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Elements of Algebra
BY
G. A. ^ENTWORTH, A.M.
AIITHOB OF A SBBUU OF TBXI-BOOKS IN MATHXMATICB
COMPLETE EDITION
BOSTON, U.S.A.
GINN & COMPANY, PUBLISHERS
1896
1
Griitis
Entered, according to Act of Congress, in the year 1881, by
G. A. WENTWORTH
in the OflBce of the Librarian of Congress, at Washington.
V
PREFACE.
THE single aim in writing this volume has been to make an
Algebra which the beginner would read with increasing in-
terest, intelligence, and power. The fact has been kept constantly
in mind that, to accomplish this object, the several parts must be
presented so distinctly that the pupil will be led to feel that he
is tnastering the subject. Originality in a text-book of this kind
is not to be expected or desired, and any claim to usefulness must
be based upon the method of treatment and upon the number
and character of the examples. About four thousand examples
have been selected, arranged, and tested in the recitation-room,
and any found too diflGlcult have been excluded from the book.
The idea has been to furnish a great number of examples for
practice, but to exclude complicated problems that consume time
and energy to little or no purpose.
In expressing the definitions, particular regard has been paid to
brevity and perspicuity. The rules have been deduced from pro-
cesses immediately preceding, and have been written, not to be
committed to memory, but to furnish aids to the student in fram-
ing for himself intelligent statements of his methods. Each prin-
ciple has been fully illustrated, and a sufficient number of problems
has been given to fix it firmly in the pupil's mind before he pro-
ceeds to another. Many examples have been worked out, in order
to exhibit the best methods of dealing with different classes of
problems and the best arrangement of the work ; and such aid has
been given in the statement of problems as experience has shown
to be necessary for the attainment of the best results. General
demonstrations have been avoided whenever a particular illustra-
tion would serve the purpose, and the application of the principle
to similar cases was obvious. The reason for this course is, that
IV PRBFACB.
the pupil must become familiar with the separate steps from par-
ticular examples, before he is able to follow them iu a general
demonstration, and to understand their logical connection.
It is presumed that pupils will have a fair acquaintance with
Arithmetic before beginning the study of Algebra ; and that suffi-
cient time will be afforded to learn the language of Algebra, and
to settle the principles on which the ordinary processes of Algebra
are conducted, before attacking the harder parts of the book.
"Make haste slowly" should be the watchword for the early
chapters.
It has been found by actual trial that a class can accomplish
the whole work of this Algebra in a school year, with one recita-
tion a day ; and that the student will not find it so difficult as to
discourage him, nor yet so easy as to deprive him of the rewards
of patient and successful labor. At least one-fourth of the year
is required to reach the chapter on Fractions; but, if the first
hundred pages are thoroughly mastered, rapid and satisfactory
progress will be made in the rest of the book.
Particular attention should be paid to the chapter on Factoring ;
for a thorough knowledge of this subject is requisite to success in
common algebraic work.
Attention is called to the method of presenting Choice and
Chance. The accomplished mathematician may miss the elegance
of the general method usually adopted in Algebras ; but it is be-
lieved that this mode of treatment will furnish to the average
student the only way by which he can arrive at an understanding
of the principles underlying these difficult subjects. In the
preparation of these chapters the author has had the assistance
and cooperation of G. A. Hill, A.M., of Cambridge, Mass., to
whom he gratefully acknowledges his obligation.
The chapter on the General Theory of Equations has been con-
tributed by Professor H. A. Howe of Denver University, Colorado.
The materials for this Algebra have been obtained from Eng-
lish, G^erman, and French sources. To avoid trespassing upon
the works of recent American authors, no American textrbook
has been consulted.
PBEFAGB. T
The author retnms his sincere thanks for assistance to Rev^ Dr.
Thomas Hill ; to Professors Samuel Hart of Hartford, Ct. ; C. H.
Judson of Greenville, S.C. ; O. S. Westcott of Racine, Wis. ; G. B.
Halsted of Princeton, N.J. ; M. W. Humphreys of Nashville,
Tenn. ; W. LeConte Stevens of New York, N.Y. ; G. W. BaOey
of New York, N.Y. ; Robert A. Benton of Concord, N.H. ; and to
Dr. D. F. Wells of Exeter.
There will be two editions of the Algebra : one of 360 pages,
designed for high schools and academies, will contain an ample
amount to meet the requirements for admission to any college ;
the other will consist of the Elementary part and about 150 pages
more, and will include the subjects usually taught in colleges.
Answers to the problems are bound separately in paper covers,
and will be furnished free to pupils when teachers apply to the
publishers for them.
Any corrections or suggestions relating to the work will be
thankfully received.
G. A. WENTWORTH.
EXBTBR, N.H.,
May, 1881.
CONTENTS.
CHAPTER I. Definitions:
Quantity and number, 1 ; numbers, 2 ; algebraic numbers, 4 ;
factors and powers, 7 ; algebraic symbols, 9 ; algebraic expres-
sions, 10 ; axioms, 11 ; exercise in algebraic notation, 18 ; simple
problems, 14.
CHAPTER II. Addition and Subtbaction:
Addition of algebraic numbers, 16 ; addition of monomials, 17 ;
addition of polynomials, 19; subtraction of algebraic numbers,
20 J subtraction of monomials, 21 ; subtraction of polynomials, 22 ;
parentheses, 25 ; simplifying algebraic expressions by removing
parentheses, 26 ; introduction of parentheses, 27.
CHAPTER ni. Multiplication:
Multiplication of algebraic numbers, 28 ; law of signs in mul-
tiplication, 29; multiplication of monomials by monomials, 30;
the product of two or more powers of a number, 30 ; multiplica-
tion of polynomials by monomials, 31 ; multiplication of polyno-
mials by polynomials, 32 ; special cases of multiplication ; square
of the sum of two numbers ; the square of the difference of two
numbers ; the product of the sum and difference of two numbers,
37 ; square of a trinomial, 39 ; the product of two binomials of the
form x + a and x + 6, 40 ; the powers of binomials of the form
a ±6, 42.
CHAPTER IV. Division :
Division of algebraic numbers, 44 ; division of monomials by
monomials, 46 ; division of powers of a number, 46 ; division of
I)olynomials by monomials, 48 ; division of polynomials by poly-
nomials, 49 ; use of parentheses in division, 63 ; special cases of
division ; the difference of two equal odd powers of two numbers
divisible by the difference of the numbers, 64 ; the sum of two
VIU ALGEBRA.
equal odd powers of two numbers divisible by the sum of the num-
bers, 66 ; the difference of two equal even powers of two numbers
divisible by the difference and by the sum of the numbers, 66 ;
the sum of two equal even powers of two numbers when each
exponent is the product of an odd and an even factor divisible by
the sum of the powers expressed by the even factor, 67 ; general
definitions of addition, subtraction, multiplication, and division, 68.
CHAPTER V. SiMPLB Equations :
Definitions, 69 ; transi)osition of terms, 60 ; solution of simple
equations, 60 ; problems in simple equations, 62.
CHAPTER VI. Factors :
Case in which all the terms of an expression have a common
simple factor, 68 ; case in which the terms of an expression can be
so arranged as to show a common compound factor, 69 ; resolution
into binomial factors of trinomials of the form of x^ + (a-\-h) x + db,
70 ; of the form of z'^—(a-\-b)x+db, 71 ; of the form of x^+(a—b)z
— ab, 72 ; of the form of x^ — (a — b)x — a6, 73 ; of the form of
x2 + 2ax + a2, 74 ; of the form of x^ — 2ax + a^, 76 ; resolution
of expressions of the form of two squares with the negative sign
between them, 76 ; resolution of the difference of two equal odd
powers, 78 ; resolution of the sum of two equal odd powers, 78 ;
resolution of the sum of two equal even powers, when possible,
79 ; resolution of trinomials of the form of x* + x'^y^-\- y*, 80 ; of
the form of 2x'^ + 6ax + 2a^, 81 ; resolution of polynomials which
are perfect powers, 82 ; resolution of polynomials composed of two
trinomial factors, 83 ; resolution of polynomials when a compound
factor of the first three terms is also a factor of the remaining
terms, 84.
CHAPTER VII. Common Factors and Multiples :
Highest common factor, 88; method of finding the highest
common factor by inspection, 88 ; method of finding the highest
common factor by division, 00 ; principles upon which this method
depends, 90 ; this method of use only to determine the compound
factor of the highest common factor, 92 ; modifications of this
method required, 93 ; lowest common multiple, 98 ; method of
finding the lowest common multiple by inspection, 98 ; method
of finding the lowest common multiple by division, 100.
CONTENTS, IX
CHAPTER YUL PaACTiONs:
Keduction of fractions to their lowest terms, 103; reduction
of fractions to integral or mixed expressions, 106 ; reduction of
mixed expressions to the form of fractioi^, 107 ; reduction of frac-
tions to the lowest common denominator, 110 ; addition and
subtraction of fractions, 112 ; multiplication of fractions, 120 ;
division of fractions, 122 ; complex fractions, 124.
CHAPTER IX. Practional Equations:
Reduction of fractional equations, 130 ; reduction of literal
equations, 184.
CHAPTER X. PsoBLEMS Probucing Fractional Equations, 137.
CHAPTER XL Simultaneous Equations of the First Degree,
161 ; elimination by addition or subtraction, 152 ; elimination by
substitution, 154 ; elimination by comparison, 155 ; literal simul-
taneous equations, 159.
CHAPTER XII. Problems Producing Simultaneous Equations,
166.
CHAPTER Xm. Involution and Evolution:
Powers of simple expressions, 181 ; law of exponents, 181 ;
powers of a binomial when the terms of the binomial have coeffi-
cients or exponents, 182 ; powers of polynomials by the binomial
method, 182 ; roots of simple expressions, 184 ; imaginary roots,
184 ; square roots of compound expressions, 186 ; square roots of
arithmetical numbers, 188 ; cube roots of compound expressions,
190 ; cube roots of arithmetical numbers, 193.
CHAPTER XIV. Quadratic Equations:
Pure quadratic equations, 196 j affected quadratic equations,
198; literal quadratic equations, 203 ; resolution of quadratic
equations by inspection, 206 ; number of roots of an equation,
208 ; formation of equations when the roots are known, 209 ;
determination of the character of the roots of an equation by in-
spection, 209 ; determination of the maximum or minimum value
Z AIiGEBBA.
of a quadratic expression, 211 ; higher equations which can be
solved by completing the square, 212 ; problems involving quad-
ratics, 214.
CHAPTER XV. SiMuETANBOus Quadratic Equations:
Solution when the value of one of the unknown numbers can
be found in terms of the other, 219 ; when each of the equations is
homogeneous and of the second degree, 222 ; when the equations
are symmetrical with respect to the unknown numbers, 223 ;
problems producing simultaneous quadratics, 226.
CHAPTER XVI. Simple Indetebminatb Equations:
The values of the unknown numbers dependent upon each
other, 228 ; method of solving an indeterminate equation in posi-
tive integers, 228.
CHAPTER XVII. Inequalities:
Fundamental proposition, 234; an inequality reversed by
changing the signs, 234.
CHAPTER XVIII. Theory of Exponents:
Fractional and negative exponents, 236; the meaning of a
fractional exponent, 237 ; the meaning of a negative exponent,
237 ; laws which apply to positive integral exponents apply also
to fractional and negative exponents, 238 ; exercise with mono-
mials having fractional and negative exponents, 239; exercises
with polynomials having fractional and negative exponents, 240 ;
radical expressions, 242 ; reduction of surds to their simplest
forms, 243 ; comparing surds of the same order, 244 ; comparing
surds of different orders, 246 ; addition and subtraction of surds,
247 ; expansion of binomials when the terms are radical expres-
sions, 249 ; rationalization of the denominator of a radical expres-
sion, 249 ; imaginary expressions, 251 ; square root of a binomial
surd, 262 ; equations containing radicals, 266 ; solution of an
equation with respect to an expression, 267 ; reciprocal equations,
268.
CHAPTER XIX. Logarithms:
Common system of logarithms, 261 ; the characteristic of a
logarithm, 263 ; the mantissa of a logarithm, 264 ; logarithm of
CONTENTS. XI
a product, 264 ; logarithm of a power and of a root, 265 ; logarithm
of a quotient, 266 ; a table of four-place logarithms, 270 ; general
proofs of the laws of logarithms, 276 ; solution of an exponential
equation by logarithms, 277.
CHAPTER XX. Ratio, Pkopobtion, and Variation:
Ratio, 278 ; commensurable and incommensurable ratios, 270 ;
theorems of ratio, 281 ; proportion, 284 ; theorems of proportion,
284 ; variation, 292 ; direct variation, 293 ; inverse variation, 293.
CHAPTER XXI. Series:
Infinite series, 299 ; finite series, 299 ; converging series, 300 ;
arithmetical series, 301 ; geometrical series, 308 ; limit of the sum
of an infinite geometrical series, 313 ; harmonical series, 314.
CHAPTER XXn. Choice. Binomial Theory:
Fundamental principle, 317; arrangements or permutations,
320 ; number of arrangements of n different elements taken all at
a time, 320 ; number of arrangements of n different elements taken
r at a time, 321 ; number of arrangements of n elements of which
p are alike, q are alike, etc. , 324 ; number of arrangements of n
different elements when repetitions are allowed, 325 ; selections or
combinations, 326 ; number of selections of r elements from n dif-
ferent elements, 326 ; the number of ways in which p elements
can be selected from p + r different elements the same as the num-
ber of ways in which r elements can be selected, 327 ; value of r
for which the number of selections of n different elements taken
r at a time is the greatest, 330 ; the number of ways of selecting
r elements from n different elements when repetitions are allowed,
336 ; the number of ways in which a selection can be made from
n different elements, 337 ; the number of ways in which a selec-
tion can be made from p + q + r elements of which p are alike,
q are alike, r are alike, 338 ; proof of the binomial theorem when
the exponent is positive and integral, 342 ; general formula for the
expansion of (a + a;)", 343 ; general term of the expansion of a
binomial, 344 ; greatest term of the expansion of a binomial, 346 ;
proof of the binomial theorem when the exponent is fractional,
346 ; when negative, 348 ; applications, 349.
Xll ALGEBRA.
CHAPTER XXnL Chance:
Theory of probabilities, 361 ; from several events of which only
one can happen, the chance that some one of them will happen,
362 ; expectation from an uncertain event, 367 ; compound events,
360; the probability that two independent events both happen,
361 ; the probability of a compound event, 361 ; the relative proba^
bility of doubtful events, 369 ; the actual happening of a doubtful
event changes the separate probabilities of the several ways of
happening in the same ratio, 370.
CHAPTER XXIV. Interest Formulas:
Simple interest, 375; compound interest, 376; sinking-funds,
377 ; annuities, 378 ; to find the amount of an unpaid annuity
when the time, interest, and rate per cent are given, 378 ; to find
the present worth of an annuity when the time it is to continue
and the rate per cent are given, 378 ; to find the present worth
of an annuity that begins in a given number of years, when the
time it is to continue and the rate per cent are given, 379; to
find the annuity when the present worth, the time, and the rate
per cent are given, 380 ; life insurance, 381 ; bonds, 381 ; to find
the rate of interest received on an investment in bonds, when the
current rate of interest, the market value of the bonds, the rate
of interest they bear, and the number of years they run are given,
381 ; to find the price that may be paid for bonds bearing a given
rate of interest, and having a given number of years to run, in
order that the purchaser may receive a given rate of interest on
his investment, the current rate of interest being known, 382.
CHAPTER XXV. Continued Fractions:
The form of a continued fraction, 386 ; any proper fraction in
its lowest terms may be converted into a terminated continued
fraction, 386 ; the successive convergents of a continued fraction
are alternately greater and less than the true value of the given
fraction, 386 ; the law for forming successive convergenta, 387 :
the difference between two consecutive convergents is equal to
the reciprocal of the product of the denominators of the two con-
vergents, 388 ; any convergent is in its lowest terms, 388 ; the
CONTENTS. XIU
successive convergents approach more and more nearly to the true
value of the fraction, 389 ; any convergent is nearer the true value
of the continued fraction than any other fraction with smaller
denominator, 389 ; a quadratic surd may be expressed in the form
of a non-terminating continued fraction, 390 ; periodic continued
fractions, 391 ; an exponential equation may be solved by con-
tinued fractions, 392 ; problems in continued fractions, 392.
CHAPTER XXVI. Theory of Limits:
Definitions of constant, variable, and limit, 394 ; a variable
may continually increase towards its limit, or continually decrease
towards its limit, or be sometimes greater and sometimes less than
its limit, 395 ; geometrical representation of the indefinite approach
of a variable to its limit, 396 ; if two variables are equal, and are
so related that a change in one produces such a change in the
other that they continue equal, and each approaches a limit, their
limits are equal, 396 ; if two variables have a fixed rUtio, and are
so related that a change in one produces such a change in the
other that they continue to have this ratio, and each approaches
a limit, their limits are in this same fixed ratio, 397 ; the limit of
the sum of two or more variables is the sum of their respective
limits, 398 ; the limit of the product of two or more variables is
the product of their respective limits, 398 ; the limit of any power
of a variable is that power of its limit, 399 ; the limit of the quo-
tient of two variables is the quotient of their limits, 399 ; the
limit of , when x approaches a as a limit, is na»—i, 400;
•C d
convergency of infinite series, 400; tests of the convergency of
series : if the terms of a series are all positive, and the limit of the
•nih term is 0, then if the limit of the ratio of the (n + 1) th term to
the rdh term is less than 1, the series is convergent, 402 ; if the
terms of a series are alternately positive and negative, and the
limit of the idh term is 0, then when the terms continually decrease
the series is convergent, 404.
CHAPTER XXVn. Indeterminate Coefficients:
Theorem of indeterminate coefi&cients, 406 ; definition of an
identical equation, 406 ; method of expanding an algebraic fraction
in the form of a series, 400 ; method of expanding an indicated
XIT AliGBBBA.
root of an expression in the form of a series, 407 ; method of
finding a fraction which when expanded will produce a given
series, 407 ; recurring series, 408 ; reversion of series, 408 ; resolu-
tion of a fraction into partial fractions, 409.
CHAPTER XXVIII. Thb Exponential Theokbm:
To expand a* in a series of ascending powers of x, 412 ; approx-
imate value of the constant e, 413 ; computation of logarithms to
the base e, 414 ; modulus of the common system of logarithms, 416 ;
computation of common logarithms, 416 ; when the difference of
two numbers is small in comparison with either of them, the
difference of their logarithms is proportional to the difference of
the numbers, 416.
CHAPTER XXIX. The Differential Method:
To express the (n + 1)^^ term of a given series in the first terms
of the successive orders of differences, 417 ; to express the sum of
n terms of a given series in the first terms of the successive orders
of differences, 418 ; piles of spherical shot : in the form of a tri-
angular pyramid, 421 ; in the form of a pyramid with square base,
421 ; in the form of a pyramid with a base rectangular but not
square, 422 ; series consisting of separable terms, 424 ; interpola-
tion of series, 426.
CHAPTER XXX. The Theory of Numbers:
Systems of notation, 428 ; to express any integral number in
the scale of r, 429 ; the common system of notation, 431 ; if a
prime number p is a factor of a6, and is not a factor of a, it is a
factor of 6, 432 ; a composite number can be separated into only
one set of prime factors, 432 ; a common fraction in its lowest
terms will not produce a terminating decimal if its denominator
contains any prime factors except 2 and 5, 433; divisibility of
numbers, 434.
CHAPTER XXXI. Imaginary Numbers:
Geometrical representation of real numbers, 436 ; geometrical
representation of imaginary numbers, 437 ; operations with imagi-
nary numbers, 488 ; complex numbers, 440 ; operations with
complex numbers, 441 ; imaginary exponent, 444 ; formulas
derived by expanding e«S 446.
C0KTENT8. XV
CHAPTER XXXn. Ixkji of Equatioks:
Geometrical representations of equations, 447; form of the
equation whose locus is a straight line, 460 ; an equation whose
locus is a circle, 450 ; an equation whose locus is an ellipse, 461 ;
an equation whose locus is a parabola, 461 ; an equation whose
locus is not a regular geometrical figure, 462 ; solution of two
simultaneous equations by constructing their loci, 462 ; method of
finding the roots of equations by constructing their loci, 464.
CHAPTER XXXni. Equations in General:
General equation of the idh degree, 467 ; F (a;) is exactly divi-
sible byx — aifoisa root of F (x) = 0, 468 ; if x — o is an exact
divisor of F{x), a is a root of F(x) = 0, 468; if a is a root of
F(x) = 0, it is an exact divisor of the absolute term, 460; syn-
thetic division, 460 ; il F{x) = (x — a) (x — b) (x — c), etc., a, 5, c,
etc. , are the roots of F{x) = 0, 461 ; if F (x) = 0 is of the nth degree,
it has n roots and no more, 461 ; the relations between the coeffi-
cients and the roots of an equation, 463 ; solutions by factoring,
464 ; Descartes' rule of signs, 467 ; when all the roots of a com-
plete equation are real, the number of positive roots equals the
number of variations of sign, and the number of negative roots
equals the number of permanences, 468 ; a rational fraction can-
not be a root of f(x) = 0, 470 ; imaginary roots enter equations in
conjugate pairs, 471; if F(x) = 0 is an odd degree, it has at least
one real root, 471 ; definition of a derivative, 472 ; to find the
derivative of a simple expression, 473 ; to find the derivative of a
compound expression, 474 ; the derivative of F{x) is the coefficient
of the first power of A in JP(x + ^), 476 ; the method of detecting
equal roots of F (x) = 0, 476 ; transformation of equations, 478 ;
the signs of the roots of /(x) = 0 are changed by changing the signs
of the alternate terms, beginning with the second, 479 ; the roots
of /(x) = 0 are multiplied by m by multiplying the second term by
w, the third by m^, and so on, 480 ; an equation is obtained with
roots the reciprocals of those of f{x) = 0 by reversing the order
of the coefficients, 480 ; method of transforming f(x) = 0 into an
equation whose roots are greater or less by h, 481 ; method of
transforming /(x) = 0 into an equation whose second term is
wanting, 482.
XVI AIiGEBBA.
CHAPTER XXXIV. Hiohbk Numbkical Equations:
The commensurable roots of /(x) = 0, 484 ; situation of incom-
mensurable roots, 486 ; if two values of z substituted in /(x) = 0
give results with unlike signs, at least one root lies between the
values ; but if the results have like signs, no root or an even num-
ber of roots lies between the values, 486 ; the value of /(x), when
X = m, is the remainder left from dividing /(x) by (x — m), 486 ;
method of finding figures of the roots by the principles of differ-
ences, 489 ; Sturm's theorem, 491 ; Horner's method of finding
other figures of the root when one or more figures have been
determined, 492 ; method of solving an equation of which the
root contains several figures in its integral part, 497 ; extraction
of roots by Homer's method, 498 ; general directions for solving
higher numerical equations, 499 ; recurring equations, 601 ; any
recurring equation is reducible to an equation of half its degree,
604 ; exponential equations, 606 ; Cardan's method for solving
cubic equations, 606 ; trigonometric solution of cubic equations,
608.
ELEMENTS OF ALGEBRA.
CHAPTER I.
Quantity and Number.
1. Whatever may be regarded as being made up of
parts like the whole is called a Quantity.
2. To measure a quantity of any kind is to find how
many times it contains another knovm quantity of the
same kind,
3. A known quantity which is adopted as a standard for
measuring quantities of the same kind is called a XTnit
Thus, the foot, the pound, the dollar, the day, are units
for measuring distance, weight, money, time.
4. A Number arises from the repetitions of the unit of
measure, and shows how many times the unit is contained
in the quantity measured.
5. When a quantity is measured, the result obtained is
expressed by prefixing to the name of the unit the number
which shows how many times the unit is contained in the
quantity measured ; and the two combined denote a quan-
tity expressed in units. Thus, 7 feet, 8 pounds, 9 dollars,
10 dayS; are quantities expressed in their respective units.
15 AL6EBBA.
When a question about a quantity includes the unit, the
answer is a number ; when it does not include the unit,
the answer is a quantity, Thus» if a man who has fifteen
bushels of wheat be asked Jiow rnany bushels of wheat he
has, the answer is the number , fifteen ; if he be asked how
much wheat he has, the answer is the quantity, fifteen
bushels.
A number answers the question. How many ? a quantity,
the question. How much ?
Numbers.
6. The symbols which Arithmetic employs to represent
numbers are the figures 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The
natural series of numbers begins with 0 ; each succeeding
number is obtained by adding one to the preceding number,
and the series is infinite.
7. Besides figures, the chief symbols used in Arithmetic
are :
+ (read, plus), the sign of addition.
— (read, minus), the sign of subtraction.
X (read, multiplied by), the sign of multiplication.
-T- (read, divided by), the sign of division.
= (read, is equal to), the sign of equality.
Exercise. — Read :
7 + 12 = 19. 8 + 3 — 5= 20 — 15 + 1.
9-4= 5. 24 + 6= 10 X 3.
6X 4 = 24. 14-7 + 5= 6X2.
48-h 3 = 16. 9X5 = 180-r 4.
8. Any figure, or combination of figures, as 7, 28, 346,
has one, and only one, value. That is, figures represent
XUMBBBS. 3
particular numbers. But numbers possess many general
properties, which are true, not only of a particular number,
but of all numbers.
Thus, the sum of 12 and 8 is 20, and the difference
between 12 and 8 is 4. Their sum added to their differ-
ence is 24, which is twice the greater number. Their
differeuce taken from their sum is 16, which is twice the
smaller number.
9. As this is true of any two numbers, we have this gen-
eral property : The sum of two numbers added to their differ-
ence is twice the greater number; the difference of two numbers
taken from their sum is twice the smaller number. Or,
1. (greater number + smaller number) + (greater number
— smaller number) = twice greater number.
2. (greater number + smaller number) — (greater number
— smaller number) =: twice smaller number.
But these statements may be very much shorteued ; for,
as greater number and smaller number may mean any two
numbers, two letters, as a and b, may be used to represent
them; and 2a may represent twice the greater, and 2b
twice the smaller. Then these statements become :
1. {a-\-h)-\-{a — b)=2a,
2. la-\-b) — {a — b)=2b.
In studying the general properties of numbers, letters
may represent any numerical values consistent with the
conditions of the problem.
10. It is also convenient to use letters to denote numbers
which are unknoum, and which are to be found from certain
given relations to other known numbers.
* ALGEBRA.
Thus, the solution of the problem, "Pind two numbers
such that, when the greater is divided by the less, the
quotient is 4, and the remainder 3 ; and when the sum of
the two numbers is iu creased by 38, and the result divided
by the greater of the two numbers, the quotient is 2 and
the remainder 2," is much simplified by the use of letters
to represent the unknown numbers.
11. The science which employs letters in reasoning about
numbers, either to discover their general properties, or to
find the value of an unknown number from its relations
to known numbers, is called Algebra.
Algebraic Numbers.
12. There are quantities which stand to each other in
such opposite relations that, when we combine them, they
cancel each other entirely or in part. Thus, six dollars
gain and six dollars loss just cancel each other ; but ten
dollars gain and six dollars loss cancel each other only in
part. Por the six dollars loss will cancel six dollars of the
gain and will leave four dollars.
An opposition of this kind exists in assets and debts, in
income and outlay, in motion forwards and backwards, in
motion to the right and to the left, in time before and after
a fixed date, in the degrees above and below zero on a
thermometer.
From this relation of quantities a question often arises
which is not considered in Arithmetic ; namely, the sub-
tracting of a greyer number from a smaller. This cannot
be done in Arithmetic, for the real nature of subtraction
consists in counting backwards, along the natural series of
numbers. If we wish to subtract four from six, we start
at six in the natural series, count four units backwards, and
ALGEBRAIC KUMBEBS. 5
arrive at two, the difference sought. If we subtract six
from six, we start at six in the natural series, count six
units backwards, and arrive at zero. If we try to subtract
nine from six, we cannot do it, because, when we have
counted backwards as far as zero, the natural series of
numbers comes to an end,
18. In order to subtract a greater number from a smaller
it is necessary to assume a new series of numbers, beginning
at zero and extending to the left of zero. The series to the
left of zero must ascend from zero by the repetition of the
unit, precisely like the natural series to the right of zero ;
and the opposition between the right-hand series and the
left-hand series must be clearly marked. This opposition
is indicated by calling every number in the right-hand
series a positive number, and prefixing to it, when written,
the sign + ; and by calling every number in the left-hand
series a negative number, and prefixing to it the sign — .
The two series of numbers will be written thus :
-4, -3,-2.-1, 0, +1, +2, +3, +4, ,
I I 1^ I \ i J I I
If, now, we wish to subtract 9 from 6, we begin at 6 in
the positive series, count nine units in the negative direction
(to the left), and arrive at — 3 in the negative series. That
is6 — 9 = — 3.
The result obtained by subtracting a greater number
from a less, when both are positive, is always a negative
number.
If a and b represent any two numbers of the positive
series, the expression a — b will denote a positive number
when a iis greater than b ; will be equal to zero when a is
equal to b ; will denote a negative number when a is less
than b.
If we wish to add 9 to —6, we begin at —6, in the
6 ALGEBRA.
negative series, count nine units in the positive direction
(to the right), and arrive at + 3, in the positive series.
We may illustrate the use of positive and negative num-
bers as follows :
-6 0 8 20
1 — I 1 H
DA C
Suppose a person starting at A walks 20 feet to the right
oiAy and then returns 12 feet, where will he be ? Answer:
at C, a point 8 feet to the right of A, That is, 20 feet
— 12 feet = 8 feet ; or 20 — 12 = 8.
Again, suppose he walks from A to the right 20 feet, and
then returns 20 feet, where will he be ? Answer : at A,
the point from which he started. That is, 20 — 20 = 0.
Again, suppose he walks from A to the right 20 feet, and
then returns 25 feet, where will he now be ? Answer : at
D, a point 5 feet to the left of A, That is, if we consider
distance measured in feet to the left of A as forming a
negative series of numbers, beginning at ^, 20 — 25 = — 5.
Hence, the phrase, 5 feet to the left of A, is now expressed
by the negative number — 5.
14. Numbers provided with the sign + or — are called
algebraic nninbers. They are unknown in Arithmetic, but
play a very important part in Algebra. In contradistinc-
tion, numbers not affected by the signs + or — are termed
absolute numbers.
16. Every algebraic number, as +4 or —4, consists of
a sign + or — and the absolute value of the number ; in
this case 4. The sign shows whether the number belongs
to the positive or negative series of numbers ; the absolute
value shows what place the number has in the positive or
negative series.
FACTORS AND POWERS. '
16. When no sign stands before a number, the sign + is
always understood ; thus, 4 means the same as + 4, a means
the same as + «• But the sign — is never omitted,
17. Two numbers which have one the sign + and the
other the sign — , are said to have unlike signs.
18. Two numbers which have the same absolute values,
but unlike signs, always cancel each other when combined;
thus +4 — 4 = 0, +a — a = 0.
19. The use of the signs + and — , to indicate addition
and subtraction, must be carefully distinguished from their
use to indicate in which series, the positive or the negative,
a given number belongs. In the first sense, they are signs
of operations, and are common both to Arithmetic and Al-
gebra. In the second sense, they are signs of opposition,
and are employed in Algebra alone.
Factors and Powers.
20. When a number consists of the product of two or
more numbers, each of these numbers is called a factor of
the product.
When these numbers are denoted by letters, the sign X
is omitted ; thus, instead of aXb, we write ab ; instead
of a X ft X c, we write abc.
The expression abc must not be confounded with a+ft+c;
the first is a product, the second is a sum. If a = 2, 6=3,
c = 4, then
aftc = 2x3X4 = 24;
a + ft + c = 2 + 3 + 4= 9.
21. Factors expressed by letters are called literal factors ;
factors expressed by figures are called numerical factors.
8 ALGE6KA.
22. A known factor of a product which is prefixed to an-
other factor, to show how many times that factor is taken,
is called a coefficient Thus, in 7 c, 7 is the coefficient of c;
in TaXy 7 is the coefficient of (ix, or, if a be known, 7 a is
the coefficient of x. When no numerical coefficient occurs
in a product, 1 is always understood. Thus, ax means the
same as lax.
23. A product consisting of two or more equal factors
is called a power of that factor.
24. The index or exponent of a power is a small figure or
letter placed at the right of a number, to show how many
times the number is taken as a factor. Thus, a*, or simply
a, denotes that a is taken once as a factor ; a* denotes that
a is taken twice as a factor ; a^ denotes that a is taken three
times as a factor; and a" denotes that a is taken n times as
a factor. These are read : the first power of a ; the second
power of a ; the third power of a ; the nth power of a,
a^ is written instead of aaa.
a" is written instead of aaa, etc., repeated n times.
The meaning of coefficient and exponent must be care-
fully distinguished. Thus,
4:a^a-\-a-{-a-]-a]
a* = aXaXaXa,
If a = 3, 4a =3 + 3 + 3 + 3 = 12.
a* = 3X3X3X3 = 81.
26. The second power of a number is generally called
the square of that number; thus, a* is called the square
of a, because if a denote the number of units of length in
the side of a square, a* denotes the number of units of
surface in the square.
ALGEBRAIC SYMBOL^. 9
The third power of a number is generally called the cube
of that number ; thus, a* is called the cube of a, because
if a denote the number of units of length in the edge of a
cube, a* denotes the number of units of volume in the cube.
Algebraic Symbols.
26. Known numbers in Algebra are denoted by figures
and by the first lettei*s of some alphabet ; as a, b, c, etc. ;
a', b*f c\ read a primes b primes c prime, etc. ; a^, bi, Cj, read
a one, b one, c one.
. Unknown numbers are generally known by the last
letters of some alphabet ; as ar, y, z, x\ y\ z\ etc.
27. The symbols of operations are the same in Algebra
as in Arithmetic. One point of difference, however, must
be carefully observed. When a symbol of operation is omit-
ted in the notation of Arithmetic, it is always the symbol
of addition : but when a symbol of operation is omitted in
the notation of Algebra, it is always the symbol of mul-
tiplication. Thus, 456 means 400 + 50 + 6, but 4 ab means
4 X a X ft ; 4| means 4 + j, but 4- means 4 X t-
28. The symbols of relation are =, >, <, which stand
for the words « is equal to," " is greater than," and " is
less than," respectively.
29. The symbols of aggregation are the bar, | ; the vin-
culum, ; the parenthesis, (); the bracket, [ ]; and
X
the brace, { }. Thus, each of the expressions, , ,
, a: + y.
(aJ+y), \^+y\y {^+y}> signifies that x+y is to be treated
as a single number.
10 ALGEBRA.
30. The symbols of continuation are dots, , or dashes,
, and are read, " and so on."
31. The symbol of deduction is .•., and is read "hence,"
or "therefore."
Algebraic Expressions.
32. An algebraic expression is any number written in
algebraic symbols. Thus, 8 c is the algebraic expression
for 8 times the number denoted by c.
If an algebraic expression contains only integral formSy
that is, contains no letter in the denominator of any of its
terms, it is called an integral expression.
33. A term is an algebraic expression the parts of which
are not separated by the sign of addition or subtraction.
Thus, 3a^, 5 a:?/, 3ab-T-5xi/ are terms.
34. A monomial or simple expression is an expression
which contains only one term.
35. A polynomial or compound expression is an expres-
sion which contains two or more terms. A binomial is a
polynomial of two terms. A trinomial is a polynomial of
three terms.
36. Like terms are terms which have the same letters,
and the corresponding letters affected by the same expo-
nents. Thus, Ta^cx^ and — 5a^cx^ are like terms ;^ut
7 a^cx^ and — ac^a^ are unlike terms.
37. The dimensions of a term are its literal factors.
38. The degree of a term is equal to the number of its
dimensions, and is found by taking the sum of the expo-
nents of its literal factors. Thus, Sxy is of the second
degree, and 5x^i/z^ is of the sixth degree.
AXIOMS. 11
39. A polynomial is said to be homogeneous when all its
terms are of the same degree. Thus, l7? — ba?y-\'xyz\s
homogeneous, for each term is of the third degree.
40. A polynomial is said to be arranged according to the
powers of some letter when the exponents of that letter
either descend or ascend in order of magnitude. Thus, ^aa^
— 4&i^ — 6gwc + 86 is arranged according to the descend-
ing powers of x, and 8Z> — 6aa; — 4ftx*-f-3ax* is arranged
according to the ascending powers of x.
41. The numerical value of an algebraic expression is the
number obtained by giving a particular value to each letter,
and then performing the operations indicated.
42. Two numbers are reciprocals of each other when their
product is equal to unity. Thus, a and - are reciprocals.
Axioms.
43. 1. Things which are equal to the same thing are
equal to each other.
2. If equal numbers be added to equal numbers, the
sums will be equal.
3. If equal numbers be subtracted from equal numbers,
the remainders will be equal.
4. If equal numbers be multiplied into equal numbers,
the products will be equal.
5. If equal numbers be divided by equal numbers, the
quotients will be equal.
6. If the same number be both added to and subtracted
from another, the value of the latter will not be altered.
7. If a number be both multiplied and divided by an-
other, the value of the former will not be altered.
12 ALGEBRA.
Exercise 1.
If a = l, b = 2, c^=S, c?=4, e = 5, /=0, find the nu-
merical values of the following expressions :
1. 9a + 2o + 3c — 2/. 4. -^-i — ; •
0 a er '
2. 46-3a-36 + 5c. 5. Te + hcd — ^^-
3. 8a^c — ^C6? + 9ce?e — ^/. 6. abc^ + bcd^— dea^+f^.
7. e* + 6e2^»2 + 6*-4e«^> — 4e^«.
8a* + 362 . 4c2 + 6ft» c^ + e?«
8.
a^^ ' c' — b^ e*
9. — • 11. ^ —
b' * ^>2_|.^2_^
10. ^-^- : 12. —
c6 — ^« - • e^^ed + d^
In simplifying compound expressions, each term must be
reduced to its simplest form before the operations of addi-
tion and subtraction are performed.
In simplifying a term, omit the sign X when possible,
and write indicated divisions in the form of fractions.
Simplify the following expressions :
13. 100 + 80-=-4. 15. 25 + 5X4-10^5.
14. 75-25X2. 16. 24-5X4-^10 + 3.
17. (24-5) X (4-^10 + 3).
Find the numerical value of the following expressions, in
which a = 2y 6 = 10, a; = 3, y = 5 :
18. xy + 4:aX2. 20. 3a; + 7y-^7 + a X y.
19. xy-15b^B. 21. 6b--^y'^2yXby2b. L ^' ^
ALGEBRAIC NOTATION. 13
22. (eb — Sy)'h2yXb + 2b.
23. (eb-S]/)'ir(2yXb) + 2b.
24. eb'-(S]/^2y)Xb — 2b.
y^ 25. 6b'i-(b — i/) — 3x + bxy'^10a.
Algebraic Notation.
26. Express the sum of a and b,
27. Express the double of x,
28. By how much is a greater than 5 ?
29. If a; is a whole number, what is the next number above
it?
30. Write five numbers in order of magnitude, so that x
shall be the middle number.
31. What is the sum of a: + a; + aj + written a times ?
32. If the product is xy and the multiplier x, what is the
multiplicand ?
33. A man who has a dollars spends b dollars ; how many
dollars has he left ?
34. A regiment of men can be drawn up in a ranks of b men
each, and there are c men over ; of how many men
does the regiment consist ?
35. Write, the sum of x and y divided by c is equal to the
product of a, b, and w, diminished by six times c,
and increased by the quotient of a divided by the
sum of x and y,
36. Write, six times the square of n, divided by m minus a,
increased by five b i^to the expression c plus d
minus a. '
37. Write, four times the fourth power of a, diminished by
five times the square of a into the square of b, and
increased by three times the fourth power of b.
14 ALGEBBA.
Exercise 2.
That the beginner may see how Algebra is employed in
the solution of problems, the following simple exercises
are introduced ;
1. John and James together have $6. James has twice
as much as John. How much has each ?
Let X denote the number of dollars John has.
Then 2x = the number of dollars James has,
and X + 2 X = the number of dollars both have.
But C = the number of dollars both have ;
.-. x+2x = 6,
or 3x=6,
and X = 2.
Therefore, John has $2, and James has $4.
2. A stick of timber 40 feet long is sawed in two, so that
one part is two-thirds as long as the other. Required
the length of each part.
Let 3x denote the number of feet in the longer part.
Then 2 x = the number of feet in the shorter part,
and 3x -h 2 X = the number of feet in both together.
But 40 = the number of feet in both together ;
.-. 3x + 2x = 40,
or 5 X = 40,
and X = 8.
Therefore, the longer part, or 3x, is 24 feet long; and
the shorter, or 2x, is 16 feet.
Note. The unit of the quantity sought is always given, and only
the number of such units is required. Therefore x must never be put
for money ^ lengthy time^ weighty etc., but always for the required
number of specified units of money, length, time, weight, etc.
The beginner should give particular attention to this caution.
PROBLEMS. 16
3. The greater of two numbers is six times the smaller,
and their sum is 35. Eequired the numbers.
4. Thomas had 75' cents. After spending a part of his
money, he found he had twice as much left as he
had spent. How much had he spent ?
v( 6. A tree 75 feet high was broken, so tlfat the part broken
off was four times the length of the part left standing.
Required the length of each part.
6. Four times the smaller of two numbers is three times
the greater, and their sum is 63. Required the
numbers.
7. A farmer sold a sheep, a cow, and a horse, for $216.
^ He sold the cow for seven times as much as the
sheep, and the horse for four times as much as the
cow. How much did he get for each ?
8. George bought some apples, pears, and oranges, for 91
cents. He paid twice as much for the pears as for
the apples, and twice as much for the oranges as
for the pears. How much money did he spend for
each?
?<
t
A man bought a horse, wagon, and harness, for $350.
He paid for the horse four times as much as for the
harness, and for the wagon one-half as much as for
the horse. What did he pay for each ?
10. Distribute $3 among Thomas, Richard, and Henry, so
that Thomas and Richard shall each have twice as
much as Henry.
11. Three men. A, B, and C, pay $1000 taxes. B paya 4
times as much as A, and C an amount equal to the.
sum of what the other two pay. How much does
each pay ?
CHAPTER II.
Addition and Subtraction.
44. An algebraic number which is to be added or sub-
tracted is often inclosed in a parenthesis, in order that the
signs + and — which are used to distinguish positive and
negative numbers may not be confounded with the + and
— signs that denote the operations of addition and subtrac-
tion. Thus, -[- 4 + (— 3) expresses the sum, and + 4 — (— 3)
expresses the difference, of the numbers + 4 and — 3.
46. In order to add two algebraic numbers, we begin at
the place in the series which the first number occupies, and
count, in the direction indicated by the sign of the second
number, as many units as are equal to the absolute value
of the second number. Thus, the sum of + 4 + (-f- 3) is
found by counting from -f-4 three units in the positive
direction, and is, therefore, -f 7 ; the sum of +4 + (—3) is
found by counting from +4 three units in the negative
direction, and is, therefore, + !•
In like manner, the sum of — 4 + (+ 3) is — 1, and the
sum of — 4 + (— 3) is —7. That is,
(1) +4 + (+3^ = 7; (3) -4 + (+3) = -l;
(2) +4 + (-3) = l; (4) -4 + (-3) = -7.
I. Therefore, to add two numbers with like signs, find
the sum of their ahsolute values, and prefix the common sign
to the sum,
II. To add two numbers with unlike signs, find the
diflerence of their absolute values, and prefix the sign of the
greater number to the difference.
addition. 17
Exercise 3.
1. +16 + (— 11)= 3. +68 + (-79) =
2. — 15 + (— 25)= 4. -7 + (+4) =
6. +33 + (+18) =
6. +378 + (+709) + (-692) =
7. A man has $5242 and owes $2758. How much is he
worth ?
8. The Pirst Punic War began b.c. 264, and lasted 23
years. When did it end ?
9. Augustus Caesar was born b.c, 63, and lived 77 years.
When did he die ?
10. A man goes 65 steps forwards, then 37 steps backwards,
then again 48 steps forwards. How many steps did
he take in all ? How many steps is he from where
he started?
Addition op Monomials.
46. If a and h denote the absolute values of any two
numbers, 1, 2, 3, 4 (§ 45) become :
(1) +a + (+&) = a + ft; (3) -a + (+ft) = -a + ft ;
(2) +a + (— 6) = a — ft; (4) — a + (— ft) = — a — ft.
Therefore, to add two terms, write thfim one after the other
with unchanged signs, '
It should be noticed that the order of the terms is im-
material. Thus, +a — ft = — ft + a. If a = 8 and ft = 12,
the result in either case is — 4.
47. 3a + 5a + 2a + 6a + a = 17a.
— 2c — 3c — c — 4c — 8c = — 18c.
Therefore, to add several like terms which have the same
18 ALGEBBA.
sign, add the coefficients, prefix the common sign, and annex
the common symbols,
48. 7a — 6a + lla + a — 6a — 2a = 19a — 13a = 6a.
— 3a — 15a — 7a + 14a — 2a = 14a — 27a = — 13a.
Therefore, to add several like terms which have not all
the same sign, find the difference between the sum of ths
positive coefficients and the sum of the negative coefficients^
prefix the sign of the greater sum, and annex the common
symbols,
49. 6a — 26 + 3a = 8a — 2ft.
— 3ax + 8y + 9aa; — 4c = 6aaj + 8y — 4c.
Therefore, to add terms which are not all like terms,
combine the like termSy and write down the other terms, each
preceded by its proper sign.
Exercise 4.
1. 5aft + (-5aft)= 6. 7aft + (-6aft) =
2. ^mx + {-'2mx)= 7. 120wy + (— 95my) =
3. —13mng + (—7mng)=^ 8. — 33a^»* + (+lla^ =
4. -6a:2-f(+8x2)= 9. —75xy + (+20xy) =
5. 25my'+(—lSmf)= 10. +16aV + (— a«ic«) =
11. — ft«m» + (+7ft*m«) =
12. 6a + (-3ft) + (+4a) + (-7ft) =
13. 4a^c + (—10xyz) + (+ea^c) + (—9xyz)
+ (-lla*c) + (+20xyz) =
14. 3a^y + (—4:ab) + ('-2mn) + (+5a^y)
addition, 19
Addition of Polynomials.
50. Two or more polynomials are added by adding their
separate terms.
It is conyenient to arrange the terms in columns, so that
like terms shall stand in the same column. Thus,
(1) 2a»— 3a26 + 4a62+ 58 (2) — 2a?y +63^ — 1
a« + 4a«6 — 7a^»^-26« — 4aj*y + 2V +5
— 3a«+ aH — ^ah^ — 4:h^ %a?y +2
2a« + 2a^^ + 6g^-3y a^y —7/
2a^ + 4:a^b — 86« —2xhj --5
- x'y + 2xy' + 5j/' + l
Exercise 5.
Add:
1. 5a + Sb + c, 3a + 36 + 3c, a + Sb + 5c.
2. Ta — U + c, ea + 3b — 5c, — 12a + 4c.
/3. a-{-b — c, ft + c — a, c-\-a — 6, a-^-b — c,
4. a + 2& + 3c, 2a — b — 2c, b — a — c, c — a — b,
5. a — 2b + 3c — 4:d, 3b — 4:C + od-'2a,
5c — ed+Sa — 4:b, 7(^-4a + 55-4c.
6. a^ — 4a^ + 5a; — 3, 2a^ — 7a^ — 7x^ — Ux + 5,
— a^ + dx' + x + S,
7. «* — 2aj» + 3a^, a^ + a^ + x, 4x* + 5x^,
2x'+3x — 4:, — 3a« — 2a; — 6.
8. a» + 3a^« — 3a*6 — 6^ 2a» + 6a*ft — 6aZ»* — 7^^
a^ — a^>2^25«.
^. 2a^ — 3aa;* + 2a2aj, 12a5 — Ga^aj + lOoa^,
aa;^ — 8aft — Sa'aj.
20
ALGEBRA.
10. d*-3c»-Mc«-4c + 7, 2c*+3c»+2c«+6c + 6,
\ — 4(J*-4c«— 6.
11. 3aj*— a:y + a»-^3y*+4y« — «■, — 6aj*— a;y — a;« + 6y«
ex^ — ey — ez, 4y« — 5y« + 3«*,
-4a^+y* + 3y« + 3«*.
12. m'— 3m% — 6mV, +w*^' + ^*^* — 5^*^j
7wV+4mV — 3w< — 2mV — 3wn*+4n',
2mn*+2nl^+3m\ — »«+2m« + 7m%.
SUBTBACTION.
61. In order to find the difference between two algebraic
numbers, we begin at the place in the series which the minu-
end occupies^ and count in the direction opposite to that indi-
cated by the sign of the subtrahend as many units as are
equal to the absolute value of the subtrahend.
Thus, the difference between +4 and +3 is found by
counting from +4 three units in the negative direction,
and is, therefore, +!> the difference between +4 and — 3
is found by counting from + 4 three units in the positive
direction, and is, therefore, + 7.
In like manner, the difference between — 4 and + 3 is
— 7 ; the difference between — 4 and — 3 is — 1.
Compare these results with results obtained in addition :
+4 + (-3) = l.
+ 4 + (+3)=7.
-4 + (-3)=-7.
-4 + (+3)=-l.
(l)+4-(+3) = l
(2) +4-(-3) = 7
(3) -4- (+3) = -7
(4) -4-(-3)=-l
Or, (1) +4-(+3) = + 4+(-3).
(2) +4-(-3)= + 4 + (+3).
(8) -4-(+3) = -4 + (-3).
(4) -4-(-3)=-4 + (+3).
SUBTRACTION. 21
62. From (1) and (3), it is evident that stibtracting a
positive number is equivalent to adding an equal negaMve
number.
From (2) and (4), it is evident that subtracting a negor
tive number is equivalent to adding an equal positive number.
To subtract, therefore, one algebraic number from an-
other, change the sign of the subtrahend, and then add the
subtrahend to the minuend.
EXEBCISE 6.
1. +26-(+16)= 3. - 31-(+68) =
2. -50 — (—25)= 4. +107-(— 93) =
5. Itome was ruled by emperors from b.c. 30 to its fall,
A.D. 476. How long did the empire last ?
Ck Baie contin^BA of Europe lies between 36® and 71® north
latitude, and between 12® west and 63® east longitude
(from Paris). How many degrees does it extend in
latitude, aud how many in longitude ?
StTBx%^CTioN or Monomials,
If a and h denote the absolute value« of any two num-
yxxsy 1, 2y 3, and 4 (§ 51) become :
(X) +a--(+^) = a~J. (8) ««-.^-f j)=-^^J.
(a) +a-(-5)=a + 5. (4) -a — (—&)=— a+&.
7a subtract, therefore, one term from another, change
th» «%m qfth^ term to he subtraeM, and unite the terms one
22 ALOBB&A.
Exercise 7.
1. 6a; — (— 4a;)= 6. 17aa»— (— 24aa;«) =
2. — 3a* — (+5a^)= 7. 6a«a;— (— 3a*aj) =
3. 3ay— (+10aZr«)= 8. — 4a;^— (— 5icy) =
4. 15mV — (— 7mV)= 9. 8aa; — (— 3a3^) =
6. — 7ay — (— 3ay)= 10. 2a6>— (+a*y) =
11. 9a^+{5x^-{+Sx^ =
12. 6ic*y-(-18a*3^) + (-10ic*y) =
13. 17aa:«— (— oaj^ — (+24flMc») =
14. —3ab+(2nix) — (—Amx) =
16. 3a — (+2*) — (—4c) =
Subtraction of Polynomials.
63. When one polynomial is to be subtracted from an-
other, place its terms under the like terms of the other,
if they have like terms, change the signs of the subtrahend,
and add.
From 4a» — 3ajV-- xi/^+2t/^
take 2aj»— x'y + Bxy' — Sy'
Change the signs of the subtrahend and add :
4a» — 3aj*y— xy*+2f
— 203+ a^y — Bxy^ + Sj/^
2«» — 2a:«y — 6V + 52/»
From a^a^ + 2a^x^'-4tax*
take a* + 4a«a:« — 3aV — 4aa;*
— a«— Sa^x^+SaV
SUBTRACTION. 28
In the last example we have conceived the signs to be
changed without actually changing them. The beginner
should do the examples by both methods until he has ac-
quired sufficient practice, when he should use the second
method only.
EXEBGISE 8.
1. Prom 6a — 2^ — c take 2a — 26 — 3c.
2. PromSa — 26 + 3c take 2a — 7Z» — c — ft.
3. PromTa:* — 8a; — 1 takeSa^ — 6a; + 3.
4. Erom4x< — 3a^ — 2a:* — 7x + 9
take a;* — 2a;« — 2x* + 7x — 9.
5. Erom 2a^ — 2aa; + 3a* take a?^ax'\'a*,
6. Prom a:* — 3xy — y^-^]/z — 2«*
take x* + 2ajy + 5aj« — 3y» — 2««.
7. Proma^ — 3a2Z» + 3aft« — ft»
take — a« + 3a2ft — 3aft»+^.
8. ¥Tomx^ — 6xi/ + xz — 7/ + 7yz + 2z*
take x^ — xt/ — xz-{'2yz-\-3s^.
