the fourth root of the expressions :
1. 16 a* - 96 a»aj + 216 aV- 216 aa^ + 81 a;*.
2. x^-%a?a + 24.a?a'-^2xa^ + Ua\
3. 1 + 4 a 4- 4 a^-f- 10 a«+ a«+ 10 a^-\- 16 a^-f 16 a*+ 19 a\
By inspection find the fifth root of the expressions :
4. 80 a^a^ - 80 ax^ + 32 a.-* - 40 a^x - a^ + 10 a^x.
6. 90aV~16aa^ + a^-270aV + 405a*aj-243a«.
244 ELEMENTS OF ALGEBRA
By inspection find the sixth root of the expressions :
6. 192 a; + 64 + 240 a^^ _^ ^.e _^ i2a^ + 60 a;^ + 160 a:^.
7. 1215 a* - 1458 a' - 540 a^ + 135 a^ - 18 a + 1 + 729 a«.
8. 60 a'x'- 16 a^x^-{- 64 a^-^ x^- 12 aa^+ 240 aV- 192 a'x.
ROOTS OF DECIMAL NUMBERS.
233. Square roots. ^1 = 1, and ^10^ = 10 ; hence the
square root of any number between 1 and 100 lies between
1 and 10 ; that is, if a number contains one or two integral
figures, its square root contains one integral figure.
Again, ^100 = 10, and V^O^OO = 100 ; hence the square
root of any number between 100 and 10000 lies between 10
and 100 ; that is, if a number contains three or four integral
figures, its square root contains two figures ; and so on.
Hence, in finding the square root of a decimal number,
the first step is to divide its integral figures into groups of
two figures each, beginning at units' place.
We thus determine the number of integral figures in the
root, and indicate the part of the number from which each
figure of the root is to be obtained.
The group to the left may contain only one figure.
E.g., in the square root of 5 38 24 there are hundreds, tens, and
units ; and the hundreds' figure is the square root of the greatest per-
fect square in 5 ; that is, the hundreds' figure is 2.
234. From § 233, we have the following principle :
(i) The first figure in the square root of a decimal number
is the square root of the greatest perfect square in the first, or
left-hand, group of figures.
Let A stand for the number denoted by one or more of
the first figures of the root, and B stand for the number
denoted by the rest ; then the root is A-}- B.
Hence the radicand = A^-\-(2A + B)B. (1)
Subtract A^ and then divide by 2 ^ + J5,
(radicand -A^^{2A-^B)=B. (2)
EVOLUTION 245
Let h stand for the number denoted by the first figure in
B, i.e., the figure whose order of units is the next lower to
the lowest in A ; then from (2)
(radicand -A^-i-2A>b. (3)
From inequality (3), we have the following principle :
(ii) If the square of the first part of the root is subtracted
from the radicand, and the remainder is divided by twice this
part of the root, the quotient will be greater than the next figure
of the root.
E.g., by (i), W\Q first, or hundreds, figure in the square root of
5 47 56
is 2 ; since 4 is the greatest perfect square in 5.
The radicand less (200)2 is 14756.
Hence, by (ii), the tens figure of the root cannot exceed
14756 -2(200), or 3 tens.
The radicand less (230)2 is 1856 ; hence the root is greater than 230,
and 3 is the tens figure of the root.
By (ii), the units figure of the root cannot exceed
1856 H- 2(230), or 4.
The radicand less (234)2 jg zero.
Hence the square root of 54756 is 234.
Instead of finding each square independently, much labor
can be saved by using the relation (§ 229)
A'-\-{2A-^b)b={A^by,
and thus making use of the previous square.
The work in the example above is usually written as
below :
5 47 56(234
^2 =(200)2= 4 00 00
2^ + 6 = 2(200) + 30 = 430)14756
(2 ^ + 6)& = 430x30= 129 00
2^ + 6=2(230)4- 4= 464)18 56
(2^ + 6)6 = 464x 4= 18 56
246 ELEMENTS OF ALGEBRA
At first A = 200. Subtracting A^, or 200^, from the radicand, and
dividing the remainder, 14756, by 2 A, or 2(200), we find that the tens
figure of the root cannot exceed 3.
Multiply 2 A + b, or 430, by 6, or 30, and subtract the product ;
then in all we have subtracted A- + (2 A+b)b, or (A-\-by; that is, 230"'2.
Now let A = 230, the part of the root already found,
and b = the next figure of the root.
Dividing the remainder 1856 by 2 A, or 460, we find that the units
figure of the root cannot exceed 4.
Multiply 2 ^ + 6, or 464, by b, or 4, and subtract the product ; then
in all we have subtracted A^ + (2A + b)b, or (A + 6)2 ; that is, (234)-^.
Omitting the ciphers and explanation, and in each remainder writing
the next group of figures only, the work will stand as below :
5 47 56(234
4
43)147
129
464)18 56
18 56
235. If a number has decimal places, its square will have
twice as many. E.g., O.S^ = 0.64 ; 0.25^ = 0.0625.
Hence to determine how many decimal figures there will
be in the square root of a number, we divide its decimal
figures into groups of two figures each, beginning at the
decimal point. If the group to the right does not contain
two figures, a cipher must be annexed.
Ex. Find the square root of 5727.2976.
Formula, A^ +(2 A + b)b = (A + by.
75 27.29 76(86.76
64
166)1127
9 96
1727)13129
120 89
17346)10 40 76
10 40 76
Here at first A = SO, b = 6 ; next ^ = 86, 6 = 0.7 , next A = 86.7,
b = 0.06.
EVOLUTION
247
Exercise 01.
Find the square
! root of the numbers
:
1. 2916.
9. 29376400.
17.
0.0022448644.
2. 2601.
10. 52.2729.
18.
0.68112009.
3. 17956.
11. 53.7289.
19.
25/49.
4. 33489.
12. 883.2784.
20.
64/81.
5. 119025.
13. 1.97262025.
21.
121/36.
6. 15129.
14. 3080.25.
22.
144/49.
7. 103041.
15. 41.2164.
23.
169/196.
8. 835396.
16. 384524.01.
24.
225/289.
236. Cube root. Since ^1 = 1, and ^1000 = 10, it fol-
lows that the cube root of any number between 1 and 1000
lies between 1 and 10 ; that is, if a number contains one,
two, or three integral figures, its cube root contains one
integral figure. Again, ^1000 = 10, and ^1000000 = 100;
hence, if a number contains four, five, or six integral figures,
its cube root contains two integral figures ; and so on.
Hence, to determine how many integral figures there are
in the cube root of a number, we divide its integral figures
into groups of three figures each, beginning at units' place.
The last group to the left may contain only one or two
figures.
When the figures of a number have been divided into
groups of three figures each, from what precedes it follows
that,
(i) The first figure in the cube root of a decimal number is
the cube root of the greatest cube in the first, or left-hand^
grouj) of figures.
Using a notation analogous to that in § 234, we have
radicand - A'' = (3 A' -^ 3 AB + B^ B.
'. (radicand - .1'') - (3 A- -\- 3 AB -^ B^ = B.
.-. (radicand -yl^) --3. 42 > 6. (1)
248 ELEMENTS OF ALGEBRA
From inequality (1) it follows that,
(ii) If the cube of the first part of the root is subtracted
from the radicand and the remainder is divided by^ three times
the square of this part of the root, the quotient will be greater
than the next figure of the root.
E.g., by (i), the Jirst or tens^ figure in the cube root of
614 125
is 8, since 8^, or 512, is the greatest perfect cube in 614.
The radicand less (80)^ is 102125.
Hence, by (ii) , the units' figure of the root cannot exceed
102125 -- 3(80)2, or 5.
The radicand less (85) ^ is zero.
Hence, the required root is 85.
Instead of finding each cube independently, much labor
can be saved by using the relation
A' -{- {S A' + 3 Ab -\- b')b = (A + by,
and thus making use of the previous cubes.
Thus, the work in the example above is usually written
as below, without the explanations to the left :
614 125(85
43 = 512 000
3^2^3(80)2 =19200)
3^6 = 3.80-5= 1200
62 = 52 = 25
20425
102 125
102 125
237. If a number has decimal places, its cube will have
three times as many. Thus 0.2^ = 0.008 ; 0.12^ = 0.001728.
Hence, to determine how many decimal figures there will
be in the cube root of a number, we divide its decimal
figures into groups of three figures each, beginning at the
decimal point.
If the group to the right does not contain three figures,
ciphers must be annexed.
lEBATIONAL NUMBERS
249
Ex. Find the cube root of 129554.216.
Formula, A^ -\-(iS A^ + S Ab + b'^)b = (A + by.
129 554.216(50.6
125
750000
9000
36
759036
4 554 216
4 554 216
Here at first ^ = 50, 6 = 0; next A = 50.0, b = 0.6.
Exercise 92.
Find the cube root of the numbers :
1. 74088.
2. 15625.
3. 32768.
4. 110592.
5. 262144.
6. 1481544.
7. 103.823.
8. 884.736.
9. 1953125.
10. 7077888.
11. 2.803221.
12. 12.812904.
13. 56.623104.
14. 264.609288.
15. 1076890625.
16. 8/27.
17. 64/125.
18. 343/1728.
INCOMMENSURABLE ROOTS, OR IRRATIONAL NUMBERS.
238. The nth power of a whole number is evidently a
whole number which is a perfect ?ith power ; and the ?ith
power of a fraction (whose numerator and denominator are
prime to each other) is a fraction whose numerator and de-
nominator are perfect nth. powers prime to each other.
Hence, it follows that
(i) The nth root of a tvhole mimber which is not the nth
poicer of another whole number is not a commensurable
number.
(ii) The nth root of a fraction whose numerator and ck^
nominator {prime to each other) are not the nth powers oj
whole numbers, is not a commensurable number.
250 ELEMENTS OF ALGEBRA
E.g., as 2 is not the square of any whole number, y/2 is not a com-
mensurable number, and therefore is not as yet included in our number
system. The same is true of ^3, y/b, ^1 •••.
Again, as the terms of the fraction 2/3 are prime to each other and
are not the squares of whole numbers, ^(2/3) is not a commensurable
number.
239. To enlarge our number concept so as to give mean-
ing to such expressions as ^2, ^5, etc., we assume the
identity
to hold when the radicand u is not a perfect /ith power.
E.g.., yJ2 is the number whose square is 2, i.e. ( \/2)^ = 2.
Again, ^5 is the number whose cube is 5, i.e. {^b)'^ = 5.
240. The ?ith root of a number which is not a perfect nth
power is called an incommensurable root or an irrational
number; as, V2, V3.
241. An irrational number, or any other number which is
not a whole or a fractional number, is called an incommen-
surable number ; as ^3, -^6, or the ratio of the circum-
ference of a circle to its diameter.
242. Approximate values of incommensurable roots.
If to 2 we add ciphers and apply the method of finding
the square root, we obtain the result below :
2.00000000)1.4142 ...
1_
24)100 1st remainder
90
281)400 2d remainder
281
2824)11900 3d remainder
11290
28282)60400 4th remainder
56564
0.00003836 5th remainder
IRRATIONAL NUMBERS 251
Each remainder in the above process is the difference be-
tween 2 and the square of the corresponding part of the root.
This remainder decreases rapidly as we increase the num-
ber of figures in the root; hence the square of the root
found approaches nearer and continually nearer 2; and
therefore the root itself approaches nearer and continually
nearer ^2.
By continuing the operation indefinitely we obtain a com-
mensurable number which approaches indefinitely near and
continually nearer ^2, but which, by § 238, can never reach
^2. This increasing commensurable number is said to
approach the incommensurable root ^2 as its limit.
In like manner we can find a commensurable number
which shall differ from any incommensurable root by as
little as we please.
Exercise 03.
Obtain to three places of decimals the value of the roots :
1. V3- 4. V^- 7. V^-3- 10. V^-^^-*- 13. ^2.
2. V^- 5. Vll- 8- V^>-^- 11- V^-^^- 14. ^4.
3. V^. 6. V13. 9. V^-^^- 12- V^-^- 15- v^2.5.
243. The quality-unit + 1 or — 1 multiplied by an arith-
metic incommensurable number is a positive or a negative
incommensurable number ; as, + ^2, — ^3, — ^5.
244. The fundamental laws which have been proved for
commensurable numbers hold also for incommensurable
numbers. The proof of these laws for incommensurable
numbers will be found in the chapter on the theory of
limits.
245. An irrational expression is one which involves the nth
root of an expression which is not a perfect nth power ; as,
^^i V(^ + ^)- ^^y i^'^'^lional numeral expression denotes
an irrational number. But, just as a fractional literal
252 ELEMENTS OF ALGEBRA
expression denotes both integral and fractional numbers,
so an irrational literal expression denotes both rational and
irrational numbers.
E.g., the irrational literal expression ^a denotes a commensurable
or rational number, when a = 1, 4, 9, 1/4, 4/9, ••• and an incommen-
surable or irrational number when a = 2, 3, 5, •••.
Observe that commensurable and incommensurable apply
to numbers only ; while rational and irrational apply to
either numbers or expressions.
Exercise 94.
1. Is the number -^jA commensurable or incommensura-
ble? sjQ? V9? V12? V14? V(4/9)? ^(8/27)?
-3/3? -^216?
2. Is the expression -^a* rational or irrational ? -^x ?
^a^7 V(^ + ^)^ ^{a + xy? ^(a/x)?
3. Give ten sets of values of a and x for which ^(a/x)
denotes a commensurable, or rational, number.
CHAPTER XVII
SURDS
246. A surd number is an irratioual number in which the
radicand is a rational number; in other words, it is an
incommensurable root of a commensurable number.
E.g., y/b, ^7, ^(2/3) are surd numbers; so also is ^a when a
denotes a commensurable number which is not a perfect square.
The incommensurable root V(^ + \/2) i^ "^^ ^ surd number, since
the radicand 3 + ^^2 is not a commensurable number.
247. A surd expression is an irrational expression in
which each radicand is a rational expression ; as -yja^ VS/G,
248. Surds of different orders. A surd of the second order,
or a quadratic surd, is a surd with the index 2 ; as ^5, ^a.
A surd of the nth order is a surd with the index n ; as -y/a.
Observe that -^ ^5 is a surd of the 6th order.
249. A rational number or expression can be written in
the form of a surd of any order.
E.g., 3 = v^, ^7, {/81, or ^43
and «= V«^ \/«^ v^«S ^^ v^«^-
250. A surd expression is in its simplest form when each
radicand is integral, its numeral factor being as small as
possible, and its literal factor of as low degree as possible.
E.g., the simplest form of the surd y/9> is 2y/2.
The simplest form of the surd ^(16 x*y5) is 2 xy^(2 x^).
263
254 ELEMENTS OF ALGEBRA
251. Reduction of surds to their simplest form. The cases
which most frequently occur are the three following :
I. Radkand integral. Eesolve the radicand into two
factors, one of which is a perfect power of a degree equal to
the order of the surd, and apply the law, ^{ab) = y/a • ^b.
Ex. 1. \/l35 = v'PTs = 3^5.
Ex. 2. 7 V50 x'y<' = 7 \/(5 x'^y)^ • 2 y
= 7x5 x'^yV2y = 35 x^yV2y,
Ex. 3. 5 </l28xV = 5 V (4 xy)'^ - 2 x^y
= 5 X 4xy \/2 x'^y = 20 xy y/2 x^y.
II. Radicand fractional. Multiply both the numerator
and denominator by such a number as will make the de-
nominator a perfect power of a degree equal to the order
of the surd, and apply the law, ^{a/h) =^a/^h.
35 V35
Ex.1. :\^=;f^=^
\49 \ 73 7
sHba^x </lbcfibx
Ex 2 ^/ 3a; _ ^n^a%x ^ V7^
\5a&5- \(5a6-0^ 5
ab^
III. Factor common to exponent of radicand and index
of root.
In this case we apply the following principle :
The index of the root and the exponent of the radicand can
he multiplied or divided by the same number.
That is, "{/'a'"'' = W«""' = V^""- §§ 223, 227
Ex. 1. ^(27 a^x^) = V(3 ax^ = V(3 ax).
Ex. 2. ^(IG a%i2) = ^x» . (2 axy
^x8. </(i2axy = x^(2ax).
SURDS 255
Exercise 95.
Reduce each oi
nhet
15.
Allowing surds 1
:o its
%5.
simplest form ;
1. ^U7.
■\/a^+2a^b+ab
a jb'-+^
6"\ a'
2. V288.
16.
Wh
3. 3V150.
17.
ivf
26.
(«^^/:;:-
4. 2V720.
5. ^256.
18.
19.
3^|.
27.
6. ^432.
7. 5V245.
20.
3xy j5z'
z \9a^y
28.
29.
</25.
</(8/:«0-
8. ^1029.
9. ^3125.
10. v'-2187.
21.
22.
Sx 3/27 a*
2a\ oj* •
26 J a*
a \Sb'
30.
31.
32.
^(9/36).
11. V27a-^6^
12. A/-108a.V-
23.
"aJ^3-
33.
13. VarV""'.
24.
34.
35.
JJ/(32/x'»).
14. V.t-^-'Y^
•V(2V«"')-
36. '%/(S'"/x
-).
38.
^c-
-54x'"+'y').
37. ^/{b-c
W-
c'*^). 39.
IJ/(64aV").
252. Surds which are rational multiples of the same
monomial surd are said to be like, or similar, surds.
E.g., 2y/3 and y/'S/7 are like surds, so also are by/a and vTa, or
2y/a.
253, To adci o?* subtract surds, we reduce them to their
simplest form, and unite those that are similar.
Ex.1. ^135 + ^40 = 3^5 + 2^5 = 5^5.
Ex. 2. 4\/T28 + 4V76 - 5^/1(32 = 32^/2 + 20^/3 - 45 V2
= 20v3- 13 V2.
256 ELEMENTS OF ALGEBRA
Exercise 96,
Simplify each of the following surd expressions :
1. V27+V48. 11. V252-V294-48Vi.
2. 2V180-V405. 12. 4V63 + 5V7-8V28.
3. 2V28-V63. 13. V^' + i^a'-3i^2U?
4. 5V208-3V325. 14. ^54+Vi-fVf
3/orT A. 3/^7-7 . 3
5. V^12-V50-V98. 15. V27c^-V8c4+V125c.
6. 3V12-V27 4-2VT5. 16. "v/o^ - ^^« + -V^326.
7. V^4-^V176 + 2V99. 17. Va^+Vb^-^/Umc.
8. 2V363-5V243 + V192. 18. 3 V147 - J Vi - V2V
9. 2^189+3^875-7^56. 19. 3Vl + 3V|-A/|f
10. ^81-7^192+4-^648. 20. j\-^72 -^^^ + 6^211.
21 . ^(9 aPf) + V(27 0^2/) + ^ a/(729 xV).
22. 2 ^(3 a^fe) - -^(9 a%^ + ^(125 a^d).
23. V(4<^'+4a'6)+V(9«&' + 9&').
24. Vit*^ — a^2/ — \^f—^ — V (a; + 2/) (^ — 2/0-
254, Surds of different orders can be reduced to identical
surds of the same order. This order can be any common
multiple of each of the given orders, but it is usually
most convenient to choose the least common multiple.
Ex. 1. Reduce v^o^, -v^P, Vc^ to identical surds of the same order.
The L. C. M. of the indices 3, 4, and 6 is 12. By III. of § 251,
V^^='<f^^; </b^^'V¥; V?='</g^o,
SURDS 257
Ex. 2. Which is the greater ^6 or ^10 ?
Reducing these surds to the same order, we have
^6 = ?2/6* = 1^1296, (1)
and ^10 = 12/103 = I^IOOO. (2)
From their values in (1) and (2), it follows that ^6 > ^10.
255. The product of two or more surds is found by ap-
plying the law
■;ya^b = ^{ab). § 222
Ex.- 1. V7 X v'28 = V7 X 2 V7 = 2 X 7 = 14.
Ex. 2. 2^14 XV21 =2V(14 x 21)= 2V(72 x 6)= 14^6.
When the surds are of different orders, they should be
reduced to the same order.
Ex. 3. V3 X ^2 = ^38 X ^2 = ^(38 x 22) = ^108.
Conversely to § 251, the coefficient of a surd can be
brought under the radical sign by reducing it to the form
of a surd of the same order.
Ex. 4. 5 V3 = V25 X V3 = >/75.
Ex. 6. X ^x^ = {/7^ X ^a;8 = ^v?.
Ex. 6. Multiply 2^3 + 3^2 by 4v3 - 5^2.
The work can be arranged as below :
2\/3 + 3V^
4\/3-5V2
24 + 2V6-30 = 2V6-6.
256. In finding powers of monomial surds we often make
use of the law, {-^aj = </ar (§ 226).
Ex.1. (3\/^)2 = 32(v/^)2 §119
= 9^(ax)2 = 9^(a2x2).
Ex. 2. (2y/xy = 2\y/xy = Sx^x.
258 ELEMENTS OF ALGEBRA
When applicable, the identities in Chap. IX. should be
used in finding the products of polynomial surds.
Ex. 3. ( V3 - \/5)2 = ( V3)2 - 2 V3 . Vs + (V5)2
= 3 - 2\/l5 + 5 = 8- 2V'l5.
257. Two binomial quadratic surds which differ only in
the quality of a surd term are called conjugate surds.
E.g., 3 + V^ 3,nd 3 — y/2 are conjugate surds ; so also are y/a-\- y/b
and ^a — y/b, or ^/a + y/b and - ^/a + y/b.
The product of two conjugate surds is rational.
E.g., (3 4- V2) (3 - V2) = 32 - ( ^2)2 = 7.
( V« + \/^)( v« - \/ft) = ( V«)^ -( V&)^ = « - ^'
Exercise 97.
Eeduce to surds of the same order :
1. V3, ^'^' 6- V»? ^^» a/«-
2. V(V2). ^(2/3). 7. 2, ^3, V4.
3. -^2, -</3. 8. ^3, 2, ^7.
4. -^8, V3, -^6. 9. V«^ ^^ Vc-
Bring the coefficient under the radical sign :
10. 11 V2. 11. 14 V5. 12. 6^4. 13. 5^6.
n^ 4 177 _ 3a5 /20 c^ _ 2a 3'27 a^^
Which is the greater :
17. V^ or -3/10 ? 19. 5 V2 or ^344 ?
18. V^oi^-v/H^ 20. ^5 or ^10?
21. -^a'^ or ya, when a<l? when a>l?
SURDS 259
Obtain the simplest form for each of the following
products :
22. 2 V15 X 3 V5. 29. ^168 x ^147.
23. 8V12 X 3V24. 30. 5^128 x 2^'432.
24. Vi2 X V^T X V75- 31. ViO X ^200.
25. ^16 X ^6 X ^9. 32. ^4 x V^'-
26. ^12 +-^75x^30. 33. (V^-V3)(V^ + V3)-
27. ^0x^12x^18. 34. (V^)-V7)(V^' + V)"-
28. v^a; + 2 X -^x - 2. 35. ( V^ - V-*^) ( V^' + V^)-
36. ( - v'c - V«) ( - V^ + V«)-
37. (-V^+V^)(V^+V»')-
Find each of the following powers :
38. (y/2f. 42. (^aby. 46. (v'^)«.
39.. (2V3)*. 43. (^ay. 47. (2V^^)*.
40. (Va;)'. 44. (-^'ay. 48. (3^/^2^)*.
41. (-^ay. 45. (-^63)^^ 49. (2^/^::rP)6^
50. (V3-V5)^ 54. (^2-</4)2.
51. (4-2V3/. 55. (V6-^2)^
52. (vr^ 4- 2 V3)=^. 56. (V2+V3+V^)'-
53. (V3-V2)^ 57. (1+^2 +V^)'-
Find each of the following products, and simplify :
58. (2V5 + 3V3)(3V5-4V3).
59. ( V2 + V^ + V^) (2 V2 + 3 V3 + V^)-
60. (5+-^4)(V3+V2).
260 ELEMENTS OF ALGEBBA
61. (2V3+^2)(2V3-^4).
62. (8-3V7)(8 + 3V7).
63. (H-V2-V3)'-
64. ( V2 + V3 - V^) ( V2 + V3 + V5).
258. Division of surds.
Suppose it is required to compute the value of y/6/^7. We might
find ^6, which is 2.236.-.; then find ^7, wliich is 2.645...; and
finally divide 2.236 ... by 2.645 ....
Of these three long operations two will be avoided if we first multi-
ply both dividend and divisor by ^7, as below :
V5/V7=V3o/7
= 5.916 •../7 = 0.845....
Observe that the new divisor is a rational number.
This example illustrates the following principle :
The quotient of one surd divided by another is put in the
simplest form for computation by multiplying both divi-
dend and divisor by such a factor as will render the divisor
rational.
This process is called rationalizing the divisor^ or rational-
izing the denominator.
The factor by which we multiply the divisor to obtain a
rational divisor is called the rationalizing factor; as, -^7
above.
The cases which most frequently occur are the three
following :
I. When the divisor is a monomial surd ; as, ^a;™.
Ex. 1. _J_^ 2XV5 ^A
3V5 3V5 X V5 15
Ex. 2. _^ = -A-^<J^=A>3.
Here the rationalizing factor is ^a*.
8UBDS 261
The simplest rationalizing factor of -s/x"^ is evidently
II. When the divisor is a binomial quadratic surd, the
simplest rationalizing factor is the conjugate of the divisor.
Ex 1 5+V7^(5+V7)(3+v7)^22 + 8v7^. ^
• 3-V7 (3-V7)(3 + V7) 9-7 ^'
Ex. 2. a/Q^ + a/^ - C a/« + a/^)(V^ + a/^) -- c< + 2 Vo^ + 6
V<^-V^ (V« — \/^)(v'« + V^) a-b
III. When the divisor is of the form (^a -f- -y/b) + -y/c,
first multiply by the expression ( y/'a + ^b) — ^c.
The divisor thus becomes
( V« + V^)' - ( V^)'. or (a + 6 - c) + 2 ^(ab). (1)
Next we multiply by the conjugate surd
(a 4. 6 _ c) - 2 ^(ab).
The divisor thus becomes the rational expression
(a-^-b-cf- (2 Vaby.
Ex. ^^2 ^ V2(V2-fV3 4-V5)
V2+V3-V5 [(V2+V3)-V'5][CV2+V3) + V5]
^ 2+V6 + V/10 ^2 + a/6+V10
^ (2 + yC + ylO) X V6
_ 2^/6 + 6 + 2^15 ^ v/6 + 3 + v/15
12 6
259. When applicable, the identities in Chapter IX. should
be used in writing the quotient of two binomial surds.
Ex. V^^ + Vy«_Cv/x)8 + (Vy)'»
=x-y/xy -\-y.
262
ELEMENTS OF ALGEBRA
Exercise 98.
Compute to three places of decimals :
1. 14--V2. 3. 48--V6- 5. 144 --V6.
2. 25 -V^- 4. V2-V^- 6. 4--V243.
Rationalize the denominator and simplify :
7. 3V3/(2V2). 11. 12/ V"^- 15- V«/^«-
8. Vl5/V(3/5). 12. 2^6/V2. 16. ^a/^a.
9. V21/V(V3).
10. 10/^5.
19.
2V5
V5 + V3
20. l^+li^^
15 - 2 V3
21 V5 + 3V3.
' 2V5-V3
22.
23.
24.
25.
26.
V6-3V12
2V<3 + V12
2 V3 + 3 V2
5 + 2V6
V9 +
V9 4- x' + 3
r
£c -f- Vi^^ — y'^
1
1 + V2 + V^*
13. 3V2/^9.
17.
V
x'/^af.
14. -^20/(3^16). 18. ^(ax)/^x.
3
27.
2+V3 4-V5
28. ^+V^+V^.
1+V2-V3
29.
30.
31.
32.
33.
34.
V3
V2+v^+V5
3
V2-V6-V'^*
■Vx — 2-\-Vx
Va? — 2 — V ic
a— Va^ + 3
Va — 6 — Va + b
Va — & 4- Va + b
3 + 4V3
V6+V2-V5'
SUBDS 268
Write each of the following quotients :
35. (a — x)-i- (y'a — ^x).
36. {ax-ay)/(^x-^^y).
37. (l-l/x)-(l4-l/V'^).
38. (a/b - x/y) h- [ V(« A) + V(V^)]-
39. (v«'-V^')^(V«-V^)-
40. (arV^J- W2/)-^(V^+V2/)•
Rationalize the denominator of each fraction :
41.
42.
x"
44
Vl + ar^-
-VI-
X
45.
Vl + ar' + Vl-
2 Va + 6 — 3 Va
a;-
6
Va^ H- a* H- a 2 Va + 6 — Va — 6
43 V10+V5 4-V3. 46 (V3+ V5)(Vo + V2)
V3+V1<>-V5 * V-+V3 + V5
260. A root of a monomial surd is found by applying the
^^^^ V V« = V«- § 227
Ex. 1. »/ V7 = ^7.
Observe that {/{fa= {/{/a, since each member = *^a.
Ex.2. Vv^(4x2) = {/V(4a;2)= */(2x).
Exercise 09.
Simplify each of the following expressions :
1. -^V(27a'). 5. V(^V^)- 9- ^WV^)-
2. V^(9ajO- 6- VA/(25iry/16). 10. ^^--^(x/^x).
3. ^v(«0. 7. V(^V^)- 11- "-V(^/V^O-
4. ^^(A^)- 8. V(2/V2) 12. V(^V</^c)-
264 ELEMENTS OF ALGEBRA
PROPERTIES OF QUADRATIC SURDS.
261. // x+^y = a^^b, (1)
where x and a are rational numbers, and ^y and -^b are surd
numbers; then,
X = a and y — b. (2)
Proof. Transposing a and squaring, from (1) we obtain
(x — ay-\-2(x — a)-y^y-\-y = b.
.'. 2(x - a)V2/ = (b-y)- (^ - af. (3)
Since a surd number cannot equal a rational number,
equation (3) is satisfied when and only when x = a and
y = ^•
262. A quadratic surd number cannot be equal to the sum
of a rational number, other than zero, and another quadratic
surd number.
Proof Let ^b = x-^^y, (1)
where cc is a rational number and -^^b and -y/y are surd
numbers; then, by § 261, we have
x = 0 Siiid y = b. (2)
263. Square root of the binomial surd a -\--^b.
Suppose Va ± V^ = V^ ± V2/- (1)
Square, a ± Vb = x + y ± 2Vxy.
Hence, by § 261, we have
x + y = a, 2-\/xy = -\/b ;
or x-\-y = a, Axy = b. (c)
Solving system (c) for x and y, and substituting their
values in (1), we obtain the value of y'(a ± V^)-
We shall here consider only those cases in which system
(c) can be solved by inspection.
SURDS 265
Ex. 1. Find the square root of 18 + 8^5.
Assume V18 + 8V5 = V^ + V'^- (1)
Square IS -^S^o = x + y + 2y/xy.
.-. a; + 2/ = 18, and 2>/^ = 8V5; §261
or X + y = 18, and xy = 80. (a)
By inspection we see that one solution of system (a) is
X = 8, y = 10.
.'. V18 + 8^6 = V8 + VIO = 2 V2 + VIO.
Ex. 2. Extract the square root of 83 — r2y'35.
Assume V83 - 12 ^35 =^x- y/y.
Square 83 - 12^35 = x + y - 2Vxy.
.-. X + y = 83, and 2Vxy = 12 V35,
or x + y = 83, and xy = 1260.
By inspection, x = 63, and y = 20.
.-. V83 - 12V35 = V63 - V20 = 3 V7 - 2y/5.
By taking x = 20 and y = 03, we would obtoin the negative root
of the given number.
Exercise lOO.
Find the square root of the binomial surds :
1. 6 + V20. 6. 11-2V30. 11. 4J-IV3.
2. 12-6V3. 7. 7-2V10. 12. 17 - 2 V66.
3. 16 + 6V7. 8. 17-12v'2. 13. 19+8V3.
4. 13-2V42. 9. 47-4V33. 14. 11+4V6.
5. 28-5V12. 10. 19 + 4V22. 15. 15-4V14.
CHAPTER XVIII
IMAGINARY AND COMPLEX NUMBERS
264. Quality-units V— 1 and — V— 1. A.s we have seen
in § 219, an even root of a negative number, as V—2,
cannot be a positive or a negative number, and therefore
is not as yet included in our number system.
To give a meaning to such expressions as V— 1 and
V— 2, we assume the identity
{</uy = u (1)
to hold when u is negative and n is even (§§ 218, 239).
Thus, V— 2 denotes that number whose square is —2.
An important particular case of (1) is
(V~l)^ = -l. (2)
Since any power or root of -f 1 or — 1, heretofore ob-
tained, is a quality-unit, Ave call V— 1 a quality-unit.
That is, V — 1 is a quality-unit ivhose square is ~ 1.
Squaring both members of (2), we obtain
(V^iy^ + l; (3)
that is, the fourth j^oiver o/ V— 1 is equal to -f 1.
Again, the opposite of the quality-unit V— 1 is —V— 1.
Also, (V"-^f-(V^'-V^l = -V^=^; (4)
that is, the cube o/ V— 1 is equal to its opposite, —V— 1.
The quality-units V— 1 and —V— 1 involve the idea of
the arithmetic one and that of oppositeness to each other.
266
IMAGINARY NUMBERS 267
Observe carefully the above relations of the quality -unit
V^ to — 1, + 1, and — V^^.
The quality-units V— 1 and — V — 1 are called imaginary
units.
The units V— 1 and — V— 1 are for brevity often de-
noted by i and — i, i being used as a numeral.
Note. The word imaginary, as here used, must not be understood
as implying that the units i and — i are any less real than -f 1 and
— 1. 'J'he expression V— 1 was called imaginary when it first made
its appearance in Algebra before its meaning and uses were under-
stood. The name is unfortunate, but with the above explanation we
shall use it.
265. Imaginary numbers. Any arithmetic multiple of the
imaginary unit i or — i is called an imaginary number.
An imaginary number is commensurable or incommensura-
ble, according as its arithmetic factor is commensurable or
incommensurable.
E.g., t3, t(3/5), — i7 are commensurable, while i^y2 and
— 1^(2/3) are incommensurable imaginary numbers.
266. Multiplication by the imaginary unit V~ 1' ^^ ^ ^^
defined by assuming the commutative law; that is, a being
any number, we assume that
a X V— 1 = V— 1 X a, or a x i = ia.
267. Since /•« = «• /, the imaginary number t • ci or — / • a
can be written ai or — ai.
Imaginary whole numbers form the following series :
..., -3i, -2i, -i, 0, i, 2i, 3*, •••.
Observe that the one and only number which is common
to the series of real and imaginary numbers is 0.
268
ELEMENTS OF ALGEBRA
268. Geometric representation of quality-units.
A directed line is a line whose direction and length are botli consid-
ered. To represent geometrically quality-numbers we use directed
lines. Of the directed line OA',
■g O is the origin and A the end.
Let 0^= + l; then 0^'=-l.
Substituting these values in
(+1). V^ri.^/3-i = _i,
we obtain
OA . \/^ . v"^ = OA'.
That is, multiplying OA by
V— 1 twice in succession re-
verses its direction ; hence we
can assume that multiplying OA
by V— i twice in succession re-
volves OA through two right
angles in the plane ABA' and in a direction opposite to that of the
hands of a clock.
Hence multiplying OA by yf^l once would revolve OA through
one right angle in this direction ; that is,
/ +
/
/
f ~1
N
" \
S
\
\
\
\
+ ' 1
\ ^
\
N
O > 1
/
/
/
/
/
y
B'
OA ' V^l = OB.
But OA-V^l=(+l).V^^=V'^.
From (1), (2), OB = V^^, or i.
.'. OB' =- OB = -i.
(1)
(2)
Hence, if the primary quality-unit + 1 is represented by the
directed line OA, the quality-units — 1, i, and — i will be represented
by the directed lines OA', OB, and OB', respectively.
As the lines OB and OB' are just as real as the lines OA and OA',
so the quality-units i and — i are just as real as -|- 1 and — 1.
Arithmetic multiples of i and — i can be represented by distances
along the lines OB and OB or their extensions, just as multiples of
-f 1 and — 1 are represented by distances along the lines OA and
OA' or their extensions.
Again, if in a football game we denote the forces exerted in the
direction OA by positive real numbers ; then negative real numbers
will denote the forces exerted in the opposite direction OA', positive
IMAGINARY NUMBERS 269
imaginary numbers will denote the forces exerted in the direction OB,
and negative imaginary numbers will denote the forces exerted in the
direction OB'.
To express by numbers the magnitudes and directions of the many
other forces in the game we need still further to enlarge our concept
of quality-numbers, as is done in § 285.
269. Since imaginary numbers are simply arithmetic mul-
tiples of the units i and — /, they are added and subtracted
the same as real numbers.
That is, at ± bi = {a± b)i, (1)
which is the converse of the distributive law.
Ex. 1. 4 I + C i = (4 4- 6) t = 10 I.
Ex. 2. (7/3) i - (0/3) i = (7/3 - 5/3) i = (2/3) i.
270. When the imaginary unit is a factor of a product, the
distributive law follows from its converse in § 269, and the
associative law follows from the commutative law in § 266 ;
that is, (fl ±b)i= ai ± bi, § 269
and ai • bi = i-ab = -ab. § 267
271. Powers of i. From § 264, we have
,^ = -1, i' = ~i, i* = -\-l. (1)
Ex. 1. 2-7 = 1-4 . 1-3 = (+ 1) ( _ i) = _ f. l)y (1)
Ex. 2. I'lo =(i4)--2i2 =(+ 1)2(- 1) = - 1. by (2)
Ex. 3. zi8 =(^iiyi=(^\)Si:^ i^ by ( 1 )
If n is any positive integer including zero, we have
,.4n =(i4y = ^_^iy^_^i. (2)
/4«+i_^-4u^- =/. by (2)
/4n+2 = ,;4«^-2 = _ 1 ; by (1), (2)
/4n+3 = ^•4.^-3 ^ _ / by (1)^ (2)
270 ELEMENTS OF ALGEBRA
Hence^ any eve7i power of i is +1 or — 1, and any odd
power of i is i or — /.
272. The square root of any yiegative number is an imagi-
nary number;
that is, V^ = V« • '• (1)
Proof By the commutative and associative laws we have
( V« • ^/^l)' = {^a)\-y/^^f = ~a; § 271
hence, y^V^^ = V~a, or conversely (1). § 221
E.g., V- 16 = i . vie = 4 i, or W^l.
and y/— ai=i ' Va^ ^ a^, or aV— 1.
273. To add or subtract imaginary numbers given in the
form Va, we first reduce them to the typeform, -^a • i.
Ex.1. V^l^ -i-y/-Sl-V-3{J = 7i + 9i-ei
= (7 + 9_6)i = 10i, orlOV^l.
Ex. 2. V-9a2 4. V- 4 &-^ - V- 7 c'^ = 3 a • i + 2 6 • i - c V7 • i
= (3a + 26 -Cv/7)i.
274. From the commutative and associative laws we have
the following principle :
The product of two or more quality-numbers is equal to the
product of their quality-units into the product of their arith-
metic values.
Ex. 1. V- 5(- Vll)= i •-! • V^' \/ll =- ^'V^^i or -V- 55.
Ex.2. V^8. \A^ = iV3- V7 =-V21.
Ex. 3. V^ . V^ . \/^ = i3y'2 . y'3 . V5= - iy/^^, or - V^M
Ex. 4. V^ • V"^ • v^ • v/^ = /I V^ V^^ V^ V' = v^«^-
IMAGINARY NUMBERS
271
Exercise 101.
Simplify each of the following expressions :
1. V^^+V^=^-V^=loo.
2. V^^-V^^-hV^^^16.
3. v^^^-v^^-v^^n^.
4. V-l)a^-V-4a2_V-l(5a='.
5. V - 36 6^ _ V - 49 6^ _f- V - 81 6^
6. V — (a; + af.+V— {x — af.
7. 3 V^ + 7 V-c - 11 V-6 + 2 V-(m-?r)-
8. V2V^^. 12. V^^-V^=^.
9. V3-V^^. 13. V^^Vie.
10. V^^ • V^ni. 14. V^=l^ • V^3 .
11. v^^V^. 15. Ve-V^^-V
16. V5 . V^=^ • V^=^ . V^=^.
-5.
17. V^=^' • V^=^' • V~^^.
18. V— 3 ax . V— 3 6u; • V— 4 aft.
19. (V:r2 + v^35)(V^^+V^.
20. (2 V"=^ + 3 \/^=^) (5 V^^ - 2 V^=^).
21. (V— a; + V— y)(V— a?— V^^).
24. V^^ 25. (V^)^
22. V-rt'-^. 23. (
6)^
275. Quotient of one quality-unit by another.
/^/=l- /-^(+l)^/; /^ (_!)=_/; §84
+ 1 ^/=/^--/ = r^ = -/; § 2G4
— 1 -i- / = i^ -^ ij = i.
272 ELEMENTS OF ALGEBRA
E.g., 1 ^ i5 = 1 ^ i = _ i ; _ 1 ^ i6 = _ 1 ^ (_ 1) = 1 ;
i ^ f = i -^ (— i) = — I.
276. From the commutative and associative laws we "have
the following principle :
The quotient of two quality-numbers is equal to the quotient
of their quality-units into the quotient of their arithmetic
values.
Ex 1 V^5 _i V5 _ V2.
Ex.2. ^^.A.^ = ,^, or^^^^.
V7 +1 v/7 7 ' 7
Ex. 3. y/~a/ \/~b = (i/i) ( y/a/y/h) = \/a/b.
Ex.4. y/^/V^ = (+l/i)i^a/^b) = -iVa7b, or -\/-(a/6).
Exercise 102.
Perform the operation of division in
1. i^ -r- i. S. 1 -r- i^ 5. i -J- ^^
2. ^5^^•. 4. - 1 -m'^. 6. V-14-V-2.
7. V^^ie^V^l. 11. V^'-^V^'.
8. V^=n^^V^^. 12. (V=T2-V-15)-r-V-3.
9. V^^^V~a. 13. (V~a+V--6) ^V~c.
10. V^(^)-^V£c. 14. (Vl6-V8)-=-V^^.
15. i-v-i^l 16. -i-~i''. 17. ^3-^F. 18. i*^ _j. jT-ss^
COMPLEX NUMBERS.
277. The sum of a real number and an imaginary number
is called a complex number, as 4: ± 5 i, 7 ± 3 i.
The general expression for a complex number is evidently
a + bi, where a and b are any real numbers.
COMPLEX NUMBERS 273
When 6 = 0, a-\-hi = a, a real number.
When a = 0, a-\-hi = hi, an imaginary number.
278. We define addition of complex numbers by assum-
ing the commutative, and therefore the associative, law of
addition for real and imaginary numbers.
Hence, in adding or subtracting complex numbers the real
parts can be added or subtracted by themselves and the
imroginary parts by themselves.
That is, (a + bi) ± (c + di) = (a ± c) -\- (b ± d) L (1)
Ex. When is the second member of ( 1 ) a complex number ? When
an imaginary number ? When a real number ?
279. If two complex numbers are equal, their reed parts are
equal, and their imaginary parts are equal.
Proof. Let a-{-bi = c-{-di, (1)
where a, b, c, d are all real numbers.
Transposing, a — c = di — bi. (2)
But, if a real number is equal to an imaginary number,
each is zero (§ 267) ; hence,
a — c = 0, or a = c,
and di — bi = 0, or d = b.
An important case of this theorem is the following :
If a + bi = 0, then a = 0 and 6 = 0.
280. Two complex numbers which differ only in the
signs before their imaginary terms are called conjugate com-
plex numbers, as a -f bi and a — bi.
Since (a + bi) -\- {a — bi) = 2 a,
the sum of two conjugate complex numbers is real.
274 ELEMENTS OF ALGEBRA
281. Multiplication by a complex number is defined by
assuming the distributive law ; tliat is,
(a H- bi) (c + di) = ac -\- adi -f bci + bdv^
= (ac — bd) 4- (ad -f- be) i. (1)
Before multiplying one complex number by another it is
convenient to reduce each to the type-form a -f- bi.
Ex. (3 + V^^)(4 - V^S) = (3 + v^ • 0(4 - V''^ • 0
= 12 +(4^5- 3^3)1 -fVlS-
282. From (1) in § 281, it follows that the product of two
complex numbers is, in general, a complex number.
But, the produet of two eonjugate eomplex numbers is real
and jjositive.
Proof. (a + bi) (a - bi) = a^ - {bif = a^ + b\
E.g., (-3 + V^^)(-3-V^^) = (-3)2-(V32)2:=ll.
283. The quotient of one complex number by another is, in
general, a complex number.
Proof a + bi_(a + bi)(c-di)
c -f di"~~ (c H- di) (c — di)
_ac + bd be — ad . ^-js
~ c^ + d^ -^ c2 + d=^** ^^
284. From (1) in § 283, it follows that when the divisor
is a complex number, the quotient can be expressed as a
complex number by multiplying both the dividend and divi-
sor by the conjugate of the divisor.
Ex 4 + 3 I ^ (4 + 3 0(3 + 20
S-'2i (3-2 0(3 + 20
^6 + 17j^A + iIj-
9 + 4 13 13 '
COMPLEX NUMBERS
275
Exercise 103.
Find each of the following sums and products :
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19. (V^^4-V2)
20. (2 + 3aO'-
2+V-4) + (3-V-l).
3+V-9)-(7-V-16).
3-2 0 + (6 + 0 1).
8+V-36)-(o+V-25).
4 + 3 0 - (3 - 4 1).
2+V^ + (2-V^:r3).
2-hV^^)(3-V^.
4+V-16)(3-V-25).
5 + 2 V^ (2 - 3 V^l).
l+V-7)(2-V-T6).
3-hv'^'5)(2-V^.
V2 + V^2) ( V3 - V^.
2+V^(2-V^.
_4-V^(-4+V^.
_7-V^^ll)(-7+V^ll).
.rV- X + 2/V^) (xV^^ - 2/V^).
V^T 4. 5V^3)(V^^ + 3V^2).
21. (2-3V^2)3.
22. (a -f ci)l
276
ELEMENTS OF ALGEBBA
Reduce each of the following expressions to the typical
form a -\-hi:
23.
24.
25.
1-V-l
26.
27.
28.
a —V— X
1
3 4-4V^5
2_3V^^
29.
30.
31.
3+2V-1
2 - 3V^=~1
3-4^'
2-V-2
Show that
3)/2]» = + 1 ; also [(- 1 - V=3)/2]-' = + 1 ;
and that therefore there are at least three cube roots of +1.
33. In § 281, when is the product a complex number ?
When an imaginary number ? When a real number ?
34. In § 283, when is the quotient a complex number ?
When an imaginary number ? When a real number ?
285. Geometric representation of a complex number.
To find the sum of + 4 and — 1 geometrically, we lay off Oil/ equal
to 4 in the positive direction ; then from the end of OM we lay off MB
equal to 1 in the nega-
tive direction. The
straight line OB^ which
extends from the origin
0 of the line 03/ to the
end B of the line MB,
is the sum of + 4 and
— 1 ; that is, the sum
is the directed line
dravm from the origin
of the first line to the
end of the second.
Similarly, to con-
struct the sum 4 + 3 1,
we lay off OM equal to
+ 4 ; at il/, the end of OM, we erect a perpendicular and on it, in the
COMPLEX NUMBERS 277
direction of positive imaginaries, lay off MP three units long ; then
OP, i.e. the directed line extending from the origin O of the first line
to the end P of the second line, will represent the complex number
4 + 3 i.
From the right-angled triangle OMP we have
the length of OP = \AP + 3^ = 5.
If we take 011= one unit long and draw HR parallel to MP, then
from the similar triangles Olill and OMP we have OB = 4/5 and
RH= (3/5)i, that is, the directed unit-line 0^ represents 4/5+ (3/5)2,
and OP, which represents 4 + 3 j, is 5 times this unit-line.
Hence the arithmetic value of 4 + 3 i is 5, and its quality-unit is
4/5 +(3/5)/, which illustrates § 286.
Observe that the quality-unit 4/5 +(3/5)i is obtained by dividing
4 + 3 I by 5, and that 5 = \/42 + 32.
In like manner we can represent any other complex number and its
quality-unit.
286. The arithmetic value, or modulus, of the complex num-
ber a +- hi is the square root of the sum of the squares of a
aud 6, or Va^ + ^^; and its quality-unit is (a -f hi)/-\/a--\-h^.
That is, a-irU = [(a +- hi)/^a? -f 6^] . Va' + h\
where the complex nnmber a -f hi is written as the product
of a quality-unit and an arithme:tic number.
E.g. the arithmetic value, or modulus, of 4 + 3i is v42 + 3^, or 5.
The modulus of 5 - 3 i is V62+(-3)2, or y/ZA.
The modulus of - 1/2 + \/^/2 is V(l/2)2 + (V3/2)2, or 1.
CHAPTER XIX
QUADRATIC EQUATIONS IN ONE UNKNOWN
287. By the principles of equivalence of equations in
Chapter VII. we can derive from any quadratic equation in
one unknown, as x, an equivalent equation of the type-form
ax' + bx + c = 0. (A)
Observe that in (A), ax- is the sum of all the terms in .^^,
bx is the sum of all the terms in x, and c is the sum of all
the terms free from x.
E.g., from the quadratic equation
6 x^ - 3 _ Gx^-S ^ x2- lOx
2 4 8
we derive tlie equivalent equation
13 x2 - 40 X - 9 = 0, (1)
which is in the type-form.
Comparing (1) with (A) we liave
a = 13, 6 = - 40, c = - 9.
If a = 0, (A) ceases to be a quadratic equation ; hence in
what follows we shall assume that a is not zero.
If neither b nor c is zero, (^1) is called a complete quad-
ratic equation.
If either b or c is zero, or if both are zero, {A) is called
an incomplete quadratic equation. When 6 = 0, the incom-
plete equation is often called a pure or binomial quadratic
equation.
E.g., equation (1) is a complete quadratic equation; while
3 x2 + 4 X — 0, 8 ic2 + 9 = 0, and -5 x^ = 0 are incomplete, the last two
being pure.
278
QUADRATIC EQUATIONS 279
Ex. 1. In examples 20-30 of exercise 105 reduce each equation to
an equivalent equation of the type-form (J.), and state the values
of a, 6, and c in each.
Ex. 2. Solve the incomplete quadratic equation
ax:^-^bx = 0. (1)
Factor x (ax + 6) = 0. (2)
Equation (2) is equivalent to the two linear equations,
X = 0, ax -\- b = 0.
Hence the roots of (2) or (1) are 0 and — b/a.
Remember that to solve any quadratic or higher equation we must
first find its equivalent linear equations. Reread §§ 148 and 149.
288. Since w^ + mu + {m/2f = (u + m/2)2, § 137
TJie expression u^ + mu is made a perfect square by adding
{m/2y, or the square of one-half the coefficient of u.
The addition of (m/2y is called completing the square.
E.g. J x2-7x is made a perfect square by adding (—7/2)"^ or
(7/2)-^
that is, x'^-lx-h 49/4 = (x- 7/2)2.
4a;2 + 8x, or (2x)^ + 4(2x), is made a perfect square by adding
(4/2)2, or 4 ;
that is, (2 x)2 + 4 (2 x) + 4 = (2 x + 2)2.
289. Any quadratic equation can be solved by transposing
all its Jtenns to one member, factoring that member by
vmting it as the difference of two squares, and then putting
each factor equal to zero.
The. following examples will illustrate the method.
Ex. 1. Solve the pure quadratic equation
«X2 + C = 0. (1)
Divide by a, a:2 _(- c/a) = 0. (2)
Factor, (x-V- c/a)(a; + v -c/a)= 0.
By § 149, oj = V— c/a, X = — V— c/a.
280 ELEMENTS OF ALGEBRA
Writing these two linear equations together, we have
x = ± V- c/a. (3)
Ex. 2. Solve the complete quadratic equation
x^ + 4x-2 = 0. (1)
Add 4-4, a;2 + 4a; + 4-6 = 0;
or (x + 2)2-(V6)2 = 0. (2)
Factor, (a; + 2 - VC) (x + 2 + V6) = 0.
By § 149, x + 2 = V6, x + 2=-V6.
Writing these two linear equations together, we have
x + 2 = ±V6. (3)
,', x = -2± V6.
Observe that in each example the two linear equations in (3) can
be obtained from (2) by transposing the known term and then
extracting the square root of both members, writing the double sign ±
with one member. The principle of equivalence of equations which
this illustrates is proved in the next article.
290. Square root. If the square root of both members of an
equation is extracted, and the double sign ± is written before
one member, the two derived equations (when rational in the
unknown) will together be equivalent to the given equation.
Proof Let the given equation be
A' = B^, (1)
Avhere A and B are rational in the unknown.
Transpose, A^-B^ = 0.
Factor, (A - B) {A + 5) = 0. (2)
By §§ 149 and 106, (2) is equivalent to the two equations
A = ±B. (3)
QUADRATIC EQUATIONS 281
Equations (3) can be obtained from (1) by extracting the
square root of both members and writing the double sign ±
with the second member.; hence the theorem.
The following examples illustrate how this principle,
which is proved by factoring, abbreviates the work of
finding the two linear equations, which are equivalent to
a given quadratic equation.
Ex. 1. Solve x2 + 32 - lOx = 0. (1)
Transpose 32, a;2 - 10 x = - 32.
Add (10/2)-2, x2 - 10 « + 25 = 25 - 32 = - 7.
Extract square root, x — 5 = ± V— 7.
.-. X = 5 ± V^. (2)
By §§ 106 and 290, no root is either introduced or lost in passing
from (1) to (2); hence the roots of (1) are 5 + V— 7 and 5— V— 7-
Ex.2. Solve xM:_9^xM:J^ ,j^
4 5
Multiply by 20, 5 x* + 45 = 4 x^ + 4.
Transpose, x^ = — 41.
Extract square root, x = ± V— 41. (2)
By §§ 106, 108, and 290, no root is either introduced or lost in
passing from (1) to (2); hence the roots of (1) are + V— 41 and
Ex.3. Solve 2x2 = 7x + ll.
Transpose 7 x, 2 x^ — 7 x = 11.
Multiply by 2, (2 x)2 - 7 (2 x) = 22.
Complete square, (2 x)2 - 7 (2 x) + 49/4 = 22 + 49/4 = 137/4.
Extract square root, 2 x - 7/2 = ± ^1^7/2.
: .-. x=(7±Vl37)/4.
Hence, to solve a quadratic equation we can proceed as
follows :
Reduce the equation to the form aa^ -\-bx = — c.
282 ELEMENTS OF ALGEBRA
If the tei'm in xr is not a perfect square, multiply (or
divide) both members by a number which will make it a
perfect square.
Add to both members what is necessary to complete the
square of the unknown member.
Extract the square root of each member, writing the
double sign ± before the known member.
Solve the two derived linear equations.
Exercise 104.
Solve each of the following equations :
1. x'-Jrl = 4.x. 9. 3x'-6x-\-2 = 0.
2. x^-2x = 4:. 10. 5x--6x-\-ll = 0.
3. ^'-^ + 5 = 80;. 11. 3^2 + 4 a; + 7 = 0.
4. 'x''-{-2x = 2. 12. 2x'-6x-{-10 = 0.
5. x'^-\-6x = -S. 13. 5x'-{-Sx + 21 = 0.
6. 4j^ + 4ic = ll. 14. 2x'-5x-\-15 = 0.
7. 9x^-\-6x = 17. 15. 2x'-3ax^2a' = 0.
8. Ax^-4.x-7 = 0. 16. {x-7y = i9(x + 2y.
When both members are perfect squares in the unknown, as in
example 16 (or can be made so, as in some of the examples which
follow), the first step is to extract the square root of both members.
17. {x-{-2y = 4.(x-lf. 21. x''-\-2ax = b^-}-2ab.
18. (x+6y = 16(x-6y. 22. x' -{-2 ab = b^ + 2ax.
19. (ic + 8f = 9i»l 23. 4.x^-\-4:ax = b'-a\
20. x^-3ax-\-2a^ = 0. 24. x'^ + 3 a^ = 4. ax.
291. To solve the general quadratic equation,
ajr^ + 6jr + c = 0, {A)
we proceed just as with the particular equations above.
Transpose c, ax^ -\-bx = — c.
QUADRATIC EQUATIONS 283
Multiply by 4 a instead of a, to avoid fractions in (1)
and (2),
(2 axf -\-2b(2ax) = -4: ac.
Add b', (2 axy -\-2b(2ax)-l-b' = b'-4.ac. (1)
Extract square root, 2ax-\-b = ± V6- — 4 ac. (2)
Hence, x = (-b± Vb'-^ac) /(2 a). (B)
By §§ 106, 108, and 290, no root is either introduced or
lost in passing from (A) to (B) ; hence the roots of (A) are
given in (B).
Let b' and c' denote the values of b and c when a = l.
/Ihen when a = 1, equations (^1) and (B) become
and x = -b '/2 ± V{b'/2f - c. (B')
By § 287, any quadratic equation can be reduced to an
equivalent equation of the form (A)-, hence, a quadratic
equation in one unknown has ttvo, and only two, roots.
292. Solution by formula. Instead of repeating the pro-
cess in § 291 with every quadratic equation, we should here-
after find the values of a, b, and c when the equation is
reduced to the type-form {A), and substitute these values in
the two equations (5),
x = {-b± V6^-4ac)/(2 a). {B)
Ex. 1. Solve 2 X'^ - 3 X -I- 5 = 0.
Here a = 2, 6 = — 3, c = 5.
Substituting these values in equations ( B) , we obtain
X = (3 ± V9 - 40) /4 = (3 ± V- 31)/4.
Ex. 2. Solve - 3 x2 = 3 A: - 2 ax.
Here a= — 3, 6 = 2 a, c = — ^k.
Substituting these values in equations {B), we obtain
X =(- 2 a ± \/4 a^ - 3(5 k) / {- 6)
= (aTVa2^^^9lfc)/3.
284 ELEMENTS OF ALGEBRA
293. Equations (JB') of § 291 afford the following simple
rule for writing out the two roots of an equation in the form
Tlie two roots are equal to minus one-half the coefficient of x
plus and minus the square root of the binomial, the square of
one-half the coefficient of x minus the known term.
Ex. 3. Solve ic2 + 4 a; + 7 = 0, by the rule given above.
x = -2± \/22 - 7 = - 2 ± V^.
Ex. 4. Solve x2 - 6 a; - 8 = 0,
a: = 3 ± V(- 3)2 -(- 8) = 3 ± V17.
Exercise 105.
Solve each of the following equations by § 293 :
1. cc2-2a; = l. 7. a^ + 31 = 10aj.
2. a^ + 8a; + 5 = 0. 8. «2 + 6a; + ll=0.
3. a;2_^4a; = l. 9. aj^ + 10 a? + 32 = 0.
4. a;2 + 18 = 10a;. 10. x'-\-52 = Ux.
5. x'-{-3 = 2x. 11. x'-\-2x = l.
6. a^ + ll = 4a;. 12. x^ = 4:X-lS.
Solve each of the following equations by § 292 :
13. 3a^ + 121 = 44x. 19. 21 + 0^ = 2 a^^
14. 25x = 6x^ + 21. 20. 9a^-U3 = 6x.
15. 8af-\-x = S0. 21. 12 a;^ = 29 a; - 14.
16. 3a;2 + 35 = 22a;. 22. 20a;2 = 12-a;.
17. x-{-22 = 6x\ 23. 15 aj^ - 2 aa; = a^.
18. 15 = 17 a; + 4a;2. 24. 21 x^ = 2 ax -\- 3 a'.
QUADRATIC EQUATIONS 285
Solve each of the following equations by the method best
suited to it :
25. 9x'-eax = a~-b'. 26. aix" -{-!) ^x^a" + 1).
27. a(pc^-l)-\-x(a^-l)=0.
28. x'-2(a-b)x-]-h^ = 2ab.
29. (b — c)x^+(c — a)x = b — a.
30. (a 4- 6) a^ 4- c.r = a + & + c.
31. abx^-(a^-{-b^x + ab = 0.
32. (a^ - 62) (a:2_ 1)^4^52.^
33. {b''-a'^(x' + l) = 2{a'-hb^x,
34. (a-a;)3 + (a;-6)» = («-6)l
35. (a;-a + 2&)3-(a;-2a + 6)' = (a4-&)^
294. Discussion of the roots, (— b ± VA^ — 4 ac)/(2 a),
wlien a, 6, c are real.
(i) If 6^ — 4 ac > 0, the two roots will be real and unequal.
(ii) If 6^ — 4 oc = 0, the two roots will be real and egzmZ.
(iii) If 6^ — 4 ac < 0, the two roots will be imaginary or
complex.
(iv) If 6 = 0, the two roots will be both real or both imagi-
nary, but opposite in quality and arithmetically equal.
(v) If c = 0, one root will be zero and the other — b/a.
(vi) If 6 = c = 0, both roots will be zero.
(vii) Both roots will be real, both imaginary, or both
complex.
(viii) If br — 4 ftc is a perfect square, the two roots will
be rational when a and b are rational.
The pupil should give the reasons for each of the above
statements.
286 ELEMENTS OF ALGEBRA
Ex. 1. What kind of numbers are the roots of the equation
Sx'^ -2x^-1? (1)
Here a = 3, 6 = - 2, c = 7 ;
... 62 _4(^c =(-2)2-4.3.7<0.
Hence the roots of (1) are complex and unequal.
295. Sum and product of roots of jr- -h 6'jr + c' = 0. (A')
Representing the two roots of (A') by Xi and x^, we have
x,^-b'/2 + V(b'/2y-c, (1)
X, = - b'/2 - ■\/(b'/2y - c. (2)
Adding (1) and (2) to find the sum, we obtain
x, + x, = -b'. (3)
Multiplying (1) by (2) to find the product, we obtain
<c,.x,= (- 672)^ - [(672)^ - c] = c. (4)
Hence, if a quadratic equation is in the form
x'-\-b'x-\-c' = 0, (A')
the sum of its roots is equal to minus the coefficient of x, and
the product of its roots is equal to the known term.
E.g., the equation 3 x^ = 7 x + 5 put in the form of {A') becomes
x^-^x-^ = 0.
Hence the sum of the roots is 7/3, and the product is — 5/3.
Note that this principle agrees with § 139 in factoring, and that it
is in reality only another form of stating the principle in that article.
Exercise 106.
1. By § 294, what kind of numbers are the roots of each
of the equations from 5 to 14 in exercise 104 ?
2. By § 295, what is the sum and what the product of
the roots of each of the equations from 7 to 18 in exercise
105?
QUADRATIC EQUATIONS 287
Solve each of the following fractional equations :
3. ^^^±l=Sx + 2. 11. -^4._l-=3.
x-1 x-2 a; + l
4. ^^-^==g^. 12. J:4-^-+ ^ =0.
ic + 1 2 2 3 + a; 2 + 3ic
. 3x-8 5a;-2 ,- 5 10 2
ic — 2 x-\-5 x-\-l ic+lO 3ic— 3
5 a; — 7 _ a; — 5 ^^ 1 4 1
7a;-5 2x-13 3-a; 5 9-2a;
X— 1 a;4-2 x l-{-x S — x 35
a; — 2 X x-\-(i x — 4 a; — 3
x— 1 a; 6 x + 1 X— 1
10. ^±1+^+2^29. jg l + _j_ + ^_ = 0.
j; + 2 a; + l 10 3 x + 3 x + U
19. _i^+ 4 15
20.
x—\ x — Z a; + 3
2 a! - 3 ^ a; - 5 5 a; - 16
4 12 a;-l '
^, 2a;-2^3-3a; 5
2a;-3 3a;-2 8a;- 12
22. -^ 6_^9^_()
a;- 3 a; + 2 3-a;
23. 2a:-l ^ 13 ^ 3 a; + 5
2a; + l 11 3 a; -5
3.r 4 4 _
.« - 2 a; + 3 2 - a; ■
288 ELEMENTS OF ALGEBRA
AlO. — -j-
26.
27.
a; + l x—1 x^ — 1
1 1 ^ 13
x^-3x d-x^ 16 aj*
1 17 1
a^-1 1-x 8 a;-f-l
28. ^^+ 1
29. -^-+ ^^ ^
3x-6 72(aj + 2) x"-^
07^ — 4 2 — ic 'dx-\-^
X g; 4- 1 _ a; — 2 a; — 1
*a.' + l x-\-2 X — 1 X
32. 1 + ^=1 + 1
xx-\-4:X-{-lx-\-2
33 ^ _ ^ — 3 , ^ _ ^ + ^^ ^ 2^
i» — 3 ic aj + 3 a? 3
34. ^ + l = a + l. 38. 5 + « = ^ + «.
ax a X a b
35. ^L_ + ^ = 4. 39. 1 = 1_1_1.
a + a; a — x x — a — b x a b
36. ^ + ^ = 1 + 1. 40. ^ + ^ = a + ..
a? — a a? — 6 a b x — b x — a
37. _^+ A_ = « + ft. 41. ^^ = i + i + i-
a? — o X — b b a x -\- a -\- b x a b
42. ^_ + ^_=_i-+ 1
a; + a a; + 6 c + a c + b
43. _^_ + -^_ = _^_+ ^
X -]- a X -\- b c + a c-\-b
QUADRATIC EQUATIONS 289
.. X -{- a , X -{-b X — a , X — h
44. \ — — 1 ^.
x — a x—o x^a x -\- o
a , b a — c , b -\- c
45. 1 -— = — ; 1-
46.
x-\-a x + b x-\-a — c jc + 6 -f- c
1 1
x-i-a-\ x — a-\-
x-\-b x — b
^^ a-{-b a + c ^ 2(a-\-b + c)
X -{- b X -\- c x^-\- b -\- c
296. The following examples illustrate how any quadratic
expression can be factored by writing it as the difference of
two squares.
Ex.1. a;2 + 4x + 0 = a;2 + 4x + 4 -(-5)
= (x + 2)2-(V:^)2
= (^x + 2 +^/^=rl)ix + 2 -V^l).
Ex.2. 3at2 + 2a;-^ = [(3x)24.2(3x)-44]--3
= [(3a; + l)2-45]-3
= (3 a; + 1 + 3\/6) (3 a; + 1 - 3V6) -4- 3
= (x + i+V5)(3x + l-3 V6).
Exercise 107.
Factor each of the quadratic expressions:
1. x'-^Gx + T.
2. x^-\-Sx-\-5.
3. iB^-lOoj + Sl.
4. 9a^- 6a; -26.
5. Sx^-{-6x-3.
6. a^-Ux + 52.
7.
ar2 4-10x + 40.
8.
a;2_8a; + 32.
9.
x^-ix-i.
10.
a,.2_^_2a;_3.
11.
x^-ix-{-5.
12.
3ar^_8a: + 7.
290 ELEMENTS OF ALGEBRA
13. Put each of the twelve foregoing trinomials equal to
0, and determine, (1) the sum and the product of the roots
of each resulting equation, (2) the character of the roots
as real, imaginary, or complex.
Factor each of the following expressions and then find
the roots of the equation formed by putting it equal to 0 :
14. a^-4a; + 16. 17. 4 x^ -|- 8 a; + 10.
15. x'-Qx^ll. 18. 9a^ + 18a:-|-18.
16. a^-8a; + 20. * 19. 16 a^ + 32 a; + 27.
CHAPTER XX
PROBLEMS
297. The solving of a problem by equations consists of
three distinct parts :
(i) The statement of the conditions of the problem by one
or more equations.
(ii) The solving of these equations.
(iii) The discussion. A problem may require for an
answer a wliole number, an arithmetic number, a real num-
ber, or numbers having some relation that is not expressed
by the equations.
To state these and other such conditions of a problem,
and to determine what solutions of the equations give
answers to the problem, is called the discussion of the
problem.
Prob. 1. Eleven times the number of persons in a room is equal to
twice the square of that number increased by 12. How many pei*sons
are in the room ?
Statement. Let x = the number of persons ;
then lla; = 2x2 + 12. (1)
Solving (1), we obtain a; = 4, x = 3/2. (2)
Equations (2) are together equivalent to (1).
Discussion. The number of persons must be an arithmetic xchole
number which satisfies one of the equations in (2) ; but 4 is the only
such number. Hence the one, and only, answer is 4 persons.
291
292 ELEMENTS OF ALGEBRA
Prob. 2. A train travels 300 miles at a uniform rate ; if the rate
had been 5 miles an hour more, the journey would have taken 2 hours
less. Find the rate of the train.
Statement. Let x = the number of miles travelled per hour ;
then 300 -^ x = the number of hours required for the journey,
and 300 h- (ic + 5) = the number of hours the journey would have
taken if the rate had been increased 5 miles
an hour.
Hence, by the conditions of the problem, we have
300 ^ 300 2 m
X X + 5 '
Solving (1), we obtain x = 25, x = — 30. (2)
Discussion. The number of miles per hour must be an arithmetic
number which satisfies one of the equations in (2) ; but 25 is the only
such number. Hence the one and only answer is 25 miles an hour.
Prob. 3. The square of the number of dollars a man is worth exceeds
by 300 twenty times that number. How much is the man worth ?
Statement. Let x = the number of dollars the man is worth ;
then x2 = 20 X + 300. (1)
Solving (1), we obtain x = 30, x = — 10.
Discussion. If a debt is regarded as a negative possession, both of
these roots give answers ; that is, the man either has $ 30 or owes $ 10.
Prob. 4. The sum of the ages of a father and son is 100 years ; and
one-tenth of the product of their ages, in years, exceeds the father's
age by 180. How old is each ?
Statement. Let x = the number of years in the father's age ;
then 100 —x = the number of years in the son's age.
Hence 0.1 a:(100 - a;)= x + 180. (1)
Solving (1), we obtain x — 60, and 100 — x = 40,
or X = 30, and 100 - x = 70.
Discussion. The father must be older than the son ; hence the
father must be 60, and the son 40, years old.
Both of the solutions of (1) would give answers if the jjroblem read
as follows : The sum of the ages of tioo persons is 100 years ; and one-
tenth of the product of their ages, in years, exceeds the age of one of
them by 180. How old is each ?
PROBLEMS 293
Prob. 5. Find a real number whose square increased by 13 is equal
to 4 times the number.
Statement. Let x = the number ;
then a;2 + 13=4a;. (1)
Solving (1), we obtain x = 2± SV^. (2)
Discussion. Since there is no real number which satisfies (2), the
problem is impossible.
If the word '■'•reaV were omitted in the problem, both the values
of X in (2) would be answers.
Prob. 6. A cistern can be filled by two pipes running together in
22^ minutes ; the larger pipe alone would fill the cistern in 24 minutes
less than the smaller one. Find in what time each would fill it.
Statement. Suppose the larger pipe to fill the cistern in x minutes ;
then the smaller pipe will fill it in x + 24 minutes. Also, 1 /x and
l/(x + 24) are the portions of the cistern which each pipe will fill in
one minute, and 1 /22 J is the portion that both together will fill in one
minute.
Hence - + — — = — • (1)
X a; + 24 22^ ^ ^
Solving (1), we obtain x = S6, x = — 15. (2)
Discussion. The answer must be an arithmetic number, but 36 is
the only such number which will satisfy either equation in (2).
Hence the larger pipe would fill the cistern in 36 minutes, and the
smallei* one in 36 + 24, or 60, minutes.
Exercise 108.
1. Find two arithmetic numbers one of which is 4 times
the other, and whose product is 196.
2. Find two arithmetic numbers whose sum is 25, and
whose product is 144.
3. Find two numbers whose sum is 15, and whose pro-
duct is - 250.
4. Divide 71 into two parts, the sum of the squares of
which is 2561.
294 ELEMENTS OF ALGEBRA
5. A rectangular court is 5 yards longer than it is
broad; its area is 1886 square yards. Find its length
and breadth.
6. The sum of the squares of two consecutive whole
numbers is 1013. Find the numbers.
7. The sum of the reciprocals of two consecutive whole
numbers is Jf . Find the numbers.
8. If a train travelled 5 miles an hour faster, it would
take 1 hour less to travel 210 miles. Find the rate of the
train.
9. The perimeter of a rectangular field is 500 yards, and
its area is 14,400 square yards. Find the length of the
sides.
10. The perimeter of one square exceeds that of another
by 100 feet; and the area of the larger square exceeds 3
times the area of the smaller by 325 square feet. Find the
length of their sides.
11. A lawn 50 feet long and 34 feet broad has a path of
uniform width round it ; the area of the path is 540 square
feet. Find its width.
12. A man travels 108 miles, and finds that he could
have made the journey in 4^ hours less had he travelled 2
miles an hour faster. At what rate did he travel ?
13. The product of the sum and difference of an arith-
metic number and its reciprocal is 3|. Find the number.
14. A cistern can be filled by 2 pipes in 33 J minutes.
To fill the cistern, the larger pipe takes 15 minutes less
than the smaller. Find in what time it will be filled by
each pipe singly.
15. A hall can be paved with 200 square tiles of a certain
size; if each tile were one inch longer each way, it would
take 128 tiles. Find the size of the tile.
PROBLEMS 295
16. There are two square buildings paved with stones
each a foot square. The side of one building exceeds that
of the other by 12 feet, and the two pavements together
contain 2120 stones. Find the sides of the buildings.
17. Find the number such that the product of the num-
bers obtained by adding to it 3 and 5 respectively is less by
1 than the square of its double.
18. The plate of a mirror is 18 inches by 12, and it is
to be framed with a frame of uniform width, whose area is
to be equal to that of the glass. Find the width of the
frame.
19. A and B distribute $ 100 each in charity ; A relieves
5 persons more than B, and B gives to each $ 1 more than
A. How many did they each relieve ?
20. The difference between the hypotenuse and two sides
of a right-angled triangle is 3 and 6 respectively. Find the
sides.
21. In the centre of a square garden is a square lawn;
outside this is a gravel walk 4 feet wide, and then a flower
border 6 feet wide. If the flower border and lawn together
contain 721 square feet, find the area of the lawn.
22. What is the property of a person whose income is
$ 2150, when he has | of it invested at 4 per cent, ^ at 3
per cent, and the remainder at 2 per cent ?
23. A person bought a certain number of oxen for $ 1200,
and, after losing 3, sold the rest for $ 20 a head more than
they cost him, thus gaining $ 35 by the bargain. How
many oxen did he buy ?
24. A can do a piece of work in 10 days ; but after he
has been upon it 4 days, B is sent to help him, and they
finish it together in 2 days. In what time would B have
done the whole work ?
296 ELEMENTS OF ALGEBRA
,25. A and B can reap a field together in 7 days^ which A
alone could reap in 10 days. In what time could B alone
reap it ?
26. A can build a wall in 8 days, which A and B can do
together in 5 days. How long would B take to do it alone ?
27. A does f of a piece of work in 10 days, when B
comes to help him, and they take 3 days more to finish it.
How long would B take to do it alone ?
28. The tens' digit of a certain number exceeds the units'
digit by 4, and when the number is divided by the sum of
the digits, the quotient is 7. Find the number.
29. Find a number of three digits, each greater by 1 than
that which follows it, so that its excess above ^ of the num-
ber formed by inverting the digits shall be 36 times the
sum of the digits.
30. A detachment from an army was marching in regular
column, with 5 men more in depth than in front ; but on
the enemy coming in sight, the front was increased by 845
men, and the whole was thus drawn up in 5 lines. Find
.the number of men.
31. The sum of two numbers is 14, and the quotient of
the less divided by the greater is -^^ of the quotient of the
greater divided by the less.
32. Find two fractions whose sum is f, and whose differ-
ence is equal to their product.
33. Two men start at the same time to meet each other
from towns which are 25 miles apart. One takes 18 minutes
longer than the other to walk a mile, and they meet in 5
hours. How fast does each walk ?
Let X = the number of minutes it takes the first man to walk a mile.
34. A and B together can do a piece of work in a certain
time. If they each did one-half of the work separately, A
PROBLEMS 297
would have to work one day less, and B 2 days more than
before. Find the time in which A and B together do the
work.
35. A man bought a certain number of railway shares
for $ 1875 ; he sold all but 15 of them for $ 1740, gaining
$ 4 per share on their cost price. How many shares did
he buy ?
36. The denominator of a fraction exceeds the numerator
by 4 ; and if 5 is taken from each, the sum of the reciprocal
of the new fraction and 4 times the original fraction is 5.
Find the original fraction.
37. A person swimming in a stream which runs Ij miles
per hour finds that it takes him 4 times as long to swim a
mile up the stream as it does to swim the same distance
down. At what rate does he swim ?
38. AVhat is the property of a person whose income is
$ 1140, when one-twelfth of it is invested at 2 per cent,
one-half at 3 per cent, one-third at 4.^ per cent, and the
remainder pays him no dividend ?
39. A person having 7 miles to walk increases his speed
one mile an hour after the first mile, and is half an hour
less on the road than he would have been had he not altered
his rate. How long did it take to walk the 7 miles ?
Let X miles an hour be his rate at first.
40. The diagonal and the longer side of a rectangle are
together 5 times the shorter side, and the longer side exceeds
the shorter by 35 yards. Find the area of the rectangle.
41. The price of photographs is raised 50 cents per
dozen ; and, in consequence, 4 less than before are sold for
$ 5. Find the original price.
42. A boat's crew can row 8 miles an hour in still water.
What is the speed of a river's current if it takes them 2
298 ELEMENTS OF ALGEBRA
hours and 40 minutes to row 8 miles up and 8 miles down
the river ?
Let X = the number of miles the current runs in an hour ;
then — ^^— + —2— = -.
8 + X 8 - X 3
43. At a concert $ 300 was received for reserved seats,
and the same amount for unreserved seats. A reserved seat
cost 75 cents more than an unreserved seat, but 360 more
tickets were sold for unreserved than for reserved seats.
How many tickets were sold all together ?
44. Out of a cask containing GO gallons of alcohol a
certain quantity is drawn off and replaced by water. Of
the mixture a second quantity, 14 gallons more than the
first, is drawn off and replaced by water. The cask then
contains as much water as alcohol. How much was drawn
off the first time ?
Let X — the number of gallons drawn off the first time ; then, in
the first mixture,
60 — X = the number of gallons of alcohol,
and X = the number of gallons of water.
60 - X xgQ _ -^4>^^ ^ ^gQ _ ^ _ ^4^^_^ ^ _^ 14.
60 ^ ^60
45. A cyclist rode 180 miles at a uniform rate. If he
had ridden 3 miles an hour slower than he did, it would
have taken him 3 hours longer. How many miles an hour
did he ride ?
46. A man drives to a certain place at the rate of 8 miles
an hour. Returning by a road 3 miles longer at the rate of
9 miles an hour, he takes 7| minutes longer than in going.
How long is each road ?
47. A father's age is equal to the united ages of his
5 children, and 5 years ago his age was double their united
ages. How old is the father ?
PROBLEMS 299
48. A and B are two stations 300 miles apart. Two
trains start simultaneously from A and B, each to the oppo-
site station. The train from A reaches B 9 hours, the train
from B reaches A 4 hours, after they met. When did they
meet, and what was the rate of each train ?
49. If a carriage wheel 14| feet in circumference takes
one second more to revolve, the rate of the carriage per hour
will be 2J miles less. How fast is the carriage travelling ?
Let X = number of miles travelled per hour ; then
X 10 X - 2f
50. The number of square inches in the surface of a
cubical block exceeds the number of inches in the sum of
its edges by 288. Find its edge and volume.
51. A cistern can be filled by 2 pipes running together
in 2 hours 55 minutes. The larger pipe by itself will till it
sooner than the smaller one by 2 hours. Find the time in
which each pipe separately will fill it.
52. My gross income is $ 3000. After paying the income
tax, and then deducting from the remainder a percentage
less by 1 than that of the income tax, the income is reduced
to $ 2736. Find the rate per cent of the income tax.
53. A set out from C toward D at the rate of 5 miles
an hour. After he had gone 45 miles, B set out from D
toward C, and went every hour -^^ of the entire distance.
After travelling as many hours as he went miles in an hour,
he met A. Find the distance from C to D.