9. Prom2aa^ + 3afta; — 4ft2a; + 12i»»
take ax^ ^4tabx + bx^ — 5b^x — a^.
10. From 6x« — 7aj2y + 4a;?/^ — 22^ — 5a;* + a;?/ — 4y* + 2
take Sx^ — 7x^y + xy^ — i/^ + 9x^ — xy + 6j/*—4,:
n. From a*'-'b* take 4a«ft — 6a*^»* + 4aft«, and from the
result take 2a* — 4a36 + 6a*ft2 + 4aft« — 2ft*.
12. From x^i^ — 3a^j^ + 4:Xi/^'-y^ take — a;* + 2a;*«/ — 4a;y*
— 4ty^. Add the same two expressions, and subtract
the former result from the latter.
13. From a'ft* — a^ftc — 8 aft*c — a V + aftc* — 6 ft V
take 2a"ftG — 6aft"c + 2aftc2 — 5ftV.
24 ALGEBRA.
«
14. From 12a + 3^> — 5c — 2rf take 10a — ^ + 4c — 3rf,
and show that the result is numerically correct
when a = 6, ^ = 4, c = l, d=5.
16. What number must be added to a to make b ; and what
number must be taken from 2a^ — 6a% + 6a^' — 2b^
to leave a^ — la^b — Sb^?
16. From2a;« — 2/* — 2a:^ + «*takea:* — y* + 2ajy— «*.
17. Froml2ac + 8cc? — 9 take — 7ac — 9(^ + 8.
18. From— 6a2 + 2a^ — 3c*take4a* + 6aZ» — 4c*.
19. From 9ajy — 4aj — 3y + 7 take Sxi/—2x + 3y + 6.
20. From — a^bc — a^*c + a^c* — abc
take a*6c + al^c — ab(^ + a^.
21. From7x* — 2x + 4 take2aj* + 3x— 1.
22. From3a^ + 2a;y— 2^ take — a:* — 3a;y + 3^, and from
the remainder take 3a:*+4a:y — 5y*.
23. From ax^ — btf take cic* — dj^,
24. From oo; + ^^ + ^y + ^y take ctx — bx — ^y + c?/.
26. From 5x2 + 4x — 4y + 3y' take 5a;2 — 3x + 3y + 2^.
26. From a*^» + 12 o^c — 9 oic* take 4tab^ — 6acx + 3 a^x,
27. Froma* — 2a& + c2 — 3^take2a«-2a^ + 3^*^
28. From the sum of the first four of the following expres-
sions, a^ + b^ + c'+d^, d^ + b^-^<^, a^-(^ + b^ — d^,
a^—b'+c' + d^, b^ + c^ + d^'-'o' take the sum of
the last four.
29. From 2x^ — 2i/ — s^ take 3y'+2x^ — «*, and from the
remainder take 3 «* — 2y' — a?*.
30. From a^ — 2 a*c -f- 3 ac* take the sum of a^c — 2a' + 2ac*
and d^ — ai? — d^c.
PABENTHESBS. 26
PARENTHESES.
64. From (§ 52), it appears that
(1) a + (+b) = a + b
(2) a+(—b) = a — b,
(3) a-'(+b) = a — b.
(4) a — (— ^) = a
The same laws respecting the removaiNof parentheses
hold true whether one or more terms are inclosed. Hence,
when an expression within a parenthesis is preceded by a
plus sign, the parenthesis may be removed.
When an expression within a parenthesis is preceded by
a minus sign, the parenthesis may be removed if the sign
of every term within the parenthesis be changed. Thus :
(1) a + (ft — c) = a + ^ — c.
(2) a — (^ — c) = a -— ^ + c.
66. Expressions may occur with more than one paren-
thesis. These parentheses may be removed in succession
by removing first, the innermost parenthesis; next, the
innermost of all that remain, and so on. Thus :
(1) a-{b-{c-d)\
=a — {b — c-\-d\,
= a — b-^ c — d.
(2) a-lh-\c-^{d-e-f)Xl
= a-[b-\c-\-{d-e+f)\1,
= a-\b-{c + d-e-\-f}-\,
= a — [J — c — rf + e— /],
= a — 6 + c + d — «+/.
26 ALGBBRiU
EXBBCISE 9.
Simplify the following expressions by removing the
parentheses and combining like terms :
(a + b) + (b + c)^{a + c). IJ-
2. (2a'-b — c)'-(a — 2b + c).
3. (2x-j/)-(2j/-z)-(2z-x).
4. (a — x — y) — (b — x + y) + (c + 2y),
6. (2x-y + Sz) + {-x-i/-4.z)-(Sx-2y-z).
6. (3a~^ + 7c)-(2a + 3^) — (5^-4c) + (3c— a).
7. 1 — (l-a) + (l — a + a^ — (1 — a + a*-a«).
8. a'-\2b'-(3c + 2b)—a\.
9. 2a— 1^ — (a — 2^)|.
10. 3a—\b + (2a — b) — (a'-b)}.
11. 7a — [3a — |4a— (5a — 2a)}].
12. 2a; + (2/ — 3«)—^3ic — 2 jf)+«}+5x — (4y — 3«).
13. |(3a-2^) + (4c-a)|-|a-(2^>-3a)— c|
+ |a-(/^-5c-a)|.
14. a— [2a+(3a — 4a)] — 5a— |6a — [(7a+8a)— 9a](.
X15. 2a— (36 + 2c)-[5^-(6c — 6^) + 5c
-|2a-(c + 26)n.
16. a — [2^+|3c-3a — (a + ^)| + j2a-(6 + c)|].
17. 16 — cc — [7aj— |8a;— (9x — 3a; — 6ir)|].
©
Ni8. 2a — [36 + (2^ — c)— 4c+J2a — (3^— 7=^)}].
•^^•19. a-[2i» + |3c — 3a-(a + ^)|+2a-(/^ + 3c)].
> 20. a — [55— |a— (3c — 36)+2c— (a — 2^ — c)|]. ^
PARENTHESES. 27
66. The rules for introducing parentheses follow direotly
from the rules for removing them :
1. Any number of terms of an expression may be put
within a parenthesis^ and the sign plus placed before the
whole.
2. Any number of terms of an expression may be put
within a parenthesis, and the sign minus placed before the
whole ; provided the sign of every term within the parenr
thesis be changed.
It is usual to prefix to the parenthesis the sign of the
first term that is to be inclosed within it.
EXEBGISE 10.
Express in binomials, and also in trinomials :
1. 2a — 36 — 4c + rf+36 — 2/.
2. a — 2x + 4:y'-3z — 2b + c.
3. a* + 3a* — 2a« — 4a2-|-a — 1.
4. — 3a — 26 + 2c — 5c^ — e — 2/.
5. (IX — by — cz — bx-{- cy-\- az,
6. 2a^ — 3x*y + 4:0^1/ -5a^l/' + x^-'2y'.
7. Express each of the above in trinomials, having the
last two terms inclosed by inner parentheses.
Collect in parentheses the coefficients of x, y, z, in
8. 2ax — 6ay-\-4ibz — 4ibx — 2cx — Scy,
9. ax'—bx'{'2ay-j-3y-\-4:az — Sbz — 2z.
^ 10. ax — 2by-\-bcz — Aibx — 3cy-^az — 2cx — ay + 45«.
11. 12aa; + 12a2/ + 46«/ — 12«>« — 15ca; + 6cy + 3c«.
12. 2ax'-3hy'-lcz — 2bx + 2cx + ^cz'-2cx--cy'-cx.
CHAPTEE III.
Multiplication op Algebraic Numbers.
67. The operation of finding the sum of 3 numbers, each
equal to 5, is symbolized by the expression, 3 X 5 = 16,
read, " three times five is equal to fifteen " ; or, by the
expression 5 X 3 = 15, read, " five multiplied by three is
equal to fifteen.''
68. With reference to this operation, this sum is called
the product ; one of the equal numbers is called the multi-
plicand ; and the number which shows how many times the
multiplicand is to be taken is called the multiplier.
69. The multiplier means so many times. The multipli-
cand can be a positive or a negative number; but the mul-
tiplier, when integral, means that the multiplicand is taken
80 many times to he added, or to he svhtracted.
60. If we have to multiply 867 by 98, we may put the
multiplier in the form 100 — 2. The 100 will mean that
the multiplicand is taken 100 times to he added; the —2
will mean that the multiplicand is taken twice to he svS-
tracted.
In general, an integral multiplier with + before it, ex-
pressed or understood, means that the multiplicand is taken
so many times to he added; and a multiplier with — before
it means that the multiplicand is taken so many times to
be subtracted. Thus,
MULTIPLICATION. 29
(1) +3 X (+5) = (+5) + (+6)+ (+6), or (+15).
(2) + 3 X (-5) = (-5) + (-6) + (-6), or (-15).
(3) -3X(+5)==-(+5)-(+5)-(+5),or(-15).
(4) -.3X(-5)=-(-5)~(-5)-(-6),or(+15).
From these four cases it follows, that, in finding the
product of two numbers,
61. Like signs produce plus; unlike signs, minium
Exercise 11.
1. —17X8= 4. — 18X— 5 =
2. —12.8X26= 5. 43 X — 6 =
3. 3.29X5.49= 6. 457X100 =
7. (-358-417) X-79 =
8. (7.512-1-2.894}) X (-6.037 + ^13.9630 =
62. The product of more than two factors, each preceded
by — , will be positive or negative, according as the number
of such factors is even or odd. Thus,
-2X-3X-4= +6X-4= -24.
— 2X-3X-4X-5 = -24X-5 = + 120.
9. 13X8 X — 7 =
10. — 38X9X— 6 =
11. — 20.9 X— 1.1X8 =
12. — 78.3 X— 0.57 X+ 1.38 X — 27.9 =
18. — 2.906 X — 2.076 X - 1.49 X 0.89 =
30 AlfGEBBA.
Multiplication of Monomials.
63. The product of numerical factors is a new number
in which no trace of the original factors is found. Thus,
4 X 9 = 36. But the product of literal factors can only be
expressed by writing them one after the other. Thus, the
product of a and h is expressed by ah ; the product of ah
and cd is expressed by ahcd,
64. If we have to multiply 5 a by — 4 ft, the factors will
give the same result in whatever order they are taken.
Thus, 5aX — 4ft = 6X— 4XaXft = — 20Xa6 = — 20aft.
65. Hence, to find the product of monomials, annex the
literal factors to the product of the numerical fa^^tors.
66. a* X a^ = aa X aaa = a^aaaa = a*.
a^X a^ X a^ = aaX aaa X a^iaa = aaaaaaaaxi = a^
It is evident that the exponent of the product is equal to
the sum of the exponents of the factors. Hence,
67. The product of two or more powers of any numher is
that numher tvith an exponent eqtial to the sum of the expo-
nents of the factors,
EXEBGISE 12.
1. +aX+b=+ab. 6. — 3^? X 8 m = — 24jm».
2. +aX—b = — ah. 7. 3a' X — a«= — 3a*.
3. —aX + b = — ab. 8. — 3a X 2a* = — 6a«.
4. —aX— ft = + a^. 9. 6aX— 2a=«
6. 7aX66 = 35aft. 10. 5wnX9m—
MULTIFLIOATION. 31
11. 3aajX— 4*y= 16. 5a'^X'-2a*=a
12. ^ScmXdn= 16. 3aVX7aV =
13. — 7a^X2ac= 17. 7aX— 46X--8o=
14. 5m^xXSm^= 18. So^ X 3ac X — 4c*=
19. 27a^X — 39m^Xl8ap =
20. 6a5VX2^YX — 5a^y =
21. 7m*a;X37?iic*X— 2m2' =
22. —Spq^X^p^qXSjfq^^
23. 2a*m«a;*X3am*a:*X4a«ma^=
24. 6a:*2^;53X— 9ic*2/*«*X — 3xV =
25. 3aa;X2amX— 4ma;X^* =
^26. 7a7w.* X 36*71* X — 4a5 X a*^^n X — 2^w X — mn*=
Of Polynomials by Monomials.
68. If we have to multiply a + b by w, that is, to take
(a + b)n times to be added, we have,
(a + b) X n = (a + b) + (a + b) + (a + b) ... .n times,
= a-\-a-\-a. ..n times 4-6 + 6 + 6 n times,
= aXn + bXn,
= an'-\- bn.
As it is immaterial in what order the factors are taken,
nX (a + 6) ^ aw + bn.
In like manner,
(a + 6 + c) Xn = an-\-bn-\'Cn^
or, w (a + 6 + c) = an + 6/1 + en.
32 ALOEBRA.
Hence to multiply a polynomial by a monomial,
69. Multiply each term of ths jpolynamial by the monomial^
and add the partial products,
EXEBGISE 13.
1. (6a — 5^)X3c = 18ac — 15^c.
2. (2 + 3a — 4a* — 5a»)6a*=12a» + 18a» — 24a*-30a«.
3. 5a{Sb + Ac — cr) = 15ab + 20ac'-5ad.
4. — Sax(;—by'—2cz'-{'5) = 3ahxy-^6a>cxz — 15ax.
5. (4a» — 36)X3a^ =
. 6. (8a» — 9a*)X3a* =
7. (3a« — 4y» + 6««)X2a«y =
8. (a^x — 5a^a^ + ax^ + 2x*)Xaa^y==
9. (— 9a« + 3a«Z»« — 4a«6« — ^X — 3a5*=
10. (3a^ — 2ic^y'-7xy' + r/^X-'5a^y =
11. (—4:Xy^+5x^y + Sa^)X—3ic^y =
12. (— 3 + 2a^> + a2^X — a* =
-4^3. (— « — 2ic«« + 5a;V* — 6a;V + 3ic«y*«) X— 3a»y« =
Op Polynomials by Polynomials.
70. If we have a + & + c to be multiplied by m + » +^,
we may represent the multiplicand a-\-b'\-chy M, Then
M(m + n+p)=MXm + MXn + MXp.
If now we substitute for J!f its value,
(a-}- b -^ c)(m + n +i?) = (a + ft + <^) X wi
+ (a + b + c)Xn
+ (a + b + c)Xp]
MULTIPLICATION. 38
or, (a + ^ + c) (m + ^ +i?) = awt -{-bm-i-cm
-{-an -{-bn -{-en
+ ap +bp +qp.
That is, to find the product of two polynomials,
71. Multiply the multiplicand by each term of the mtiltiplier
and add the partial products; or, multiply ea^h term of one
factor hy ea^h term of the other and add the partial products.
72. In multiplying polynomials, it is a convenient ar-
rangement to write the multiplier imder the multiplicand,
and place like terms of the partial products in columns.
Thus : (1) 5a — 6 b
3a— 4 b
IBa' — lSab
(2) Multiply 4a; + 3 + 5a^ — 6a« by 4 — 6a^-5a.
Arrange both multiplicand and multiplier according to
the ascending powers of x.
3+ 4tx+ 5a^— 6a?
4 — 5x — 6x^
12 + 16x + 20a?-24:x^
-15a;-20aj« — 25a:» + 30aj*
-18x'--24a;« — 30a;* + 36a^
12+ X — 18a;2 — 73a^ +36a^
(8) iMultiply l + 2a; + ic*— ]3ic«by a»— 2 — 2x.
Arrange according to the (descending powers of x.
x^—3x' + 2x +1
a?—2x-2
a?'-Sa? + 2x^+ a?
— 2a* +6a^ — 4a«— 2a:
—2a;* +6a;» — 4.r~2
af^&a? +7a^ + 2a^— 6a — 2
84 ALOEBSA.
(4) Multiply a* + ^* + ^— <*^ — ^ — ^^ by a+^ + o.
Arrange according to the descending powers of a.
a' — ah — ac+ ^ — hc-^- (?
g -|- ^+ ^
of — a% — dfc •\'ah^ — dbc + oc*
+ a*6 — o^^*— o^ +^* — **<J + ^
-|-a'c ~ g&c — OA? '\-l?c — h<?-\-(?
The student should observe that, with a view to bringing
like terms of the partial products in columns, the terms of
the multiplicand and multiplier are arranged in the same
order.
In order to test the accuracy of the work, interchange
the multiplicand and multiplier. The result should be the
same in both operations.
Exercise 14.
Multiply :
1. ir* — 4bya^ + 5. 3. a^ + aV + a;* by a« — ic*.
2. y — 6byy+13. 4. a^ + ajy + y* by cc — y.
5. 2x — y by 05 + 2y.
6. 2ic» + 4a^ + 8a; + 16by 3aj — 6.
7. a^ + a:* + a;--l by x — 1. 8. a* -— 3 oo; by a; + 3 a.
9. 2^»^ + 3a^ — a*by— 5& + 7a.
10. 2a + ^'bya + 26. 12. a* — a^ + ^»2 by a + ft.
11. a^ + oZ^ + Z^^by a — ^. 13. 2aZ>-5Z>2 by 3 a*— 4a6.
14. — a«+2a«* — Z»»by 4a2 + 8a5.
16. a« + a^ + ^by a* — a^ + ^.
MULTIPLICATIOK. S5
16. a» — 3a** + 3a^—^« by a*— 2o^ + ^.
17. x + 2i/'-3zhy x — 2y + 3z.
18. 2a^ + 3a;y + 4y*by 3aj* — 4xy + y«.
19. oc^ + X7/ + yhy a^ + xz + s^,
20. a^ + b^ + c^ — ab — ac — bc by a + b + e.
21. oi^ — xi/ + i^+x + i/ + lhjx + j/—l.
Arrange the multiplicand and multiplier according to
the descending powers of a common letter, and multiply :
^22. 5x + 4x^ + a^ — 24thy x'+ll-^x.
23. x'+llx — 4.x'--24.hyx^ + 5 + 4:X.
24. a;< + a;* — 4a: — ll+2ar»bya^ — 2a: + 3.
25. — 5a;* — x^ — a: + a:* + 13a:' by a* — 2 — 2aj.
26. 3ar + ar^ — 2x2— 4by2a; + 4a:» + 3x*+l.
^27. 5a^ + 2a^b^ + ab^'-'3a^ by 5a^'-'2ab^ + Sa^b* + b\
/yc 28. 4a'y — 32ay* — 8aV + 16ay by aV+4aV' + 4«y-
29. 3m« + 3w' + 9m7i* + 9w*7iby 6mV — 2m7i*
— 6m»»* + 2??t*n.
30. ea'^^ + Sa*^'* — 2aZ^* + &«by 4a* — 20^^* — 3Z»*
Find the products of :
^31. X — 3, a; — 1, a; + 1, and a; + 3.
.32. «* — x + l, a;«+a; + l, andx* — x^ + l.
\S 33. a^+ab + P, a^--ab + b^, and a^ — aH^ + b\
34. ^a^ — Aan + ab^ Aa^+Sa^ + b^, and 2a^ + b\
35. ap+a, a; + 2a, a; — 3a, a; — 4a, and a; + 5a.
36. 9a*+^i«,27a» — ^, 27a2+ft», and 81a* — 9a«^ + 5*
36 AXGEBSA.
rA 37. Fromtheproductof y*— 2y«— «*andy" + 2y«— «■
take the product of y*— y« — 2«*andy"+ yz'-2f?.
38. Find the dividend when the divisor =3a*— a^--3^',
the quotient =a*^ — 2^ the remaiyder = — 2a^* v^
The multiplication of polynomials may be indicated by
inclosing each in a parenthesis and writing them one after
the other. When the operations indicated are actually per-
formed^ the expression is said to be simplified.
Simplify :
39. {a + h-'C)(a + c — b)(h + c-'a){a + h + c).
40. (a + *)(^ + c) — (c + rf)(rf + a)-(a+c)(6-c?).
41. {a + h + c + df+{a'-h'-'C + dy
42. (a + «» + c)*— a(^ + c — a) — 6(a + c — 6) — c(a + 6 — c).
43. (a — 6)x— (6 — c)a-|(/^ — aj)(^> — a) — (ft-c)(6 + c)}.
Ax 44. (m + «)wi — \(m — rif — (n — m)n\,
I 46. (a-^ + c)*— {a(c — a — Z>) — [^(a + ^ + c)
46. (l>^+^y— (i>+?)(pjr— 3^1— yfr-^^l).
\ 47. (9aj«y«-4^(««-j^-{3a!y-2y»nM«^+y)
-'4ft*|-(» + (>)»,
49. {ac-^(a— i)(6+«)J--i^6— <a.-H^)^.
60. 5|(a— ft)rc-*cy| — 2Ja(ic--y)--*a6|
— |aam— ^d— 2a)34.
61. (a;-l)(«--2)-3»(«+3)+2^(a:-h8)(» + l)--a^.
MULTIPLICATION. 37
62. \(2a + by+(a-2by\X\(Sa-2by-'(2a'-3by\.
63. 4(a — 3^»)(a + 36)-2(a-65)«-2(a« + 6ft«). ^^
64. arV + 2/^*-2ay(a: + y)(a:-y)-(a^-y»)«.
66. 16(a* + «>«)(a*-5«)-(2a-3)(2a + 3)(4a« + 9)
+ (25-3)(26 + 3)(4^+9).
78. There are some examples iu multiplication which
occur so often in algebraical operations that they should
be carefully noticed and remembered. The three which
follow are of great importance :
(1) «+ b
(2) a— h
(8) a + ft
a+ b
a — 6
a — b
«'+ ab
a*— ah
a» + oA
ab + l^
— ai + 6«
— ab — V
a*-\-2ab-\-1f'
a*~2ab + b*
a' -*»
From (1) we have (a + by = a^ + 2 ab + b^. That is,
74. The sqitare of the sum of two numbers is equal to the
sum of their squares + tunee their product,
From (2) we have (a -'by=a^—2ab + b\ That is,
76. The square of the difference of two numbers is equal
to the sum of their squares — twice their product.
From (3) we have (a -f 5) (a — 5) = a* — V, That is,
76. The product of the sum and difference of two numbers
is equal to the difference of their squares,
77. A general truth expressed by symbols is called a
formula.
38 ALQEBRA.
78. By using the double sign =b, read plus or minus, we
may represent (1) and (2) by a single formula ; thus,
(aztby=a^:h2ab + h^;
in which expression the upper signs correspond with each
other, and the lower signs with each other.
By remembering these formulas the square of any bino-
mial, or the product of the sum and difference of any two
numbers, may be written by inspection ; thus :
Exercise 16.
1. (127)*-(123)»=(127 + 123)(127-123)
= 260X4 = 1000,
2. (29)*=(30-l)« = 900 -60 + 1 = 841.
3. (63)8 = (50 4- 3)2 = 2600 + 300 + 9 = 2809.
6. (2a»a;-6aj2^)* = 4a*aj2-20aV2^ + 26a;V-
6. (3a^c + 2a*c^(3a5*c — 2a*c^ = 9a8ft*c» — 4a*c*.
7. (x + yy= 15. (ab + cdy =
8. (y — «)*= 18- (3mn — 4)8=
9. (2a; + 1)8= 17. (12 + 6a;)8 =
..10. (2a + 5by= 18. (4a;2/8-y«8)8=
11. (1 — ic8)8= 19. (3abc — bedy:=
12. (3a« — 4ary= 20. (4:a^ — xiy)^=
13. (1 — 7a)8= 21. (x + yXx — y) =
14. (5xy + 2y= 22. (2a + 5)(2a — 5)==
MULTIPLICATION.
23. (3— aj)(3 + a5) =
24. (Sab + 2b^(3ab-2lj^^
25. (4a:«-32^(4«> + 33/«) =
26. (aV-fty)(aV + ft/) =
27. (6ay-52^(6a;2/ + 63/«) =
28. (4x«-l)(4x« + l) =
29. (l + 3ab^(l'-3ab'^ =
30. (oaj + 5y) (oo; — by) (a*ix^ + ft'y*) =
79. Also the square of a trinomial should be carefully
noticed.
a 4" b-\- G
a6 +^4- ^0
dc + 6c 4" g*
= a2 + ^ + c« + 2a6 + 2ac4-26c.
It IS evident that this result is composed of two sets of
numbers :
I. The squares of a, 6, and c ;
II. Twice the products of a, 5, and c taken two and two.
Again,
a — •
b —
c
a —
b-
c
a»-
ab-
dC
—
ab
+ 6' +
be
—
ac
+
bc + c'
a^ — 2ab-'2ac + b^ + 2bc + c'
=^a* + b^ + (^'-2ab — 2ac + 2bc.
40 ALGEBRA.
The law of formation is the same as before :
I. The squares of a, h, and c;
II. Twice the products of a, h, and c taken two and two.
The sign of each double product is + or — according as
the signs of the factors composing it are like or unlike.
The same law holds good for the square of expressions
containing more than three terms, and may be stated thus :
80. To the stim of the squares of the several terms add tunce
the prodv^ct of each term by each of the terms that follow it.
By remembering this formula, the square of any poly-
nomial may be written by inspection; thus :
Exercise 16.
2. (a; — y4-«)2= 10. (a^ — f — zy^
3. (m + n—p — qy= 11. (x + 2y — 3zy =
^ 4. (x^+2x — 3y= 12. (x^-'2f + 5^^ =
5. (x* — 6a: + 7)2= 13. (ic* + 2a: — 2)^ =
6. (2x^-'7x'\'9y= n/>J 14. (oi^-'5x + 7y =
"^ 7. (x^ + y' — zy^ 15. (2x^-Sx-'4:y =
8. (aj*-4a:V + yT= l^. (x + 2y + Szy =
81. Likewise, the product of two binomials of the form
x-^-a, x-^b should be carefully noticed and remembered.
(1) x+5 (2) X -5
x +3 X —3
x^-{'5x a^— 5aj
+ 3a; + 15 —3a; + 15
a^ + 8a; + 15 aj2_8^^15
MULTIPLICATION. 41
(8) X +5 (4) X -5
aj — 3 aj +3
-3g — 15 +3a; — 15
a^+2a; — 16 aj« — 2a;-15
It will be observed that :
I. In all the results the first term is a;^ and the last term
is the product of 5 and 3.
II. From (1) and (2), when the second terms of the
binomials have like signs, the product has
the last term positive;
the coefficient of the middle term = the sum of 3 and 5 ;
the sign of the middle term is the same as that of the
3 and 5.
III. From (3) and (4), when the second terms of the
binomials have unlike signs, the product has
the last term negative;
the coefficient of the middle term = the diflerenoe of
3 and 5;
the sign of the middle term is that of the greater of the
two numbers.
82. These results may be deduced from the general
formula,
{x + a)(x + b)=x^-j-(a + b)x + ab,
by supposing for (1) a and b both positive;
(2) a and b both negative;
(3) a positive, b negative, and a>b;
(4) a negative, b positive, and a > ft.
By remembering this formula the product of two bino-
mials may be written by inspection ; thus :
42 ALaSBRA.
Exercise 17.
1. (x + 2)(x + S)=^ 11. (aj — c)(aj — d^ =
r^2. (x + l)(x + 5)= 12. (x-4.y)(x + y) =
3. (aj — 3)(x — 6)= >rJ 13. {a'-'2b)(a — 5b) =
4. (a:-8)(a;-l)= 14. (a^ + 2f)(a^ + f) =
^5, (x — S)(x + l)= 16. (a»-3a;y)(ic« + a;y) =
6. (x — 2)(x + 5)= J 16. (aaj — 9)(aaj + 6) =
7. (x-'S)(x + 7)= 17. (a; + a)(a; — *) =
>N 8. (a: — 2)(x — 4)= 18. (a;-ll)(aj + 4) =
9. (a + l)(a + ll)= >r^l9. (a; + 12)(a-ll) =
10. (aj — 2a)(x + 3a)= >f* 20. (x — 10) (aj — 5) =
83. The second, third, aaid fourth powers oi a-\-b are
found in the following manner :
a + h
a + b
a»+ ah
g^ + y
a +5
(a + 5)8 = a» + 3a*5 + 3a^«+6»
a 4-^
(a + 5)* = a* + 4a% + 6a*6« + 4a^» + **
MULTIPLICATION. 43
From these results it will be observed that :
I. The number of terms is greater by one than the ex-
ponent of the power to which the binomial is raised.
II. In the first term, the exponent of a is the same as
the exponent of the power to which the binomial is raised;*
and it decreases by one in each succeeding term.
III. b appears in the second term with one for an expo-
nent, and its exponent increases by one in each succeeding
term.
IV. The coefficient of the first term is 1.
V. The coefficient of the second term is the same as
the exponent of the power to which the binomial is raised.
VI. The coefficient of each succeeding term is found
from the next preceding term by multiplying its coefficient
by the exponent of a, and dividing the product by a num-
ber greater by one than the exponent of b,
84. If b is negative, the terms in which the odd powers
of b occur are negative. Thus :
(a'-'by==a*-4:a^b + 6a''P — ^ab^ + b\
Exercise 18.
Write by inspection the results :
1. (x + ay= 5. (x + ay= 9. (x + yy =
2. (x — ay= 6. (x — a)*= ^0. (x — yy =
3. (x+iy= 7. (x+iy=: 11. (x+i/=
4. (a;-l)»= 8. (x — iy=^ 12. (aj~l}i=
A'h
CHAPTER IV.
Division.
86. Dmsion is the operation by wMch, when a product
and one of its factors are given, the other factor is deter-
mined.
86. With reference to this operation the product is called
the dividend ; the given factor the divisor ; and the required
factor the quotient.
87. The operation of division is indicated by the sign -s- ;
by the colon :, or by writing the dividend over the divisor
12
with a line drawn between them. Thus, 12 -j- 4, 12 : 4, — ,
each means that 12 is to be divided by 4.
88. + 12 divided by + 4 gives the quotient + 3 ; since
only a positive number, + 3, when multiplied by + 4, can
give the positive product, + 12. § 61.
+ 12 divided by — 4 gives the quotient — 3 ; since only
a negative number, — 3, when multiplied by — 4, can give
the positive product, + 12. § 61.
— 12 divided by + 4 gives the quotient — 3 ; since only
a negative number, — 3, when multiplied by + 4, can give
the negative product, — 12. § 61.
— 12 divided by —4 gives the quotient +3 ; since only
a positive number, + 3, when multiplied by — 4, can give
the negative product, —12. § 61.
DIVISION. 46
(1) ^ = + 3. (3) =^ = -3,
(2) ±f = -3. (4) ^ = + 3.
Trom (1) and (4) it follows that
89. The quotient is positiye when the dividend and
divisor have like signs.
From (2) and (3) it follows that
The quotient is negative when the dividend and divisor
have unlike signs.
90. The absolute value of the quotient is equal to the
quotient of the absolute values of the dividend and divisor.
Exercise 19.
+ 264 ^ +3840 ^ 106.33
3. rrr-- =^ 5.
+ 4
— 4648 =
— 8
7.
-264
+24
ft
— 3670
-85
9.
+ 6.8603
— 30 —4.9
2568 -42.435
4. . ^^ = 6.
+ 12 +34.5
-7.1560
10.
11.
+ 324
-1
-3.14159
— 0.31831
— 61 ~" ^^' —31.4159
46 AIiGEBBA.
Division of Monomials.
91. If we have to divide dbc by be, aahx by ahyj 12 abe
by — 4 a^, we write them as follows :
abc aabx ax 12 abc
be aby y — 4ao
Hence, to divide one monomial by another,
92. Write the dividend over the divisor with a line between
them; if the expressions have common factors, remove the
common factors.
If we have to divide a* by a^, a^ by a*, a* by a, we write
them as follows :
a^
aaaaa
= aaa
= a^
a'
aa
a^
a*
aaxiaaa
aaaa
= aa
= a'
a
aaaa
a
aaa =
= a«.
93. That is, if a power of a number is divided by a
lower power of the same number, the quotient is that power
of the number whose exponent is equal to the exponent of the
dividend minus that of the divisor.
Again,
€? axi 11
a* aaaaa axia a*'
a^ aaa 1 1
a^ aaaaa aa a*'
a* aaaa 1
1
DIVISION. 47
^ 94. That is, if any power of a number is divided by a
higher power of the same number, the quotient is expressed
by 1 divided by the number ttdth an exponent equal to the
exponent of the divisor minus that of the dividend.
^
Exercise 20.
+ a
7.
10 ab
2bc
13.
— 3b7nx
4ax"
— a
8.
7?
14.
abc
+ a
9.
— 12 am
— 2m
= 16.
mp^a^
-«*-+s
10.
35 abed
5bd
16.
- 51 abdy^
-a ' *•
3bdy
6vtx
2x
11.
abx
5aby
17.
225mh/_
25 my^
12 a*
— 3a
12.
27 a'
-3a'
18.
30xy_
— 57^y
Sa'm'x'
=
21.
- a'b^cd' '
=
20. ^'/t^
=
22.
12 am^nYq^
4??iVpy
=
23. (4 a%^ X 10 a^'^z) -r- 5 a«Z> V =
24. (21 xY^ -^ 3 xy^z) (— 2 a^fz) =
25. 104a^V-5-(91a*^»V-f-7a*^>*x) =
26. (24a«6«aj-5-3a25^ + (35a%V-5- — 5a8&c) =
27. 85a**+^-«-5a*"-*= ^ 28. 84a'^-^12a2 =
48 ALGEBRA.
Of Polynomials by Monomials.
96. The product of (a + b-\-c) Xp = ap-{'bp-\'ep.
If the product of two factors is divided by one of the
factors, the quotient is the other factor. Therefore,
(ap -\-bp-\- <yp) -^ p=^a'\'h'\'C.
But a, h, and c are the quotients obtained by dividing
each term, ap, bp, and cp by p.
Therefore, to divide a polynomial by a monomial,
96. Divide each term of the polynomial by the m^anamial.
Exercise 21.
1. (Sab — 12ac)-^4a = 2b — Sc,
2. (15am — 10bm + 20cm)-^—5m = — 3a + 2b — 4:C,
3. (lSamy — 27bny + S6cpy)-^—9y =
4. (21ax — lSbx + 15cx)-^ — 3x =
6. (12a^ — Sa^ + 4:x)-T-4^x =
6. (Sar' — 6x^ + 9a;^ — 12x^-^3x*=
7. (35m^y + 2Smy-'Umf)-^—7my=z
8. (4:a^b — 6a^b^+12a^b^-^2a^b =
9. (12ar^/ — 15ary — 24a;V)-f-— 3x*y =
10. (12a;y-24a:y + 36a;V — 12 x«y«)-^ 12x^2/'=
^11. (3a*-2a^6-a«Z^-4-a* =
-> 12. (Sx'yii^+6x^s^ — 15x^y'z^ + lSxyz)'^ — Sa^yz:=^
13. (— 16a8W + 8a**V — 12a«&«c»)-^— 4a«^c« =
DIVISION. 49
Op Polynomials by Polynomials.
97. If the divisor (one factor) = a-^-b + c,
and the quotient (other factor) = »*+i>+2'>
an-\'hn'\' en
then the dividend (product) =-^ + a/? + 5^ + <5p
(iq-\'hq'{'cq.
The first term of the dividend is an ; that is, the product
of a, the first term of the divisor, by n^ the first term of the
quotient. The first term n of the quotient is therefore
found by dividing an, the first term of the dividend, by a,
the first term of the divisor.
If the partial product formed by multiplying the entire
divisor by n is subtracted from the dividend, the first term
of the remainder ap is the product of a, the first term of
the divisor, by p, the second term of the quotient. That is,
the second term of the quotient is obtained by dividing the
first term of the remainder by the first term of the divisor.
In like manner, the third term of the quotient is obtained
by dividing the first term of the new remainder by the first
term of the divisor, and so on.
Therefore, to divide one polynomial by another,
98. Divide the first term of the dividend by the first term
of the divisor.
Write the result a^ the first term of the quotient.
Multiply all the terms of the divisor by the first term of
the quotient,
Subtra^yt the product from the dividend.
If there is a remainder, consider it a^ a new dividend and
proceed cw before.
50 IXGEBBA.
99. It is of great importance to arrange both dividend
and divisor a,ccording to the ascending or descending powers
of some common letter ^ and to keep this order throughoitt the
operation.
Exercise 22.
Divide
(1) a« + 2a^ + **bya + 5; (2) a«-^bya + 5;
a» + 2a^ + y|a4-^ a'- l^\a±h
a^-\- db g 4" ^ a^-^ah a — h
ab-^V" —ah — b^
ab + b^ -ah-V
(3) a^'-2ab-\-}?\f^ a--b\
a^'-'2ah-\-l^\a — b
€? — ah a — b
— ah-^b^
— ab-^b^
(4) 4aV-4aV4-a;«~a«byx» — a«;
a;«-4aV4-4aV-a«|aj2— a*
r.« —
aj*-3aV+a*
--3aV + 4aV-a«
(6) 22aV+155* + 3a*-10a»5-22a^«bya« + 35«-2a^;
3a*-10a'6 + 22a*y — 22gj^« + 15^*| a«-2a6 + 36'
3a^— 6a^^+ 9a'^>' 3a« — 4a5 + 6^
— 4a«6 + 13a26*-22a*»
— 4a»^+ 8a«^« — 12aZ^
piYisiosr. 51
Divide
6. 5c*--7a; + 12by a: — 3.
7. aj* + x — 72by a; + 9.
8. Sfa^^ — aj' + Sx — 9 by 2a; — 3.
9. 6x' + Ux'-Ax + 24^hy2x + 6.
10. 3aj* + a; + 9a;« — Iby 3a; — 1.
11. 7a» + 58a;-24x* — 21by 7a; — 3.
12. x^ — 1 by a; — 1.
13. a»--2a5« + ^«by a — 5.
14. a;* — 81y*by a; — 32^.
15. a;* — y^bya; — y.
16. a« + 326*bya4-26.
17. 2a* + 27a^« — 815*by a + 35.
18. a;* + lla;« — 12a;-6a;« + 6by3 + a;«-3a;.
19. «* — 9a;« + a;« — 16a; — 4by a;*4-4 + 4a;.
""20. 36+a;*-13a;« by 6 + a;'+5a;.
21. a;*4-64bya;*+4a;4-8.
22. a;* + x8 + 57 — 36a; — 24a;«bya;'-3 + 2x.
23. 1 — X — 3a;* — a;'by l+2a;4-aj'.
24. a;« — 2a;» + lby a;* — 2a;4-l.
26. a* + 2aV + 9h^hj a^ — 2ab + Sb^.
26. 4a;« — ar' + 4a;by2 + 2a;« + 3a;.
27. a« — 243 by a — 3.
28. 18a;* + 82a;« + 40-67a-45a;«by3a;« + 6-4a;.
29. a;* — 6a;2^— 9a;* — y*by a;* + y + 3a;.
52 ALGBB&A.
SO. a5* + 9«»y» — 6a^y — 42/* by ar* — 3x^ + 2/.
31. a;* + a;*^ + 2/*by ic* — a;y + y«.
32. a^ + ^ + «V + 2^~'2a;^-—a'^ by a^ + a; — y.
33. 20?* — 3y"+a;y — «« — 4y« — «*by 2a;+3y + «-
34. 12 + 82ar« + 106a;*~70ar« — 112x» — 38 a;
by 3 — 5a; + 7a:*
36. a;'+2^ ^y x^ — a^i/ + a^i/^ — x^ + y\
36. 2a* + 2a;y — 2jry' — 7x«y — 2/*by 2ic* + 2/*--a;y.
37. 16a;* + 4icy + 2/*by 4a;*— 2«y + y'.
"38. 32a«i + 8a«^>« — ai« — 4a'*i* — 56a*i«
byi« — 4a% + 6ay.
39. l + 5a;»-6a;*byl— a; + 3a;*.
>40. l-52a*^>* — 51aWby 4a«^ + 3d^ — 1.
41. oj'y — xy' by x*y + ^^J^ ~~ 2a;'y^ — ^.
42. a;«+15a:y+15a;y + 2/* — 6a;*y — 6a;2/^ — 20a;*y'
^ bya;«-3a;V+3a;3^— y».
bya^— 2a«^ — ai«.
44. 81a;V + 18a;y — 54a;*y' — 18a;»3/* — 180;^' — 9/
by3a;* + ay + y*
46. a^ + 2a^b + Sa^b^ + Sab^+16b^hj a^ + UK
46. 82/* — a;«+21a;'y» — 24a;y^by 3a;y — «« — 3^.
47. 16a*+9^* + 8a«^«by4a« + 3^»«— 4a*.
48. a«+^»« + c» — 3a*c by a + 6 + c.
49. a« + 8*«:f c» — Go^cby a^+U^+c' — ac — 2ab'-2^,
60. a« + &* + c* + 3a** + 3a*«bya + * + c.
DIVISION. 53
100. The operation of division may be shortened in some
cases by the use of parentheses. Thus :
^'{•((i + b + c)x^+(ab-^ac-\'be)x '\-abe\x +h
a^+( +b )x^ x^+(a+c)x+ae
(a + c) a^ + (ab + ac + ic) 05
(a +c)x^'\-(a^ -{-bc)x
CLCX -{-dbc
EXEBCISE 23.
Divide
1. a'»(^ + c) + ^»«(a — c) + ca(a — i) + a^ by a + ft + c.
2. aj*— (a + ft + c)a;*4-(a^ + ac + ^c)a; — aic
hy x^—{a-\'h)X'\-(ib,
3. a^'-2a3i^ + {a^ + db'--h'^X'-a% + ah^hy X'-a-\'h.
4. x^ — (a* — b — c)3t^—(b — c) ax '\- be hy x^ — ax -}• c,
5. j^ — (m + n +i?)y* + (wiw + mj) + nj))y — mnp by y — p:
6. a^+(5 + a)x»— (4 — 5a + ^»)a;«-(4a + 5i)a; + 4ft
by a* + 5a; — 4.
7. x*''(a+b+c+d)a^'\'(ab+ac+ad+bc'\'bd+cd)x^
— (abc + a^c? + acd + ^ccQa; + al)cd
by a:^ — (« + c)x + ac.
8. «■ — (m — c)x^+(n — cm-\'d)a^-^(r-\'cn — dm)x*
+ (cr + dn) x-{-drhj a^ -;- mx^ -\- nx + ^.
9. as" — mx^ + waj* — nx^-\'mx — Ibyo; — 1.
10. (x + yy + 3(x + yyz + 3(x + y)z' + ;^
hj (x + yy + 2(x + y)z + z?^.
X
54 ALGEBRA.
101. There are some cases in Division which occur so
often in algebraic operations that they should be carefully
noticed and remembered.
Case I.
The student may easily verify the following results :
(1) ^—^ = a^+ab + b\
a — 0
(3) ^^^ = a^-\-a% + a%^ + ah'^ + h\
^ a — 0
r,5 OO 7^
(4) ^ = a* + 2a% + 'ia%'' + 8ab''+16b\
\ Erom these results it may be assumed that :
\ 102. The difference of two equal odd powers of any two
numbers is divisible by the difference of the numbers.
It will also be seen that :
I. The number of terms in the quotient is equal to the
exponent of the powers.
II. The signs of the quotient are all positive.
III. The first term of the quotient is obtained, as usual,
by dividing the first term of the dividend by the first term
of the divisor. .
IV. Each succeeding term of the quotient may be ob-
tained by dividing the preceding term of the quotient by
the first term of the divisor, and multiplying the result by
the second term of the divisor (disregarding the sign).
DIVISION. 56
EXSRGISB 24.
Write by inspection the results in the following exam-
ples :
1. (3^ — l)-^(y-l). 6. (a^^^)^(x--y).
2. .(6» — 125)-^(i-5). 6. (a'-l)-^(a~l).
3. (a» — 216)-^(a-6). 7. (1-8 «»)-§- (1—2 a).
4. (x» — 343)-5-(a;-7). 8. (x* — 32^-5- (a — 2^).
9. (8aV-l)-5-(2aa; — 1).
10. (1— 27a^y^-5-(l-3ajy).
11. (64a»6» — 27a:»)-5-(4ai^ — 3a;).
12. (243a« — l)-^(3a — 1).
13. (32a' — 243^) -5- (2a — 3ft).
Casb II.
^ "^ a + h
(4) ^^g^^^^^^ = 81a;^-54a;«y + 36a;y-24a;/ + 16y^
\From these results it may be assumed that :
4.03. The sum of two equal odd powers of two number's is
divisible by the sum of tJie numbers.
The quotient may be found as in Case I., but the signs
are alternately plus and minus.
66 ALGEBBA.
Exercise 2B.
Write by inspection the results in the following exam-
ples :
1. (a^ + f)-^(x + y). 6. (8aV + l)-^(2aaj + l).
2. (a^ + y')-^(x + y). 6. (a^ + 27f)-i-(x + Sy).
3. (l + Sa^^(l+2a). 7. (a' + 32b^ ^ (a + 2b).
4. (27a^ + b^-^(3a + b). 8. (512a^/+^-f-(8a;y+«).
9. (729a» + 2166^-r-(9a + 6i).
10. (64 a« + 1000 ^8) -^ (4a + 10 i).
11. (64:a%^ + 27a^-i'(AaI> + Sx),
12. (a^ + 343)-^(a; + 7).
13. (27a^f+S^-^(3xt/ + 2z).
14. (1024a« + 243^^«)-5-(4a + 3^).
Case III.
. Erom these results it may be assumed that ;
104. T?ie difference of two equal even powers of two num-
bers is divisible by the difference and also by the sum of the
numbers.
When the divisor is the difference of the numbers, the
quotient is found as in Case I.
When the divisor is the sum of the numbers, the quo-
tient is found as in Case II.
/:
r.
DIYISION. 57
EXEBCISE 26.
Write by inspectioii the results in the following exam-
ples :
1. (oj* — /)-f-(x-y)- 8. (16a;*-l)-^(2« + l).
2. (a;* — y*)-^(a + y)- «• (81aV--l)-5-(3aaj — 1).
3. (a« — a«)-5-(a — «). 10. (81aV— l)-f-(3aaj+l).
4. (d« — a;«)-f-(a + a;). 11. (64a« — ^»«) -^ (2 a — ^).
5. (a;* — 81y^)-f-(a; — 3y). 12. (64a« — ^»«) -f- (2a + ^).
6. (ar*-8l2/*)--(a; + 3y). 13. (a^-7292^) - (x-3y).
7. (16aj* — l)-f-(2a; — 1). 14. (a;«-729y^ -5- (a; + 3y).
16. (81a* — 16c*) -f- (3a — 2c).
16. (81a* — 16c*) -5- (3a + 2c).
17. (256 a* — 10,000) -f- (4 a - 10).
18. (256 a* — 10,000) -^ (4 a + 10).
19. (625a;* — 1) -^(505- 1).
Case IV.
It may be easily verified that :
106. The sum of two equal even powers of two numbers
is not divisible by either the sum or the difference of the
numbers.
But when the exponent of each of the two equal powers
is composed of an odd and an even factor, the sum of
the given poWers is divisible by the sum of the powers
expressed by the even factor.
Thus, «• + y" is ^ot divisible by a; + y or by a; — y, but is
divisible by o^ + y^.
The quotient may be found as in Case II.
58 ALGEBRA.
Exercise 27.
Write by inspection the resiQts in the following exam-
ples :
1. ix' + f)^{^ + }/^. 6. {x^ + l)-^{x' + l).
2. (a« + l)--(a»+l). 7. (%4.x' + f)^(4.a?-\-f).
3. (a^^ + y'^-5-(a' + 2^. 8. (64 + «^ -^ (4 + a«).
4. (i^« + l)-^(^' + l). 9. (729a« + ^')-^(9a«+^.
6. (a" + ^'^-^(a*+^*). 10. (729c« + l)-f-(9c« + l).
Note. The introduction of negative numbers requires an exten-
sion of the meaning of some terms common to arithmetic and
algebra. But every such extension of meaning must be consistent
with the sense previously attached to the term and with general laws
already established.
Addition in algebra does not necessarily imply augmenJbationy ajs it
does in arithmetic. Thus, 7 + (— 6) = 2. The word sum, however,
is used to denote the result.
Such a result is called the algebraic sum, when it is necessary to
distinguish it from the arithmetical sum, which would be obtained by
adding the absolute values of the numbers.
The general definition of Addition is, the operation of uniting
two or more numbers in a sirigle ea^ession written in its simplest
form.
The general definition of Subtraction is, the operation of finding
from two given numbers, called minuend and subtrahend, a third
number, called dijff'erence, which added to the subtrahend will give the
minuend.
The general definition of Multiplication is, the operation of finding
from two given numbers, called multiplicand and muttipHer, a third
number, called produ^, which may be formed from the multiplicand
as the multiplier is formed from unity.
The general definition of 'Division is, the operation of finding the
other factor when the product of two factors and one factor are
given.
CHAPTER V.
Simple Equations.
106. An equation is a statement that two expressions
are equal. Thus, 4 a; — 12 = 8.
107. Every equation consists of two parts, called the
first and second sides, or members, of the equation.
108. An identical equation is one in which the two sides
are equal, whatever numbers the letters stand for. Thus,
^109. An equation of condition is one which is true only
when the letters stand for particular values. Thus, a + 5
:=8 is true only when a; =3.
110. A letter to which a particular value must be given
in order that the statement contained in an equation may
be true is called an unknovm number,
111. The value of the unknown number is the number
which substituted for it will satisfy the equation, and is
called a root of the equation.
112. To solve an equation is to find the unknown num-
ber.
113. A simple equation is one which contains only the
first power of the symbol for the unknown number, and is
also called an equation'©^ the yirs^ degree.
ALGEBRA.
114. If equal changes he made in both sides of an equor
tian, tJie results tvill be equal, § 43.
(1) To find the value of a; in a: + 6 = a.
x + 6 = a;
Subtract 6 from each side, x + 6 — 6 = a — 6;
Cancel +6 — 6, x = a — h,
(2) To find the value of a; in a; — ft = a.
ac — 6= a;
Subtract — 6 from each side, x — 6+6=a + 6;
Cancel — 6 + 6, x = a + 6.
The result in each case is the same as if h were trans-
posed to the other side of the equation with its sign
changed. Therefore^
^ - 116. Any term m^y he transposed from, one side of an
equation to the other provided its sign he changed.
For, in this transposition, the same number is subtracted
from each side of the equation.
116. The signs of all the terms on each side of an
equation may be changed ; for this is in effect transposing
every term.
117. When the symbols for the known and unknown
numbers of an equation are connected by the sign + or — ,
they may be separated by transposing the symbols for
the known numbers to one side and the symbols for the
unknown to the other.
118. Hence, to solve an equation with one unknown
number,
Transpose all the terms involving the unknoum numher to
the left side, and all the other terms to the right side; combine
SIMPLE EQUATIONS. 61
the Uke termsy and divide both sides by the coeffioietU of the
symbol for the unknown number,
119. To verify the resiilt, substitute the value of the
symbol for the unknown number in the original equation.
EXEBCISE 28.
Find the value of a; in
1. 5aj — 1 = 19. ' 8. 16a: -11 = 7* + 70.
2. 3a: + 6 = 12. 9. 24a; — 49 = 19a-- 14.
3. 24a; = 7a; + 34. 10. 3a; + 23 = 78 — 2a;.
4. 8a; — 29 = 26 — 3a;. 11. 26 — 8a; = 80 — 14a;.
6. 12— 5a; = 19 — 12a;. 12. 13 — 3a; = 6a; — 3.
6. 3a; + 6 — 2a; = 7a;. 13. 3a; — 22 = 7a; + 6.
7. 5a; + 50 = 4a; + 56. 14. 8 + 4a; = 12 a; — 16.
,^ 5a5-(3a;-7) = 4a;-(6a5-35). ^v-^^^- V^ 'r x-^^'
16. 6a; — 2(9— 4a;) + 3 (5 a; — 7) = 10a; -(4+ 16a; + 35).^
17. 9a; — 3(5a; — 6) + 30 = 0.
18. aj— 7(4a;-ll) = 14(a;-5)-19(8-a;) — 61.
19. (a; + 7)(a; — 3) = (a;-5)(a;-15).
20. (a; — 8)(a; + 12) = (a; + l)(a; — 6).
21. (a; — 2)(7 — a;) + (a; — 5)(a; + 3)— 2(a; — 1)+12 = 0.
22. (2a;-7)(a; + 5) = (9-2ar)(4-a;) + 229.
23. 14-a; — 5(a;— 3)(a; + 2) + (5 — a;)(4-5a;)=45a;— 76.
24. (a; + 5)*— (4-a;)2 = 21a;.
25. 5(a;-2)«+7(a?-3)«=(3a;-7)(4x-19)+42.
62 ALGSBBJu
Exercise 29.
problems.
1. Find a number such that when 12 is added to its double
the sum shall be 28.
Let X = the number.
Then 2 x = its double,
and 2 z + 12 = doable the number increased by 12.
But 28 = double the number increased by 12.
.•.2xH-12 = 28.
2x = 28-12.
2x=16,
x = 8.
2. A farmer had two flocks of sheep^ each containing the
same number. He sold 21 sheep from one flock and
70 from the other, and then found that he had left in
one flock twice as many as in the other. How many
had he in each?
Let X = the number of sheep in each flock.
Then x — 21 = the number of sheep left in one flock,
and X — 70 = the number of sheep left in the other.
.•.x-21 = 2(x-70),
x-21 = 2x — 140.
x-2x= -140 + 21,
-x=-119,
X = 119.