CHAPTER XXI
IRRATIONAL EQUATIONS
298. An irrational equation is an equation one or both of
whose members is irrational in an unknown.
In this chapter, as heretofore, the radical sign will denote
only the principal root of a number or expression.
E.g.^ Vx^ — 2 = X — 7 is an irrational equation, and Vx^ - 2 de-
notes only the principal square root. Note that we cannot speak of
the degree of this or any other irrational equation.
In solving irrational equations we use the following
principle ;
299. If both members of an irrational equation are raised
to the same integral power, the derived equation will have all
the roots of the given one and often others in addition.
Proof Let A = B (1)
be the given irrational equation.
Squaring, A^ = B\ (2)
By §§ 105 and 106, (2) is equivalent to the equation
{A-B){A + B) = (>. (3)
By § 74, the roots of (3) include those of A — B=0, or
(1) ; hence no root is lost by squaring (1).
But the roots of (3) include also those of A + B = 0,ov
A = — B; hence any root of A = — B which is not a root
of ^ = J5 must be introduced by squaring (1).
In like manner the principle can be proved for any other
positive integral power.
300
IRRATIONAL EQUATIONS 301
Ex. 1. Solve the equation x — 6 = — Vx — 6. (1)
Square, x2 - 12 x + 36 = x - 6. (2)
Transpose, x^-lSx-{-4t2 = 0.
Factor, (x - 6) (x - 7) = 0. (3)
Now (3) , or (2) , is satisfied when x = 6 and x = 7 ; but (1) is satisfied
only when x = 6. Hence by squaring (1) the root 7 was introduced.
By § 299, the roots of (2) include those of (1) and also those of
A = - B, or X - 6 = Vx-e. (4)
Equation (4) is satisfied both when x = 6 and when x = 7.
Hence if it had been required to solve (4), by squaring we would
have obtained (2), and no root would have been introduced.
Notice that we cannot say that (2) is equivalent to (1) and (4)
jointly (as would be the case, by § 290, were (1) and (4) rational
equations); for (2) has only two roots, while (1) and (4) together have
three roots, 6 being a root of each.
Observe that, since we cannot speak of the degree of an irrational
equation, we do not know how many roots it has until we have solved it.
Ex. 2. Solve 2 - V2 X + 8 + 2 Vx + 6 = 0. (1)
Our purpose being to obtain a rational equation, it is better before
squaring to put the more complex surd in one member by itself, as
below.
Transpose, 2 + 2 Vx + 6 = a/2 x + 8.
Square, 4 4- 8 Vx + 6 + 4x + 20 = 2x + 8.
Transpose, x + 8 = - 4Vx + 5. (2)
Square, x2 + 16 x + 64 = 16 x + 80.
Transpose, x^ - 16 = 0. (3)
Hence, by § 299, if (1) has any root, it is 4 or — 4. But neither
X = 4 nor x = — 4 satisfies (1) ; hence (1) has no root, i.e., it is impos-
sible, and therefore both roots of (3) were introduced by squaring (1)
and (2).
If we use both the positive and the negative values of V2x + 8 and
vx + 6, we obtain in addition to (1) the three equations.
2-V2x4-8-2Vx + 5 = 0, (4)
2+V2x + 8-2\/x+5 = 0, (5)
2+V2x + 8 + 2Vx-f5=0. (6)
302 ELEMENTS OF ALGEBRA
By treating (4), (5), or (6) as we did (1), we would obtain (3).
Hence the roots of (3) include the roots of each of the four equations,
(1), (4), (5), and (6).
By trial we find that — 4 is a root of (4), — 4 and + 4 are both
roots of (5), and neither — 4 nor + 4 is a root of (6) ; that is, (6)
expresses an impossible condition as well as (1).
Hence in rationalizing (1) or (6), i.e., in deriving the rational
equation (3) from (1) or (6), two roots are introduced.
In rationalizing (4), one root is introduced.
In rationalizing (5), no root is introduced.
Exercise 109. i^
Solve each of the following irrational equations :
1. Vaj — 5 = 3. 7. Va;4-25 = 1+V^.
2. 7 - Va; - 4 = 3. 8. Vi» + 3 -f ^x = 5.
3. V5a;-l=2Vaj + 3. 9. V8a; + 55 - 3 = 2V2^.
4. 2V3-7a;=3V8fl7-12. 10. 10 -V25 -\-'i) x = 3^x.
5. .V9x'-llx-o = 3x-2. 11. V9ic-8 = 3Va.'-h4-2.
6. V4.x^-7 x-^l = 2x-l^. 12. ■Vx-4.-\-3=Vx-\-ll.
13. In each of the foregoing examples, from what other
irrational equation or equations would we have derived the
same rational equation ?
14. ■\/Sx-{-17 -V2x = -V2x-\-9.
15. V3a;-ll+V3^=Vl2x-23.
16. -Vi2x-5+VSx-l=^27x--2.
17. Vx + 3 + Va; + 8 = V4 a; 4- 21.
18. Va5 + 2+V4a7 + l=V9aj + 7.
19. ■\/x -i- 4: ab = 2 a -\- ^x.
20. ■^x+V4:a-^x=2Vb + x.
21. Vx — l-\-^x = 2^^x.
IRRATIONAL EQUATIONS 303
22. ViC + o + V^ = 10 -=- Y^ic.
23. V^- vaJ-8 = 2H-V.T-8.
24. Vl + a; + V^ =2 -J- VI + a;.
25. 2 VaJ — V4ic— 3 = l^V4a; — 3.
26. V^ - '^ = 1 -^ (V^ + ")•
27. -^Hi- = 3 + ^^ + 1.
V« — 1 2
Simplify the first member in example 27.
28.
^
- 2Va; 2
-1
2 =
-2 + 1
29.
V2 + « + V2 - a;
V2 + a;-V2-a!
30.
1
1-
+ 1 + ^
a; ^x -h 1 V^ ~
- = 0.
In the next seven examples, first reduce the
improper fractions to
lixed expressions :
31.
4- 3 _ S^x - 5
- 2 3 V-x- - 13
32.
9V^
3v
r-23_6V«-17
'X - 8 2 V-» - <3
V^ + - V-'« + '^
34 6Va?-7 g^7Vx-26
V^-l 7V.T-21
35 2Va;-l^Vx-2 3^ 12v^-ll ^ 6^^4-5
* Va' + t V^-i ' 4va.'-4f 2V^ + t'
36.
304 ELEMENTS OF ALGEBRA
39. ^(a-x)+^(b-x)=^(a-\-b-2x).
40. ^(ax + b^) — -^(bx i- a^) = a — b.
41. ^(a-{-x)-\--^(b + x)=^(a + b-^2x).
42. ^(a-x)+^(b~-x)=-y/(2a-{-2b).
300. Equations in quadratic form. If an equation has only
two unknown terms, and if the unknown factor of one of
these terms is the square of the unknown factor of the
other, the equation is in quadratic form.
E.g., since x^ + Sx is the square of vx^ + 3x, the equation
(x2 + 3x)+ 5vxM-3x = 7 is in quadratic form.
The following examples illustrate how the principles of
quadratic equations can be applied to irrational equations
which are, or can be put, in quadratic form.
Ex.1. Solve 2x2 + 3x-5\/2x'^ + 3x + 9=-3. (1)
Add 9, (2x2 + 3x + 9)-5V2x2 + 3x + 9 = 6. (2)
Since 2 x^ + 3 x + 9 is the square of V2 x^ + 3 x + 9, equation (2)
is in quadratic form. Transposing 6 and factoring, we have
(\/2 x2 + 3 X + 9 - 6) ( V2 x2 + 3 X + 9 + 1) = 0. (3)
The roots of (3) include the roots of
V2 x2 + 3 X + 9 = 6, (4)
and of \/2 x2 + 3 X + 9 = - 1, (5)
but no others.
The roots of (4) are 3 and — 4J ; while (5) is an impossible equa-
tion, since a principal square root cannot be a negative number.
What would be the roots of (1), if the sign before the radical
were + ?
Ex. 2. Solve 3 x2 - 7 + 3 V3 x2 - 16 X + 21 = 16 X. (1)
Transposing 16 x and adding 28 — 28, we obtain
(3x2 - 16x + 21) + 3V3x2 - 16x + 21 - 28 = 0. (2)
IRRATIONAL EQUATIONS 305
Factor, (VSx^- 16 a: + 21 - 4) ( V3 x^ - 16 x + 21 + 7) = 0. (3)
The roots of (3) include the roots of
V3 x2 - 16 a; + 21 = 4, (4)
and of V3 x2 - 16 X + 21 = - 7, (5)
but no others.
The roots of (4) are 5 and 1/3, and (5) is impossible.
What would be the roots of (1), if the sign before the radical
were — ?
If we could not factor (2) by inspection, by § 293 we would have
V3x2-16x + 21 = - f ± Vf + 28 = + 4 or - 7.
Exercise 110.
Solve each of the following irrational equations :
1.
Sx^-4:X-\-V3x'-4:X-6 = lS.
2.
a^ _ a; -f 4 4- Vur - a? -h 4 = 2.
3.
a^ + 2a; - Var* H- 2a; - 6 = 12.
4
1 1 1 Var* \ X \ 5 —
Va;2 + a; -f 5
5.
a^ 4. V4ar^ H- 24 a; = 24 - 6 a;.
6. 2a^-\-6x = l-Va^-]-3x + l.
7. 2 (2 a; - 3) (a; - 4) - V2 ar^ - 11 a; + 15 = 60.
8. ^4.0^ -{-2 X -^ 7 = 12 x'-^Gx- 119.
9. 2a;2_2a;_17 + 2V2a;2_3^_l_7^^^
10. 3a;(3-a;) = ll-4Var'-3a; + 5.
11. 2ar2-4a;-Va;^-2a;-3 = 9.
CHAPTER XXII
HIGHER EQUATIONS
301. The following examples illustrate how the princi-
ples of quadratic equations are applied to higher equations
which are, or can be put, in quadratic form.
Ex. 1. Solve (x2 + 2a;)2-5(a;2 + 2a;)-14 = 0. (1)
Factor, (a:2 + 2 x - 7)(a;2 + 2 a; + 2) = 0. - (2)
Equation (2) is equivalent to the two equations
x2 + 2x-7 = 0, x2 + 2x + 2 = 0,
each of which is readily solved.
Ex. 2. Solve x* - 8 x3 + 10 ic2 + 24 X + 5 = 0. (1)
Adding C x2 — 6 x^ to the first member, we have
(x* - 8 a:3 + 16 a;2) - 6 x2 + 24 X + 5 = 0,
or (a:2-4x)2-6(x2-4x) + 5 = 0. (2)
Factor, (a;2 _ 4 x - 5) (x2 _ 4 x - 1) = 0. (3)
Equation (3) is equivalent to the two equations
ic2-4x-5 = 0, x2-4x-l=0,
whose roots are 5, — 1, 2 i VS.
Ex. 3. Solve -^^ + ^^ = ^. (1)
X — 1 x2 4
Here the second term is the reciprocal of the first.
Putting y for the first term, and therefore the reciprocal of y for
the second, (1) becomes
y 4
306
HIGHER EQUATIONS 307
Multiply by 4 y, 4 y^ - 17 ?/ + 4 = 0.
Factor, (y - 4) (4 2/ - 1) = 0.
.-. 2/ = 4, or 1/4.
Hence (1) is equivalent to the two equations
^^ = 4 and -^^ = 1. (2)
x-l x-1 4
The roots of equations (2) are 2, 2, (1 ± V— 16)/8.
Exercise 111.
Solve the following equations :
1. a:^-5ar' + 4 = 0. 3. a;* - 7 ar^ - 18 = 0.
2. a;* -10 0^2.^9^ Q 4^ (oF -l)/9 -i-l/a^ = 1.
5. a^ + lOO/ar^ = 29.
6. (a^ -h a!)2 - 22 (ar' + a-) =-40. •
7. (ar^ - a^)- - 8 (a-2 - .1-) = - 12.
9. 2»2 + 3a;H-l= 30/(2 ar' + 3 x).
10. ar^ + 3a;-20/(ar + 3a;) = 8.
11. a^ + a; + 1 = 42/(a^ + a;).
12. a;^ - 8 ar'^- 12^2 4- 112 a; = 128.
13. aj* + 2ar'-3ar-4.r-96 = 0.
14. a;*-10.t'3-}-30ar'-25a; + 4 = 0.
15. a;^-14ar^-hGla:2_g4^_^20 = 0.
16. ^_ + ^ + l_o
17.
a; + l x"
X a.-^+1^5
X- + 1 a; 2
18 - ar^-f-2 ar^ + 4.r4-1^5
ar^ + 4a;-fl ar'4-2 2*
308 ELEMENTS OF ALGEBBA
302. A binomial equation is an equation of the form a;" = a,
where n is a positive integer.
The binomial quadratic equation x^ = a has already been
solved. Certain binomial higher equations are readily solved
by previous principles.
Ex. 1. Solve the binomial cubic equation x^ — 1 = 0. (1)
Factor, (a; - 1) (a;2 + x + 1) = 0. (2)
Equation (2) is equivalent to the two equations
X - 1 = 0, x2 + X + 1 = 0. (3)
The solutions of equations (8) are 1 and (— 1 ± v'— 3)/2.
Hence, the cubic equation (1) has one real and two complex
solutions.
Since by (1), x^ = + 1, tlie cube of each solution of (1) is equal to
+1 ; that is, +1 has the three cube roots +1 (— 1+V— 8)/J, and
(—1 — V— 3)/2. See example 32, exercise 103.
Since + 27 =(+ 1) x 27, the three cube roots of +27, or the three
solutions of the cubic equation x^=^21, can be obtained by multiply-
ing the three cube roots of +1 by the cube root of the arithmetic num-
ber 27.
Thus the three solutions of x^ =+27 are +3 and 3(- 1 ± V^^)/2.
Ex. 2. Solve the binomial biquadratic equation x^ — 1 = 0. (1)
Factor, (x^ - l)(x2 + 1) = 0. • (2)
The solutions of (2) are ± 1 and ± V— 1.
Hence +1 has four fourth roots, two real and two imaginary.
The four solutions of x* = 81, or the four fourth roots of +81, are
±3and±3V^^.
Ex. 3. Solve x5 = 1, or x^ - 1 = 0. (1)
Factor, (x - 1) (x* + x^ + x^ + x + 1) = 0. (2)
One solution of (2) is 1, and the other solutions are those of the
equation
x4 -f x3 + x2 + X + 1 = 0. (3)
Divide by x^, a;2 + x + 1 + - + ^ = 0.
HIGHER EQUj^TIONS 309
Addl, a;2 + 2+l+x + - = l,
(..i)%(..l)=x.
.-. x2+l = K-l±V5)x. (4)
Solving the two equations in (4), we obtain four solutions, all of
which are complex. Hence 1, or any other positive number, has five
fifth roots, one real and four complex.
Ex. 4. Solve x6 = 1^ or x5 - 1 = 0. (1)
Factor, (x^ - 1) (x^ + 1) = 0,
or (a;-l)(a;2 + x+l)(x+l)(x2-x + l) = 0. (2)
Equation (2) is equivalent to the four equations
X - 1 = 0, x' + X + 1 = 0, X + 1 = 0, x^ - X + 1 = 0. (3)
Solving equations (3), we obtain six solutions, two real and four
complex.
Hence 1, or any oXhQV positive number, has six sixth roots, two real
and four complex.
Ei. 5. Solve x^ = 1, or x8 - 1 = 0. (1)
•Factor, (x* + 1) (x^ + 1 ) (x^ - 1 ) = 0. (2)
The roots of (2) are ±1, ± V- 1, and the roots of
X* + 1 = 0. (3)
Add 2 x2 - 2 x2, x4 + 2 x2 + 1 - 2 x2 = 0.
Factor, (^2 + 1 + Xy/'l) (x2 + 1 - Xy/I) = 0. (4)
Equation (4) is equivalent to the two equations
X2 + 1 + Xy/2 = 0,
and x2 + ,l-xV2 = 0,
each of which has two complex roots.
310 ELEMENTS OF ALGEBRA
Hence, any positive number has eight eighth roots, two real, two
imaginary, and four complex.
Observe that any root of + 1 or — 1 is a quality-unit.
Exercise 112.
Solve each of the following binomial equations :
1. ar'' + l = 0. 5. a^ + l = 0. 9. a^^ _ ^4 ^ 0^
2. a^ + 27 = 0. 6. ar' + 32 = 0. 10. a:* -625 = 0.
3. .T^ + 1 = 0. 7. x^-\-l = 0. 11. af- 243 = 0.
4. x'-\-16 = 0. 8. a.'«H-64 = 0. 12. x«- 729 = 0.
CHAPTER XXIII
SYSTEMS INVOLVING QUADRATIC AND HIGHER
EQUATIONS
303. As in linear systems, so in any other determinate
system tliere must be as many independent consistent equa-
tions as there are unknowns.
In solving systems which involve quadratic or higher
equations we have frequent use for the following principle
of equivalent systems:
304. If M, N, P, Q denote any integral unknown expres
sions, then system (a)
PxQ
0,1
(a)
is equivalent to the four systems (6), (c), (d), (e).
M=0,
P=0.
M=0,] , ^=0,1 , iV^=0,l
("). Q^oM^^ p=oM'^ e=o:}«
Proof Any solution of system (a) must reduce the factor
M or N (or both) to 0, and at the same time must reduce
P or Q (or both) to 0.
Now any solution of system (a) which reduces M to 0
and P to 0 is a solution of system (6) ; any solution of (a)
which reduces M to 0 and Q to 0 is a solution of (c) ; and
so on. Hence any solution of system (a) is a solution of
system (6), (c), (d), or (e).
('onversely, any solution of system (b) reduces M to 0 and
P to 0, and therefore reduces M x iV to 0 and P x Q to 0;
311
312 ELEMENTS OF ALGEBRA
hence, any solution of system (b) is a solution of system
(a), and so on. Hence, any solution of system (6), (c), (d),
or (e) is a solution of system (a).
Whence system (o) is equivalent to the four systems (6),
(c), (d), (e).
Ex. 1. Solve thie system
a;2 - a:?/ - 2 ?/2 = 0, (1)
3y2_i0y + 8 = 0. (2)
By § 201, system (a) is equivalent to (&).
Factor (1), (^^ _ 2 ?/) (x + 2/) = 0. (3)
Factor (2), (Zy - ^){y -2)=Q. (4)
By § 304, (6) is equivalent to the four linear systems (c),
a;-2?/ = 0, 1 x-2y = 0, I »: + ?/ = 0, 1 x + ?/ = 0, '
3 2/-4 = 0. J ?/-2 = 0. J 3?/-4 = 0. J ?/-2 = 0. ,
(a)
(&)
(^)
The solutions of the four systems (c) are |, f ; 4,2; — -f, | ; — 2, 2
which are therefore the four solutions of (a).
Ex. 2. Solve the system
x^ + 2xy + y'^ = 36, (1)
a:2-2a;i/ = 0. (2)
System (a) is equivalent to system (6) .
From(l), x^y=±A. ^^H (5)
From (2), x{x -2y)=0. (4) J
By § 304, (&) is equivalent to the four linear systems (c).
x + ?/ = 4, 1 a; + ?/ = 4, 1 a; + !/ = -4, 1 x + y = -4
4, 1 a; + !/ = -4, 1 a; + y = -
0. J x = 0. J ic - 2 ?/ = 0.
cc = 0. J a; — 2?/ =
In applying the principle of this article to system (6), observe that
the two equations in (3) are equivalent to the equation
{x-\-y-i)(ix + y + i)=0.
The solations of (a) are therefore 0, 4 ; f , f ; 0, — 4 ; — |, — f .
SYSTEMS OF QUADRATIC EQUATIONS 313
Whenever one or each of the equations of a system can
be resolved into two or more equivalent equations, the Jlrst
step in solving the system is to apply the principle of this
article.
305. The two examples in § 304 illustrate the theorem :
A system of two quadratic equations in two unknowns has,
in general, four, and only four, solutions.
Exercise 113.
Solve each of the following systems of equations :
1. (x-2y)(x-l) = 0,] 6. (x-\-y){x-y + l) = 0,]
x-\.y-i = 0. 1 (x-\-2)(y + S) = 0. J
2. (x-3)(y-2) = 0,) 7. (x-hyf = 16,\
a; + 2/ = 7. J (p-yY
3. x'-^xy + Zy'-^OA 8. cr + 2 a^y + 2/' = 144, |
X -\-y = \. J Q(? — 2 xy -\- y"^ = A:. J
4. a-?/ -7 2/ + 3 a; = 21,1 9. a?-{-xy = x-\-y.
.1
x-\-y = 2. ) y^ — 2xy = Sy — 6x.
5.4:a^ — xy = 0,] 10. a^ — y^ = x -\- y, 1
2a?-32/ = 6. J ix?-Sxy = 5x-15y.)
306. A system of two equations, one linear and the other
quadratic, can be solved by first eliminating one unknown by
substitution.
Ex. 1. Solve the system
aj + 2 y = 5,
S!<«
«2 + 2y2 = 9.
Solve (1) for x,
x = 5-2y.
(3)
From (2) and (8), (5 -
-2?/)2 + 2?/-^ = 9.
(6)
Factor, (3 y -
-4)(2 2,-4)=0.
(4).
314 ELEMENTS OF ALGEBRA
By § 201, (a) is equivalent to the system, (3) and (4), or (6).
By § 304, (b) is equivalent to the two systems (c) and (d).
X = 5 — 2y,] x = 6 — 2y,]
■(c) (d)
3?/ -4 = 0. r ' y-2 = 0. J ' '
The solution of (c) is 7/3, 4/3 ; and that of (d) is 1, 2.
After the theory is clearly understood, the work after equation (4)
can be abridged as below :
From (4), y = 4/3, or 2.
When y = 4/3, from (3), x = 6- 8/3 = 7/3 ;
When y = 2, from (3), a; = 5 - 4 = 1.
This example illustrates the following theorem :
A syste7n of one linear and one quadratic equation in two
unknowns has, in general, two, and only two, solutions.
Exercise 114.
Solve each of the following systems :
1. x-\-y = 15,] S. x — y = 3.
}
xy = 36. J iK^ + 19 + 2/- = 3 xy
2. x-\-y = bl,\ d. 2x — y = 5,
xy = 518. J x-\-3y = 2xy
3. Sx-4.y = -12, I 10. 3x-\-2y = 5, 1
3ic2 4- 2 2/2 - 2/ = 48. J x' - ixy -\- 5y' = 2.)
4. x-y = 10, 1 11. 3x'-2xy = 15,]
0^ + 2/2 = 58.1 2x-{-3y = 12. J
5. 3x-\-3y = 10,) 12
xy = 1. J
6. 2x-5y = 0, 1 13. x^ + 3xy-y^ = 23,
3)2-3 2/2 = 13.1 x + 2y = 7.
7. 2x-\-3y = 0, I 14. x'-{-if=lS5,
. x + y = 15, I
a^2 -f 2/' = 125. J
SYSTEMS OF QUADRATIC EQUATIONS 815
15. 2x-7y = 2o, 1 17. x-\-y = 2, 1
5x'-{-4:xy+Sy-=2S.} 2x + 3y = 6xy.}
le. 3x — 31 = 5y, 1 18. x-^2y = 7, 1
a^ -\- 5 xy + 2o = y^. } 3y + 6x = 5xy. }
19. x-y = l, I
^-f = (^/6)xy.)
20. x^-2xy=^0, (1)|
4a^ + 92/2 ^225. (2) I
Factor (1), a;(x - 2 y) = 0. (3)
System, (2) and (3), which by § 201 is equivalent to (a), is equiva-
lent to the two systems (6) and (c).
4a;2 4-9?/2 = 225, 1 4^2 + 9y2 = 225, 1
a; = 0. J a;-2y = 0.
21. a:2-3a:?/ = 0, 1 23. Q!^-2xy + o = 0,
5ar' + 3/ = 48.J (a;-2/y = 4. J
22. 2ar-3a;.y = 0,l 24. a^ + 4/ = 4a^ + 16, 1
2/2 + 5an/ = 34. J a^ 4- .v' = 5. J
307. If each of two quadratic equations has onSf and only
one J temi below tlie 2d degree^ and these two terms are similar;
the system can be solved by first eliminating the term below
the second degree by addition or subtra^ion.
Ex. 1. Solve the system x"^ + xy + 2 y- = 44, (1) |
2a;2-xy + 2/2 = 16. (2)}^''^
Each equation in system (a) has one, and only one, term below
the 2d degree, 44 and 1(3, respectively ; and these terms are similar.
We proceed to eliminate the term below the 2d degree.
Multiply (1) by 4, 4^2 + 4 a;.v + 8 y2 = ne. (3)
Multiply (2) by 11 , 22 x"^ - 11 xy -\- 11 y"- = 176. (4)
Subtract (3) from (4), 18 x"^ - Ibxy + Sy^ = 0. (5)
Factor, (y - 3 x) (y-2x) = 0. (6)
316 ELEMENTS OF ALGEBRA
System, (6) and (1), which is equivalent to (a), is equivalent also
to the two systems (&).
x2 + ce?/ + 2 y2 = 44, I x2 + a;?/ + 2 2/2 = 44, |
y-^x = 0. J y-'2x = 0. J ^^^
The solutions of systems (&) are y/2, Zy/2; —y/2, —3^2; 2, 4;
and — 2, — 4 ; which are therefore all the solutions of (a).
Ex. 2. Solve the system ?/2 _ 2 a:2 = 4 x, (1) ]
3 ?/2 + xz/ - 2 a;2 = 16 x. (2) J ^^^
The terms below the 2d degree, 4 x and 16 x, are similar.
We proceed to eliminate the term in x.
Multiply (1) by 4, 4 ?/2 _ 8 a;2 = 16 x. (3)
Subtract (2) from (3), y^ - xy - Qx'^ = 0.
Factor, (y + 2 x) (y - 3 x) = 0. (4)
System, (4) and (1), which is equivalent to system (a), is equiva-
lent also to the two systems (6) and (c).
?/2 _ 2 a:2 = 4 X, ] ?/2 - 2 ^2 = 4 x.
] 2/'-2x2 = 4x, 1
?/ + 2x = 0. ' ^"^ .. o^ A 1 (<^)
The two solutions of system (h) are 0, 0 and 2, — 4 ; those of (c)
are 0, 0 and 4/7, 12/7 ; which are therefore the four solutions of (a).
Observe that by eliminating the term below the second
degree in each of the systems above, we obtained a homo-
geneous equation in x and y, which we resolved into two
equivalent equations.
Instead of eliminating the term below the second degree,
it is sometimes better to eliminate one of the terms of the
second degree.
Ex. 3. Solve the system 9 x2 - 8 y2 ^ 28, (1) ]
7x2 + 3 2/2 = 31. (2) J '^^'^
Multiplying (1) by 3 and (2) by 8, and adding, we eliminate y"^
and obtain
83 x2 = 332, or X = ± 2. (3)
When X = 2, from (2) we obtain y = ±\.
When X = — 2, from (2) we obtain y =±l.
SYSTEMS OF QUADRATIC EQUATIONS 317
Hence the four solutions of (a) are 2, 1 ; 2, — 1 ; — 2, 1 ; —2,-1.
Ex. 4. Solve the system xy + x = 25, (1) |
2xy-Zy = 2S. (2) J
Eliminating the product xy we obtain
2 X + 3 2/ = 22. (3)
Solving system, (1) and (3), which is equivalent to system (a), we
obtain the two solutions 5, 4 ; 16/2, 7/3.
Ex. 5. Solve the system x^ — Sxy = 10, (1) ]
4y2-xy = -l. (2)
|(«:
Sometimes by adding or subtracting the given equations we obtain
an equation which can be resolved into equivalent equations.
Add (1) and (2) , x:^ - 4xy + iy^ = 9,
or x-2y=±S. (3)
System, (1) and (3), which is equivalent to system (a), is equiva-
lent to the two systems (6) and (c) .
x^-Sxy = lO,] x^-Sxy = lO,
x-2y = S. >^'^ - "- '^ ^^'^
^-Sxy = lO, I
x-2y=-3. J
Exercise 115.
Solve each of the following systems of equations :
1.
x^ + xy=12,
x!/-if = 2. J
6.
ar^ + 52/2 = 84,
3a^ + 17a'?/ + 84 = 2/^
2.
x^-{-xy = 24,
2f-{-3xy = S2.
7.
ar-7xy-9y- = 9,
x'^5xy-\-lly' = 5
3.
ic2 + 3a«/ = 7,
f + xy = 6. .
8.
x(x-^y) = AO,
\
y(x-y)=:6. J
4.
2,x?-hf = 2%\
Sxy-4:f = S. J
9.
x^ + xy + f = 7,
6x'-2xy-\-y' = i5..
5.
x^-Sxy-^2f = 3
y
10.
x'-^3xy = 2S,\
2x' + y' = 6.
xy-\-4:y' = S. >
818
ELEMENTS OF ALGEBRA
In example 10, add the two equations.
= 40,1
= 9. J
11. a^ + 3aj?/ = 40,
4 ?/2 + 0^2/
12. a^ + 3fl7?/ = 54,
xy -\-4:y^ = 115.
13. x^-\-xy-\-A4: = 2y^,)
xy + 3 / = 80. J
14. 3xy-{-x'^ = 10,
5xy — 2 x'^ = 2..
In example 14, eliminate x^ or the product xy.
= 301. J
15. Ax'-Sy^^-ll,
11 a^ + 5 2/'
16. 2x^-\-y^ = 9,
5aP-{-(yy' = 26.}
17. 20 a^- 16/ = 179,
5a;2-336/ = 24.
18. 2a^-2xy-3y^ = lS,
3x'-2y' = 19.
19. i»2 4-3a; — 2?/ = 4
2x'-5x-{-3y
u\
20. (a; + l)(2/ + l) = 10,
a?2/ = 3.
21. 4:x'-3xy = 10,
y^-xy = 6.
22. x^ — 2xy = 3y,
2x'-9y'' = 9y.
23. 2x^-xy-^y^ = 2y;
2 a^ -]- 4: xy = 5 y.
24. a.'3 + l = 9?/,|
a^ 4- a; = 6 2/. J
308. Systems of symmetrical equations. A symmetrical
equation is one which is not changed by interchanging its
unknowns.
E.g., x-\- y = 12, xy = 35, x^ + y^ _ 74^ x^±2 xy + y^ = 16 arc
symmetrical equations. The equations x — y = 2, x^ — y^ = 4,
x^ — y^ = IQ are symmetrical except for sign.
The methods given below for solving systems of sym-
metrical equations can usually be employed when the equa-
tions are symmetrical except for sign.
(a)
Ex. 1. Solve the system,
X^ + y2 = 74,
(1)
xy = 35,
(2)
Multiply (2) by 2,
2xy = 70.
(3)
SYSTEMS OF QUADRATIC EQUATIONS
319
Add (3) to (1), x^-\-2xy + y^ = 144,
x + y = ±l2. (4)
Subtract (3) from (1), x-y = ±2. (6)
By § 301, system (6) is equivalent to the four systems (c) .
a; + y = 12,
x-y = 2.
X 4- y = 12,
x-y=-2.
+ i/=-12, I x + y = -l2,
-y = 2. J x-y = -2.
The solutions of systems (c) are 7, 5 ; 5, 7 ; — 5,
Ex. 2. Solve the system, x^ - xy -\- ij^ = 49,
x + y=lS.
or
Square (2),
Subtract (1) from (3),
Subtract (4) from (1),
x2 + 2 xy + 1/2 = 169.
3 xy = 120,
xy = 40.
x-y=±3.
System, (2) and (5), is equivalent to the two systems (6).
a; + y=13, 1
(1)
(2)
(3)
(&)
(c)
5.
)
|(«
(4)
(5)
a: + y = 13, 1
x-y = 3. J
a; _ y = _ 3. J
(6)
The solutions of systems (6) are 8, 6, and 5, 8.
The four solutions of system, (1) and (5), must include the two
solutions of (a), since no solution was lost by squaring (2).
Hence the two solutions of (a) must satisfy (2) and also (5).
Therefore the solutions of systems (b) are the two solutions of (a).
Observe that each of tlie above systems was solved by first finding
the values of x +y and x —y.
Ex. 3. Solve the system x* + y< = 82,
x-y = 2.
Let x — v + w^
and y = v — w.
From (2), (3), and (4), w=\.
From (1), (3), (4), and (6),
(v+l)*+ (tj-l)* = 82,
or (r2 + i0)(t;2_ 4)3,0.
.•.v = ±2, or±^^
10.
(1)
(2)
(3)
(4)
(5)
(6)
1"
(ft)
320 ELEMENTS OF ALGEBRA
From (3), (5), and (6), x = S, -1, \ ± V^^TO.
From (4), (5), and (6), ^ = i, _ 3, _ i ^ V^TTo.
(c)
System (a) with (3) and (4) forms a system equivalent to (6),
which is equivalent to (c) with (5) and (6).
Hence the four solutions of (a) are given in (c).
Exercise 116.
Solve each of the following systems of equations by first
finding the values oi x -]- y and x — y :
1. a^2_^/ = 89, 1 5. x'^l+y' = Zxy,
xy = 4.0. J Ss(f — xy-{-3y^ = 13.
2. x'-\-y' = 170y
xy = 13.
6. ^
^ - ^2/ + r = "6, 1
a; + .?/ = 14. J
3. x' + y^=Qo,\ 7. x' + xy + f^^QlA
iC2/ = 28. J .'c + ?/ = 9. J
4. a^ + ic2/ + 2/^ = 67, 1 8. x" - 4.xy + y~ = b2,\
a^-^^ + 2/' = 39.J _i^(ar_2/) = l. J
9. Solve the systems in examples 1, 2, 4, 5, 8, 12, and 14
in exercise 113, by first finding the values oi x -\- y and
x-y.
Solve each of the following systems of equations :
10. x-.y = 3, 1 13. x^ + y^ = 212A
a;2 _ 3 ic2/ + / = - 19. J x-y = 2. J
11. x^ — xy-\-y'-=12A 14:. x — y = 2, 1
x-{-y = lA. J x' — jf = 242. J
12. a; + 2/ = 4, 1 15. aj^ 4- ^Z'' = 706,
J
a;4 _j_ ^4 ^ 82. j a; + 2/ = 8.
16. Solve system (2) in § 263, and observe that x and y
are rational only when a^ — 6 is a perfect square.
SYSTEMS OF QUADRATIC EQUATIONS 321
309. Division. If the members Qf one equation (1) are
divided hy the corresponding memhers of another equation
(2), and the derived equation (3) is integral in the unknowns;
then the system (a) is equivalent to the two systems (6) and (c).
AB = A'B', (1)1 A = A\ (3)1 5 = 0, (5)1
B = B'. (2)r ^ B = B'. (4) J ^ ^ B' = 0. (6) J ^ ^
Observe that (3) is the derived equation, that (4) is the
same as (2), and that (5) and (6) are formed by equating
to 0 the members of (2).
Proof. Substituting B for B' in (1) we obtain the sys-
tem (d).
B(A-A') = 0,
B = B'.
)W
By § 202, system (d) is equivalent to system (a).
By § 304, system (cl) is equivalent to the two systems (6)
and (e).
■8 = 0, \
By substitution (§ 202), system (e) is equivalent to (c).
Hence (a) is equivalent to the two systems ip) and (c).
E.g.^ dividing (1') by (2') we obtain the integral equation (3');
y^ = x{x^y), (1')
y^ = X + y. (2
Hence system (a') is equivalent to the two systems (6') and (c')
;;}-
V = -, (30 I .^ = 0, (50 1
y^ = ic + y. (40J j; + y = 0. (60 J
Whenever the equation B = 0 or JB' = 0 is impossible, sys-
tem (c) will be impossible, and system (a) will be equivalent
to system (6),
(a)
5) J
822 ELEMENTS OF ALGEBRA
Ex. 1. Solve the system a;^ -y^ = 21, ^^^ \
X - y = 3, (2) J
Dividing (1) by (2) we obtain the integral equation (3); hence,
as B' =0, or 3 = 0, is impossible, system (a) is equivalent to (6).
x2 + x^ + if = 9, (3)
x-y = S. (4)
Ex. 2. Solve the system x* + x^ -\. i/ = lii7\, (1)
x2 _ a;i/ + 1/^ = 03. (2)
■ Divide (1) by (2), x^ + xy + y^ = 117. (3)
Add (2) and (3), x'^ + y- = CO, (4) ]
Subtract (2) from (3), •2xy = 54. (5
Since 03 = 0 is impossible, by division (§ 309) and addition (§ 204)
system (&) is equivalent to (a).
Ex. 3. Solve the system x-y + xy- = 30, (1)
x-\-y = 5. (2)
Divide (1) by (2), xy = 0. (3)
Equations (2) and (3) form a system equivalent to (a).
Exercise 117.
Solve each of the following systems :
1. a^ + 2/' = 3473, 1 6. x' + x'y' + y' = 2^3
x-y = 4.. i a^-xy-^y' = S7
(a)
a; + ?/ = 23. J x'-xy+y' = 9. J
2. a.-^ - 2/' = 218, 1 7. x* + x^ -h y' = 91, )
x-y = 2. J aj2+ a.'^ + 2/' = 13. J
3. x'-7f=^9SS,\ 8. a;^-fa^/ + ^' = 2923,1
^3 - 7/3 ^ 2197, 1 9. x^-\-a^y^-{-y' = 7371,
x-y = 13. J a.-2-j'?/ + 2/2 = 63.
a..4 + ^2^2 _^ 2/^ ^ 2128, 1 10. x'-f=56, 1
a;2 4_ 0^2/ + 2/' = "6. J x" -\- xy -{- y' = 2S. J
SYSTEMS OF QUADRATIC EQUATIONS
823
11. a^ + f = 126, I 12. x-^y-Vxy = 7, 1
ay'-xy + f = 21.S ^^fj^xy=^ 133. J
In the next four systems apply § 304 first.
13. a; + 2/ = 5, 1 15. a; + 2/ = l,
4 icy = 12 - x'y\ J x'y^ + 13 rt^ + 12 = 0. J
14. a?y-\-xf = imA 16. Sa^ - 5/ = a; + ?/, 1
a2/=400. J 3a;2_32^2^3._2, J
In the next four systems let the unknowns be the reciprocals of x
and y, and let v = \/x and w = \/y.
17. 2/.^ + 1/?/ = 1,
^ + ^ + i=5.
y? xy y"^
18. l/a; + l/7/ = 2,
1/a^ 4- 1// = 20. J
19. 3/ar^-l// = l,
^-1 + 1 = 3.
a? xy f
20.
21.
l/x2-l/(4y^)
i_l + J_
ar* a^ 42/^
x_-j-j x—ji
x-y x+y
x' + f = 20.
3,
= 9.
22. a^A + /A=9/2,
a; + 2/ = 3.
310. It should be observed that the methods given in this
chapter are applicable only to special systems of quadratic
and higher equations, and do not enable us to solve a sys-
tem of any two quadratic equations ; for the equation de-
rived by eliminating one unknown will, in general, be above
the second degree in the other unknown, and we have not
yet learned how to solve an equation of a higher degree
than the second, except in very special cases.
E.g.^ consider the system
a;2 + x + y = 3, x2 + y2 = 5. („)
Solving the first equation for y and substituting its value in the
second, we have
a;-2 + (3 - y - x)2 = 5,
or x4 + 2a;8-4x-^ -6a; + 4 = 0. (1)
Equation (1), which is of the fourth degree, cannot be solved by
any methods which have been given in the previous chapters.
324 ELEMENTS OF ALGEBRA
Exercise 118.
1. The difference of two numbers is 7, and the sum of
their squares is 169. Find the numbers.
2. The sum of the squares of two numbers is 130, and
the difference of their squares is 32. Find the numbers.
3. The sum of two numbers is 39, and the sum of their
cubes is 17,199. Find the numbers.
4. A person bought some fine sheep for $ 360, and found
that if he had bought 6 more for the same money, he would
have paid $ 5 less for each. How many did he buy, and
what was the i)rice of each ?
5. If the length and breadth of a rectangle were each
increased by 1 yard, the area would be 48 square yards ; if
they were each diminished by 1 yard, the area would be 24
square yards. Find the length and breadth.