3. A and B had equal sums of money ; B gave A $5, and
then 3 times A's money was equal to 11 times B's
money. What had each at first?
Let X = the number of dollars each had.
Then x + 5 = the number of dollars A had after receiving
$5 from B,
aud X — 5 = the number of dollars B had after giving A ^6.
SIMPUEB BQUATIONS. 68
.•.8(aj + 6) = ll(x-6);
8«+16=Hx-66;
8« — llx=-66-16;
-8x=-70;
x = 8i.
Therefore^ each had f 8.75.
4. Pind a number whose treble exceeds 50 by as much as
its double falls short of 40.
Let X = the number.
Then 3 x = its treble,
and 3 X — 60 = the excess of its treble over 60 ;
also, 40 — 2 X = the number its double lacks of 40.
.•.3x — 60 = 40 — 2x;
8x + 2x = 40 + 60;
6x = 90;
x=18.
6. What two numbers are those whose difference is 14,
and whose sum is 48?
Let X = the larger number.
Then 48 — x = the smaller number,
and X — (48 — x) = the difference of the numbers.
But 14 = the difference of the numbers.
.•.x-(48-x) = 14;
x-48H-x=14;
2x = 62j
x=31.
Therefore, the two numbers are 31 and 17.
6. To the double of a certain number I add 14, and obtain
as a result 154. What is the number?
7. To four times a certain number I add 16, and obtain as
a result 188. What is the number?
8. By adding 46 to a certain number, I obtain as a result
a number three times as large as the original number.
Find the original number.
ALGEBRA.
One number is three times as large as another. If I
take the smaller from 16 and the greater from 30,
the remainders are equal What are the numbers ?
10. Divide the number 92 into four parts, such that the
first exceeds the second by 10, the third by 18, and
the fourth by 24.
11. The sum of two numbers is 20 ; and if three times the
smaller number is added to five times the greater,
the sum is 84. What are the numbers ?
12. The joint ages of a father and son are 80 years. If the
age of the son were doubled, he would be 10 years
older than his father. What is the age of each?
13. A man has 6 sons, each 4 years older than the next
younger. The eldest is three times as old as the
youngest. What is the age of each ?
14. Add 11^24 to a certain sum and the amount will be as
much above $80 as the sum* is below $80. What
is the sum?
15. Thirty yards of cloth and 40 yards of silk together
cost $330; and the silk cost twice as much a yard
as the cloth. How much did each cost a yard ?
16. Find the number whose double increased by 24 exceeds
80 by as much as the number itself is less than 100.
17. The sum of $500 is divided among A, B, C, and D. A
and B have together $280, A and C $260, and A
and D $220. How much does each receive?
18. In a company of 266 persons composed of men, women,
and children, there are twice as many men as women,
and twice as many women as children. How many
are there of each?
SIMPLE EQUATIONS. 65
19. Find two numbers differing by 8, such that four times
the less may exceed twice the greater by 10.
A is 58 years older than B, and A's age is as much
above 60 as B's age is below 50. Find the age of
each.
A man leaves his property, amounting to 11^7500; to be
divided among his wife, his two sons, and three
daughters, as follows: a son is to have twice as
much as a daughter, and the wife l|^500 more than
all the children together. How much is the share
of each?
22. A vessel containing some water was filled by pouring
in 42 gallons, and there was then in the vessel seven
times as much as at first. How much did the vessel
hold?
23. A has $72 and B has $52. B gives A a certain sum ;
then A has three times as much as B. How much
did A receive from B ?
24. Divide 90 into two such parts that four times one part
may be equal to five times the other.
25. Divide 60 into two such parts that one part exceeds
the other by 24.
26. Divide 84 into two such parts that one part may be
less than the other by 36.
Note I. When we have to compare the ages of two persons at a
given time, and also a number of years after or before the given time,
we must remember that both persons will be so many years older or
younger.
Thus, if z represent A's age, and 2 x B^s age, at the present time,
A's age five years ago will be represented by x ~ 6 ; and B's by
2 a! — 6. A's age five years hence will be represented by x + 5 ; and
B*6ageby2x + 6.
66 ALaSBRA.
27. A is twice as old as B^ and 22 years ago he was three
times as old as B. What is A's age ?
28. A father is 30 and his son 6 years old. In how many
years will the father be just twice as old as the son ?
29. A is twice as old as B, and 20 years ago he was three
times as old. What is B's age ?
30. A is three (imes as old as B, and 19 years hence he
will be only twice as old as B. What is the age of
each?
31. A man has three nephews ; his age is 50^ and the joint
ages of the nephews is 42. How long will it be
before the joint ages of the nephews will be equal
to that of the uncle ?
Note II. In problems involving quantities of the same kind
expressed in different units, we must be careful to reduce all the
quantities to the same unit.
Thus, if X denotes a number of inches, all the quantities of the same
kind involved in the problem must be reduced to inches.
Q 32. A sum of money consists of dollars and twenty-five-
cent pieces, and amounts to $20. The number of
coins is 60. How many are there of each sort ?
33. A person bought 30 pounds of sugar of two different
kinds, and paid for the whole $2.94. The bette^
kind cost 10 cents a pound and the poorer kind 7
cents a pound. How many pounds were there, of
each kind?
34. A workman was hired for 40 days, at $1 for every day
he worked, but with the condition that for every
day he did not work he was to pay 45 cents for his
board. At the end of the time he received $22.60.
How many days did he work?
SIMPLE EQUATIONS. 67
35. A wine merchant has two kinds of wine; one worth 60
cents a quart, and the other 75 cents a quart. From
these he wishes to make a mixture of 100 gallons,
worth $2.40 a gallon. How many gallons must he
take of each kind?
36. A gentleman gave some children 10 cents each, and
had a dollar left. He found that he would have
required one dollar more to enable him to give them
15 cents each. How many children were there ?
37. Two casks contain equal quantities of vinegar ; from
the first cask 34 quarts are drawn, from the second,
20 gallons ; the quantity remaining in one vessel is
now twice that in the other. How much did each
cask contain at first?
38. A gentleman hired a man for 12 months, at the wages
of $90 and a suit of clothes. At the end of 7
months the man quits his service and receives
$33.75 and the suit of clothes. What was the price
of the suit of clothes ?
39. A man has three times as many quarters as half-
dollars, four times as many dimes as quarters, and
twice as many half -dimes as dimes. The whole sum
is $7.30. How many coins has he in all ?
40. A person paid a bill of $15.25 with quarters and half-
dollars, and gave 51 pieces of money all together.
How many of each kind were there ?
41. A bill of 100 pounds was paid with guineas (21 shil- .
lings) and half-crowns (2J shillings), and 48 more
half-crowns than guineas were used. How many of
each were paid ? <^:p PS
■ ' I r.
CHAPTER VI.
Factobs.
150. An expression is rational if none of its terms con-
tain square or other roots. It is often important to
determine the rational and integral factors of a given
expression.
151. Case I. The simplest case is that in which all the
terms of an expression have one common factor. ThuS;
(1) a^ + xy = x(x + y).
(2) 6a« + 4a«+8a = 2a(3a« + 2a + 4).
(3) 18a«^-27aW + 36a^ = 9ad(2a«— 3aft + 4).
Exercise 30.
Besolve into factors :
1. 5a*— 15a. 4. Aafy — 12 a^i^+Sxt^.
2. 6a»+18a* — 12a. 6. t/^ -- aj^ + bf + cy.
3. 49a«— 21a;4-14. 6. 6a«^-21a*^ + 27aV.
7. 54aj«/+108a:V-243a:y.
8. 46ajy<»— 90a:*/ — 360xy.
9. 70 ay -140 ay + 210 ay
10. 32a»«»« + 96a«^— 128aW.
FACTORS. 69
122. Case IL Frequently the terms of an expression
can be so arranged as to show a common factor. Thus,
(1) a:^ + ax + bx + ab=^(a^+ax) + (bx + ab),
= x (x -{- a) -{- b (x-^ a),
=:(x+bXx + a),
(2) ac — ad — bC'{-bd=(ac — ad)'—(bc — bd),
=^a(c — d) — b(c'—d),
= (a — ^)(c — (^.
Exercise 31.
Resolve into factors :
1,0^ — ax — bx-\-ab, 6. abx — aby-^pqx — pqy,
2. ab-^-ay — by — ^. 7. cdx^-^adxy — bcxy — ab^.
3. bc-^bx — ex — x\ 8. obey — l^dy — acdx-^bdh:.
4. mx-{'mn'\'ax-\-an. 9. ax — ay — bx-{-by.
6. cdoc^ — cxy-^dxy — ^. 10. cdz^ — cyz-\-dyz — j^,
123. The square root of a number is one of the two equal
factors of that number. Thus, the square root of 25 is 5 ;
for, 25 = 5X5.
The square root of a* is a* ; for, a*=^a^X a^.
The square root of a%V is abc ; for, a^b\^ = abc X abc.
In general, the square root of a power of an expression
is found by taking the square root of the numerical co-
efficient and writing each letter with an exponent equal to
one-half the exponent of the letter in the given power.
The square root of a product may be found by taking
the square root of each factor, and finding the product of
the roots.
70 ALGEBRA.
The square root of a positive number may be either
positive or negative; for,
a*=aX a,
or, a^^-^aX—a'y
but throughout this chapter only the positive value of the
square root will be taken.
124. Case III. From § 73 it is seen that a trinomial is
often the product of two binomials. Conversely, a trino-
mial may, in certain cases, be resolved into two binomial
factors. Thus,
To find the factors of
aj« + 7a+12.
The first term of each binomial factor will obviously be x.
The second terms of the two binomial factors must be
two numbers
whose prodtict is 12,
and whose sum is 7.
The only two numbers whose product is 12 and whose
sum is 7 are 4 and 3.
.-. a* + 7a; +12 = (a; + 4) (a; + 3).
Again, to find the factors of a;* + 5a;y-f 6y*.
The first term of each binomial fa<jtor will obviously be x.
The second terms of the two binomial factors must be
two numbers
whose prodiLct is 6y*,
and whose sum is by.
The only two numbers whose product is 6y* and whose
sum is by are Sy and 2y,
.•.a^' + 5a:y + 63^=(a; + 3y)(a; + 2y).
FACTOBS.
71
Exercise
Find the factors of :
1. x'+llx + 2L
2. aj«+llaj + 30.
^. 2^+172^ + 60.
4. z'+13z + 12.
5. x' + 21x + 110.
6. 2^ + 35y + 300.
7. ^ + 23^ + 102.
8. x' + Sx + 2.
9. it' + Tx + e,
10. a« + 9aft + 86».
21. aV + 14a*aj + 33*l 24.
22. a\^ + 7acx + 10x\ 26.
23. xyz^ + 19xyz + ^S. 26.
32.
11. a*+13aa: + 36a*.
12. y' + ldpy + ^Sp^
13. ««+29^» + 100g«.
14. a* + &a^+e.
15. «« + 4««+3.
16. a2** + 18a^ + 32.
17. a^i/' + 7xY + 12.
18. «»" + 10««+16.
19. a^ + 9ah + 20b\
20. a;« + 9a;» + 20.
^^V+18a6c+65a».
A2 + 23rs5J + 90«^
m^n^ + 20 m^n^pq + Sl^?*!/*.
186. Case IV. To find the factors of
aj2— 9a; + 20.
The second terms of the two binomial factors must be
two numbers
whose product is 20,
and whose sum is — 9.
The only two numbers whose product is 20 and whose
sum is — 9 are — 6 and — 4.
.•.aj»-9a; + 20 = (x~5)(x~4).
72 ALGEBRA.
EXEBCISE 33.
Resolve into factors :
1. a;2 — 7a; + 10. 13. a%V— 13a^c + 22.
2. a:^ — 29a; + 190. 14. a;^ — 15a; + 50.
3. a2_23a + 132. 15. a;2__20a; + 100.
4. 5* — 30^ + 200. 16. aV-21aa; + 54.
5. z^ — ^3z + 460. 17. aV — 16 a5a; + 395*.
6. a;* — 7a; + 6. 18. aV — 24 <w;« + 143 «l
7. a;* — 4aV + 3a*. 19. a;2_20a; + 91.
8. a;* — 8a; + 12. 20. a;* — 23a; + 120.
9. z^ — 57^ + 56. 21. ;5:2-53« + 360.
10. 2/« — 7^^+12. 22. x^—(a + c)x + ac.
11. a;*y* — 27a;y + 26. 23. y'z^ — 2Sabyz + lSTaV.
12. a*5« — lla%» + 30. 24. c'd^- 30 abed + 221 a^b\
126. CaseV. To find the factors of
a;* + 2a; — 3.
The second terms of the two binomial factors must be
two numbers
whose product is — 3,
and whose sum is + 2. '
The only two numbers whose product is — 3 and whose
sum is + 2 are + 3 and — 1.
.•.a;« + 2a;-3 = (a; + 3)(a; — 1).
FACTORS. 73
Exercise 34.
Kesolve into factors :
1. x' + ex — T. 8. a«4-25a — 160.
2. a^ + 6a;--84. 9. ft« + 3ft*-4.
3. y' + Ty — eO. 10. ftV + S^c — 164.
4. 3/2_^i2y — 46. 11. c^«+16c*--100.
5. «« + ll« — 12. 12. c2 + 17c-390.
6. «* + 13«-140. 13. a^ + a-132.
7. a*+13tf — 300. 14. 3i^f«^ + 9xyz — 22.
187. Case VI. To find the factors of
a* — 6a; — 66.
The second terms of the two binomial factors must be
two numbers
whose prodtcct is — 66,
and whose sum is — 6.
The only two numbers whose product is — 66 and whose
sum is — 6 are — 11 and + 6.
.•.a* — 6x — 66 = (aj-ll)(a; + 6).
Exercise 36.
Besolve into factors :
1. a* — 3a;— .28. 6. a^ — 15wl00.
2. y«-7y-i8. 7. c^o-Oc* — 10.
3. a;* — 9a; — 36. 8. a;^ — 8a; — 20.
4. :^ — llz — 60. 9. 2/2- 5ay — 50a«.
6. «« — 13«-14. 10. a^b^ — Sab — 'L
74 ALGEBBA.
11. aV— 3aaj — 64. 14. 3^«*-53^«»— 84.
12. c»cP — 24(?rf — 180. 16. a«^ — 16a^ — 36.
13. a^(? — ah — 2. 16. a^ — (a ■— b) x — ab.
We now proceed to the consideration of trinomials which
are perfect squares. These are only particular forms of
Gases III. and TV., but from their importance demand
special attention.
188. Case VII. To find the factors of
aj«+18» + 81.
The second terms of the two binomial factors must be
two numbers
whose prodicct is 81,
and whose sum is 18.
The only two numbers whose product is 81 and whose
sum is 18 are 9 and 9.
.\a^+lSx + Sl=^(x + 9)(x + 9) = (x + 9y.
Exercise 36.
Resolve into factors :
1. x' + nx + Se. 8. 2/* + 162/2«* + 64«*.
2. a;* + 28ir + 196. 9. 2/^ + 242^+144.
3. ic» + 34a; + 289. 10. aV + 162 ar« + 6561.
4. z' + 2z + l. 11. 4a» + 12aJ« + 95*.
6. 2^ + 2002^+10,000. 12. dx'y* + SOxf z + 25^.
6. «* + 14«« + 49. 13. 9a^ + 12xy+4:f.
7. aj* + 36a2/ + 3242/*. 14. 4aV + 20aV2/ + 25ajy.
FACTORS. 75
1S9. OabbVIII. To find the factors of
0^-1805 + 81.
The second terms of the two binomials must be two
numbers
whose product is 81,
and whose sum is — 18.
The only two numbers whose product is 81 and whose
sum is — 18 are —9 and — 9.
.•.a« — 18a: + 81=(a;-9)(a — 9) = (a;-9)».
Exercise 37.
Resolve into factors :
1. a^ — 8a + 16. 10. 4a;y — 20aV« + 26y*«*.
2. a^ — 30a + 225. 11. lea^y^ — Sxy'z' + i/z*.
3. a2_33a._^35i^ 12. 9a%V — 6a^V(£ + WcP.
4. x* — 40aj + 400. 13. 16a;«— 8ajy + a;y.
5. 2/2 _ioOy + 2500. 14. a^x* — 2a%xy + by.
6. j/* — 20f + 100. 15. 36a;y — 60a;y»+252/*.
7. 2^ — 50y« + 625«l 16. l-eab^ + 9a^b\
8. x^ — S2xy + 25et/'. 17. 9mV — 24WH + 16.
9. «« — 34»» + 289. 18. Al^a^ -• 12 bxhj + 9 xy.
19. 49a« — 112aJ + 64ftl
20. 64a;y — 160a;y;sj + 100»V.
21. 49a26V — 28a5cx + 4x2.
22. 121a;*-286a:V + 169y2.
23. 2S9a^y'^'-102xf^d + 9y'z^(P.
24. 361x^fz^ — 76abcxyz + 4:a^b'(^.
76 ALAEBRA.
130. Case IX. An expression in the form of two
squares, with the negative sign between them, is the prod-
uct of two factors which may be determined as follows :
Take the square root of the first term, and the square
root of the second term.
The sum of these roots will form the first factor ;
The difference of these roots will form the second factor.
Thus:
(2) a^^(b^cy=\a+(h-c)\\a--{h-c)l
= |a + ^ — <^\\^ — ^ + <'|«
= |a — 5 + c — d\\a — h — c-^-dl,
181. The terms of an expression may often be arranged
so as to form two squares with the negative sign between
them, and the expression can then be resolved into factors.
Thus:
a^ + b^'-(^ — €^-{-2ab-{-2cd,
= a^ + 2ab + b^ — (^ + 2cd — d^,
= (a^ + 2ah+h')-{(^'-2cd + d^,
= (a + hy-{c^dy,
= \{a + h) + {c^d)\\{a + b)^{c--^l
=^\a + h + c — d\\a-^h — c + d\,
132. An expression may often be resolved into three or
four factors. Thus :
(1) x^'-y''=^(??+f)(7?^f)
= {^ + f){x' + y'){x' + f){x + y){x--y).
FAOTORS. 77
(2) 4((^ + cd)«-(a«4-6«-c*-<£«)»,
\2ab + 2cd-'a''-i^ + c'+d%
= \(a^+2ab + h^ — (c''-2cd + d*)l
\((^ + 2cd + d^'-(a^-'2ab + b^\,
= l(a+by-(c-^dyi\(c+dy^(a^byi
= \a + b+(c-d)\\a + b^(c^d)l
\c + d+(a^b)\\o + d^(a--b)l
= \a + b + c--d\\a + b—c + d\
le + d+a-^bWo + d — a + bl
Exercise 38.
Resolve into factors; -
1. o'—bK /l4. (a + by'-(c + dy.
2. a« — 16. / 15. (x + yy—(x — t/y.
3. 4a« — 25. 16. 2ab — a^'-b^ + l.
4. a* — b*, 17. Q^ — 2yz'-%y — ^.
5. a* — 1. 18. x^-'2xy\-if—:^.
y 6. c?'-b\ "'N.l^- a' + 126c — 4i»2 — 9c«.
7. a«— 1. ^'20. a«— 2ay+y*— V— 2aj« — «».
// 8. 36a:* — 49y». 21. 2xy — a;* — 2^ + «*.
. 9. lOOx^y* — 121aW 22. a;*+y*— ;^— rf*— 2a;y — 2(£«.
,10. 1 — 49a:*. 23. a:*— y*+«*— a»— 2x« + 2ay.
^1. a* — 25^. ' 24. 2a5 + a* + «»* — cl ;
'/12. (a — ft)* — c*. ^^26. 2a;y — a:*— 2^ + a*+ft*— 2aft. '
'^13. a:*— (a — 6)*. 26. (ox + fty)*- 1. y
78 ALGEB&A.
27. l-aj*-2^ + 2ajy. 31. (oj + 1)« - (y - 1)« y
28. (6a — 2)2— (a-4)l 32. (£« — a:* + 4xy — 42/«. ^/
"^ 29. a^ — 2ai + 5* — ir*. 33. a* — 5* — 26c — c».
30. (x+l)*-(y + l)l 34. 4x*-9x*+6a;-l.
183. CaseX.
a^ — V*
Since — 3^=a^ + a^yH-y*,
a? y
X* — V*
and ^ = a;* + a:*y + aj2y* + ac/ + y,
aj — y
and so on^ it follows that the difference between two equal
odd powers of two numbers is divisible by the difference
between the numbers.
EXEBCISE
39.
Besolve into factors:
1. a* -I?.
e. 8a:»-27y».
2. a* — 8.
7. 64y»— 1000«».
3. x»-343.
8. 729ie«-512y».
4, y«-125.
9. 27 a»- 1728.
6. y»-216.
10. 1000a'- 13316'.
184, Case XL
«» + «'
Since — — =
= x'-
-ax-{-
<
«* + '/
and — p-=^ =
x + i/
= x*-
-^i/-^
■a;y — a!y» + y*,
and so on, it follows that the sum of two equal odd powers
of two numbers is divisible by the sum of the numbers.
FACTORS. T9
EXERGISS 40.
Besolve into factors:
1. aj' + y'. 6. 216a» + 612o«.
2. af+S. 7. 729x«+1728y«.
3. ic»4-216. 8. a* + y«.
4. 2^ + 64««. 9. aj' + y'.
6. 64y + 125(j«. 10. 32y+243A
136. Case XII. The sum of any two powers of two
numbers, whose exponents contain the same odd factor, is
diyisible by the sum of the powers obtained by dividing
the exponents of the given powers by this odd factor.
Thus,
In like manner, x*® + 32y', which is equal to 05**+ (2y)',
is divisible by a:^ + 2 y ; but x* + t/*f whose exponents do not
contain an odd factor, and a^ + y^^ whose exponents do
not contain the same odd factor, cannot be resolved into
rational factors.
Exercise 41.
Resolve into factors :
1. a^ + b\ 3. a^ + y'^ 5. a«+l. 7. 64a« + ««.
2. a^^ + b^^. 4. ^« + 64c«. 6. a"+l. 8. 729 + c«.
\
80 AliGEBRA.
136. Case XIII. For a trinomial to be a perfect square,
the middle term must be twice t?ie product of the square roots
of the first and Uist terms.
The expression a* + ic*y* + y* will become a perfect
square if icV ^® added to the middle term. And if the
subtraction of o^^ from the expression thus obtained be
indicated, the result will be the difference of two squares.
Thus :
= (a* + ^ + «y) (a;* + 3^— »y),
or, (a:* + ajy + 3/*)(a;* — ay+y^.
Exercise 42.
Eesolve into factors:
1. a* + a«5« + **. 8. 49m* + 110mV + 81n*.
2. 9x* + 3aj*y«+4y*. 9. 9a* + 21aV+25c*.
3. IGx^-lTx'y' + y*. 10. 49a*-15aV + 121ft*. \
4. 81a* + 23a*6* + 16&*. 11. 64ir*+128a;y + 81y*. ^
6. 81a*-28aV + 16**. 12.*4a;*-37ay + 9y*.
6. 9a* + 38xy+492/*. 13. 25 a* - 41 aj«2^ + 16 2^ ^-
7. 25a* — 9a*** + 165*. 14. 81a;*-34«y + y*.
• If, in Example 12, 0 y* = (— 3 y^, then 26 x^^ should be added to
4 x* — 37 x^ + 0 y*, in order to make the expression a perfect square.
That is', we should have : ,
(4aj* - 12 x2y2 + 0 y4) - 25ajV,
= (2a;2-3y2)2-26x2y8,
= (2x2 - 3y2 + 5a;y) (2x2- 3y2 - 6xy),
or, (2x2 + 6xy — 3y2) (2x2 — 5xy- 3y2),
FACTORS.
81
187. Case XIV. To find the factors of
6a2 + «-12.
It is evident that the first terms of the two factors
might be 6 a; and x, or 2 a; and 3x, since the product of
either of these pairs is Cx'.
Likewise, the last terms of the two factors might be 12
and 1, 6 and 2, or 4 and 3 (if we disregard the signs).
From these it is necessary to select such as will produce
the middle ter^ of the trinomial. And they are found by
trial to be 3a; and 2x, and —4 and + 3.
.•.6x*+a;-12 = (3a;-4)(2a; + 3).
Exercise 43.
/
!Resolye into factors:
1. 12a;* — 5a; — 2.
2. 12x^ — 1 x + 1.
3. 12ar2-a; — 1.
4. 3a;« — 2a; — 5.
5. 3a;* + 4a;-4.
6. 6a;* + 5a; — 4.
7. 4a;* + 13a; + 3.
8. 4a;« + lla; — 3.
9. 4a;* — 4a; — 3. /
10. a;* — 3aa; + 2a2.^
11. 12a*+aV — a;*.
12. 2x^ + 5xi/ + 2i/.
13. 6aV + aaj — 1.
14. 6b^ — 7bx — 3a^.
15. 4a;* + 8a; + 3.
16. a* — aa; — 6a:*.
17. 8a* + 14a^-156*
18. 6a* — 19ac + 10c*.
19. 8a;* + 34a;y + 2l3^.
20. 8a;* — 22a;y — 21y*.
21. 6aj* + 19a;y — 7y*.
22. lla*-23aJ + 25*.
23. 2(^'-13cd + ed\
24. 62/*+7y« — 3«*.
&2 ALGEBRA.
188. Case XV. The factors, if any exist, of a polyno-
mial of more than three terms can often be found by the
application of principles already explained. Thus it is
seen at a glance that the expression
a'^ — Sa^b + Sab^ — V
fulfills, both in respect to exponents and coefficients, the
laws stated in § 83 for writing the power of a binomial ;
and it is known at once that
a^-Sa^b + Sab^-'b^=(a-by.
Again, it is seen that the expression
a^ — 2xy-j-i^-^2xz — 2yz-\-s^
consists of three squares and three double products, and
from § 79, is the square of a trinomial which has for terms
X, y, z.
It is also seen from the double product —2xyy that x and
y have unlike signs ;
and from the double product 2xz, that x and z have like
signs. Hence,
Qi? — 2xy-\-f + 2xz — 2yz + ^=^(x-'y+z)\
EXEBCISE 44.
Resolve into factors:
1. a? + Za^b-\-^a^ + b\ 4. x* + 4ajV + 6ajy+4V+2^V
2. a» + 3a« + 3a+l. 5. ic*-4a;« + 6a:»--4a; + l.
3. a» — 3a* + 3a — 1. 6. a* — 4a«c + 6aV — 4ac»+c*.
7. x^ + 2xy + f + 2xz-\-2yz + s^.
8. x^ — 2xy + f — 2xz + 2yz + s^.
9. a^-\'b^ + (^'\-2ab'-2ao--2bc.
FACTORS. 83
139. Case XVI. Multiply 2ar — 2/ + 3bya; + 2y — 3.
2x- y + 3
X + 2y 3
2a;2- xy +3ar
4.xy-2f +6y
~6a- + 3?/ — 9
2a:* + 3iry — 2^* — 3a; + 9y — 9
It is to be observed that 2x2 4- 3xy — 2y', of the product, is obtained
from (2x — y) X (X + 2y);
that — 9 is obtained from 3 X — 3 ;
that — 3x is the sum of 2x x — 3 and x x 3 ;
thatOy isthesum of 2y X3 and — 2^x — 3. '
From this result may be deduced a method of resolving
into its factors a polynomial which is composed of two
trinomial factors. Thus:
Find the factors of
6x2 — To;?^— 3^" — 9x-f30y — 27.
The factors of the first three terms are (by Case XIV.)
3x4-2/ aJ^d 2x — 3y
Now — 27 must be resolved into two factors such that the sum of
the products obtained by multiplying one of these factors by 3x and
the other by 2x shall be — 9x.
These two factors evidently are — • 9 and -f 8.
That is, {^Q^—lxy-'3y^ — 9x + Z0y — 21)
= (3aj + 2^~9)(2aj~3y + 3).
140. The following method is often most convenient for
separating a polynomial into its factors :
Find the factors of
2x^ — 5xy + 2f-\'lxz — ^yz-\-Z9?.
1. Reject the terms that contain z,
2. Reject the terms that contain y,
3. Reject the terms that contain x.
84 ALOEBBA.
Factor the expression that remains in each case.
1. 2x«-6xy4-2y2 = (x-22/) (2x-y).
2. 2x« + 7x«4-3z2=(x4-32)(2«+«).
3. 2y2-5yz4-322=(-2y + 32)(-y + «).
Arrange these three pairs of factors in two rows of three factors
each, so that any two factors of each row may have a common term.
Thus:
x — 2y, x + Sz, — 2y4-3«;
2x — y, 2x + z, —y + z.
From the first row, select the terms common to two factors for one
trinomial factor:
X — 2y + 3z.
From the second row, select the tenns common to tu)0 factors fbr
the other trinomial factor :
2x — y + «.
Then, 2a:* — 5xy + 2y^ + 7xz — 5yz + 3z^
= (x-22/ + Sz)(2x-7/ + z).
141. When a factor obtained from the first three terms
is also a factor of the remaining terms, the expression is
easily resolved. Thus :
a^-3xy+2y'-Sx + 6y,
= (x-2y)(x-y)-3(x-2y),
= (x^2y)(x-y-S).
Exercise 45.
Resolve into factors:
1. 2x^—5xy + 2y' — 17x + 13y+2i:
2. ea^ — STxy + ef — Bx — By — l.
3. 6x^ — 5xy — 6y^ — x — 5y — l.
<h 4. 5a^ — Sxy + 3f + 7x-5y + 2.
^^^,^5. 2x'-xy-3i/-Sx + 7y+e.
FAGTOB8. 85
^6. aj«— 252^-lOa; — 20y + 21.
,-^"'^7. 2a^ — 5x^ + 2^ — xz — yz — «*.
U^8. 6x' + xy-'f—Sxz + 6yz — 9!^.
, 9. 6a^— 7a;y + 3^ + 35aj« — 6y« — 6«".
- 10. 5iK*— Sajy + Sy* — 3a;« + y«--2«*.
11. 2iK* — a?y — 3y*— 5y« — 2«*.
W~ 12. 6a;« — 135cy + 6z/« + 12a:« — 13y« + 6«»
13. x^ — 2xy + y^ + 5x — 5y,
JL 14. 2a^ + 5a;y — Sy* — 4a;« + 2y«.
Exercise 46.
miscellaneous examples.
The following expressions are to be resolved into fac-
tors by the principles already explained. The student
should first carefully remove all monomial factors from
the expressions.
f 1. 5x2— 15x — 20. 9 a^^a^ + 1.
L 2. 2x* — 16x* + 24a^. 10. a^—f — xz + yz.
a 3. 3a^b^ — 9ab — 12. 11. ab — ac — b^ + be. ''
y/^^. a* + 2aa: + x*+4a + 4a;. . 12. 3a^-— 3aj« — ^y-^ry^*
6. a^ — 2(ib'\-l^'-i». 13. a^ — Q^ — ab — bx,
6. Qi?—2xy-^f-~<?-^2cd — d'^. 14. a^ — 2aa; + a^ + a — a.
y^ 7. 4 — a:^ — 2a:* — X*. 15. Sx'-'3y^ — 2x + 2y.
^ 8. aJ' — l^ — a — b, 16. a* + x* + aJ* + «. ^
17. aV — aV-aV + 1. ^=^
18. 3a*— 2aj«y-27xy« + 18y».
86
ALGEBRA.
19.
20.
21.
22.
23.
24.
b 25.
26.
27.
37.
38.
X.42.
43.
44.
^ 45.
46.
47.
' '-r 48.
^49.
^ 69,
61.
62.
63.
b
u
0
4a:*-aj«+2a;— 1. 28. 4a2 — 4a^ + 6«. ^
a* — 2/^. 29. 16a;2 — 80iC2/ + 100y».
aj« + y^. 30. ^^a^7^f — 26hVf.
729— ««. 31. 9ay — 30V« + 26»«.
x^y\-y^. 82. 16ic«— a.
a*c — c«. 33. :x^-'2xy — 2xz + f + 2yz + ^.
cc2 + 4«-21. 34. a«-a5-6**-4a + 125.
3a2— 21a5+305*. 35. 7^ + 2xy + f — X'-y — ^.
2x^—4:X^y—6xy, 36. (a + 5)* — c*. C
x^ — xy — ef — 4:x + 12y, 39. Scc^ — lla^z + Gy*. /'
1 — x-f-a^^ — a^.
40. a:2 + 20ic + 91.O r,l7
41. (x-y)(x'-z^^(x-z)(a^-f).
a:*— 5aj — 24.
(x^ — y'--zy — 4:fz\
Si^f + 5a^yz-'60xz'.
Sa^ — a^ + Sx — 1.
7? — 2 ?/ia5 + m* — w^
4a«^-(a2+«>2_^a^
1 — 14a8x + 49aV.
68. a^-^h^ — aia?
N]^
50. 2/^ — 4^ — 117.
51. ^ + 6^-135. ^
52. 4a2— 12a5+9^'-4c
63. (a + 3^)2-9(^^-0)2.
54. 9a;2— 42/' + 4y« — «2.
65. e^^ica — 7^^— 3«*. G
66. d^ — h^ — Zahia — hy-f
57. aJ*+2^+3xy(x + 2/).
■52) + ft(a-^)*.
60. 6aj2 + 135cy + 6y2. ^>
6a%2-a5« — 125*.
a2 + 2ac^ + rf2_4^2^125c-9c2.
«»— 2a;V + 4a:y2 — 8/. 64. 4aW — 8a5x + 3ftl
FACTORS. 87
I \
\ 66. ISic^ — 24a;y+8y»+9aj — 6y. 74. 16a»a; — 2aj*.
V 66. 2sc' + 2xy-'12y' + 6xz + lSyz. 76. 32&b» — 4V-
/ 67. (x + yy — l—xy^x + y + l). 76. a; — 27a;*.
/ 68. a^'-y^ — z^ + 2yz + x + y'-'Z. 77. a:" — y".
( ^^9. 2x^ + 4:xy + 2f + 2ax + 2ay. 78. 49w« — 121n«.
"!:> 70. 16a^b + 32abc + 12bc'. 79. 16 — 81^.
y*' 71. m^j) — m^q — n^p + n^q. SO. 12«* — «* — 6. J
^72. 12ax^'-14:axy — 6ay'. 81. a* — aj^ + a; — 1.
G 73. 2a;« + 4aj2 — 70x. 82. a;2+2a;+l — y».
83. 49(a-*)*-64(m-7i)l
<'
85. a;2~53a; + 360.
86. a'' — 2x^ + 2?^ — 4a; + 82/ — 4. ^
87. 2ab — 2bc'-ae-^ce + 2b^ — be.
• /
^ f
88. 125a;* + 350a:«2/2 + 245 a;y*. ^ f, j^
f 89, a^ + a'^b + a^b^+a^b^ + a^b^ + al/'.) ""
,90. 2a*a; — 2a*ca; + 2ac^a; — 2c*a;.
91. 6x2 — 5a;z/ — 6y* + 3a;« + 15y« — O^*. ^ ^ ^
92. 4:X^ — ^xy + 2i/ — Zxz'-vz^z\ . \.^ v »
93. 3a2-7a5 + 2*2^5ac-5^c + 2o2. ' "^
.--^--7 94. a;* — 2a;» + a;2 — 8x + 8.
96. 6a;2 — 8a;2/ + 32^-5a; + 3y. ^^
96. a^ — 2arf + ei» — 45* + 126c — 9A
97. (a;2-a;-6)(ar^ — a;-20). .f
\
CHAPTER VII.
Common Factors and Multiples.
142. A common factor of two or more expressions is an
expression which is contained in each of them without a
remainder. Thus,
5 a is a common factor of 20 a and 25 a;
3a;V is a common factor of 12icy and 15icV.
148. Two expressions which have no common factor
except 1, are said to be prime to each other.
144. The Highest Common Factor of two or more ex-
pressions is the product of all the factors common to the
expressions.
Thus, 3 a* is the highest common factor of 3 a*, 6a^; and
12 a*.
63?}^ is the highest common factor of lOicy and 15icy.
For brevity, H. C. F. will be used for Highest Common
Factor.
(1) Find the H. C. F. of ^2a%^x and 21a*^)V.
A2a^b^x =^2X^X1 XaJ^Xh^Xx]
21a*6V = 3 X 7 X a* X 5' X a^.
.-. the H. C. F. =3 X 7 X a* X 52 X a;
= 21a%^x,
(2) Find the H. C. F. of 2 a*ic + 2 ax^ and 3 ahxy + 3 hQi?y.
2 a^x + 2 aaj* = 2 oaj (a + a;) ;
Sdbxy + Sboc^y = Sbxi/ (a + x).
.\theIl.G,^.=x(a + x),
COMMON FAGTOBS AND MULTIPLES. 89
(3) Find the H. C. F. of
8aV — 24a*aj + 16a«and 12aa:«y — 12 ooy — 24 ay.
8aV— 24a«x + 16a*=8a»(aj« — 3aj + 2),
= 2V(a;-l)(a; — 2);
12aa^y — 12aajy — 24ay = 12ay(a;* — ic — 2),
= 2«X3ay(x + l)(a — 2).
/. the H. C. F. =22a (a: — 2),
= 4a(a: — 2).
Hence, to find the H. 0. F. of two or more expressions :
Resolve each expression into its lowest factors.
Select from these the lowest power of each common factor^
and find the product of these powers.
EXEBCISB 47.
Find the H. C. F. of :
1. ISal^i^d and 36a%crf» 2. 17^ 34^^^, and 51j?V.
3, 8«>/«^ 12a^2^«», and 20aV«'-
4. 30a;V, 90iB»/, and 120a^3/*.
6. a*— ^anda»--ft» 7. a«+ic» and (a + «)«
6, a*— a:* and (a — aj)* 8. 9a^ — 1 and (3aj+l)*
9. 7a^— 4aj and 7a^ — 4a*.
10. 12 aVy — 4 a'ajy* and 30 aVy* — 10 a'aV-
11. Sa^V'c — 12a^h<? and 6a^*c+4a^A
12. a:*— 2ic — 3 and»»+a; — 12,
13. 2a« — 2a^and4ft(a + 5)*.
^14. 12aj»y(a; — y)(aj — 3y) and 18a:*(aj — y)(3aj — y).
16. 3aj»+6a:2 — 24»and6a^— 96a5.
90 ALGEBRA.
16. ac(a — ^)(a — c) and 5c(5 — a)(5 — c).
17. 10a^y — e0a^y*-^5xi/^SLTid5a^y^'--5xy^'--100t/^.
18. a;(a; + l)«, iB«(x* — 1), and 2a;(a:* — a; — 2).
19. 3ic*— 6aj + 3, 6a^ + 6« — 12, andl2a^ — 12.
20. 6 (a — by, 8 (a* — b^, and 10 (a* - b').
21. a^-3/«, (a: + y/, and a^ + Sxi/ + 2y'.
22. x* — 3/*, a^ — ^, and aj*— 7a;yH-62/*.
23. «2— 1, «« — 1, ando^ + a^ — 2.
146. When it is required to find the H. C. F. of two or
more expressions which cannot readily be resolved into
their factors, the method to be employed is similar to that
of the corresponding case in arithmetic. And as that
method consists in obtaining pairs of continually decreas-
ing numbers which contain as a factor the H. C. F. required ;
so in algebra, pairs of expressions of continually decreasing
degrees are obtained, which contain as a factor the H. C. F.
required.
The method depends upon two principles :
1. Any factor of an expression is a factor also of any
multiple of that expression.
Thus, if F represent a factor of an expression A^ so that A = nF,
then mA = mnF, That is, mA contains the factor F.
2. Any common factor of two expressions is a factor of the
sum or difference of any multiples of the expressions.
Thus, if F represent a common factor of the ejqpressions A and B
so that
A = mF, and B=nF;
then pA = pmF, and qB = qnF.
Hence, pA ±qB== pmF ± qnF,
= (pm ± qn)F.
That is, pA ± qB contains the factor F.
COMMON FACTORS AND MULTIPLES. 91
146. The general proof of this method as applied to
numbers is as follows:
Let a and b be two numbers, of which a is the greater.
' The operation may be represented by :
b)a(p
42)154(3
nF)mF(j)
pb
, 126
jmF
c)b(3
28)42(1
cF)nF(q
qe
28
qcF
d)c(T
14)28(2
F)eF(c
rd
28
cF
Pj 9, and r represent the several quotients,
c and d represent the remainders,
and d is supposed to be contained exactly in c.
The numbers represented are all integral.
Then c = rd,
5 = grc + d = grrd + (J = (gr + 1) (J,
a = i>6 + c = pqrd +pd 4- rd,
= (pqr + p + r)d,
.*. d is a common factor of a and b.
It remains to show that d is the highest common factor of a and b.
Let / represent the highest common factor of a and 6.
Now c = a — j?6, and / is a common factor of a and b,
.-. by (2) / is a factor of c.
Also, d = b—qc, and / is a common factor of b and c.
.*. by (2) / is a factor of d.
That Is. d contains the highest common factor of a and 6.
But It has been shown that d is a common factor of a and b,
.'. d is the highest common factor of a and 6.
NoTB. The second operation represents the application of the
method to a particular case. The third operation is intended to rep-
resent clearly that every remainder in the course of the operation
contains as a factor the H. C. F. sought, and that this is the highest
factor commxm to that remainder and the preceding divisor.
92 ALGEBRA.
147. By the same method, find the H. C. F. of
2a^+x — 3 and 4.x^ + Sx^—x — 6.
2a^+x-'S)4:X^ + Sx*— x — 6(2x + S
4:0t^+2a^ — 6x
6x^ + 5x — 6
2x + 3)2x'+ x — Six — l
— 2ic — 3
.-. the H. F. C. = 2a; + 3. — 2a;T-3
The given expressions are arranged according to the descending
powers of x.
The expression whose first term is of the lower degree is taken for
the divisor ; and each division is continued until the first term of the
remainder is of lower degree than that of the divisor.
148. This method is of use only to determine the com-
pound factor of the H. C. F. Simple factors of the given
expressions must first be separated from them, and the
highest common factor of these must be reserved to be
multiplied into the compound factor obtained.
Find the H. C. F. of
12x^ + 30a^ — 72x^a.nd32a^ + S4tx^-176x.
12x^ + 30x^'-72x' = 6x'(2x' + 5x'-12).
32i^ + S4:X^ — 176x'^ix(Sx^ + 21x — U).
6a^ and 4 a; have 2x eommon,
2a^+5a; — 12)8a;2 + 21a; — 44(4
8a^ + 20a; — 48
x+ 4)2a^ + 5a; — 12(2a;-3
'" 27^+Sx
— 3a;-12
.•.theH.{F/a = 2ar(aj + 4). -3a;~12
COMMON FACTORS AND MULTIPLES. 93
149. Modifications of this method are sometimes needed.
(1) Find the H. C. F. of 4x'' — Sx — 5 and 12x^ — 4:X — 65.
4x« — 8a; — 5)12x2— 4a; — 65(3
12a;^ — 24 a; — 15
20a; -50
The first division ends here, for 20 x is of lower degree than 4 as*.
But if 20x — 60 be made the divisor, 4x2 will not contain 20 x an
integrcU number of times.
Now, it is to be remembered that the H. C. F. sought is contained
in the remainder 20 x — 50, and that it is a compound factor. Hence
if the simple fo/dtor 10 be removed, the H. C. F. must still be con-
tained in 2 X — 5, and therefore the process may be continued with
2 X — 6 for a divisor.
2a;-5)4a;*- 8a; — 5(2a; + l
4a;«-10a;
2a; — 5
2a;-5
.-.theH. C. F. =2a; — 5.
(2) Find the H. C. F. of
21a;» — 4a;« — 15a;-2and21a;« — 32a;2 — 54a;-7.
21a;« — 4a;* — 15a;-2)21a;« — 32a;* — 54a;-7(l
21a;«— 4a;* — 15a; — 2
-28x* — 39a; — 5
The difficulty here cannot be obviated by removing a simple factor
from the remainder, for — 28x2 — 39x — 5 has no simple factor. In
this case, the expression 21x8 — 4x2 — 15x--2 must be multiplied by
the simple factor 4 to make its first term divisible by — 28 x^.
The introduction of such a factor can in no way affect the H. C. F.
sought ; for the H. C. F. contains only factors common to the remain-
der and the last divisor, and 4 is not a factor of the remainder.
The signs of all the terms of the remainder may be changed ; for
if an expression A is divisible by — F, it is divisible by +F.
94 AL6EBBA.
The process then is continued by changing the signs of the remain-
der and multiplying the divisor by 4.
2Sx^ + 39x + 5)S4:X^— 16x*— 60a;— 8(3a;.
S^a^ + 117x'+ 15x
— 133ic*— 75a;- 8
Multiply by — 4, 3^4
632a;« + 300a; + 32(19
532a;« + 741a; + 95
Divide by - 63, — 63 )— 441 a; — 63
7a;+ 1
7a; + l)28a« + 39aj + 5(4x + 5
28a:^+ 4:x\
S5x + 5
.•.theH.C. F. = 7a; + 1. 35a; + 5
(3) Find the H. C. F. of
8a;* + 2a; — 3 and 60^ + 5x^ — 2.
6a^+ Bx"— 2
Multiply by 4, 4
8a;« + 2a;-3)24a;» + 20a;*— 8 (3a; + 7
24a;«+ 6x^— 9x
Ux^+ 9a;— 8
Multiply by 4, _4
56 a;* + 36a; — 32
56a;^ + 14x — 21
Divide by 11, H )22x — 11
2x- l)8a;> + 2a;-3(4a; + 3
8 a;' — 4a;
6a; — 3
.-. the H. C. F. =2a; — 1. ex-^^
In this case it is necessary to multiply by 4 the given expression
6a;* + 6a* — 2 to make its first term divisible by Sx^, 4 being obvi-
ously not a common factor.
COMMON FACTOBS AND MULTIPLES.
96
The following arrangement of the work will be found
most convenient :
•
8aJ + 2x — 3. ,
6«»+ 6jb*- 2.
4
6x — 3
6x — 3
2.4x»+20«»- 8
24a^+ 6a;»— 9?;
3a;
14a;'+ 9a;- 8
4
66x* + 36a;-32
56x'+Ux-21
11) 22 a; — 11
2a;- 1
+ 7
4a; + 3
150. From the foregoing examples it will be seen that,
in the algebraic process of finding the highest common
factor, the following steps, in the order here given, must
be caref uUy observed :
I. Simple factors of the given expressions are to be re-
moved from them, and the highest common factor of these
is to be reserved as a factor of the H. C. F. sought.
II. The resulting compound expressions are to be ar-
ranged according to the descending powers of a common
letter ; and that expression which is of the lower degree is
to be taken for the divisor ; or, if both are of the same
degree, that whose first term has the smaller coefficient.
III. Each division is to be continued until the remainder
is of lower degree than the divisor.
IV. If the final remainder of any division is found to
contain a factor that is not a common factor of the given
expressions, this factor is to be removed ; and the resulting
expression is to be used as the next divisor.
V. A dividend whose first term is not exactly divisible
by the first term of the divisor, is to be multiplied by such
an expression as will make it thus divisible.
96 AliOEBRA.
EXEBCISE 48.
Find the H. C. F. of :
•^ 1. 5x^+4:X'-l, 20x'+21x — B.
2. 2«»- 4a:«— 13a; — 7, 6ar8— 11a:*— 37a; — 20.
8. 6a*+25a»-21a* + 4a, 24a* + 112a« — 94a2 + 18a.
r' 4. 9a;» + 9a:* — 4a; — 4, 45a;« + 64a;« — 20a; — 24.
5. 27a;« — 3a;* + 6ar» — 3a;*, 162a;« + 48a;»— 18a;* + 6a;.
/6. 20a;»-60a;* + 50a; — 20, 32a;*-92a;» + 68a;* — 24a;.
^•- 7. 4a;« — 8a; — 5, 12a;* — 4a; — 65.
8. 3a* — 5a*a; — 2aa;*, 9a*— 8 a*a; — 20 aa;*.
9. 10a;* + a;* -9a; + 24, 20a;^ — 17a;*+48aj-3.
\10. 8a;* — 4a;* -32 a; -182, 36a;* — 84a;* — Ilia; — 126.
11. 5a;*(12a;*+4a;*+17a;-3), 10a;(24a;*— 52a;*+14a;— 1).
/ 12. 9a;V — a;*y^ — 20a;^, 18ar^y — 18a;*y* — 2a;2^ — 8y*.
13. 6a;* — a; — 15, 9a;*— 3a; — 20.
14. 12a;*-9a;*+5a; + 2, 24a;* + 10a; + l.
15. 6a;*+15a;*-6a; + 9, 9a;* + 6a;* — 51a; + 36.
16. 4a;* — a;*y — ary*- 5^^, 7a;* + 4a;*y + 4ajy* — 3^.
17. 2a* — 2a* — 3a — 2, 3a* — a* — 2a — 16.
/IB. 12y* + 2y*— 94y-60, 48y*-24y*-348y + 30.
19. 9a;(2a;* — 6a;*-a;* + 15a; — 10),
6a;*(4a;* + 6a;* — 4a;* — 15a; — 15).
^^0. 15a;*+2a;*-75a;*+5a; + 2, 35a;*+a;*-175a;*+30a;+l.
/ 21. 21a;* — 4a;* — 15a;* — 2a;, 21 a;* -32a;* — 54a; -7.
22. 9a;*y— 22a;*/— 3a;^+10y', 9a;«y — 6a;y+a;*y*— 26a;y*.
COMMON FACTORS AND MULTIPLES. 97
.— 23. 6aj* — 4a*— llaj« — 3x2 — 3a — 1,
4aj*+2a^— 18a» + 3« — 5.
^24. «* — aaj« — aV — a^x- 2a*, Sa^'-7aa^ + 3a^X'-2a\
151. The H. C. F. of three expressions will be obtained
by finding the H. C. F. of two of them, and then of that
and the third expression.
For, if A, B, and C are three expressions,
and D the highest common factor of A and B,
and liJ the highest common factor of D and (7,
then 2> contains every factor common to A and B,
and U contains every factor common to D and C.
.'. U contains every factor common to A, B, and C.
EXEBCISB 49.
FindtheH. C. F. of:
1. 2x^-\-x — l, a? + bx+^, 7? + l,
2. 1^-f-y + l, 3/-2y-l, ,/^f + y^l,
3. a;«— 4a2+9a-10, aj«+2a2— 3x-+2(), x^^bx^—^x+3^.
4. a^ — 7a^ + 16x — 12, Sx' — Ux^ + iex,
Ba^ — lOx^+Tx — U. )
22^-72^ + 16y-15.
6. 2x* + 3x — 5y3a^—x — 2,2a^ + x — 3.
7. a« — 1, a^ — a*— a — 2, 2a;^— a^ — x — 3.
8. ar« — 3a; — 2, 2a:«+3a« — 1, a'+l.
9. 12(a*-2^), 10(a«-y^, 8(aV + a^2/*>
10. 05* + a;^, oi?y-\-t/, a;* + ^V + 2^*-
11. 2(a^y — xf), 3(a^y-x2/), 4(aV-a2^), 5(a*y-x2^).
(
W AL6BBKA.
Lowest Common Multiple.
162. A common multiple of two or more expressions is
an expression which is exactly divisible by each of them.
163. The Lowest Common Mnltiple of two or more ex-
pressions is the product of all the factors of the expres-
sions, each factor being written with its highest exponent.
164. The lowest common multiple of two expressions
which have no common factor will be their product.
For brevity L. C. M. will be used for Lowest Common
Multiple.
(1) Find the L. C. M. of 12 a% Ubc", 36 a^.
12a^c = 2^X3a%
Ubc^ = 2 Xlh<?,
36aZ^ = 22X3«a5«
.-. The L. C. M. =2« X 32 X 7aWc2 = 252a»ftV.
(2) Find the L. C. M. of
2a^ + 2ax, Ga^ — 6a;^ 3a*— 6aaj + 3x*.
2a^ + 2ax =2a(a + aj),
6a2 — 6aj« =2 X 3(a + «) (« — «),
3a* — 6aaj + 3aj«=3(a — «)«.
.-.theL. C. M.-=6a(a + aj)(a — a;)*
Exercise 50.
Find the L. C. M. of
1. 4:a\ ^ah^, 2a7?, 4. a:*— 1, cc* — aj.
2. 18aaj*, 72 ay*, 12 ay. 6. a* — 5*, a^-\-aJb.
3. a^, ax-^T?. 6. 2a: — 1, 4a:* — 1.
COMMON FACTOSS AND MULTIPLES. 99
7. a + b, a^ + b\ 9. a* — oj, «« — 1, a^ + 1.
8. 05*— 1, aj' + l, «*— 1. 10. x^ — 1, x* — a, «« — 1.
11. 2a + l, 4a2-.l, 8a«+l.
12. (a + by, a^ — V.
13. 4(l + aj), 4(l-aj), 2(l-jr«).
14. « — 1, a^ + oj + l, a;« — 1.
16. x«-y«, (x + y)*, (x-y)*.
16. ^-f, S(x-yy, 12(x«+y»).
17. 6(a;«+xy), S(xy-f), lOix'-f).
18. a^+5x + 6, x2+6aj + 8.
19. a*— a — 20, a^ + a — 12.
20. x« + lla; + 30, a^+12x + 35.
21. x' — 9x — 22, X* — 13a; + 22.
22. 4:ab(a^ — 3ab + 2b^, 6a«(a« + a* — 66^.
23. 20(x« — 1), 24(x» — X — 2), 16(x2 + x — 2).
24. 12xy(a^-y^, 2a^(x + yy, 3f{x-yy.
25. (a — b)(b — c),(b'-c)(c — a), (c — a)(a — b).
26. (a — b)(a — c), (b — a)(b — 6), (c — a)(c — b).
27. x« — 4x« + 3x, x* + x« — 12x«, x* + 3x* — 4x».
28. x'y — xy', 3x(x — yy, 4y(x — y)«.
29. (a + ^)2^(o + c^)^ (a+c)2-(i + c0', (a+^^-(5+c)^
30. (2x— 4)(3x — 6), (x — 3)(4x — 8), (2x — 6)(5x — 10).
166. When the expressions cannot be readily resolved
into their factors, the expressions may be resolved by find-
ing their H. C. F.
100
ALGEBRA.
Find the L. C. M. of
6aj» — lla^ + 2y« and 9x» — 22 V — 83^.
3a^!/ + 4.xf+2f
3a^y + 4.xy' + 2j^
9aj» — 22aj2^-8y*
2
ISa^ — Uxt^ — ief
lly)33a^y~44V~22y«
3a* — 4ajy — 2f 2x—y
Hence, 6aj« — llcr«y + 2y*= (2aj — y) (3a:* — 4a;y — 2y«),
and 9ar»-22a:y' — 8y^=(3x+4y)(3a:* — 4ajy — 2y*).
.-. the L. C. M. =(2a; — y)(3x + 4y)(3a^ — 4a;y — 2y»).
In this example we find the H. C. F. of the given expres-
sions, and divide each of them by the H. C. F.
166. It will be observed that the product of the H. C. F.
and the L. C. M. of two expressions is equal to the product
of the given expressions. For,
Let A and B denote the two expressions, and D their
H. C. F.
Suppose A = aDf and B = bD;
Since D consists of all the factors common to A and B,
a and b have no common factor.
.'. L. C. M. of a and b is ah.
Hence, the L. C. M. of aD and bD is ohD.
Now, A = aD, and B = bD;
.\AB = abDXD,
.*. -jr- = abD=the lowest common multiple. That is,
iTie L. C. M. of two expressions can be found by dividing
their product by their H. C. F.
Or, by dividing one of the expressions by the H. C. F., and
multiplying the result by the other expressidn.
COMMON FACTORS AND MULTIPLES. 101
157. To find the L. C. M. of three expressions -4, Bj C.
Find M, the L. C. M. of ^ and ^ ; then the L. C. M. of M
and C is the L. G. M. required.
EXEBCISE 51.
Find the L. CM. of:
1. 6x* — X — 2, 21x'—17x + 2, 14a^+6aj-l.
2. «*— 1, a^ + 2a; — 3, 6a^— a — 2.
3. x8_27, x^-15x + 3e, ar8-3a^-2aj + 6.
4. 6a^+19ic — 4, 10«» + 13a; — 3.
6. 12^ + aji^-62^, lSa^+lSxi/ — 20y'.
6. a:* — 2x« + a:, 2x*— 2a^ — 2ic — 2.
7^ 12a:« + 2« — 4, 12a:«-42« — 24, 12a^— 28aj — 24.
8. a»-6a^ + llx-6, a»-9x*+26»~24,
x» — 8a;«+19a; — 12.
9. a^ — 4a^ aJ» + 2aa* + 4a*aJ + 8a^
0^ — 2aa*+4a*aj — 8a«.
10. a»+2a^y — ay* — 23^, aj« — 2a^y — ajy«+23^.
11. l+i>+i>*, l-i>+i?*, l+y+i>*.
12. (1-a), (l-a)«, (l-a)»
,^ 13. (a + c)« — «»«, (a + ft)« — c*, (b + cy — a^
rl4. 3c»-3c«y + cy«-3^, 4c»-c«y-3c2^.
16. m» — 8w + 3, m« + 3w* + m + 3.
16. 207i* + w« — 1, 267i* + 57i« — n — 1.
17. b^-2b^ + b^Sb + S, Ab^-12h^ + 9b'-l.
18. 2r«—8r* + 12r» — 8r« + 2r, 3r'-6r» + 3r.
CHAPTEE VIIL
Fractions.
4
168. In Arithmetic, an expression of the form of - is
employed to indicate that 4 units are divided into 5 equal
parts, and that one of these parts is taken ; or, that one
unit is divided into 6 equal parts, and that 4 of these parts
are taken.
The expression is also used to indicate the qiiotient of 4
divided by 5,
4
159. The expression - is called a fraction. 4 is the nn-
merator, and 5 the denominator.
160. The numerator and denominator together are called
the terms of the fraction.
161. The denominator shows into how many equal parts
the unit is divided and therefore names the part ; and the
numerator shows the number of the parts taken.
It will be observed that a figure written above the line
in a fraction serves a very different purpose from that of a
figure written below the line.
A figure written above the line denotes nnmber;
A figure written below the line denotes name.
162. Every whole number may be written in the form
4
of a fraction with unity for its denominator ; thus, 4 = --
FBAGTION8. 103
To Beduce a Ebagtion to its Lowest Tebms.
163. Let the line AB be divided into 5 equal paxts^ at
the points C, D, E, F.
a\ \ \ \ \ \ \ \ \ \ \ t \ \ \ \b
C B E F
Then AF is f of AB. (1)
Now let each of the parts be subdivided into 3 equal parts.
Then AB contains 15 of these subdivisions, and AF con-
tains 12 of these subdivisions.
.-. AF is \i of AB. (2)
Comparing (1) and (2), it is evident that f = |f.
In general :
If we suppose AB to be divided into b equal parts, and
that AF contains a of these parts, where a and h signify
positive whole numbers,
Then ^i^ is ~ of ^15. (3)
Now, if we suppose each of the parts to be subdivided
into c equal parts, e being a positive whole number.
Then AB contains be of these subdivisions, and AF con-
tains ae of these subdivisions.
.-.^L^^'is^^of^^. (4)
uC
Comparing (3) and (4), it is evident that
a ao
I'^Te
Since — is obtained by multiplying by e both terms of
the fraction 7>
0
and, conversely, - is obtained by dividing by c both terms
of the fraction -;-, it follows that
be
104 ALGEBRA.
If the numerator and denominator of a fraction he muUi-
plied by the same number, or divided by the same number,
the value of the fraction is not altered.
In Algebra, a fraction simply indicates the quotient of
the numerator divided by the denominator, but the laws
that apply to fractions in Arithmetic apply also to frac-
tions in Algebra.
164. To reduce a fraction to lower terms.
Divide the numerator and denominator by any common
factor.
A fraction is expressed in its lowest terms when both
numerator and denominator are divided by their H. C. F.
Eeduce the following fractions to their lowest terms :
. c? — 7? {a — 7^{a^-\'ajx-\'7?) €?-\'ax-\'7f
^ ^ a* — 7? (a — «)(« + «) a + aj
a»+7a + 10^(a + 5)(a + 2)^a + 5
^^ a^-^-ha-^^ (a + 3)(a + 2) a + 3
6x^->5a;-6 ^ (2a; -3) (3a; + 2) ^ 3a; + 2.
W
8a;2 — 2a; — 16 (2a; — 3) (4a; + 6) 4a; + 5
a« — 7tt'+16<^ — 12
3a« — 14a2 + 16a '
Since in Ex. (4) no common factor can be determined
by inspection, it is necessary to find the H. C. F. of the
numerator and denominator by the method of division.
FBACTIOKS.
105
Sappress the factor a of the denominator and proceed to diyide :
a»- 7a»+16a- 12
3
3a»-14a + 16
3a*— 6o
o — 7
3a»— 21a"+48o— 36
3a» — 14a* + 16a
— 7a* + 32a— 36
3
— 8a+16
— 8aH-16
3a — 8
— 21a''+96a-108
— 21a' + 98a-112
-2)- 2a+ 4
a— 2
.-. the H. C.
F. = a — 2.
Now, if a« — 7 a* +16 a — 12 be divided by a-'2f the
result isa* — 5a + 6; and if 3a' — 14 a* +16 a be divided
by a — 2, the result is 3 a* — 8 a.
. a^-7a* + 16a-12^a'-6a + 6
•• 3a»-14a« + 16a 3a« — 8a '
166. If the common factors cannot be determined by
inspection, the H. G. F. must be found by division.
EXEBCISE 52.
Seduce to lowest terms :
a^-1
1.
a» + l
4aj(a- + l)
«« — 7aj + 12'
a:* — 2a; — 3
a«-10a; + 2l'
x' + x^+l
a^ + x + 1
g«+2a^+y«.
aJ* — y*
7.
8.
9.
a«+2a« + 2a + l
g* — a — 20
a«+a-12'
g» — 4x*+9a; — 10
a» + 2a;^-3aj + 20'
g»-5a;*+lla; — 15
a^-x' + Sx + B
a;* — a'y — ajy* — 3/*
106 ALGEBBA.
a''-3a + 2 a* - a*b - a'b' + atf' ,
ic' + a^ — aj — 1 a^—ah — 2V
^"^ aj» — a:* — 2a; + 2 4(a^b-'ab^^
_ 4x* — 12aa; + 9a3 a« + 2a^ + ft« — c»
14 • 24 ■ . •
a« + 2(^ + 5*-c2 /' (a-ft)2-(c + rf)2
a;^ — x^ — 2a; + 2 / 6a;« — 5a; — 6
2a;« — a; — 1 ( 8a;^ — 2a;-15
\ x«-6x^+lla;-6 .; ^^ + ^V + 2/*
«» — 2x«-a; + 2 \ {x — y){7?'-f)
_ 6a;« — 23a;«+16a; — 3 ^ aj« + y»
6ar»-17a;«+lla;-2 \ a;*-«»y' + 2/*
-^ aj*-2a;«-x*-2a; + l v"" (a»-5«)(a^-a5 + 5»)
To Reduce a Fraction to an Integral or Mixed
Expression.
aj^ + 1
Change ——7 to a mixed expression,
a; "~~ J.
{a?^l)-^{x—l)=a?->rx-\-l-\ — ^- Hence,
166. ij^ ^Ae degree of the numerator of a fraction equals or
exceeds that of the denominator, ths fraction may be changed
to the form of a mixed or integral expression by dividing the
numerator by the denominator.
FRACTIOlTfl. 107
If there is a remainder, this remainder must be written
as the numerator of a fraction of which the divisor is the
denominator, and this fraction with its proper sign must
be annexed to the integral part of the quotient.
EXEBCISE 63.
Change to integral or mixed expressions :
1. - • 6. z •
X — 1 oa — aj
^ Sx' + 2x + l ^ 16(3a;^ + l)
x + 4 ' 4x — 1
05 + 4 * X — 4
a* — 0x4- x^ ' a^4-b^
4. -j--' 9. ~
a-\-x a — 0
2a:^ + 5 5a^-a;^ + 5
aj-3 ' 5x^ + 4:x-l
To Reduce a Mixed Expression to the Eobm op a
Fraction.
167. In arithmetic 5f means 5 + f .
But in algebra the fraction connected with the integral
expression, as well as the integral expression, may be posi-
tive or negative ; so that a mixed expression may occur in
any one of the following forms :
.a a , a a
108 ALGEBRA.
Change n+-r to a fractional form.
Since when 6 is a positive integral number there are
b bths in 1, in n there will be n times b bths^ that is, nb bths,
which, with the additional a bths, make (nh + a) bths.
.a nb-\-a
In like manner :
n — T
a nb — a .
, a — nh-^ra
-"+*=— j— 5
J a — nh — a
and — n — -= r •
0 0
The same law holds whatever be the value of b.
168. Hence, to reduce a mixed expression to a fraction,
Multiply the integral expression by the denominator^ to
the product annex the numerator^ and under the result urrite
the denominator.
169. It will be seen that the sign before the fraction is
transferred to the numerator when the mixed expression is
reduced to the fractional form, for the denominator shows
only what part of the numerator is to be added or sub-
tracted.
The dividing line has the force of a vinculum or paren-
thesis affecting the numerator ; therefore if a minus sign
precede the dividing line, and this line be removed, the
sign of every term of the numerator mu^t be changed.
Thus,
a — b C7i — (a — b) en — a-^b
c c c ■
(1) Change to fractional form x — 1
x — 1
FBACTIONS. 109
X
x-1
X
^ — x-\-(x — 1)
X
x^-^x + x^l
X
x^ — l
X
aj-1
X
(2) Change to fractional form a; — 1 -
. x-1
05— -1 9
X
g^ — a; — (g — l)
X
7? — X — x-\-l
= — }
x
^x^ — 2x-{-l
X
Exercise 54.
Change to fractional form :
2. 1 + ^^^
3, Zx
x-\-y
X
4. «-«=+^z:^-
o.
oa — jii
oa —
6b
6.
a^b-
a^ + b'
a + b
7.
la ^
-3a + 4
a'
(a ■
5 — 6a
8.
3.-^
ax — 3
n
2a
110 ALGEBRA.
9. — ^ + 1. 16. 2a — ft r-T-
a — 0 a + 6
10. ^J-l. /l6. 3X-10+ ^
a + ft ' / aj + 4
12. 6^^ + 6a + 3x. 18. ^_3a,-5£fc^.
4 aj — 2
1 6a^
13. a — l-\ r-rr- 19. a'— 2aa;+4x2
1 a-\-2x
r— 20. x — a + y] f-L-^.
Lowest Common Denominatob.
170. To reduce fractions to equivalent fractions haying
the lowest common denominator :
Reduce -]—i> zr^y and -r—i to equivalent fractions having
4a^ 3a 6a®
the lowest common denominator.
The L. C. M. of 4a», 3a, and 6a8 = 12a^
If both terms of j-^ be multiplied by 3 a, the value of
the fraction will not be altered, but the form will be
9aa; 2?/
changed to t^t— g ; if both terms of ^ be multiplied by 4 a*,
the equivalent fraction tk^ is obtained ; and, if both terms
5 10
of —^ be multiplied by 2, the equivalent fraction To~8 is
obtained.
Hence,
FRACTIONS. Ill
3x 2y ^
. ^ 9ax Sa'y 10 ^ ,
are equal to _j, -_j, __, respectively.
The multipliers 3 a, 4 a*, and 2 are obtained by dividing
12 a', the L. C. M. of the denominators, by the respective
denominators of the given fractions.
171. Therefore, to reduce fractions to equivalent frac-
tions having the lowest common denominator,
JF^ind the L. C. M. of the denominators.
Divide the L. C. M. hy the denominator of each fraction.
Multiply the first numerator by the first quotient, the
second by the second quotient, and so on.
The products will be the numerators of the equivalent
fractions.
The L. C. M. of the given denominators unll be the
deriominator of each of the equivalent fractions.
Exercise 55.
Reduce to equivalent fractions with the lowest common
denominator :
, 3aj— 7 4a;-9 ^ 1 1
1. -z > -r-T — • 5.
18 * ia — b){b'-c) (a — b){a — c)
2 a; — 4y 3x--8y ^ 4a;' xy
Bx" lOaj 3(a + by 6(a* — Z»«)
4a — 5c 3a — 2c ^ 8a; + 2 2aj — 1 3a; + 2
* • TT ' 7i ;; '
5ac 12a^e a;-2 3a;-6 5aj — 10
5 6 a — bm i c — bn
8. 9 ^9
1 — a; 1 — a* " 7»a;
112 ALOEBBA.
Addition and Subtbaction of Fbactioks.
172. To add fractions:
Reduce the fractions to equivalent fractions having the
lowest common denominator.
Add the numerators of the equivalent fractions.
Write the result over the lowest common denominator.
178. To subtract one fraction from another:
Bedu^e the fraMions to equivalent fractions Tiaving the
lowest comm^on denominator.
Subtract the numerator of the subtrahend from the numer-
ator of the minuend.
Write the result over the lowest common denominator.
(1) Simplify, _E_ + __.
The lowest common denominator (L. C. D.) = 15.
The multipliers are 3 and 1 respectively.
12 a; + 21 = Ist numerator,
Sx — 4 = 2d numerator,
16 a; + 17 = sum of numerators.
. 4a; + 7 3a; — 4^15a; + 17
•'' 6 "^ 15 16 "
(2) Simplify, 1 j^
TheL. C. D. = 84.
The multipliers are 12, 28, and 7 respectively.
36 a — 48 ft = 1st numerator,
— 56a + 286 — 28c = 2d numerator,
91a — 28c = 3d numerator.
71a — 20 b — 66 c = sum of numerators.
. 3a — 4ft 2a — b + c lSa — ^c71a — 20b — 56c
•'•7 3 "^ 12 "" 84
FRACTIONS. 113
Note. Since the minm »ign precedes the second fraction, the
signs of all the terms of the numerator of this fraction are changed
after being multiplied by 28.
Exercise 56.
Simplify :
Sx — 2y 5x — 7y Sx + 2y
5x "^ lOaj "^ 26
3a:> "^ 6aj "^ 12 *
Aa^ + 5b^ , Sa + 2b , 7~2a
^' 2b' "^~5^+~9~'
4fl; + 5 3a; — 7 9_
3 5a; '^12a;^'
4a;~3y 3a; + 7 5a; — 2y 9a; + 2y
7 "•" 14 21 ^ 4.2 *
/><e 3a;y — 4 5y^+7 6a;' — 11
x^ xx^ ix^y
a''-2ac + (^ b^'-2bc + c'
5a« — 2 3a^ — dt
8a« 8
a — 5 5 — c c — a . ab' -{- bc^ -^ ca^
e a b abc
_! 1 1 2x — z y — 2z
'^ 2x*y Qfz 2xz^'^ 4xV "•" 4x»y« '
. \
114 ALGBBBA.
Simplify ^+5±y.
TlieL.C.D.=a:> — y*.
The multipliers are x — y and x+y, i^si>ectiyel7.
x^ — 2xy'\- y* = Ist Dumerator.
x^'\-2xy'\- y^ = 2d numerator.
2a? + 2y* = sum of numeratoi&
or, 2(a?+f)
u u u
''x + y x — y a*— -y*
Exercise 67.
Simplify :
^ ■ 1 e. 1
* x — 6 aj + 5 * 2a(a + aj) 2a (a — a;)
_J 1_ __a h__
x — 1 x — S (a + h)b (a — b)a
3. T-^ h:: 8.
1+x ' 1— aj • 2aj(aj — 1) 4a;(a; — 2)
_J 2_ 1 + a? 1 — x
1-aj 1-aj*' l + aj + a* l-a; + a«'
1 X 2aX'-Sby 2ax + 3by ^
x — y"^ (x — yy ' 2xy(x-'y) 2xy(x+y)
(1) Simplify
a — b a-\-b a* — b^
The L. C. D. = (a — ^) (a + b).
The multipliers are a + &, a — &, and 1, respectively.'
FILiCTIONS. 115
2 a* + 3 oft + ^* = 1st numerator,
— 2a^'{-3ab — i*=2d numerator,
— Qab =3d numerator,
0 = sum of numerators.
2a + b 2a — 5 6ab _
' ' a — b a-\-b a^ — b^
The L. C. D. = (x + y)(x — y)(x^ + f).
The multipliers are, respectively:
x2 4- y2, (X - y) (x2 4- y2), (x 4- y) (z — y) (x* 4- y% (x + y) (x - y)
a^y + ^ = Iflt numerator,
— a;^ + 2ic*^ — 2a;y + 2xy' — y*=2d numerator,
«* — y* = 3d numerator,
2 a:^^ — 2 x^ = 4th numerator,
4 Qi^y — xy — y* = sum of numerators.
.-. Sum of fractions = — y-'^lr'-'y ,
iC* — /
Exercise 58.
Simplify :
1 . 1 . 2a « « x^ , X
1 + a^l — a^l — a« l-x 1 — x^l+x*
1 1 , 2x ^ x'y A _j^__
1 — X 1+a; l + x* ' y x + y os' + xy
g — 1 . X — 2 a; — 3
a: — 2'^aj — 3'*"x — 4'
3 , 4a 5a»
X — a (x — of (x — of
116
ALOEBBA.
3
X — 1 x + 2 (x + l)(x + 2)
a — b , h — c
e — a
(b + e)(cJra) ' (c + a)ia-\-b) ' (a + J)(*+c)
X — « , 33 — h (g — by
x — h X — a (x — a)(x — b)
10.
11.
12.
13.
14.
16.
c-\'a
(6 — c)(c — a)"^(c— a)(a — 5) ' {a'-b)(b — c)
a^ — bc , ^' — ac , (^'\-<Lb
(a + b){a + cy (b-\-a){b + cy {c + b)(c+a)
a x a^-^-a?
a — X a-\-2x (a — x)(a']-2x)
3 4.6
(a — b)(b — c) (a — ^)(a — c)"^(a — c)(^ — c)
X — 2y 2x-\-y 2x
a — b a — b (a — ^)(^H~y).
x(a + b) y(a + b) xy(a + b)
/17 3x x + 2y 3y .
/■''• (x + yy a?-^^(x^yy
18.
a — b
(a + ^)2— c« (a + cy — V"
19 tt + ^ fl^ — ^ . ab(x — y)
ax -{-by ax — by aV — b^j^
FBACTI0K8. 117
174. Since t* = «> and — - = a,
0 — 0
it is evident that if the signs of both numerator and
denominator are changed^ the value of the fraction is not
altered.
A • g — b — (g — b) — a-^b b — a
Hence^ if the numerator or denominator be a compound
expression^ or if both be compound expressions, the sign of
every term in the denominator may be changed, provided
the sign of every term in the numerator be changed.
Since the change of the sign before the fraction is equiv-
alent to the change of the sign before every term of the
numerator of the fraction, the sign before every term of
the denominator may be changed, provided the sign before
the fraction be changed.
Since, also, the product of + a multiplied by + ^ is ah,
and the product of — a multiplied by — i is ab, the signs
of two factors, or of any even number of factors, of the
denominator of a fraction may be changed without altering
the value of the fraction.
By the application of these principles, fractions may
often be changed to a more simple form for addition or
subtraction.
(1) Simplify --^^^-j + fni^-
Change the signs before the terms of the denominator of the third
fraction, and change the sign before the fraction.
The result is,
2 3 2x — S
X 2x — l 4.x' — 1
in which the several denominators are arranged in the same ord£T vjUh
respect to X.
118 ALaSBBA.
The L. C. D. = a(2aj-l)(2aj+l).
8 aj* — 2 = 1st numerator,
— Gx* — 3a: = 2d numerator,
— 2a:* + 3a5 = 3d numerator.
— 2 = sum of numerators.
-2
Sum of the fractions =
a;(2aj-l)(2aj + l)
(2) Simplify:
Cliange the sign of the factor {h — a) m the denominator of the
second fraction, and change the sign before the fraction.
Then change the signs of the factors (c — a) and (c — 6) in the
denominator of the third fraction.
The result is,
1 I + 1
a (a — b)(a — c) b(a — b)(b — c) c(a — c)(b — c)
in which the factors of the several denominators are written in
alphaJbeticcU order,
TheL. C. 'D. = abc(a — b)(a — c)(b — c).
bc(b — c) i=b^c — ftc* =: 1st numerator,
— ac(a. — c)=— a^c-^-ac^ = 2d numerator,
ab (a — b) = (j?b — aft* ^=3d numerator.
a^ft — a*c — aft* + a<? + ft*c — ftc* = sum of numerators.
= a*(ft — c) — a(ft* — c*) + ftc(ft — c),
= [a*-a(ft + c) + ftc][ft-c],
= \c? — aft — atf + ftc] [ft — - c],
= [(a* — ac) — (aft — ftc)] [ft - c],
= [a (a — c) — ft (a — c)] [ft — c],
= (a — ft)(a — c)(ft — c).
FRACTIONS. 119
.-. Sum of the fractionB= /'*r^\t~'^t7'\
ahc (a -^o^^a — c)(b — e)
abc
EXEBCISE 59.
Simplify :
3 + 2a; Sx — 2 16a? — g*
2-x "^ 2 + x "•■ a;«— 4
x« — 1 ' x + 1 1 — x
-i . 1 . 1
1 2
6.
(2 — m)(3 — w) (w — l)(w — 3) * (m — l)(m — 2)
1 . 1
(b — a)(x + a)^ (a — b)(x + b)
&»+y 2ay 2a%
^ — a g — 2^ Sx(a — b)
x—b b + x V — 7?
Z + 2x 2 — 3a; l^x—a?
^' 2—x 2 + x "^ a;«-4
3 7 4 — 20a?
^^' 1 — 2x l+2a; 4a?2 — l'
120 ALGEBRA.
11 <^ + ^ , b + c , c + a
(b — c)(c-d)^(b'-a)(a — c)^(a — b)(b — c)
N" ' (a'-b)(a-c)'^ (b + c)(b-a)'^ {c-a)(c + b)
(a;-y)(a;-«) (y- x) (y-«) "^ («-a:)(«-y)*
- 14 3 ^ 6
16.
(a — b)(b'-c) (b'-a)(c — a) (a'-c)(c — b)
1.1 1
^- ' ' x(x — y)(x — z) y{y — x){y — z) xyz
Multiplication of Fractions.
175. In Arithmetic it is often necessary to take equal
parts of fractions of units.
Suppose it is required to take J of f of a unit.
Let the line AB represent the unit of length.
^1 I i I I I I I I I I I I I I Ib
C D E F
Suppose AB divided into 5 equal parts, at C, D, JBJ, and
F, and each of these parts to be subdivided into 3 equal
subdivisions.
Then one of the parts, 2ls AC, will contain 3 of these
subdivisions, and the whole line AB will contain 16 of
these subdivisions.
That is, J of J of the line will be ^ of the line ;
J of f will be tV + tV + tV + tV» ot ^ of the line; and
J of i will be twice ^, or ^ of the line.
FRACTIONS. 121
Suppose it is required to take - of - of the line AB,
where a, h, c, d represent positive whole numbers.
^1 I I I I I I I I I I I I I I u
C D E F
Let the line AB be divided into h equal parts, and let each
of these parts be subdivided into d equal subdivisions.
Then the whole line will contain hd of these subdivisions^
and one of these subdivisions will be 7-; of the line.
od
If one of the subdivisions be taken from each of a parts,
they will together be — of the line. That is,
od
l«a 1,1,1 ,. ,. a
d °* i = w+ jrf+M "^^ " *^"^^«' = hd'
and -; of 7 will be e times r-i, or — of the line.
do od od
Therefore, to find a fraction of a fraction,
Find the prodtict of the numerators for the numerator of
the product, and of the denominators for the denominator
of the 'product,
^^_ __ c a c ^ a
176. JNow, - X r means - of 7-
do do
Therefore, to find the product of two fractions.
Find the product of the numerators for the numerator of
the product, and of the denominators for the denominator
of the product.
The same rule will hold whatever be the values of
a, b, e, d,
11 a factor exist in both a numerator and a denominator
it may be cancelled ; for the cancelling of a common factor
before the multiplication is evidently equivalent to cancel-
ling it after the multiplication ; and this may be done by
§163.
122 ALGEBBA.
Division of Fractions. '
177. Multiplying by the reciprocal of a number is equiv-
alent to dividing by the number. Thus, multiplying by \
is equivalent to dividing by 4.
The reciprocal of a fraction is the fraction with its terms
interchanged.
Thus, the reciprocal of f is }, for } X f =1. § 42.
Therefore, to divide by a fraction,
Interchange the terms of the fiuction, and multiply by the
resulting fraction. Thus,
2a ^ 1 __3a; 2a _2a
^^ S2? • 3«"~ 1 30*"^ x'
The common factor cancelled is 8 a;.
14a;«^7g_9y 14a^_2a;
^^ 2iy ' 9y""7a;^273^""3y'
The common factors cancelled are Oy and 1x,
/«\ ^^ ^ g^ {a-\-x)(a^x) ax
^ ^ (a — xy ' af — T? ah (a^x)(a^x)
x{a'\-x)
b(a — x)
The common factors cancelled are a and a — x.
If the divisor be an integral expression, it may be
changed to the fractional form. § 162.
Simplify ;
Exercise 60.
1 ^v-. 3 -J£_^_?£_.
hx d 2p-2 • p-1
2x 3a^ 3ac 8xV _ 2cf
' a ^ c ^ 2b' • 16a^ * Sai^'
F&ACTI0K8.
123
9aV 15xy^
20a^lj^e
ISxy'z
4:xs^ 6xy 2xy*
9mV 5p'q 24a^y«
25fe°/;t» 70w*y 3pm
XO.
a — ^ a» — 6»
a* + ^ »* — ^
11.
a^ — ly" • a + 6*
12.
«*+» — 2 ^a* — 13aj + 42
ic* — 7a;
a;«H-2aj
18.
g»— lla; + 30 a^'-Sa;
x'—ex + d a^ — 5x
16. ^-(^-t^\
cx
14 ^n^x^^i^'-
(x — y){x + y)*
a*-{-2abah — 2V
18. ^±5^x^=4*.
X — y x* — y^
17.
aJ_4 a;«-26
«»+5a; a:« + 2aj
19.
L* OT* VL
m* — 71^
m
c« + c? • c + d
-k
./-
g'— 4a + 3 a» — 9a + 20 a^ — 7a
a«-5a + 4 a2_i0a + 21 a«~5a
y-7^> + 6,,^>' + 10^ + 24 . V + U
21.
22.
a;«~y«
a^ — 3xy + 2y^ x^ + xy {x — yf
a»~3a'5 + 3ay-^>» 2a5 — 2^ a' + o^
124 AL6EBBA.
26.
i
\ 27.
(a + cy—(b + d)^ ' {a — by—id — cy
0^— 2xy + y' — g^ g + y — a;
^^ a5* + 2a3y + y* — «* as — y + «
Complex Eragtions.
178. A complex fraction is one which has a fraction in
the numerator or in the denominator, or in lx)th.
To simplify a complex fraction,
Divide tJie numerator by the denominator.
(1) SimpUfyi-
i-
i
i-^i =
=!xi=*.
(2) SimpUfyg-
?l = ± =
5f V
f-^V- =
= AXf =
= TVr.
(3) Simplify/,!^.
3% 3x
3x ,
~ 1 *
4« — 1
4
X^'-
x — i 4a; — 1
4 ~4x-l
^ 1
4
12a
2—
4a: — 1
FBACTIOirS. 125
179. It is often shorter to multiply both terms of the
fraction by the L. G. D. of the fractions contained in the
numerator and denominator.
Thus, in § 178, (1), multiply both terms by 6 ; (2), both
terms by 24 ; (3), both terms by 4. The results obtained
are f , ^, ^^^^, respectively.
Simplify by first simplifying the denominator :
1 + x-
l-x+a?
X
1-
x(l—x + xF)
11^1 (l + x)(l^x + a^ + x
1 — « + «*
1
x-
:?T^
l + x + a?
x(l + x + a^
■^l + aj + a:*— (aj — «* + aj^
x+oc^+x*
^ l+x"
The expression is reduced to the form
X I 1 "^ X ~T* X^ X ^^ X^ *^~ Cr
.^ , 0/i . Jv I > and this is equal to . , , ,♦
The expression __^ , . is reduced to the form
"^~l+a; + a?»
i-a- l^!3"^^."^"^.a,^> ^d this is equal to ^g^^'
126 ALGSBSA.
EZEB0I8E 61.
Simplify :
3a? . g — 1
2 ■*■ 3 ^1
1. To 8. 1 T
«-l+rZ^ 9. 1-f
x-2-
3- ^+^ + 1^
aj — 6
10.
3 2a:-y 1
«+i «.+-'_• 1+1
2 2
05 — a
11.
« — ^^ ^r ^ l-hx + -^^
(a x\
X a)
Hy , '^^ fxtz — x^ x+y^
'— y* g+yl(x— y)' x—yj
1-
1
X — a
x + a
X
X— y
x«-y«
X
y
x«+xy
x-1
x-l '
x + 1
x + l
x-1
18. —
7. — T-r; ^ 14. -
xy
aj — 1 aj + i « + y
FBAOnOKS. 127
ah otf
15, *
«* + (* + c)a: + 6o
a + 6
16. -^+1-4- 17. \ ;+^
1 + i '^ ^ + F
2m-3 + - 1+i + i
^ *-. «^ «^ ^ «^ 3
18. — 5 3 — 19. -T — Tr-r-TT 20. J,
2m — 1 or — (^ + c)'
1 +
m ah 1 J. ^
1— a
EXEBCISB 62.
MISCELLANEOUS EXAMPLES.
7 ^- S^P^^^ ^+7«^-9.:»-7x + 8'
2. Find the value of , ^ — ^ ^ . when a = 4, 6 = *,
c = l.
3. !Fmd the value of 3a' H rj when a = 4, 6 = ^,
c = l.
^ «. ,.- 2 1 1
4. Simplify
(a^-l)« 2a^-4a; + 2 l-aj«
a; a;
6. Find the value of I ; ) ; 777 when x=—^.
\x — b/ x + a — zb 2
7. Simplify |^^_^-^^_p^ + -,-^.|.
2^ 1 o-^
2b *
128 ALGEBRA.
..Si.pHI,(S±^-S=^).(|±f-|^>
9. Simplify
10. Simplify
fa*-ab\fa*+ab+l>'\ ( 2a* _ A A 2ah \
i+iz:£ i+«'-=^
/ 11. Simplify 7—! 5 5-^^ — 5.
1 ^""^ < or — or
a + ^ a* + ^
12. Divide «^ + ^-3(|-a!»)+4(x + i)bya! + ^-
1-
2 ay r;£_y"l*
X
1+!
/ 13. Simplify ('«+y)V
^ / 14. Find the value of ^rr + ttt— i tts \ when a;
X/ , 2o — x 2b + x 4 6* — ar
15. Find the value of ^ , ^ when a;= - . v and
ab + a '^-2' + ^ '^^^
16. Simplify
+
a (a — b) (a — c) h(b — c)(h — a) c(c — d)(c — b)
17. Simplify
a—1 b — 1 . c — 1
Sahc a b c
bc-\-ca — ab ^ jl.^ ^
a b c
FKA.CTI0ir8. 129
^ is. Simnlifv -Jt Z V ^Ilif
» m
1 . 1
a 6 + c
20. Simplify 3a-[6+ {2«-(6-c)n + ^ + |7^
21. Simplify «-» g-y (»-»)' («-y)»
(a — y)(a — xy (a— «)(« — y)»
22. Simplify Lr- 23. (»*-yO(2^-2xy)
//
14
tfl 4(x-,)«.^-^^
3 — a:
26. Simplify ^ —^ — ^ ^
26. Simplify
1 1 > i_
a (a — ft) (a — c) b(b — a) (b — c) abc
,_4+ 6 ^. x + 5
27. Simplify 1±1 X ^ ^
^ 6_ '^(*-l)(*-2)
CHAPTER IX.
FRACTIONAL Equations.
TO REDUCE EQUATIONS CONTAINING FRACTIONS.
180. (1) | + i = 12.
Multiply both sides by 4, the L. C. M. of the denominators.
Then, 2x + x = 48,
3x=48.
.•.x=16.
VJ
6 '•
— ^"X
8
Multiply both sides by 24, the L. 0. M. of the denominators.
Then,
4x-96 = 576-3x,
4x + 3x=676 + 96,
7x=672.
.•.x=96.
X X
— 1
(3)
3
11
aj — 9.
Mull
biply by
33, the L. C. M. of the denommators.
Then,
Ua
-8x + 3 = 33x-297,
llx-
3x-33x= -297-3,
-26x=-300.
.-. X = 12.
Since the minus sign precedes the second fraction, in removing
the denominator, the + (understood) before x, the first term of the
numerator, is changed to — , and the — before 1, the second term
of the numerator, is changed to +.
181. Therefore, to clear an equation of fractions^
Multiply each term by the L. C. M. of the denominators.
FKAGTIOKAI. SQUATI0N8. 131
If a fraction is preceded by a miniu sign, the sign of
every term of the numerator must be changed when the
denominator is removed.
Exercise 63.
Solve the equations:
^ ^ x + 2 _ ^ 6a; 5a; 9 3 — a;
1. 6x— ^ = 71. 4. _-_ = -— ^.
3 — x 17 , - 60! — 4 _ 1— 2x
2. X —=^ 6. 2x -g-=7 g-.
5—2x , „ ■ 6a;— 8 x + 2 14 3 + 6aj
-^- + 2=0, 2— «• ^- = -9 T-
5a; + 3 3 — 4aj , a; 31 9 — 5a;
7.
8 3 ' 2 2 6
,„ 7a!-|-5 6a: — 6 8 — 6a!
3 5 ^ 15
1 o o
7 flZ. i(3a;~4)+|(5a; + 3) = 43-5a^
;| 14. i(27-2x) = |-^(7a!-54).
1S2 ALOBBBA.
16. 6a!— |8a!-3[16 — 6a! — (4 — 6a;)]} =6.
5a!~3
7
9 — a! 6a: , 19. ..
- 3 =2+6^* *>•
2a!+7
9a:-8 a!-ll
7
11 2
8a! — 16
lla; — 1 7a: + 2
3
7 13
7a; + 9
3a: + l 9a:-13 249 — 9a!
8 7 4 14
18lL If the denominators contain both simple and com-
pound expressions, it is best to remove the simple expres-
sions first, and then each compound expression in turn.
After each multiplication the result should be reduced to
the simplest form.
. Sx + 5 7a;--3_4a; + 6
^^ 14 '^6x + 2'^ 7
Multiply both sides by 14.
Then, 8aj+ 6 + 1.1 = 8a + 12.
OX "T 1
Transpose and combine, ^ , .. = 7.
Multiply by 3 X + 1, 49a - 21 = 21 x + 7,
28 a = 28.
.a=l.
(2)
q 4a; 7a;
'4 10
Simplify the complex fractions by multiplying both terms of each
fraction by 9.
^ 27-4a 1 7a-27
^^^' ~36- = 4 90r-
Multiply both sides by 180.
136-20a=46-14a + 64,
-ea=-86.
FKAGTIONAL EQUATIONS. ISS
EXEBCISB 64
Solve the equations:
36 5a;-4 "^4
9(2g — 3) lla;-l_9a; + ll.
14 ■^3a: + l~ 7
IQg + lT 12a? + 2 ^5g--4
/• 18 13a:-16"" 9
6a; + 13 3a; + 5 _2a;
15 6a; — 25"" 5 '
"J' 18^^^^"22r . ^ ' l + 16aj , ^ 101-64*
0^«- ■39=:6^ + ^^+^^=^A 24
) / 6--5g 7-2*^ ^l + 3g IQg-ll . 1
/ 16 14(aj-l) 21 30 "^106
9a; + 5 , Sx-1 ^ZQx + lb 41
14 "^6a; + 2 66 "^66*
6g + 7 2a; — 2^2g + l
15 7a; — 6 5
6a; + l 2a; — 4 _2a; — 1
15 7a;-16"" 5
7a; — 6 x — b x
10.
36 6«-101 6
18S. Literal equations are equations in which some or
all of the knavm numbers are represented by letters ; the
numbers regarded as known numbers are usually repre-
sented by the first letters of the alphabet.
184 ALGEBBA.
(1) (a — x)(a + x) = 2a* + 2ax'-a^.
Then, a^ — aja = 2a2 + 2ajc — a^,
— 2 ax = a^.
a
(2) (x'-a)(x — b) — (x — b)(x'-c) = 2(x — a)(a'-c).
(x^ — oas — te + a6) — (x* — 6x — ca 4- 6c) = 2 (ox — ex — a^ 4- ac)^
x2 — ox — 6x4-a6 — x2 4-6x4-cx — 6c = 2ax — 2cx — 2a2 + 2<ic.
That is, — 3ax + 3cx= — 2a2 + 2ac — a6 + 6c,
— 3 (a — c) X = — 2 a (a — c) — 6 (a — c),
— 3x=— 2a— 6.
2a + 6
...X=-g--
Exercise 65.
Solve the equations:
1. ax + 6c = 6a; + ac. 2. 2a — cx = 3c — 5bx,
3. <i^X'-{'bx — c=^b^X'^cx — d.
4. — ac* + ^*c + <3t^<^^ = <3t^^ + ^'wa;--ac*a; + ^'c — mc.
5. (a + iJ5 + ft) (a + 6 — x) := (a -{' x) (b — x) — ab.
6. (a* + xy=x'+4:a^ + a\
7. (a*— a;)(a* + aj) = a* + 2aaj— x".
aa — b , xA-ac ^ Sa — bx 1
8. -^+a = -_. 10. ox 2— = 2-
a(^x + a:«) , oo* ^ 4flMC — 25
bx b 3
^ a^-a a — a; 2aj a
bx b b X
. ^ 3 ab — x^ 4:X — ae
13. r =
c bx ex
14. am—'b r- H = 0.
16.
16.
FBAOTIOKAL XQUATIOKS.
Sax — 2b ax — a aa5 2
3^ 2r"""T""3' ^
aJ>'-{'X b^ — X X — b ab — x
136
l\ ^"* 6« a*b "" a« l^
bx + 1 a(aj2-l) ^^ ab , . ,T 1
V
18. T h«H — = 0.
b — ex c
EZERCISK 66.
Solve the eqtiations :
a; — 3
x-5 , 1
4(a!-l) 6(x-l) ' 9
„ , X (x-2)(x+4:)
2. x-\ T = i JlT^*
as — 1 x + 1
7 6a! + l 3(l + 2a;^
a; — 1 SB + l «» — 1
1 1 x — 1
4.
2(a! — 3) 3(x — 2) (x — 2)(a;— 3)
2(2x + 3) 6 5x + l
9(7 — 3!) 7-x 4(7 — x)'
7.
x — 7 2a; — 15
aj + 7 2a; — 6 2(aj + 7)
8. ,£±^ + ^=§^«.
3x+5
2x + 3
136 ALGEBRA.
/ 132X + 1 8«-}-6_-.„| ^, 3a!-l 4x-2 1 ^
'•O ^"- 2a;-3^x-2 3a;+2 «-! x-1 1-a?
36 — 4 a; — 6 X — 7 x — 8
Id*
X--Z a — 6 a — 8 x— 9
14. (x — a)(a; — «►) = (« — a — ^)*
16. (a — 6)(aj— c) — (^— c)(a; — a) — (c— a)(aj— 6)=0
/ le. ^z:£±l + ^±£±l=2..
aj — 1 a; + l
j^ 4 . 7 37
a! + 2 ' x + Z a^ + 5« + 6
18. (a: + l)' = a:[6 — (1 — «)] — 2.
/ 19 25-ix 16a; + 4i_ 23
V « — 8^2a;— 16 24 3*— 24
\2 a;y 2 4
^23.
5 a; — 1 3
24. -^Ili- + -^ni- = i4.
|(x-l)^4(x+l) ^ +
16
0-^).
CHAPTEE X.
Problems.
Exercise 67.
Ex. Find tlie number the sum of whose third and fourth
parts is equal to 12.
I/et X = the number.
X
Then « = ^® third part of the number,
and -r = the fourth part of the number.
.-. o + 7 = the sum of the two parts.
But 12 = the sum of the two parts.
.-.- + - = 12
Multiply both sides by 12 :
4x + 3aj=144,
7x = 144.
.-. X = 20f
1. Find the number whose third and fourth parts together
make 14.
2. Find the number whose third part exceeds its fourth
part by 14.
3. The half, fourth, and fifth of a certain number are
together equal to 76; find the number.
4. Find the number whose double exceeds its half by 12.
6. Divide 60 into two such parts that a seventh of one
part may be equal to an eighth of the other.
138 ALGEBBA.
6. Divide 50 into two such parts that a fourth of one
part increased by five-sixths of the other part may
be equal to 40.
7. Divide 100 into two such parts that a fourth of one
part diminished by a third of the other part may be
equal to 11.
The sum of the fourth, fifth, and sixth parts of a
certain number exceeds the half of the number by
112. What is the number?
9. The sum of two numbers is 5760, and their difference
is equal to one-third of the greater. What are the
numbers ?
10. Divide 45 into two such parts that the first part
^.-^"flmded by 2 shall be equal to the second part
y multiplied by 2.
11. Find a number such that the sum of its fifth and its
seventh parts shall exceed the difEerence of its fourth
and its seventh parts by 99.
12. In a mixture of wine and water, the wine was 25
gallons more than half of the mixture, and the
water 5 gallons less than one-third of the mixture.
How many gallons' were there of each ?
13. In a certain weight of gunpowder the saltpetre was 6
pounds more than half of the weight, the sulphur
5 pounds less than the third, and the charcoal 3
pounds less than the fourth of the weight. How
many pounds were there of each?
14. Divide 46 into two parts such tiat if one part be
divided by 7, and the other by 3, the sum of the
quotients shall be 10.
PBOBLBMS. 139
16. A house and garden cost $850, and five times the price
of the house was equal to twelve times the price of
the garden. What was the price of each ?
16. A man left the half of his property to his wife, a
sixth to each of his two children, a twelfth to his
brother, and the remainder, amounting to $600, to
his sister. What was the amount of his property ?
17. The sum of two numbers is a and their difference is h\
find the numbers.
18. Find two numbers of which the sum is 70, such that
the first divided by the second gives 2 as a quotient
and 1 as a remainder.
19. Find two numbers of which the difference is 25, such
that the second divided by the first gives 4 as a
quotient and 4 as a remainder.
20. Divide the number 208 into two parts such that the
sum of the fourth of the greater and the third of
the smaller is less by 4 than four times the difference
of the two parts.
21. Find four consecutive numbers whose sum is 82.
KoTE I. It is to be remembered that if x represents a person^s age
at the present time, his age a years ago will be represented by x — a,
and a years hence by a + a.
Ex. In eight years a boy will be three times as old as he
was eight years ago. How old is he ?
Let X = the number of years of his age.
Then x — 8 = the number of years of his age eight years ago,
and X + 8 = the number of years of his age eight years hence.
.•.x + 8 = 3(x-8),
x+8 = 3x-24,
X— 3x= —24 — 8,
-2x=-32,
x=16.
4-
140 AL6EBBA.
-H
22. A is 72 years old, and B's age is two-thirds of A's.
How long is it since A was five times as old as B ?
23. A mother is 70 years old, her daughter is half that
age. How long is it since the mother was three and
one-third times as old as the daughter ?
24. A father is three times as old as the son ; four years
ago the father was four times as old as the son then
was. What is the age of each ?
26. A is twice as old as B, and seven years ago their
united ages amounted to as many years as now
represent the age of A. Find the ages of A and B.
26. The sum of the ages of a father and son is half what it
will be in 25 years ; the difference is one-third what
the sum will be in 20 years. What is the age of each?
Note IL If A can do a piece of work in x days, the part of the
work that he can do in one day will be represented by 1. Thus, if he
can do the work in 6 days, in 1 day he can do ^ of the work.
Ex. A can do a piece of work in 5 days, and B can do it
in 4 days. How long will it take A and B together
to do the work?
Let X = the number of days it will take A and B together.
Then i = the part they can do in one day.
Now, I = the part A can do in one day,
and { = the part B can do in one day.
.'. J + J = the part A and B can do in one day together.
4x+5x=20,
9x=20,
x = 2}.
Therefore they will do the work in 2 J days.
27. A can do a piece of work in 5 days, B in 6 days, and
C in 7 J days; in what time will they do it, all
working together?
PBOBLEMS. 141
28. A can do a piece of work in 2^ days, B in 3^ days, and
G in 3} days; in what time will they do it, all
working together?
29. Two men who can separately do a piece of work in 15
days and 16 days, can, with the help of another, do
it in 6 days. How long would it take the third man
to do it alone?
30. A can do half as much work as B, B can do half as
much as C, and together they can complete a piece
of work in 24 days. In what time can each alone
complete the work?
31. A does I of a piece of work in 10 days, when B comes
to help him, and they finish the work in 3 days
more. How long would it have taken B alone to
do the whole work?
32. A and B tojgether can reap a field in 12 hours, A and
C in 16 hours, and A by himself in 20 hours. In
what time can B and C together reap it ? In what
time can A, B, and C together reap it?
33. A and B together can do a piece of work in 12 days,
A and C in 15 days, B and C in 20 days. In what
time can they do it, all working together ?
Note III. If a pipe can fill a vessel in x hours, the part of the
vessel filled by it in one hour will be represented by 1. Thus, if a
pipe will fill a vessel in 3 hours, in 1 hour it will fill ^ of the vessel.
34. A tank can be filled by two pipes in 24 minutes and
30 minutes, respectively, and emptied by a third in
20\ minutes. In what time will it be filled if all
three are running together ?
36. A tank can be filled in 15 minutes by two pipes, A and
B, running together. After A has been running by
142 ALGEBBA.
itself for 6 minutes, B is also turned on, and the
tank is filled in 13 minutes more. In what time
may it be filled by each pipe separately ?
86. A cistern could be filled by two pipes in 6 hours and 8
hours, respectively, and could be emptied by a third
in 12 hours. In what time would the cistern be
filled if the pipes were all running together?
87. A tank can be filled by three pipes in 1 hour and
20 minutes, 3 hours and 20 minutes, and 6 hours,
respectively. In what time will the tank be filled
when all three pipes are running together?
88. If three pipes can fill a cistern in a, b, and c minutes,
respectively, in what time will it be filled by all
three running together?
39. The capacity of a cistern is 755^ gallons. The cistern
has three pipes, of which the first lets in 12 gallons
in 3^ minutes, the second 15^ gallons in 2^ minutes,
the third 17 gallons in 3 minutes. In what time
will the cistern be filled by the three pipes running
together?
NoTB IV. In questions involving distance, time, and rate :
Distance ^.
—5-- — = Time.
Rate
Thus, if a man travels 40 miles at the rate of 4 miles an hour,
~ = the number of hours required.
Ex. A courier who goes at the rate of 31^ miles in 5
hours, is followed, after 8 hours, by another who
goes at the rate of 22^ miles in 3 hours. In how
many hours will the second overtake the first?
Since the first goes 31} miles in 6 hours, his rate per hour is 6^
miles.
PBOBLEMS. 143
Since the second goes 22} miles in 3 hours, his rate per hour is 7^
miles.
Let X = the number of hours the first is travelling.
Then x — 8 = the number of hours the second is travelling.
Then ^A* = the number of miles the first travels ;
(x — 8) 7^ = the number of miles the second travels.
They both travel the same distance.
... eAx=(x-8)7}.
The solution of which gives 42 hours.
40. A sets out and travels at the rate of 7 miles in 6
hours. Eight hours afterwards, B sets out from the
same place and travels in the same direction, at the
rate of 5 miles in 3 hours. In how many hours will
B overtake A?
41. A person walks to the top of a mountain at the rate
of 2^ miles an hour, and down the same way at the
rate of 3^ miles an hour, and is out 5 hours. How
far is it to the top of the mountain?
42. A person has a hours at his disposal. How far may
he ride in a coach which travels b miles an hour, so
as to return home in time, walking back at the rate
of c miles an hour ?
43. The distance between London and Edinburgh is 360
miles. One traveller starts from Edinburgh and
travels at the rate of 10 miles an hour; another
starts at the same time from London, and travels at
the rate of 8 miles an hour. How far from London
will they meet?
44. Two persons set out from the same place in opposite
directions. The rate of one of them per hour is a
mile less than double that of the other, and in 4
hours they are 32 miles apart. Determine their
rates.
144
▲L6EBBA.
45. In going a certain distance, a train travelling 35 miles
an hour takes 2 hours less than one travelling 26
miles an hour. Determine the distance.
Note V. In problems relating to clocks, it is to be observed that
the minute-hand moves twelve times as fast as the hour-hand.
Ex. Find the time between two and three o'clock when
the hands of a clock are :
I. Together.
II. At right angles to each other.
III. Opposite to each other.
Fig. 1.
Fig. 2.
Fig. 3.
I. Let CH and CM (Fig. 1) denote the positions of the hour and
minute hands at 2 o'clock, and CB the position of both hands when
together.
Then arc HB = one4welfth of arc MB,
Let X = the number of minute-flpaces In arc MB,
X
Then
12
= the number of minute-spaces in arc HB,
and 10 = the number of minute-spaces in arc MH,
Now arc MB = the arc MH + arc HB,
« = io + -.
That is,
The solution of this equation gives x = 10}^}.
Hence, the time is 10|f minutes past 2 o'clock.
IL Let CB and CD (Fig. 2) denote the positions of the hour and
minute hands when at right angles to each other.
PBOBLEM8. 145
Let X = the number of minute-spaces in arc MBBD.
Then — = the number of minute-spaces in arc ITB,
and 10 = the number of minute-spaces in arc JfH.
15 = the number of minute-spaces in arc BD.
Now arc MHBD = the arcs MB + HB + BD.
That is, X = 10 + ^ + 16.
1a
The solution of this equation gives x = 27^.
Hence, the time is 27^ minutes past 2 o'clock.
in. Let CB and CD (Fig. 3) denote the positions of the hour and
minute hands when opposite to each other.