6. The numerator and denominator of one fraction are
each greater by 1 than those of another, and the sum of the
two is 1 j^ ; if the numerators were interchanged, the sum
of the fractions would be 1^. Find the fractions.
7. For a journey of 108 miles, 6 hours less would have
sufficed, had the traveller gone 3 miles an hour faster. At
what rate did he travel ?
8. The hypotenuse of a right-angled triangle is 20 feet,
and its area is 96 square feet. Find the length of the other
two sides.
9. A number is divided into two parts such that the
sum of the first and the square of the second is twice the
sum of the second and the square of the first ; and the sum
of the number and the first part is 4 more than twice the
second. Find the number.
10. The small wheel of a bicycle makes 135 revolutions
more than the large wheel in a distance of 260 yards; if
SYSTEMS OF QUADRATIC EQUATIONS 325
the circumference of each were one foot more, the small
wheel would make 27 revolutions more than the large wheel
in a distance of 70 yards. Find the circumference of each
wheel.
11. A man bought 6 ducks and 2 turkeys for $ 15. For
$ 14 he could buy 4 more ducks than he could turkeys for
$ 9. Find the price of each.
12. The sum of the cubes of two numbers is 407, and the
sum of their squares exceeds their product by 37. Find
the numbers.
13. A rectangular field contains 160 square rods. If its
length be increased by 4 rods, and its breadth by 3 rods, its
area will be increased by 100 square rods. Find the length
and breadth of the field.
14. A man rows down stream 12 miles in 4 hours' less
time than it takes him to return. Should he row at twice
his ordinary rate, his rate down stream would be 10 miles
an hour. Find his rate in still water, and the rate of the
stream.
15. The sum of two numbers is 7, and the sum of their
fourth powers is 641. Find the numbers.
16. A gentleman left $ 210 to 3 servants to be divided in
continued proportion, so that the first should have $ 90
more than the last. Find the legacy of each.
17. From a sheet of paper 14 inches long, a border of
uniform width is cut away 'all round it, and the area is
thereby reduced |; but had the sheet been 3 inches nar-
rower, and a border of the same width had been cut away,
the area would have been reduced ^. Find the breadth of
the paper, and the width of the border cut away.
18. A and B set out from the same place, and travel in
the same direction at uniform rates. B starts 5 hours after
326 ELEMENTS OF ALGEBRA
A, and overtakes him after travelling 100 miles. Had their
rates of travelling been a mile per hour less, B would have
overtaken A after travelling 60 miles. Find their rates.
19. A man has to travel a certain distance, and, when he
has travelled 40 miles, he increases his speed 2 miles per
hour. If he had travelled with his increased speed during
the whole journey, he would have arrived 40 minutes
earlier; but if he had continued at his original speed, he
would have arrived 20 minutes later. Find the whole dis-
tance he had to travel, and his original speed.
20. A cubical tank contains 512 cubic feet of water. It
is required to enlarge the tank, the depth remaining the
same, so that it shall contain 7 times as much water as
before, subject to the condition that the length added to
one side of the base shall be 4 times that added to the
other. Find the sides of the new rectangular base.
CHAPTER XXIV
INEQUALITIES
311. An inequality is the statement that one number is
greater or less than another, as 6 > 4, — 3 < — 2. See
§§ 7 and 31.
312. When a and h are real, in § 31 we agreed to say
that :
a > 6, when a — 6 is positive;
and a<b, when a — b is negative.
The statement 'a — b is positive ' is expressed in symbols
by (I — ft > 0 ; and 'a — b is negative ' by a — 6 < 0.
In this chapter we shall not consider imaginary or complex
numbers.
313. Two inequalities are said to be like or unlike in
species according as they do or do not have the same sign of
inequality.
E.g.y the inequalities 8>4 and a>b are like in species; while
2 < 3 and « > 6 are unlike in species.
If a > 6 ; then, conversely, b<a.
The inequality a > 6 and its converse b<a are unlike.
314. Principles of inequalities.
(i) If one number > a second, and this second number > a
third, then the first number > the third number.
That is, if a > 6 and 6 > c, then a'^c.
(ii) If the same number is added to both members or sub-
tracted from both members of an inequality, the derived in-
equality will be like the given one.
That is. if a > ft, then a ±7)i>b ± m.
327
328 ELEMENTS OF ALGEBRA
(iii) If the corresjyonding members of two or more like
inequalities are added, the derived inequality will he like the
given ones.
That is, if a > 6 and c > <?, then a -f c> 6 + d.
(iv) If both members of an inequality are multiplied, or
divided, by the same positive number, the derived inequality
will be like the given one.
That is, if a>b, then a (+n) > b (^w), or a-r-'^n^b-h '^n.
(v) If both members of an inequality are multiplied, or
divided, by the same negative number, the derived inequality
will be unlike the given one.
That is, if a > 6, then a (~n) < b (~n), or a-i-~n<b-r- ~n.
(vi) If all the members of two or more like inequalities are
positive, and if the corresponding members are multiplied
together, the derived inequality will be like the given ones.
That is, if +%>+&!, ^a2>+bz, ••♦,
then +ai • +a2--->+6i • +62"-.
(vii) If both members of an inequality are positive, and they
are raised to the same positive integral power, the derived
inequality ivill be like the given one.
That is, if ^a >+6, then ("^a)** > (+6)% where w is a positive
integer.
(viii) If the same principal roots of both members of an
inequality are taken, the derived inequality will be like the
given one.
That is, if a > 6, then ^a >-^&.
Proof of {}). (a — 6) + (6 — c)=a — c.
Hence, if a — 6 > 0 and 6 — c> 0, then a — c > 0 ;
that is, if a>h and & > c, then a>c.
Proof of (it), a — b = (a ± m) — (b ± m).
Hence, if a > 6, then a ±m>b ± m.
The proof of the other principles is left as an exercise for the pupil.
INEQ UA LITIS S 329
316. The following principle is often useful in proving
inequalities :
Jfa and b are unequal and realy a- + 6- > 2 ab.
Proof. (a-by>0,
or a^-2ab-\-b'>0. (1)
Adding 2 a6 to each member, by (ii) of § 314 we obtain
a^ + b'>2ab, when a ^ b. (2)
Observe that a^ and b' are both positive.
Ex. 1. Prove (x + y)/2>y/xy, if a; > 0, y > 0, and x # y.
If in (2) we put x for a* and y for 6*, we obtain
X -\-y>2 y/xy.
Hence, by (iv), (x + y)/2 >\/xy, where x > 0, y > 0, and x :^ y.
Ex. 2. a« + 6' > a^b + ab\ if « + 6 > 0 and a ^ 6.
From (1), a^ -ab + 6^ > ah. by (ii)
Multiply by a + 6, a^ ■¥b^> a'^b + aft-^. by (iv)
Ex. 3. The sum of any positive number, except 1, and its recipro-
cal is greater than 2.
Let the number be n ; then in (2), putting n for a^ and 1/n for
62 we obtain
n + l/«>2.
316. The following examples illustrate some of the uses
of the principles of inequalities :
Ex. 1. For what values of x is (6x - 7)/3 >(2 - 3 x)/5 ? (1)
Multiply by 15, 25 x - 35 > 6 - 9 x. by (iv)
Transpose, 34x>41. by (ii)
Divide by 34, »> 41/34. by (iv)
Hence (1) is satisfied for any value of x gi-eater than 41/34.
Ex. 2. For what values of x is x^ - 4 x + 3 > - 1 ? (1)
Addl, x2-4x + 4>0, or (x-2)2>0. by (ii)
Hence (1) is satisfied when (x — 2)2 >0, i.e.., when x has any real
value except 2.
330 ELEMENTS OF ALGEBRA
Ex. 3. Find what values of x satisfy the inequalities
4iK-6<2x + 4, (1) \
and 2 X + 4 > 16 - 2 ic. (2) J
From (1), 2x<10, or x<5. by (ii), (iv)
From (2), 4ic>12, or ic>3. by (ii), (iv)
Hence (1) and (2) are satisfied by any value of x between 3 and 5.
Ex. 4. Find what values of x satisfy the inequality
x^-lx<S. (1)
Subtract 8, x2 - 7 ic - 8 < 0, or (x - 8) (x + 1)< 0. by (ii)
The product (x - 8) (x + 1) will be negative, when, and only when,
one factor is positive and the other negative.
One of these factors will be positive and the other negative when x
has any value between —1 and 8, and only then.
Hence (1) is satisfied by any value of x between — 1 and 8.
Ex. 5. Find what values of x and y satisfy the inequality
3x + 2?/>5, (1)|
and the equation 6 x -\- 7 y = 12. (2) J
Multiply (1) by 5, 15 x + 10 2/ > 25. (3)
Multiply (2) by 3, 15 x + 21 y = 36. (4)
Subtract (4) from (3), - 11 y >- 11, or ?/ < 1. by (v)
Multiply (1) by 7, 1 x + 14 y > 35. (5)
Multiply (2) by 2, 10 x+Uy = 24. (6)
Subtract (6) from (5), 11 x > 11, or x > 1.
Hence any solution of equation (2) in which x > 1 and y < 1 will
satisfy both (1) and (2).
Exercise 119.
If the letters denote unequal positive numbers, prove :
1. a^-\-b'^-\-c^>ah + ac + bc. (1)
Use the relation a^ 4. 52 -> 2 ab.
2. a^ + b''>d'b-{-ab\ (2)
INEQUALITIES 331
3. ^>^; % + -A+-- <3)
4. a^j^ly' + (?> (a'b + ab^ + a-c + ac" + b^c + 5c2)/2.
5. am + 67i+cr<l, if a^-{-b^-]-(^=zl, and m^+7i^+)'^=l.
Find the limits between which the values of x must lie
to satisfy each of the following inequalities :
6. 6 a; > fa; 4- 18. 10. a;^ _^ a; > 12.
7. ^a;-|a;>|a;-3. 11. (a; + 2)/(a; - 3) > 0.
8. -2(a; + 7)>-16. 12. (a; - 7) /(a; + 4) < 0.
9. x^ — ox> — 4:. 13. 3(a;+7)/5>5(a;-3)/7.
14. If 5 a; — 6 < 3 a; + 8 and 2a; + l<3a; — 3, show that
the values of x lie between 4 and 7.
15. If 3 a; — 2 > I a; — I and J — fa;<8 — 2 a;, show that
the values of x lie between 12/25 and 82/9.
Find what values of x and y will satisfy each of the
following systems :
16. 2x-\-3y=4:,] 17. Sx-y=Q,) 18. 4:X-2y=e,
x—y>2.) 2x+y>4:.) 2x—Sy>5.
19. Show that (1) in example 1 holds, if a, 6, and c are
real and either a ^ b, or a ^ c, or b ^ c.
20. Show that (2) and (3) in examples 2 and 3 hold, if a
and b are real and unequal and a-\-b>0.
CHAPTER XXV
RATIO AND PROPORTION
317. The ratio of one number to another is the quotient
of the first divided by the second.
The dividend is called the first term, or the antecedent, of
the ratio ; and the divisor, the second term, or consequent.
The ratio of a to ft is written -, a/b, a -^ b, or a: b, each of which
b
forms can be read ' a is to 6 ' or ' a by &.'
The ratio of 8 to 2 is 8/2, or 4 ; the ratio of 7 to 5 is 7/5.
It is clear that a ratio is arithmetically greater than, equal
to, or less than 1, according as its first term is arithmetically
greater than, equal to, or less than, the second.
318. Since a ratio is a fraction, all the properties of frac-
tions belong to ratios in whatever form the ratios are written.
Thus a:b = am:bm, or a/b = am/(bm) ; §172
and a : b=(a -^ m) : (b -^ m), or a/b={a -^ m)/(b -=- m). § 173
Two ratios can be compared by reducing them as fractions
to a common denominator.
Ex. 1. Which is the greater, 3:11 or 5 : 19 ?
3:11= 3/11 = 57/209, and 5 : 19 = 5/19 = 55/209 ;
hence the ratio 3 : 11 > the ratio 5 : 9.
Ex. 2. (a : bf = a^ : b^ ; V^76 = V« - V&- §§ 186, 225.
319. By § 91, (a : &) (c : d) (e :/) = ace : hdf.
The ratio ace : hdf is said to be compounded of the ratios
a:b, c: d, and e : /.
RATIO AND PROPORTION
320. The inverse of a ratio is its reciprocal.
Hence the inverse of the ratio a : 6 is the ratio b : a
(§ 182).
321. By § 183, a/b :c/d = ad: be.
Hence the ratio of any two fractions can be expressed by
the ratio of two integers.
322. Two numbers are said to be commensurdble or incoiii-
mensurable with each other according as their ratio can or
cannot be expressed by the ratio of two integers.
E.g., y/2 and 5 are incommensurable with each other, so also are
^3 and ^5. The incommensurable numbers 3^2 and 7v^2 are com-
mensurable with each other; for their ratio is 3/7. Compare § 224.
323. Ratio of concrete quantities. If A and B are two con-
crete quantities of the same kind, whose numerical measures
in terms of the same unit are the numbers a and 6, then the
ratio of ^ to J5 is defined to be the ratio of a to b.
Exercise 120.
Find the simplest expressions for the following ratios :
1. 6 a to 12 a'*. 4. a/x to c/y.
2. 3 a^x/5 to 6 aa^/J. 5. a/{x - 2) to S/(x - 2y.
3. 1/a to 1/6. 6. 9/(a-by to 6/(a-b).
7. Write as a ratio (2x:Syy; (2a: bf ; (« : c)« ; i^oTb.
Find the ratio compounded of :
8. The ratio 25 : 8 and the square of the ratio 4 : 3.
9. The ratio 32 : 27 and the cube of the ratio 3 : 2.
10. The ratio 6 : 7 and the square root of the ratio 25 : 36.
11. Arrange the ratios 5 : 6, 7 : 8, 41 : 48, and 31 : 36 in
descending order of jnagnitude
334 ELEMENTS OF ALGEBRA
12. For what value of x will the ratio 15 + a;: 17 + a?
be equal to 1/2 ?
13. What number must be added to each of the terms of
the ratio 3 : 4 to make it equal to the ratio 25 : 32 ?
Let X = the number to be added ; then
(3 + a;)/(4 + a;)=25/32.
14. Find two numbers in the ratio of 5 to 6, whose sum
is 121.
15. Which is the greater ratio, 5:7 or 5 + 2:7 + 2?
16. Which is the greater ratio, 7:5 or 7 + 2:5 + 2?
PROPORTION.
324. Four quantities are said to be in proportion when the
ratio of the first to the second is equal to the ratio of the
third to the fourth.
An equality whose members are two equal ratios is called
a proportion. Thus, if
a:b = c: d, (1)
then a, b, c, and d are in proportion, or are proportional, and
equation (1) is 2^ proportion.
A proportion can be written in the form
a/h = c/d, a:b = c: d, or a:b : : c: d,
each of which is read ^ a by 6 is equal to c by d,' or ^ a is to
6 as c is to d.'
The four numbers in a proportion are called the propor-
tionals, or the terms, of the proportion.
The first and fourth terms are called the extremes, and
the second and third the means.
E.g., a and d are the extremes, and b and c are the means in the
proportion
a :h = c : d.
In (1), d is called the fourth proportional to a, b, and c.
RATIO AND PROPORTION 335
325. The following theorem aud its converse in the next
article are the two fundamental principles in proportion.
In any propoHion the product of the extremes is equal to the
product of the means.
That is, if a\b = c\d, (1)
then ad — be. (^2)
Proof. Clearing (1) of fractions, we obtain (2).
Ex. The first, second, and fourth terms of a proportion are c^, 2 a,
and 5 h respectively ; find the third term.
Let x = the third term of the proportion ;
then c2 : 2 a = X : 5 6.
.-. 2 ax = 5 6c2, or X = 5 6cV(2 a).
326. Conversely, if the product of one set of tivo numbers
is equal to the product of another set of two numbers, either set
can be made the extremes and the other set the means of a
proportion.
Proof. Let ad = bc. (1)
Divide (l)hj db, a:b = c:d, ot c: d = a:b.
Divide (1) by dc, a: c = b : d, or b : d = a: c.
Divide (1) by a6, d:b = c: a, or c: a — d\b.
Divide (1) by ac, d : c = 6 : a, or 6 : a = d : c.
From this principle it follows that —
(i) A proportion is proved ivhen it is proved that the product
of its extremes is equal to the product of its means.
(ii) In a given proportion, we can interchange the means,
or the extremes, or we can take the means as extremes and the
extremes as means.
§172
6,91
327. If
a: b = c: d,
then
ma : mb = no : nd,
and
ma : nb = mo : nd.
336 ' ELEMENTS OF ALGEBRA
328. Any proportion, as a : 6 = c : (/, (1)
can he taken by
(i) inversion ; that is, b : a = d : c, (2)
(ii) alternation ; that is, a: c — b . d, (3)
(ill) addition ; that is, a -\- b . a = c -\- d . c, (4)
or a -{- b : b = c + d : d, (5)
(iv) subtraction; that is, a — b : a = c — d : c, (6)
or a - b : b = c - d : d, (7)
(v) addition and subtraction ; that is,
a + b: a-b = c + d\c-d. (8)
Proof. From (1), ad = he. (1')
Add hd to (1'), (a -f 6) c/ = (c + d) h. (2')
Add - hd to (1'), (a - ?>) d = (c - d) h. (3')
By § 326, from (1'), we have (2) and (3); from (2'), (5);
and from (3'), (7).
Dividing (2') by (1'), we obtain (4).
Dividing (3') by (1'), we obtain (6).
Dividing (2') by (3'), we obtain (8).
Observe that (2) and (3) can be obtained from (1) by (ii) of § 326.
329. The products or the quotients of the corresponding
terms of two proportions are x^^^ojyortional.
That is, if a: b = c: d, (1)
and a':b' = c''. d\ (2)
then aa^ : 66' = c& : d&, (3)
and ala! : bjb^ = c/c' : d/d'. (4)
Proof Multiplying (1) by (2), by §§ 6 and 91 we ob-
tain (3).
Dividing (1) by (2), since ^ = ^, we obtain (4).
a 10 010
330. Like powers or like principal roots of proportionals are
proportional.
RATIO AND PROPORTION 337
That is, if a:b = c:d, (1)
then a" : 6" = c" : d", (2)
and {/a:^b=^c:^d. (3)
Proof. By §§ 128 and 186, from (1) we obtain (2).
Hy §§ 221 and 225, from (1) we obtain (3).
331. In a series of equal ratios the sum of the antecedents
is to the sum of the consequents as any one antecedent is to its
consequent.
That is, if a: b = c d = e .f= ", (1)
then a + c + e -{-•' .b + d ^f+ •' = a:b = c:d= •.-.
Proof Let a/b = r; then c/d = r, e//= ?', ••• ;
hence a = br, c = dr, e =fr, •••.
Adding the members of these equations, by § 6, we obtain
a + c + e -f- ••• = (& + <« +/+•••) »••
a4- c 4- e + ••• _ _a_c_
** b-hd-\-f-\--'~~^~b~d •
332. A general and easy method for proving a proportion
is to represent the value of one of the equal ratios in the
given proportion by a single letter, as was done in the last
section.
Ex. 1. Given a:b = c:d, prove that
a2 + a6 : c2 + cd = 62 - 2 a& : (P _ 2 cd. (1)
Let a/b = r ; then c/d = r ;
then a = br, and c = dr.
Substituting these values of a and c in each ratio of (1), we have
a^ + ab^ b^r^ -h 6^r ^ b^(r^ + r) ^ &«
c2 + cd d^r^ + dh' d%r^ + r) d^'
and b^-2ab ^ 6^-2 fc^r ^ b'^(l-2r) ^ 6^
d2-2cd d^-2d^r d\l-2r) <P*
Hence the ratios in (1) are equal.
338 ELEMENTS OF ALGEBRA
Ex. 2. Given a : & = c : d = e :/, prove that
a^ + c^-^ e^'.h^ + d^+P = ace:'bdf. (1)
Let a/h = r ; then c/d = r, and e//= r.
Hence a = br, c = dr, and e = fr.
Substituting the values of «, c, and e first in the product of the
extremes, and then in the product of the means, we obtain
(«3 + c3 + e^)bdf = (&3 + # ^fs^r^Mf,
and (63 + (?3 + j3)«ce = (b^ + # j^ p)'fiMf.
That is, the product of the extremes in (1) is equal to the product
of the means ; hence, by (1) of § 326, (1) is proved.
333. A continued proportion is a proportion in which, the
consequent of each ratio is the antecedent of the following
ratio. Thus a, b, c, d -•' are in continued proportion if
a:b = b:c = c: d= •••.
li a:b = b : c, then b is called a mean proportional between
a and c, and c is called a third proportional to a and b.
It a:b = b : c = c: d, then b and c are called the two mean
proportionals between a and d.
334. The mean proportional between two numbers is equal
to the square root of their product.
Proof. If a:b = b:Cf then b^ = ac, or 6 = Vac.
Exercise 121.
From each of the following products form four different
proportions and their converses :
1. xy = mn. 2. 6x3 = 2x9. 3. a^-b^ = a^-y\
Find the fourth proportional to the three numbers :
4. a, ab, c. 5. a^, 2ab, 3b^. 6. a^, xy, 5 ay^y.
Find the third proportional to the two numbers :
7. a%ab. 8. Q?,2iii?. 9. Sx,%xy. 10. 1, x.
RATIO AND PROPORTION
Find a mean proportional between the two numbers :
11. d',b\ 13. 12ax',Sa^
12. 2a^, 8 a;. 14. 27 a^b% 3 b.
li a : b = c : d, show that
15. ac:bd = c-:(l'.
16. d':<^ = a^-b-:c'-(r-.
17. 2a-\-3c:3a-\-2c = 2b-{-Sd:3b + 2d.
18 . la-\- mb : ;)« -f ^6 = /c + md : pc + ^d.
19. a: a-\-c = a-\-b:a-\-b -\-c + d.
20. a- + a^ -f- ^'- : «- - «^ + ^- = c^ 4- ccZ + d' : c- - cd -{■ d?.
21. a + & : c + ^Z = Va^ + 6' : Vc2 + d^.
22. V^M=^: V?+^ = ^c?T^:-v/^M^.
23. a^c + (M? : 6^d + ftd^ = (« + c)"' : (6 + d)\
24. Va" + 6" : ^c" + d" = </ar - b"- : </(f —d\
25. If a : ^ = 6 : c, prove that a:c = a- -. b\
26. If a : 6 = 6 : c = c : d, prove that a:d = a^:b\
Let r = a -4- 6 ; then a = br, b = cr, c = dr.
.'. abc = bcdr^. .-. a -^ d = r^ = a^ -^ b^.
27. If a, 6, c, (i be any four numbers, find what numbei
must be added to each to make the results proportional.
28. Two numbers are in the ratio of 3 to 8, and the sum
of their squares is 3577 ; find them.
29. The ages of two persons are as 3:4, and 30 years
ago they were as 1 : 3 ; find their present ages.
30. The sides of a triangle are as 3 : 4 : 5, and the perim-
eter is 480 yards ; find the sides.
340 ELEMENTS OF ALGEBRA
31. Divide the number 14 into two such parts that the
quotient of the greater divided by the less shall be to the
quotient of the less divided by the greater as 16 to 9.
32. Show that the ratio of any two fractions, not involv-
ing surds, can be expressed by the ratio of two whole
numbers.
33. Express the ratio of 5^ to 7y\ by the ratio of two
whole numbers.
34. Express the ratio of 17 J to 14| by the ratio of two
whole numbers.
35. The sum of two numbers is 8, and their product is
to the sum of their squares as 3 to 10. What are the
numbers ?
36. The sum of two numbers is 10, and the sum of their
squares is to the square of their sum as 13 to 2^. What
are the numbers ?
37. A hare is pursued by a greyhound, and is 60 of her
own leaps before him. The hare takes 3 leaps in the time
that the greyhound takes 2 ; but the greyhound goes as far
in 3 leaps as the hare does in 7. In how many leaps will
the greyhound catch the hare ?
Let X = the number of leaps taken by the greyhound,
and y = the number of leaps taken by the hare in the same time ;
then aj : y = 2 : 3,
and af-f60:y = 7:3.
CHAPTER XXVI
THEORY OF EXPONENTS
335. Hitherto we have defined and used only positive
integers as exponents. It is, however, found convenient to
extend the meaning of an exponent so that we can use zero,
a fraction, or a negative number, as an exponent.
As it is desirable that all exponents should obey the same
laws, we shall fix the meaning of (i.e., define) any new expo-
nent by imposing the restriction that all exponents must
obey the fundamental law,
a"* xa" = a'"-^". (1)
E.g., to find the meaning of a', we have by law (1),
that is, (a' )3 = a ; .*. a» = the cube root of a. § 213
Again, to find the meaning of a', we have by law (1),
J-J = J-^^ = J = a^;
that is, (a*)2 = «« . .^ ^t = t^e square root of a\ § 213
336. Meaning of a positive fractional exponent.
Let r and s denote any positive integers; then by the
fundamental law of exponents, we have
n , ■< 1- — to « terms
a' ' a* ••• to s factors = a* *
that is, (a')' = a*"; .-. a' = the sth root of a'' ; § 213
r
that is, a' is only another way of writing the sth root of the
rth power of a.
S41
342 ELEMENTS OF ALGEBRA
Hence a positive fractional exponent denotes a root of a
power of its base. The denominator indicates the root, and
the numerator the power.
1
Thus a' denotes the sth root of a.
r
337. Using a* to denote only the principal sth root of a*",
we have
r
a' = W={-^a)% §§220,226
and a« — ^a.
Ex. 1. 83=^8 = 2; 4^ = ( ^4)^ = 2^ = 8.
Ex. 2. 8^ = (^8)2 = 22 = 4.
Ex.3. (-32)^ = (\/^r32)4=(_2)4=16.
338. Meaning of zero as exponent.
By the fundamental law of exponents, we have
a"* • a^ = cr+^ = «'" ;
.-. a^ = qT'/oT' — 1.
That is, any base, except zero, with zero as an exponent is
equal to positive one.
Observe that a° is only another way of writing a'^/a'^, or 1.
E.g., a'i = a/a = a'^/a^- = a'^/a^="' = l.
339. Meaning of a negative exponent.
Let n denote any positive integer or fraction; then by
the fundamental law of exponents, we have
a" • a-'^ = a"+(-'^) = a^=l.
.: a-" = l/a".
That is, a~" is only another way of writing the reciprocal
of a", or of denoting that a'* is to be used as a divisor.
E.g., 3-2 = 1/32 = 1/9 ; (- 2)-3 = l/(_ 2)3 = - 1/8.
THEORY OF EXPONENTS 343
Note. The arithmetic value of an exponent denotes a power, or
a root of a power, of its base ; and its quality denotes whether this
power or root is to be used as a factor or divisor.
Fractional and negative exponents express no new ideas, and are
not necessary to the notation of algebra ; but they are very con-
venient, and greatly facilitate many operations. Fractional ex-
ponents simply afford another way of writing a root of a power ; and
negative exponents, another way of writing a divisor.
r
340. Hereafter in this chapter we shall use a' to denote
only the pnncipcd sth root of a*", or, what is the same thing,
the rth power of the principal sth. root of a.
Observe that r can be either positive or negative, but
that s is always positive.
Ex. 1. 8-i = 2-2 = 1/22 = 1/4.
Ex. 2. (- 27)-^ = (- 3)-* = l/(- 3)4 = 1/81.
Ex.3. (-32)-f = (-2)-« = l/64.
Note the advantage of tirst extracting the root in these examples.
341. A base with any exponent is called an exponential
expression ; as, 3^, a', (x -f- ?/)", (f^'.
342. The quality of an exponent can be changed if the
sign before the exponential expression is changed from x
to -^, or from ^ to x .
Proof. axb-" = ax (1/6") = a -t- 6". (1)
Also a -h b-» = a-T- (1/6") = a x 6". (2)
343. Any exponential factor can be transfeired from the
dividend to the divisor, or from the divisor to the dividend, if
the quality of its exponent is changed from -^ to —, or from
- to -h.
Proof This is the converse of (1) and (2) in § 342.
Or this operation is simply multiplying both dividend
and divisor by the same exponential expression.
344 ELEMENTS OF ALGEBRA
Ex. 1. a-Vc2 = c-2/a2. a-^/b-^ = bya^
Ex. 2. — — - = — - = cH^a^y^ =
Exercise 122.
Eind the value of each of the following expressions :
1. 4-1 5. (2/3)-^ 9. 8t 13. (27/8)1
2. 5-^ 6. 1/5-3. ^Q 4-f. 14. (81/16)"l
3. (3/4)-\ 7. l/.Z-\ 11. 9-1 15. (-27)1
4. (2i)-2. 8. 35". 12. (4/9)-l 16. (- 125)-^
17. 8-^.4-2. 18. (l/64)-3 . (l/9)i 19. (1/25)"^ . 27~l
Write each of the following expressions without using
fractional or negative exponents :
20. ah. 22. 3i»i 24. 2ax~^. 26. {x/y)^.
21. x^. 23. 2a;V^- 25. (a/6)i 27. (a;/?/) ".
28. ^— 31. ^. 34. "-^^'" .
5 a-%^ x-^y-h-^ (a + hyx-^y
' lyh-' ' ab-'c-' ' (a-\-b)-'
^^ 5x-'y-h\ ^^ b(x-y)-\ ^^ x--(x + y)-^
7 c-'db-
(x-\-y)-^ a--{x + yy'
37. Write each of the expressions in examples 28 to 36
in the integral form, i.e.y transfer all the factors from each
denominator to the numerator.
38. Using fractional exponents, write V^^5 -\fa^''> '\/^\
39. Using negative fractional exponents, write Vi/^;
^17^; -^17^'; ViT^.
THEORY OF EXPONENTS 346
344. The meaning of any real commensurable exponent
having been determined, it remains to prove that any such
exponent obeys all the laws of positive integral exponents.
For convenience in stating these laws we shall enlarge the
meaning of the word power so as to include whatever is
denoted by an exponent.
345. Exponents were so defined as to obey the following
fundamental law :
Tlie product of the mth and the nth power of any ba^e is
equal to the (m + ri)th power of that Inxse.
That is, a^^a" = «*"+", I
where m and n denote any commensurable real numbers.
Ex. 1. x^ . a;^ = x^'^^ = x^.
Ex. 2. x~^ . a;"^ = x~^^"^) = x~^.
Ex. 3. x^a~^ ' x-hi^ = x-^a~^ = l/(xa^).
346. Tfie quotient of the mth power of any base divided by
the nth power of the same base is equal to the (m — n)th power
of that base.
That is, a'^/a" = a'"-", II
where m and n denote any commensurable numbers.
Proof a*" H- a" = a"* x a"" = a"-". §§ 342, 345
Ex. 1. x-8 ^ X-* = x-3-(-'«) = X. §§ 342, 345
Ex. 2. J-^ oT^ = a^-(-^) = al
Ex. 3. a~^ -r a"^ = a~^~^~^^ = a^.
347. By §251, V^=Va^*; .-. a' = a^.
r^p r»^p8 rq+pt
Hence a' ' = a*« *« = a '" .
Ex. 1. x^ . x^ = x^* = x^2" = x"^.
Ex. 2. x^/x^ = x*"^ = x^ = xT^.
846 ELEMENTS OF ALGEBRA
Exercise 123.
Simplify each of the following expressions :
1. x'x\ 5. x^3fx-\ 9. a^6-2 X a-^h\
2. a-W. 6. a-^a-V. 10. ^ x~^^y^ x 2 x~^y^ .
3. rt'V. 7. «y. 11. 4 a;~^a~^ X 3 a^' ai
4. a;-2a;-l 8. x'^^-x^. 12. 2A^x5:K"V^r.
13. a/a-\ 15. 4-74-^ 17. a^-yiu-^ 19. x-'^/x-l
14. .'^Va;-^ 6. 7-77-^ 18. x-'/x-^\ 20. t/V^-I
21. (8x--)/(6a.-0. 29. ^-^^"1
22. .^'"-70?-"*.
30. ic^/a; ^.
31. a^^^.aM.
23. x-^/x^-'^.
24. ar'-ya?2-n^
25. x''^/y?-'\ 32. a'^aj"^^ • a^ici
26. o^-Yr--. 33^ a.V/(^-Vb.
27. a.-^-a^"l 34. a/^^^ v'^^.
28. aWa"^. 35. .J/a?'^ • ^ic'" -j- a^^.
348. If m and n denote any commensurable real numbers,
the other three laws of exponents are :
The nth poicer of the mth power of any base is equal to the
mnth power of that base.
That is, (a'")" = a'"". Ill
The nth power of the product of any number of factors is
equal to the product of the nth powers of the factors.
That is, (ab.'.y = a"b" '". IV
THEORY OF EXPONENTS 347
TJie nth power of the quotient of one number by another is
equal to the quotient of the nth power of the first by the nth
power of the second.
That is, {a/by = a"/b". V
Ex. 1. (a:2)-3 = x2(-3) = x-e = l/x*. By IH
Ex. 2. (8-2)-i = 8-2H) = 8^ = 16. By HI
Ex. 3. (a"^)~^ = a~^^"^^ = a^. By III
Ex. 4. (x-iy-8)-2 = (a;-i)-2(y-8)-2 = x^^. By III, IV
Ex. 5. (9"^ x-6)"^ = (9"*)~^(x-6)"^ = 9^ X* = 3 X*.
Ex. 6. (x2/a-8)-2 = (a;2a3)-2 = a;-*a-«.
Ex. 7. {x^/y^y^ = (x^y"^)-» = x'V.
Ex.8. (^_£l\-''^(l^\-'^^:!^
\^y'l \zy^l 3-S2,-t
= 27 2/^x8/8.
Ex 9 Q^v^^ rT>/R_aV^ . /a6-2\^
6\/a^ '^^v^ 6a"^ ' \ba-^l
= aV^^(a26-8)^
349. Proof of laws III, IV, V when m and n are positive
fractions.
Let p, g, r, and s denote any positive integers.
To prove III, (J)' = (V^)' § 337
= (V^y § 227
= [(V«W §226
= (^a)^ = a". Ill
348 ELEMENTS OF ALGEBRA
r
To prove IV, (aby = -</{aby § 337
= ^tf^ § 119
= </a''<JW § 224
r r
= a'b\ IV
To prove V, (a/by = </(a/by § 337
= -</ar/b'' § 186
= </a^/</F - ' §225
r r
= a'/b^. V
350. Proof of laws III, IV, V wJien m and n are negative.
Let h and k denote any positive integers or fractions.
To prove III, (a"^)"* = 1 -- (1/a*)* § 339
= 1 -f- (l/a'^*) §§ 186,118
= a^\ III
To prove IV, (ab)-^ = l/{aby § 339
= l/a'^^* § 119
= (1/a*) (1/6*) = a-^b-\ IV
To prove V, (a/b)-^ = 1 -- {a/by § 339
= 1 -f- (ay^'^) § 186
= by a'' = a-'^/b-K V
The verification of laws III, IV, and V when m or n is
zero is left as an exercise for the pupil.
Exercise 124.
Simplify each of the following expressions :
!• (O'- 4. {x-^\ 1- («-")-*•
2- W"'- 5. {x-i)-i. 8. {aSl
3. (a;')-*. 6. («")-« 9. (-5^*^^'.
THEORY OF EXPONENTS 349
10. (VaryK 14. (4.a-yK 18. (x~^a^)-^^.
11. (a/^)-'*. 15. (Sa'')i 19. (a-2c-t)-8.
12. (VO'" ^^- (^"'2/"')"'- 20. (a-y32)-i.
13. (a;V2)-3. 17. (8a~V^)"^. 21. (2a-Vaj)"'
22. (3ar^^2)-l 33. </^^yV^.
23. (aj-"c-)"'. 34. ^(a + &)' • (a + bp.
24. (3v^^o-«. 35. f^^Y^f^^y.
25. (ay.-)-.^ y^"^ ,^«"^'^^
26. [-2-^a-y(4a;-i)]-2. 36. J^./^^Y^^
27. (32a-i/^)-i U^-U-.-i
28. ^^«/^. 37. (1^ .(^^ '.
29. V^?6^/^5/S^. ^„ »^-| ^„-ii,h^-
38
30. V^rY^'/VsrY^\
31. V^Fy^^=^. ^^Y^«/^
32. ^(a + 6)«.(a2-62)-i. * V^-^a/ " Xa;-^*
351. The following examples are applications of the
methods of multiplication, division, and evolution, to poly-
nomials whose terms involve fractional and negative expo-
nents :
Ex. 1. Multiply > + 1 + 1/ ^a by ^a + 1/ ^a - 1.
The terms + 1 and — 1 may be regarded as the coefllcients of a^ ;
hence arranging both expressions in descending powers of a, we have
a^ + 1 + a~^
g^ - 1 + q~^
a^ + a^ + 1
- a3 - 1 - a 3
+ 1 + a~^ + g"^
af +1 + a~^
350
ELEMENTS OF ALGEBRA
Ex. 2. Divide 16 a-^ + 5 a-i - 6 a-2 + 6 by 1 + 2 a-K
Arrange in ascending powers of a.
16 a-3 - 6 a-2 4- 5 a-i + 6 [ 2 «-i + 1
16a-3+ 8q-2 8 a'^ - 7 a"! + 6
- 14 a-2 + 5 a-i
- 14 a-2 - 7 g-i
12 a-i + 6
12a-i + 6
Ex. 3. Eind the square root of
4 X + 2 x^' - 4 x^ - 4 x^ + x^ + xi
Arrange in descending powers of x.
x^ - 4 x^ + 2 x^ + 4 X - 4x^ 4- x^ | x^ - 2 x^ + x «"
J
2 x^ - 2 x^
xt
-4x^ + 2x^ + 4x
-4x^ +4x
-4
x^ + x^
2x^ -4x^ + x^
2x« -4x^ + x^
Exercise 125.
Multiply :
1. a^-\-ah^-^b^ by a^-bi-
2. x^ — ic^2/^ + y^ by a?^ + y^.
3. 3aj^ — 5 + 8a;~^ by 4 a;^ + 3 aj~i
4. 3a^-4a*-a~^ by 3 a^ -^ oT^ - 6 oT^
5. a^ + a^6^ + 6^ by a^ — ah^-{-bK
6. ic^ — a;2 + a?^ — aj by a;^ + x^.
5 3 1 -1 1 3 1
7. a;^ — a?^ + a?^ — a? * by a;^ + ^^.
8. a3 + &^ 4- c^ — bh^ — c^a^ — a^6^ by a^ + b^ -\- ci
THEORY OF EXPONENTS 351
n n
9. x'^ + x^-j-l by X-" -\- x~~^ -}- 1.
10. iJ-^i^a^^i-^i^ah-^b^ by ia^ + ^bi
11. (f + 2c-'-7 by 5-3c-'= + 2(f.
Divide :
12. 21x + a;^4-a;^4-l by 3a;^4-l.
13. 15a-3a^-2a"* + 8a-i by 5a*4-4.
14. 55^-66^-4fe-^-46~*-5 by 6^-26"^.
15. 21a'^ + 20-27a^-26a2' by 3a*-5.
16. a;^y~^ + 2 H- ic~^y^ by x^2/~^ — 1 + x~^yK
17. a^ + a-6^ — ah^ — ab + ah^ + &^ by a^ 4- &i
18. x^y~^ + y^x~^ by a;'?/"^ + y^a;~».
19. «' — 2 -h a~3 by a^ — a '.
20. 8 c-" - 8 C + 5 c^" - 3 c-3" by 5 c" - 3 c"".
Find the square root of the following expressions :
21. 25a* + 16-30a-24a.^ + 49a*.
22. 9x-12x^-\-10-4:X~^-{-x-\
23. 4 ay^a-' + 12 a;a-^ + 25 + 24 x-^a + 16 aj-^a^
24. 25 arV' + i y^x-^ - 20 ic^/"^ - ^yx'^ + 9.