Let X — the number of minute-spaces in arc MHBD,
Then — = the number of minute-spaces in arc HB,
iZ
and 10 = the number of minute-spaces in arc MH,
30 = the number of minute-spaces in arc BD,
Now arc MHBD = the arcs 21H+HB + BD,
That is, X = 10 + ~ + 30.
The solution of this equation gives x = 43^3^.
Hence, the time is 43t7j minutes past 2 o'clock.
46. At what time are the hands of a watch together :
I. Between 3 and 4 ?
II. Between 6 and 7 ?
III. Between 9 and 10 ?
47. At what time are the hands of a watch at right-angles :
I. Between 3 and 4 ?
II. Between 4 and 6 ?
III. Between 7 and 8?
48. At what time are the hands of a watch opposite to
each other:
I. Between 1 and 2?
IT. Between 4 and 5?
III. Between 8 and 9?
146 ALGEBRA.
49. It is between 2 and 3 o'clock ; but a person looking
at bis watcb and mistaking the bour-hand for the
minute-band, fancies that the time of day is 55
minutes earlier tban it really is. What is the true
time?
Note VI. It is to be observed that if a represent the nuiqber of
feet in the length of a step or leap, and x the number of steps or leaps
taken, then ax will represent the number of feet in the distance
made.
Ex. A bare takes 4 leaps to a greybound's 3 ; but 2 of the
greyhound's leaps are equivalent to 3 of the hare's.
The hare has a start of 50 leaps. How many leaps
must the greyhound take to catch the bare ?
Let 3x = the number of leaps taken by the greyhound.
Then 4x = the number of leaps of the hare in the same time.
Also, let a denote the number of feet in one leap of the hare.
Then --- will denote the number of feet in one leap of the grey-
bound.
That is, 3x X -^ = the whole distance,
and (50 + 4 a;) a = the whole distance.
... ^=(60 + 4a;)a.
Divide by a, ^=50 + 4x,
0x=100 + 8x.
X = 100.
.-. Sx = 300.
Thus the greyhound must take 300 leaps.
60. A hare takes 6 leaps to a dog's 5, and 7 of the dog^s
leaps are equivalent to 9 of the hare's. The hare
has a start of 60 of her own leaps. How many
leaps will the hare take before she is caught?
PROBLEMS. 147
51. A greyhound makes 3 leaps while a hare makes 4; bat
2 of the greyhound's leaps are equivalent to 3 of
the hare's. The hare has a start of 50 of the grey*
hound's leaps. How many leaps does each take
before the hare is caught?
52. A greyhound makes 2 leaps while a hare makes 3; but
1 leap of the greyhound is equivalent to 2 of the
hare's. The hare has a start of 80 of her own leaps.
How many leaps will the hare take before she is
caught?
Note Vll. It is to be observed that if the number of units in the
breadth and length of a rectangle be represented by x and x + a,
respectively, then x (x + a) wiU represent the number of surface units
in the rectangle, the imlt of surface having the same name as the
linear unit in which the sides of the rectangle are expressed.
63. A rectangle whose length is 5 feet more than its
breadth woxdd have its area increased by 22 feet if
its length and breadth were each made a foot more.
Find its dimensions.
54. A rectangle has its length and breadth respectively 5
feet longer and 3 feet shorter than the side of the
equivalent square. Eind its area.
55. The length of a rectangle is an inch less than double
its breadth ; and when a strip 3 inches wide is cut
off all round, the area is diminished by 210 inches.
Find the size of the rectangle at first.
56. The length of a floor exceeds the breadth by 4 feet ; if
each dimension were increased by 1 foot, the area
of the room would be increased by 27 square feet.
Eind its dimensions.
NoTB VUL It is to be observed that if h pounds of metal lose a
pounds when weighed ia water, 1 pound wiU lose i of a pounds,
or I of a pound.
148 ALGEBBA.
57. A mass of tin and lead weighing 180 pounds loses 21
pounds when weighed in water ; and it is known that
37 pounds of tin lose 5 pounds, and 23 pounds of
lead lose 2 pounds, when weighed in water. How
many pounds of tin and of lead in the mass ?
68. If 19 pounds of gold lose 1 pound, and 10 pounds of
silver lose 1 pound, when weighed in water, find the
amount of each in a mass of gold and silver weighing
106 pounds in air and 99 pounds in water.
69. Fifteen sovereigns should weigh 77 pennyweights ; but
a parcel of light sovereigns, having been weighed and
counted, was found to contain 9 more than was sup-
posed from the weight ; audit appeared that 21 of
these coins weighed the same as 20 true sovereigns.
How many were there all together ?
60. There are two silver cups, and one cover for both. The
first weighs 12 ounces, and with the cover weighs
twice as much as the other without it; but the sec-
ond with the cover weighs one-third more than the
first without it. Find the weight of the cover.
61. A man wishes to enclose a circular piece of ground with
palisades, and finds that if he sets them a foot apart
he will have too few by 150 ; but if he sets them a
yard apart ho will have too many by 70. What is the
circuit of the piece of ground?
62. A horse was sold at a loss for $200 ; but if it had been
sold for $250, the gain would have been three-fourths
of the loss when sold for $200. Find the value of
the horse.
63. A and B shoot by turns at a target. A puts 7 bullets
out of 12, and B 9 out of 12, into the centre. Be-
^ tween them they put in 32 bullets. How many shots
did each fire ?
PROBLEMS. 149
64. A boy buys a number of apples at the rate of 5 for 2
pence. He sells half of them at 2 a penny and the
rest at 3 a penny^ and clears a penny by the trans-
action. How many does he buy ?
65. A person bought a piece of land for $6760, of which
he kept f for himself. At the cost of $260 he made
a road which took -j^^ of the remainder, and then sold
the rest at 12^ cents a square yard more than double
the price it cost him, thus clearing his outlay and
$500 besides. How much land did he buy, and
what was the cost price per yard ?
66. A boy who runs at the rate of 12 yards per second
starts 20 yards behind another whose rate is 10^-
yarcis per second. How soon will the first boy be
10 yards ahead of the second ?
67. A merchant adds yearly to his capital one-third of it,
but takes from it at the end of each year, $6000 for
expenses. At the end of the third year, after de-
ducting the last $5000, he has twice his original
capital. How much had he at first ?
68. A shepherd lost a number of sheep equal to one-fourth
of his flock and one-fourth of a sheep ; then, he lost
a number equal to one-third of what he had left and
one-third of a sheep ; finally he lost a number equal
to one-half of what now remained and one-half a
sheep, after which he had but 25 sheep left. How
many had he at first ?
69. A trader maintained himself for three years at an ex-
pense of $250 a year ; and each year increased that
part of his stock which was not so expended by one-
third of it. At the end of the third year his original
stock was doubled. What was his original stock ?
/
150 ALGEBBA.
70. A cask contains 12 gallons of wine and 18 gallons
of water; another cask contains 9 gallons of wine
and 3 gallons of water. How many gallons must
be drawn from each cask to produce a mixture
containing 7 gallons of wine and 7 gallons of water?
71. The members of a club subscribe each as many dollars
as there are members. If there had been 12 more
members, the subscription from each would have
been $10 less, to amount to the same sum. How
many members were there ?
72. A number of troops being formed into a solid square,
it was found there were 60 men over; but when
formed in a column with 5 men more in front than
before, and 3 men less in depth, there was lacking
one man to complete it. Pind the numb^ of troops.
73. An officer can form the men of his regiment into a
hollow square twelve deep. The number of men in
the regiment is 1296. Find the number of men in
front of the hollow square.
74. A person starts from P and walks towards Q at the
rate of 3 miles an hour ; 20 minutes later another
person starts from Q and walks towards P at the
rate of 4 miles an hour. The distance from P to Q
is 20 miles. How far from P will they meet ?
75. A person engaged to work a days on these conditions :
for each day he worked he was to receive h cents,
and for each day he was idle he was to forfeit c
cents. At the end of a days he received d cents.
How many days was he idle ?
76. A banker has two kinds of coins : it takes a pieces of
the first to make a dollar, and "J pieces of the second
/^ to make a dollar. A person wishes to obtain c
pieces for a dollar. How many pieces of each kind
must the banker give him ?
CHAPTER XI.
SiMULTAKEOUS EQUATIONS OF THE EiRST DSORBB.
184. If one equation contain two unknown numbers, an
indefinite number of pairs of values may be found that will
satisfy the equation.
Thus, in the equation aj + y = 10, ani/ values may be
given to x, and corresponding values for y may be found.
Ant/ pair of these values substituted for x and y will
satisfy the equation.
186. But if a second equation be given, expressing dif-
ferent relations between the unknown numbers, only oTie
pair of values of x and y can be found that will satisfy
both equations.
Thus, if besides the equation X'\-y = 10, another equa-
tion, X — y = 2, be given, it is evident that the values of
X and y which will satisfy both equations are
x = 6
y=Ar
for 6 + 4 = 10, and 6 — 4 = 2; and these are the only
values of x and y that will satisfy both equations.
186. Equations that express different relations between
the unknown numbers are called independent equations.
Thus, X'\-y = 10 and x — y = 2 are independent equa-
tions ; they express different relations between x and y,
BuIj a: + y=10 and 3a; + 3y = 30 are not independent
152 ALGEBRA.
equations ; one is derived immediately from the other,
and both express the same relation between the unknown
numbers.
187. Equations that are to be satisfied by the same
values of the unknown numbers are called simultaneous
equations.
188. Simultaneous equations are solved by combining
the equations so as to obtain a single equation containing
only one unknown number ; and this process is called
elimination.
Three methods of elimination are generally given :
I. By Addition or Subtraction.
II. By Substitution.
III. By Comparison.
Elimination by Addition or Subtraction.
(1) Solve: 2aj — 3y= 41 (1)
3x + 22/ = 32j (2)
Multiply (1) by 2 and (2) by 3,
4x-6y= 8 (3)
9g+6y= 96 (4)
Add (3) and (4), 13 a: =104
.-. X = 8.
Substitute the value of x in (2),
24 + 2y=32.
•• y = 4.
In this solution y is eliminated by addition.
(2) Solve: 6aj + 35y = 1771 (1)
8aj — 21y= 33J (2)
Multiply (1) by 4 and (2) by 3,
24x+140y=708 (3)
24x- 63y= 99 (4)
Subtract (4) from (3), 203 y = 609
.•.y = 3.
SIMULTANEOUS EQUATIONS. 153
Substitute the yaJue of y in (2).
8x— 63 = 33,
.-. X = 12.
In this solution x is eliminated by subtraction,
189. Hence, to eliminate an unknown number by addi-
tion or subtraction,
Multiply the equations by stick numbers as wUl make ths
coefficients of the syrnbol for this unknovm number equal in
the resulting equations.
Add the resulting equations, or subtract one from the other,
ax^ording as these equ^l coefficients have unlike or like signs.
Note. It is generally best to select that unknown number to be
eliminated which requires the smallest multipliers to make its co-
efficients equal ; and the smallest multiplier for each equation is found
by dividing the L. C. M. of the coefficients of this unknown number
by the given coefficient in that equation. Thus, in example (2), the
L. C. M. of 6 and 8 (the coefficients of x), is 24, and hence the smallest
multipliers of the two equations are 4 and 3 respectively.
Sometimes the solution is simplified by first adding the
given equations, or by subtracting one from the other.
(3)
x + 49y= 61
(1)
49a + y= 99
(2)
Add (1) and (2),
50 i
c + 60y=160
(3)
Divide (3) by 60,
X + y = 3.
(4)
Subtract (4) from (1),
48y = 48.
.•.y=l.
Subtract (4) from (2),
48x = 96.
.-. X = 2.
Exercise 68.
Solve by addition or subtraction :
1. 2x + Sy = T\ 3. 7x + 2y = 30^ 5. 5aj-f 4y = 68\
4aj — 5y = 3j y-3x= 2J 3x + 7y = e7j
2. a — 2y = 41 4. 3x — 5y = 511 6. 3ic-f2y = 391
2x— y = 5J 2x + 7y= 3j 32^-2a; = 13J
154 ALGBBBA.
7. 3a; — 4y = — 6\ 11. 12a:+ 72^ = 1761
4:x — 5y= IJ 32^ — 19a;= 3J
8. llaj + 32/ = 1001 12. 2x — 7y= 81
4a; — 72^= 4j 4:y — 9x = 19)
9. a; + 49y = 693\ 13. 69y-17a;= 103\
49a;+ 2^ = 357j 14a; — 132/ = — 41 J
10. 17a; + 32/ = 57\ 14. 17a; + 302/ = 59\
162/-3a; = 23j 19a; + 282/ = 77j
Elimination by Substitution.
Solve: 2a; + 32/ = 8
3x + 7y
=?}
2x + 32/=8 (1)
3a+72/=7 (2)
Transpose 3 y iii (1), 2 x = 8 — 3 y. (3)
Divide by coefficient of x, x = — r— *^ (4)
2
Substitute the value of x in (2), 3 ( — r-^^ + 7 y =7,
24-9y+14y=14,
6y=-10.
.-. 2/ = - 2.
Substitute the value of 2/ in (1), 2 x — 6 = 8.
.•.x=7.
190. Hence to eliminate an unknown number by sub-
stitution,
From one of the equations obtain the value of one of the
unknown numbers in terms of the other.
Substitute for this unknown number its value in the other
equation, and reduce the resulting equation.
simultaneous equations. 155
Exercise 69.
Solve by substitution :
1. 3a; — 4y = 2\ 8. 3a; — 4y = 181
7a; — 9y = 7j > 3g + 2y= OJ
2. 7a; — 52/ = 241 9. 9a; — 5y = 521
4a; — 3y = llJ 8y — 3a;= 8J
3. 3a; + 2y = 321 10. 5a; — 3y= 41
20a; — 32^= IJ 12^^ — 7a; = 10j
4. 11a; — 72^ = 371 11. 9^/- 7a; = 13\
8a; + 9y = 41j 15a; — 7^= 9j
5. Tx+ 52^ = 601 12. 5a; — 2y= 51 1
13a; — lly = 10j 19a; — 32/ = 180 J
6. 6a; — 72^ = 42\ 13. 4a;+ 9y=106\
7x — 6ij = 75J 8a; + 172/ = 198J
7. 10a;+ 9y = 290\ 14. Sx + 3y = S^
12a;- 11 2/ = 130 J 12a; + 92/ = 3j
Elimination by Comparison.
Solve: 2a; — 9y = ll
-92^=in
-42/= 7J
3a; — 42/
2x-9y = ll, (1)
3x-4y=7. (2)
Transpose 9 y in (1) and 4 y in (2), 2x=ll + 9y, (3)
3a=7 + 4y. (4)
Divide (3) by 2 and (4) by 3, x = li±li^, (5)
7 + 4i/
,.11. 11 + Qy 7-h4y ,^,
Equate the values of a, ^ — ~ — 3 (')
166 ALGEBBA.
Beduce (7), 83 + 27y = 14 + 8y,
19y=-19.
.•.y=-l.
Subfltitate the value of y in (1), 2 x + 9 = 11.
.-. X = 1.
191. Hence, to eliminate an unknown number by com-
parison,
From each equation obtain the value of one of the unknown
numbers in terms of the other.
Form an equation from these eqtuil values and reduce the
eqtiatian.
Note. If, in the last example, (3) be divided by (4), the resulting
2 11 + 9 v
equation, - = » would, when reduced, give the value of y.
This is the shortest method, and therefore to be preferred.
Exercise 70.
Solve by comparison :
1. a; + 16y=531 8. 3y — 7x= 41
Sx+ y = 27j 2y + 5x=:22j
2. 4:x+ 9y = 611 / 9. 2l2/ + 20aj = 1651
8a: — 13^^= 9j / 77y — 30a; = 295 J
3. 4:X + 3t/ = ^S\ I 10. llx — 10y = 141
5y--3x = 22j. \ 5x+ 7y = 41J
4. 2a: + 32^ = 431 / 11. 7y— 3x = 1391
10a— y= 7J 2a; + 5y= 91J
6. bx— ly= 331 ; 12. 17a; + 12y= 591
lla; + 122/ = 100j \ 19x- 4y = 153J
6. 6a; + 72/ = 431 \ 13. 24x+ 7y= 271
lla; + 9y = 69J \ 8a; — 33y = 115J
7. 8aj — 2l2/= 331 \ !*• « = 32/ — 191
6a; + 35y = 177j \ y = 3x — 23)
SIMULTANEOUS EQUATIONS. 157
192. Each equation must be simplified, if necessary,
before the elimination is performed.
Solve: (x-l)(y + 2) = (x-3)(y-l) + 81
2x-l 3(y-2)_ y
6 4 J
(x-l)(y + 2)=(x-3)(y-l) + 8 (1)
gx-l 3(y--2)_
6 4
(2)
Simplify (1), «y + 2x-y-2 = xy — x-3y + 8 + 8.
Transpose and combine, 3 x + 2 ^ = 13. (3)
Simplify (2), 8x - 4 - 16y + 30 = 20.
Transpose and combine, 8 x — 16 y = — 6. (4)
Multiply (3) by 8, 24 x + 16 y = 104. (6)
Multiply (4) by 3, 24 x — 46 y = - 18. (6)
Subtract (6) from (6), 61 y = 122.
.-. y = 2.
Substitute the value of y in (3), 3 x + 4 = 13.
.-. X = 3.
Exercise 71.
Solve :
1. x(y + 7) = i/(x + l)l ^ __!_
2« + 20 = 3y + l J ' ^ + 3 y — 2
5(x + 3) = 3(y-2)+2
)
5 *"l *5 10 "I
=^-2 ^=oJ 1 + ^ = 3 J
2. 2a!
6 ' 4
■2)(y
3(a! + 3)-4(2/+4) = -8
6. (x + l)(y + 2)-(x + 2)(y+l)=-l^
x — 2 10—x_i/ — 10
*• ~5 3~-~T~
2y + 4 2a! + y^a; + 13
3 8
}
158
ALGEBBA.
x + 1 y + 2^2(x-y)
3 4 6
x-3 y-3 _
—. ^-^- = 2y-x
16.
-4_y + 2-i
5
10
6+ 4 -^
2x-3y_^^x-Sy^^^^
1 — 3a;_ll-3y
2x-y + 3 a!-2y + 3
3 4
3ar — 4y + 3 4a; — 2y — 9_
4 + 3 ~
/ 11
10. lix = liy + 4:^-]
Hx = iy-21^^j
13 3
12.
a;+2y+3 ix-5y+6
3 ^ 19
6a:— 5y+4~3a;4-2y+l ,
^j-y_16
y — x 8
17. 6a; — i(52/ + 2) = 32'
3y+4(« + 2)=9
18. 3a;— 0.26y=28
0.12a;+0.72/=2.64
19. 7(a;-l) = 3(y + 8)-|
4x + 2 5y±9 I
9 ~ 2 J
13. 3^^^3^2x±y
^ 5
a;-2y_a; y
^ ^-2 + 3
20. 7a; + i(2y+4)=16-.
3y-i(a;+2) = 8 J
/
14.
4a; — 3?/ — 7 _ 3^ __ 2y 5
5 10 15 6
/
3 ' 2 20
y — 1 a; 3y -1—2/ — ^ i ^ t 1_
X
V-
^
SIMULTANEOUS EQUATIONS.
21. 5^ + 3. = 4y-2
ga' + ey 3£--2y_
535 — 3 3a; — 19 3y — a;
2 2 ~ 3
2a5 + y 9a; — 7_3(y + 3) 4x + 5y
2 8 ~ 4 16
23. 3y+ll = '^'"'-y(^+/y> + 31-4a» '
^ is — y + 4
(a; + 7)(y-2) + 3 = 2a!y-(y-l)(x + l).
6H-9 3« + 5y_ 3« + 4i
169
4a;- 6 *.' 2
8y + 7 ■ 6a;-3y 4y-9
10 "^ 2y — 8 "^ 5
1^
25. X ^ =
23— a;
20-
'+i^=*'
59 -2a;
2
73 -3y
Literal Simultaneous Equations.
193. The method of solving literal simultaneous equa-
tions is as follows:
Solve:
ax'\-hy = m'\
cx'-\-di/= n )
Multiply (1) by c,
Multiply (2) by o,
Subtract (4) from (3),
Divide by coefficient of y,
ax-hby = m
CX + dy = n
acz-i- bey = cm
acx+ ady=an
(6c — ad)y = cm — an
y- be-ad
(1)
(2)
(3)
(4)
160
ALGEBRA.
To find the value of x :
Multiply (1) by d,
Multiply (2) by b,
Subtract (6) from (5),
Divide by coefficient of z,
adx + bdy = dm
bcx -H bdy = bn
{ad — be) X = dm — bn
_dm — bn
~ ad — bc
(6)
(6)
Solve : \
■ ir^ x-{-y = a'\ 3. 7nx -{-ni/ = a
: — y=ib)
ix + bij-
px + qy-
EXERCISE 72.
• + 7iy = a^
px'\-qy--
►Kc — ny = r "^
6. mas — ny = r
Ji. ax-{-by = c'\ 4. aa; + % = e^
/=r J ax--{'€y=d J
6. oa; + ^y = ^J 1
^ * a
8. abx + cdy = 2 '
d — b
ax — cy=--
bd
ttx-\-2by=^d ,
a-\-b a — b a-\-b
a-\-b a — b a — b,
11. a(a — x)^=b(x-\-y—a)^
a(y — b'-x)=b(y — b)J
12. ^-^=a^
x — yl I
x + y — l
a^ + b^
13. aa; = fti/H ^
(a — ^) a; = (a -|- i)
14. ax-{-by = c^
a b
= 0
16.
2a
a-\- b a — b
x — y x-^-y
2ab ""^"T^
16. bx — bc = ay — aC)
A—y^a — b J
V
SIMULTANEOUS EQUATIONS.
161
X — a
17. - — -^ = e
a(x — a)'^b(y — ft) + dbc = 0 ,
18. (a + b)x — (a--b)y=^^ab 1
la'-b)x + (a + b)y = 2a^ — 2b^}
19. (x + a)(y + b)-(x-a)(y-b)=2(a''by^
x-y + 2(a-b) = 0
20. (a + b){x + y)-(a^b)(x''y) = a*^
(a-b)(x + y) + (a + b)(x'-y)=^b' S
194. ^Fractional simultaneous equations, of which the
denominators are simple expressions and contain the
unknown numbers^ may be solved as follows:
(1) Solve:
a , b
X y
e , d
X y
- + - = m.
X y
c , d
X y
ac , be
1 — = cm.
X y
ac , ad
— I = an.
X y
he — ad
Multiply (1) by c,
Multiply (2) by o,
Subtract (4) from (3),
Multiply both sides by y, 6c — od = (cm — arC) y,
be — ad
y
= cm — an.
(1)
(2)
(3)
(4)
Multiply (1) by d,
Multiply (2) by 6,
ad , bd
h — = dm.
X y
bc.bd ,
X y
(5)
(6)
162
ALGEBRA.
Subtract (6) from (5),
ad — be
• dm — 6fi,
Multiply both sides by x, ad — bc= (dm — 6n) as.
ad — bc
dm — bn
(2) Solve:
3x^ 5y
1_
6x
Multiply (2) by 4,
Add (1) and (3),
Divide both sides by 19,
1
3x by
-^-J- = 3
6x lOy
14 2
— — — = 12.
3x 5y
19
3x
= 19.
Substitute the value of x in (1), 6 + ^— = 7.
Transpose,
Divide both sides by 2,
by
\^
by
= 2
= 1
Solve :
Exercise 73.
6.
(1)
(2)
(8)
5. ?-^ = 5 •
X y
^ y
a h ac^
X y h
^ , a he
X y a ,
SIMULTANEOUS EQUATIONS.
163
7. h7-=o
ax by
ax by
8. = m + n
nx my
n m 2 12
X y
a.
, b
9.
+-=
m
X
y
b
a
n
X
y
195. If three simultaneous equations are given, involv-
ing three unknown numbers, one of the unknown numbers
must be eliminated between two pairs of the equations;
then a second between the resulting equations.
196. Likewise, if four or more equations are given, in-
volving four or more unknown numbers, one of the
unknown numbers must be eliminated between three or
more pairs of the equations; then a second between the
pairs that can be formed of the resulting equations ; and
so on.
Solve :
2a; — 32^ + 4«= 4^
3x + 52/ — 7;^ = 12
5a; — y — 8«= 5
Eliminate z between two pairs of these equations.
Multiply (1) by 2,
(3) is
Add,
Multiply (1) by 7,
Multiply (2) by 4,
Add,
Multiply (6) by 7, -
(5) is
Subtract (6) from (7),
4x — 6y + 8z= 8
5x-- y~-8z= 6
173 X =619
.•.x=3.
(1)
(2)
(3)
(4)
9x-7y =13
(6)
14x-2l2/ + 28z = 28
12x + 20y-28« = 48
26x- y =70
(6)
182x-72/ = 632
(7)
9x-7y= 13
Substitute the value of x in (6), 78 -- y = 76.
.-. y = 2.
Substitute the values of x and y in (1), 6 — 6 + 4 « = 4.
.-. « = 1.
164
ALGEBBA.
Exercise 74.
Solve :
1. 5a; + 3y — 6« = 4^
X'-2y + 2z=^2 J
7x-4y + 3« = 35j
3. x + ?/ + « = 6
bx + ^y + 3z=^22
15a; + 10y + 6^ = 53.
4. 4a; — 3y + « = 9
9a; + ?/ — 5;2: = 16
a; — 42^ + 3« = 2
5. 8x + 4y — 3« = 6^
a; + 3y-^ = 7 I
4aj — 52/ + 4« = 8j
6. 12x + 5y—4.z = 29'
13a;-2y + 5« = 58
17a: — y — « = 15
7. 3/ — x-{-z = — 5
« — y — x = — 25
x + y + z = 35
8. a; + 2/ + « = 30 -j
8a: + 42/ + 2« = 50 I
27a; + 9y + 3« = 64j
9. 15i/ = 24;s! — 10a: + 41 '
15a; = 12y — IGis + lO
18a;— (7;3f — 13) = 14y.
10. 3x — 2/ + « = 17
5x + 3ij-'2z = 10
7x + 4:y — 5z = S J
11. a; + y + ;2; = 5
3a; — 52/ + 7« = 75
95-ll« + 10 = 0.
12. a; + 2y + 3« = 6
2x + Ay + 2z = S
3a; + 2y + 8« = 101J
13. x — 3y — 2z = l
2x'-3y+5z = — 19
5x + 2y — z^l2
.14. 3x — 2y = 5 1
4:x — 3y+2z = ll I
a; — 2y — 5« = — 7 J
15. a; + y=l^
y + ;. = 9l -^
a; + « = 5 J
16. 2a; — 3y = 3l
32/-4« = 7l
4« — 5a; = 2j
17. 3x — 4y + 6z = l'
2x + 2y — z = l
7x — 6y + 7« = 2.
18. 7a; — 32/ = 30
9y — 5« = 34
x + y + z=:33}
/
/
y
SIMULTAKEOUS EQUATIONS.
166
4 14
V »U
20. ^ + ? = 6
Z X
24. 2-^ + ^=2.9
« y «
_5_6_I=_io.4
y « a;
21. -H = a
X y z
X y z
1,1 1
y ;s a;
25. ? + i-^=0
35 2/ «
2-2-2=0
bz'^cy = a '
a« + ex = i
ay + &c = c ,
/26. aa; + fty + c« = a ^
/ ax — by — cz=^b
ttx-\'Cy-\'bz=^c ,
3 4 5
x—y y — z x-^rz x — a — b
a b c a + ft + <?
* In example 21, subtract from the sum of the three equations each
equation separately.
t In example 22, multiply the equations by a, 6, and c, respectively,
and from the sum of the results subtract the double of each equation
separately.
CHAPTER XII.
Problems producing Simultaneous Equations.
197. It is often necessary in the solution of problems
to employ two or more letters to represent the numbers to
be found. In all cases the conditions must be sufficient to
give just as many equations as there are unknown numbers
employed.
If there are more equations than unknown numbers,
some of them are superfluous or contradictory ; if there
are less equations than unknown numbers, the problem is
indeterminate or impossible.
(1) When the greater of two numbers is divided by the
less the quotient is 4 and the remainder 3; and
when the sum of the two numbers is increased by
38, and the result divided by the greater of the
two numbers, the quotient is 2 and the remainder 2.
Find the numbers.
Let
X = the greater number,
and
y = the smaller number.
Then
y
and
x +
X
From the solution of these equations, x = 47, and y = 11.
(2) If A give B $10, B will have three times as much
money as A. If B give A $10, A will have twice
as much money as B. How much has each ?
/
P&0BLSM8. 167
Let £ sc the number of dollan A has,
and y = the number of dollars B has.
Then y + 10 = the number of dollars B has, and « — 10 « the
number of dollars A has after A gives $10 to B.
.-. y + 10 = 3 (X - 10), and « + 10 = 2 (y - 10).
From the solution of these equations, x = 22, and y = 26.
Therefore, A has $22 and B $26.
Exercise 75.
1. The sum of two numbers divided by 2 gives as a quo-
tient 24, and the difference between them divided by
2 gives as a quotient 17. What are the numbers ?
2. The number 144 is divided into three numbers. When
the first is divided by the second, the quotient is 3
and the remainder 2 ; and when the third is divided
by the sum of the other two numbers, the quotient
is 2 and the remainder 6. Find the numbers.
3. Three times the greater of two numbers exceeds twice
the less by 10 ; and twice the greater together with
three times the less is 24. Find the numbers.
4. If the smaller of two numbers is divided by the greater,
the quotient is 0.21 and the remainder 0.0057 ; but
if the greater is divided by the smaller, the quo-
tient is 4 and the remainder 0.742. What are the
numbers ?
6. Seven years ago the age of a father was four times that
of his son ; seven years hence the age of the father
will be double that of the son. What are their ages ?
6. The sum of the ages of a father and son is half what
it will be in 25 years ; the difference between their
ages is one-third of what the sum will be in 20 years.
What are their ages ?
168 AXiOEB&A.
7. If B give A f 25, they will have equal sums of money ;
but if A give B $22, B's money will be double that
of A. How much has each ?
A
>^
y
8. A farmer sold to one person 30 bushels of wheat and
40 bushels of barley for $67.50 ; to another person
he sold 50 bushels of wheat and 30 bushels of barley
for $85. What was the price of the wheat and of
the barley per bushel ?
9. If A give B $5, he will then have $6 less than B ; but
if he receive $5 from B, three times his money will
be $20 more than four times B's. How much has
each?
/lb. The cost of 12 horses and 14 cows is $1900 ; the cost
of 5 horses and 3 cows is $650. What is the cost
of a horse and a cow respectively ?
Note I. A fraction of which the terms are unknown may be
represented by —
Ex. A certain fraction becomes equal to J if 3 is added to
its numerator, and equal to f if 3 is added to its
denominator. Determine the fraction.
Let
2C
- = the required
fraction.
By the conditions
x + 3 .
and ^^3=f.
From the solution of these equations it is found that
x = 6,
y=18.
Therefore the fraction = y\.
.11. A certain fraction becomes equal to 2 when 7 is added
to its numerator, and equal to 1 when 1 is subtracted
from its denominator. Determine the fraction.
X
X
PROBLEMS. 169
12. A certain fraction becomes equal to J when 7 is added
to its denominator^ and equal to 2 when 13 is added
to its numerator. Determine the fraction.
13. A certain fraction becomes equal to | when the
denominator is increased by 4, and equal to ^f
when the numerator is diminished by 15. Deter-
mine the fraction.
14. A certain fraction becomes equal to § if 7 is added to
the numerator, and equal to | if 7 is subtracted
from the denominator. Determine the fraction.
15. Find two fractions with numerators 2 and 5 respec-
tively, such that their sum is Ij; and if their
demominators are interchanged their sum is 2.
16. A fraction which is equal to f is increased to -^ when
a certain number is added to both its numerator
and denominator, and is diminished to | when one
more than the same number is subtracted from
each. Determine the fraction.
Note II. A number consisting of two digits which are unknown
may be represented by 10 x + y, in which z and y represent the digits
of the number. Likewise, a number consisting of three digits which
are unknown may be represented by 100 a; + 10 y + 2, in which x, y,
and z represent the digits of the number.
For example, consider any number expressed by three digits, as
364. The expression 364 means 300 + 60 + 4 ; or, 100 times 3+10
times 6 + 4.
Ex. The sum of the two digits of a- number is 8, and if
36 be added to the number the digits will be
interchanged. What is the number ?
Let X = the digit in the tens' place,
and y = the digit in the units' place.
Then 10 x + y = the number.
By the conditions, x + y = 8, (I)
and 10x + y + 36 = 10y + «. (2)
170 ALGEBBA.
From (2),
9x — 9y = — 36
Divide by 9,
X — y= —4.
Add (1) and (3);
2x = 4.
.-. X = 2.
Subtract (3) from (1),
2y= 12.
r,y = 6.
Hence, the number is 26.
17. The sum of the two digits of a number is 10, and if
54 be added to the number the digits will be
interchanged. What is the number ?
18. The sum of the two digits of a number is 6, and if
the number be divided by the sum of the digits
the quotient will be 4. What is the number?
19. A certain number is expressed by two digits, of which
the first is the greater. If the number is divided by
the sum of its digits the quotient is 7 ; if the digits
are interchanged, and the resulting number dimin-
ished by 12 is divided by the difference between the
two digits, the quotient is 9. What is the number ?
20. If a certain number is divided by the sum of its two
digits the quotient is 6 and the remainder 3 ; if the
digits are interchanged, and the resulting number is
divided by the sum of the digits, the quotient is 4
and the remainder 9. What is the number ?
21. If a certain number is divided by the sum of its two
digits diminished by 2, the quotient is 5 and the
remainder 1 ; if the digits are interchanged, and the
resulting number is divided by the sum of the digits
increased by 2, the quotient is 5 and the remainder
8. Find the number.
22. The first of the two digits of a number is, when doubled,
3 more than the second, and the number itself is less
by 6 than five times the sum of the digits. What is
the number ?
PBOBLEMS. 171
23. A number is expressed by three digits, of whicli the
. first and last are alike. By interchanging the digits
in the units' and tens' places the number is increased
by 54 ; but if the digits in the tens' and hundreds'
places are interchanged, 9 must be added to four
times the resulting number to make it equal to the
original number. What is the number ?
24. A number is expressed by three digits. The sum of
the digits is 21 ; the sum of the first and second
exceeds the third by 3 ; and if 198 be added to the
number, the digits in the units' and hundreds' places
will be interchanged. Find the number.
26. A number is expressed by three digits. The sum of
the digits is 9; the number is equal to forty-two
times the sum of the first and second digits; and
the third digit is twice the sum of the other two.
Find the number.
26. A certain number, expressed by three digits, is equal
to forty-eight times the sum of its digits. If 198
be subtracted from the number, the digits in the
units' and hundreds' places will be interchanged ;
and the sum of the extreme digits is equal to twice
the middle digit. Find the number.
Note III. If a boat moves at the rate of x miles an hour in still
water, and if it is on a stream that runs at the rate of y miles an hour,
then
x-\- y represents its rate doton the stream,
x — y represents its rate wp the stream.
27. A waterman rows 30 miles and back in 12 hours. He
finds that he can row 5 miles with the stream in the
same time as 3 against it. Find the time he was
rowing up and down respectively.
172 ALGEBRA.
28. A crew which can pull at the rate of 12 miles an hour
down the stream, finds that it takes twice as long to
come up the river as to go down. At what rate does
the stream flow ?
29. A man sculls down a stream, which runs at the rate of
4 miles an hour, for a certain distance in 1 hour and
40 minutes. In returning it takes him 4 hours and
15 minutes to arrive at a point 3 miles short of his
starting-place. Find the distance he pulled down
the stream and the rate of his pulling.
30. A person rows down a stream a distance of 20 miles
and back again in 10 hours. He finds he can row
2 miles against the stream in the same time he can
row 3 miles with it. Find the time of his rowing
down and of his rowing up the stream ; and also the
rate of the stream.
Note IV. When commodities are mixed, the quantity of the
mixture is equal to the quantity of the ingredients ; the cost of the
mixture is equal to the cost of the ingredients.
Ex. A wine-merchant has two kinds of wine which cost
72 cents and 40 cents a quart respectively. How
much of each must he take to make a mixture of
50 quarts worth 60 cents a quart ?
Let X — the required number of quarts worth 72 cents a
quart,
and y = the required number of quarts worth 40 cents a
quart.
Then, 72 x = the cost in cents of the first kind,
40 2^ = the cost in cents of the second kind of wine,
and 3000 = the cost in cents of the mixture.
.-. X 4- y = 60,
72x+40y=3000.
From which equations the values of x and y may be found.
PROBLEMS. 173
31. A grocer mixed tea that cost him 42 cents a pound
with tea that cost him 54 cents a pound. He had
30 pounds of the mixture, and by selling it at the
rate of 60 cents a pound, he gained as much as 10
pounds of the cheaper tea cost him. How many-
pounds of each did he put into the mixture ?
32. A grocer mixes tea that cost him 90 cents a pound
with tea that cost him 28 cents a pound. The cost
of the mixture is f 61.20. He sells the mixture at
60 cents a pound, and gains $3.80. How many
pounds of each did he put into the mixture ?
33. A farmer has 28 bushels of barley worth 84 cents a
bushel. With his barley he wishes to mix rye
worth f 1.08 a bushel, and wheat worth $1.44 a
bushel, so that the mixture may be 100 bushels,
and be worth $1.20 a bushel. How many bushels
of rye and of wheat must he take?
Note V. It is to be remembered that if a person can do a piece
of work in x days, tfie part of the work he can do in one day will be
represented by ^.
Ex. A and B together can do a piece of work in 48 days ;
A and C together can do it in 30 days ; B and C
together can do it in 26f days. How long will it
take each to do the work ?
Let X = the number of days it wiU take A alone to do the work,
y = the number of days it will take B alone to do the work,
and z = the number of days it will take C alone to do the work.
Then, - 1 - » - » respectively, will denote the part each can do
X y z . , ^
ma day,
and - + - will denote the part A and B together can do in a day;
but — wiU denote the part A and B together can do in a day.
48
174 ALGEBRA.
Therefore, " + " * 75 (1)
Likewise, ~ + " = ^ (2)
^ 1,113
^^ y+r26f=80 <^)
Add (1), (2), and (3), l + ^ + f =^ (4)
Multiply (1) by 2, | + ? =^ (5)
Subtra^jt (5) from (4), ? = i
.-. z = 40.
2 1
Subtract the double of (2) from (4), - = t^
.-. y = 80.
2 1
Subtract the double of (3) from (4), - = ^
.-. X = 120.
34. A and B together earn f 40 in 6 days ; A and C to-
gether earn $54 in 9 days; B and C together earn
$80 in 15 days. What does each earn a day ?
35. A cistern has three pipes, A, B, and C. A and B will
fill it in 1 hour and 10 minutes ; A and C in 1 hour
and 24 minutes ; B and C in 2 hours and 20 minutes.
How long will it take each to fill it ?
36. A warehouse will hold 24 boxes and 20 bales; 6 boxes
and 14 bales will fill half of it. How many of each
alone will it hold ?
37. Two workmen together complete some work in 20 days;
but if the first had worked twice as fast, and the
second half as fast, they would have finished it in
15 days. How long would it take each alone to do
the work ?
38. A purse holds 19 crowns and 6 guineas; 4 crowns and
5 guineas fill H oi it. How many of each alone
will it hold ?
PROBLEMS. 176
39. A piece of work can be completed by A, B, and C to-
gether in 10 days ; by A and B together in 12 days;
by B and C, if B works 15 days and C 30 days.
How long will it take each alone to do the work ?
40. A cistern has three pipes, A, B, and C. A and B will
fill it in a minutes ; A and C in 6 minutes ; B and
C in c minutes. How long will it take each alone
to fill it ?
Note VI. In considering the rate of increase or decrease in quan-
tities, it is usual to take 100 as a common standard of reference, so that
the increase or decrease is calculated for every 100, and therefore
called per cent.
It is to be observed that the representative of the number resulting
after an increase has taken place is 100 + increase per cent; and
after a decrease, 100 — decrease per cent.
Interest depends upon the time for which the money is lent, as well
as upon the rate charged ; the rate charged being the rate on the
principal lor one year. Hence,
„. , . , ^ Principal X Rate X Time
Smiple mterest = -^j: »
100
where Time means number of years or fraction of a year.
Amount = Principal + Interest.
In questions relating to stocks, 100 is taken as the representative
of the stock, the price represents its market value, and the per cent
represents the interest which the stock bears. Thus, if six per cent
stocks are quoted at 108, the meaning is, that the price of $100 of the
stock is $108, and that the interest derived from |100 of the stock will
be y^ of $100, that is, $6 a year. The rate of interest on the money
invested will be JJJ of 6 per cent.
41. A man has $10,000 invested. For a part of this sum
he receives 5 per cent interest, and for the rest 4 per
cent ; the income for his 5 per cent investment is
$60 more than from his 4 per cent. How much
has he in each investment ?
176 AIiGEBBA.
42. A sum of money, at simple interest, amounted in 6
years to $26,000, and in 10 years to $30,000. Find
the sum and the rate of interest.
43. A sum of money, at simple interest, amounted in 10
months to $26,250, and in 18 months to $27,250.
Find the sum and the rate of interest.
44. A sum of money, at simple interest, amounted in m
years to a dollars, and in n years to b dollars. Find
the sum and the rate of interest.
45. A sum of money, at simple interest, amounted in a
months to c dollars, and in h months to d dollars.
Find the sum and the rate of interest.
46. A person has a certain capital invested at a certain
rate per cent. Another person has $1000 mpre
capital, and his capital invested at one ger cent
better than the first, and receives an income $80
greater. A third person has $1500 more capital,
and his capital invested at two per cent better
than .the first, and receives an income $150 greater.
Find the capital of each, and the rate at which
it is invested.
47. A person has $12,750 to invest. He can buy three
per cent bonds at 81, and five per cents at 120.
Find the amount of money he must invest in each in
order to have the same income from each investment.
48. A and B each invested $1500 in bonds ; A in three
per cents and B in four per cents. The bonds were
bought at such prices that B received $5 interest
more than A. After both classes of bonds rose 10
points, they sold out, and A received $50 more than
B. What price was paid for each class of bonds ?
PROBLEMS. 177
49. A person invests $10,000 in three per cent bonds,
$16,500 in three and one-half per cents, and has an
income from both investments of $1056.26. If his
investments had been $2750 more in the three per
cents, and less in the three and one-half per cents,
his income would have been 62^ cents greater.
What price was paid for each class of bonds ?
50. The sum of $2500 was divided into two unequal parts
and invested, the smaller part at two per cent more
than the larger. The rate of interest on the larger
sum was afterwards increased by 1, and that of the
smaller sum diminished by 1 ; and thus the interest
of the whole was increased by one-fourth of its
value. If the interest of the larger sum had been
so increased, and no change had been made in the
interest of the smaller sum, the interest of the
whole would have been increased one-third of its
value. Find the sums invested, and the rate per
cent of each.
NoTB Vn. If X represents the number of linear units in the length,
and y in the width, of a rectangle, xy will represent the number of its
units of surface ; the surface unit having the same name as the linear
unit of its sides.
51. If the sides of a rectangular field were each increased
by 2 yards, the area would be increased by 220
square yards ; if the length were increased and the
breadth were diminished each by 5 yards, the area
would be diminished by 185 square yards. What
is its area ?
52. If a given rectangular floor had been 3 feet longer and
2 feet broader it would have contained 64 square
feet more ; but if it had ])een 2 feet longer and 3
feet broader it would have contained 68 square feet
more. Find the length and breadth of the floor.
178 ALGEBRA.
63. In & certain rectangular garden there is a strawberry-
bed whose sides are one-third of the lengths of the
corresponding sides of the garden. The perimeter of
the garden exceeds that of the bed by 200 yards ;
and if the greater side of the garden be increased
by 3, and the other by 5 yards, the garden will be
enlarged by 645 square yards. Find the length
and breadth of the garden.
Note VIII. Care must be taken to express the conditions of a
problem with reference to the same principal unit.
Ex. In a mile race A gives B a start of 20 yards and beats
him by 30 seconds. At the second trial A gives B
a start of 32 seconds and beats him by 9j\ yards.
Find the rate per hour at which each runs.
Let X = the number of yards A runs a second,
and y = the number of yards B runs a second.
Since there are 1760 yards in a mile,
1760
= the number of seconds it takes A to
run a mile,
and ^ = the number of seconds B was running
^ in the first and second trials,
Hence, = 30, respectively.
y X
y X
The solution of these equations gives x = 6}f and y = 5^^.
That is, A runs ri^ ' or xtt » of a mile in one second ;
and in one hour, or 3600 seconds, runs 12 miles.
Likewise, B runs l^^ix ™iles in one hour.
54. In a mile race A gives B a start of 100 yards and
beats him by 15 seconds. In the second trial A
gives B a start of 45 seconds and is beaten by 22
yards. Find the rate of each in miles per hour.
PROBLEMS. 179
55. In a mile race A gives B a start of 44 yards and beats
him by 51 seconds. In the second trial A gives B
a start of 1 minute and 15 seconds and is beaten by
88 yards. Find the rate of each in miles per hour.
56. The time which an express train takes to go 120 miles
is ^5 of the time taken by an accommodation-train.
The slower train loses as much time in stopping at
different stations as it would take to travel 20 miles
without stopping; the express-train loses only half
as much time by stopping as the accommodation-
train, and travels 15 miles an hour faster. Find
the rate of each train in miles per hour.
57. A train moves from P towards Q, and an hour later
a second train starts from Q and moves towards P
at a rate of 10 miles an hour more than the first
train ; the trains meet half-way between P and Q.
If the train from P had started an hour after the
train from Q its rate must have been increased by
28 miles in order that the trains should meet at the
half-way point. Find the distance from P to Q.
58. A passenger-train, after travelling an hour, meets with
an accident which detains it one-half an hour ; after
which it proceeds at four-fifths of its usual rate, and
arrives an hour and a quarter late. If the accident
had happened 30 miles farther on, the train would
have been only an hour late. Determine the usual
rate of the train.
69. A passenger-train, after travelling an hour, is detained
15 minutes ; after which it proceeds at three-fourths
of its former rate, and arrives 24 minutes late. If
the detention had taken place 5 miles farther on,
the train would have been only 21 minutes late.
Determine the usual rate of the train.
180 ALGEBRA.
60. A man bought 10 oxen, 120 sheep, and 46 lambs.
The cost of 3 sheep was equal to that of 6 lambs ;
an ox, a sheep, and a lamb together cost a number
■-..^ of dollars less by 57 than the whole number of
animals bought ; and the whole sum spent was
$2341.50. Find the price of an ox, a sheep, and
a lamb, respectively.
61. A farmer sold 100 head of stock, consisting of horses,
oxen, and sheep, so that the whole realized $11.75 a
head ; while a horse, an ox, and a sheep were sold
for $110, $62.50, and $7.50, respectively. Had he
sold one-fourth of the number of oxen that he did,
and 25 more sheep, he would have received the same
sum. Find the number of horses, oxen, and sheep,
respectively, which were sold.
62. A, B, and C together subscribed $100. If A's sub-
scription had been one-tenth less, and B's one-tenth
more, C's must have been increased by $2 to make
up the sum ; but if A's had been one-eighth more;
and B's one-eighth less, C's subscription would have
been $17.50. What did each subscribe?
63. A gives to B and C as much as each of them has ; B
gives to A and C as much as each of them then has;
and C gives to A and B as much as each of them
then has. In the end each of them has $6. How
much had each at first ?
64. A pays to B and C as much as each of them has ; B
pays to A and C one-half as much as each of them
then has ; and C pays to A and B one-third of what
each of them then has. In the end A finds that he
has $1.50, B $4.16J, C $0.58f How much had
each at first ?
CHAPTER XIII.
Involution and Evolution.
198. The operation of raising an expression to any
required power is called Involution.
Every case of involution is merely an example of mvltir
plication, in which the factors are equal. Thus,
199. A power of a simple expression is found by multi-
plying the exponent of each factor by the exponent of the
required factor, and taking the product of the resulting
factors. The proof of the law of exponents, in its general
form, is :
(»"•)" = «"» X a"* X a"* X to n factors,
— a"* + "» + "*+ *<* *• tenM,
Hence, if the exponent of the required power be a com-
posite number, it may be resolved into prime factors, the
power denoted by one of these factors may be found, and
the result raised to a power denoted by another, and so on.
Thus, the fourth power may be obtained, by taking the
second power of the second power; the sixth by taking
the second power of the third power ; the eighth by taking
the second power of the second power of the second power.
200. Erom the Law of Signs in multiplication it is
evident that,
I. All even powers of a number are positive,
II. All odd powers of a number have the same sign as
the number itself.
182 ALGEBEA.
Hence, no even power of any number can be negative;
and of two compound expressions whose terms are identical
but have opposite signs, the even powers are the same.
Thus,
(i — a)2=| — (a — 6)P = (a — ^>)l
201. A method has been given, § 83, of finding, without
actual multiplication, the powers of binomials which have
the form {a±h).
The same method may be employed when the terms of a
binomial have coefficients or exponents,
(1) (a — 6)8 = a8-3a% + 3a^»2 — ft«
(2) {bx'-'2f)\
= (5xy-3(5xy(2/) + 3(5a^(22^2-(22/y,
= 125a;«-150xV+60a;y~82/».
(3) (a — &)* = a* — 4a«6 + 6a26* — 4a^»« + ^>*.
(4) (?^-W^
2. In like manner, a polynomial of three or more
terms may be raised to any power by enclosing its terms
in parentheses, so as to give the expression the form of
a binomial. Thus,
(1) {a + b + cf=^\a + {h + c)Y,
= a^-^^a\h + c) + ^a{b + cy + {b + cy,
= a8 + 3a«^ + 3a2c + 3a5* + 6aftc
INVOLUTION AND EVOLUTION, 188
(2) (a»-2a:*+3a; + 4)«,
= (0^-2x^+2 (x^-2x^(8x + A)+(3x + 4:y,
=a;«-4a;*+4a;*+6x*— 4a;«-16a;*+9a;*+24a;+16,
=x« — 4a;« + 10a*-4a;«-7a^ + 24a; + 16.
Exercise 76.
Write the second members of the following equations :
1. (ay= 11. (2a^b<^*= 21. (-3a%>cy =
2. (a^y=: 12. ('-5axhf)^= 22. (— 3a;y«)« =
3. {x^ff= 13. (;-lm^nxhff= 23. (— 5a2fta^* =
^■(f)*= »(-l2)= -(-'^7=
6. (ic + 2)»= 16. (2a; — a)*= 26. (1 — a — ay =
7. (aj-2)*= 17. (3a; + 2a/= 27. (2-32c+4a;y=
8. (a: + 3/= 18. (2a; — 2^)*= 28. (1— 2x + a;2)3 =
9. (l + 2a;)«= 19. (x^y — 2xi/)^= 29. (1 — a: + a;2)3 =
10. (2m — 1)»= 20. (ai- 3/= 30. (l+a; + a;«/ =
Evolution.
203. The operation of finding any required root of an
expression is called Evolution.
Every case of evolution is merely an example of factor-
ing, in which the required factors are all equal. Thus,
the square, cube, fourth roots of an expression are
found by taking one of the two, three, four equal factors
of the expression.
^^ jm ^L '^
184 ALGEBBA.
204. The symbol which denotes that a square root is to
be extracted is V; and for other roots the same symbol is
used, but with a figure written above to indicate the root,
thus, -^f ^y etc., signifies the third root, fourth root, etc.
205. Since the cube of a* = a", the cube root of a^ = a^.
U Since the fourth power of 2a' = 2*a*, the fourth root of
Since the square of (ibc = a^^c^y the square root of a^lj^(^
= abc.
Since the square of — = -^-5, the square root of -5-5 = —
xy a;y ^ afy' xy
Hence, the root of a simple expression is found by
dividing the exponent of each factor by the index of the
rooty and taking the product of the resulting factors,
206. It is evident from § 200 that
I. Any even root of a positive number will have the
double sign, dz .
II. There can be no even root of a negative number.
III. Any odd root of a number will have the same sign
as the number.
Thus,^^ = ±^; ^-27mV = -3mn«;
i^lf -9y
V
81a" ~ 3a*
But V— a^ is neither + x nor — x, for (+«)*=+ a^, and
(-xy= + x'.
The indicated even root of a negative number is called
an impossible, or imaginary, number.
INVOLUTION AND EVOLUTION.
185
207. If the root of a number expressed in figures is not
readily detected, it may be found by resolving the number
into its prime factors. Thus, to find the square root of
3,415,104 :
2«
3415104
2«
426888
3'
63361
7
6929
7
847
11
121
11
3,416,104 = 2« X 3» X 7' X 11*.
••• V'3,416,104 = 2»X3 X7Xll= 1848.
Simplify: Exeecisb 77.
1. V^, ^^, V4^, \/64, v'^^V, -^16a"JV, ^-Z2a^.
2. -v'— 1728c«d»V/, -^33766»V«, \-^3111696c"'«*.
-3 V633616WV« a'/-M^*^, A'/^i^.
4. V26a«6V+ -^S^W- 'l'81a*i»c« — 4'32a*S>V.
5. V27a;y X V^43pi» X Vi6^.
When a=l, J = 3, a; = 2, y = 6, find the values of:
6. 4:-\/2x — ■\/abxy + 5 ^/a'bfxy.
7. 2a V8a« + i -^12% + iabx Vtoy.
8. Va* + 2a6 + 6»X Va* + 3«»6 + 3a6« + J».
». V — 3i'a + 3ia» — a»-f-V6* + a'-2a4.
186 ALGEBRA.
Square Roots of Compound Expressions.
, Since the square of a + b is a^-\-2ab + b^y the
square root of a^ -\- 2 db -\- b* is a-\-b.
It is required to find a method of extracting the root
a + b when a^ + 2ab'\-b^ is given:
Ex. The first term, a, of the root is obviously the square root of
the first term, a^, in the expression.
a^-^.2ab + b^\a + b If the a^ is subtracted from the
^8 given expression, the remainder is
n iT] 9/172 2ab + b^, Therefore the second term,
' "^ 6, of the root is obtained when the
Zao-f-o gjg|. ^Ym of this remainder is divided
by 2 a, that is, by double the part of
the root already found. Also, since 2 a6 + 62 = (2 a + 6) 6, the divisor
w completed by adding to the trial-divisor the new term of the root.
Ex. Find the square root of 25 a:* — 20x^i/ + 4a;y.
25a^ — 20x^i/ + ^xY\5x — 2x^t/
25x^
-20x^1/ + 4:xy
"-20a;V + 4a;V
10x — 2x^y
The expression is arranged according to the ascending powers of x.
The square root of the first term is 6 x, and 5 x is placed at the
right of the given expression, for the first term of the root.
The second term of the root, — 2x2y, is obtained by dividing
— 20 a^y by 10 x, and this new term of the root is also annexed to the
divisor, 10 x, to complete the divisor.
209. The same method will apply to longer expressions,
if care be taken to obtain the trial-divisor at each stage of
the process, by doubling the part of the root already found;
and to obtain the complete divisor by annexing the new term
of the root to the triaUdivisor,
INVOLUTION AND EVOLUTION. 187
Ex. Find the squaxe root of
16aj»-24x«+25a;*-20x«+10x2-4a:+l |4a;«— 3a^+2g— 1
8a^-3aj«
-24a«+25a:*
Sx^'-6x'+2x
16aj*--20a;«+10««
16ic*-12aj8+ 4a;«
8x»-6x>+4aj-l
- 8x»+ ex'—^x+l
— Sa^+ 6x^-'4:x+l
/
The expression is arranged according to the descending powers
of X.
It will be noticed that each successive trial-divisor may be obtained
by taking the preceding complete divisor with its last term doubled.
Exercise 78.
Extract the square roots of :
1. a* + 4a3 + 2a-4a + l.
2. a;* — 2a;V + 3icy — 2V + 2/*-
3. 4a» — 12a«x + 5aV + 6aV + aV.
4. 9aj«— 12ajV + 16xy — 24ary + 4/ + 16ay*.
5. 4a«+16c«+16aV — 32aV.
6. 4x* + 9 — 30a — 20a^+37a^.
7. 16x* — 16abx' + 16b^x^ + 4a%^-Sab^ + U\
8. x'' + 25x^+10x'-4x^ — 20x^+16 -24:x,
9. x^ + Sxy — 4:0^1/ — 4:Xt/ + Sx^l/* — 10a^l/^ + t/^.
10. 4 — 12a--llaH5a'-4a« + 4a«+14a«
11. 9a^ — 6ab + 30ac + 6ad + h^'-10hc — 2bd
+ 25c'+10cd+(P,
188 AL6EBBA.
12. 25a;«--31a;y + 34a^2^-30a^?/ + 2/*— 8xy» + 10a^/..
13. m« — 47»' + 10m« — 20m* — 44m»
+ 35m* + 46m« — 40m + 25.
7
14. 0?* — ^j/ — j^j^'{'X7^-\-y\
16. a* — 4aV + 6«*3/* — 6V+52^*-^ + ^-
a; mr
t, 1,4 ,10, 20 ,25 ,24 ,16
"• 6i-T + 3-T + «'- "• * +^-12-3 + 9-
Square Roots of Arithmetical Numbers.
210. In the general method of extracting the square
root of a number expressed by figures, the first step is to
mark off the figures in groups.
Since 1 = l^, 100 = I02, 10,000 = IOO2, and so on, it is evident that
the square root of any number between 1 and 100 lies between 1 and
10 ; the square root of any number between 100 and 10,000 lies be-
tween 10 and 100. In other words, the square root of any integral
number expressed by one or two figures is a number of one figure ;
the square root of any integral number expressed by three or four
figures is a number of two figures ; and so on.
If, therefore, an integral number be divided into groups of two
figures each, from right to left, the number of figures in the root will
be equal to the number of groups of figures. The last group to the
left may have one or two figures.
Ex. Find the square root of 3249.
32 49 (57 In this case, a in the typical form a^ + 2db-\-l^
^^ represents 5 tens, or 60, and b represents 7 ones.
107)749 The 26 subtracted is really 2600, that is, a^, and
7 49 the complete divisor, 2 o + 6, is 2 X 60 + 7, or 107.
INVOLUTION AND EVOLUTION. 189
211. The same method will apply to numbers of more
than two groups by considering a in the typical form to
represent at each step the part of the root already found.
It must be observed that a represents so many tens with respect to
the next figure of the root,
Ex. Find the square root of 6,322,249.
6 32 2249(2307
4
43J132
129
4607)32249
32249
212. If the square root of a number have decimal places,
the number itself will have ttoice as many.
Thus, if 0.21 be the square root of some number, this number will
be (0.21)2 = 0.21 X 0.21 = 0.0441 ; and if 0.111 be the root, the num-
ber will be (0.111)2 = 0.111 X 0.111 = 0.012321.
Therefore, the number of decimal places in every square decimal
will be even, and the number of decimal places in the root will be
haHf as many as in the given number itself.
Hence, if the given square number contains a decimal, we divide it
into groups of two figures each, by beginning at the decimal point
and marking toward the left for the integral number, and toward
the right for the decimal. We must be careful to have the last group
on the right of the decimal point contain two figures, annexing a
cipher when necessary.
Ex. Find the square roots of 41.2164 and 965.9664.
41.21 64 (6.42 9 65.96 64) 31.08
36 9
I24J52I 61)65
496 61
1282)2564 6208)49664
2564 49664
It is seen from the grouping that the root of the first example will
have one integral and two decimal places, and that the root of the
second example will have two integral and two decimal places.
190 ALGEBRA.
218. If a number contains an odd number of decimal
places, or if any number gives a remainder when as many
figures in the root have been obtained as the given number
has groups, then its exact square root cannot be found.
We may, however, approximate to its exact root as near as
we please by annexing ciphers and continuing the operation.
Ex. Find the square roots of 3 and 357.357.
3. (1.732 3 57.35 70 (18.903..
1 1
27J2OO 28j257
189 224
343)1100 369)3335
1029 3321
3462) 7100 37803) 147000
6924 113409
Exercise 79.
Extract the square roots of :
1. 120,409; 4816.36; 1867.1041; 1435.6521; 64.128064.
2. 16,803.9369; 4.54499761; (1.24373969; 0.5687573056.
3. 0.91*^21 ;]U43; 0.00852; 17; 129; 347.259.
4. 14,23^38^,-^2.5; 2000; 0.3;(o.03; 111.
5. 0.00111; 0.004; 0.005; 2; 5;|3.25; 8.6. ^ ^"V"^
6. i; if; m; iM;\fM; m- . ^.. ' 1
Cube Roots of Compound Expressions.
214L Since the cube of a + b is a^+3a^b + 3 dl^ + V, the
cube root of a^ + Sa^^ + Sa&^ + ^s is a-\-h.
It is required to find a method for extracting the cube
root a + ^ when a^ + ^a^^ + So-^H- ^^ is given :
t ^^''
INVOLUTIOK AKD EVOLUTIOK. 191
(1) Find the cube root of a» + 3 a% + 3 a*» + h\
3a>
+ 3aM-^
3a*+3a^ + ^'
3a% + 3a**+6»
3tt^6 + 3ay + y
The first term a of the root is obyiously the cube root of the first
term a^ of the given expression.
K aS is subtracted^ the remainder is 3 a^ft + 3 aft^ + fts . therefore,
the second term 6 of the root is obtained by dividing the first term of
this remainder by three times the square of a.
Also, since SaPb-^- Sab^-{- Ij^= (Za^+Sab + b^b, the complete
divisor is obtained by adding 3 a6 + 6^ to the trial divisor 3 a^.
(2) Find the cube root of Sa^ + 36a^+54:Xi/ + 27 1/.
Sx'+3ex^i/+54:xy'+27y' \2x + Sy
12x^ 8««
(6g+3y)3y= +18xy+V
12aH-18a;y+V
3%7?y+Uxi^-\-27f
3Qx^y+Uxtf-\-21f
The cube root of the first term is 2 x, and this is therefore the first
term of the root.
The second term of the root, 3 2/, is obtained by dividing 36 x^ by
3(2x)2= 12 x^ which corresponds to Sa^ in the typical form, and is
completed by annexing to 12x2 the expression {3 (2x)4-3 y} 3 y =18xy
+ 9y2, which corresponds to 3 a6 + b\ in the typical form.
215. The same. method may be applied to longer expres-
sions by considering a in the typical form 3a® + 3a^ + 6^
to represent at each stage of the process the part of the root
already found.
Thus, if the part of the root already found be x + y, then 3 a* of
the typical form will be represented by 3 (x + yY ; and if the third
term of the root be + «, the 3 06 + 6^ will be represented by 3 (x + y)z
+ z2. So that the complete divisor, 3 a^ + 3 06 + ft^, will be repre-
sented by 3 (X + y)2 + 3 (X + y) « + z%
192 ALGEBRA.
Ex. Find the cube root of x^—Saf + Bx^—Sx — l.
3x* afi
(3X2- X)(- X)= -3X8+ g2
3x*-3x8+ x2
-3x6 + 6x8
-3x6 +3x*- x8
3 (X2 - X) 2 = 3 X* - 6 x8 + 3 x2
(3x2-3x-l)(-l)= -3x2+3x + l
3x*-0x8 +3X + 1
-3x* + 6x8-3x-l
-3x* + 6x8-3x-l
The root is placed above the given expression for convenience of
arrangement.
The first term of the root, x2, is obtained by taking the cube root
of the first term of the given expression ; and the first trial-divisor,
3 X*, is obtained by taking three times the square of this term of the
root
The first complete divisor is found by annexing to the trial-divisor
(3x2 — x)(— x), which expression corresponds to (Ha + b)b in the
typical form.
The part of the root already found (a) is now represented by x2 — x ;
therefore, 3 o2 is represented by 3 (x2 — x)2 = 3 x* — 6 x^ + 3 x2, the sec-
ond trial divisor ; and (3 a + 6) 6 by (3 x2 — 3 x — 1) (— 1); therefore,
in the second complete divisor, 3 a2 + (3 a + 6) 6 is represented by
(3x*-6x8 + 3x2) + (-3x2-3x-l)X(-l) = 3x*-6x8+3x + l.
Exercise 80.
Find the cube roots of :
1. x^-j- 6x^1/ + 12x^ + 82^ 3. x^ + 12x^ + ^Sx + 64:,
2. a« — 9a2 + 27a — 27. 4. x^—3a^+5a^a^—3a^x—a\
6. a:« + 3a;« + 6jr* + 7ir3 + 6ic2-f 3a: + l.
6. l — 9x + 39x^ — 99x^+156x^ — lUafi + 64tx\
7. a« — 6a* + 9a^ + 4a3 — 9a* — 6a — 1.
8. 64jr«+192x« + 144ir*-32a« — 36a:* + 12a; — 1.
9. l—3x+6a^-10a^+12x*-'12a^+10x^ — 6x'+Sa?-x^
INVOLUTION AND EVOLUTION. 193
10. a^ + 9a''b — lSoa^b^ + 72dab^ — 729b\
11. c^ — 12b(^ + 60 b^c' - 160 ^V + 240 ^»V — 192 V'c + 6U\
12. 8 a« + 48 a«6 + 60 a*6* - 80 a»^« - 90 a^b'' + 108 aft* - 27 6«.
Cube Eoots of Arithmetical Numbers.
216. In extracting the cube root of a number expressed
by figures, the first step is to mark it off into groups.
Since 1 = !«, 1000 = 108, i,000,000 = lOO^, and so on, it follows
that the cube root of any integral number between 1 and 1000, that is,
of any integral number which has one, two, or three figures, is a num-
ber of one figure ; and that the cube root of any integral number
between 1000 and 1,000,000, that is, of any integral number which
has four, five, or six figures, is a number of two figures ; and so on.
K, therefore, an integral number be divided into groups of three
figures each, from right to left, the number of figures in the root will
be equal to the number of groups. The last group to the left may
have one, two, or three figures.
217. If the cube root of a number contain any decimal
figures, the number itself will contain three times as many.
Thus, if 0.3 is the cube root of a number, the number is
0.3 X 0.3 X 0.3 = 0.027.
Hence, if the given cube number have decimal places, we divide
it into groups of three figures each, by beginning at the decimal point
and marking toward the left for the integral number, and toward
the right for the decimal. We must be careful to have the last group
on the right of the decimal point contain three figures, annexing
ciphers when necessary.
If the given number be not a perfect cube, ciphers may be annexed,
and a value of the root may be found as near to the true value as we
please.
218. It is to be observed that if a denotes the first term
of the root, and b the second term, the first complete divisor
is 3a^ + Sab + b^,
and the second trialrdivisor is 3 (a + ^)^ *^^* is,
3a^ + 6ab + 3b%
194
AL6EBBA.
which may be obtained from the preceding complete divisor
by adding to it its second term and twice its third term:
Sa^ + 3ab+ b^
+ Sab + 2b^
Sa^ + 6ab + Sb^
a method which will very much shorten the work in long
arithmetical examples.
219. Ex. Extract the cube root of 5 to five places of
decimals.
5.000(1.70997
1
3X10«=300
!}
3(10X7) = 210\
7» = j49
559 J
259
3X17002 = 8670000
3(1700X9)= 45900
9^ = 81
8715983.
45981
3X1709^ = 8762043
4 000
3 913
i
87 000 000
78 443 829
8 556 171 0
7 885 838 7
670 332 30
613 343 01
After the first two figures of the root are found, the next triaJ-
divisor is obtained by bringing down the sum of the 210 and 49
obtained in completing the preceding divisor ; then adding the three
lines connected by the brace, and annexing two ciphers to the result.
The last two figures of the root are found by division. The rule
in such cases is, that two less than the number of figures already
obtained may be found without error by division, the divisor to be
employed being three times the square of the part of the root already
found. /
INVOLUTION AND EVOLUTION.
196
Exercise 81.
Find the cube roots of :
1. 274,625.
2. 110,592.
3. 262,144.
4. 884.736.
5. 109,215,352.
6. 1,481,544.
7. 1601.613.
8. 1,259,712.
9. 2.803221.
10. 7,077,888.
11. 12.812904.
12. 56.623104.
13. 33,076.161.
14. 102,503.232.
16. 820.025856.
16. 8653.002877.
17. 1.371330631.
18. 20,910.518875.
19. 91.398648466125.
20. 5.340104393239.
21. Find to four figures the cube roots of 2.5; 0.2; 0.01;
4; 0.4.
220. Since the fourth power is the square of the square,
and the sixth power the square of the cube ; the fourth root
is the square root of the square root, and the sixth root is the
cube root of the square root. In like manner, the eighth,
ninth, twelfth roots may be found.
Exercise 82.
Find the fourth roots of :
1. 81 a' — 540 a^b + 1350 a'b^ - 1500 ab' + 625 b\
2. l — 4:x + 10x^—iea:^ + 19x^ — 16a^ + 10x^'--4tx^ + a^.
Find the sixth roots of :
3. 64: — 192x + 24t0x^'-ie0a^ + e0x*-12x' + x\
4. 729a:« — 1458a:« + 1215a*- 540aj»+135aj2--18a; + l.
Find the eighth root of :
6. l-8y + 282^-562^ + 70z/*-56i/« + 28y«-8/ + 2/«.
CHAPTER XIV.
Quadratic Equations.
221. An equation which contains the square of the
unknown quantity, but no higher power, is called a
quadratic equation.
If the equation contains the square onlj/, it is called
a pure quadratic ; but if it contains the first power also, it is
called an afieoted quadratic.
Pure Quadratic Equations.
(1) Solve the equation 5 x* — 48 = 2 «*.
5x* — 48 = 2 a;* It will be observed that there are two roots of
^^--.^g equal value but of opposite signs ; and there are
o 1 /, only two, for if the square root of the equation,
,aj=±4
x2 = 16, were written ± x = ±4, there would be
only two values of x ; since the equation — x
= + 4 gives X = — 4, and the equation — x = — 4 gives x = 4.
Hence, to solve a pure quadratic,
Collect the unknown numbers on one side^ and the known
numbers on the other; divide by the coefficient of the
unknown number; and extract the square root of each side
of the resulting equation. i
(2) Solve the equation 3x* — 15 = 0.
3ar — 15 = 0 j^. ^-^ y^ observed that the square root of 5
30^ = 15 cannot be found exactly, but an approximate
x^ = 5 value of it to any assigned degree of accuracy
.'. X = it V5 °^*y ^ found.
QUADRATIC EQUATIONS. 197
228. A root which is indicated, but which can be found
only approximately, is called a Surd.
Ex. Solve the equation 3 a* + 15 = 0.
3ar + 15 = 0 it will be observed that the square root
3x* = — 15 of— 6 cannot be found even approximately;
x^= — 5 for the square of any number, positive or
.". X = db V 5 ^®&*^i^®i is positive.
224. A root which is indicated, but which cannot be
found exactly or approximately, is imaginary. § 206.
Exercise 83.
Solve :
1. aj*-3 = 46. 6. 5x^-9 = 2x^ + 24l.
2. 2(a*-l)-3(x2+l)+14=0. 7. (x + 2y = ^x + 5.
• 3 "^ 6 2 5 15 25 '
4 _?_4._?--« Q 3x^-27 90 + 4x^_
l + x"^l-x • o^ + S "^ x'+9 ""
4ar 6ic' 3 a 7 X
^^ 4x^+5 2x«-5_7x2-.25
12.
13.
10 15 20
10x^+17 12x^ + 2 ^5x^--4
18 11x^-8 9
14x^+16 2x2+8 2x2
21 8x2-11 3
14. x* + 6x + a = ftx (1 — bx).
16. mx2 + w = 2'. 16. x* — ax'\'b = ax(x — 1).
198 ALGEBRA.
Affected Quadratic Equations.
225. Since (ax ±l by = aV ± 2 abx + ^^ it is evident that
the expression a V ± 2 aftaj lacks only the third term, b^, of
being a complete square.
It will be seen that this third term is the square of the
quotient obtained from dividing the second term by tunce the
square root of the first term,
226. Every affected quadratic may be made to assume
the form of a^x^ ± 2 abx = c.
The first step in the solution of such an equation is to
complete the square; that is, to add to each side the square
of the quotient obtained from dividing the second term by
twice the square root of the first term.
The second step is to extract the square root of each side
of the resulting equation.
The third and last step is to reduce the resulting simple
equation.
(1) Solve the equation 16x^ + 5^; — 3 = 7a;2 — a; + 4:5.
16x2 + 6a; - 3 = 7 x2 - X + 46.
Simplify, 9x2+6x = 48.
Complete the square, 9x2 + 6x + 1 = 49.
Extract the root, 3x + 1 = ± 7.
Reduce, 3x= — H-7or — 1 — 7,
3x = 6or — 8.
.-. X = 2 or — 2|-.
Verify by substituting 2 for x in the equation
16x2+5x-3=7x2-x + 46,
16 (2)2 + 5 (2) - 3 = 7 (2)2 - (2) + 45,
64 + 10 - 3 = 28 - 2 + 46,
71 = 71.
QUADRATIC EQUATIONS. 199
Verify by substituting — 2f for x in the equation
16 x2 + 5 X - 3 = 7 x2 - X + 46,
16(-j)2+5(_|)_3 = 7(^j)a_(-j)+45.
1024 - 120 - 27 = 448 + 24 + 406,
877 = 877.
(2) Solve the equation 3 a;^ — 4 a; = 32.
Since the exact root of 3, the coefficient of x\ cannot be found, it
is necessary to multiply or divide each term of the equation by 3 to
make the coefficient of x^ a square number.
Multiply by 3, 9 x^ - 12 x = 96.
Complete the square, 9x2 — 12 x + 4 = loo.
Extract the root, 3 x — 2 = ± 10.
Reduce, 3x = 2 4- 10 or 2 — 10 ;
3x= 12 or —8.
.-. X = 4 or — 2f
^ ^. .^ ,_ n ^ 4x 32
Or, divide by 3, * "3" "^ 3^ '
o ^ . .X. „ 4x . 4 32 , 4 100
Complete the square, x^ — —+-=-+-= - •
2 10
Extract the root, x — - = ±
.'. X
3 3
2 ±10
= 4 or - 2I-.
Verify by substituting 4 for x in the original equation,
48 - 16 = 32,
32 = 32.
Verify by substituting — 2|- for x in the original equation,
21i--(~10i) = 32,
32 = 32.
200 AL6EBBA.
(3) Solve the equation — 3a:^ + 5a; = — 2.
Since the even root of a negative number is impossible, it is neces-
saiy to change the sign of each term. The resulting equation is,
3x2 -5a; = 2.
Multiply by 3, 9 x2 - 16 x = 6.
25 49
Complete the square, 9x2 — 15x + -7 = -t-"
Extract the root.
"-I'^i-
Reduce,
s..iil.
8x = 6or — l.
.-. X = 2 or — i.
Or, divide by 3,
"" 3 -3
Complete the square,
^2-5^ + 26^49.
3 ^ 36 36
Extract the root,
^-6=^6*
5±7
...x= g .
= 2or-i.
If the equation 3x2 — 5x = 2be multiplied by four times the coeffir-
cienJt of x2, fractions will be avoided :
36x2-60x = 24.
Complete the square, 36 x2 — 60 x + 26 = 49.
Extract the root, 6 x — 5 = ±7,
6x=6±7,
6x= 12 or —2.
.-. X = 2 or — i.
It will be observed that the number added to complete the square
by this last method is the square of the co^dent ofxia the original
equation 3 x2 — 6 x = 2.
QUADRATIC EQUATIONS. 201
3 1
(4) Solve the equation ^— ^ = 2.
Simplify (as in simple equations),
4x2-23x=-30.
Multiply by four times the coefficient of x^, and add to each
side the square of the coefficient of x,
64 x2 - ( )+ (23)2 = 629 - 480 = 49.
Extract the root, 8 x — 23 = ± 7.
Reduce, 8x = 23±7;
8x = 30or 16.
.-. X = 3i or 2.
If a trinomial is a perfect square, its square root is found by taking
the square roots of the first and third terms and connecting them by
the 8ign of the middle term. It is not necessary, therefore, in com-
pleting the square, to write the middle term, but its place may be
indicated as in this example.
(6) Solve the equation 72 a;* — 30 a = — 7.
Since 72 = 2^ X 3^, if the equation be multiplied by 2 the coeffi-
cient of x^ in the resulting equation, 144 x^ — 60 x = — 14, will be a
square number, and the term required to complete the square will be
(|j)2 = (5)2 = ;y.. Hence, if the original equation be multiplied by
4X2, the coefficient of x^ in the result will be a square number, and
fractions will be avoided in the work.
Multiply the given equation by 8,
576x2- 240x=- 66.
Complete the square, 676x2 — ( ) + 25 = ~ 31_^
Extract the root, 24 x — 5 = ± V— 31.
Reduce, 24 x = 6 ± V- 31^
.•.x= 5^(5 ±^-31).
Note. In solving the following equations, care must be taken to-^
select the method best adapted to the example under consideration.
Solve: Exercise 84.^^^
/l. aj*+4a = 12. 4. a^ — 7x = S. 7. «« — x = 6.'
2. a:* — 6x = 16. 6. Sic^ — 4a; = 7. ' 8. 5a^ — 3aj = 2.
^3. 0^— 12a:+6=i) 6. 12a;«+x— 1=0. 9. 2x'—27x=U.
202 ALGEBRA.
11. J-f = 2(a; + 2). c 14.
x + 1 2x — 1
« + 4
x + 3
,„ 3a; . 4 13
« + ! 2(a; + 4) 18
2 . 3 2
/t/^
ic — 1 a? — 2 X — 4
16. 5x(a; — 3) — 2(a;«— 6) = (a: + 3)(aj + 4).
Sx 5^ 3a;^ 23
2(x + l) 8~a^ — l 4(x-l)* ^
18. (a;-2)(a;-4)-2(a;-l)(a;-3) = 0.
19. i(a,-4)-.?(a.-2) = ^(2a. + 3).
20. ?(3x2-ar-5)-i(«2-l) = 2(a;--2/.
21.
5
2x
+
3a;-50 _12a; + 70
22.
16 ' 3(10 + a;)
15 -7a;
j^ — 1 8(1 -a;)
2aj — 1 . 1 2a:-3
2
/ o ?^— 1 _L 1 _ 2a; — a
190
25. a;
26. 1-
14a;-9^a;' — 3 /
8a; — 3 " x + l' ^
x-^-h X — 6
24.
x + 2 4 — a;
a;-l
28.
T
3
27.
2a;
_2^+3 ^
2 (2a; — 1) 2 (a; + 1) "" 4 — 3a;'
2a; + l a; — 2
7 — a; ■ X
7 — x 7 — 3a;
r^'
12a;'-lla;» + 10a;-78_ ,, 1
^*- 8ar'-7a; + 6 " '-^*^~2'
l^/
30.
6
18
8
a; — 1 x-{-5 a; + l a; — 5
QUADRATIC EQUATIONS. 203
SS7. Literal quadratic eqtiations are solved as follows :
(1) Solve the equation aa^ -{-bx = c.
Multiply the equation by 4 a and add the square of 6,
4a2a;2+( ) + 62 = 4^^ + 52,
Extract the root, 2az + b = ± -sf^tac-hh^.
Reduce, 2 ox = — 6 ± V4 oc + 6^.
-6±V4ac + &g
.-. X = r •
2a
(2) Solve the equation adx — acx^ = bcx — bd.
Transpose bcz and change the signs,
acx^ + 6cx — adz = M.
Express the left member in two termSy
acx^ + (6c — od) X = W.
Multiply by 4 ac,
4 a^c^^ + 4ac(6c — a<i)x = 4 oftcd.
Complete the square,
4o2c2x2+( ) + (6c-a(f)2 = 62c2 + 2a6cd + a2d2.
Extract the root, 2 acx + (&c — ad) = ± (6c + ad).
Reduce, 2 acx = — (6c— ad) ± (6c+ ad),
= 2 ad or — 2 6c.
d 6
.-. X = - or •
c a
(3) Solve the equation px^ — px + g'x^ -J" 2^^ = ^ ■
Express the left number in two ienn«,
(p + g)x2-(i)-3)x = ^-
Multiply by four times the coefficient of x2,
" X 4(p + g)2x2-4(p2 — g2).a. = 4pg.
Complete the square,
4(p4-9)2x2-( )+(p-g)2 = ^2+2l)g4. ^ .
Extract the root, 2{p + q)x— {p — q) = ± (p + q).
Reduce, 2(p + g)x= (p — g) ± (p + g),
= 2 p or — 2 g.
.'. X = — ^ or ^ —
p + q p + g
Note. The left-hand member of the equation when simplified
must be expressed in two term^ simple or compound^ one term con-
taining x2, and the other term containing x.
204 ALGEBBA.
Exercise 85.
Solve :
1. a5' + 2aaj = a*. 14. a^-\-ax = a-\-x,
2. a:* = 4aa;+7a*. 16. x^ + ctx ^= bx + alf. ^
-8. ^=I^*_3^. le. * + ? = ? + *. ;>-
4 a X 0 X J^
^ , 571X Zn^ ^ ,^1.1 1.1
. 4. 05* ?r=0. 17. --
2 2' ' X x-\-b a a-^b
a* ft* .o ^ I ^^ ^* A
... y
(x + af {x — af *"' 3 ' 4 3a
6. cx^ax^-^-ba? -:-:• 19. - = «+—-—• ^
a-\-b X — 3 x + 3 '^-—
X ^ aV , ft* 2aa: ^^ . i a^(^* — w")
^ 7. --r- + - = 20. ma;*— 1 = — ^^ ^•
ft* c* c mn
8. (a* + l)a; = aa:*4-a. 21. (aa; — ft) (ftx — a) = c*.
a , ft 2 c ax-\-b inx-]-n
9. 7= 22. , ' = — -J- —
X — ax — ft X — c bx-\-a nx-f-m
10. ———— = - + - + - 23. — J— H =c. -r
a + ft + ^ a ft a; m + a: m — x
11. _1 L_=3+^. 24. («-l)V+2(3a-l)»^^^
'a — a; a4"^ a^-^x^ ' 4a — 1
a* + ft'^ a*+ft*
(2x — a)* , ^^ a;* — 4mwaj . .,
2x^a + 2b (m + ny ^ ^
a — ft_14a* — 5aft — 10ft* (2a — 3ft)g
^^' "^ oft* "" 18a*ft* "^ 2aft
QUADRATIC EQUATIONS. 205
28. abs[?'\ = ■ — 5 •
. ^^ ic* m* — 4a* x
-L 3m — 2a ^a — dm 2
30. 6g+^ ; ^ =5(a-ft)+-^— '
^Zl. f (x* + a* + a5) = ia;(20a + 4ft).
32. 05* — (b — a)c=^ax — bx'{'CX,
33. 05* — 2wM; = (n — p-{-in)(n — p — m).
34. 05* — (wi + ^) aj = i^ (2? + fi' + ^ + ^) ( J3 + 3' — m — n).
35. mriic* — (m-^n) (mn + 1) a; + (m + w)* = 0.
2b — X — 2a, 4^ — la x — 4a
bx ax — bx ah — 6*
37. 2x*(a*-6*)-(3a* + 6*)(a:-l) = (3i»*+a*)(x + l).
a — 2b — X 5b — X . 2a — x — 19b
a* — 46* ax-{-2bx 2bx — ax
aj + 13a + 36 ^ a — 2b
5a-36 — a; aj + 26
x + Sb 3^^ a + Sb _
Sa^ — 12ab 96*-4a* (2a + 3b) (x — 3b)
41. nx^-{-px — J9X* — ma; + m — n = 0.
42. (a + i + c)aj* — (2a + 6 + c)a; + a = 0.
43. (ax — b)(c — d)=^(a — b) (ex — d)x,
2g + l 1/1 2\ 3a; + l
x\b aj a
45.
44.
6
1,1 a 2bx + b
2x* + a; — 1^2a;* — 3x + l 2bx — b ax' — a
206 ALGEBRA.
228. An affected quadratic may be reduced to the form
7? -{-px -f- g' = 0, in which p and q represent any numbers,
positive or negative, integral or fractional.
Ex. Solve: x^+px + q = 0.
^x' + 0+p'=p'-^q,
2x + p = zt Vy — 42',
1 _.^__^—
.•.a; = -|±-yp2-4y.
By this formula, the values of x in an equation of the
form x^'{-px-{-q = 0, may be written at once. Thus, take
the equation
3x^ — ox + 2 = 0.
Divide by 3, x^ — f x + | = 0.
Here, P = - f, and g = j.
.•.x = |±iV^-J,
= 1 or |.
229. A quadratic which has been reduced to its simplest
form, and has all its terms written on one side, may often
have that side resolved by inspection into factors.
In this case, the roots are seen at once without com-
pleting the square.
(1) Solve x^ + 7 ic- 60 = 0.
Since x^ + 7x - 60 = (x + 12) (x - 5),
the equation ' x^ + 7 x — 00 = 0
may be written (x + 12) (x — 5) = 0.
It will be observed that if either of the factors x+12orx — 5 isO,
the product of the two factors is 0, and the equation is satisfied.
Hence, x + 12 = 0 and x — 5 = 0.
.-. X = — 12, and x = 6.
QUADRATIC EQUATIONS. 207
(2) Solve a^ + 7 a; = 0.
The equation z^+7x = 0
becomes a; (x + 7) — 0,
and is satisfied if a; = 0, or if x + 7 = 0.
.*. the roots are 0 and — 7.
It will be observed that this method is easily applied to an equation
aU the terms of which contain x.
(3) Solve 2ar8 — «2_g^^Q
The equation 2x« — x^ — 6x = 0
becomes x (2x2 — x — 6) = 0,
and is satisfied if x = 0, or if 2 x^ — x — 6 = 0.
By solving 2 x2 — x — 6 = 0 the two roots 2 and — | are found.
.'. the equation has three roots, 0, 2, — J.
(4) Solve ic« + cc2--4aj — 4 = 0.
The equation x8+ x2--4x — 4 = 0
becomes x^ (x + 1) — 4 (x + 1) = 0,
(x2-4)(x+l) = 0.
.-. the roots of the equation are — 1, 2, — 2.
(5) Solve a:^-2a;2 — llx + 12 = 0.
c- x8-2x2-llx+12 .
Smce = x2 — X — 12,
X— 1 *
the equation x^ — 2 x^ — 11 x + 12 = 0
may be written (x — 1) (x2 — x — 12) = 0.
The three roots are found to be 1, — 3, 4.
An equation which cannot be resolved into factors by inspection
may sometimes be solved by guessing at a root and reducing by divi-
sion. In this case, if a denote the root, the given equation (all the
terms of the equation being written on one side) may be divided by
X — o.
208 ALGEBRA.
Exercise 86.
Find the roots of :
1. (x+l)(x—2)(x^+x-2)=0, 7. a^ — x' — x + l^^O.
2. (x^—3x+2)(x^—x—12)=0. 8. 8aj» — 1=0.
3. (x + l)(x-2)(x + 3)= — 6. 9. Sa^ + 1 = 0.
4. 2x^ + 4x2 — 70x = 0. 10. a;« — 1 = 0,
6. (x^ — x — 6) (a:^ — X — 20) = 0. 11.x (x—a)(x^— b^=0.
6. x(x+l){x+2)={a+2)(a+l)a. 12. n(x»+l)+x+l=0.
230. If r and r' represent two values of x, then
cc — r = 0,
and x — r' = 0.
.'. (x — r) (oj — r*) =: 0.
This is a quadratic equation, as may be seen by performing the
indicated multiplication.
Now, r and r' are roots of this equation ; for, if either r or r' be
written for x, one of the factors, x — r, x — r', is equal to 0, and the
equation is satisfied. Also r and r' are the only roots, for no value
of X, except r and r', can make either of these factors equal to 0.
Since r and r' may represent the values of x in any quadratic
equation, it follows that every quadratic equation has two roots, and
orUy two.
Again, if r, r', r", represent three values of x,
then, (X — r) (X — r') (x — r") = 0.
This is a cubic equation, as may be seen by performing the indi-
cated multiplication. Hence, it may be inferred that a cuMc equation
has three roots, and only three; and so, for any equation, that the
number of roots is equal to the degree of the equation.
It may also be inferred that if r be a root of an equation, x — r
ivm be a fcuAor cf the equation when the equation is vnritten with all
its terms on one side.
QUADRATIC EQUATIONS. 209
If r and r^ represent the roots of the general quadratic equation,
x2 + px + g = 0.
This equation may be written (x — r) (x — r^ = 0,
or, x2 — (r + O X + rr' = 0.
A form which shows that
the sum of the roots =""!>»
and the prodiict of the roots = q,
231. It will be seen from § 230 that an equation may be
formed if its roots are known.
If the roots of an equation be — 1 and i^
the equation will be (x + 1) (x — i) = 0,
or, by multiplying by 4, 4 x^ + 3 x — 1 = 0.
If the roots of an equation be 0, 1, 5,
the equation will be (x — 0) (x — 1) (x — 6) = 0 ;
that is, X (x — 1) (x — 5) = 0,
or, X* — 6 x"^ + 5 X = 0.
If X occur in every term, the equation will be satisfied by putting
X = 0, and may be reduced to an equation of the next lower degree
by dividing every term by x.
232. By considering the roots oi x^'\-px'\-q = 0,
namely, r = — f- + -^p^ — 4:q,
and ^' =— 1~2^-^'""^^'
it will be seen that the character of the roots of an equation
may be determined without solving it :
I. As the two roots have the same expression, V// — 4y,
both roots will be real, or both will be imaginary.
If both be real, both will be rational or both surds, accord-
ing as j9^ — 4: q is or is not a perfect square.
210 ALGEBBA.
II. When ^ is greater than 4 y, the two roots will be
reaZ, for then the expression p^ — 4:q is positive, and there-
fore Vp^ — 4^' can be found exactly or approximately.
Since also its value in one root is to be added to — "^j
and in the other to be subtracted from — ^> the two roots
will be different in value,
III. When p^ is equal to 4 q, the roots will be equal in
value.
IV. When p^ is less than 4 q, the roots will be imaginary,
for then the expression p^—Aq will be negative, and there-
fore y^'—Aq represents the even root of a negative number,
and is imaginary.
V. If 2' (= r X r') be positive, the roots, if real, will have
the same sign, but opposite to that oi p (since r-|-r'= — p).
But if y be negative, the roots will have opposite signs,
233. Determine by inspection the character of the roots of:
(1) x^ — 5x + 6 = 0.
In this equation p is — 5, and q isQ.
.-. ^p^-4q = V26 - 24 = 1.
.-. the roots will be rational, and both positive,
(2) x' + Sx + l^O.
In this equation p is 3, and g^ is 1.
.-. ^p'^ — 4q = V9 — 4 = VS.
.-. the roots will be surds, and both negative,
(3) x^ + 3x + A = 0.
In this equation p is 3, and g is 4.
.-. ^p^-4tq = V9-16 = V^.
.-. the roots will be imaginary.
QUADRATIC EQUATIONS. 211
EXEBOISE 87.
Form the equations whose roots are :
1. 2,1. 6. -5,-i. 9. 0,-i, 1,-1.
2. 7,-3. 6. —J, f. 10. a — 2b,3a-{'2b.
3. i,i. 7. 3, -3, f, — i. 11. 2a — b,b — 3a.
4. i— |. 8. 0,1,2,3. 12. a(a + l), 1 — a.
Determine by inspection the character of the roots of :
13. a:*-7x + 12=0. 17. x^ + ^x + l = 0.
14. x^— 7a; — 30 = 0. 18. a:* — 2a: + 9 = 0.
16. x^ + 4:X — 5 = 0. 19. 3ic* — 4ic — 4 = 0.
16. 59^ + S = 0. 20. a;«+4aj + 4 = 0.
234. It is often useful to determine the maximum or
minimum value of a given quadratic expression.
(1) Find the maximum or minimum value ofl+X'^a^,
Let 1 + x — x2=m;
then, x^ — z = l-'m,
and 4a;2~( ) + l = 6-4m,
2x--l= ±V6--4m.
.-. x = i±iV6-4m.
Now, for all possible values of x, 5 — 4 m cannot be negative ; that
is, m cannot be greater than f ; and for this value x is i. Therefore,
f is the maximum value of the given expression.
(2) Find the maximum or minimum value of a* + 3ar + 4.
Let x2 + 3x + 4=w;
then, x2 + 3 X = m — 4,
and 4x2 + ( ) +9=4m-7.
2x + 3= ± V4 m - 7.
x= - |±iV4m-7.
For all possible values of x, 4 m — 7 cannot be negative ; that is,
TO cannot be less than } ; and for this value x = — f. Therefore, } is
the minimum value of the given expression.
212 ALOEBBA.
£XEBCISB 88.
Find the maximum or minimum value (and determine
rhich) of:
1. i + ex-x". 4. (a — x)(x-b). 7. a5»-2a;+9.
i6. •* '■■
X
3. '''+^-
6. a!« + 8«4-20.
9. fv
10. Divide a line 20 in. long into two parts so that the
sum of the squares on these two parts may be the
least possible.
11. Divide a line 20 in. long into two parts so that the
rectangle contained by the parts may be the greatest
possible.
12. Find the fraction which has the greatest excess over
its square.
S36. Two other cases of the solution of equations by
completing the square should be noticed.
I. When any two powers of x are involved, one of which
is the square of the other,
II. When the addition of a number to an equation of the
fourth degree will make both sides complete squxires.
(1) Solve8a;«+63a;«=8.
In this equation the exponent 6 is the double of 3, hence zfi is
the square of sfi.
8«« + 63x« = 8,
256jc8 + ( )+ (63)2 = 4226,
16x« + 63=±65,
16x«=2, or-128,
x« = i, or — 8.
By taking cube root, as = i, or — 2.
QUADRATIC EQUATIONS.
213
The other roots of the equation are found by finding the remaining
roots of the equations, x^ = ^, and »« = — 8.
Smce, x« = i, .-. 8 x8 — 1=0
Now, by § 230,
8x8 -l=(2x-l)(4x2 + 2x4-1)
.-. (2x-l)(4x2 + 2x+l)=0
and i8satisfiedif4x3+2x+l=0
as well as if 2x— 1=0.
The solution of4x«+2x+l=0
gives X = i (— 1 ± V^).
Smce, x8 = — 8, .-. »» + 8=0
Now, by S 230,
x8+8=(x+2)(x2-2» + 4)
.-. (x + 2)(x3-2x + 4)=0
and is satisfied if x^ — 2x + 4=0
as well as if x + 2=0.
The solution of x« — 2 x + 4=0
X = 1 ± V^.
the roots are i, -2, 1 ± V^, i (~ 1 ± V^).
(2) Solvea;*-10a:» + 35a;* — 50x + 24 = 0.
Take the square root of the left side.
g*~10a^ + 35g«— 50a; + 24|g^~5a; + 5
2a:* — 5a;
-10a;« + 25a:*
2a:* — 10a; + 5
10x'—50x + 24:
lOx* — 50a: + 26
- 1
It is now seen that if 1 were added, the square would be complete
and the equation would be
a:* — 10a:» + 35a:* — 50a; + 26 = l.
Extract the square root, and the result is,
x2-6x+6=±l.
That is, x2 — 5x = — 4, or — 6.
4x2- ( ) + 25 = 9, orl,
2x — 6= ±8, or±l,
2x=8, 2, 6, or 4.
.% » = 4, 1, 8, or 2.
214
ALGEBRA.
Exercise 89.
Find the roots of :
1. afi + 7a^ = S.
2. x*—ox' + 4: = 0.
3. 37x«-9 = 4aj*
4. 16a^=17aj*-l.
6. 32xi«-33x«4-l = 0.
6. (a;2— 2)«=i(ic2+12).
^- "^ 3 12-"-
8. (a:*-9)' = 3 + ll(a:*-2).
9. aj«+14a;» + 24 = 0.
10. 19aj* + 216aj' = a;.
11. aj8 + 22x* + 21 = 0.
12. ic^ + 3a:" — 4 = 0.
13. 4a;*-20x»+23a*+5x=6.
14. 4 + 4-20 = 0.
16. X* — 4x«-10a;«+28a;-15 = 0.
16. X* — 2ar' — 13aj2 + 14aj = -24.
17. 108aj* = 20a;(9a;* — 1) — 51a* + 7.
18. (a»-l)(ic»-2) + (aj2-3)(a:*-4) = a:* + 6.
Problems Involving Quadratics.
236. Problems which involve quadratic equations have
apparently ttw) solutions, as a quadratic has two roots.
Sometimes both will be solutions : but generally one only
will be a solution, and the other be inconsistent with the
conditions of the problem. No difficulty will be found
in selecting the result which belongs to the problem, and
sometimes a change may be made in the statement of a
problem so as to form a new problem corresponding to the
solution which was inapplicable to the original problem.
QUADRATIC EQUATIONS. 215
(1) The sum of the squares of two consecutive numbers is
481. Find the numbers.
Let
X = one number,
and
a; + 1 = the other.
Then
x2+(x+l)a = 481,
or
2x2 + 2x+ 1 = 481.
The solution of which gives, x = 15, or — 16.
The positive root 15 gives for the numbers, 15 and 16.
The negative root — 16 is inapplicable to the problem, as coraecu-
tive numbers are understood to be integers which follow one another
in the conmion scale, 1, 2, 3, 4
(2) What is the price of eggs per dozen when 2 more in a
shilling's worth lowers the price 1 penny per dozen ?
Let X = the number of eggs for a shilling.
Then, - = the cost of 1 egg in shillings,
X
12
and — = the cost of 1 dozen in shillings.
But, if X + 2 = the number of eggs for a shilling,
12
, Q = the cost of 1 dozen in shillings.
X T ^
.'. — - ^:p^ = i5 (^ P®^^y ^^g A ^^ * shillmg).
The solution of which gives x = 16, or — 18.
And, if 16 eggs cost a shilling, 1 dozen will cost j{ of a shilling,
or 9 pence.
Therefore, the price of the eggs is 9 pence per dozen.
If the problem is changed so as to read : What is the
price of eggs per dozen when two less in a shilling's worth
raises the price 1 penny per dozen ? the algebraic state-
ment will be
12 1?_ 1.
X — 2 X 12'
The solution of which gives x = 18, or — 16.
Hence, the number 18, which had a negative sign and was inappli-
cable in the original problem, is here the true result.
216 ALGEBRA.
Exercise 90.
1. The sum of the squares of three consecutive numbers
is 366. Find the numbers.
2. Three times the product of two consecutive numbers
exceeds four times their sum by 8. Find the
numbers.
3. The product of three consecutive numbers is equal to
three times the middle number. Find the numbers.
4. A boy bought a number of apples for 16 cents. Had
he bought 4 more for the same money he would have
paid ^ of a cent less for each apple. How many did
he buy ?
6. For building 108 rods of stone-wall, 6 days less would
have been required if 3 rods more a day had been
built. How many rods a day were built ?
6. A merchant bought some pieces of silk for $900.
Had he bought 3 pieces more for the same money
he would have paid $15 less for each piece. How
many did he buy ?
7. A merchant bought some pieces of cloth for $168.76.
He sold the cloth for $12 a piece and gained as
much as 1 piece cost him. How much did he pay
for each piece ?
8. Find the price of eggs per score when 10 more in 62^
cents' worth lowers the price 31^^ cents per hundred.
9. The area of a square may be doubled by increasing
its length by 6 inches and its breadth by 4 inches.
Determine its side.
10. The length of a rectangular field exceeds the breadth
by 1 yard, and the area is 3 acres. Determine its
dimensions.
M'
QUADRATIC EQUATIONS. 217
11. There are three lines of which two are each ^ of the
third, and the sum of the squares described on them
is equal to a square yard. Determine the lengths
of the lines in inches.
12. A grass plot 9 yards long and 6 yards broad has a
path round it. The area of the path is equal to
that of the plot. Determine the width of the path.
13. Find the radius of a circle the area of which would be
doubled by increasing its radius by 1 inch.
14. Divide a line 20 inches long into two parts so that the
rectangle contained by the whole and one part may
be equal to the square on the other part.
15. A can do some work in 9 hours less time than B can
do it, and together they can do it in 20 hours.
How long will it take each alone to do it ?
16. A vessel which has two pipes can be filled in two
hours less time by one than by the other, and by
both together in 2 hours 55 minutes. How long
will it take each pipe alone to fill the vessel ?
17. A vessel which has two pipes can be filled in 2 hours
less time by one than by the other, and by both
together in 1 hour 52 minutes 30 seconds. How
long will it take each pipe alone to fill the vessel.?
18. An iron bar weighs 36 pounds. If it had been 1 foot
longer each foot would have weighed ^ a pound
less, rind the length and the weight per foot.
A number is expressed by two digits, the second of
which is the square of the other, and when 54 is
added its digits are interchanged. Find the number.
20. Divide 35 into two parts so that the sum of the two
fractions formed by dividing each part by the other
may be 2^^.
218 ALGEBRA.
21. A boat's crew row 3^ miles down a river and back
again in 1 hour 40 minutes. If the current of the
river is 2 miles per hour, determine their rate of
rowing in still water.
22. A detachment from an army was marching in regular
column with 5 men more in depth than in front.
On approaching the enemy the front was increased
by 846 men, and the whole was thus drawn up in
6 lines. Find the number of men.
23. A jockey sold a horse for $144, and gained as much
per cent as the horse cost. What did the horse cost ?
24. A merchant expended a certain sum of money in goods,
which he sold again for $24, and lost as much per
cent as the goods cost him. How much did he pay
for the goods ?
25. A broker bought a number of bank shares ($100 each),
when they were at a certain per cent discount, for
$7500 ; and afterwards when they were at the same
per cent premium, sold all but 60 for $5000. How
many shares did he buy, and at what price ?
26. The thickness of a rectangular solid is f of its width,
and its length is equal to the sum of its width and
thickness; also, the number of cubic yards in its
volume added to the number of linear yards in
its edges is | of the number of square yards in
its surface. Determine its dimensions.
27. If a carriage-wheel 16J feet round took 1 second more
to revolve, the rate of the carriage per hour would
be Ij- miles less. At what rate is the carriage
travelling? f \n
H
CHAPTEE XV.
Simultaneous Quadratic Equations.
287. Quadratic equations involving two unknown quan-
tities require different methods for their solution, according
to the form of the equations.
238. Case I. When from one of the equations the value
of one of the unknown quantities can be found in terms of
the other, and this value substituted in the other equation.
Ex. Solve: Sx^-'2xy = 5\' (1)
x-y=2 J (2)
Transpose x in (2), y = » — 2.
Substitute in (1), 3 x^ — 2 x (x — 2) = 6.
The solution of which gives x = 1 or — 6.
.-. y = — 1 or — 7.
Special methods often give more elegant solutions of examples than
the general method by syJ)stitiUion.
I. When equations have the form, x ± y = a, and xy — h; x^ ± y^
= a, and xy=^h; or,x±y = a, and x^ + y^ = 6.
(1) Solve: aj+y = 401 (1)
xy = SOO S (2>
Square (1), x2 + 2 xy + y2 = 1600. (3)
Multiply (2) by 4, 4xy = 1200. (4)
Subtract (4) from (3), x2 - 2 xy + y2 = 400.
Extract root of each side, x — y = ±20. (5)
Add (1) and (5), 2 x = 60 or 20.
.-. X = 30 or 10.
Subtract (6) from (1), 2 y = 20 or 60.
... y = 10 or 30,
220
ALGEBRA.
(2) Solve:
aj — y = 4 1
a:« + y« = 40j
Square (1), x^ — 2zy + ]fi=16.
Subtract (2) from (3), — 2 xy = — 24.
Subtract (4) from (2), x^-h2xyhy^ = 64.
Extract the root, x + y = ±8.
By combining (6) and (1), x = 6 or — 2.
y = 2 or — 6.
(1)
(2)
(3)
W
(6)
iolve :
X y
9 1
''20
^ + i
41
400 J
Square (1),
1 + 1 + 1=81..
»2 xy y« 400
Subtract (2) from (3)
2 _ 40.
xy""400*
Subtract (4) from (2),
1 .2 1 _ 1 .
x« xy y2 400*
Extract the root,
1-1 = ±±.
X y 20
By combining (1)
and
(6),
X = 4 or 6.
y = 6 or 4.
(1)
(2)
(8)
(4)
(6)
II. When one equation may be simplified by dividing it by the other.
(4) Solve:
a:» + y^ = 911
Divide (1) by (2),
Square (2),
Subtract (3) from (4),
Divide by — 3,
Add (6) and (3).
Extract the root,
By combining (6) and (2),
(1)
(2)
x2-xy + y2=13. (3)
x2+2xy + y2 = 49. (4)
3xy = 36.
-xy=-12. (6)
x2 — 2xy4- 2/2=1.
x-y=±l. (6)
X = 4 or 3.
y = 3 or 4.
SIMULTANEOUS QUADBATIC EQUATIONS.
221
Exercise 91.
Solve I
1. x + y=13-
xy
2. x + y=29"
3. a; — y = 19"
a;y:
hy=13-l
= 36 J
l-y=29\
= 100 J
-y=19->
= 66 J
-y=A5\
=260 J
4. X — y^i&'
xy-
6. » — y = 10 \
' = 178j
x» + y» =
6. a; — y=14 \
ir' + 2/! = 436j
7. a; + y=12 \
x'+2/*=104/
o; y 4
ar*"*"^/* 16
9. - + - = 5
X y
10. 7a;2 — 8a-y = 169"\
5x+2i/ = 7 J
11. a; + y=49 1
a:« + 2/* = 1681J
12. a;« + 2^=341|
x + y=ll J
13. x« + 2^=1008^
x + 2/=12 J
14. X* — 2/^=98
^-2/'=981
16. x« — 2^ = 2791
3J-y = 3 J
16. X — 32^ = 1
17. 4?/ = 5a; + l \
2.T// = 33 — a;2j
18. i-i=3
i-i=21
19. i-i=2i
a; y
20. a^ — 2a;^ — y' = l\
a; + y = 2 J
222 AIiGEBBA.
239. Case II. When each of the two equations is lionuh
geneous and of the second degree.