25 . x^ — 2 oT^x^ + 2 a^ic^ + a~h^ — 2 a^a;^ + ai
26. 4 .T" + 9 «-" + 28 - 24 aj-i" - 16 a;K
27 . 9 a;-^ - 18 a;- Vy + 15 2/ -^ a^ - 6 Vp ^ a; + ^V^.
352 ELEMENTS OF ALGEBRA
352. The following examples are applications of the for-
mulas for products and quotients in Chapter IX., to binomials
whose terms involve fractional and negative exponents :
(1) (2 a^ - x~i)^ = (2 Jy + 3(2 a^)2(- x~^)
+S(2 J)(-x~^y+(-a~^y. (1)
= 8 a^ - 12 ax~^ + 6 a^x-i - a"i (2)
(2) (x^ + yi) (x^ - yi) = {x^^ - {y^y = x^- yK § 122
(3) (7 x-9 2/-1) (7 x+9 y-i) =49 a;2_8l y-\
(4) (4a;-5a;-i)(4x+3x-i) = 16a:2+(3x-i-5x-i)4ic-15ic-2 §123
= 16 x2 - 8 - 15 a;-2.
(5) (pfi - 1) - (X^ - 1) = [(X^)5 - 15] ^ (x^ _ 1)
= (x^f + (x^s ^ (a;^-)2 + x^ + 1 § 129
= a;'^ + ic + cc^ + ic^ + 1.
(6) (x^ + 27) - (x^ -t- 3) = i(xh^ + 33] - {x^ + 3)
= X - 3 a:^ + 9.
Exercise 126.
Write the value of the following expressions :
1. {ar^-\-h^f. 3. {m^ + Jf. 5. (f^ - s'^f.
2. (x^^y-^f, 4. (c2-6*)l 6. (2a^-a-^f.
7. (r-2 + 6^y. 11. (a^ + &~^/.
8. {i^'-^n-^y. 12. (a^^-2/"^)«.
9. (iVa-i-^6)^ 13. (o?^ + 1) (aJ^ - 1).
10. (a~^-2 62c^)^ 14. (x^-\.y^)(x^-y^),
15. (4 a;^ + 3 a"^) (4 a;^ - 3 a"*).
16. (3 a; - 5 a-^ (3 a; + 2 a-i).
THEORY OF EXPONENTS 353
17. (ab-c^(ab-^5x-^).
18. (x-9a)-^{x^ + 3a^).
19. (a-2x_i6)^(a-x_4).
20. (aj-3- + 8) -f- (a;-* 4- 2).
21. (c2* - c-'') -=- (c* - c-i*).
22. (l-8a-3)^(l-2a-^).
23. (a;* + ic* + 1) (a;^ - 1).
24. (a;-'*-l)^(a;-i + l).
25. (ar^+-32)H-(a;'* + 2).
26. (a,-3 - 2/S) ^ (a;^ — 2/^).
27. (ar^ + /)^(a;* + A
28. (a; - 243 2/^) ^(a;^- 3 2/^.
CHAPTER XXVII
INDETERMINATE EQUATIONS AND SYSTEMS
353. Division by zero. As a quotient, 0/0 denotes the
number which multiplied by 0 is equal to 0 (§ 85). By
§ 74 any number multiplied by 0 is equal to 0 ; hence 0/0
denotes any number whatever, or is indeterminate. That is,
when the dividend is zero, division by zero is indeterminate.
As a quotient, a/0 denotes the number which multiplied
by zero is equal to a. But any number, however large,
multiplied by zero, is zero; hence the division of a by 0^
is impossible. That is, ivhen the dividend is not zero, division
by zero is impossible in the sense that no number can express
the quotient or any part of it.
354. The forms 0/0 and a/0. As an answer to a problem
the indeterminate form 0/0 denotes that the problem is
indeterminate, i.e., has an unlimited number of answers.
As an answer to a problem, a/0 denotes that the problem
involves inconsistent conditions, and is therefore impossible,
as is illustrated by the following problem :
Prob. A and B are travelling in the direction PB at the rates of a
and h miles per hour. At 12 o'clock A is at P and B at ^, which is c
miles to the right of P. Eind when they are together.
P Q B
Let distances measured to the right from P, and periods of time
after 12 o'clock, be regarded as positive.
Let X = the number of hours from 12 o'clock to the time when A
and B are together.
Then ax = 6x + c. (1)
Hence x = — —• (2)
• a — 0
364
INDETERMINATE EQUATIONS AND SYSTEMS 355
Discussion. If c > 0 and a > 6, a; is positive ; that is, A will over-
take B at some time after 12 o'clock.
If c > 0 and a < &, x is negative ; that is, A and B were together
at some time before 12 o'clock.
If c = 0 and a =^ b^ x = 0; that is, A and B are together at 12
o'clock, but not before or after that time.
If c = 0 and a = b, x = 0/0 ; that is, A and B are always together
under the conditions; and the problem is indeterminate, i.e., has an
unlimited number of answers.
If c ^fc 0 and a = 6, a; = c/0 ; that is, A and B can never be
together as they are always at a fixed distance apart ; the problem
involves inconsistent conditions, and is therefore impossible.
Observe that the fraction — - — assumes the form - by
a-b 0 -^
reason of two indej^endent conditions; namely, c = 0 and
a = b. In any such case the form 0/0 indicates that the
given fraction can have any value under the conditions.
355. An impossible equation is one which expresses a con-
dition which cannot be satisfied.
E.g., Sx + 5 = Sx — S is an impossible etjuation ; for it expresses
the condition 0 • a; = — 13, which no value of x can satisfy.
Again, ^a; = — 3 is an impossible equation, when ^x is restricted
to its principal value.
An impossible system of equations is a system whose equa-
tions are inconsistent (§ 206).
E.g., the system ax -{- by = c, (1)
3 ax + 3 6y = 5 c, (2)
} («)
is impossible ; for its equations are evidently inconsistent (§ 206).
An impossible equation or system of equations is often a particular
case of a more general equation or system, in which the solutions
involve the form a/0.
Thus, the equation ax = b becomes impossible only when a = 0,
and then its root b/a becomes 6/0.
It will be seen in § 356 that a system of two linear equations in
X and y becomes impossible only for a certain relation between the
coefficients of its equations, which makes the values of x and y assume
the form a/0.
856 ELEMENTS OF ALGEBBA
Again the system
x + y = 9, . (1)
2x + y = lS, (2) [ (&)
x + 5y = 16, (3).
is impossible ; for the only solution common to (1) and (2) is 4, 5,
and this reduces (3) to 29 = 16.
Equation (3) cannot be obtained from (1) or (2), or by combining
(1) and (2); hence it is independent of them separately and jointly.
System (b) illustrates the principle that
When the number of independent equations in a system ea>-
ceeds the number of unknowns, the system is impossible.
356. A defective system is one which, lacks one or more of
the full number of solutions which we would expect from
the degrees of its equations.
E.g.^ the system
|(«)
a2x2 - 62y2 = c2, (1)
ax-(h-\- e)y = c, (2)
which has, in general, two solutions (§306), becomes defective when
e = 0.
For, dividing (1) by (2) when e = 0, we obtain
ax + by = c. (3)
Equations (2) and (3) form a system equivalent to system (a);
hence system (a) has but one solution when e = 0.
357. An indeterminate equation is one which has an un^
limited number of solutions. Thus any equation in two or
more unknowns is indeterminate.
An indeterminate system of equations is one which has an
unlimited number of solutions.
E.g. J the system Sx + 4:y-{-5z = 0,
x-y-2z = 0, ^^^^
Ida]
is an indeterminate system ; for, assigning any value whatever to z,
we can find a corresponding set of values of x and y. Hence, system
(a) has an unlimited number of solutions, and is indeterminate.
INDETERMINATE EQUATIONS AND SYSTEMS 357
Again, the system
2x-\-Sy-z = lo, (1)
Sx-y + 2z = 8, (2)
5 X H- 2 y + 2! = 23, (3)
is indeterminate. No two of its equations are equivalent, but any
one of them can be obtained from the other two ; thus, by adding
(1) and (2), we obtain (3). Hence the system contains but two in-
dependent equations, and therefore any solution of two of them will
be a solution of the third.
These examples illustrate the following principle :
WJien the number of independent equations in a system is
less than the number of unknowns, the system is indeterminate.
Ex. By discussing its solution, show that tlie system
ax-i- by = c,
a'x + b'y = c',
|(«)
is (i) indeterminate if ^ = _ = 2_ j (1^
a' b' c'
and (u') impossible if ^ = ^ r,^ ^. (2)
a b' c'
By § 207 the values of x and y in system (a) are
_ b'c — be' _ ac' — a'c .on
~ ab' - a'b ^ ab' - a'b ^ ^
(i) When condition (1) is satisfied, from (1) we have
ab' - a'b = 0, b'c - be' = 0, ac' - a'c = 0 ;
hence the values of x and y in (3) each assume the form 0/0 ; that is,
the system has an unlimited number of solutions, and is therefore
indeterminate.
(ii) When condition (2) is satisfied, we have ab' — a'b = 0.
But neither b'c — be' nor ac' — a'c is zero.
Hence the value of each x and y assumes the form a/0 ; that is,
the system has no solution, and is therefore impossible.
The equations in (a) are evidently equivalent when (1) is satisfied,
and inconsistent when (2) is satisfied.
358 ELEMENTS OF ALGEBEA
368. Sometimes it is required to find the positive integral
solutions of an indeterminate equation or system.
The following examples will illustrate the simplest general
method of finding such solutions.
Ex. 1. Solve 7 X + 12 y = 220 in positive integers.
Dividing by 7, the smaller coefficient, expressing improper fractions
as mixed numbers, and adding the proper fractions, we obtain
aj + y + ^pi=:31. (1)
Since x and y are integers, 31 — x — y is an integer ; hence the
fraction in (1) denotes an integer.
Multiplying this fraction by such a number as will make the coeffi-
cient of y divisible by the denominator with remainder 1 (which in this
case is 3) , we have
- — ^ = 2 ?/ — 1 + ^—^ — = an mteger.
Hence ^ = an integer = p, suppose.
:.y = lp + 1. . (2)
From (1) and (2), x = 28-12j9. (3)
Since x and y are positive integers, from (2) it follows that^ >— Ij
and from (3) it follows that p < 3 ; hence
i) = 0,1,2. (4)
From (2), (3), and (4), we obtain the three solutions
a: = 28, 16, 4;
y= 2, 9, 16.
Ex. 2. Solve in positive integers the system
a; + 2/ + ^ = 43, (1)
10 aj + 5 ?/ + 2 ;s = 229. (2) J
INDETERMINATE EQUATIONS AND SYSTEMS 359
Eliminate z, 8x-\-Sy = 143,
y + 2x-\-^-^^ = ^7. (3)
.-. i^LzA = x-l+ ?-=-i = an integer.
o o
x—1
.'. = an integer = p, suppose.
o
.-. x = 3p + l. (4)
From (3) and (4), y = io-8p. (6)
From (1), (4), and (5), z = 5p- 3. (6)
From (6), i) > 0 ; and from (o), /> < 6 ; hence
p= 1, 2, 3, 4, 5. .
Whence x = i, 7, 10, 13, 16;
y = 37, 29, 21, 13, 5 ;
z= 2, 7, 12, 17, 22.
Thus, the system has five positive integral solutions.
Exercise 127.
Solve in positive integers :
1. 3a; + 291/ = 151. 8. 12 x -lly + 4:Z = 22,
2. 3.'c + 82/ = 103. -4:X-{-5y-\-z = 17.
3. 7x-^12y = 152. 9. 20a: - 21?/ = 38, 1
4. 13a; + 72/ = 408. S y -\- 4. z = 34.. J
5. 23 a; + 25 2/ = 915. 10. 5ic- 14^ = 11.
6. 13 a; + 11 2/ = 414. H. 13 a; + 11 2; = 103, |
7. 6a;+72/+4.=122, 1 7.-52/ = 4. J
lla;+8i/-62=145.J 12. 14 a; - 11 y = 29.
13. A farmer buys horses at $ 111 a head, cows at $ 69,
and spends $ 2256. How many of each does he buy ?
360 ELEMENTS OF ALGEBRA
14. A drover buys slieep at $ 4 a head, pigs at $ 2, and
oxen at ^17. If 40 animals cost Mm $ 301, how many of
each kind does he buy ?
15. I have 27 coins, which are dollars, half-dollars, and
dimes, and they amount to $ 9.80. How many of each sort
have I ?
16. A drover buys sheep at $ 3.50 a head, turkeys at
$ 1.33i, and hens at | 0.50. If 100 animals cost him $ 100,
how many of each does he buy ?
CHAPTER XXVIII
THEORY OF LIMITS
359. A variable is a quantity which is, or is conceived to
be, continually changing in value.
E.g.^ the time since any past event is a variable ; so also is the
height of an ascending or a descending balloon.
The amount of water in a cistern which is being filled by a con-
tinuous stream is a variable ; and the number which measures this
amount is a variable number.
Variable numbers are usually represented by the final
letters of the alphabet, as x, ?/, z.
A constant is a quantity whose value is fixed or invari-
able. Constant yiumbers are usually represented by figures
or the first letters of the alphabet, as 4, 7, a, 6, c.
E.g.^ the time between any two past events is a constant ; and the
number which measures this time is a constant number.
360. Limit of a variable. When, according to its law of
change, a variable approaches indefinitely near, and con-
tinually nearer a constant, but can never reach it, the
variable is said to approach the constant as its limit.
E.g.^ let A, B, and N be three ^xed points in the straight line AN;
then AB and NB will be constant distances.
P
I I I I I. J
A O D E B N
Suppose a point P, starting from A, moves 1/2 the distance from
^to 5, or to C, the first second ; 1/2 the remaining distance, or to i),
the next second ; 1/2 the remainin<? distance, or to E, the next second ;
and so on indefinitely ; then AP, NP, and 5P will be variable distances.
361
362 ELEMENTS OF ALGEBRA
The variable distance AP will approach indefinitely near and con-
tinually nearer the constant distance AB, but can never reach it ;
that is, the variable distance AP will approach AB as its lijnit. The
variable distance NP will approach indefinitely near and continually
nearer the constant distance NB, but can never reach it ; that is, the
variable distance NP will approach NB^ as its limit. The variable
distance BP will approach indefinitely near and continually nearer
zero, but can never reach it ; that is, BP will approach zero as its
limit.
The variable AP will always be less than its limit AB ; and the
variable NP will always be greater than its limit NB.
Again, suppose the number of sides of a regular polygon inscribed
in a given circle is increased from 4 to 8 ; from 8 to 16 ; from 16 to 32,
and so on indefinitely ; then the area of the inscribed polygon will
approach the area of the circle as its limit, and the area between the
perimeter of the polygon and the circumference of the circle will
approach zero as its limit.
Some variables change according to such laws that they
approach limits, others do not. The theory of limits ap-
plies only to such variables as approach limits.
361. Notation. The sign = is read ^approaches as a
limit ' ; thus x = a \& read ^ x approaches a as its limit.'
The phrase ' as a limit ' is sometimes omitted, x = a
being read 'x approaches a.'
The expression It {x) is read ^ the limit of x.^
Thus, It (x) = a, read ' the limit of x is equal to a,^ is
only another way of writing x = a.
362. The difference between a variahle and its limit ap-
proaches zero as its limit. That is, if jr = a, then jr — a = 0.
Conversely, if the difference between a variable and a con-
stant approaches zero as its limit, the variable approaches the
constant as its limit. That is, if jr — a = 0, then x = a.
Proof. X — a approaches just as near to 0 as a; does to
a and no nearer ; hence if ic = a, then x — a = 0; and
conversely.
THEORY OF LIMITS 363
363. A variable cannot approach two unequal limits at the
same time.
Proof. In approaching as a limit the more remote of two
unequal constants, a variable would evidently reach a value
between the two constants, and thereafter while it ap-
proached the one as a limit, it would recede from the other,
which therefore is not a limit. Hence the theorem.
364. Tlie limit of the sum of a constant and a variable is
the sum of the constant and the limit of the variable.
That is, if jr = a, c 4- Jr = c + fl-
Proof The sum c -{- x approaches just as near to c + a
as X does to a, and no nearer ; hence if x = a, c -{- x = c-\- a.
365. TJie limit of the product of a constant and a variable
is the product of the constant and the limit of the variable.
That is, if x = a, ex = ca, when c^O.
Proof By § 362, if a; = a, a: - a = 0.
Choose a constant k as small as you please ; then x — a
will become arithmetically less than k/c.
Hence cx — ca will become arithmetically less than k.
But cx — ca cannot become 0, since x — a cannot.
Hence when x = a, cx = ca.
366. If two variables are always equal and one approaches
a limit, the other approaches the same limit.
That is, if / =jr and x = a, then / = a.
Proof li y = Xj then y approaches just as near to a as ic
does, and no nearer ; hence, if x = a, y = a.
367. If two variables are always equal and each approaches
a limit, their limits are equal.
That is, if / = jr, and x = a, and/ = 6, then b = a.
364 ELEMENTS OF ALGEBRA
Proof. If y = cc and x = a, then, by § 366, y ^ a. (1)
But by hypothesis, y ==b. (2)
By (1), (2), and § 363, b = a.
368. If It (i^) = 0 and It (w) = 0, then It (vw) = 0.
Proof If ?; = 0 and w = 0, vw approaches nearer to 0
than either v or w; but vw cannot equal 0, since neither v
nor w can.
Hence if v = 0 and w ^^ 0, vw = 0.
369. TJie limit of the variable sum of two or more variables
is the sum of their limits.
That is, if x = a, y = b, z = c, •-•,
then x+y + z-\ = ff + 6 + c H .
Proof Let Vi = x — a, V2 = y — b, Vs = z — c, -";
then (aj 4- 2/ + 2 + •••) - (a + 6 + c + ...) = '^1 + ^2 + ^3 + • -.
Now, however small a constant k may be, each one of the
n variables Vi, v^, v^, ••• can become arithmetically less than
k/n ; hence their sum can become arithmetically less than k.
But, since x-\-y -\- z -\- '•• is variable, '^i + i^2 + '^3 + ••• cannot
reach and remain zero. Hence I'l + Vg + % + • • • =0.
Hence x-\-y + z-\ = a + b+c-\ . §362
370. The limit of the variable product of two or more vari-
ables is the product of their limits.
That is, lt(jr/z...) = It(jr) • lt(/) • lt(z) ....
Proof. Let x = a, y = b,
and let v = x — a, w = y —b;
then x = a-\-Vj y — h -\-u).
THEORY OF LIMITS 365
Hence oey = ab -{- aw -\- bv -{- vw.
.: It (xy) = a6 + It (aw -^bv-{-vw) §§ 367, 364
= ab-{-alt(w)+blt(v)^\t(vw) §§369,365
= ab § 368
= lt(a^).lt(2/). (1)
By (1), It (xy • zu) = It (xij) • It (zu)
= lt(x) '\t(y) 'lt(z) At(u).
And so on for any number of variables.
371. TJie limit of the variable quotient of two variables is
the quotient of their limits, when the limit of the divisor is not
zero.
That is. It (jr//) = It (jr)/lt (/), when It (/) ^ 0.
Proof Let z = x/y, or x — yz-,
then \t(z) = \t{x/y), (1)
and It (x) = It {yz) = It (y) • It (z). (2)
Dividing (2) by It (2^) when It (y) ^ 0, we obtain
It (2) = It (x) /It (2/), when It {y) ^ 0. (3)
Equating the two values of It (2) in (1) and (3), we have
\t(x/y) = \t{x)/\t{y), when \t(y) ^ 0.
Ex. it(^U;i^ §371
_lt(a;).lt(y).lt(g) „ 3^^
c . It(tj) . It(Mj) ' ^
372. Lt (jr") = [It(jr)]'', where n is a positive integer.
Proof. It (af*) = It (a; • ic • a; • • • to w factors) by notation
= It (x) '\t(x)'"ton factors § 370
= [It(aj)]". by notation
E.g., if X = a, lt(x*) = a*, lt(x6) = a«, lt(a;") = a".
366 ELEMENTS OF ALGEBRA
Exercise 128.
It x = a, y = b, z = c, u = e, v = 0, find :
1. lt(aa;). 4. lt(x^-\-a^). 7. lt(a^/f),
2. \t(cx + av). 5. \t(a:^-x'). 8. It (2 a^/y').
3. lt(xy-7zu). 6. It (xy^-zu^). 9. lt(vx'^y/z*).
10. It (a.-^i/'' 4- mafz^ + wi?). 12. It (a^y + mic^;^ + wa?^^).
11.
373. TF/ien ^/ie quotient of two variables or the product or
sum of two or more variables is equal to a constant, the
quotient, product, or sum of their limits is equal to the same
constant.
Proof, (i) Let xy = m -, tlien xyz = mz.
We multiply by z to make the members variable.
.-. It (x) . It (?/) . It . (0) = m . It {z). §§ 370, 365
Divide by It (z), It (x) • It {y) = m.
(ii) Let X ^ y = m ; then x = my.
.'. It (a;) = It (wiy) = w • It («/).
.-. It (x) H- It (y) = m, when It {y) =^ 0.
(iii) Let x + y -{- z + ••• = m. (1)
y + z + •" = m — X.
.'. lt(2/)+lt(^)+...=:m-lt(a;).
.-. lt(x)+lt(2/)+lt(2;)... =m.
374. Lt (c/y) = c/lt (y), ivhen It (/) ^ 0.
Proof. Let z = c/y; then zy = c.
Hence It (0) = It (c/y), (1)
and It (2!) . It (y) = c. § 370
Hence It (z) = r/lt (y), when It (y) ^ 0. (2)
From (1) and (2), It {c/y) = c/lt (?/), when It («/) ^t 0.
THEORY OF LIMITS 367
m m
375. Lt(jr'') = [lt(x)]''.
Proof. Let
;s = x" ; then s" = x^.
(1)
From (1),
lt(0)=lt(X-),
(2)
and
[lt(e)]"=[lt(x)]'».
§§ 367, 372
Hence It (2)= [It (x)]". (3)
m m
From (2), (3), It (x«) = [It (x)] «
Ex. It (x^) = [It (x)]^ = J when x = a.
376. An infinitesimal is a variable whose limit is zero.
Thus, the difference between a variable and its limit is
an infinitesimal.
In approaching its limit zero, an infinitesimal becomes indefinitely
small and continually smaller, but it never equals zero. A small
quantity becomes an infinitesimal when it begins to approach zero as
its limit rather than when it reaches any particular degree of small-
ness. A quantity, however small, which does not approach zero as
its limit is not an infinitesimal.
377. An infinite is a variable which under its law of
change can exceed any constant however great.
Thus, the reciprocal of an infinitesimal is an infinite.
E.g.^ if x = 0, 1/x can exceed any constant number however great ;
thus, since 1/(0.1") = 10", we have
when x = .l, .l^o, .l^'^', .V-^, .\^^\ ...,
1/x = 10, 1010, 10^00, 101000, 1010000^ ....
The general symbol for an arithmetic infinite is oo ; and
a; = 00 is read ' x increases without limit,' ov ^ x is infinite.'
A positive infinite is denoted by + oo, and a negative infinite
by — 00, read ^ 00 is a negative infinite.'
An infinite does not approach a limit, but increases
arithmetically without limit.
368 ELEMENTS OF ALGEBRA
378. Any number which is neither an infinitesimal nor
an. infinite is called a finite number. All the numbers con-
sidered prior to this chapter are finite numbers.
379, Any finite number, not zero, divided by an infini-
tesimal is an infinite; and conversely, any finite number,
not zero, divided by an infinite is an infinitesimal.
That is, when x = 0, a/x = oo (where a ^ 0).
And conversely, when a; = oo, a/x = 0.
Ex. rmdltfi^i=l^i+i^ whenx =
4 x^ - 3 x2 + 5 _ 4 - 3/x + 6/x»
7x-3 + 4x-8~7 + 4/x2 - S/x^'
173
. ^ /4x3-3x2 + 5\ lt(4-3/x + 5/x3). «« 367, 371
V7a;3 + 4x-8J \t(7+4/x'^-S/x^)
= 4/7. §§ 364, 369, 379
Exercise 129.
Find the limit of each of the following expressions,
when a; = 00 :
1.
3.
Ix'-Sx ' ^ (Sx-4:)(5x-\-A)
5a;2 + 9* ' 9a^4-8a;-ll'
ax^-bx-\-e ^ (3 + 2 a;') (2 a; - 7)
mar^ + ca^'^ + na;* * (5 a;2 + 7) (7 + 9 a;)'
(3ar^-l)^ g (3 + 2a^)(a;-5)
x' + 9 ' (4a^-9)(l + ^)
If a; = a, y = b, z = c, find the value of :
7. abc, if xyz — m. 10. Lt (m/ar^ + n/i/^).
8. ab/c, if xy/z=^n. 11. Lt (?i/a;** + /i/y").
9. ayc^, li xyz'^h. 12. Lt ((xYt" -}- ^V?/"*).
THEORY OF LIMITS 369
13. Lt(2/^). 14. Lt(a;V)- 15. Lt (V^).
16. Ltf^4^^V 17. Ltf^-^^^\
\xu^ — nz^J V 2^ n^'x^J
380. Fractions which assume the form 0/0.
Substituting 1 for x in the fraction (x- — l)/(x — 1), we have
x2 - 1 ^ 1 - 1 ^ 0
x-1 1-1 0*
That is, by the method in § 12 we fail to obtain any definite value
for the fraction (x^ - l)/(x — 1), when x = 1.
But for any value of z other than 1, we have
(x2_i)/(x-l) = x + l. (1)
Hence (1) holds true when x = 1.
•••""'Uffl)- ""'!(-+!) = 2. (2)
That is, 2 is the Umit which the fraction approaches when x = 1.
The first member of (2) is read 'the limit of (x^ - l)/(x- 1),
when X = 1.'
The example above suggests the following definition :
The value of an expression for any particular value of its
variable is the Umit which the expression approaches when
the variable approaclies this particular value as its limit.
This -definition applies to any expression, but we shall
use it only when the simpler one in § 12 fails.
Ex. Find the value of (x^ + x - 2)/(x2 - 1), when x = 1.
Putting 1 for x in this fraction, we obtain the form 0/0.
Hence to find the value of this fraction, when x = 1, we must find
its limit when x = 1.
For values of x other than 1 , we have
x'^ + a;-2_(x- l)(x + 2^_x + 2
x2-l ~(x-l)(x + l)~x + l'
imit /xg + x-2\ limit /x + 2\
limit
X
_lt(x + 2)_3
~lt(x+l) 2*
370 ELEMENTS OF ALGEBBA
Hence the value of the fraction when aj = 1 is 3/2.
Note that we cannot apply § 371 to the given fraction, for the
limit of its divisor is 0 when x = \.
Observe that the indeterminate form 0/0 arises here by
reason of one condition ; viz. x — some particular value. In
any such case an indeterminate form simply indicates that
the method of evaluation by substitution (§ 12) fails, and
that the more general method by limits must be used.
Exercise 130.
Find the value of each of the following expressions :
1.
2.
^ ^, when x = l.
x — 1
^ x'-l
'• 0^-1'
when x = l.
^ + J, whenx= 1.
.T^ — a^
when x = a.
X^ — 4:
x^ — a^'
when x = a.
-^+J»f2 hen.= 3.
x' — 9
3 ^-f-a'
x^-^-a^'
when x = —
9. (^ -«')', when
x = a.
(x-a)i
^ of — ax — Sx — 3
x'-a'
— , when X -.
= — a.
381. a/0, or absolute infinity. The expression a/0, read ' a by-
zero,' frequently occurs in mathematics, and the question arises 'what
does it mean ? ' By § 353, a/0 must symbolize that of which no part
can be expressed by any number however large ; hence it symbolizes
that which transcends all number, or absolute infinity, of which we
can have no positive idea.
The expression a/0 is commonly denoted by the symbol co. When
this notation is adopted, this meaning of go must be clearly distinguished
from that in § 377, where oo denotes an infinite, or a variable which
increases without limit.
In this book x never denotes a/0.
THEORY OF LIMITS 371
382. Certain combinations of 0 and a/0 ; as
(a/0) - (a/0), (a/0) • 0, a/0 - a/0, etc.,
produce additional indeterminate forms. But any expression which
assumes any one of these forms can be reduced to an identical expres-
sion which for the same values of its variables will assume the funda-
mental form 0/0.
E.g., we have the identities,
ia/x)^Ca/y)=y/x (1)
{a/x) ' y = ay/x (2)
a/z-a/y = aiy-x)/ (xy) . (8)
If X = 0 and y = 0, (1), (2), and (3) become
(a/0) - (a/0) =0/0
(a/0) .0 = 0/0
a/0 -a/0 = 0/0.
K in the identity a;«-y = x-/xf
we put X = 0 and y = a, we obtain
00 = 0/0.
That is, 0^ is an indeterminate form.
LAWS OF INCOMMENSURABLE NUMBERS
383. The laws, already proved for commensurable num-
bers, are, by the theory of limits, easily proved for incom-
mensurable numbers.
384. Proof of the fundamental laws.
Let a, 6, c be any incommensurable constant numbers,
and let x, y, z be commensurable variable numbers such that
X = a, y = b, z = c.
Proof of (A). x-^y = y -\-x. § 36
. •. It (x) -h It (y) = It (y) + It (a?) ; § § 367, 369
that is, a + b = 1 -\-a.
372 ELEMENTS OF ALGEBRA
Proof of (^') . xy = yx. § 49
•. It (x) It (y) = It (y) It (x); § § 367, 370
that is, a-b = b -a.
Proof of (O). {x-^y)z = xz + yz. § 60
.% [It (x) + It (2/)] It (z) = It (x) It (2) + It (y) It (;2) ; § § 369, 370
that is, (a -\- b) c = ac -{- be.
Laws (B), (B'), and (C) follow from laws (A), (A'), and
(C), as in §§ 36, 49, and 88.
385. If X is commensurable and jr = 0, then a' == 1, when
a=^l or 0.
E.g., giving to x the successive values |, J, |, •••, and finding the
corresponding values of 16*, we obtain the results below :
When x = l, I, |-, j\, ^V s^f» ih^ ^h ■-i
16^ = 4, 2, 1.4, 1.19, 1.09, 1.04, 1.019, 1.009 ....
Observe that each value of 16* is the square root of the preceding
value.
From this table of values it is evident that when x = 0, 16*= will
approach indefinitely near and continually nearer 1 ; but it cannot
reach 1, since x cannot reach 0, and 16" = 1, when, and only when,
x = 0.
386. Meaning of a'", m incommensurable.
Let x<m<Zf
where x and z are commensurable, and let a>l; then in
harmony with the meaning of commensurable exponents
we assume a"* to denote a number such that
a'Ka'^K a'. (1)
Let x=:m, and 2; = m ;
then It (z-x)^ It (z) - It (x) = 0.
Hence by § 385, lt(a^-^) = 1. (2)
THEORY OF LIMITS 373
Now a'-a' = a^ (a'-' - 1).
.-. It (a* - a') = It (a') [It (a'-=^) - 1] § § 370, 364
= lt(a^)(l-l)=0. by (2)
From (1), a' — a' > a"" — a"" ',
hence, as a' — a' = 0, a"" — a' = 0,
,'. a"" = It (a') when x = m.
That is, any base a icith an incommensurable exponent m
denotes the limit of a* when x = m.
387. Proof of laws of exponents I, II, IV, V.
Let m and n be any incommensurable constant numbers,
and let x and y be commensurable variable numbers such
that x^m, y = n.
Proof of I. a'a^ = a'^". § 345
Hence It (a'a*) = It (0*+") = ar+". (1)
But It (a'a^) = It (a*) • It (a") = a'^a^ (2)
From (1) and (2), a'^a'' = a"»+". Law I
Proof of 11. By law I we have
(i^~^a!* = a"*.
Hence by § 32 ci— " = a'^/a\ Law II
Proof of IV. a'b' = (aby. § 348
Hence It (a'6*) = It [(«6)*] = (a^)- (o)
But It (a'b') = It {a') • It (6^) = a^ • &«. (6)
From (5) and (6), (aft)*" = a-^ft"*. Law IV
iVoo/ 0/ V. a'/b' = (a /by. § 348
Hence It (a'^/b^) = It [(«/6)^] = (a/ft)"*. (7)
But It (a'/b^) = It (a-) / It (6^) = a'"/ 6"*. (8)
From (7) and (8), (a/b^ = a'^/b'^. Law V
374 ELEMENTS OF ALGEBRA
388. To prove law III for incommensurable exponents
we need the following theorem of limits :
Ify and z are commensurable and m and c are incommen-
surable; tlieny when y = m and z = c, z^ = c"'.
Proof. For all values of z (except 0 and 1), and hence
for z variable, by § 386, we have
2;y _ 2;'* = 0, when y = m. (1)
By § 375, g'^ _ c"* = 0, when z = c. ' (2)
From (1) and (2), (2^ - z^) + {z^ - C') = 0 ;
that is, 2^ — c"* = 0, when y = m and 2 = 0.
Hence z^ = c"* when y = m and z = c. (3)
Proof of III. Using the same notation as in § 387, we
have
{ay = rt^^.
Hence It [(a^)^] = It (a^^ = a^^. (4)
But by (3), It [(a^)^] = [It (a^)]"^^) = {ar)\ (5)
From (4) and (5), («'")''=«""*.
From these laws for incommensurable numbers the other
laws follow by the proofs already given for commensurable
numbers.
VARIATION.
389. Two variables are often so related that the value of
one depends upon the value of the other.
E.g.^ the distance a train runs at a given speed depends upon the
time it runs, and this distance increases when the time increases.
The length of an elastic cord depends upon its tension, and this
length varies when the tension varies.
liy = b x'^, the value of y depends upon the value of x, and y varies
when X varies.
We shall here consider only the simplest kinds of variation.
THEORY OF LIMITS 375
390. Direct variation. When the ratio of two variables
is a constant, either variable is said to vary directly as the
other.
The symbol oc, read ^ varies directly as/ is called the sym-
bol of direct variation. When placed between two variables
it denotes that their ratio is some constant.
The word ' directly ' is sometimes omitted.
E.g., 2/ OCX, read ^y varies directly as a:,' denotes that y/x = m^
where m is some constant.
Again, if y = 3 x, y/x = 3 ; hence, yocx or xccy.
391. If one vanable varies directly as another, either vari-
able is a constant multiple of the other; and conversely.
Proof If yc^x,y/x = m] .-. y=mx, or x=(l/m)ys
Conversely, if y = mx, y/x = m\ .-.yocx, or xccy.
E.g., the area of a rectangle = base into altitude.
Hence if the altitude is constant, the area oc the base.
And if the base is constant, the area oc the altitude.
392. If y^x, and if x\ y' and «", ?/" are any two sets of
corresponding values of x and y, then
2/':a:' = y':a:". (1)
Proof If 2/ oca:, y'/x^ = m and y'^/x^^ = m. (2)
From the equal ratios in (2), we have the proportion (1).
Conversely, if y' : x' = y" : x", y = mx and yocx.
393. Inverse variation. One variable is said to ixiry in-
versely as another when the first varies as the reciprocal of
the second.
That is, / varies inversely as jr, when y oc 1/jr.
394. If one variable varies inversely as another, the product
of the two variables is a constant; and conversely.
Proof. If yaz 1/x, y = m (1/x) ; .-. yx = m.
Conversely, if yx = m, y = m(\/x); .-. yocl/x.
E.g., if yx = 3, y varies invereely as x.
376 ELEMENTS OF ALGEBRA
395. Joint variation. One variable is said to vary as two
others jointly when it varies as the product of the two.
That is, / varies as x and z jointly^ lohen y (jz {xz), or
y = m(xz).
E.g., it W= the amount of work done by 2lf men in Z> days ; then,
if 31 and D both vary, WazM x D ;
if M is constant, TT x D ;
if D is constant, W^M.
One variable is said to vary directly as a second, and in-
versely as a third, when it varies as the product of the
second into the reciprocal of the third.
That is, y varies directly as x, and inversely as z, when
yccx(l/z), or y = mx(l/z).
E.g., if yz = Sx, y = 3x(l z); hence yccx(\/z').
396. In each of the preceding cases of variation, the
value of the constant, m, can be found when any set of
corresponding values of the variables is known.
Ex. 1. Given y<xx, and y = Q when x = 2; find the constant ratio
of y to X.
Since y x.x, y = mx, where m is some constant. (1)
Since y = 6, when oj = 2, from (1) we have
6 = 2 m, orm = 3 = y/x.
Ex. 2. The volume, V, of a pyramid varies jointly as its height,
H, and the area of its base, B. When the area of the base is 60
square feet and the height 14 feet, the volume is 280 cubic feet. Find
the area of the base of a pyramid whose volume is 390 cubic feet, and
whose height is 26 feet.
Since VccBH, V = niBH, where m is some constant. (1)
Substituting the given values of V, B, II, in (1), we have
280 = m X 60 X 14, or m = ^.
THEORY OF LIMITS 877
Hence, when V= 390, and A = 26, we have
390 = ^ 5 X 26.
/. B = 45, the number of sq. ft. in base.
397. The simplest way to treat variations is to convert
them into equations.
Ex. If M Qc y and yccx, prove that uccx.
Since uccy and yxx, by § 391 we have
V = ay and y = 6x, where a and 6 are some constants.
.'. u = abx. .'. uccx. § 390
398. If u ccx when y is constant j and u ccy when x is conr
stantj then u ccxy when x and/ both vary.
Let x', y\ vi! be one set and a;", 2/", w" another set of cor-
responding values of a;, y^ z, when all change together.
Let X change from a;' to a;", y remaining constant, and
suppose that in consequence u changes from m' to u^ ; then
since ucnx when y is constant, by § 392 we have
v) : a;' = u^ : a;". (1)
Now let y change from 1/' to y", x remaining constant;
then u will change from u^ to w"; hence as Mocy when x
is constant, we have
u^'.y^ = v!^'.y\ (2)
Multiplying (1) by (2), and dividing the antecedents by
w,, we have
v) : x'y' = u" : x'Y.
Hence uccxy. § 392
Similarly it may be proved that, if u varies as each one
of the three variables «, y, z when the other two are con-
stant, then u oc xyz when they all change ; and so on.
E.g., let A denote the area of a triangle, B Its base, and H its
altitude ; then
AccB, when H is constant,
and AccH, when B is constant ;
hence A oc BH, when B and H both change.
378 ELEMENTS OF ALGEBRA
Exercise 131.
1 . If ic varies directly as y, and ?/ = 7 when ic = 18 ;
find X when y = 21.
2. If 2/ varies inversely as x, and y = 4: when a; =15;
find 2/ when a; = 12.
3. If aj varies jointly as y and 2;, and x = 6 when 2/ = 2
and z = 2', find ic when ?y = 5, z = 7.
4. li x^ccy and 2;^ x 2/, then 0^2; oc y.
5. If icocl^2/j ^^i<i y = ^ when a; = 15; find y when
x = 6.
6. If a; varies directly as y and inversely as z, and
a; = 10 when y = 15 and z = 6; find a; when y = S, z = 2.
7. If a; varies directly as y and inversely as z, and
a;=14 when 2/=!^ ^.nd 2; = 14; find 2; when a; =49, y=4:5.
8. If ajocl-i-2/? and yccl-i-z, prove 2; oca;.
9. If 3a5 H-72/<x3 a? + 13 2/, ^^^ y=^ when x = 5',
find the equation between x and y,
10. If the cube of x varies as the square of y, and if
a; = 3 when y = 5', find the equation between x and y.
11. If the area of a circle varies as the square of its
radius, and if the area of a circle is 154 square feet when
the radius is 7 feet ; find the area of a circle whose radius
is 10 feet 6 inches.
12. The velocity of a falling body varies directly as the
time during which it has fallen from rest, and the velocity
at the end of 2 seconds is 64. Find the velocity at the end
of 5 seconds.
13. The volume of a sphere varies directly as the cube
of its radius, and the volume of a sphere whose radius is
1 foot is 4.188 cubic feet. Find the volume of a sphere
whose radius is 3 feet.
THEORY OF LIMITS 379
14. The pressure of a gas varies jointly as its density
and its absolute temperature; also when the density is 1
and the temperature 300, the pressure is 15. Find the
pressure when the density is 3 and the temperature is 320.