Ex. Solve: 2f—4:xy + ZQ?=^ll^ (1)
y»-x2 = 16 J . (2)
Let ^ = vx, and substitute vx f or ^ in both equations.
From (1), 2t?2x2 - 4ttB2 4- 3aj2 = 17.
2c2-4t?4-3
From (2), r^^ — x^^ 16.
16
17
Equate the values of a^,
-1
17 16
2t2-4«4-3 »2_i
32i?2-64«4-48=17i?2-17,
15i?2-64c=-65.
The solution gives, t) = - or — •
Substitute the value of v In x^ =
V —
26
then, x^ = 9 or — •
8 5
16
26
9
.•.x=±3or±|»
13
and y = t)x=±5or±-^*
Exercise 92.
Solve :
1. a» + a;y + 22/^=74 "1 4. a«-4y« — 9 = 0 \
2x2_|_2x^ + 2/' = 73J ajy + 22^-3 = 0j
2. a^ + xy + 42^ = 61 6. jr^ — aj^/ — 35 = 01
^ = 14 J ajy + 2/«-18 = Oj
3a;»+8y2 = 14 J ajy + 2/^ — 18 =
ic« — ajy + 2/* = 211 6. a^ + iry + 22/* = 44'
2^«2a:2/ = -16 J 2a;«-ajy+2^ = 16.
SIMULTANEOUS QUADBATIG EQUATIONS. 223
7. a^ + xi/ — 15 = 0'\ 9. 2cc« + 3a;y + 3^=70\
«y — 2^ — 2 = 0 J 6x' + xy — i/=::50 J
8. x^—'X2/-{-i^=7 ) 10. x^ — xi/ — i/ = 5 1
3x*+13a;3/ + 82^=162j 2x* + 3a;y + 3^ = 28J
240. Case III. When the two equations are symmetrical
with respect to x and y; that is, when they have x and y
similarly involved in them.
Thus, the expressions 2x8 + 3x2^2 + 22^, 2xy — 3x--3y + 1,
X* — 3 x^y — 3 xy2 + y4 are symmetxical expressions.
(1) Solve: a^ + 2^ = 18xy1 (1)
x + y = 12 J (2)
Put u 4- « for X, and u — r? for y, in (1) and (2).
(1) becomes (u + v)^ + (u — v)^ = 18 (u 4- v) (u — c),
or w8 + 3 MW2 = 0 (u2 - 1?2). (3)
(2) becomes (w 4- r) + (u — u) = 12,
or 2 u = 12.
.*. w = 6.
Substitute 6 for u in (3).
^ (3) becomes 216 + 18 »2 = 0 (36 - 1?2),
whence, ©2=4.
.-.» = ± 2,
.-. X = li 4- 1> = 6 ± 2 = 8 or 4,
and y=u— t? = 6:f2 = 4or8.
(2) Solve: x + y==S -) (1)
x^ + y^ = 70ej (2)
Put u 4- r for X, and u — r for y in (1) and (2).
(1) becomes (m 4- ») 4- (w — «) = 8.
.'. u = 4.
(2) becomes u* 4- 6 u^^ 4- 1?* = 363. (8)
Substitute 4 for u in (3),
266 4- 96 u2 4-^4 = 353,
or, i?*4-96u2=97. (4)
The solution of (4) gives r = di 1 or ± V— 97.
Taking the possible values of t>, x = 5 or 3, and y = 3 or 5.
224 AI.OEBKA.
EXEBCISE 93.
Solve :
1. ixy=96 — x>y'\ 4. 4(a; + y)=3a^ \
x + i/ = 6 J x + y + x'+f=26j
2. a^ + y» = 18-x — y\ 5. 4:x' + x!/ + iif=58\
xy = 6 J 5x'+5y'=65 J
3. 2(x' + f) = 5xi/\ 6. X!/(x + y) = 30^
i(x-y)=xy J (r»+2/»=35 J
241. The preceding cases are general methods for the solution of
equations which belong to the kinds referred to ; often, however, in
the solution of these and other kinds of simultaneous equations in-
volving quadratics, a little ingenuity will suggest some step by which
the roots may easily be found.
Exercise 94.
Solve :
1. X — y = 7 \ 8. x—y=^l \
a^ + xy + ^f=l^i a?^f = %^\
2. x^'\-xij = 35'\ 9. ic2 + 4xy = 3 > ,
xy-f = ^ j 4.xy+if = 2i\*
3. xy — 12 = 01 10. a;2-_a.y_f_^=:48|
X — 2y = 5 J X — y — 8 = 0 )
4. xy — l = 0 1 11. x^ + Sxij+f^l I
x2 + y2 = 50 J 3x^+xy+3i/=lS )
6. 2x — 5y = 9 1 12. aj2_2xy+3?/^=lJ >
aj* — a;?/ + 2/^ = 7j x^ + xy—i/^i )
6. aj — 2/ = 9 1 13' a + ?/ = a )
a;2/ + 8 = 0) 4x2/ — a2 = — 4^2)
7. 5x — 7y=0 •) 14. x — y = l
6a^-i^
) -^ 14. X — y = l ^
SIMULTANEOUS QUADRATIC EQUATIONS.
225
15. a^ + 9xi/=S4:0}
16. a; + y = 6 \
a^ + f = 72i
17. 3xy + 2x + y = AS5\
3a; — 2y = 0 )
18. x — y = l \
0^8 — 2^ = 19)
28 \
= 124)
19. a^ + f = 272S
xl^'-xy + y^
20. x + y = a }
x' + f = b']!
21. a^ — f = 0
3a^'-4:xy + 57/
x + y x-^y 10
22.
X — y X'\-y
x' + f = 4:5
2/^ = 9)
}
23. - + - = 5
X y
17
x+1 ' y + 1 12
24. a;^ — a;y + 7/^ = 7 |
aJ* + a;y + 2/* = 133>
25. x + y = 4:
26. a«--y8=aM
X — y=a )
27. ^-x, = a'+lr'^ ^
xy — y^ = 2ab )
28. x^-'i/ = Aab\
xy = a^ — b^ )
29. xy = 0 \
x^ + y' = 16l
30. a^ + xy + y^=37 f"
x* + xy + y< = 481>
31. a^ = ax'{'by}
y'^ay + bx)
32. X — 2^ — 2 = 0 I
15(x2-2^ = 16a;y)
x — y x + y 40 ^
6x = 20y + 9 J
a 0
^ + ^ = 4
35. x' + y' = 7 + xy \
x^-\'y^ = Qxy—l )
36, a^ — y' = 3093|
x—y=3 )
r+y=4 I
ic* + y' = S2)
37. f(x-l)-t(aj + l)(y-l) = -in
i(2/ + 2)=i(a; + 2) I
38. 10x^+15xy = 3ab — 2an ^
102/^ + 15x// = 36^ — 2^>M ""^
226 AliOEBBA.
Exercise 95.
1. If the length and breadth of a rectangle were each in-
creased by 1, the area would be 48 ; if they were
each diminished by 1, the area would be 24. Find
the length and breadth.
2. The sum of the squares of the two digits of a number
is 25, and the product of the digits is 12. Find the
number.
3. The sum, the product, and the difference of the squares
of two numbers are all equal. Eind the numbers.
Note. Represent the numbers hy x + y and x — y, respectiyely.
4. The difference of two numbers is f of the greater, and
the sum of their squares is 356. What are the
numbers ?
6. The numerator and denominator of one fraction are
each greater by 1 than those of another, and the
sum of the two fractions is 1^ ; if the numerators
were interchanged the sum of the fractions would
be 1^. Find the fractions.
6. A man starts from the foot of a mountain to walk to its
summit. His rate of walking during the second half
of the distance is i mile per hour less than his rate
during the first half, and he reaches the summit in
5^ hours. He descends in 3f hours, by walking 1
mile more per hour than during the first half of the
ascent. Find the distance to the top and the rates
of walking.
Note. Let 2 x = the distance, and y miles per hour = the rate at first.
Then - H = 6^ hours, and — r-; = 3f hours.
y y-^ ^ y+1 ^
SIMULTANEOUS QUADRATIC EQUATIONS. 227
7. The sum of two numbers which are formed by the
same two digits in reverse order is ff of their
difference; and the difference of the squares of the
numbers is 3960. Determine the numbers.
8. The hypotenuse of a right triangle is 20, and the area
of the triangle is 96. Determine the legs.
Note. The square on the hyx)otenu8e = sum of the squares on
the legs ; and the area of a right triangle = | product of legs.
9. Two boys run in opposite directions round a rectan-
gular field the area of which is an acre ; they start
from one corner and meet 13 yards from the oppo-
site corner; and the rate of one is | of the rate of
the other. Determine the dimensions of the field.
10. A, in running a race with B to a post and back, met
him 10 yards from the post. To make it a dead
heat, B must have increased his rate from this point
41^ yards per minute; and if, without changing his
pace, he had turned back on meeting A, he would
have come in 4 seconds after him. How far was it
to the post ?
11. The fore wheel of a carriage turns in a mile 132 times
more than the hind wheel ; but if the circumferences
were each increased by 2 feet, it would turn only
88 times more. Mnd the circumference of each.
12. A person has $6500, which he divides into two parts
and loans at different rates of interest, so that the
two parts produce equal returns. If the first part
had been loaned at the second rate of interest, it
would have produced $180 ; and if the second part
had been loaned at the first rate of interest, it would
have produced $245. Pind the rates of interest.
CHAPTEE XVI.
Simple Indeterminate Equations.
242. Ip a single equation be given which contains two
unknown numbers, and no other condition be imposed,
the number of its solutions is unlimited; for, if any value
be assigned to one of the unknown numbers, a correspond-
ing value may be found for the other. Such an equation
is said to be indeterminate.
243. The values of the unknown numbers in an inde-
terminate equation are dependent upon each other; so that,
though they are unlimited in number, they are confined to
a particular range.
This range may be still further limited by requiring
these values to satisfy some given condition; as, for
instance, that they shall be positive integers.
244. The method of solving an indeterminate equation
in positive integers is as follows:
(1) Solve 3 a: + 41/ = 22, in positive integers.
Transpose, 3 x = 22 — 4 y,
.•.x = 7-y\-^^
the quotient being written as a mixed expression.
...x + y-7 = i^.
Since the values of x and y are to be integral, x + y — 7 will be
tegral, and henc<
form of a fraction.
1 — 2/
integral, and hence, — ^-^ will be integral, though written in the
Let — ^ = m, an integer ;
SIMPLE INDETERMINATE EQUATIONS. 229
Then l--y = 3m,
... y=l — Sm.
Substitute this yalue of ^ in the original equation,
3a; + 4-12m = 22,
.-. X = 6 4- 4 m.
The equation y = 1 — 3 m shows that m in respect to y may be 0,
or have any negative value, but cannot have a positive value.
The equation x = 6 + 4 m shows that m in respect to x may be 0,
hut cannot have a negative value greater than 1.
.*. m may be 0 or — 1,
and then x = 6, |^ = 1 ;
or X — 2, y = 4.
(2) Solve 6 aj — 14y = 11, in positive integers.
Transpose, 6x = ll + 14y,
...x = 2 + 2y + i±il^.
...x-2y-2 = i^.
l + 4y ,, . , ,
.•. — ^— ^ must be mtegral.
Now, if — -r-^ be put = m, then y = — r — , a fraction in form.
To avoid this difficulty, it is necessary in some way to make
14-42/
the coefficient of y equal to unity. Since — r— ^ is integral, any
multiple of — r-^ is integral. Multiply, then, by such a number as
wiU make the coefficient of y greater by 1 than some multiple of the
denominator. In this case, multiply by 4. Then
4+16 y ^ . 4 + y . _ ,
— - or 3 y H ^ is mtegraL
o o
4 + y
/. — r-^ = m, an mteger ;
.*. y = 5 m — 4.
Since x = J (11 + 14 y), from the original equation,
.-. x= 14m — 9.
Here it is obvious that m may have any positive valtte^ and
x=5, 19, 33
y=l, 6, 11
230 ALGEBRA.
The required multiplier can always be found when the coefficients
are prime to each other, and it is best to divide the original equation
by the smaller of the two coefficients, in order to have the multiplier
as small as possible.
246. The necessity for a multiplier may often be obviated
by a little ingenuity. Thus,
The equation 4 y = 29 — 7 « may be put in the form of
4y=29 — 8x+a:,
.•.y=7-2x4-^^»
f 4
in which the fraction is of the required form.
The equation 6 x = 18 4- 13 y
gives x= 3 + 2y + -^-"^-^^»
14- « o
in which — -z-^ is of the required form.
246. It will be seen from (1) and (2), § 244, that when only
positive integers are required, the number of solutions will
be limited or unlimited according as the sign connecting
X and y is positive or negative,
(1) Find the least number that when divided by 14 and 5
will give remainders 1 and 3 respectively.
If N represents the number, then
-^j- = x,and~^=y.
.•.JV= 14x4-1, and^=6y+3.
.-. 14x+l = 5y+3.
6y = 14x — 2,
5y= 16x — 2 — X.
o 2 + x
,.y=3^ — r"
2 4- X
Let — — = wi, an integer ;
.*. x = 6m--2,
y = ^ (14 X — 2), from original equation.
.'. y = 14 m — 6.
If m = 1, X = 3, and y = 8.
.-. JV= 14x4- l = 5y+ 3 = 43. Ana.
SIMPLE INDETEBMINATE EQUATIONS. 231
(2) Solve 5x + 6y = S0y so that x may be a multiple of y,
and both positive.
Let X = my.
Then (6m4-6)y = 30.
and
5m + 6
30 m
6m4-6
If m = 2, X = 3|, y = If
Km = 3, x = ^,y=^.
(3) Solve 14a5 + 22^^ = 71, in positive integers.
x=6~y + -^'
If we multiply the fraction by 7 and reduce,
the result is -" 4 y 4- i,
a form which shows there can be no integral solution.
There can be no integral solution of ox ± &y = c, if a and b have a
common factor not common also to c ; for, if d be a factor of a and
also of 6, but not of c, the equation may be written,
mdx ± ndy = c,
or wx ± ny = - » a fraction.
Exercise 96.
Solve in positive integers :
1. 2x + lly = 4t9, 5. Sx + Sy = 61.
2. 7aj + 32/ = 40. 6. 8aj + 5y = 97.
3. 5x + 7y = 53, 7. 16a;+ 72^ = 110.
4. x + 10y = 29. 8. 7ic + 102^ = 206.
Solve in least positive integers :
9. 12x — Ty = l. 12. 23a; — 9^ = 929.
10. 5x — 17y = 23, 13. 23aj — 332/ = 43.
11. 23y — 13a; = 3. 14. 555ic — 22y = 73.
232 ALGEBBA.
16. How many fractions are there with denominators 12
and 18 whose sum is §| ?
16. What is the least number which, when divided by 3
and 5, leaves remainders 2 and 3 respectively ?
17. A person counting a basket of eggs, which he knows
are between 50 and 60, finds that when he counts
them 3 at a time there are 2 over; but when he
counts them 5 at a time there are 4 over. How
many are there in all ?
18. A person bought 40 animals, consisting of pigs, geese,
and chickens, for $40. The pigs cost $5 apiece, the
geese $1, and the chickens 25 cents each. Find the
number he bought of each.
19. Find the least multiple of 7 which, when divided by 2,
3, 4, 5, 6, leaves in each case 1 for a remainder.
20. In how many ways may 100 be divided into two parts,
one of which shall be a multiple of 7 and the other
of 9?
21. Solve 18a; — 5y = 70 so that y may be a multiple of x,
and both positive.
22. Solve 8a; + 12y = 23 so that x and y may be positive,
and their sum an integer.
23. Divide 70 into three parts which shall give integral
quotients when divided by 6, 7, 8, respectively, and
the sum of the quotients shall be 10.
24. Divide 200 into three parts which shall give integral
quotients when divided by 5, 7, 11, respectively, and
the sum of the quotients shall be 20.
26. A number consisting of three digits, of which the
middle one is 4, has the digits in the units' and
hundreds' places interchanged by adding 792. Find
the number.
SIMPLE INDETERMINATE EQUATIONS. 233
26. Some men earning each $2.50 a day, and some women
earning each $1.76 a day, receive all together for
their daily wages $44.76. Determine the number
of men and the number of women.
27. A wishes to pay B a debt of £1 125., but has only half-
crowns in his pocket, while B has only 4-penny pieces.
How may they settle the matter most simply ?
28. Show that 323 a; — 627 y= 1000 cannot be satisfied by
integral values of x and y,
29. A farmer buys oxen, sheep, and hens. The whole
number bought is 100, and the whole price £100.
If the oxen cost £6, the sheep £1, and the hens
Is, each, how many of each did he buy?
30. A number of lengths 3 feet, 6 feet, and 8 feet are cut;
how may 48 of them be taken so as to measure 176
feet all together ?
31. A field containing an integral number of acres less
than 10 is divided into 8 lots of one size, and 7 of 4
times that size, and has also a road passing through
it containing 1300 square yards. Find the size of
the lots in square yards.
32. Two wheels are to be made, the circumference of one
of which is to be a multiple of the other. What
circumferences may be taken so that when the first
has gone round three times and the other five, the
difference in the length of rope coiled on them may
be 17 feet ?
33. In how many ways can a person pay a sum of £16
in half-crowns, shillings, and sixpences, so that the
number of shillings and sixpences together shall be
equal to the number of half-crowns ?
CHAPTEE XVII.
Inequalities.
247. Expressions containing any given letter will have
their values changed when different values are assigned to
that letter ; and of two such expressions, one may be for
some values of the letter larger than the other, for other
values of the letter smaller than the other.
Thus, 1 + X + x2 will be greater than 1 — x + x^ for all positive
values of x, but less for all negative values of x.
248. One expression, however, may be so related to
another that, whatever values may be given to the letter,
it cannot be greater than the other.
Thus, 2x cannot be greater than x^ + 1, whatever value be given
to X.
249. For finding whether this relation holds between two
expressions, the following is a fundamental proposition :
If a and h are unequalj a^-\-h^>2db.
For, (a — 6)2 must be positive, whatever the values of a and h.
That is (a— &)2>0,
or a2 — 2a&+62>0;
.-. a2 + 62^2a6.
260. The principles applied to the solution of equations
may be applied to inequalities, except that if each side of
an equality have its sign changed^ the inequality will be
reversed.
Thus, if a>6, then — a will be < — 6.
INEQUALITIES. 236
(1) If a and h be positive, show that a^ + J* is > a^h + ai*.
a8 + 68> 026+062,
if (dividing each side by a + 6),
a2 — a6 + 62>a6,
if a2 + 62>2a6.
But a2 + 62 is>2 06. < 249.
.-. a8 + 68>o26 + a62.
(2) Show that a^ + ft^ + c^ is >ab-\-ac + hc.
Now, a2+62is>2a6,
a2+ c2is>2 oc, §249.
62 + c2is>26c.
By adding, 2a2 + 262 + 2c2 is>2a6 + 2ac + 26c.
.-. a2 + 62 + c2 is>a6 + ac + 6c.
Exercise 97.
Show that, the letters being unequal and positive :
1. a^-\-Wis>2h{a + h). 2. a»6 + o^Ms > 2 a^^l
3. (a^ + *')(a* + **)is>(a'rh^'T
5. The sum of any fraction and its reciprocal is > 2.
6. If a?=a^-{-b^, 2^=c2-[-6^^ xy is not < ac-{-hdy or ad-\-hc,
7. ab + ac + hc<(a + b — cy + {a + c--hy+{b + C'-'ay.
8. Which is the greater, (a^ + b^ ((^ + ^) or {ac + bdf?
9. Which is the greater, m^-\-m or m^+l?
10. Which is the greater, a* — b^ or 4 a^ (a — b) when a \%>b ?
11. Which is the greater, \— + \— or Va + V^?
^« TTT1-* i-'xi ^ a + ft 2a^^
12. Which IS the greater, ' or — tt ?
2 a + 6
13. Which is the greater, T2 + -2<^^7 + ~^
CHAPTEE XVIII.
Theory of Exponents.
261. The expression a", when tj is a positive integer, has
been defined as the product of n equal factors each equal
to a. § 24.
And it has been shown that a"* X a" = a"***"*. § 66.
That «"• -5- a" = a"-", if m is greater than w ; § 93.
or y if m is less than n, § 94.
And that («"•)» = oT". § 199.
Also, it is true that a" X ^* = {ahy ; for
(^ahy =^ ab taken n times as a factor,
= a taken n times as a factor X b taken ?» times as a
factor = a" X 5".
262. Likewise, v'a, when n is a positive integer, has been
defined as one of the n equal factors of a (§ 203) ; so that if
Va be taken n times as a factor, the resulting product is a;
that is, i^/aT = a.
Again, the expression ^^o" means that a is to be raised
to the mth power, and the nth root of the result obtained.
And the expression (-Ta)"* means that the nth root of a
is to be taken, and the result raised to the mth power.
It will thus be seen that any proposition relating to roots
and powers may be expressed by this method of notation.
It is, however, found convenient to adopt another method of
notation, in which fractional and negative exponents are
used.
THEORY OP EXPONENTS. 237
853. The meaning of a fractional exponent is at once sug-
gested, by observing that the division of an exponent, when
the resulting quotient is integral, is equivalent to extracting
a root. Thus, a* is the square root of a*, and 3, the expo-
nent of a*, is obtained by dividing the exponent of a* by 2.
If this division be indicated only, the square root of a*
will be denoted by at, in which the denominator denotes the
root, and the numerator the power. If the same meaning
be given to an exponent when the division does not give an
integral quotient, a* will represent the square root of the
m
cube of a; and, in general, a", the nth root of the mth power
of a. This, then, is the meaning that will be assigned to a
fractional exponent, so that in a fractional exponent
254. The numerator taill indicate a power, and the denom-
inator a root,
255. The meaning of a negative exponent is suggested
by observing that in a series of descending powers of a.
z*, a*, a', a*, a^.
the subtraction of 1 from the exponent is equivalent to
dividing by a; and if the operation be continued, the
result is
a®, a~^, a~*, a~^, a~^ a~*.
Then a« = - = l; a-' = l-=-a = i;
a ' a
This, then, is the meaning that will be assigned to a
negative exponent, so that,
266. A number unth a negative exponent will denote the
reciprocal of the number unth the corresponding positive
exponent.
238 ALGEBRA.
It may be easily shown that the laws which apply to
positive integral exponents apply also to fractional and
negative exponents. , , t
257. To show that a* X J"= (o^)* :
— — n
a»x 6"= Va^ X V5",
= V(a6)'»,
m
= (a6) » (by definition).
Ill 1
Likewise a" X 6» X c« = (a6c)«, and so on.
2 1 j_
258. To show that (a«)* = a"«:
1 1
Let
X = (a"«)».
1
x» = a*», and ««» = a.
Then
1
j[ 1
But
X = (a"»)" (by supposition).
1 i ^
269. To show that a*" X a"** = a"^ :
Now
a"» X a-» = a"» X — >
= ? = «'^^'^>^J
<03.
or
= a-<»-«). (by definition),
§04.
260. In like manner the same laws may be shown to
apply in every case.
THEORY OF KXPONENTS. 239
261. Hence, whether w and n are integral or fractionaly
positive or negative :
,,j| ^I. a"Xa* = a"'-^. III. (ary = a'^.
II. a" -^ a* = «'»-». IV. a'^Xlr^iaby.
EXBBCISE 98.
Express with fractional exponents :
1. V5^; ^^; (V^)*; ^^;te; (^^a)'; ^\
2. -^i^; ^yiy?; ^/^WJ '5y/^M^.
Express with radical signs :
3. al; aW; 4aji^i; Sx^yi.
Express with positive exponents :
2a'h:
4. a-«j Sor-ijr*; 6aj-^y; aV"*;
Write in the form of integral expressions :
Sxy z a (? a5~i or^ v
Simplify :
6. aiXai; JiX&*; c»XcA; diXd^.
7. miXm-J; wfXw-A; a<*Xa*; a^ X a""*.
8. aixVa; c'^xVc; y^X^i xiX-s/cF\
/^ 9. a^icXa-iftci; aWc-k X a^b'iM.
10. aiyW X aj-§y-4«-4 f.x«yi«4 X x-^jT^zri,
11. aiXa-»Xa-iXa-»\ (^yx (^^X (0-
240 ALaEBRA.
13. (a«)i-f-(a«)l; (c"*)!; (m-*)*; (n^; (^)*y/^
14. 0?-t)-l; (yl)-*;(ar-lyl)-l;^a§X «♦)-».
16. (4a-l)-f; (27 &-«)-§; (64ci«)-l;)(32c-»<)l.
262. The laws that apply to the exponents of simple
expressions also apply to the^ exponents of compound ex-
pressions.
(1) Multiply yl+y* + y* + l by yi-1.
yf + yi + yi+1
yi-1
y +yl + yi + yi
— yl — yh — yk—l
y —1 y — 1. Ans.
(2) Divide «§ + a* — 12 by a:i — 3.
X'
x*-3a;i
4x*-12
4x*-12 x* + 4. AnB.
,,,,., Exercise 99.
Multiply :
1. a^ + x'*y'* + y^hy a:^ — xPyP + ^.
2. «"«-" — y" by ic" + y""*~**.
3. id — 2a* + l by a;i — 1.
^•4. 8af + 4aW + 5aW + 9^f by 2af — W.
THEORY OF EXPONENTS. 241
6. l + ab-^ + a^b-^hjl-ab-^ + a^b-^
^ 6. a^b-^ + 2 + a-^h" by a^b-^ — 2 — a" V.
7. 4ar« + 3aj-« + 2x-i + lbyx-* — ar-» + l.
Divide :
8. oj** — ^ by 05* — y*.
9. x-\-y-\-z — 3a;i^«i by ai + yi-|-«i
1^ 10. X + y by a* — «*yi + ajly* — a*y* + yt.
11. a:*^* + 2 + a;-ybyx2r* + x'y
> 12. a-^ + a-^'b-^ + b-^hja-^ — a-^b-^ + b-K '
Find the squares of :
13. 4a*-^; a* — 64; a + a~M 2ai6i — a-W.
/
/
If a = 4, 6 = 2, c = l, find the values of :
14. a46; SaJ-^ 2(a6)4; a-JJ-^; 12a-%-^.
16. Expand (ai-64)»; (2ar-i + a)*; {ab-^ -- bij-y.
Extract the square root of :
(;i6. 9x-^ — 18x-^?^ + 15a-V — 6a;-y + 2^.
Extract the cube root of :
17. 8aj» + 12a:2-30a-35 + 46x-^ + 27«-2-27ar^.
Eesolve into prime factors with fractional exponents :
18. ^12y ^12^ \^> \/64 ; and find their product.
Simplify :
r 19. |(a^'*)«X(a^*)-'|8«-». ^20. (a:^«« Xic-^8«-«.
21.3 (a4 + b\y - 4 (a4 + 64) (a4 — 64) + (a4 - 2 64)1
/ 22. {{0^ "•{"•+1. 24. [K^"")"*W-^[K^"')"i'^]~
242 ALGEBRA.
Radical Expressions.
268. An indicated root that cannot be exactly obtained is
called a surd. An indicated root that can be exactly ob-
tained is said to have the form of a surd.
264. The required root shows the order of a surd ; and
surds are named quadratic, cuMe, biquadratic^ according as
the second, third, or fourth roots are required.
266. The product of a rational factor and a surd factor
is called a mixed surd; as^ 3 V2, h ^la,
266. When there is no rational factor outside of the
radical sign, the surd is said to be entire; as, V2, Va.
267. Since \^ X Vft X Vc = "^abc, the product of two or
more surds of the same order will be a radical expression
of the same order consisting of the product of the numbers
under the radical signs.
268. In like manner, Va*6 = Vo^ X V^ = a Vft. That is,
A foMor under the radical sign whose root can he taken,
may, by having the root taken, be removed from under the
radical sign.
269. Conversely, since a V^ = Va*^,
A factor outside the radical sign may be raised to the
corresponding power and placed under it.
Again: ^^ = ^a X^2 = -^ V^;
RADICAL EXPBESSIOKS. 243
270. A surd is in its simplest form when the expression
tinder the radical sign is integral and as sm^ll as possible.
271. Surds which, when reduced to the simplest form,
have the same surd factor, are said to be similar.
Simplify :
V50; -^108; ^/TV/; yj^', yl^', ^296352.
(1) V60 = V26X 2 = 6\^.
(2) 'V^='V^27X4 = 3\^.
(3) VTxV = V7a;2y2 xy^ = yV7xV.
._ 4/ 6a 4/40a6cg_ 1 </.^^, »
(6) \/'296342.
Hence, 296362 = 2* X 38 x ?»,
.-. \^296862 = -^8 X \^ X \^,
= 7 X 3X2V22,
= 42>/i. ^Irw.
In simplifying numerical expressions under the radical sign, the
method employed in (6) may be used with advantage when the factor
whose root can be taken is not readily determined by inspection.
Exercise 100.
Express as entire surds :
1. 3V5; 3V2I; 5V32; a^b-^b^-, x\^.
2. Sy'^/^; 2x^xy) a^^/a?b^', Sc'^c, babe^ahcrK
3. ^V^; i6Vh?; (^+yW^+2?y+7^
28
296362
22
37044
32
9261
3
1029
7'
343
7
49
244 ALGEBKA.
Express as miKed surds :
4. Viy^; V8^; \^54aVy»; V24; Vl26a*^.
Simplify :
7. 2\^80^W; 7V396^; 9\^81^; 5V726.
12 2 /xV\ / ^ \ t /^\/^\*
11. (aa;) X (b^x)h ; (2 a«^>*) X (b^s^i ; 5 (3 a'^^V) X (a*^-y)*-
12. Show that V2O, V46, V| are similar surds.
13. Show that 2 Va'^*, VS^, iX/T- are similar surds.
14. If V2 = 1.414213, find the values of
272. Surds of t?ie same order may be compared by
expressing them as entire surds.
Ex. Compare f V7 and jVlO.
iVt" =Vv,
|Vio = V^.
VV = '^^» and V^ = V^.
As V^ is greater than V^, fVlO is greater than f V?.
BADICAL EXPRESSIONS. 245
878. The product or quotient of two surds of the sa7ne
order may be obtained by taking the product or quotient
of the rational factors and the surd factors separately.
(1) 2V5X5V7=10V35.
(2) dV5-5-3V7 = 3V^ = 3V|| = fV35.
Exercise 101.
1. Which is the greater SVT or 2Vi5 ?
2. Arrange in order of magnitude 9 V3, 6V7, 6\^.
3. Arrange in order of magnitude 4v4, 3v5, 6 V3.
4. Multiply 3 V2 by 4V6 ; f VlO by t^^ Vi5.
^5. Multiply SVf by f Vi62 ; ^\^ by 2^/2,
6. Divide 2V5 by 3 VlS ; f V2T by ^^^/^.
7. Simplify fV3XtV5-T-fV2.
8 Sim lif ^^ X '^^^ --- i^^
^^^^^3V27 5Vi4 ' I5V2I
9. Simplify 2-^4 X5\/32-^\^i08.
274. The order of a surd may be changed by changing
the power of the expression under the radical sign. Thus,
V5 = \^; ^c = ^.
Conversely, V25 = V5 ; Vc^ = Vc ;
or, in general, Vc" = Vc.
In this way, surds of different orders may be reduced to
the same order, and may then be compared, multiplied, or
divided.
246 ALGEBBA.
(1) To compare V2 and V3.
^^ = 2* = 2* = y/2» = \^;
,', Vs is greater than \^.
(2) To multiply Via by V6ic.
V4a=(4a)*= (4 a)^ = V{4 a)2 = V 16 a2 ;
Ve^ = (6a)* = (6x)8 = V(6x)3 = V216a8.
/. Via X V6x = Vl6 a2 X V216x8,
= Vl6a2x 216x8,
= -^2^2 X 28 X 38x8.
= •v^2« X 2 X 38a2x8,
= 2V64a2x8. ^ns.
(3) To divide \^ by V6^.
-v^=(3a)*= (3a)»=\^)2=\^;
V66 = (6&)* = (66)8 = ^^(66)8 = \^21668.
... -^3^-V66 = ^i^^--^^l6F8=^g»
_ G/~a2"_ e/ a2.
~V24&8 \28X3&8'
26 X 366« 66
Exercise 102.
Arrange in order of magnitude :
1. 2-^,3V2, 4^4. 3. 2^, 3^, 4V2.
2. Vj, V^. 4. 3Vi9, 5\^2, 3-^.
BADIGAL EXPRESSIONS. 247
Simplify :
6. 3 (4 aJ^l -I- (2 a«6)i ; (2 a^V^i X {al'V)^ -s- (a«^i
7. (2a3)iX(3a^^i-^(5a^>«)i;4Vi2-^2V3.
9. (7V2-5V6-3V8 + 4V20)X3V2.
10- VOf/xVCfl/; V(4^xV(2a26)*.
11. ( \/^)« X ( \^^%^* ; aiMcirf-i -^- a«3-^c-tyH.
276. In the addition or subtraction of surds, each surd
must be reduced to its simplest form ; and, if the resulting
surds are similar,
Add the rational factors, and to their sum annex the
common surd factov.
If the resulting surds are not similar.
Connect them with their proper signs,
276. Operations with surds will be more easily per-
formed if the arithmetical numbers contained in the surds
are exjjressed in their prime factors, and if fractional expo-
nents are used instead of radical signs.
(1) Simplify V27 + V48 + V147.
V27 = (38)i = 3X3^ = 3V3 ;
Vis = (24 X 3)* = 22 X 3* = 4 X 3* = 4^3 ;
Vl47 = (72 X 3)i = 7 X 3* = 7^/3.
.-. V27 4-V48 + Vi47=(3 4-4 4-7)V3=14V3. An8.
248 ALGEBBA.
(2) Simplify 2-^320 — 3-^^40.
2-^320 = 2 (2«X 6)4 = 2X2ax 6*=8\^;
8\^ = 3(28X5)i = 3x 2X 6* = 6\^
.•.2\^-3'V^=8'^-6'^ = 2'^. Ans,
(3) Find the square root of VSl.
The square root of ^/Sl = (81^)* = 81* = (3*)*
= 3* = (32)i = ^/d.
(4) Find the cube of I \^.
Thecubeof i\/2= (i)« X (2*)« = ^ X 2* = ^ V5.
Exercise 103.
Simplify :
1. V27 4- 2 V48 + 3 VlOS; Is V1000 + 4 V50 + 12 Vm
2. \^128 + \^686+\^16j/\^54 + 3v^l6 + \^
3. 12V72 — 3Vi28;<W81 — 3\^i029.
4. 2V3 + 3Vli-V5jrXy4 + '^"-^^-"^-
,■'■ >f -Vf3VS)3Vi+2VA-4VS.
^^X8^ 2-^ + 3^^108 + ^^500 — -v^320-- 2 \^1372. ^
9. (2^3^)^ (3^)1 10. (^V^)*; (V27)i j
11. (-V^)i; (\/5i2)i; (-^256)"^; ^; '-^27.
^. \/45 ^-^; '-5^32; ^-5^243; ^125; ^49. D
13. -V^«; ^/9^*; \/i6^; -5^32^.
RADICAL £XPB£SSIONS« - ^ 249
- 14. (^y; {^Y; {^f; {^f. ^ )
16. (a-^)-»; (a;\^)-t; {j^^^; (a-'-v'O"*-
^ Expand by the method explained in § 201 :
16. (Va + VJ)*; (V^+V«»)»; (Va — 2Vi)».
17. (2a«-iVa)«; {2^*-iy^*; (^-^^^.
■•■ (IV^V?)" (»■--•)•; (^-*'^-)*
- {&m^ (;|-4 (f -I)*
Find the square root of :
21. ic^ + 6x*''y + lla:^y^ + 6a^3^ + 2^.
22. l + 4a;-i-2a;-§-4a;-i + 25a;-i-24ari + 16aj-«.
3
277. If we wish to find the approximate value of ^^'
it will be less labor to multiply first both numerator and
denominator by a factor that will render the denominator
rational; in this case by V2. Thus,
3 3V2 3V2
V2 V2 X V2 '
878. It is easy to rationalize the denominator of a frac-
tion when that denominator is a binomial involving only
quadratic surds. The factor required will consist of the
same terms as the given denominator, but with a different
250 ALaBBRA.
7 — ^ \fK
sign between them. Thus, _ , ^ ._ will have its denomi-
nator rationalized by multiplying both terms of the fraction
by6 — 2V5. For,
^7 — 3V5^(7 — 3V5)(6~2V5)
6 + 2V5""(6 + 2V5)(6 — 2V5)
279. By two operations the denominator of a fraction
may be rationalized when that denominator consists of
three quadratic surds.
Thus, if the denominator is V6 + V3 — V2, both terms of
the fraction may be multiplied by V6 — V3 + V2. The
resulting denominator will be 6 — 5 + 2 Vg = 1 + 2 V6 ;
and if both terms of the resulting fraction be multiplied
by 1 — 2V6, the denominator will become 1 — 24 = — 23.
Exercise 104.
Find equivalent fractions with rational denominators, for
the following :
3 7 /'4— V2 - 6
2.
V7 + V5'2V5-V6'l+V2'5-2V6'
a + 5 2x — yfxy
yfh — Vc a — \lb \lxy — 2y
Find the approximate values of :
g ^.__JL 7V5 7 + 2ViO
V3' V5- V2' V7 + V3' 7-2V10'
radical expressions. 251
Imaginary Expressions.
280. All imaginary square roots may be reduced to one
form.
V^ = VaX(— 1) = a* V^.
281. V — 1 means an expression which, when multiplied
by itself, produces — 1. Therefore,
(V=l)'=-ij_ _ _ _
(V-l)='=(V-l)*xV-l=-lV-l=-V-l;
(V^)*=(V^)>X(V^)«=(-1)X(-1) = 1;
and so on. So that the successive powers of V— 1 form
the repeating series, + V— 1, — 1, — V — 1, + 1.
(1) Multiply 1 + V^ by 1 — V^.
(1 + 2 V^) (1-2 V^) = 1 - 4 (- 1) = 5.
(2) Divide V— ah by V^.
and V^ =6iV^.
~ = Vo.
V-6 6*V^
Exercise 105.
Multiply :
1. 4 + V^by4 — V^5 V3 — 2V^by V3 + 2V^.
2. V54byV— 2j aV— ^by a;V— y.
252 ALOEBRA.
8. V— a + V^ by V— a — V^; aV— «*^* by V— a***.
4. V^^byV^; 2V3 — 6V^ by 4V3 — V^.
Divide :
6. a; V^ by y\f^ ; 1 by V^^ ; a by a^^T^.
6. V=^by V^^; Vi5by V^; V^by V^^.
Square Root op a Binomial Surd.
282. The product or quotient of two dissimilar quadratio
surds will be a quadratic surd. Thus^
'slab X ^/abc = ab^;
Vabc -h- Va^ = Vc.
For eyery quadratic surd, when simplified, will have under the
radical sign one or more factors raised only to the first power ; and
two surds which are dissimilar cannot have all these factors alike.
Hence, their product or quotient will have at least one factor
raised only to the first power, and will therefore be a surd.
283. The sum or difference of two dissimilar quadratic
surds cannot be a rational number, nor can it be expressed
as a single surd.
For if Va ± Vft could equal a rational number c, we should have,
by squaring,
a±2Va6+6=c«;
thatis, ±2Va6 = c2 — a — 6.
Now, as the right side of this equation is rational, the left side
would be rational ; but, by § 282, Va6 cannot be rational. Therefore,
Va ± V6 cannot be rational.
In like manner, it may be shown that Va ± VS cannot be ex-
pressed as a single surd Vc.
RADICAL EXPRESSIONS. 253
884. A quadratic surd cannot equal the sum of a rational
number and a surd.
For, if Va could equal c + V5J we should have, by squaring,
a = c« + 2cV6'+6,
and, by transposing, 2cV6=a — 6 — c^.
That is, a surd equal to a rational number, which is impossible.
885. If a + V^ = a; + Vy, then a ttnll equal x and b unll
equal y.
FOr, by transposing, Vft — ^fy^ x — a ; and if 6 were not equal to
y, the difference of two unequal surds would be rational, which by
§ 283 is impossible.
.% 6 = y, and a = x.
In like manner, if a — V6 = x — 'V^, a will equal x and h will
equal y.
To extranet the square root of a binomial surd a •\' yb.
Let Va+ V6 = Vx + Vy.
Squaring, a + Vft = x + 2 Vxy + y.
.% X + y = a, and 2 \Sy = Vf. § 285.
From these two equations the values of x and y may be found.
This method may be shortened by observing that, since
V^ = 2Vay,
a — \b = x — 2yxy-^y,
By taking the root, \/a — Vb = ^/x~- Vy.
-'»(VQ^ + V^)(Vfl^^)==(V^+Vy)(V^-Vy)»
And, as a = ic + y,
the values of x and y may be found by addition and
subtraction.
254 AIiOBBBA.
Ex. Extract the square root of 7 + 4 y/S.
Let
>/J+V^= V7 + 4V3.
Then
>/s"->^= V7-4V3
By miUtiplying,
x--y= V49-48.
.-. aj — y = 1.
But
« + y=7.
.♦. X = 4, and y = 8^
.'.'^+'^=2 + yfl
,'
•.V7 + 4V3=2 + V5.
Exercise 106.
Extract the square roots of:
1. 14 + 6V6. 6. 20 — 8V6. 11. 14 — 4V6.
2. 17 + 4Vi5. 7. 9-6V2. 12. 38 — 12ViO.
3. 10 + 2V2i. 8. 94 — 42V5. 13. 103 — I2VII.
4. I6 + 2V55. 9. 13 — 2V3O. 14. 57 — 12 Vis.
5. 9-2Vi4. 10. 11— 6V2. 15. 3i — Vio.
16. 2a + 2Va' — ^l 18. 87 — 12V42.
17. a^ — 2^Va2 — Z>2. 19. (a + 6/--4 (a — ft) V^.
287. A root may often be obtained by inspection. For
this purpose, write the given expression in the form a+2 Vft,
and determine what two numbers have their sum equal to
a, and their product equal to b,
(1) Find by inspection the square root of 18 + 2 V77.
It is required to find two numbers whose sum is 18 and whose
product is 77 ; and these are evidently 11 and 7.
Then 18 + 2 Vtt" = 11 +_7 + 2 Vll + 7,
= (Vll + V7)2.
That is, VlT + Vt" = square root of 18 + 2 V77.
RADICAL EXPRESSIONS. 256
(2) Find by inspection the square root of 76 — 12 V^T.
It is necessary that the coefficient of the surd be 2 ; therefore,
76 — 12 V2T must be put in the form of
76-2V62X21 = 75-2V766.
The two numbers whose sum is 75 and whose product is 766
are 63 and 12.
Then 76-2V766 = 63 + 12 -2 V63+ 12,
= (V63-Vi2)«.
That is, V63 — VT2 = the square root of 76 — 12 VST;
or, 8 V7 — 2 V3 = the square root of 76 - 12 V2T.
Equations containing Eadicals.
888. An equation containing a single radical may be
solved by arranging the terms so as to have the radical
alone on one side, and then raising both sides to a power
corresponding to the order of the radical.
Ex. Va:^ — 9 + aj = 9.
Vx2 — 9 = 9 — X.
By squaring, x2 — 9 = 81 — 18 x + x^,
18x = 90.
/. X = 6.
289. If two radicals are involved, two steps may be
necessary.
Ex. Va; + 15 + V^=15.
Vx+16 + >^= 15.
By squaring,
X + 16 + 2Vx2+15x + X = 226.
By transposing, 2 Vx2 + 16x = 210 — 2 x.
By dividing by 2, Vx2+15x = 106 — x.
By squaring, x^ + 16 x = 11026 — 210 x + »«.
226 x= 11026.
.-. X = 49.
256 ALGEBRA.
Some of the following radical equations will redace to
simple and others to quadratic equations.
gQlyg. Exercise 107.
1. Va; — 5 = 2. 6. Va; + 4 + V2ic — 1=6.
2. 2V3a; + 4 — a: = 4. 7. VlScc— 1 — V2a— 1=5.
3. 3 — ^a^-l = 2x. / 8. V4+^ + Vi = 3.
4. \/3x — 2 = 2(x-^). '9. ^25 + x + ^25 — x = S,
6. 4a;-12Va; = 16. 10. a^ = 21 + Va^ — 9.
11. 2aj — V8a^ + 26 + 2 = 0.
/"12. Va + l + Vx + 16 = Va: + 25.
i 13. V2aj + 1 — Va; + 4 = iVa;-3.
y 14. Vx + 3 + Va; + 8 = 5>^.
Va; + 1 Va:-1 Vx^ — 1
Va; + 2a — Vic — 2a
18.
Vx — 2a + Vic + 2a 2a
^^ 3x + V4x^^^ g^ V7V+4 + 2V3x~l^^
3a;-V4a; — x2 * ' 77x^ + 4 — 2 V3a; — 1
21. V(x-a)2 + 2aft + i>« = x-a + ^.
22. V(a; + a)2 + 2a^> + &2^5 — a — «.
RADICAL EXPRESSIONS. 257
24. 4a:J — 3(x* + l)(aji — 2) = a;*(10 — 3»i).
25. (aj§ — 2)(aji — 4) = al(aj§ — 1)« — 12.
26. «« — 4aji = 96. 28. a* + 2 a^af* = 3 a.
s 81
27. a; + ari==2.9. 29. 8ivi + -f^ = 52aj.
290. Equations may be solved with respect to an expres-
sion in the same manner as with respect to a letter.
(1) Solve(x2 — x)*-8(aj«— aj) + 12 = 0.
Consider (x^ — x) as the unknown number.
Then (x^ -x)a- 8(xa-x) = - 12.
Complete the square, (x^ — x)2 — ( ) + 16 = 4.
Extract the root, (x^ — x) — 4 = ±2.
x2 - aj = 6 or 2.
Complete the square, 4 x^ — ( ) + 1 = 26 or 9.
Extract the root, 2x~l=±5or±3.
2x=6, -4,4, -2.
.-. X = 3, - 2, 2, - 1.
(2) Solve5x-7x2 — 8Vfx2-5a; + l = 8.
Change the signs and annex + 1 to both sides.
7xa-6x+l + 8V7x2-6x+l=-7.
Solve with respect to VTx^— 5x+ 1.
(7x2- 5x 4- 1) + 8 (7a.2_ 5a. + i)4 + 16 = 9.
(7x2~5x+l)* + 4=±3.
(7x2-5x4- !)*=-! or -7.
Square, 7 x2 — 5x + 1 = 1 or 49.
Transpose, 7 x2 — 5 x = 0 or 48.
From 7x2 — 6x = 0, ^ = 0 or f ;
From 7x2 — 5x = 48, x = 3 or — 2f
Note. In verifying the values of x in the original equation, it is
seen that the value of V7 x2 — 6 x 4-1 is negative. Thus, by putting
0 for X the equation becomes 0 — 8 Vl = 8 ; and by taking — 1 f or Vl
we have (— 8) (- 1) = 8 j that is, 8 = 8.
258 AI^GEBRA.
(3) Solve a:« + x + l + i + f3 = l.
Arrange as follows : fx^ "^ "a) "^ (^ "^ x) ~ ^*
By adding 2U> (x-h ^ V
1 / 1\*
there is obtained x^ + 2 + -^^ ix + -y
Multiply by 4 and complete the square,
Extract the root, 2 (x + -) 4- 1 = ± 8.
i'^D'
2 or — 4.
X + - = 1 or - 2.
X
Multiply by X, x^ — x = — 1, and aj2 + 2 x = — 1.
.-. 4x2 - ( ) + 1 = - 3, .-.xs + 2x + 1 = 0.
2x-l=±V^, a; + l = 0.
.-. a; = i(l±V=^). .•.x=-l.
291. An equation like that of (3) which will remain
unaltered when - is substituted for x, is called a reciprocal
equation.
It will be found that every reciprocal equation of odd degree will
be divisible by » — 1 or x + 1 according as the last term is negative
or positive ; and every reciprocal equation of even degree vjUh its last
term negative will be divisible by sc* — 1. In every case the equation
resulting from the division will be reciprocal.
RADICAL EXPRESSIONS. 259
(4) Solve ar^ + 2a;^ — 3aj«-3a;2 + 2a; + l==0.
This is a reciprocal equation, for, if xr-^ is put for x, the equation
becomes flr« + 2 x-^ — 3x-« — 3x-2 + 2 x-i + 1 = 0, which multiplied
by a^ gives 1 + 2x — 3x2 — Sx^ 4- 2x* + x^ = 0, the same as the
original equation.
The equation may be written (x^ + 1)+ 2 x(x8+ 1) ~ 3x2(x + 1) = 0,
which is obviously divisible by x + 1. The result from dividing by
x+1 is x* + x8-4x2 + x + l = 0, or (x* + 1) + x(x2+ 1) = 4x2.
By adding 2 x2 to (x* + 1) it becomes (x* + 2 x2 4- 1) = (x2 + 1)2.
Then (x2 + 1)2 + x (x2 + 1) = 6x2.
Multiply by 4 and complete the square,
4(x2 4-l)24-() + x2 = 25x2.
Extract the root, 2 (x2 + 1) + x = ± 6x
Hence, 2 x2 + 2 = 4 x or — 6 x.
By simplifymg, x2 — 2x= — 1; andx2 + 3x= — 1,
whence, x = 1 and 1; whence, x = i(— -3 ± Vs).
Therefore, including the root — 1 obtained from the factor
x+1, the five roots are — 1, 1, 1, i (— 3 ± Vs).
By this process a reciprocal cvMc equation may be reduced to a
quadratic, and one of the fifth or sixth degree to a biquadratic^ the
solution of which may be easily effected.
Solve ;
Exercise 108.
1. ic2_3a.__(57a;2__3^__3^2 = 0.
3. (2aj«—3cB)2 — 2(2x2 — 3£c) = 15.
9a*
4. (ax — by-{-4:a(ax — ^)=^~7"
6. 3(2x' — x)'-(2x''-x)h = 2,
6. 15x — 3aj» + 4(a;2— 55c + 5)i = 16.
7. ««+cc-2 + a: + a;-i = 4. 9. x' + x + i (x' + x)h=^i.
«. «>+V^^=^ = 19. 10. (x + l)i+(x'-l)i = 5.
260 ALGEBRA.
11. aj — l = 2 + 2ar*. 12. V3aj + 5 — V3a; — 5=4.
13. (x' + l)'-x(x^ + l) = '-'2x^
14. 2a^'-2^2x^'-5x = 5(x + S).
15. a: + 2 — 4Wa; + 2 = 12x«.
16. V2a; + a4- V2a; — a = 5.
.17. ^/9x^ + 21x + l — ^9a^ + 6x + l = Sx.
18. a;i--4aji + cc-4 + 4a;""5 = --J.
19. (2aj + 3y)'-2(2aj + 3y) = 8^
a^-y' = 21 )
20. g + y+V^+y = a) 22. a^ + y^ + a; + y = 48|
x--y+^x — y = b) xj/ = 12 J
21. a;* — a^^2 + y* = 13^ 23. «' + xy + 2/* = a»|
x« — icy + 2/2=3 J x+Vxy + y = bi
24. (a;-y)«-3(aj-2^) = 10^
x*2/^-3«y=54 J
25. V^ — V^ = «*(^* + y*)\
- fe)'+(^)*-|
xy — (ir + y) = 54 J
2Ta*
28. --3aaj = V4a^ + 9aa:2+^
a 4
29. (aj + l + 0(^ — l + 0 = ^i- ^r'^
30. 2(aji — l)-i-2(a;i — 4)-^ = 3(ici — 2)-». ;,
CHAPTER XIX.
Logarithms.
292. I^ the common system of notation the expression
of numbers is founded on their relation to ten.
Thus, 3854 indicates that this number contains IC three times, 10^
eight times, 10 five times, and four units.
293. In this system a number is represented by a series
of different powers of ten, the exponent of each power
being integral. But, by employing fractional exponents,
any number may be represented (approximately) as a single
power of 10.
294. When numbers are referred in this way to 10, the
exponents of the powers corresponding to them are called
their logarithmB to the base 10.
For brevity the word " logarithm " is written log.
From § 256 it appears that :
ib''= 1, io->(=iV) =0.1, . ,
io»= 10, io-»(=Ti^) =0.01,.— • ; '■'
io«=ioo, 10-* (=T^^)= 0.001, '_.:-..
and so on. Hence,
log 1=0, log 0.1 =-1,
log 10 = 1, log 0.01 =—2,
log 100 = 2, '■ log 0.001 = — 3,
and so on.
262 ALGEBRA.
It is evident that the logarithms of all numbers between
1 and 10 will be 0 + a fraction,
10 and 100 will be 1 + a fraction,
100 and 1000 will be 2 + a fraction,
1 and 0.1 will be — 1 + a fraction,
0.1 and 0.01 will be — 2 + a fraction,
0.01 and 0.001 will be — 3 + a fraction.
295. The fractional part of a logarithm cannot be
expressed exactly either by common or by decimal frac-
tions : but decimals may be obtained for these fractional
parts, true to as many places as may be desired.
If, for instance, the logarithm of 2 be required ; log 2 may be
supposed to be ^,
Then 10* = 2 ; or, by raising both sides to the third power, 10 = 8,
a result which shows that \ is too large.