15. The volume of gas varies directly as the absolute
temperature and inversely as the pressure. Also when the
pressure is 15 and the temperature 280, the volume is 1
cubic foot. Find the volume when the pressure is 20 and
the temperature 300.
16. The distance through which a heavy body will fall
from rest varies directly as the square of the time, and a
body will fall through 144 feet in 3 seconds. Find how far
it will fall in 2 seconds.
17. The pressure of wind on a plane surface varies
jointly as the area of the surface and the square of the
wind's velocity. The pressure on a square foot is 1 pound
when the wind is moving at the rate of 15 miles per hour.
Find the velocity of the wind when the pressure on a
square yard is 16 pounds.
18. The volume of a right circular cone varies jointly as
its height and the square of the radius of its base ; and the
volume of a cone 7 feet high with a base whose radius is
3 feet is 66 cubic feet. Find the volume of a cone 9 feet
high with a base whose radius is 14 feet.
CHAPTER XXIX
THE PROGRESSIONS
399. A series is a succession of terms whose values are
determined by some one law.
A series is said to be finite or infinite according as the
number of its terms is limited or unlimited.
In this chapter we shall consider only the three forms of
series which are called the arithmetic, the geometric, and the
harmonic progressions.
ARITHMETIC PROGRESSIONS.
400. An arithmetic progression (A. P.) is a series in which
the difference between any term (after the first) and the pre-
ceding term is the same throughout the series.
The difference, which can be either positive or negative,
is called the common difference.
E.g., the series
2, 6, 8, 11, 14, 17, 20, 23, ..., (1)
and 7, 5, 3, 1, -1, -3, -5, -7, .-., (2)
are arithmetic progressions.
In series (1) the common difference is 3, and in (2) it is — 2.
If in (2) we add — 2 to any term, we obtain the next term.
401. The nth term. Let d denote the common difference
in an A. P., and a the first term ; then, by definition,
the second terln = a -\- d,
the third term =a + 2d,
and the nth term = a + (w — l)d (1)
380
THE PROGRESSIONS 381
E.g., a the first term of an A. P. is 4, and the common difference
is 5,
the ninth term = 4 + ( 9 - 1)5 = 44,
and the twenty-first term = 4 + (21 - 1)5 = 104.
Ex. The fourth and fifty -fourth terms of an A. P. are, respectively,
64 and — 61. Find the twenty-seventh term.
I (a
Here 64 = the fourth term = a -\- Sd<,
and - 61 = the fifty-fourth term = a + 53 d.
Solving system (a), we find a = 71J, d = — 5/2.
.*. the twenty-seventh term = a -\- 2Qd = 6 J.
402. When three numbers, «, b, c, are in A. P., the middle
term b is called the arithmetic mean of the other two terms
a and c.
403. If a, 6, c are in A. P., by definition we have
b — a = c — b.
.-. b = (a-\-c)/2.
That is, the arithmetic mean of any two numbers is half
their smn.
404. All the terms between any two terms of an A. P.
may be called the arithmetic means of the two terms.
The following example illustrates how any number of
arithmetic means can be inserted between any two numbers.
Ex. Insert 9 arithmetic means between 50 and 80.
Since there are 9 arithmetic means, 80 must be the eleventh term.
60 being the first ; hence, by definition, we have
the eleventh term = 50 + 10 cZ = 80.
Hence d = 3, and the required series is
60, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80.
382 ELEMENTS OF ALGEBRA
405. Sum of n terms. Let I denote the nth term, and S
the sum of n terms of an A. P. ; then
^ = a + (a + d) + (a + 2 fZ) H i.(i-d)-\-l,
or ^ = / 4- (Z - a^) + (Z _ 2 d) + ... -f (a + (^) + a.
Adding the corresponding terms, we have
2 S = (a + I) -\- (a -^ I) -{- (a -\- 1) -{- '" to 71 terms.
.: S = in(a + l). (1)
From § 401, i = a-\- {n -l)d. (2)
.-. >S = inl2a-{-(n-l)dl. (3)
If any three of the five numbers a, d, I, 71, S are given,
the other two can be found from equations (1) and (2), or
from (3) and (2).
Ex. 1. Find the sum of 20 terras of the A. P.
-5-1 + 8 + 7+ 11 + -.
Here a = - 5, d = 4, n = 20.
.'. S=^n{;2a+(n- l)d}
= 10{-10 + 19 X 4}
= 660.
Ex. 2. Find the sum of the first n consecutive odd numbers,
1, 3, 5 ....
Here a = 1, d = 2, n = n.
.-. S=^n{2a-\-{n-l)d}
= iw{2+(>i-l)2}
Hence the sum of n consecutive odd numbers, beginning with 1,
is w2.
Ex. 3. The first term of an A. P. is 6, and the sum of 25 terms is
25. Find the common difference.
THE PROGRESSIONS 383
Here a = 6, /S' = 25, w = 25 ; hence from (3) of § 405 we have
25 = ^ X 25(12 + 24 d}
.-. d = -6/12.
Ex. 4. How many terms must be taken of the series 11, 12, 13,
... to make 410?
Here a=ll, (Z = l, S = 410 ; hence from (3) of § 405 we have
410 = in{22+(n-l)}. (1)
.-. w = 20, or -41.
Since the number of terms must be an arithmetic whole number,
the number of terms is 20. See § 297.
Ex. 5. How many terms must be taken of the series — 16, — 15,
-14, ... to make - 100?
Here a = — 16, d = 1, S = — 100 ; hence we have
-100 = in{-32+(n-l)}.
.-. n = 8, or 25.
Hence the number of terms is 8 or 25.
The sum of the 17 terms following the first 8 must therefore be
zero. These 17 terms are — 8, — 7, — •••, 7, 8, and their sum is
evidently zero.
Exercise 132.
1. Find the twenty-seventh and forty -first terras in the
series 5, 11, 17, •••.
2. Find the seventeenth and fifty-fourth terms in the
series 10, 11^, 13, •••.
3. Find the twentieth and thirteenth terms in the series
_.S _2 —1 ...
4. If the twelfth term of an A. P. is 15, and the twen-
tieth term is 25, what is the common difference ?
5. The seventh term of an A. P. is 5, and the twelfth
term is 30. Find the common difference.
6. The first term of an A. P. is 7, and its third term is
13. Find the tenth term.
384 ELEMENTS OF ALGEBRA
7. The first term of an A. P. is 20, and its sixth term is
10. Find the twelfth term.
8. The seventh term of an A. P. is 5, and the fifth term
is 7. Find the twelfth term.
9. Which term of the series 5, 8, 11, ••• is 65 ?
10. Which term of the series J, f, f , ♦•• is 18 ?
11. Insert 6 arithmetical means between 8 and 29.
12. Insert 7 arithmetical means between 269 and 295.
13. Insert 15 arithmetical means between 67 and 43.
14. If a, b, c, d are in A. P., prove that a -\- d = b -^ c.
15. The sum of the second and fifth terms of an A. P. is
32, and the sum of the third and eighth is 48. Find the
first term.
16. The sum of the third and fourth terms of an A. P.
is 187, and the sum of the seventh and eighth terms is 147.
Find the second term.
Find the sum of each of the following series :
17. 5, 9, 13, •.. to 19 terms.
18. 1, 21,31 ... to 12 terms.
19. — 5, — 1, 3, ••• to 20 terms.
20. i, I, i, ... to 7 terms.
21. 10, -%S ^8-, ... to 7 terms.
22. I, 1, |, ... to 15 terms.
How many terms must be taken of :
23. The series 42, 39, 36, ... to make 315?
24. The series 15, 12, 9, ... to make 45 ?
25. The series — 8, — 7, - 6, ... to make 42 ?
TUE PROGRESSIONS 385
26. Find the sum of all the numbers between 100 and
500 which are divisible by 3.
27. Find the sum of all the odd numbers between 100
and 200.
28. The sum of 10 terms of an arithmetical series is 145,
and the sum of its fourth and ninth terms is 5 times the
third term. Determine the series.
29. Divide 80 into 4 parts which are in A. P., and which
are such that the product of the first and fourth parts is |
of the product of the second and third.
30. Find 4 numbers in A. P., such that the sum of their
squares shall be 120, and that the product of the first and
last shall be less than the product of the other two by 8.
31. If a body falling to the earth descends a feet the
first second, 3 a the second, 5 a the third, and so on ; (1) how
far will it fall during the tth. second ? (2) how far will it
fall in t seconds ? Ans. (2t — l) a, at-.
32. How many strokes does a common clock make in 12
hours ?
33. A debt can be discharged in a year by paying $ 1 the
first week, $ 3 the second week, $ 5 the third, and so on.
Find the last payment and the amount of the debt.
34. One hundred apples are placed on the ground at the
distance of a yard from one another. How far will a person
travel, who shall bring them, one by one, to a basket, placed
at a distance of a yard from the first apple ?
35. Two boys A and B set out at the same time, to meet
each other, from two places 343 miles apart, their daily
journeys being in A. P. ; A's common difference being an
increase of two miles, and B's a decrease of 5 miles. On
the day at the end of which they met, each travelled exactly
20 miles. Find the duration of each journey.
386 ELEMENTS OF ALGEBRA
GEOMETRIC PROGRESSIONS.
406. A geometric progression (G. P.) is a series in which
the ratio of any term (after the first) to the preceding term
is the same throughout the series.
This ratio, which can be either positive or negative, is
called the common ratio.
•-g-,
, the series
2,
6,
18,
54,
162,
8,
4,
2,
1,
h
1,
-1,
h
-i
^,
or
is a geometric progression (G. P.). In the first series, the common
ratio is 3 ; in the second series it is 1/2 ; and in the last it is — 2/3.
If we multiply any term in either series by the common ratio, the
product will be the next term of that series.
407. The nth. term. Let r denote the common ratio, and
a the first term of any G. P. ; then by definition
the second term = ar^
the third term = ar^,
and the nth. term = ar'*"^ (1)
E.g.^ if the first term of a G. P. is 8, and the common ratio is 1/2,
the fifth term = 8 x (l/2)5-i = 1/2,
and the ninth term = 8 x (l/2)9-i = 1/32.
Ex. The sixth term of a G. P. is 156, and the eighth term is 7644.
Find the seventh term.
Here 156 = the sixth term = ar^, (1)
and 7644 = eighth term = ar^. (2)
Divide (2) by (1), 49 = r2. (3)
.-. r = ± 7. (4)
But the seventh term = sixth term x r
= 156 (±7; = ±1092.
THE PROGRESSIONS 387
408. Sum of n terms. Let S denote the sum of n terms ;
then
S = a-{- ar + a?*^ + ••• -|- ar**-^ + «?•"-*
= a (1 -\- r -{- 7^ + r^ ■] h ^'""^ + r**-^)
1 — r"
= a-
1 — 7*
_a(l — r")
§ 129
Hence S = ^^^^^ ^. (1)
1 — r
Let Z denote the ?ith term ; then from § 407
I = ar'^-K (2)
From (1) and (2)
c, a — rl
(3)
If any three of the five numbers a, I, n, r, s, are known,
the other two may be found from equations (1) and (2),
or from (2) and (3).
Ex. Sum the series 6, — 18, 64, ••• to 6 terms.
Here a = C, r = - 18 -=- G = - 3, n = 6.
= f{l-36}
= - 1092.
409. When three numbers, a, b, c, are in G. P., the middle
term b is called the geometric mean of the other two terms
a and c.
410. If a, 6, c are in G. P., by § 406, we have
c: b = b: a. .-. 6 = Vac.
That is, the geometric mean of any two numbers is the mean
proportional between them.
388 ELEMENTS OF ALGEBBA
411. All the terms between any two terms of a G. P. may
be called the geometric means of the two terms.
412. To insert m geometric means between a and b.
Calling a the first term, h will be the (m + 2) th term ;
hence by (1) of § 407, we have
h = ar^+\
•»+!/
r= ^/bTa. (1)
Hence the m means required are ar, ar^, ••• ar"^, in which
r has the value given in (1).
Ex. Insert 6 geometric means between 56 and — 7/16.
Here a = 56, 6 = - 7/16, and w + 1 = 7.
^ IQ > 2* X 23 2
Hence r = — 1/2, and the 6 means required are
- 28, 14, - 7, 7/2, - 7/4, 7/8.
Exercise 133.
Find the last term in the following series :
1. 2, 4, 8, ••• to 9 terms. 2. 2, 3, 4^, ••• to 6 terms.
3. 3, -3'^, 3^ ••• to 2 7i terms.
4. X, 1, 1/x, ••• to 30 terms.
5. The first term of a G. P. is 3, and the third term is 4.
Find the fifth term.
6. The third term of a G. P. is 1, and the sixth term
is — 1/8. Find the tenth term.
7. The fourth term of a G. P. is 0.016, and the seventh
term is 0.000128. Find the first term.
8. The fourth term of a G. P. is 1/18, and the seventh
term is — 1/486. Find the sixth term.
9. Insert 3 geometric means between 486 and 6.
THE PROGBESSIONS 389
10. Insert 4 geometric means between 1/8 and 128.
11. Insert 5 geometric means between 3 and 0.000192.
12. Insert 4 geometric means between a^b~^ and a~^b^.
Find the sum of the following series :
13. 64, 32, 16, ••• to 10 terms.
14. 8.1,2.7,0.9, ..-to? terms.
15. 3, - 1, 1/3, ••• to 6 terms.
16. 1/2, 1/3, 2/9, ... to 7 terms.
17. - 2/5, 1/2, - 5/8, ... to 6 terms.
18. 2, - 4, 8, ... to 2p terms.
413. When r < 1 arithmetically, the successive terms of
a G. P. become smaller and smaller arithmetically, and the
G. P. is said to be a decreasing progression.
414. TJie limit of the sum of an infinite number of terms of
a decreasing G. P. is
Proof Prom (1) of § 412, we have
1-r 1
(1)
Now if r < 1 arithmetically, and the number of terms,
or n, is increased without limit, then
1 — r
Hence from (1), by § 364, we obtain
The limit of the sum of an infinite number of terms of a
series is often called the sum of the series.
390 ELEMENTS OF ALGEBRA
E.g., if a = 2 and r = 1/2, we have the decreasing G. P.,
5, 1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128. .... (1)
The sum of an infinite number of terms of this series approaches 4
as its limit. For suppose that we bisect a line four inches long, and
take away one of the parts ; then bisect the remainder, and take away-
one of the parts ; and continue this process without limit. It is evi-
dent that the part remaining will approach zero as its limit, and the
sum of the successive parts taken away will approach four inches as
its limit. But the numbers of inches in the successive parts taken
away will be the terms of series (1). Hence the sum of an infinite
number of terms of that series approaches 4 as its limit.
Ex. 1. Eind the sum of the series 1, 1/2, 1/4, •••.
Here a = 1, r = 1/2.
From (2), it (6') = ^— ^ = 2.
Ex. 2. Find the sum of the series 9, —3, 1, •••.
Here a = 9, r = - 1/3.
From (2), ,t (^) = j_^ = f =6i.
Ex. 3. Express 0.423 as a common fraction.
0.423 = 0.4232323 ...= — + ^ + H + .".
10 103 105
23
103
H-
"l02J
23
103
«s^
23
990
23 , 23 , 23 ,
^^^' 10-3 + 10-5 + 10^ +
.-. 0.423 = ^ + ^'^ = IM-
Ex. 4. Find the infinite G. P. whose sum is 18, and whose second
term is — 8.
Here ar = -8, (1)
and -^ = 18. (2)
1 — r
Divide (1) by (2), r(l - r) = - 4/9.
... ,.2 _ y _ 4/9 = 0.
.'. r = - 1/3 or 4/3.
THE PROGRESSIONS 391
Only the value — 1/3 is admissible for r, since the series is a
decreasing one.
From(l), a = -8 -(-1/3) = 24.
Hence the series is 24, -8, 8/3, -8/9, ....
Exercise 134.
Find the sum of each of the following series :
1. 9, 6, 4, .... 4. f, -1, I, ....
2. 1 -J, I, .... 5. 0.9, 0.03, 0.001, ....
3. h h yV •••• 6. 0.8, -0.4, 0.2, ....
Express as a common fraction :
7. 0.3. 8. 0.16. 9. 0.24. 10. 0.378. 11. 0.037.
12. Find the infinite G. P. whose sum is 4, and whose
second term is f.
13. Find the infinite G. P. whose sum is 9, and whose
second term is — 4.
14. If every alternate term of a G. P. is taken away, the
remaining terms will be in G. P.
15. If all the terms of a G. P. are multiplied by the same
number, the products will be in G. P.
16. Show that the reciprocals of the terms of a G. P. are
in G. P.
17. By saving 1 cent the first day, 2 cents the second
day, 4 cents the third day, and so on, doubling the amount
every day, how much would be saved in a month of 30
days?
18. Suppose a body to move eternally as follows : 20 feet
the first minute, 19 feet the second minute, 18 J^ feet the
third minute, and so on. Find the limit of the distance
passed over.
392 ELEMENTS OF ALGEBRA
19. A ball falling from the height of 100 feet rebounds
one-fourth the distance, then falling, it rebounds one-fourth
the distance, and so on. Find the distance passed through
by the ball before it comes to rest.
20. If in problem 31 of exercise 132, a = 16 ^J^"? ^^^ ^^ng
will it be before the ball in problem 19 comes to rest ?
To fall 100 feet, it takes Vl00TT6^, or 10\/J^, seconds; to
rebound, or to fall, 25 feet, it takes •\/25 -^ 10^^, or 5\/J^2_^ seconds ;
to rebound, or to fall, Q\ feet, it takes \/6|~n(3^, or | V^^, seconds ;
and so on.
Hence the time = 10 vTS + 2(5 VJS + f ^^ + -)
= 30 V^ = 1%% V579 = 7.4805 +.
HAEMONIC PROGRESSIONS.
415. An harmonic progression is a series of numbers whose
reciprocals form an A. P.
E.g.^ the series
1, h h 7' •••> or 4, -4,
is an harmonic progression ; for the reciprocals of their terms
1, 3, 5, 7,...., or \, -\, -f, ...
are in A. P.
416. When three numbers are in harmonic progression
(H. P.), the middle term is called the harmonic mean of the
other two.
417. Let H be the harmonic mean of a and h ; then by
§415,
-' 77' 7 ^^® ^^ ^- ^•
a H h
" H a b h'
2 1,1 rj 2 ah
= --{--, or H
Hah a + 6
THE PROGRESSIONS 393
418. If A and G denote respectively the arithmetic and
the geometric mean of a and 6, then (§§ 403, 409)
2ab
2 a + 6
r.AxH = '^±^X^^==ab = 0\
2 a + b
Hence A:G=G:H.
That is, the geometric mean of any two numbers is also the
geometric mean of their arithmetic and harmonic means.
419. Problems in H. P. are generally solved by inverting
the terms, and making use of the properties of the resulting
A. P.
Ex. The fifteenth term of an H. P. is 1/25, and the twenty-third
term is 1/41. Find the series.
Let a be the first term, and d the common difEerence of the corre-
sponding A. P. ; then
26 = the fifteenth term = a 4- 14 d,
and 41 = the twenty-third term = a + 22 d.
.-. (Z = 2, a = - 3.
Hence the A. P. is — 3, — 1, 1, 3, 5, •••,
and the H. P. is - \, -1, 1, \, \, ....
Exercise 135.
1. Find the sixth term of the series 4, 2, IJ, ..•.
2. Find the eighth term of the series IJ, 1||, 2^, •••.
Find the series in which
3. The second term is 2, and the thirty-first term is ^\.
4. The thirty-ninth term is ^, and the fifty-fourth term
394 ELEMENTS OF ALGEBRA
Find the harmonic mean between
5. 2 and 4. 6. 1 and 13. 7. { and J^.
8. Insert 2 harmonic means between 4 and 12.
9. Insert 3 harmonic means between 2|- and 12.
10. Insert 4 harmonic means between 1 and 6.
11. If a, b, c are in harmonic progression, prove that
a — b:b — c = a:c.
CHAPTER XXX
PERMUTATIONS AND COMBINATIONS
420. Fundamental principle. If one thing can he done in m
ivays, and (after it has been done in any one of these ways)
a second thing can he done in n ways; then the two things can
he done in m x n ways.
Ex. 1. If there are 11 steamers plying between New York and
Havana, in how many ways can a man go from New York to Havana
and return by a different steamer ?
He can make the first passage in 11 ways, with each of which he
has the choice of 10 ways of returning ; hence he can make the two
jourueys in 11 x 10, or 110, ways.
Ex. 2. In how many ways can 3 prizes be given to a class of 10
boys, without giving more than one to the same boy ?
The first prize can be given in 10 ways, with each of which the
second prize can be given in 9 ways ; hence the first two prizes can be
given in 10 X 9 ways. With each of these ways of giving the first two
prizes, the third prize can be given in 8 ways ; hence the three prizes
can be given in 10 x 9 x 8, or 720, ways.
Proof. After the first thing has been done in any one of
the m ways, the second thing can be done in n different
ways ; hence there are n ways of doing the two things for
each of the m ways of doing the first ; therefore in all there
are mn ways of doing the two things.
This principle is readily extended to the case in which
there are three or more things, each of which can be done in
a given number of ways.
421. The different ways in which r things can be taken
from n things, the order of selection or arrangement being
395
396 ELEMENTS OF ALGEBRA
considered, are called the permutations of the n things taken
r at a time.
Thus, two permutations will be different unless they con-
tain the same things arranged in the same order.
E.g.^ of the four letters a, 6, c, d, taken one at a time, we have the
four permutations
a, &, c, d.
Of these four letters taken two at a time, we have the twelve permu-
tations
a&, ac^ ad, ha, be, bd, ca, cb, cd, da, db, dc.
If after each of these permutations we place in turn each of the
letters which it does not contain, we shall obtain 24 permutations of
the four letters taken three at a time.
The number of permutations of n different things taken r
at a time is denoted by the symbol T^. Thus ^P^, ^P^, ^P^
denote respectively the numbers of permutations of 9 things
taken 2, 3, 4 at a time.
422. To find the number of permutations of n dissimilar
things taken r at a time.
The number required is the same as the number of ways
of filling r places with n different things.
The first place can be filled by any one of the n things,
and after this has been filled in* any one of these n ways,
the second place can be filled in {ii — 1) ways ; hence with
n things two places can be filled in n(n — 1) ways ; that is,
-P, = n(n~-1). (1)
After the first two places have been filled in any one of
these n (n — 1) ways, the third place can be filled in (n — 2)
ways ; hence three places can be filled in n(n — l)(n — 2)
ways ; that is,
-P, = n(n-l)(n-2). (2)
PERMUTATIONS AND COMBINATIONS 397
For like reason, we have
^P, = n(n - l)(7i -2)(n- 3); (3)
and so on.
From (1), (2), (3), •••, we see that in "P^ there are r fac-
tors, of which the rth is n — r -\-l; hence
"P, = n(n-l) (n _ 2) ... (?i - r -f 1). (A)
If all the n things are to be taken at a time, r = n, and
(A) becomes
«P„ = w(n-l)(n-2)...3.2.1. (B)
423. The continued product ri (?i — 1) (n — 2) . . . 3 . 2 . 1 is
denoted by the symbol [n, or w!, either of which is read
* factorial ri/
Thus [4 = 4.3.2.1; [9 = 9. 8. 7- 6. 5- [4.
With this notation (B) in § 422 can be written
""Pn^ln. (B)
That is, the number of permutations of n different things
taken all at a time is factorial n.
Ex. 1. In how many different ways can 7 boys stand in a row?
The number =7P7 = 7.6.5-4.3.2.1= 5040. by (B)
Ex. 2. How many different numbers can be formed with the figures
1, 2, 3, 4, 5, 6, taken four at a time ?
The number required = ep^ = 6 . 6 • 4 • 3 = 360. by (A)
424, If N denote the number of jyermutations of n things
taken all at a time, of which r things are alike, s others alike,
and i others alike; then
Proof Suppose that in any one of the N permutations
the r like things were replaced by r dissimilar things ; then,
from this single permutation, without changing in it the
398 ELEMENTS OF ALGEBRA
position of any one of the other n — r things, we could form
|_r new permutations. Hence from the N original permuta-
tions we could obtain N\r permutations, in each of which
s things would be alike and t others alike.
Similarly, if the s like things were replaced by s dissimilar
things, the number of permutations would be N\r\s, each
having t things alike.
Finally, if the t like things were replaced by t dissimilar
things, we should obtain iV^[r|s|^ permutations, in which all
the things would be dissimilar.
But the number of permutations of n dissimilar things
taken all at a time is [ri.
Hence iV|r[s[^ = [n.
\n
Therefore N =
r sit
Ex. 1. How many different numbers can be formed by the figures
2, 2, 3, 4, 4, 4, 5, 5, 5, 5 ?
110
The number = ,-^==— = 12600.
[2 [3 [4
Exercise 136.
1. A cabinet maker has 12 patterns of chairs and 7 pat-
terns of tables. In how many ways can he make a chair
and a table ? Ans. 84.
2. There are 9 candidates for a classical, 8 for a mathe-
matical, and 5 for a natural science scholarship. In how
many ways can the scholarships be awarded ?
3. In how many ways can 2 prizes be awarded to a class
of 10 boys, if both prizes may be given to the same boy ?
4. Find the number of permutations of the letters in
the word numbers. How many of these begin with n and
end with s ?
PERMUTATIONS AND COMBINATIONS 399
5. If no digit occur more than once in the same number,
how many different numbers can be represented by the 9
digits, taken 2 at a time ? 3 at a time ? 4 at a time ?
6. How many changes can be rung with 5 bells out of
8 ? How many with the whole peal ?
7. How many changes can be rung with 6 bells, the
same bell always being last ?
8. In how many ways can 15 books be arranged on a
shelf, the places of 2 being fixed ?
9. Given "P4 = 12 • "Pg ; find n.
10. Given n : "Pg : : 1 : 20 ; find n.
11. Given "Pg : "+2p^ : : 5 : 12 ; find w.
12. How many different arrangements can be made of
the letters of the word commencement ?
Of the 12 letters, 2 are c's, 3 are m's, 3 are e's, and 2 are n's ;
112
.-. JVr=. ^— =3326400.
[2 [3 [3 [2
13. Find the number of permutations of the letters of the
words mammalia, caravansera, 3fississi'p2n.
14. In how many ways can 17 balls be arranged, if 7 of
them are black, 6 red, and 4 white ?
Prove each of the following relations :
15. n(n — l)(n — 2)"'(n — r -\- l)\n — r = \n.
16. 9-8.7.6/[3 = |9/([3|5).
17. n(n — l)(n-2)"'(n — r-\- l)/\r = \n/(\r\n — r).
18. [5[5(6/5)=[6|4; . •. [5 [5 < ^6 [4.
19. \a\a(a-^T)-i-a = \a-\-l\a — l}, .: |ft|a<|a + l|a— 1.
20. |a|a<|a+l|a-l<|a-l-2|a-2<|a + 3|a-3<..».
21. |18 — x\x is least when x = 9.
400 ELEMENTS OF ALGEBRA
425. The different ways in which r things can be taken
from n things, without regard to the order of selection or
arrangement, are called the combinations of the n things taken
r at a time.
Thus, two combinations will be different unless they both
contain precisely the same things.
E.g.^ of the four letters a, 6, c, (Z, taken two at a time, there are
the six combinations
a6, ac^ ad, be, bd, cd.
Taken three at a time, there are the four combinations
abc, abd, acd, bed.
Taken four at a time, there is one combination only.
The number of combinations of n things taken r at a time
is denoted by the symbol ""C^.
426. To fiyid the number of combinations of n different
things taken r at a time.
Every combination of r different things has |_r permuta-
tions ; hence, ""Cr \r_ will denote "P^; that is,
^Cr\r_=''Pr
= n(n — l){n — 2) '" (n — r + 1).
Hence „o^^«(n - l)(n- g-(.- r + 1)_ ^^^
In applying this formula, it is useful to note that the
suffix r in the symbol "(7^ denotes the number of the factors
in both the numerator and denominator of the formula.
Ex. How many groups of 4 boys are there in a class of 17 ?
The number = i7(74 = iLJ^jiidi = 2380.
4.3.2.1
PERMUTATIONS AND COMBINATIONS 401
427. In (C) of § 426, multiplying the numerator and
denominator of the fraction by | n — r, we obtain
"Gr=r-T^-- (D)
„(„_l)(„_2)...(n-r + l)l!Lz
- r
\r\n — r
fifi — L_
^-|r|7i-r
)
lituting n — r for r in (D), we obtain
-0 - ^ .
^-'-|»-r|r
(1)
From (D) and (1), "O, = "a_,. (E)
The relation in (E) follows also from the consideration that for
each group of r things which is selected, there is left a corresponding
group of n — r things.
The relation in (E) often enables us to abridge computation.
E.g., i«Ci3 = "Ca = iL2<Ji = 105.
428. Value of r which renders "Cr greatest.
"Cv, or |n/(|2] |w — ?•)? is greatest when [r \n — r is least.
[a [a (a + 1) -7- a = [a + 1 |a — 1, etc. ;
.-. \a\a< [g + l |a — 1 < |a + 2 |a — 2 < .-•.
Hence, ic/ien n is even, \r \n — r is least, and therefore **Cr
is greatest, when r = n — r, or r = n/2.
Again [6 |6 + 1 = \b-\-l 1 6,
and |6 + l|6<|6 + 2|6-l<|6 + 3|6-2<...
Hence when n is odd, \r \n — r is least, and therefore **C,
is greatest, when r = n — r ± 1, or r = (n ± l)/2.
i/.gf., [r |18 — r is least and ^^Cr is greatest when r = 9.
Again |r [16 — r is least and ^^Cr is greatest when r = 7 or 8.
402 ELEMENTS OF ALGEBRA
Exercise 137.
1. How many combinations can be made of 9 things
,taken 4 at a time ? taken 6 at a time ? taken 7 at a time ?
The last number = ^C7=^C2 = 36.
2. How many combinations can be made of 11 things
taken 4 at a time? taken 7 at a time?
3. Out of 10 persons 4 are to be chosen by lot. In how
many ways can this be done ? In all the ways, how often
would any one person be chosen?
4. From 14 books in how many ways can a selection of
5 be made, when one specified book is always included?
when one specified book is always excluded ?
5. On how many days might a person having 15 friends
invite a different party of 10 ? of 12 ?
n
6. Given ^C.2 = 15, to find n.
7. Given «+i(74 = 9 x "Cg, to find n.
8. In a certain district there are 4 representatives to be
elected, and there are 7 candidates. How many different
tickets can be made up?
9. Of 8 chemical elements that will unite one with
another, how many ternary compounds can be formed ?
How many binary ?
10. There are 15 points in a plane, no 3 of which lie in
the same straight line. Find how many straight lines there
are, each containing 2 of the points.
11. In a town council there are 25 councillors and 10
aldermen ; how many committees can be formed, each con-
sisting of 5 councillors and 3 aldermen ?
12. Find the sum of the products of the numbers 1, 3, 5,
2, taken 2 at a time ; taken 3 at a time.
13. Find the number of combinations of 55 things taken
50 at a time.
PERMUTATIONS AND COMBINATIONS 403
14. If 2«(73:'*(72 = 44:3; find n.
15. If ^Cu = "Cg; find n ; find "d, ; find ""C^.
16. In a library there are 20 Latin and 6 Greek books ;
in how many ways can a group of 5 consisting of 3 Latin
and 2 Greek books be placed on a shelf ?
17. From 3 capitals, 5 other consonants, and 4 other
vowels, how many permutations can be made, each begin-
ning with a capital and containing in addition 3 consonants
and 2 vowels ?
18. If i«a = ''Cr+2 ; find r ; find ^C,.
19. From 7 Englishmen and 4 Americans a committee of
6 is to be formed; in how many ways can this be done
when the committee contains, (1) exactly 2 Americans,
(2) at least 2 Americans ?
20. Of 7 consonants and 4 vowels, how many permutations
can be made, each containing 3 consonants and 2 vowels ?
21. When repetitions are allowed, "P^ = 7i*", and "P„ = n".
When repetitions are allowed after the first place has been filled in
any one of n ways, the second place can be filled in n ways ; hence
"Pa = n2, etc.
22. In how many ways can 4 prizes be awarded to 10
boys, each boy being eligible for all the prizes ?
23. There are 25 points in space, no 4 of which lie in the
same plane. Find how many planes there are, each con-
taining 3 of the points.
24. For what value of r is [r \ 18 — r least ? [r | 21 — r ?
|r|45-r?
25. For what value of r is ^^C, greatest? "C,? ^^C,?
CHAPTER XXXI
BINOMIAL THEOREM
429. In § 126 the laws of exponents and coefficients of
the binomial theorem were proved for positive integral
exponents up to 7. These laws hold for all exponents,
integral or fractional, positive or negative.
In this chapter we shall prove these laws for any positive
integral exponent, and apply them to all exponents.
430. From the distributive law for multiplication, it fol-
lows that if we take one term from each of any number of
binomials and multiply these terms together, we shall
obtain a term of the continued product of these binomials ;
and if we do this in every possible way, we shall obtain all
the terms of the continued product of these binomials.
E.g., if we take a letter from each of the three binomials,
(a + 6)(a + &)(a + 6),
and multiply the three letters together, we shall obtain a term of the
continued product ; and if we do this in every possible way, we shall
obtain all the terms of this product.
We can take the a's from the three binomials, and we can do this
in one, and only one, way ; hence a^ is a term of the product.
We can take the b from one binomial and the a's from the other
two, and we can do this in three ways ; for the b can be taken from
any one of the three binomials ; hence 3 a^b is a term of the product.
We can take the 6's from two binomials and a from the third, and
we can do this in three ways ; hence 3 ab^ is a term of the product.
Finally, we can take the &'s from the three binomials in one, and
only one, way ; hence h^ is a term of the product.
Hence (a + 6) (a + b) (a -f ?>) = «» + 3 a'^b + 3 ab^ + &».
404
BINOMIAL THEOREM 405
431. Binomial theorem. Suppose we have
(a -\- b) {a -\- b) (a -\- b) ' • ' to n factors. (1)
If we take a letter from each of the ?i binomials, and
multiply these letters together, we shall obtain a term of
the continued product ; and if we do this in every possible
way, we shall obtain all the terms of this product.
We can take the a's from all the binomials in one, and
only one, way ; hence a'* is one term of the product.
We can take b from one binomial and the a's from the
remaining (n — 1) binomials, and we can do this in as many
ways as one b can be taken from the n binomials, i.e., n,
or "Ci, ways ; hence "C\ • a"~^b is a term of the product.
Again, we can take the 6's from two binomials, and the
a's from the remaining (n — 2) binomials, and we can do this
in as many ways as two 6's can be taken from the n binomials,
i.e., "(72 ways ; hence "C2 • a"~-6^ is a term of the product.
And, in general, we can take the 6's from r binomials
(where r is any positive integer not greater than w), and the
a's from the remaining (n — r) binomials, and we can do
this in as many ways as r 6's can be taken from the n bi-
nomials, i.e., "(7, ways ; hence "C^ • a"~''6'' is the (r -f l)th, or
general, term of the product.
The 6's can be taken from the n binomials in one, and
only one, way ; hence we have the term 6", and this is what
the general term '•C^a'*~''6'" becomes when r = n.
Hence (a -\- b) (a -\- b) (a -\- b) -•• to n factors
= a" + ^da^'-'b + " C^"" -2^2 _^ ... 4- ^Cror-'b' + . . . + 6". (2)
If we substitute for ^'Ci, "C2, etc., their values as given in
§ 426, we obtain (?i denoting any positive integer)
(a H- by = a" + na'^-^b + '^(^~^) a^-''b^ -f • • •
if
_^ n(n - 1) (n - 2) ... (n - r + 1) ^n-.^. + ... + 6". (3)
\r_
406 ELEMENTS OF ALGEBRA
Identity (2) or (3) is tlie symbolic statement of the bi-
nomial theorem.
The second member of either is called the expansion of
(a -{-by.
Observe that (3) states in symbols the laws in § 126, and
that therefore (3) can be written out by these laws.
Note that the sum of the exponents of a and b in any
term is ii.
Ex. Expand (x-^ - i/y.
= (X-^y + 4 (x-=^)3(- 2/3) + 6 (a;-2)2(_ ^3)2 + 4 (a;-2) ( _ y3)8
= x-s - 4 x--y^ 4- 6 x-^f - 4 x'V + y^^.
432. In the expansion of (a + by, the general term
n(n-l)(n-2)..^(n-r + l) ^„_.^. _ ^^^ ^^ _^ ^^^^ ^^^^
\r
Observe that there are r factors in both the numerator
and denominator of the coefficient of the (r + l)th term.
By giving to r the proper value, we can find any term in
the expansion of (a + by.
When n is a positive integer, the coefficient of the (r+l)th
term becomes zero for any value of r greater than n ; hence
there are /? + 1 tei^ms in the expansion of (a + by.
Thus, when r = n + 1,
n(yi-l)(n-2) -'(n-r-\- 1) ^ n(?i-l)(n-2) •-(n-n)_Q^
[r |_r
Ex. Find the seventh term of the expansion of (4 x/5 — 5/2 xy.
Here a = 4 x/5, b=-5/(2x), w = 9, r = 6.
Substituting these values in the formula, we have
the seventh term = — )
1.2.3.4.5.t3V 5 / \2x I
= 10500x-3.
BINOMIAL THEOREM 407
433. The coefficients of the expansion in (2) of § 431 are
i nri nri nri nip nfl nfl .
X, L/1, U2, L/3, •••, L/„_2, ^n-\1 Wj
hence the (r4-l)th term from the beginning is ''Cra^~''b''j
and the (r 4- l)th term from the end is "C^.^a'^^""''.
But "O, = "(7„_, for all values of r (§ 427).
Hence, in the expansion of (a + by, the coefficients of any
two terms equidistant from the beginning and the end are the
same, so that the coefficients of the last half of the expansion
can be written from those of the first half
434. If, in identity (3) of § 431, we put a = 1 and b = x,
we have
(l+a;)- = l + nx + ''^V^'^^+-+.-7^^— ^"+-+^"
2 \r\n — r
This is a convenient form of the binomial theorem^ and
one which is often used.
Observe that this form includes all cases ; e.gr., if we want to find
(a 4- ?>)", we have
435. In (1) of § 434 the coefficients of x, x^, x^, •••, x'^ are
the values of "Ci, "Cg, "C3, •••, "(7„; hence (t) can be written
(1 + a;)" = 1 + "C,x + ^C^ + ... + "Oa' + ••• + ^'C^a;". (1)
Putting x = l, we obtain
2« = 1+ "Ci + '^C^ + - + "C, + - + "a. (2)
That is, the sum of the coefficients in the expansion of
(1 + xy, or (a + by, is 2\
From (2) it follows also that the sn7n of all the combinor
tions that can be made of n things, taken 1, 2, •", n at a time,
is 2" - 1.
■408 ELEMENTS OF ALGEBRA
Exercise 138.
By the laws in § 126 write the expansion of :
1. {3x'-2y)\ Q (r-^_^h)\ n. {x-'^-2c^)\
2. (2a2-3 6y. 7. {f-^n^)\ 12. (1 - 1 A)«.
3. (c^ + hy. 8. (2 x/^ - 3/xy. 13. (^2/ _ «-f)5.
4. (3a^ + 2/)'- 9. (x-^-afy. 14. (2 a;/3 - a/c)«.
5. (2-3i«y. 10. (a-'--x-y. 15. (a;-^-2?/-^)^
16. Find the 3d term in the expansion of (a — 3 by^.
17. Find the 7th term in the expansion of (1 — xy^.
18. Find the middle term in the expansion of (1 + xy^.
19. Find the middle term in the expansion of (2 a? — 3 yy.
20. Find the 18th term in the expansion of (1 + x)^.
21. Find the 7th term in the expansion of
[4:x/5-5/{2x)J.
22. Find the 17th term in the expansion of (xF — 1/x)^.
436. Binomial theorem, exponent fractional or negative.
When the exponent of a binomial is fractional or nega-
tive, the laws in § 126, or, what is the same thing, the
formula
(a + by = a" + na-~^b + <^lz1} a^-^^b' + -
_^n(n -l)(n- 2) ... (n - r + 1) ^._,^. ^ _^ ^^^
\l
gives an infinite series ; for in this case no one of the factors
n, n — 1, n~ 2, etc., in the (r + l)th term can ever be zero.
When, however, r increases ivithout limit, the sum of r terms
of this series will approach (a -\- by as its limit, provided the
first term of the binomial is arithmetically greater than the
BINOMIAL THEOREM 409
second term. That is, when 7i is fractional or negative, the
infinite series in (1) is the expansion of (a + by provided
a>b arithmetically.
A proof of this theorem is too difficult to be given here.
For a rigorous proof, see Taylor's " Calculus," § 98.
Ex. 1. Expand (c-i — (P)~^ and find the general term.
Applying the laws in § 126 we obtain
(c-i _ cf^)-f = [(c-i) + (- d2)]-t
-i¥.(0"^'(-^^)« + - (1)
= c^ + f cV + ^ c'^V .+ ^^ c^(P + -. ' (2)
The two distinct steps, that of applying the laws to obtain (1) and
that of performing the indicated operations in (1) to obtain (2), must
be taken separately.
In performing the operations indicated in (1), first note the number
of negative numeral factors in a term to determine the quality of its
numeral coefficient. Thus in the fourth term there are four negative
factors, — ^^\ and (— l)^.
Substituting in the general term for n, a, and b their values — |,
c-i, and — ^2^ ^e obtain
the (r+l)th term = (~i)^~^)(~Y^ "• ^~^~^"^^^(c-0"^""(-<f^/
[r
~ 5'"[r ^ ^
Since there are r factors in the numerator in (3), the term involves
the 2 rth power of — 1, which is + 1.
In (2), by this article d^ must be arithmetically less than c-i.
Ex. 2. Expand 1/(1 + x) and find the general term.
Applying the laws in § 120, we obtain
(1 + x)-i = 1-1 - 1 . 1-2 . x + 1 . 1-3 a;2 - 1 . 1-* x^-{- "'
= 1 -x-\-x^-x^-hx* . (1)
The (r + l)th term = (- ^^^-^ "'(- '^^ l-i-'-x'-=(- lyxr. (2)
In (1), by this article x is limited to values between — 1 and + 1.
410 ELEMENTS OF ALGEBRA
Ex. 3. (1 + a;)-2= 1-2 - 2 . 1-3 . x + 3 . 1-4 . a;2 - 4 . 1-5x3 + ...
= l-2x + 3x2-4x3 + ....
The (r+l)th tevm = ^~^^^~^^"'^~''~'^^l-'-'xr={-iyir-\-l)xr.
Ex. 4. Expand 1/ Vl - x, or (1 - x) 2.
(1 _ x)-^=r^ - i . r*(- X)+^- H(- X)2 _ ^% . l-^(- X)3+ ...
=l+lX+|x2+3:V:«^+•••.
The (r + l)th tern ^(-i)(-t)(-g- (-^-^'+l)i-i-(-xy
_1.3.5...(2r-l)^,
Ex. 5, Find the cube root of 127.
127 = 125 4- 2 = 53 + 2.
.-. v/l27=(53 + 2)i
= OT^ + K5')~^2 - i(53)"^' 22 + ^(53)-t23 + ...
3 52 9 55^81 58
= 5 + 0.0266666 - 0.0001422 + 0.0000012
= 5.0265255 -.
The smaller the ratio of the second term of the binomial to the first,
the more rapidly the successive terms of the expansion decrease, and
therefore the fewer the terms it is necessary to find.
Here we put 127 = 125 + 2, because 125 is the perfect cube which
makes the ratio of the second term to the first the smallest.
Exercise 139.
Expand to four terms :
1. (l-x)-\ 5. {l-\-x)-\ 9. (l-5a;)l
2. {l-x)-\ 6. (l + 2a;)-^ 10. (6^ _ c-*)"^.
3. (l-x)-\ 7. (2-x)-^ 11. a/-Vx^-f.
4. (l-x)-\ 8. (l-3a;)"^. 12. b/(J-b-^).
BINOMIAL THEOREM 411
Find the general term in the expansion of :
13. (l-x)-'. 15. (1-xyK 17. (l-2x)-^.
14. (l-x)-\ 16. (l + x)~i 18. (l4-3a;)~i
In its expansion find the :
19. Sixth term and eleventh term of (i a — bVb)^.
20. Fifth term of (1 - a^)"!
21. Seventh term of {x~^—y^y.
22. Third term and eleventh term of (1 -f 2a;)^^.
23. Fifth term of (c-2+ e-^)-'».
24. Sixth term of (a;"^-ci26f)-f.
Find to four places of decimals the value of :
25. ^31. 27. ^29. 29. </620. 31. ^/998.
26. </l7. 28. ^122. 30. ^31. 32. ^3128.
Expand to four terms :
33. (8 + 12 a)*. 38. (9 + 2»)*.
34. (l-Sx)^. 39. (4a- 8a;)"^.
35. (1-3 a;)-*. 40. (c^a"* - 6 V^)"^.
36. (a^ + c^)*. ^^ a
37. (c-d^l ' (cb-^-^y'¥
42. Find the general term in each of the examples from
33 to 39 inclusive.
CHAPTER XXXII
LOGARITHMS
437. The exponent which the base a must have in order
to equal the number N is called the logarithm of N to the
base a.
That is, if a^ = N, (1)
X is the logarithm of N to the base a, which is written
X^IOQaN. (2)
Equations (1) and (2) are equivalent; (2) is the logarith-
mic form of writing the relation between a, x, and N,
given in (1).
U.g.i since 3^ = 9, 2 is the logarithm of 9 to the base 3, or 2 = logs 9.
Since 2* = 16, 4 is the logarithm of 16 to the base 2, or 4 = log2 16.
Since 2-3 = 1/8, - 3 = logs (1/8).
Since 4^ = 8, 3/2 = log4 8.
Eeview §§ 52, 336, 338, 339, 386 on exponents.
Exercise 140.
1. Express each of the following relations in the loga-
rithmic form :
2^ = 8, 3^ = 81, 4^^ = 64, 12^ = 144, 6^ = 216, n' = b.
2. Express each of the following relations in the expo-
nential form :
logs 125 = 3, log2 32 = 5, log4 64 = 3, log3 81 = 4, \og,M=b.
412
LOGARITHMS 413
3. When fclie base is 3, what are the logarithms of 1, 3, 9,
27, 81, 243, 729 ?
4. When the base is 4, what are the logarithms of 1, 4,
16, 64, 256, 1024 ?
5. When the base is 2, what are the logarithms of 1, 1/2,
1/4, 1/8, 1/16, 1/32, 1/64, 1/128, 1/256 ?
6. When the base is 10, what are the logarithms of 1, 10,
100, 1000, 10000, 100000, 0.1, 0.01, 0.001, 0.0001, 0.00001 ?
7. When the base is 3, and the logarithms are 0, 1, 2, 3,
4, — 1, — 2, — 3, — 4, what are the numbers ?
438. The logarithms of all arithmetic numbers to any given
base constitute a system of logarithms.
Since 1* = 1, 1 cannot be the base of a system of log-
arithms. Any arithmetic number except 1 can evidently
be taken as the base of a system of logarithms.
Since logarithms are exponents, from the general proper-
ties of exponents, we obtain the general
PROPERTIES OF LOGARITHMS TO ANY BASE.
439. The logarithm of 1 is zero.
Proof. a« = l, .-. log«l = 0.
440. The logarithm of the hOjSe itself is 1.
Proof a^ = a, .-. \o^^a = l.
441. The logarithm of a product is equal to the sum of the
logarithms of its factors.
Proof Let M=a% N=a^',
then MxN=a'+«. §345
Hence log« (M x N) = x-\-y = log„ M -f log^ N.
E.g. , log4 (16 X 64) = log4 16 + log4 64 = 2 + 3 = 5.
414 ELEMENTS OF ALGEBRA
442. The logarithm of a quotient is equal to the logarithm
of the dividend minus the logarithm of the divisor.
Proof. Let M=a% N=aV',
then M-i-]Sr= a^ ^ § 346
Hence log« (M -i- N) = x — y = log^ M — log„ N.
E.g., logs (243 - 27) = logs 243 - logs 27
= 5-3 = 2.
443. The logarithm of any power of a number is equal to
the logarithm of the number multiplied by the ex2)onent of the
power.
Proof Let M=a'\
then, for all real values of p, we have
Jf p = aP\ § 348
Hence log„( Jf^) =px=p log„ M.
E.g., log4 (163) = 3 . log4 16 = 3 X 2 = 6 ;
log8(81^)=|log381 = f x4 = 3;
and logs (25"^) = - f logs 25=-fx2=-3.
444. By § 443, the logarithm of any positive integral
power of a number is equal to the logarithm of the number
multiplied by the exponent of the power ; and the logarithm
of any root of a number is equal to the logarithm of the
number divided by the index of the root.
Ex. 1. Given
logio 2 = .30103 and logio 3 = .47712 ; find logio ^720.
logio ^720 = ^ logio (23 X 32 X 10) § 443
= K'3 logio 2 + 2 logio 3 + logio 10) §§ 441, 443
= K-90309 + .95424 + 1) = .95244.
LOGARITHMS 415
Ex. 2. log„ [ V^ - (63c?) ] = loga x^ - loga {b^c^) § 442
= loga x^ - (log« 63 + log^ c^) § 441
= I log«a; - 3 loga & - f loga c. § 443
445. If a series of numbers are in geometric progression^
their logarithms are in arithmetic progression.
E.g., if N=\/21, 1/9, 1/3, 1, 3, 9, 27,-.
log3iV=-3, -2, -1, 0, 1, 2, 3, ....
Proof The logarithms of the terms of the Gr. P.
Ny .JSTr, ..., JVr",
are log^iV, log« .V + log„ r, ••., log„ JV^ + »i loga r,
which is an A. P. whose common difference is log^ r.
Exercise 141.
Express log^ y in terms of log^ b, log„ c, log, ic, and log, z,
having given the following equations :
1. y = T'b^<^. 4. y=W^-
2. 2/=^2'-Vc«. 5. 2/=V^-VW.
3. y = '^. 6. , = ^.
C^ ^^^&
Given logjo 2 = .3010, log^ 3 = .4771, find :
7. Iogio4; logio5; logioO; logioS; logio9; logwlO.
logio 5 = logio 10 - logio 2 = 1- .3010 = .6990.
8. Iog,ol2. 10. logio 30. 12. logio (3/2).
9. logio 16. .11. logio 50. 13. logio (6/5).
14. logioV^OO. 15. logio ^120.
16. Between what integral numbers does logio -^ lie?
when N lies between 10 and 100? Between 1 and 10?
Between .1 and 1 ? Between .01 and .1 ? Between .001
and .01?
416 ELEMENTS OF ALGEBRA
446. If the base of logarithms is greater than 1,
(i) The logarithm of a number is positive or negative,
according as the number is greater or less than 1.
(ii) The logarithm of an infinite is infinite ; and the
logarithm of an infinitesimal is a negative infinite, or, as it
is often stated, the logarithm of zero is negative infijiity.
Proof By § 437 x is the logarithm of a"" to the base a.
Let a > 1 ; then, by the principles of exponents, we know
that
if a* > 1, x>0; il a'' <1, x<0; hence (i).
If a* = 00, a; = 00 ; if a" = 0, x = — :/:>; hence (ii).
COMMON LOGARITHMS.
447. The logarithms used for abridging arithmetic compu-
tations are those to the base 10 ; for this reason logarithms
to the base 10 are called common logarithms.
Thus the common logarithm of a number answers the
question, ^ What power of 10 is the number ? '
Most numbers are incommensurable powers of 10; hence
most common logarithms are incommensurable numbers,
whose approximate values we express decimally.
Hereafter in this chapter when no base is written, the
base 10 is to be understood.
When a logarithm is negative, for convenience it is ex-
pressed as a negative integer plus a positive decimal.
E.g., the conmion logarithm of any number
between 10 and 100 is +1 + a positive fraction ;
between 1 and 10 is 0 + a positive fraction ;
between 0.1 and 1 is -1 + a positive fraction;
between 0.01 and 0.1 is -2 4- a positive fraction.
LOGARITHMS 417
448. The integral part of a logarithm is called the
characteristic, and the positive decimal part the mantissa.
A negative characteristic is usually written in the form
I, or 9 - 10; 2, or 8 - 10; 3, or 7 - 10; etc.
E.g., log 434.1 = 2.63759 ; +2 is the characteristic and .+63759 is the
mantissa : log 0.0769 = 2.88593, or 8.88593 - 10 ; 2, or 8 - 10, is the
negative characteristic, and .+88593 is the mantissa. The sign - is
written over the 2 to show that it affects the characteristic alone.
449. The characteristic of the common logarithm of any
number is found by the following simple rule:
Wlien the number is greater than 1, the characteristic is
positive and arithmetically one less than the number of digits
to the left of the decimal point ; when the number is less than
1, the characteristic is negative and arithmetically one greater
than the number of zeros between the decimal point and the
first significant figure.
E.g., 785 lies between IO2 and 10' ;
hence log 785 = 2 + a mantissa.
Again 0.0078 lies between lO-^ and IO-2 ;
hence log 0.0078 = — 3 + a mantissa.
Proof. Let N denote a number which has m digits to the
left of the decimal point ; then N lies between 10*""^ and
10"*;
that is, N = 10<'"-^) + * f™^"°°.
.-. log -^= (m — 1) + a mantissa.
Again let. ^denote a decimal which has m zeros between
the decimal point and the first significant figure; then iV
lies between lO-^""^^) and lO""* ;
that is, N= lO-^'^+i) + * <^'=«°".
.-. log N=— (m -h 1) -f a mantissa.
418 ELEMENTS OF ALGEBBA
450. The common logarithms of numbers which differ only
in the positio7i of the decimal point have the same mantissa.
Proof. When a change is made in the position of the
decimal point, the number is multiplied or divided by some
integral power of 10 ; that is, an integer is added to, or
subtracted from, the logarithm, and therefore its mantissa
is not changed.
E.g., log 1054.3 = 3.02296,
log 1.0543 = 0.02296,
log .010543 = 8.02296 - 10, or 2.02296.
451. When a negative logarithm is to be divided by a
number, and its negative characteristic is not exactly divis-
ible by that number, the logarithm must be so modified in
form that the negative integral part will be exactly divisible
by the number.
Ex. Given log 0.0785 = 2.8949; find log \/0. 0785.
Log v/0.0785 = I log 0.0785 = } (2.8949)
= }(7. +6.8949)= 1.8421.
Adding —5 + 5 to the logarithm does not change its value and
makes its negative part divisible by 7.
Exercise 142.
1. Log 427.32 = 2.6307. Find log 42732, log 42.732.
2. Log 23.95 = 1.3793. Find log 23950, log 239.5,
log 239500, log 0.002395, log 0.0002395, log 2395.
3. Log 4398 = 3.64326. Find log V0.4398, log -^0.4398,
log ^439.8, log ^0.04398, log ^0.004398.
4. Log 674.8 = 2.82918. Find log ^0.6748, log ^0.6748,
log -^0.06748, log ■{/0.06748, log ^0.006748.
452. Tables of logarithms. Common logarithms have two
great practical advantages : (i) Characteristics are known
LOGABITHMS 419
by § 449, so that only mantissas are tabulated ; (ii) mantis-
sas are determined by the sequence of digits (§ 450), so that
the mantissas of integral numbers only are tabulated.
At the close of this chapter will be found a table which
contains the mantissas of the common logarithms of all
numbers from 1 to 999 correct to four decimal places.
Note. Tables are published which give the logarithms of all num-
bers from 1 to 99999 calculated to seven places of decimals ; these are
called ' seven-place ' logarithms. For many purposes, however, the
four-place or five-place logarithms are sufficiently accurate.
From a table of logarithms we can obtain :
(i) The logarithm of a given number ;
(ii) The number corresponding to a given logarithm.
453. To find the logarithm of a given number.
Ex. 1. Find log 7.85.
By § 450, the required mantissa is the mantissa of log 785.
Look in column headed "N" for 78. Passing along this line to
the column headed 5, we find .8949, the required mantissa.
Prefixing the characteristic, we have
log 7.85 = 0.8949.
Ex. 2. Find log 4273. 2.
When the number contains more than three significant figures, we
must use the principle that when the difference of two numbers is
small compared with either of them, the difference of the numbers
is approximately proportional to the difference of their logarithms.
By § 450, the required mantissa is that of log 427.32.
The mantissa of log 427 = .6304.
The mantissa of log 428 = .6314.
That is, an increase of 1 in the number causes an increase of .0010
in the mantissa ; hence an increase of .32 in the number will cause an
increase of .32 of approximately .0010, or .0003, in the mantissa.
Adding ,0003 to the mantissa of log 427, and prefixing the character-
istic, we have
log4273.2 = 3.6307.
420 ELEMENTS OF ALGEBBA
Ex.3. Find log 0.0006049.
By § 450, the required mantissa is that of log 604.9.
The mantissa of log 604 = .7810.
Also, an increase of 1 in the number causes an increase of .0008 in
the mantissa; hence .9 of .0008, or .0007, must be added to .7810.
.-. log 0.0006049 = 4.7817, or 6.7817 - 10.
To find log 30 or log 3, find mantissa of log 300.
Exercise 143.
Find, from the table, the logarithm of the numbers :
1. 8. 5. 703. 9. 0.05307. 13. 7.4803.
2. 50. 6. 7.89. 10. 78542. 14. 2063.4.
3. 6.3. 7. 0.178. 11. 0.50438. 15. 0.0087741.
4. 374. 8. 3.476. 12. 0.00716. 16. 0.017423.
454. To find a number when its logarithm is given.
Ex. 1. Find the number of which the logarithm is 3.8954.
Look in the table for the mantissa .8954. It is found in line 78 and
in column 6 ; hence
.8954 = the mantissa of log 786.
.-. 3.8954 = log 7860;
or 7860 is the number whose logarithm is 3.8964.
Ex. 2. Find the number of which the logarithm is 1.6290.
Look in the table for the mantissa .6290. It cannot be found ; but
the next less mantissa is .6284, and the next greater is .6294.
Also, .6284 = mantissa of log 425,
and .6294 = mantissa of log 426.
That is, an increase of .0010 in the mantissa causes an increase of 1
in the number ; hence an increase of .0006 in the mantissa will cause
an increase of approximately 3% of 1, or .6, in the number; hence
I .6290 = the mantissa of log 425.6 ;
.-. 1.6290 = log 42.56.
LOGARITHMS 421
Ex. 3. Find the number of which the logarithm is 3.8418,
Look in the table for the mantissa .8418. It cannot be found ; but
the next less mantissa is .8414, and
.8414 = mantissa of log 694.
Also, an increase of .0006 in the mantissa causes an increase of 1 in
the number ; hence an increase of .0004 in the mantissa will cause an
increase of f of 1, or .66 in the number ; hence
.8418 = the mantissa of log 694.66.
.-. 3.8418 = log 0.0069466.
Exercise 144.
Find the number of which the logarithm is :
1. 1.8797. 6. 8.1648-10. 11. 3.7425.
2. 7.6284-10. 7. 9.3178-10. 12. 7.1342-10.
3. 0.2165. 8. 1.6482. 13. 3.7045.
4. 2.7364. 9. 8.5209-10. 14. 8.7982-10.
5. 4.0095. 10. 3.8016. 15. 3.4793.
455. The cologarithm of a number is the logarithm of its
reciprocal.
That is, colog N=\og(l -^N) = — log N.
To make the fractional part of the cologarithm positive, if
log iV > 0 and < 10, colog N is written
(lO-logiV)-lO;
if log ^ > 10 and < 20, colog N is written
(20-logiV^)-20.
E.g., cologO.0674 =-(2.7589)= 1.2411;
colog 432 = (10 - 2.6263) - 10 = 7.3737 - 10 ;
colog 345000000000 = (20 - 11.5378) - 20 = 8.4622 - 20.
Instead of subtracting the logarithm of a divisor, we can,
by § 87, add its cologarithm.
422 ELEMENTS OF ALGEBRA
Ex.1. Find the value of 1M^AM723.
0.0534 X 7.238
log 15.08 = 1.1784
log0.0723 = 8.8591 -10
colog0.0534 = 1.2725
colog 7.238 = 9.1404-10
Add, log (fraction) = 0.4504 = log 2.8213.
Hence the fraction = 2.8213.
Ex. 2. Find the value of 0.0543 x 6.34 x (- 5.178).
Iog0.0543 = 8.7348 -10
log 6.34 = 0.8021
loa: 5.178 = 0.7141
Add, log (product) = 0. 2510 = log 1 . 7824.
Hence the product is — 1.7824.
By logarithms we obtain simply the arithmetic value of the result ;
its quality must be determined by the laws of quality.
J 5.42 X 427.2
Ex. 3. Find the value of Ji
\3.244 X 0.0231^
log5.42 = 0.7340 =0.7340
2 log 427. = (2.6304) x 2 = 5.2608
4 colog 3.24 = (9.4895 - 10) x 4 = 7.9580 - 10
^ colog 0.0231 = (1.6364) -2 = 0.8182
5)4.7710
.-. log (root) = 0.9542
.-. root = 9.00
456. An exponential equation is one in which the unknown
appears in an exponent ; as 2* = 5, af = 10.
Such equations are solved by the aid of logarithms,
Ex. 1. Solve 32* - 14 x 3^ + 45 = 0. (1)
Factor (1), (3* - 9) (3^ - 5) = 0. (2)
LOGARITHMS 42^
(4)
Equation (2)
is equivalent to the two equations
3' = 9, (3) 3- = 5.
From (3),
x = 2.
From (4),
a; log 3 = log 5.
log 5^ 0.6990 _g49
log 3 0.4771
Hence the roots of (1) are 2 and 1.4649.
Exercise 145.
Find by logarithms the value of :
1. 742.8x0.02374. 7. 4743-5-327.4.
2. 0.3527x0.00572. g 9.345^ (_ 0.0765).
3. 78.42x0.000437.
_ 2.476 x(- 0.742)
4. 5234 X (-0.03671). 9. ^3 ^^ ^ (_o.ooi21)*
5. 3.246 X (-0.0746). ^^ 321 x(- 48.1) x (357)
6. - 4.278 X (- 0.357). * 421 x (- 741) x (4.21)
11. 5l 14. («)«. 17. (iiyi
12. 0.02li 15. 714.2i 18. (3|)^27
13. 0.5328. 16. (Iff)^ 19. 4.71=^
/0.035'* X 54.2 X 785^ x 0.0742
206
20.
21.
22.
10.035'^
^ 4.72^ X 7.14^ X 8.47^
3/ 0.0427^ X 5.27 x 0.875^
V7.42I* X Vl-'^4 X V0.00215
5/0.714^x0.1371^x0.0718^
^' 0.5242x0.742^ x 0.0527^
424 ELEMENTS OF ALGEBRA
Solve each of the following equations :
23. 31^ = 23. 25. 5^ = 800. 27. 5^-3 = 8^*+^
24. 0.3^ = 0.8. 26. 12^ = 3528. 28. aH^'' = (^,
29. 23*52^-1 = 4^^3^+1. 30. 42=^-15 (4^) +56 = 0.
COMPOUND INTEREST AND ANNUITIES.
457. To find the compound interest, $ /, and amount, $ M,
of a given principal, $ P, in n years, $ r being the interest on
$ 1 for 1 year.
Let ^R = the amount of $ 1 in 1 year ; then R = l-\-r,
and the amount of f P at the end of the first year is $ PR ;
and since this is the principal for the second year, the
amount at the end of the second year is $ PR x R, or
$ PR^. For like reason the amount at the end of the third
year is $PR^, and so on; hence the amount in n years is
^PR""', that is,
M=PR'', or P(l + r)^ (1)
Hence I=P(R--1). (2)
If the interest is payable semi-annually, the amount of
$P in 1/2 a year will be $P(l-\-r/2); hence, as n years
equals 2 n half-years,
M=P(l + r/2y\ (3)
Similarly, if the interest is payable quarterly,
M=P{l+r/4:y\ (4)
Ex. Find the time in which a sum of money will double itself at
ten per cent compound interest, interest to be "converted into prin-
cipal" semi-annually.
Here 1 + r/2 = 1.05. Let P = 1 ; then M=2.
Substituting these values in (3), we obtain
2 = (1.05)2«
.-. log2 = 2w .log 1.05.
2 log
^^gl- = ^^^ = 7.1 years, ^ns.
02 1.05 0.0424 ^
LOGARITHMS 425
458. Present value and discount. Let f P denote the pres-
ent value of the sum $M due in n years, at the rate ?-; then
evidently in n years at the rate r, $P will amount to $M-j
hence
M=PB^, or P = MR-\
Let $ D be the discount ; then
D = M- P= 3/(1 - Z2-**).
459. An annuity is a fixed sum of money that is payable
once a year, or at more frequent regular intervals, under
certain stated conditions. An Annuity Certain is one pay-
able for a fixed number of years. A Life Annuity is one
payable during the lifetime of a person. A Perpetual An-
nuity, or Pei'petuity, is one that is to continue forever, as,
for instance, the rent of a freehold estate.
460. To find the amount of an annuity left unpaid for a
given number of years, allowing compound interest.
Let ^ J. be the annuity, n the number of years, $ R the
amount of one dollar in one year, $ JHf the required amount.
Then evidently the number of dollars due at the end of the
First year = A ;
Second year = AR + A ;
Third year = AR' + AR -{- A -,
nth year = AR'-^ + AR" ---\ [-AR-\-A
^A(R^-l)
R-1
That is, M=-(R''-1). • (1)
Ex. 1. Find the amount of an annuity of 1 100 in 20 years, allow-
ing compound interest at 4 J per cent.
r ^ 0.046
426 ELEMENTS OF ALGEBRA
By logarithms, 1.04520 = 2.4117.
... 3/ ^l^lill^ 3137.11.
0.045
Hence the amount of the annuity is $3137.11.
Ex. 2. What sum nmst be set aside annually that it may amount
to $ 50,000 in 10 years at 6 per cent compound interest ?
rrom(l), .4 = -^!^=.^M00Aa06^ 3793.37.
Hence the required sum is $3793.37.
461. To find the preseiit value of an annuity of ^A pay-
able at the end of each of n successive years.
Let $P denote the present value; then the amount of
^P in n years will equal the amount of the annuity in the
same time : that is,
PR^ = A{R'-l)r-\ (1)
.-. P=A{l-R--)r-\ (2)
If the annuity is perpetual, then n = oo, i2~" = 0, and
(2) becomes
P = Ar-\
Exercise 146.
1. Write out the logarithmic equations for finding each
of the four numbers, M, R, P, n.
2. In what time, at 5 per cent compound interest, will
$100 amount to f 1000?
3. Find the time in which a sum will double itself at
4 per cent compound interest.
4. Find in how many years f 1000 will become f 2500
at 10 per cent compound interest.
5. Find the present value of $10,000 due 8 years hence
at 5 per cent compound interest.
LOGARITHMS 427
6. Find the amount of $1 at 5 per cent compound in-
terest in a century.
7. Show that money will increase more than thirteen-
thousand-fold in a century at 10 per cent compound interest.
8. If A leaves B $1000 a year to accumulate for 3
years at 4 per cent compound interest, find what amount B
should receive.
9. Find the present value of the legacy in example 8.
10. Find the present value, at 5 per cent, of an estate of
$ 1000 a year to be entered on immediately.
11. A freehold estate worth f 120 a year is sold for
$ 4000 ; find the rate of interest.
12. A man has a capital of $20,000, for which he re-
ceives interest at 5 per cent; if he spends $1800 every
year, show that he will be ruined before the end of the
17th year.
428
ELEMENTS OF ALGEBRA
N
0
1
2
3
4
5
6
7
8
9
10
0000
0043
0086
0128
0170
0212
0263
0294
0334
0374
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
13
1139
1173
1206
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
1969
1987
2014
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2465
2480
2604
2529
18
2563
2577
2601
2625
2648
2672
2695
2718
2742
2766
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
3679
3598
23
3617
3636
3655
3674
3692
3711
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
4266
4281
4298
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4466
28
4472
4487
4502
4518
4633
4548
4664
4679
4694
4609
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4767
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4942
4955
4969
4983
4997
6011
6024
5038
32
5051
5065
5079
5092
5105
6119
6132
6145
6169
5172
33
5185
5198
5211
5224
5237
6250
5263
5276
6289
6302
34
5315
5328
5340
5353
6366
5378
5391
6403
6416
5428
35
5441
5453
5465
5478
5490
5602
6514
6627
5639
5561
36
5563
5575
5587
5599
5611
6623
6636
6647
6658
6670
37
5682
5694
5705
5717
6729
5740
5752
5763
5776
5786
38
5798
5809
5821
5832
6843
6865
6866
5877
6888
5899
39
5911
5922
5933
5944
5955
5966
6977
5988
6999
6010
40
6021
6031
6042
6053
6064
6075
6086
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
6376
6385
6395
6406
6416
6425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
45
6532
6542
6551
6561
6671
6580
6590
6699
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
6876
6884
6893
49
6902
6911
6920
6928
6937
6946
6955
6964
6972
6981
50
6990
6998
7007
7016
7024
7033
7042
7060
7069
7067
51
7076
7084
7093
7101
7110
7118
7126
7136
7143
7162
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7236
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
54
7324
7332
7340
7348
7356
7364
7372 7380
7388
7396
TABLE OF MANTISSAS
429
N
0
1
2
3
4
5
6
7
8
9
55
7404
7412
7419
7427
7436
7443
7451
7459
7466
7474
56
7482
7490
7497
7505
7613
7520
7528
7536
7543
7661
67
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
7762
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7826
7832
7839
7846
61
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
8036
8041
8048
8056
64
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
65
8129
8136
8142
8149
8166
8162
8169
8176
8182
8189
66
8196
8202
8209
8216
8222
8228
8236
8241
8248
8264
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
68
8325
8331
8388
8344
8351
8357
8363
8370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8446
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8637
8643
8549
8566
8561
8567
72
8573
8579
8585
8591
8597
8603
8609
8616
8621
8627
73
8633
8639
8645
8651
8667
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8746
75
8751
8756
8762
8768
8774
8779
8786
8791
8797
8802
76
8808
8814
8820
8826
8831
8837
8842
8848
886^4
8869
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8964
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
9176
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
9316
9320
9326
9330
9335
9340
86
9345
9360
9355
9360
9365
9370
9375
9380
9386
9390
87
9395
9400
9405
9410
9416
9420
9426
9430
9436
9440
88
9445
9450
9465
9460
9466
9469
9474
9479
9484
9489
89
9494
9499
9504
9509
9613
9518
9523
9528
9633
9638
90
9542
9547
9552
9567
9562
9566
9571
9576
9681
9686
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
92
9638
9643
9647
9652
9657
9661
9666
9671
9676
9680
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
9731
9736
9741
9745
9750
9764
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9806
9809
9814
9818
96
9823
9827
9832
9836
9841
9846
9860
9864
9869
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
CHAPTER XXXIII
GRAPHIC SOLUTION OF EQUATIONS AND SYSTEMS
M'
4--4-
M
P
462. Let XX' and YY' be any two fixed straight lines
at right angles to each other at 0. Let the directions
OX and OF be positive directions; then the directions
^Y . OX' and 0Y> will
be negative direc-
tions.
The lines XX
and YY' are called
axes of reference,
__ and their intersec-
tion 0, the origin.
From P, any
point in the plane
of the axes, draw
PM parallel to
IP
M
Fig. 1
YY' ; then the po-
sition of P will be
determined when
we know both the lengths and the directions of the lines
OMsindMP.
The line OM, or its numerical measure, is called the
abscissa of the point P; and MP, or its numerical measure,
is called the ordinate of P. The abscissa and ordinate
together are called the coordinates of P.
E.g., OM' and M'P' are the coordinates of P' ; the abscissa, OW,
is negative, and the ordinate, M'P', is positive. OM'", the abscissa
of P"', is positive, and M'"P"', its ordinate, is negative.
430
SOLUTION OF EQUATIONS AND SYSTEMS 431
An abscissa is usually deaoted by the letter x, and an
ordinate by y.
Observe that the numerical measure of OM or MP is a
positive number, if it extends in the direction OX or OF;
and a negative number if it extends in the direction OX'
or or.
The axis XX' is called the axis of abscissas, or the jr-axis ;
and YY\ the axis of ordinates, or the /-axis.
The point whose coordinates are x and y is denoted by
{^y y)'
E.g., (2, — 3) denotes the point of which the abscissa is 2, and the
ordinate — 3.
We use a system of coordinates analogous to that explained above
whenever we locate a city by giving its latitude and longitude ; the
equator ie one axis, and the assumed meridian the other.
Ex. Plot the point (- 2, 3) ; (-3,-4).
In the figure lay off OJf = - 2, and on MT* parallel to YT lay off
JiPP' = + 3 ; then P' is the point (-2, 3).
To plot (-3, - 4), lay off OM" = - 3, and on M"P" parallel to
YY' lay ofE M"P" = - 4 ; then P" is the point (-3, - 4).
The lines XX' and YY divide the plane into four equal
parts called quadrants, which are numbered as follows :
XOY is the Jirst quadrant j YOX' the secondy X'OY the
third, and YOX the fouHh.
Exercise 147.
1. Plot the point (2, 3); (4, 7); (3, -5); (-2, +3);
(-3, +5); (4, -2); (-2, -3); (-5, -3); (-2, 4);
(-4,-1); (0,0).
2. In which quadrant is (+a, +b)? (+a, "6)? ('a, +b)?
(-a,-b)?
3. What is the quality of x and of y, when the point
(Xy y) is in the first quadrant ? Second quadrant ? Third
quadrant? Fourth quadrant?
432 ELEMENTS OF ALGEBRA
4. In which quadrants can the point (x, y) be, when x is
positive ? X negative ? y positive ? y negative ?
5. In what line is the point (a;, 0) ? (0, y) ?
6. Where is the point (0, 0) ? (4, 0) ? (- 3, 0) ? (0, 2) ?
(0,-5)?
463. Graphic solution of equations in x and /.
The locus, or graph, of an equation in a; and y is the line
or lines which include all the points, and only those, whose
coordinates satisfy the equation.
Ex. 1. Draw the locus of y = x"^ — x — 6. (1)
If in (1) we put a; = - 3, - 2, - 1, ..., we obtain
when X = - 3, - 2, - 1, 0, 1/2, 1, 2, 3, 4, ...,
y = 6, 0, - 4, - 6, - 6|, - 6, - 4, 0, 6, ....
Drawing the axes XX' and TY' in fig. 2, and assuming 01 as the
linear unit, we plot the points
(-3,6), (-2, 0), (-1, -4), (0, -6), ....
The relative positions of these points indicate the form of a curve
through them.
Whenever there is any doubt about the form of this curve between
any two plotted points, as between (0, —6) and (1, —6), one or
more intermediate points should be found and plotted.
As X increases indefinitely from 3, y (^or x^ — x — 6) continues posi-
tive and increases indefinitely ; hence the locus has an infinite branch
in the first quadrant. As x decreases indefinitely from — 2, y con-
tinues positive and increases indefinitely ; hence the locus has an
infinite branch in the second quadrant.
Drawing a smooth curve through the plotted points we obtain the
curve ABC in fig. 2, which, with its infinite branches, is the locus
of equation (1).
This curve is called the locus of the equation because each and
every real solution of equation (1) is the coordinates of some point
on the curve.
SOLUTION OF EQUATIONS AND SYSTE3IS 433
Imaginary or complex solutions of an equation cannot be
represented by the coordinates of any points in the plane
XO Y, since by definition the coordinates of every point in
this plane are real.
Note. The pupil should use coordinate or cross-section paper, and
with a hard pencil draw the loci of equations neatly and accurately.
X X-
t—^
Ex. 2. Draw the locus ot y = 3? — 2x. (1)
When x = -2, -y^, -1, -0.8, 0, 0.8, 1, y/2, 2, ...,
y=-4, 0, 1, 1.1, 0, -1.1, -1, 0, 4, ....
As X increases indefinitely from 2, y (or x^ — 2x) continues positive
and increases indefinitely ; hence the locus has an infinite branch in
the first quadrant.
As X decreases indefinitely from —2,y continues negative and
arithmetically increases without limit ; hence there is an infinite
branch in the third quadrant.
Plotting these points, as in fig. 3, and tracing a smooth curve
through them, we obtain the curve ABCD, which, with its infinite
branches, is the locus of (1).
434
ELEMENTS OF ALGEBRA
Ex. 3. Draw the locus oi y = x^ — S x"^ + i.
When x = -3/2, - 1, - ^, 0, 1/2, 1, 2, 3, ...,
y=- 6.1, 0, 3.1, 4, 3.4, 2, 0, 4, ....
As X increases from 2, y increases indefinitely from 0 ; and, as
X decreases from —!,'</ decreases indefinitely from 0.
The locus is the curve ABCD in fig. 4, which has one infinite
branch in the first quadrant and another in the third.
2/ = a:* + x3 - 3 x2 - ic + 2.
A \Y E
Fig. 5
Ex. 4. Draw the locus of y = x* + a;^ - 3 x^ — a: + 2.
When a; = -|, -2, _ |, - 1, - ^, - .2, 0, J, 1, f, ..-,
y = 9.2, 0, -1.6, 0, 1.7, 2.08, 2, 0.9, 0, 2.2, ••..
The locus is the curve in fig. 5, which has one infinite branch in
the first quadrant and another in the second.
The foregoing examples illustrate how each real solution of an
equation in x and y is the coordinates of some point in the locus ;
hence, by the coordinates of its points, the locus of an equation in
X and y giveg all its infinite number of real solutions.
SOLUTION OF EQUATIONS AND SYSTEMS 435
Exercise 148.
Draw the locus of each of the following equations, stating
in which quadrants the infinite branches lie :
2. y = x — a^ + 6. 5. y = x^ — a^ — 6x.
3. y = oc^ — 4:X. G. y = x^ — 5x^ + 4:.
7. Using one set of axes, draw the loci of
y = x, y = x + 2, y = x-2.
Observe that these loci are parallel straight lines.
8. Using one set of axes, draw the loci of
y = 2x, y = 2x + 3, y = 2x-3.
9. Draw the locus of
y = 3x-2,oiy = -2x-\-l,of2y = 4:X-6.
These examples illustrate the truth that the locus of any linear
equation in x and y is a straight line.
Hence to draw the locus of any linear equation we can plot two of
its points and draw a straight line through them.
10. Draw the locus of
y = 4, of 2/ = — 3, of X = 3, of a; = — 4.
Observe that each of these loci is parallel to one of the axes.
11. From the origin 0 as a centre and with a radius 5,
draw the circumference of a circle. Draw the ordinate PM
of any point P on this circumference and the radius OP.
Denote the coordinates of P hy x and y. Then from the
right-angled triangle OMP, we obtain a?^ + ^^ _ 52^
What then is the locus of a^ -f 2/^ = 2^ ?
436 ELEMENTS OF ALGEBRA
12. Draw the locus of or -|- 3/- = 9, of x- -\- y- = 16.
Examples 11 and 12 illustrate the truth that the lucus of any equa-
tion of the form x^ -\- y^ = r^ is the circumference of a circle whose
centre is at the origin and whose radius is r.
13. Draw the locus of 4tx^ + ^y' = 36. (1)
Here y =± |V9 - xK
Evidently — 3 is the least value of x which will render y real ; hence
no part of the locus can lie to the left of the line a; = — 3. For like
reason no part of the locus can lie to the right of the line x = 3-
When a; =-3, -2.5, -2, -1, 0, 1, 2,3,
2/ = 0, ±1.1, ±1.5, ±1.9, ±2, ±1.9, ±1.5,0.
The locus is the ellipse RASB (fig. 8, page 439), the semi-axes
being 3 and 2.
Observe that in (1) the coefficients of x^ and ?/2 are unequal, while
in examples 11 and 12 they are equal.
14. Draw the locus of a;- -f 4?/- = 4.
15. Draw the locus of a.-^ — 4 ?/^ = 4.
Here y =±\/2V^^~^^. (1)
When X > 2 or <— 2, the values of y in (1) are real ; when x lies
between — 2 and +2, y is imaginary ; hence there is an infinite
branch in each of the four quadrants, but no point of the locus lies
between the lines x = — 2 and x = 2.
16. Draw the locus of 2/^ = 4 x.
17. Using one set of axes, draw the loci of
X — y = — 3 and a; -f 1/ = 1.
Observe that these equations are independent and consistent, and
that the one and only point common to their loci is (— 1, 2).
18. Using one set of axes, draw the loci of
2 ic — 2/ = — 1 and 2x — y = — 3.
Observe that these equations are inconsistent, and that their loci
are parallel and hence have no point in common.
SOLUTION OF EQUATIONS AND SYSTEMS 437
19. Using one set of axes, draw the loci of
2 a; + 2/ = 1 and 6x-\-3y = 3.
Observe that these equations are equivalent, and that their loci
coincide and hence have all points in common.
20. What is the greatest number of points in which a
straight line can cut the locus in fig. 2 ? In fig. 3 ? In
fig. 4 ? In fig. 5 ? Compare each answer with the degree
of the equation of each locus.
464. Graphic solution of systems of equations.
Ex. 1. By the aid of loci discuss the system
ax + by = c, (1)
ax + b'y = c'. (2)
(a)
Let the locus of (1) be the straight line MN, and that of (2)
the line liP. Then the coor-
dinates of the point P, which is
common to both loci, will be
the solution common to (1)
and (2), or the solution of the
system (a). By measuring
the coordinates OA and AP,
the numerical solution of the
system could be obtained.
This example illustrates
graphically the theorem in
§207.
The loci will have one, and only one, point in common,
if a/a' ^ b/h>,
i.e., if (1) and (2) are independent and consistent (§ 207).
The loci will coincide throvghotit their whole extent,
if a/a' ^ b/V = c/c',
i.e., if (1) and (2) are equivalent (§§ 207, 357).
The loci will be parallel and have no point in common,
if a/a' = b/b' and a/a^ =^ c/c',
i.e., if (1) and (2) are inconsistent (§§ 207, 357).
438
ELEMENTS OF ALGEBBA
Ex. 2. By the aid of loci, discuss the system
for different values of c.
x2 + ?/2 = 25,
(1)
(2)
(a)
The locus of (1) is the circle PP'BP" ; and, if c = 1, the graph of
(2) is the straight line MN; hence the coordinates of the two points
P and B are the two solutions of system (a).
By measurement we find the two solutions to be 3, 4 and — 4, — 3.
As c increases, the locus MN moves upward parallel to itself, and
P and B approach P'.
When c = 5\/2, the locus of (2) is the tangent N'3I', and the two
solutions of the system are equal.
Similarly, when c = — 5\/2, the locus of (2) is M"N".
When c < 5\/2 and > - 5V2, the locus of (2) lies between N'M'
and N"M", and the two solutions of the system are real and unequal.
When c > 5\/2 or < — 5V2, the locus of (2) does not cut the circle,
and both solutions of the system are imaginary or complex.
SOLUTION OF EQUATIONS AND SYSTEMS 489
Ex. 3. By aid of loci discuss the system
x^+ y^ = r^
(1)
}(«)
C2 + ^2 _ ^2 (2)
for different values of r.
The locus of (1) is the ellipse ABBS, in which OA=S and 0B=2.
If r = 5/2, the locus of (2) is the circle PP'P"P"', and the four
solutions of the system are the coordinates of the four points P, P',
P", P"', and thus are real and unequal.
If r = 3, the circle will be tangent to the ellipse at A and B ; hence
two solutions of the system will be 3, 0, and the other two — 3, 0.
If r = 2, the circle will be tangent to the ellipse at B and S.
If r < 2 or > 3, the two loci will have no common points, and all
four solutions of the system will be imaginary or complex.
When r = 5/2, by clearing (2) of fractions and then subtracting it
from (1) we obtain 5]/^ = 11, the locus of which is the parallel lines
PP and P"'P'. These lines cut either the ellipse or the circle in all
the points which are common to these curves, and only in these points.
This illustrates the equivalency of system (a) to the system
4x2 + 9y2 = 36,
6y2
= 36, I x'^-hy^ = ^,]
= 11.1 °^ 5?/2=ll. J
440
ELEMENTS OF ALGEBRA
(a)
Ex. 4. By aid of loci discuss the system
xy = 12, (1)
y = mx + n, (2) ,
for different values of m and n.
The locus of (1) is the curves AB and 02), whose infinite branches
approach the axes.
When n = 0 and m = 3/4, the locus of (2) is the line PP', and the
two solutions of system (a) are the coordinates of the points P and P'.
Let m = 0 ; then P will move out along the infinite branch PB, and
P' along the infinite branch P'C ; that is, y = 0 and x = + go or — oo.
y lA
Again, when n = 0, the two solutions of system (a) are
2V:i/m, 2\/3hi, and -2\/^i/m, - 2V8to.
For m = 0, the solutions in (.3) assume the forms
a/0, 0 and - a/0, 0 ;
(3)
hence equation (1) and y = 0 are inconsistent, and system (a) is then
impossible. This agrees with the figure, for the locus oiy = 0 coincides
with X'OX, and does not intersect the locus of (1).
When m is negative, the solutions in (3) become imaginary. This
agrees with the figure ; for when m is negative, x and y in y = mx are
SOLUTION OF EQUATIONS AND SYSTEMS 441
opposite in quality, and hence the locus of y = mx will lie in the
second and fourth quadrants, and will not cut the locus of (1).
If »i = 0 and n ^ 0, the locus of (2) will be parallel to XOX', and
will cut the locus of (1) in only one point ; hence system (a) will be
defective in one solution.
Exercise 149.
1. By aid of loci show that system (a) is equivalent to
the four systems in (6).
a^ -h y- = 25
xy
r = 25
x?/ = 12 J
x + y = 7] x + y = 7 ) x-^y = -7) x-\-y = -7]
a; — l = lJ x — y = — lj x — y = \ i x — y = — l^
2. By aid of loci show that the following six systems are
equivalent :
0^ + 2/2 = 25
ar'=16
0^ + 2/^ = 25
7?-f = 7 J
2/2 = 9
x'-f = 7
1 ^-f=7
a^ = 16
a^ = 16
1 / = 9 i
/ = 9
I
1-
GRAPHIC SOLUTION OF EQUATIONS IN X.
465. A variable whose value depends upon one or more
other variables is called a dependent variable, or a function of
those variables. A variable which does not depend upon any
other variable for its value is called an independent variable.
E.g., x^, 2 a;"2 — .3 X + 7, or x"* — 7 x^ + 9, is a function of the inde-
pendent variable x.
Again, y in each of the equations in this chapter is a function of the
independent variable x.
The symbol f{x), read ^function x,' is used to denote any
function of x.
442 ELEMENTS OF ALGEBRA
The symbols f(a), f(2), /(I) represent the values of f(x)
when x = a, 2, 1, respectively.
E.g., iif(x) = ^3 + X, then
/(a) =a^^a, /(2) = 2^ + 2 = 10, /(I) = 2.
Since /(a;) denotes any function of x, y =f{x) denotes any
equation in x and y, when the equation is solved for y.
Thus, any one of the equations in the first ten exsrcsples in
exercise 146 is a particular case of y =f(x).
466. A continuous variable is a variable which in passing
from one value to another passes successively through all
intermediate values.
A function, as /(a;), is said to be continuous between x = a
and ic = 6, if when x changes continuously from a to h, f(x)
varies continuously from /(a) to f{b). In other words, /(a:)
is continuous between x = a and x=b, when the locus of
y =f(x) is an unbroken curve between the lines x = a and
x = b.
E.g., the time since any past event varies continuously. The veloc-
ity acquired by a falling body and the distance fallen are continuous
functions of the time of falling.
In each of the four examples in § 463, y is a continuous function of
X for all real values of x.
In example 2 of § 464, y in equation (1) is real and a continuous
function of x between x = — 6 and x = + 5.
The examples in § 463 illustrate the fact that
Any rational integral function of x is a continuous function.
In what follows we shall use f(x) to denote a rational
integral function of x.
467. The ordinates of the points in the locus ot y = a^
— x — 6 in fig. 1, of § 463, are the successive values of
a^ — x — 6 corresponding to successive values of x ; hence,
the locus of y =f{x) is often caUed the graph of f{x).
SOLUTION OF EQUATIONS AND SYSTEMS 443
E.g., in fig. 2, while x increases continuously from —3 to zero,
the function x^ — x — 6 decreases continuously from + 6 though zero
to — 6 ; and while x increases from zero to +4, x^ — x — Q first de-
creases from — 6 and then increases to -f 6.
Again, in fig. 3, while x increases continuously from — 2 to — 0.8,
the function x^ — 2x increases continuously from — 4 to 4- 1.1 ; while
X increases from —0.8 to +0.8, x^ — 2x decreases from +1.1 to
— 1.1 ; while x increases from + 0.8 to + 2, x' — 2 x increases from
-1.1 to 4.
In like manner, in the other figures, the pupil should follow the
changes in f{x) as x increases.
468. The abscissas of the points in which the graph of
f(x) cuts or touches the axis of x are the real vahies of x
for which f(x) is zero ; that is, they are the real roots of
the equation f{x) = 0.
At a point of tangency the graph is properly said to touch
the axis of x in two coincident j)oints.
E.g., from the graph in fig. 2, we learn that one root of the equa-
tion z^ — x — Q = 0 is — 2 and the other is 3.
From the graph in fig. 3, we learn that the three roots of the equa-
tion ar^ — 2 X = 0 are — ^^2, 0, and y/2.
In fig. 4, the graph cuts the axis of x at (— 1, 0) and touches it at
(2, 0) ; hence, one root ofx'' — 3x2 + 4 = 0is — 1 and the other two
roots are 2 each.
Hence, to find the real roots of f(x) = 0, we can draw
the graph of f{x), or the locus of y =f(x), and measure the
abscissas of the points of intersection and tangency with
the ar-axis.
Exercise 150.
Construct the graph of f(x), and find approximately the
real roots of each of the following equations :
1. 3^^x-2 = 0. 4. x^-Sar-Ax-\-ll = 0.
2. ar'-f 2a;-5 = 0. 5. a^ - 4.0(^ - 6x - S = 0.
3. .r^-3a;+-4 = 0. 6. a;^- 4 a^-3a;+- 2 = 0.
CHAPTER XXXIV
THEORY OF EQUATIONS
489. Horner's method of synthetic division.
Let it be required to divide
Ax^ + Bx^+ Cx-^ D hj X- a.
If for convenience we write the divisor to the right of the dividend
and the quotient below it, by the usual method we have :
Ax^ + Bx"^ + Cx -\- D
Ax^ — Aax^
Ax' + (Aa + B)x
+ (Aa^ + Ba+ O)
(Aa + B)x:^
(Aa + B)x^ - (Aa^ + Ba)x
(Aa^ + Ba + C)x
{Acfi + Ba+ C)x - {Aa^ + ga^ + Ca)
Aa^ + Ba^ + Ca + D
In the shorter or synthetic method, we write only the coefficients of
the dividend and place a at their right, as below :
A B G D\a
_ Aa Aa^ + Ba Aa^ + Bii^ + Ca
A Aa-\- B Aa'^-h Ba+C Aa^ + Ba"^ + Ca -\- D
Multiplying A by a, writing the product under J?, and adding, we
obtain Aa + B. Multiplying this sum by a, writing the product under
O, and adding, we obtain Aa'^ + Ba + C. In like manner the last
sum is obtained.
Now A and the first two sums are respectively the coefficients of
x"^, X, and x'' in the quotient obtained above by the ordinary method,
and the last sum is the remainder.
In like manner any rational integral function of x can be divided
by X — a. If any power of x is missing, its coefficient is zero, and
must be written in its place with the other coefficients.
Observe that the shorter or synthetic method of division
includes only that part of the usual method given above
which is in black-faced type.
444
THEORY OF EQUATIONS 445
Since we omit the sign — before the second term of the
divisor, we must omit also that sign before the second term
of each product, and then add that term to the dividend, as
in the shorter method above.
Here the remainder Aa^ + Bar -{■ Ca -\- D is the value of the divi-
dend Ax^ + Bx:^ + Cx-t D for x = A, which affords a second proof
of § 131.
Ex. 1. Divide 2 x* + a;^ - 29 x2 - 9 x + 180 by x - 4.
"Write the coefiBcients with 4 at their right and proceed as below :
2 +1 -29 - 9 +180|4
4-8 +36 +28 +76
2 +9 +7 +19 +256
Hence the quotient = 2 x** + 9 x^ + 7 x + 19,
and the remainder, or /(4),= 256.
Ex. 2. Divide 2 x* + x^ - 29 x^ - 9x + 180 by x + 5.
2 +1 -29 - 9 + 180 1 - 5
_ 10 +45 - 80 +445
2 - 9 +16 -89 +626
Hence the quotient = 2 x^ - 9 x^ + 10 x - 89,
and the remainder, or /( - 5), = 625.
Ex. 3. Divide x^ + 21 x + 342 by x + 6.
1 +0 +21 +342
_ 0 +36 - 342
1 _6 +57 0
Hence the quotient = x^ — 6 x + 57,
and the remainder, or /(— 6),= 0.
Hence the division is exact, and x + 6 is a factor of /(x).
Exercise 151.
By Horner's method
1. Divide x^ -2x^ - 4.X + 8 by x-3; by x - 2.
2. Divide 2a;^ H-4a^-ar^- 16 a; - 12 bya;-|-4; by x + 3.
446 ELEMENTS OF ALGEBRA
3. Divide 3 a;* - 27 aj^ + 14 a? + 120 by x-%; by a; + 5.
4. Find the value of 2 a;* — 3 a^^ + 3 a? — 1 when a; = 4 ;
when a; = — 3 ; when a; = 3 ; when x = 5.
5. Show that one factor of a;^ -f 8 x-^ + 20 a; + 16 is x -{-2,
and from the quotient find the others.
6. Show that two factors of a^* + a^ - 29 a;^ - 9 a; + 180
are x — 3 and a; -f 3, and find the others.
7. Show that two factors of a?^— 4 ar^— 8 a; + 32 are x — 2
and a; — 4, and find the others.
INTEGRAL RATIONAL EQUATIONS IN ONE UNKNOWN.
470. If all the terms of an integral rational equation in x
are transposed to the first member and arranged in descend-
ing powers of x, we shall obtain an equivalent equation of
the form
Aa^" + ^1^"-' + ^l2^"-' + • • • + ^n-l^ + -4n = 0, (B)
where Aq, Ai, A2, •••, A^-i, A^ denote any known numbers,
real, imaginary, or complex, and n denotes the degree of the
equation.
Denoting the first member of (B) by f(x), (B) can be
written
471. To solve equation (B), or f(x) = 0, by § 149 we
need to factor its first member, equate each factor to zero,
and solve the resulting equations. But when (JB) is above
the second degree in x, the first member cannot be factored
by inspection except in certain special cases.
The methods which follow should be used when, and only
when, f(x) cannot be factored by inspection.
472. If a is a root of the equation f(x) = 0, that is, if
f{a) = 0, theii f{x) is divisible by x—a (§ 131).
THEORY OF EQUATIONS 447
Conversely, if f(x) is divisible by x — a, then f(a) = 0 ; tJiat
is, a is a root of the equation f{x) — 0.
E.g.^ if 2 is a root of the equation
a;3_2x-2-4a; + 8 = 0, (1)
then its first member is divisible by x — 2 (§ 132).
Conversely, if the first member of (1) is divisible by a; — 2, then 2 is
a root of this equation.
473. It was proved in § 148 that n linear equations in x
are jointly equivalent to an equation of the ?ith degree in x.
In proving the converse of this theorem in the next article
we assume the following theorem :
Any integral rational equation in one unknown has at least
one root, real, imaginary, or complex.
Note. The proof of this theorem is too long and difficult to be
given here.
474. Any equation of the nth degree in one unknown has n,
and only n, roots.
Pi'oof By § 473, the equation f{x) — 0 has a root.
Let rtj denote this root ; then, by § 472, f{x) is divisible
by a; — Oj, so that
f{x) = {:x-a,)f{x), (1)
in which, by the laws of division, f (x) has the form of f{x),
and is of the {n — l)th degree.
Now the equation f (x) = 0 has a root.
Denote this root by a^ ; then
f{x) = (x-a.;)f(x), (2)
in which f2(x) is of the (?i — 2)th degree.
Repeating this process n — 1 times, we finally obtain
fn-l(x) = (^-Ctn)A^, (n)
where A^ is the coefficient of x'' in f(x).
448 ELEMENTS OF ALGEBBA
From (1), (2), •••, {n), we obtain
f(x) = {x-a,)f^{x)
= (x-a,)(x-ao)f2(x)
= (x~ ai) (x — as) (^* — ttg) • • • (a? — a„) A- (3)
Hence the equation f(x) = 0 is equivalent to the n linear
equations
X — ai = Oy a; — 0-2 = 0, • • •, x — «„ = 0,
and therefore has n and only n roots.
From (3), it follows that any expression of the 71th. degree
in X can be resolved into n linear factors in x.
475. Equal roots. If two or more of the factors x — a^,
X — a2, '•', X — a^ are equal, the equation /(x) = 0 has two
or more equal roots.
E. g. , of the equation
(x-4)3(a; + 5)2(a;-7) = 0,
three roots are 4 each, and two are — 5 each.
Ex. One root of 2 ic^ - 5 x^ - 37 x + 60 = 0 is 5. Find the others.
One root being 5, one factor of /(oj) is x — 5 (§ 472).
By division the other factor is found to be 2 x^ + 5 x — 12.
Hence the two roots required are those of the equation
2x2 + 5x-12^0. (1)
The roots of (1) are evidently — 4 and 3/2.
Exercise 152.
Solve each of the following equations :
1 . cc^ — 6 £c^ + 10 a; — 8 = 0, one root being 4.
2. 3 a^ — 25 .^•2 + 42 ic 4- 40 = 0, one root being 5.
3. 2 ic^ 4- a;^ — 15 a; — 18 = 0, one root being — 2.
4. 3 a^ - 8 a;2 - 31 a; + 60 = 0, one root being - 3.
THEORY OF EQUATIONS 449
5. 4 a^ - 9 x^ — 3 a; -h 10 = 0, one root being — 1.
6. a;^ + a^ - 29 x2 - 9 X 4- 180 = 0, two roots being 3 and
-3.
7. a;'' — 4a;^ — 8x4-32 = 0, two roots being 2 and 4.
8. 2 x^ — 15 x^ 4- 35 x^ — 30 a; + 8 = 0, two roots being 1
and 2.
9. 3 a;^ — 5 a;'' - 17 ^2 4- 13 X 4- 6 = 0, two roots being —2
and 3.
By § 148, form the equation whose roots are :
10. The two numbers, ± V— 2.
11. The four numbers, ±V— 3, ± V— 5.
12. The four numbers, 3 ± V^^, 5, — 2/3.
13. 3/4, liV^Ts, l-t-V^=^.
14. 2, ±V^^, 3±V^^.
15. 3, -4, V^^.
In each of the last six examples, observe that the coefficients of the
equation obtained are all real when, and only when, the imaginary or
complex roots occur in conjugate pairs. This illustrates the converse
of the next article.
476. In any integral rational equation having only real
coefficients^ imaginary or complex roots occur in conjugate
pairs; that is, if a 4- bi is a root, then a — bi is also a root.
Proof. If the coefficients in f(x) are all real, then all the
terms of the expression obtained by substituting a 4- hi for
X in fix) will be real except those containing odd powers of
hi, which will be imaginary.
Representing the sum of all the real terms by A, and the
sum of all the imaginary terms by Bi, we have
f{a + hi) = AJrBi. (1)
450 ELEMENTS OF ALGEBRA
Now f(a — bi) will evidently differ from f(a + bi) only in
the signs before the terms containing the odd powers of bi ;
that is, in the sign before Bi ; hence
f(a - bi) = A- Bi. (2)
Since a + bi is a root of f(x) — 0, from (1) we have
A + Bi = 0.
Therefore ^ = 0 and ^ = 0. § 279
Hence by (2), f(a - bi) = 0.
That is, when a -\- bi is a root of f(x) = 0, a — bi is also a
root.
Ex. One root oix^ - 4x^ + 4x - S = 0 (1)
is (1 + V^3)/2 ; find the others.
Since 1/2 + aA^/2 is a root, 1/2 - V^^/2 is also a root (§ 476).
Hence two factors of the first member of (1) are
X - 1/2 - V^/2 and X-1/2+ V^^/2,
whose product is (x — 1/2)^ + 3/4, or x^ — x+1.
But cc3 - 4ic-2 + 4rB - 3 = (x2 - ic + 1) (a; - 3); (2)
hence the third root of (1) is 3.
Identity (2) illustrates the following principle :
477. Any rational integral function of x wliose coefficients
are real can be resolved into real factors, linear or quadratic
in X.
Proof. If the coefficients of f{x) are real, the imaginary
or complex roots of f{x) = 0 occur in conjugate pairs, as
a 4- bi and a — bi-^ hence the complex factors of f{x) occur
in conjugate pairs, as x — a — bi and x — a-{- bi, whose
product is a real quadratic expression in x-, that is
(x — a—bi) (x - a H- bi) = {x - ay -|- b^.
THEORY OF EQUATIONS 451
Exercise 153.
Solve each of the following equations, and find the real
factors of the first member :
1. a^ - 6 ar' + 57 a; - 196 = 0, one root being 1 - 4 V^3.
2. a:^ — 6 a; + 9 = 0, one root being (3 + V^^)/2.
3. ar' - 2 a^ + 2 a; - 1 = 0, one root being (1 + V^r3)/2.
4. a;*-f4aj^ + 5ar^-|-2a; — 2 = 0, one root being — 1 + ^.
5. a;*4-4a^4-6a^ + 4a; + 5 = 0, one root being i.
6. a^ — ar* 4- a;'' — a.*^ + a; — 1 = 0, two roots being — i and
(l + V^/2.
7. Show that in an equation with commensurable real
coefficients, surd roots occur in conjugate pairs; that is, if
a + -y/b is a root of f(x) = 0, a — ->/6 is a root also, -y/b being
a surd number.
All the terms in /(a + ^/b) will be rational except those containing
odd powers of -^6, whiph are surd. Denote the sum of all the rational
terms by A and the sum of all the surd terms hy B^b ; then
/(a+ y/b) = A + By/b.
Hence /(a — y/b) = A — By/b; and so on as in § 476.
8. Solve 6 a^ — 13a;^ — 35 ar^ — a; + 3 = 0, one root being
2-V3.
9. Solve X*- 36 a^+ 72 a; - 36 = 0, one root being 3 - V3.
478. The graph of /(») illustrates the fact that equal real
roots form the connecting link between unequal real roots
and imaginary or complex roots, and that imaginary or com-
plex roots occur in pairs.
E.g., by slightly diminishing the term 4 of the function x^—Sx^-\-i,
its graph in fig. 4 of § 46.3 would be moved downward, and would
then cut the axis of x in three points ; by slightly increasing the term
452 ELEMENTS OF ALGEBRA
4, the graph would be moved upward, and would then cut the axis of x
in hut one point.
That is, the two equal real roots of the equation
x3 _ 3 x2 + 4 = 0
would become unequal real roots or complex roots according as the
known term 4 were diminished or increased.
From fig. 5 in § 463 the pupil should follow the changes in the
roots of the equation
ic4 + x3 - 3 x2 - a; + 2 = 0,
(i) when the term 2 is decreased continuously to - 1 ;
(ii) when the term 2 is increased continuously to 4.
479. An equation of the form (B) in § 470, is said to be
in the type-form when the coefficient of ic" is 1.
E.g., ic4 _ I a;3 + 3 ^.2 _^ 4 _ 0 is in the type-form.
480. If an equation of the nth degree is in the type form, then
— the coefficient of x""'^ = the sum of the roots;
the coefficient of x""'^ = the sum of the products of the
roots taken two at a time;
— the coefficient of jr"~^ = the sum of the products of the
roots taken three at a time.
(— 1)** (the coefficient of x^) = the product of the n roots.
Proof. Let a^, a2, a^, ••• a„ denote the n roots; then, by
§ 148, the equation can be written in the form
(x — tti) (x — ag) (x — as)"'{x— a„) = 0. (1)
When n = 2, by multiplication (1) becomes
a^ — (ai + ttg) ^ + cti<^2 = 0,
which proves the theorem when n = 2.
When 71 = 3, by multiplication (1) becomes
^ — («i + «2 4- cis)^ + (<^i«2 + «i0t3 + a2a3)x — aia^^ = 0, (2)
which proves the theorem when ?i = 3.
THEORY OF EQUATIONS 45B
From the laws of multiplication it is evident that the
same relation holds when w = 4, 5, 6, •••.
Observe that, if the term in a;""^ is wanting, the sum of
the roots is 0, and ?.f the kno^n term is wanting, at least
one root is 0.
E.g., in the equation
x4 + 6«2_iix-6 = 0,
the sum of the roots is 0 ; the sum of their products taken two at a
time is 6 ; the sum of their products taken three at a time is 11 ; and
their product is — 6.
Note. The coeflBcients in any equation are functions of the roots ;
and conversely, the roots are functions of the coefficients. The roots
of a literal quadratic equation have been expressed in terms of the
coefficients (§ 291). The roots of a literal cubic or biquadratic equa-
tion can also be expressed in terms of the coefficients, as is shown in
college algebra. But the roots of a literal equation of the fifth or
higher degree cannot be so expressed, as was proved by Abel in 1825.
Ex. Its roots being in arithmetic progression, solve
4 x3 - 24 x=« + 23 X + 18 = 0. (1)
Let a denote the second term in the A. P. and b the difference ;
then the three roots are a — b, a, a + b. Hence their sum is 3 a ;
the sum of their products taken two at a time is 3 a^ — b'^ ; and their
product is a{a^ — b^).
Divide (1) by 4 to reduce it to the type-form; then, by §480,
we have
3 a = 6, 3 a2 - 62 = 23/4, a(a^ -b^) = - 9/2. (2)
Solving the first two equations in (2), we obtain a = 2, b = ± 5/2 j
and these values are found to satisfy the third equation in (2).
Hence the roots are — 1/2, 2, and 9/2.
Exercise 154.
1. The sum of two of its roots being zero, solve
The sum of the three roots is — 4 ; hence the third root is — 4.
2. Its roots being in arithmetic progression, solve
454 ELEMENTS OF ALGEBRA
3. Its roots being in geometric progression, solve
3 0^3 _ 26 a^ + 52 a: - 24 = 0.
4. One root being 1 —V— 3, solve
aj3-4«2^g^_3^Q^
One root being 1 — V— 3, a second root is 1 + V— 3.
The sum of these two roots is 2, and the sum of the three roots
is 4 ; hence the third root is 2.
5. By § 480, solve each of the first five examples in
exercise 153.
481. If the coefficients of /(a?) are all +, /(a;)> 0 when
a;>0; hence, if the coefficients off{x) are all positive, f(x) = 0
has no positive real root.
If the coefficients of f(x) are alternately + and — ; then,
when a;<0, /(aj)>0 or <0 according as n is even or odd;
hence, if the coefficients of f{x) are alternately 4- a7id — ,
f (jr) = 0 has no negative real root.
If the sum of the coefficients of f(x) is zero, /(I) = 0 ;
hence, when the sum of the coefficients of f(x) is zero, one root
of f(x) = Ois -\- 1.
E.g., x^ + 6 x^ -\- llx -\- 6 = 0 has no positive root, since /(x)> 0
when OS > 0.
o;^ — 6 a;2 _|- 10 X — 8 = 0 has no negative root ; since /(x)< 0 when
aj<0.
/k4 + 2 ic3 - 13 ic2 - 14 X + 24 = 0 has + 1 as a root ; since /(I) = 0.
482. If all the coefficients of an equation in the typeform
are whole numbers, any commensurable real root of the
equation is an integral factor of its known term.
E.g., any commensurable real root of the equation
a;3 _6a;2 + i0a;-8 = 0
is an integral factor of its known term — 8 ; that is, any such root is
±1, ±2, ±4, or i8.
THEORY OF EQUATIONS 455
Proof. Let all the coefficients of the equation
a-" + Aix""-^ + A^x""-^ -i \-A^ = 0 (1)
be whole numbers, and suppose that s/t, a fractional num-
ber in its lowest terms, is one of its roots.
Substituting s/t for x, we obtain
oft ©n-l ©n— 2
Multiplying by r~^, and transposing, we obtain
s^^/t = - (Ais^-^ + A^s""-^ + • • • 4- A„r-^) (2)
Now (2) is impossible, for its first member is a fractional
number in its lowest terms, and its second member is a
whole number.
Hence a fractional number cannot be a root, and there
fore any commensurable root must be a whole number.
Next, let a be an integral root of (1).
Substituting a for ic, transposing A^, and dividing by a,
we have
a"-i +A,a--'' + ^,a-« + • • • + A,,_, = - AJa. (3)
The first member of (3) is a whole number; hence the
quotient AJa is a whole number, ^.e., a is an integral
factor of A^.
Ex.1. Solve a;8 - 6x2 + 10x- 8 = 0. (1)
By § 481, (1) has no negative root ; hence, by § 482, any commen-
surable real root of (1) is + 1, + 2, + 4, or + 8, i.e. it is one of the
positive integral factors of 8.
The work of determining whether -)- 4 is a root can be arranged
as below :
1 _6 +10 -S[±
+4 - 8 +8
1-2+2 0
The division is exact, and the quotient is x^ — 2 a; + 2.
Hence the roots of (1) are 4 and the roots of
x2-2x + 2 = 0. (2)
456 ELEMENTS OF ALGEBRA
Solving (2), x = l ± V^.
Hence the roots of (1) are 4 and 1 ± V— 1.
Ex.2. Solve a;* + 2 ic3 - 13x2- 14^ + 24 = 0. (1)
By § 481, one root of (1) is + 1, and by § 482 any other commen-
surable real root is
±1, ±2, ±3, ±4, ±6, ±8, ± 12, or ± 24,
i.e. it is one of the integral factors of 24.
1
+ 2
+ 1
-13
+ 3
-14
-10
+ 24|J_
-24
1
+ 3
-2
-10
- 2
-24L
+ 24
-2
1 +1 - 12
Hence the roots of (1) are 1, — 2, and the roots of
a;2 + a; - 12 = 0.
Hence the roots of (1) are 1, — 2, 3, and — 4.
Usually it is better to try the smaller factors of An first.
Exercise 155.
Solve each of the following equations :
1. i^ + 2x' + 9x-{-lS = 0.
2. a^-6a^-{-llx-6 = 0.
3. a^-4.x^-6x-{-9 = 0.
4. x^-3a^-{-x'-^2x = 0.
5. x^-Sa^-\-13x-6 = 0.
6. aj3 + 6a;2-f-9aj + 2 = 0.
7. a^ + 5a^-9a;-45 = 0.
8. x^-Aa^-Sx-^32 = 0.
9. x*-6x^ + 2Ax-16 = 0.
10. x^-SiK^-Ux'-^4Sx-32 = 0.
11. a^-3x*-9a^-\-21x'-10x-\-24. = 0.
12. a^-\-2x^-23x-60 = 0.
THEOBY OF EQUATIONS 467
483. Limits of real roots. Superior limit. In evaluating
/(4) in example 1 of § 469, the sums are all positive, and
they evidently would all be greater for x > 4.
Hence f{x) can vanish only for x < 4 ; and therefore all
the roots of f(x) = 0 are less tlian 4.
Hence, if in computing the value of f('^c) all the sums are
positive, the real roots off(x) = 0 are all less than +(?.
The least integral value of +c which fulfils this condition
is called the superior limit of the real roots of f(x) = 0.
Inferior limit. In evaluating /(— 5) in example 2 of
§ 469, the sums are ^ternately — and -f, and they evidently
would all be greater arithmetically for x< — 5. Therefore
all the real roots of f(x) = 0 are greater than — 5.
Hence, if in computing the value of f{~b) the sums are alter-
nately — and -}-, all the real roots of f{x) = 0 are greater
than ~b.
The greatest integral value of ~h which fulfils this condi-
tion is called the inferior limit of the real roots of f(x) — 0.
Observe that the above reasoning holds when we regard a
zero sum as either positive or negative, and that when the
last sum is zero, the limit obtained is itself a root.
E.g., if /(x) = x* + 2x8- 13a:2-14a; + 24 = 0; (1)
then in evahiating/(4), the sums are all + ; and in evaluating/(— 5),
the sums are alternately — and + ; hence the real roots of f(x) = 0 lie
between — 5 and 4.
Hence, by § 482, any commensurable roots of (1) must be
±1, ±2, ±3, or -4.
Compare this result with example 2 in § 482.
Exercise 156.
1. Show that any commensurable real root of
a:3_2a:_50 = 0
lies between — 2 and 4 ; and hence is ± 1 or 2.
458 ELEMENTS OF ALGEBRA
2. Show that any commensurable real root of
is ± 1, ± 2, ± 4, ± 5, ± 8, or 10.
3. Show that any commensurable real root of
a;4_i5a^-|.10a; + 24 = 0 (1)
is ±1, ±2, ±3, or -4.
4. Eind the roots of equation (1) in example 3.
Solve each of the following equations:
5. x*-9ic^ + lTa;'^+27a;-60 = 0.
6. «^-45a^-40x + 84 = 0.
7. a^_4a;*-16i»« + 112a;2-208a; + 128 = 0.
8. ic4-ar^~39a;2 + 24ic + 180 = 0.
9. a^ + 5ar^ - 81 a;^ - 85 ar^ + 964a^ -f 780aj - 1584 = 0.
10 . ic^ + aj« - 14 a^ - 14 a;^ 4- 49 aj^ + 49 a^ - 36 a; = 36.
11. aj6-10x4-3a.-2 + 108 = 0.
12. a;«-2ar^-7a;* + 20aj3-21a^-18a;-f-27 = 0.
484. To transform an equation into another whose roots
shall be some multiple of those of the given one.
Proof. If in the equation
of" + ^ix"-i + A^x^-^ + ^3X"-3 H h ^„ = 0, (B)
we put X = Xi/a, and multiply by a", we obtain
a;i« + ^laa^i"-^ + ^gOt V~' + ^sO^V"^ + • • • + -4^ = 0. (2)
Since Xi = aa;, the roots of (2) are a times those of (1).
Hence, to effect the required transformation, multiply the
second term of (B) by the given factor, the third term by its
square, and so on.
Observe that before the rule is applied the equation must
THEORY OF EQUATIONS 459
be put in the type-form, and any missing power of x must
be written with zero as its coefficient.
This theorem becomes evident also when we observe that
if in equation (2) in § 480 each root is multiplied by a, the
second term will be multiplied by a, the third term by a^,
and the fourth term by a\
The chief use of this transformation is to clear an equa-
tion of fractional coefficients.
Ex. Solve the equation
x3-.i^x2 + fx-x^^ = 0, (1)
first transforming it into another with integral coefficients.
Multiplying the second term by a, the third by a-, the fourth by a^,
we obtain
x3 _ J^ ax2 -f f a^x - i^« a' = 0. (2)
By inspection we discover that 4 is the least value of a which will
render the coefficients of (2) integral. Putting a = 4, we obtain
a;8 _ 11 x2 + 36 X - 36 = 0. (3)
The roots of (3) are found to be 2, 3, and 6.
But the roots of (3) are four times the roots of (1); hence the roots
of (1) are 1/2, 3/4, and 3/2.
Exercise 157.
Solve the following equations by transforming them into
others whose commensurable real roots are whole numbers :
1. a^-^x^-^^ix + ^\ = ().
2. a^-x'/4:-x/2-\-l/S = 0.
3. 8a^-26a;2-hllx-hl0 = 0.
4. a^- x^/3 - x/m + 1/108 = 0.
5. 24:3^ -520^ + 26x-S=-0.
6. 9x*-9a^ + 5x'-3x-^2/S = 0.
7. x' - ^76 - ar^/12 - 13 a^/24 + 1/4 = 0.
8. 2a;^-12ar^ + 19a:'-6a;4-9 = 0.
460 ELEMENTS OF ALGEBRA
485. If f(a) and f(b) are opposite in quality, an odd number
of real roots off(x) = 0 lies between a and b.
Iffip) and fib) are like in quality, no real root, or an even
number of real roots of f(x) = 0 lies between a and b.
Proof If the ordinates of two points in the graph of fix)
are opposite in quality, the points are on opposite sides of
the a^axis, and the part of the graph between these points
must cross that axis an odd number of times (§ 466) ; that
is, f{x) is zero for an odd number of values of x between a
and 6.
If the ordinates of two points are like in quality, the
points are on the same side of the avaxis, and the part of
the graph between these points either does not cross that
axis or crosses it an even number of times, touching it being
regarded as crossing it twice.
E.g., in fig. 3 of § 463, the graph cuts XJT an odd number of
times between A and B or A and Z>, and an even number of times
between A and C or B and D.
In fig. 5 of § 463, the graph cuts XXf an odd number of times
between A and B or B and E, and an even number of times between
A and (7, C and E, or A and E.
Ex. Find the first figure of eacli real root of the equation
ic3 _ 4 a;2 _ 6 ic + 8 = 0. (1)
By §§ 474 and 476, (1) has either three or only one real root.
By Horner's method we find that : -
when x = -2, -1, 0, 1, 2, 3, 4, 5,
f(x)=-4, +9, +8, -1, -12, -19, -16, +3.
Since /(— 2) and /(— 1) are opposite in quality, at least one root
of (1) lies between — 2 and — 1. For like reason a second root lies
between 0 and 1, and a third between 4 and 6.
Hence two roots are -(!• + ) and 4- + , and, since /(0.9) is +
and /(I) is -, the third root is 0.9 +.
THEORY OF EQUATIONS 461
486. Any equation of an odd degree in which Aq is positive
has at least one real root whose quality is opposite to that of
its known term A^.
Proof If J-o > 0 and f{x) is of an odd degree, then
/(-^) is -, /(O) = A, /(+^) is +.
Hence if A^ is positive, one root of f(x) = 0 lies between
—00 and 0 (§ 485) ; and if A^ is negative, one root lies
between 0 and +qo.
487. Any equation of an even degree in which A^ is positive
and the known term A^ is negative has at least one positive
and one negative real root.
Proof If A > 0 and f{x) is of an even degree, then
/(-oo) is +, /(O) is -, X+^) is +.
Hence one root of f{x) = 0 lies between — oo and 0, and
another between 0 and +qo.
Exercise 158.
Find the first figure of each real root of the equations :
1. a^-3x2-4a:4-ll = 0. 5. a^-2aj-5 = 0.
2. a^ + a^-2x-l = 0. 6. a:^ ^_ a; _ 50O = 0.
3. x*-4:2i^-3x-\-2S = 0. 7. x^ -\- 10 x" + 5 x = 260.
4. x^-4:X--6x = -S. 8. a;*-12a.-2+12a:-3 = a
48 FRENCH.
The Qa Ira Series of French Plays.
Edited by Professor B. W. Wells, of the University of the South.
6 volumes, i6mo, cloth. Each, 36 cents.
THE plays selected have not heretofore been edited for use in
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reading of these plays will help the student not only in the study
of French, but also in the development of a literary taste.
The following works are contained in the series : —
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SCIENCE. 51
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56 SCIENCE.
Herbarium and Plant Descriptions.
Designed by Professor EDWARD T. NELSON, late of Ohio Wesleyan
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The preliminary matter gives full directions for collecting,
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The portfolio is strong, durable, and attractive in appearance.
In the class-room and in the field this work has been found
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Professor D. P. Penhallow, McGill University, Montreal, Can.: The idea
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SCIENCE. hi
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58 MA THEM A TICS.
Elements of Algebra.
By Professor JAMES M. TAYLOR, Colgate University, Hamilton, N.Y.
At Press.
IN this book Professor Taylor aims primarily at simplicity in
method and statement, and at a natural and logical sequence
in the series of steps which lead the pupil from his arithmetic
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An introductory chapter explains the meaning and object of
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