Suppose, then, log 2 = 3^. Then 10^ = 2, or by raising both sides
to the terdh power, lO^ = 2io. That is, 1000 = 1024, a result which
shows that ^ is too small.
Since \ is too large and ^ too small, log 2 lies between \ and ^^ ;
that is, between 0.33333 and 0.30000.
In supposing log 2 to be \^ the error of the result is if^* = 3% = 0.2.
In supposing log 2 to be 3^, the error of the result is *^^^¥5nr\?~^ ~
fj^ = — 0.024 ; log 2, therefore, is nearer to ^j^ than to \,
The difference between the errors is 0.2 — (— 0.024) = 0.224, and the
difference between the supposed logarithms is 0.33333 — 0.3 = 0.03333.
The last error, therefore, in the supposed logarithm may be con-
sidered to be approximately ^ of 0.03333 = 0.0036 nearly, and this
added to 0.3000 gives 0.3035, a result a little too large.
By shorter methods of higher mathematics, the logarithm of 2 is
known to be 0.3010300, true to the seventh place.
296. The logarithm of a number consists of two parts,
an integral part and a fractional part.
Thus, log 2 = 0.30103, in which the integral part is 0, and the frac-
tional part is .30103 ; log 20 = 1.80103, in which the integral part is 1,
and the fractional part is .30103.
LOGARITHMS. 263
297. The integral part of a logarithm is called the ohar-
acteristic ; and the fractional part is called the mantisia.
298. The mantissa is always made positive, Hence^ in
the case of numbers less than 1 whose logarithms are nega-
tive, the logarithm is made to consist of a negative charac-
teristic and a positive mantissa.
299. When a logarithm consists of a negative character-
istic and a positive mantissa, it is usual to write the minus
sign over the characteristic, or else to add 10 to the char-
acteristic and to indicate the subtraction of 10 from the
resulting logarithm.
Thus, log. 0.2 = 1.30103, and this may be written 9.30103 — 10.
300. The characteristic of a logarithm of an integral
number, or of a mixed number, is one less than the number
of integral digits.
Thus, from § 294, log 1=0, log 10 = 1, log 100 = 2. Hence, the
logarithms of all numbers from 1 to 10 (that is, of all numbers con-
sisting of one integral digit), will have 0 for a characteristic ; and the
logarithms of all numbers from 10 to 100 (that is, of all numbers con-
sisting of two integral digits), will have 1 for a characteristic ; and so
on, the characteristic increasing by 1 for each increase in the number
of digits, and therefore always being 1 lessjthan that number.
301. The characteristic of a logarithm of a decimal fra^
tion is negative, and is equal to the number of the place occu-
pied by the first significant figxire of the decimal.
Thus, from § 294, log 0.1 = — 1, log. 0.01 = — 2, log. 0.001 = — 3.
Hence, the logarithms of all numbers from 0.1 to 1 will have — 1 for
a characteristic (the mantissa being ylus) \ the logarithms of all num-
bers from 0.01 to O.l will have — 2 for a characteristic ; the logarithms
of all numbers from 0.001 to 0.01 will have — 3 for a characteristic ;
and so on, the characteristic always being negative and equal to the
nuniber of the place occupied by the first significant figure of the decimal.
264 ALGEBRA.
302. The mantissa of a logarithm of any integral numr
her or decimal fraction depends only upon the digits of the
number, and is unchanged so long as the seqv^ence of the
digits remains the same.
For, changing the position of the decimal point in a number is
equivalent to multiplying or dividing the number by a power of 10.
Its logarithm, therefore, will be increased or diminished by the expo-
nent of that power of 10 ; and, since this exponent is vntegrdL^ the
mantissa of the logarithm will be unaffected.
Thus, if 27196 = W««,
then 2719.6 = 108-««,
27.196 = 10i««,
2.7196 = 100««,
0.27196= 109««-io,
0.0027196 = 10''««-io.
808. The advantage of using the number 10 as the base
of a system of logarithms consists in the fact that the
mantissa depends only on the sequence of digits, and the
characteristic on the position of the decimal point
804. As logarithms are simply exponents (§ 294), there-
fore,
The logarithm of a product is the sum of the logarithms
of the factors.
Thus, log 20 = log (2 X 10) = log 2 + log 10,
= 0.3010 + 1.0000 = 1.3010 ;
log 2000 = log (2 X 1000) = log 2 + log 1000.
= 0.3010 + 3.0000 = 3.3010 ;
log 0.2 = log (2 X 0.1) = log 2 + log 0.1,
= 0.3010 + 9.0000 - 10 = 9.3010 - 10,
log 0.02 = log (2 X 0.01) = log 2 + log 0.01,
= 0.3010 + 8.0000 - 10 = 8.3010 - 10.
Exercise 109.
Given : log 2 = 0.3010 ; log 3 = 0.4771 ; log 5 = 0.6990 ;
log 7 = 0.8451.
Find the logarithms of the following numbers by resolv-
LOGABITHMS. 266
ing the numbers into factors, and taking
the
sum of the
logarithms
of the factors
:
1. log 6.
9. log 25.
17. log 0.021.
26.
log 2.1.
2. log 16.
10. log 30.
18. log 0.36.
26.
log 16.
3. log 21.
11. log 42.
19. log 0.0036.
27.
log 0.066.
4. log 14.
12. log 420.
20. log 0.004.
28.
log 0.63.
5. log 36.
13. log 12.
21. log 0.06.
29.
log 1.75.
6. log 9.
14. log 60.
22. log 12.6.
30.
log 106.
7. log 8.
15. log 75.
23. log 1.25.
31.
log 0.0106.
8. log 49.
16. log 7.5.
24. log 37.6.
32.
log 1.06.
805. As logarithms are simply exponents (§ 294), there-
/ The logarithm of a power of a number is equal to the '
/ logarithm of the numher multiplied by the exponent of the
'. power.
Thus, log 67 = 7 X log 6 = 7 X 0.6990 = 4.8930.
log 3" = 11 X log 3 = 11 X 0.4771 = 6.2481.
806. As logarithms are simply exponents (§ 294), there-
fore, when roots are expressed by fractional indices.
The logarithm of a root of a number is equal to the loga-
rithm of the numher multiplied by the index of the root.
Thus, log 2* = i of log 2 = i X 0.3010 = 0.0763.
log 0.002* = i of (7.3010 - 10).
The expression i of (7.3010 — 10)
may be put in the form of i of (27.3010 — 30) which = 9.1003 — 10 ;
for, since 20 — 20 = 0, the addition of 20 to the 7, and of -- 20 to the
— 10, produces no change in the loaXue of the logarithm.
807. In simplifying the logarithm of a root the equal pos-
itive and negative numbers to be added to the logarithm must
be such that the resulting negative number, when divided by
the index of the root, shall give a quotient of — 10.
266 ALGEBRA.
Exercise 110.
Given : log 2 = 0.3010 ; log 3 = 0.4771 ; log 5=0.6990 ;
lo
g7 =
0.8461.
Find
logarithms of the following :
1.
2».
6. 5».
11. 61.
16.
7».
21.
6i
2.
6*
7. 21.
12. 7A.
17.
61.
22.
2V.
3.
7*.
8. 5».
13. 21
18.
3A.
23.
51
4.
3«.
9. 31.
14. 61.
19.
71.
24.
7V.
5.
7».
10. 71.
15. 3K
20.
31.
26.
211.
308. Since logarithms are simply exponents (§ 294),
therefore,
The logarithm of a quotient is the logarithm of the divi-
dend minus the logarithm, of the divisor.
Thus, log f = log 3 - log 2 = 0.4771 - 0.3010 = 0.1761.
log I = log 2 - log 3 = 0.3010 - 0.4771 = - 0.1761.
To avoid the negative logarithm —0.1761, we subtract the entire
logarithm 0.1761 from 10, and then indicate the subtraction of 10
from the result.
Thus, - 0.1761 = 9.8239 - 10.
Hence, log f = 9.8239 — 10.
309. The remainder obtained by subtracting the loga-
rithm of a number from 10 is called the cologarithm of
the number, or arithmetical complement of the logarithm
of the number.
Cologarithm is usually denoted by colog, and is most
easily found by beginning with the characteristic of the loga-
rithm and subtracting each figure from 9 down to the last
significant figure, and subtracting that figure from 10.
Thus, log 7 = 0.8461 ; and colog 7 = 9.1549. Colog 7 is readily
found by subtracting, mentally, 0 from 9, 8 from 9, 4 from 9, 6 from
9, 1 from 10, and writing the resulting figure at each step.
LOGARITHMS. 267
810. Since colog 7 = 9.1649,
and log ;^ = log 1 - log 7 = 0 - 0.8451 = 9.1549 - 10,
it is evident that,
If 10 be subtracted from the cologarithm of a number, the
result is the logarithm of the reciprocal of that number.
811. Since log J = log 7 — log 5,
= 0.8451 - 0.6990 = 0.1461,
and log 7 + colog 5 - 10 = 0.8461 + 9.3010 - 10,
= 0.1461,
it is evident that.
The addition of a cologarithm — 10 is equivalent to the
subtraction of a logarithm.
The steps that lead to this result are :
therefore, log i = log (7 X J) = log 7 + log J. § 304.
But log I = colog 5—10. § 309.
Hence, log J = log 7 + colog 6 — 10.
Therefore,
812. The logarithm of a quotient may be found by add-
ing together the logarithm of the dividend and the cologa-
rithm of the divisor, and subtracting 10 from the result.
In finding a cologarithm when the characteristic of the logarithm
is a negalive number, it must be observed that the subtraction of a
negative number is equivalent to the addition of an equal positive
number.
^^' ^°^ Crlo2 = ^^S 6 4- colog 0.002 - 10,
= 0.6990 + 12.6990 - 10,
= 3.3980.
Here log. 0.002 = 3.3010, and in subtracting — 3 from 9 the result
is the same as adding + 3 to 9.
2
Again, log g-j^ = log 2 + colog 0.07 — 10,
= 0.3010 + 11.1549-10,
= 1.4659.
268 ALGEBRA.
Alao, log ^ = 8.8461 - 10 + 9.0970 - 10,
= 17.9421-20,
= 7.9421 - 10.
Here, log 2« = 3 log 2 = 3 x 0.3010 = 0.9030.
Hence, colog 2* = 10 — 0.9030 = 9.0970.
EXEBCISE 111.
Given : log 2 = 0.3010 ; log 3 = 0.4771 ; log 5 = 0.6990 ;
log 7 = 0.8451.
Find logarithms for the following quotients :
■•I-
7.
5
3'
13.
0.05
3
19 ^•^^.
^^' 0.003 .
25- 3»
-1-
a
6,
2'
14.
0.005
2
^' 0.02
2a ^ .
0,02a
-!■
9.
7
3'
16.
0.07
6
21 ^•^.
^- 0.007
^•oS-
^f
10.
7
2'
la
5
0.07
0.005.
0.07
2& °°^'-
0.003»
-f
11.
3
2'
17.
3
0.007
23. ^-^^^
0.005»
*^' 78
-i
12.
7 .
0.6*
la
0.003
7
0.0007.
^ 0.2
0.005*
813. A table oi four-place logarithms is here given, which
contains logarithms of all numbers under 1000, t?ie decimal
point and characteristic being omitted. The logarithms of
single digits 1, 8, etc., will be found at 10, 80, etc.
Tables containing logarithms of more places can be pro-
cured, but this table will serve for many practical uses,
and will enable the student to use tables of six-place,
seven-place, and ten-place logarithms in work that requires
greater accuracy.
LOGARITHMS. 269
814; In working with a four-place table, the numbers
corresponding to the logarithms, that is, the antilogarithms,
as they are called, may be carried to four significant digits.
To Find the Logabithm of a Number in this Table.
815. Suppose it is required to find the logarithm of
66.7. In the column headed " N " look for the first two
significant figures, and at the top of the table for the third
significant figure. In the line with 65, and in the column
headed 7, is seen 8176. To this number prefix the charac-
teristic and insert the decimal point. Thus,
log 66.7 = 1.8176.
Suppose it is required to find the logarithm of 203k7.
In the line with 20, and in the column headed 3, is seen
3076 ; also in the line with 20, and in the 4 column, is
seen 3096, and the difference between these two is 21.
The difference between 20300 and 20400 is 100, and the
difference between 20300 and 20347 is 47. Hence, ^V^ of
21 = 10, nearly, must be added to 3076. That is,
log 20347 = 4.3086.
Suppose it is required to find the logarithm of 0.0006076.
In the line with 60, and in the 7 column, is seen 7060 ; in
the 8 column, 7069: the difference is 9. The difference
between 6070 and 6080 is 10, and the difference between
6070 and 6076 is 6. Hence, ^ of 9 = 6 must be added to
7060. That is, ^^^ 0.0006076 = 6.7056 - 10.
To Find a Number when its Logarithm is Given.
816. Suppose it is required to find the number of which
the logarithm is 1.9736.
Look for 9736 in the table. In the column headed " N,"
and in the line with 9736, is seen 94, and at the head of
270
ALOEBBA.
N
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
0000
0414
0792
1139
1461
0043
0453
0828
1173
1492
0086
0492
0864
1206
1623
0128
0531
0899
1239
1563
0170
0669
0934
1271
1584
0212
0607
0969
1303
1614
0263
0646
1004
1336
1644
0294
0682
1038
1367
1673
0334
0719
1072
1399
1703
0374
0755
1106
1430
1732
15
16
17
18
19
1761
2041
2304
2663
2788
1790
2068
2330
2677
2810
1818
2096
2366
2601
2833
1847
2122
2380
2626
2866
1875
2148
2405
2648
2878
1903
2175
2430
2672
2900
1931
2201
2465
2695
2923
1969
2227
2480
2718
2945
1987
2253
2604
2742
2967
2014
2279
2529
2765
2989
20
21
22
23
24
3010
3222
3424
3617
3802
3032
3243
3444
3636
3820
3064
3263
3464
3666
3838
3076
3284
3483
3674
3856
3096
3304
3502
8692
3874
3118
3324
3622
3711
3892
3139
3345
3641
3729
3909
3160
3366
3660
3747
3927
3181
3386
3679
3766
3945
3201
3404
3698
3784
3962
25
26
27
28
29
3979
4160
4314
4472
4624
3997
4166
4330
4487
4639
4014
4183
4346
4502
4664
4031
4200
4362
4618
4669
4048
4216
4378
4533
4683
4065
4232
4393
4548
4698
4082
4249
4409
4564
4713
4099
4266
4426
4579
4728
4116
4281
4440
4694
4742
4133
4298
4456
4609
4757
30
31
32
33
34
4771
4914
5061
6186
6316
4786
4928
5066
6198
5328
4800
4942
5079
6211
6340
4814
4955
6092
6224
5363
4829
4969
5105
6237
5366
4843
4983
5119
5250
5378
4867
4997
5132
5263
5391
4871
6011
5145
5276
5403
4886
5024
5169
6289
6416
4900
6038
5172
5302
5428
35
36
37
38
39
6441
6663
6682
6798
6911
6463
6576
6694
6809
6922
6466
5587
6705
6821
5933
5478
6699
6717
5832
6944
5490
5611
5729
5843
6966
6602
5623
5740
5856
5966
5514
5635
5762
6866
6977
6527
6647
6763
6877
6988
5539
5658
6775
5888
5599
5551
5670
6786
5899
6010
40
41
42
43
44
6021
6128
6232
6336
6436
6031
6138
6243
6346
6444
6042
6149
6253
6356
6464
6053
6160
6263
6365
6464
6064
6170
6274
6376
6474
6076
6180
6284
6386
6484
6086
6191
6294
6396
6493
6096
6201
6304
6405
6503
6107
6212
6314
6416
6613
6117
6222
6325
6426
6622
45
46
47
48
49
6632
6628
6721
6812
6902
6542
6637
6730
6821
6911
6551
6646
6739
6830
6920
6661
6666
6749
6839
6928
6671
6666
6768
6848
6937
6580
6676
6767
6867
6946
6590
6684
6776
6866
6955
6599
6693
6786.
6876
6964
6609
6702
6794
6884
6972
6618
6712
6803
6893
6981
50
51
52
63
54
6990
7076
7160
7243
7324
6998
7084
7168
7261
7332
7007
7093
7177
7269
7340
7016
7101
7186
7267
7348
7024
7110
7193
7276
7366
7033
7118
7202
7284
7364
7042
7126
7210
7292
7872
7060
7135
7218
7300
7380
7059
7143
7226
7308
7388
7067
7162
7235
7316
7396
LOGABITHMS.
/'
271
N
0
1
2
3
4
6
6
7
8
0
55
56
57
58
59
7404
7482
7669
7634
7709
7412
7490
7666
7642
7716
7419
7497
7674
7649
7723
7427
7506
7582
7667
7731
7435
7513
7689
7664
7738
7443
7620
7597
7672
7745
7451
7528
7604
7679
7752
7459
7536
7612
7686
7760
7466
7643
7619
7694
7767
7474
7551
7627
7701
7774
60
61
62
63
64
7782
7863
7924
7993
8062
7789
7860
7931
8000
8069
7796
7868
7938
8007
8076
7803
7876
7945
8014
8082
7810
7882
7962
8021
8089
7818
7889
7959
8028
8096
7825
7896
7966
8036
8102
7832
7903
7973
8041
8109
7839
7910
7980
8048
8116
7846
7917
7987
8065
8122
6«
66
67
68
69
8129
8195
8261
8326
8388
8136
8202
8267
8331
8396
8142
8209
8274
8338
8401
8149
8215
8280
8344
8407
8156
8222
8287
8361
8414
8162
8228
8293
8357
8420
8169
8235
8299
8363
8426
8176
8241
8306
8370
8432
8182
8248
8312
8376
8439
8189
8264
8319
8382
8445
70
71
72
73
74
8461
8613
8673
8633
8692
8467
8619
8679
8639
8698
8463
8626
8685
8646
8704
8470
8531
8591
8661
8710
8476
8537
8697
8057
8716
8482
8643
8603
8663
8722
8488
8649
8609
8669
8727
8494
8665
8615
8675
8733
8500
8561
8621
8681
8739
8606
8567
8627
8686
8745
75
76
77
78
79
8761
8808
8866
8921
8976
8766
8814
8871
8927
8982
8762
8820
8876
8932
8987
8768
8826
8882
8938
8993
8774
8831
8887
8943
8998
8779
8837
8893
8949
9004
8785
8842
8899
8954
9009
8791
8848
8904
8960
9015
8797
8854
8910
8965
9020
8802
8869
8915
8971
9025
80
81
82
83
84
9031
9086
9138
9191
9243
9036
9090
9143
9196
9248
9042
9096
9149
9201
9263
9047
9101
9154
9206
9268
9063
9106
9169
9212
9263
9058
9112
9166
9217
9269
9063
9117
9170
9222
9274
9069
9122
9176
9227
9279
9074
9128
9180
9232
9284
9079
9133
9186
9238
9289
85
86
87
88
89
9294
9346
9396
9446
9494
9299
9360
9400
9460
9499
9304
9356
9405
9465
9504
9309
9360
9410
9460
9509
9315
9365
9416
9465
9513
9320
9370
9420
9469
9518
9325
9375
9425
9474
9623
9330
9380
9430
9479
9528
9336
9385
9435
9484
9633
9340
9390
9440
9489
9538
90
91
92
93
94
9642
9690
9638
9686
9731
9647
9596
9643
9689
9736
9562
9600
9647
9694
9741
9557
9605
9652
9699
9745
9562
9609
9667
9703
9750
9566
9614
9661
9708
9754
9571
9619
9666
9713
9759
9576
9624
9671
9717
9763
9581
9628
9676
9722
9768
9586
9633
9680
9727
9773
95
96
97
96
99
9777
9823
9868
9912
9966
978^
9827
9872
9917
9961
9786
9832
9877
9921
9966
9791
9836
9881
9926
9969
9795
9841
9886
9930
9974
9800
9846
9890
9934
9978
9806
9850
9894
9939
9983
9809
9864
9899
9943
9987
9814
9869
9903
9948
9991
9818
9863
9908
9952
9996
272 ALGEBRA.
the column in which 9736 stands is seen 1. Therefore,
write 941, and insert the decimal point as the character-
istic directs. That is, the number required is 94.1.
Suppose it is required to find the number of which the
logarithm is 3.7936.
Look for 7936 in the table. It cannot be found, but the
two adjacent mantissas between which it lies are seen to
be 7931 and 7938 ; their difference is 7, and the difference
between 7931 and 7936 is 6. Therefore, f of the differ-
ence between the numbers corresponding to the mantissas,
7931 and 7938, must be added to the number corresponding
to the mantissa 7931.
The number corresponding to the mantissa 7938 is 6220.
The number corresponding to the mantissa 7931 is 6210.
The difference between these numbers is 10,
and 6210 + ^ of 10 = 6217.
Therefore, the number required is 6217.
Suppose it is required to find the number of which the
logarithm is 7.3882 — 10.
Look for 3882 in the table. It cannot be found, but the
two adjacent mantissas between which it lies are seen to
be 3874 and 3892 ; their difference is 18, and the difference
between 3874 and 3882 is 8. Therefore, ^ of the differ-
ence between the numbers corresponding to the mantissas,
3874 and 3892, must be added to the number corresponding
to the mantissa 3874.
The number corresponding to the mantissa 3892 is 2450.
The number corresponding to the mantissa 3874 is 2446.
The difference between these numbers is 10,
and 2440 + 1 of 10 = 2444.
Therefore, the number required is 0.002444.
l06abithms. 273
Exercise 112.
Find logarithms of the following numbers ?
1. 60. 6. 3780. 11. 70633. 16. 877.08.
2. 101. 7. 54327. 12. 12028. 17. 73.896.
3. 999. 8. 90801. 13. 0.00987. 18. 7.0699.
4. 9901. 9. 10001. 14. 0.87701. 19. 0.0897.
6. 5406. 10. 10010. 16. 1.0001. 20. 99.778.
Find antilogarithms to the following logarithms :
21. 4.2488. 26. 4.7317. 29. 9.0410 — 10.
22. 3.6330. 26. 1.9730. 30. 9.8420 — 10.
23. 2.5310. 27. 9,8800 — 10. 31, 7.0216-10.
24. 1.9484. 28. 0.2787. 32. 8.6580 — 10.
Ex. Find the product of 908.4 X 0.05392 X 2.117.
log 908.4 = 2.9583
log 0.06392 = 8.7318 -10
log 2.117=0.3267
2.0168 = log 103.7. Ans.
Find by logarithms the following products :
33. 948.76X0.043875. 35. 830.75x0.0003769.
34. 3.4097X0.0087634. 36. 8.4395x0.98274.
317. When any of the factors are negative, find their
logarithms without regard to the signs ; write the letter n
after the logarithm that corresponds to a negative number.
K the number of logarithms so marked is odd, the product
is negative; if even, the product is positive.
274
ALGEBRA.
Find the products of :
37. 7564 X(— 0.003764). 39. —5.840359 X (—0.00178).
38. 3.7648 X(- 0.083497). 40. —8945.07X73.846.
, 8.3709 X 834.637
Ex. Find the quotient of
7308.946
log 8.3709 = 0.9227
log 834.637 = 2.9215
colog 7308. 946 = 6.1362-10 J .
9.9804 - 10 = log 0.9668. Aim.
Find the quotients of :
. 70654
•iX.
54013
42.
58706
93078
43.
8.32165
0.07891
44.
65039
90761
AK
7.652
— 0.06875
46.
47.
48.
49.
50.
0.07654
83.947 X 0.8395
7564 X 0.07643
8093 X 0.09817'
89 X 753 X 0.0097
36709X0.08497
413 X 8.17 X 3182
915 X 728 X 2.315*
21^ X (- 6.12) X(— 2008)
365 X(- 531) X 2.576
Ex. Find the cube of 0.0497.
log 0.0497 = 8.6964 - 10
3
6.0892 - 10 = log 0.0001228. Am,
Find by logarithms :
51. 6.05^ 55. 0.78765«
52. 1.051^ 56. 0.691'*^.
53. 1.1768*. 57. (^f.)".
54. 1.3178^«. 68. {\\y. '
59. (lOf)^
60. (H)«.
61- (Mi)'.
62. (Tj^y"^.
63. (3§J)*".
64. (1A)H
65. (8J)H
66. (5f|)«-^^
LOGABITHMS. 275
Ex. Find the fourth root of 0.008^62.
log 0.00862= 7.9365-10
30. -30
4)37.9365 -- 40
9.4839 - 10 = log 0.3047. Ana.
Find by logarithms :
67. 7i 70. 8379A. 73. 0.176438. 76. (j^;^^)*.
68. 11*. 71. 906.80*. 74. 2.5637A. 77. (9Ji)i
69. 783*. 72. 8.1904*. 75. (^H)*. 78. (11^)*-
Find by logarithms the values of :
JO.OO754332 X 78.343 X 8172.4* X 0.00052
79.
80
81
82
83.
84.
85.
4
■4
64285^ 154.27^ X 0.001 X 586.79*
15.832^ X 5793.6* X 0.78426
0.000327* X 768.942 X 3015.3 X 0.007*'
'7.1895 X 4764.22 X 0.00326*
' \0.00048953 X 457» X 5764.4^
5/3.1416 X 4771.21 X 2.7183*
■4.
4
30.103* X 0.4343* X 69.897*
7 /0.032712 X 53.429 X 0.77542*
32.769 X 0.000371*
3/732.0562 X 0.0003572* X 89793
42.2798^ X 3.4574 X 0.0026518*
8/7932 X 0.00657 X 0.80464
4
0.03274 X 0.6428
gg . . . .1206 X VO.13274 X 0.057389
VO.43468 X 17.385 X VO.0096372
r 3.075526^ X 5771.2* X 0.0036984* X 7.74 -j I
U 72258X327.93^X86.97* J '
276 ALGEBBA.
318. Since any positive number other than 1 may be
taken as the base of a system of logarithms, the following
general proofs to the base a should be noticed.
I. The logarithm of the product oftv>o or more numbers is
equal to the sum of the logarithms of the numbers.
For, let m and n be two Dumbers, and x and y their logarithms.
Then, by the definition of a logarithm, m = a^, and n — aff.
Hence, my. n^^o^xw^a^+y.
.-. log (m X n) = X + y,
= log m + log u.
In like manner, the proposition may be extended to any number
of factors.
II. The logarithm of a quotient is equul to the logarithm
of the dividend minus the logarithm of the divisor.
For, let m and n be two numbers, and x and y their logarithms.
Then m = a*, and n = a}f.
Hence, m-^n = a'^-^aM = a*-^.
.-. log (m -f n) = X — y,
= log m — log n.
From this it follows that log — = log 1 — log m.
But, since log 1 = 0, log — = — log m.
III. The logarithm of a power of a number is equal to
the logarithm of the number multiplied by the exponent of
the power.
For, let X be the logarithm of m.
Then m^a^,
and mp = (a'')p = ai«.
.'. log mP = px,
= p log m.
LOGARITHMS. 277
ly. The logarithm of the root of a number is eqtud to the
logarithm of the number divided by the index of the root.
For, let X be the logarithm of m.
Then m = a*,
1 1 X
and m»" = (a*)^ = a^
, 1 X log m
••. log m*- = - = —2 — •
r r
319. An exponential equation, that is, an equation in
which the exponent is the unknown quantity, is easily
solved by logarithms.
For, let a'^ = m.
Then log a^ = log w.
.'. X log a = log m,
log m
log a
Ex. Mnd the value of x in 81^ = 10.
81» = 10,
_ log 10
* log 81*
.-. log X = log log 10 4- colog log 81,
= 0 + 0.7103 - 10.
.•.x = 0.624.
320. Logarithms of numbers to any base a may be con-
verted into logarithms to any other base b by dividing the
computed logarithms by the logarithm of b to the base a.
For, let log m = ^ to the base b,
and log & = X to the base a.
Then m = bv, and b = a*.
.-. m = (a*)y = a*y.
.•. log m (to base a) = xy = log b (to base a) X log m (to base b),
.-. log m (to base 6) = '^|f^^^-
This is usually written, log6 m = . y ^'
CHAPTEE XX.
Katio, Proportion, and Variation.
321. The relative magnitude of two numbers is called
their ratio, when expressed by the fraction which the first
is of the second.
Thus, the ratio of 6 to 3 is indicated by the fraction f , which is
sometimes written 6 : 3.
322. The first term of a ratio is called the antecedent,
and the second term the consequent When the antecedent
is equal to the consequent, the ratio is called a ra^io of
equality; when the antecedent is greater than the conse-
quent, the ratio is called a ratio of greater inequalitt/; when
less, a ratio of less inequality,
323. When the antecedent and consequent are inter-
changed, the resulting ratio is called the inverse of the
given ratio.
Thus, the ratio 3 : 6 is the inverse of the ratio 6 : 3.
324. The ratio of two quantities that can be expressed
in integers in terms of a common unit is equal to the ratio
of the two numbers by which they are expressed.
Thus, the ratio of $9 to $11 is equal to the ratio of 9 : 11 ; and the
ratio of a line 2f inches long to a line 3f inches long, when both are
expressed in terms of a unit ^^ of an inch long, is equal to the ratio
of 32 to 45.
325. Two quantities different in kind can have no ratio,
for then one cannot be a fraction of the other.
RATIO. 279
326. Two quantities that can be expressed in integers
in terms of a common unit are said to be commensurable.
The common unit is called a common measure, and each
quantity is called a multiple of this common measure.
Thus, a common measure of 2^ feet and 3f feet is J of a foot, which
is contained 15 times in 2^ feet, and 22 times in 3f feet. Hence, 2^
feet and 3|- feet are multiples of ^ of a foot, 2^ feet being obtained
by taking ^ of a foot 16 times, and 3f by taking } of a foot 22 times.
327. When two quantities are incommensurable, that is,
have no common unit in terms of which both quantities
can be expressed in integers, it is impossible to find a frac-
tion that will indicate the exact value of the ratio of the
given quantities. It is possible, however, by taking the
unit sufficiently small, to find a fraction that shall differ
from the true value of the ratio by as little as we please.
Thus, if a and h denote the diagonal and side of a square,
Now^^= 1.41421356 , a value greater than 1.414213, but less
than 1.414214.
If, then, a milliorUh part of h be taken as the unit, the value of the
ratio ^ lies between IJJHH ^^^ rHMJJ» ^^^ therefore differs from
either of these fractions by less than y^T^i^^^.
By carrying the decimal farther, a,fraction may be found that will
differ from the true value of the ratio by less than a billionth, trU-
lionth, or any other assigned value whatever.
328. Expressed generally, when a and b are incommen-
surable, and b is divided into any integral number (n) of
equal parts, if one of these parts be contained in a more
than m times, but less than m + 1 times, then
7 > — > but < :
b n n
- ^ a , . , ^ m - ??t + 1
that IS, the value of - lies between — and •
' b n n
280 ALGEBRA.
The error, therefore, in taking either of these values for
a . 1 1
7- IS < — But by increasing n indefinitely, - can be made
to decrease indefinitely, and to become less than any as-
signed value, however small, though it cannot be made
absolutely equal to zero.
329. The ratio between two incommensurable quantities
is called an incommensurable ratio.
330. As the treatment of Proportion in Algebra depends
upon the assumption that it is possible to find fractions
which will represent the ratios, and as it appears that no
fraction can be found to represent the exact value of an
incommensurable ratio, it is necessary to show that two
incommensurable ratios are equal if their true values always
lie between the same limits, however little these limits differ
from each other.
Let a : b and cdhe two incommensurable ratios.
Suppose the true values of the ratios a : h and c.d lie between
— and . Then the difference between the true values of these
n n J 1
ratios is less than -, however small the value of - may be. § 328.
^ n ** 1
But since - can be made to approach zero at pleasure, - can be
n n
made less than any assumed difference between the ratios.
Therefore, to assume any difference between the ratios is to assume
it possible to find a quantity that for the same value of - shall be both
1 ""
greaJter and less than - ; which is a manifest absurdity.
Hence, a\h^ cid.
331. It will be well to notice that the word limit means
a fixed value from which another and variable value may
be made to differ by as little as we please ; it being impos-
sible, however, for the difference between the variable
value and the limit to become absolutely zero.
BATIO. 281
2. A rath vnll not he altered if both its terms be muUir
plied by the same number.
For the ratio a : 6 is represented by - » the ratio maimh ]& repre-
sented by — r ; and since — r = t » ^'.maimb^aiK
"" mb' rnb b
333. A ratio will be altered if different multipliers of its
terms be taken; and vnll be increased or diminished accord-
ing as the multiplier of the antecedent is greater or less than
that of the consequent.
For, ma : nb will be > or < a : 6
,. ma . ^ ^a / na\
i«cordmga8 lift » > »' < &(= ■^)
as
ma is > or < na,
as
m is > or < n.
334. A ratio of greater inequality will be diminished^
and a ratio of less inequality increased by adding the sams
number to both its terms.
For, a4-x:6+iBis>or<a:6
a + X . a
according as . , is > or < -,
0 I X 0
as a6 4- 6x is > or < a& + ox,
as 6x is > or < ax,
as 6 is > or < a.
335. A ratio of greater inequality will be increased, and
a ratio of less inequality diminished, by subtracting the same
number from, both its terms.
For, a—X',h—x will be > or < a : 6
according as
as
a — x . ^ ^a
^_^is>or<^,
as
a6 — 6x is > or < a& —
ox,
as
ox is > or < 6x,
as
a is > or < 5.
282 AL6EBBA.
336. Ratios are compounded by taking the product of
the fractions that represent them.
Thus, the ratio compounded oiaih and c : d is found by taking the
productof^and- = — •
The ratio compounded of a : 6 and a : 5 is the duplicate ratio a^ : &^,
and the ratio compounded of a : 6, a:b, and a : 6 is the triplicate ratio
337. Eatios are compared by comparing the fractions
that represent them.
Thus, a:6is>or<c:d
according as - is > or < - »
ad . ^ ^bc
-w>or<^.
as od is >• or < 6c.
Exercise 113.
1. Write down the ratio compounded of 3:5 and 8:7.
Which of these ratios is increased, and which is
diminished by the composition?
2. Compound the duplicate ratio of 4 : 15 with the tripli-
cate of 5 : 2.
3. Show that a duplicate ratio is greater or less than its
simple ratio according as it is a ratio of greater or
less inequality.
4. Arrange in order of magnitude the ratios 3:4; 23 : 25 ;
10 : 11 ; and 15 : 16.
5. Arrange in order of magnitude
a-\-b:a — b and a^ + b^ : a^ — b^, if a>b,
Pind the ratios compounded of :
6. 3:5; 10:21; 14:15. 7. 7:9; 102:105; 15:17.
RATIO. 283
a^+ax + a? , a^ — ax + a:^
«• ^3-,., + ^.,^ and ^^^ -
0^ — 6a; x^ — 5x
10. a + *:a-5; a«+ft« : (a + *)'; (a* - ^* : a* - ft*.
11. Two numbers are in the ratio 2 : 3, and if 9 is added to
- each, they are in the ratio 3 : 4. Find the numbers.
(Let 2x and Sx represent the numbers.)
12. Show that the ratio a : ft is the duplicate of the ratio
a + c : ft + c, if c^= aft.
13. Find two numbers in the ratio 3 : 4, of which the sum
is to the sum of their squares in the ratio of 7 to 50.
14. If five gold coins and four silver ones are worth as much
as three gold coins and twelve silver ones, find the
ratio of the value of a gold coin to that of a silver
one.
16. If eight gold and nine silver coins are worth as much
as six gold and nineteen silver coins, find the ratio
of the value of a silver coin to that of a gold one.
16. There are two roads from A to B, one of them 14
miles longer than the other ; and two roads from
B to C, one of them 8 miles longer than the other.
The distance from A to B is to the distance from B
to C, by the shorter roads, as 1 to 2 ; by the longer
roads, as 2 to 3. Find the distances.
17. What must be added to each of the terms of the ratio
m : n, that it may become equal to the ratio piq?
18. A rectangular field contains 5270 acres, and its length
is to its breadth in the ratio of 31 : 17. Find its
dimensions.
284 ALGEBKA.
Proportion.
338. An equation consisting of two equal ratios is called
a proportion ; and the terms of the ratios are called pro-
portionals.
339. The algebraic test of a proportion is that the two
fractions which represent the ratios shall be equal.
A C
Thus, the ratio a : h will be equal to the ratio c:d if - = - ; and
the four numbers a, b, c, d are called proportionals, or are said to be
in proportion.
340. If the ratios a : b and c : d form a proportion, the
proportion is written a:b=^c:d, and is read
the ratio of a to ft is equal to the ratio oi do d;
or a:b::c:d, and is read
a is to ft in the same ratio as c is to d.
The first and last terms, a and d, are called the extremes.
The two middle terms, ft and c, are called the means.
341. When four numbers are in proportion^ the product
of the extremes is equal to the product of the means.
For, if a : 6 : : c : d,
XV a c
then r = J •
0 a
By multiplying by M, ad = 6c.
342. If the product of two numbers is equal to the prodvM
of two others, either two may be made the extremes of a pro-
portion a7id the other two the means.
For, if ad = 6c,
by dividing by 6d, i:j = i^i '
a c
.'. a:6: :c:d.
PBOPORTION. 285
343. The equation ad=bc gives
be - ad
*=^' *=T'
SO that an extreme may be found by dividing the product
of the means by the other extreme ; and a mean may be
found by dividing the product of the extremes by the other
mean.
344. If four quantities, a, b^ c, d, be in proportion, they
will be in proportion by :
I. Inversion.
That iS; b will be to a as (2 is to c.
For,
if
a:b::c:dy
then
a c
and
or
b_d^
a c
.•.b;a::d:c.
345. II. Composition.
That is, a-^b will be to ^ as c + rf is to c?.
For, if a:b::c:d,
then
a_ c
b" d'
and ^■^1 = 5+1»
a + 6 c + d
or
b d
, a + b:b::c-\- did.
346. III. Division.
That is, a — b will be to J as c — c? is to d.
For, if aibr.cidj
« <^
then ^=^»
ALGEBBA.
a — b c — d
»'. a ^ b : b :: c — d : d.
347. lY. CompoBition and Division.
That is, a + ft will be to a — ft as c + ef is to c — cL
For, from II., — r— = —^ >
,, TTT a — b c — d
and from III., —7 — = — r—
0 a
By dividmg, : = --
a — bc — d
,\a+b:a^b::c-\-d: c — d.
348. When the four quantities a, ft, c, e? are all of the
same kindy they will be in proportion by :
V.
Alternation.
That is,
a will be to c as ft is
tod.
For, if a:l
:: C
:d,
then
a _
b~
c
'd
By multiplying by - »
c
ab__
be
be
or
.6
d
.-. a
: ci:
bid.
349. From the proportion a:c::b:d may be obtained
by:
VL Composition. a-\-c:c::b-\-d:d.
VII. Division. a — c:c::h--did.
VIII. Composition and Division, a+c.a— c::ft-j-<f;<f— df.
PBOPOBTION. 287
350. In a series of equal ratios, the sum of the antecedents
is to the sum of the consequents as any antecedent is to its
consequent,
r may be put for each of these ratios.
Then -=r, ^=r,y = r,;^ = r.
/. a = 6r, c = dr, e — fr^ gr = hr,
.-. a + c + c + fir = (6 4- d +/+ h) r.
a4- c + e + flr __ _ a
*'*6 + d+/ + A""*'""6*
.'.a+c + e + gxb^d+f + hiiaih.
In like manner it may be shown that
ma + nc + pe ■\- qg : mh + nd -\- pf + qh \i aih.
351. If a, b, c, d be in continued proportion, that is, if
a:h-=hic=^cidy then will a : c = a^ : ^^ and a : e?= a' ; 6^
or
For,
a_6__ c
6 c"5*
Hence,
6 c 6 6
a a2
.\aic-a^: 62.
So
6 c d 6 6 6
or
a_a^
d" 68*
.•.a:d=a8:68.
352. If a, ft, c are proportionals, so that a : 5 : : ft : c, then
ft is called a mean proportional between a and c, and c is
called a third proportional to a and ft.
288 AL6EBBA.
It a:b: :b:c, then b = y/ae.
For, if
a:&::&:e,
then
a b
6 = c'
and
bfi=ac.
.-. b = y/ac.
aib:
:e:d,
e:f:
:g:h,
k:l:
imirij
b~~ d' f
A 2 n
353. TAe products of the corresponding terms of two or
more proportions are in proportion.
For, if
and
then
Hence, by finding the product of the left members, and also of the
right members of these equations,
aek __ cgm
bfl" dhn
.'. aek : bfl :: cgm : dhn.
354. Like powers, or like roots, of the term^ of a propor-
tion are in proportion.
For, if aibiicidy
then 7 = 1'
b d
By raising both sides to the nth power.
£11.— £!!.
By extracting the nth root,
1 1
a«_ ^^
1 ~" i'
1111
.•. flp*:6»:: c«:(f».
PBOPORTION. 289
366. If two qtutrUUies be increased or diminished by like
parts of each, the results toiU be in the same ratio as the
quantities themselves.
a __\ n/
For,
h
(>*?)
b
a±'-a
n
,', a :h :: a ± — a :b ±^ b.
n n
366. The laws that have been established for ratios
should be remembered when ratios are expressed in their
fractional form.
« Solve: -,__-^ = -,-j_^
and this equation is satisfied, when x = 0 ;
2x^ 1 1
or, dividing by—. ^^ipj = ^3^-
.•.x = h
(2) li a:b::c:dy show that
a^-^-ab : b^ — ab::(^-]-cd:d* — cd.
If
a c
b=r
^x, a+b c + d . o^«
then 7 = -^ § 347.
a— 6 c—d
and
that is,
a _ c
-6" -d*
, « x^=^X^-±^; §363.
—6 a—b — d c—d
a^-hab d^ + cd
l^-ab d^-cd
or a^+aJbi^-aJbiic^ + cdiO^-cd.
290 ALGEBRA.
(3) When a:b::c:d, and a is the greatest term, show that
a + d\s greater than b + c.
Since
a,
h'
= -1 and a>c,
then
Also,
since
b>d,
a—b c—d
b = d '
and
then
By adding to this,
h>d,
a — b':>c — d.
6 + d = 6 4- d,
a + d>6 + c.
Exercise 114.
lia:b::c:
rf, prove
that:
1.
ma :nb::
mc : nd.
4.
2.
Sa + b:b
:3c + ^
',d.
5.
3.
a + 2b:b
:c + 2(i
:d.
6.
§346.
:b^::(^:d^
6. aia-j-b: :c:c-\'d,
6. a: a — b::c:c — d.
7. ma-\-nb:ma — rib : : mc -\- nd : mc — nd.
8. 2a + 3b:Sa — U::2c + 3d:Sc — 4:d.
9. m^^ + ^ • mb^-\- nd^ \\a^: b^,
10. ma^ + riaft + J9&^ : mc^ + ?icc? +i3c^' : : ft* : cP.
If a : ft : : 5 : c, prove that :
11. a + ft:ft + c::a:ft. 12. a^ + aft : ft* + ftc: :a:c.
13. a;c::(a + ft)2:(ft + c)2.
14. When a, ft, and c are proportionals, and a the greatest,
show that a + c > 2ft.
16. If — j-^ = = , and x, y, z are unequal, show
that Z + w + n = 0.
tBOPORTIOK. 291
16. Findx wlienaj + 6:2aj — 3::6aj + l:3a; — 3.
17. Find x when x + a:2x — biiSx + b-Ax — a.
18. Find x when Va + V^: Va — V^; :a:b.
19. Find cc and y when x:27 ::y:9, and a; : 27 ; ; 2 : a — y.
20. Find « and y when aj + y + 1 :aj + y+2 : : 6 : 7, and
wheny + 2«:y — 2a;; :12ic + 6y — 3: 6y — 12a; — 1.
21. Find x when a:^^4a;+2: a;^*— 2a;— 1 ; : a;*— 4a;: a;*— 2a;— 2.
22. A railway passenger observes that a train passes him,
moving in the opposite direction, in 2 seconds ; but
moving in the same direction with him, it passes him
in 30 seconds. Compare the rates of the two trains.
23. A and B trade with different sums. A gains $200 and
B loses $60, and now A's stock : B's : : 2 : ^. But,
if A had gained $100 and B lost $85, their stocks
would have been as 15 : 3^. Find the original stock
of each.
24. A quantity of milk is increased by watering in the
ratio 4 : 5, and then three gallons are sold ; the
remainder is mixed with three quarts of water, and
is increased in the ratio 6 : 7. How many gallons
of milk were there at first ?
26. In a mile race between a bicycle and a tricycle their
rates were as 5 : 4. The tricycle had half a minute
start, but was beaten by 176 yards. Find the rate
of each.
26. The time which an express-train takes to travel 180
miles is to that taken by an ordinary train as 9 : 14.
The ordinary train loses as much time from stopping
as it would take to travel 30 miles ; the express-train
loses only half as much time as the other by stop-
ping, and travels 15 miles an hour faster. What
are their respective rates ?
292 ALGEBRA.
27. A line is divided into two parts in the ratio 2 : 3; and
into two parts in the ratio 3:4; the distance be-
tween the points of section is 2. Find the length
of the line.
28. A railway consists of two sections ; the annual expendi-
ture on one is increased this year 5^, and on the
other 4^, producing on the whole an increase of
^A%* Compare the amount expended on the two
sections last year, and also this year.
29. When a, b, c, d are proportional and unequal, show
that no number x can be found such that a-\-Xy
h-^rXyC-^rXyd-^rx shall be proportionals.
Variation.
367. Two quantities may be so related that, when one
has its value changed, the other will, in consequence, have
its value changed.
Thus, the distance travelled in a certain time will be doubled if the
rate be doubled. The time required for doing a certain quantity of
work will be doubled if only haV the number of u)orhnen be employed.
368. Whenever it becomes necessary to express the
general relations of certain kinds of quantities to each
other, without confining the inquiry to any particular values
of these quantities, it will usually be sufficient to mention
two of the terms of a proportion. In all such cases, how-
ever, four terms are always implied.
Thus, if it be said that the weight of water is proportional to its
volume, or varies as its volume, the meaning is, that one gallon of
water is to any number of gallons afi the weight of one gallon is to the
weight of the given number of gallons.
VARIATION. 293
359. Quantities used in a general sense, as distance, time,
weight, volume, to which particular values may be assigned,
are denoted by capital letters, -4, By C, etc. ; while assigned
values of these quantities may be denoted by small letters,
a, b, c, etc. The letters A, B, (7 will be understood to
represent any numerical values that may be assigned to
the quantities ; and when two such letters occur in an
expression they will be understood to represent any corre-
sponding numerical values that may be assigned to the two
quantities.
360. When two quantities A and B are so connected
that their ratio is constant, that is, remains the same for
all corresponding values of A and J5, the one is said to vary
as the other ; and this relation is expressed hy A <^ B (read
A varies as B),
Thus, the area of a triangle with a given base varies as its altitude ;
for, if the altitude be changed, the area will be change^ in the same
ratio.
If this constant ratio be denoted by m, then — = m, or
A = mB,
From this equation m may be found when two corre-
sponding values of A and B are known.
361. When two quantities are so connected that if one
be changed in any ratio, the other will be changed in the
inverse ratio, the one is said to vary inversely as the other.
Thus, the time required to do a certain amount of work varies
inversely as the number of workmen employed ; for, if the number
of workmen be doubled, halved, or changed in any ratio, the time
required will be halved, doubled, or changed in the inverse ratio.
362. If A varies inversely as J5, two values of A have to
each other the inverse ratio of the two corresponding values
of J5 ; OT a:a' ::b' :b, that is, ab = a'b\
294 ALGEBRA.
Hence, the product AB is constant, and may be denoted
by m. That is, AB = m.
If any two corresponding values of A and B are known,
the constant m may be found.
The equation AB=m may be written -4 = it» and as m
is constant, A is said to vary as the reciprocal of B, or
"^ B
363. The two equations,
A = mB (for direct variation),
A = rr (for inverse variation),
furnish the simplest method of treating Variation.
If A = mBC, A is said to y2ltj jointly as B and C,
If ^ = -r7? A is said to vary directly as B and inversely
as a ^
364. The following results are to be observed :
L If ^oc B and ^oc (7, then A^ C.
For A = mJB, where m is constant,
and B = nC, where n is constant
••. ^ = mnC,
.*. A cc C, since 7nn is constant.
In like manner, it Ace B and JB « 7=j» then Acc-r^*
11. If ^oc (7andJ5oc (7, then ^ ± J5 oc (7,and ^sTABcc a
For ^ = mC, where m is constant,
and B=nC, where n is constant.
.\A±B=(m±n) C,
.*. ^ ± B « C, since m ± n is constant.
Also, Vab= yJmCx nC- ^TiimC^ = C Vmn.
.*. VAB « C, since '^mn is constant.
VARIATION. 295
III. If ^ oc ^ and (7 oc 2), then AC « BD.
For A = mJ5, where m is constant,
C = nD, where n is constant.
,\ AC = cc BD, since mn is constant.
IV. If ^oc^then^»oc^«.
For A = mB, where m is constant.
.*. -4» X J5», since m» is constant.
V. If ^ Qc ^ when C is unchanged, and A<^ C when B
is unchanged, then A<x: BC when both J5 and C change.
For A = mJB, when B varies and C is constant.
Here, m is constant and cannot contain the variable B.
.'. ^ mM«< contain B, but no other power of B.
Again, A = nC, when C varies and J5 is constant.
Here, n is constant and cannot contain the variahle C,
.'. A must contain C, but no other power of C.
Hence, A contains both B and C, but no other powers of B and
C, and therefore,
■^tt: = p, or -4 = pJBC, where p is constant.
••. Ace BC, since i> is constant.
In like manner, it may be shown that if A varies as each
of any number of quantities B, C, D, etc., when the rest
are unchanged, then when they all change, A QC BCD, etc.
Thus, the area of a rectangle varies as the base when the altitude
is constant, and as the altitude when the base is constant, but as the
product of the base and altitude when both vary.
The volume of a rectangular solid varies as the length when the
width and thickness remain constant ; as the width when the length
and thickness remain constant ; as the thickness when the length and
width remain constant ; but as the product of length, breadth, and
thickness when all three vary.
296 ALGEBRA.
(1) If A varies inversely as By and when A = 2 the corre-
sponding value of B is 36, find the corresponding
value of B when .4 = 9.
Here -4 = — i
or m = AB,
.-. m = 2 X 36 = 72.
And if 9 and 72 are substituted for ^ and m respectively in
72
the result is 9 = — •
.-. 9J5= 72.
.-. J5 = 8. ^7W.
(2) The weight of a sphere of given material varies as its
volume, and its volume varies as the cube of its
diameter. If a sphere 4 inches in diameter weighs
20 pounds, find the weight of a sphere 5 inches
in diameter.
Let W represent the weight,
V represent the volume,
D represent the diameter.
Then Wca V and Foe L^,
.: TToc i)3
Put W= ml)»,
then, since 20 and 4 are corresponding volumes of W and D,
20 = m X 64,
.•.m = |J = ^^.
.-. when D = 6, >r = i«\j of 126 = 39^^.
EXBRCISE 115.
1. If ^ oc By and A = 4: when B = 5y find A when jB = 12.
2. If -4 Qc By and when B = iy A = -^, find ^ when ^ = i.
3. If A varies jointly as B and (7, and 3, 4, 5 are simultane-
ous values of A, B, C, find A when ^= C = 10,
VARIATION. 297
4. It Aoc —y and B = 2, when .4 = 10, find the value of
X)
B when ^ = 4.
TO
5. If ^ oc — , and ^ = 4, and (7=3, when ^ = 6, find the
c
value of A when ^ = 5 and (7=7.
6. If the square of X varies as the cube of Y, and X=^S
when Y= 4, find the equation between X and F.
7. If the square of X varies inversely as the cube of Y,
and X=2 when F=3, find the equation between
Xand r.
8. If Z varies as X directly and F inversely, and if X= 3,
and Y= 4, when Z= 2, find the value of Z when
X=15and F=8.
9. If Ace B-{-c where c is constant, and if A = 2 when
^ = 1, and if ^=5 when ^=2, find A when J5=3.
10. The velocity acquired by a stone falling from rest
varies as the time of falling ; and the distance fallen
varies as the square of the time. If it is found
that in 3 seconds a stone has fallen 145 feet, and
acquired a velocity of 96f feet per second, find the
velocity and distance at the end of 5 seconds.
11. If a heavier weight draw up a lighter one by means
of a string passing over a fixed wheel, the space
described in a given time will vary directly as the
difference between the weights, and inversely as
their sum. If 9 ounces draw 7 ounces through 8
feet in 2 seconds, how high will 12 ounces draw 9
ounces in the same time ?
12. The space will vary also as the square of the time.
Find the space in Example 11, if the time in the
latter case is 3 seconds.
298 AL6EBBA.
13. Equal volumes of iron and copper are found to weigh
77 and 89 ounces respectively. Find the weight of
10^ feet of round copper rod when 9 inches of iron
rod of the same diameter weigh Sl^fty ounces.
14. The square of the time of a planet's revolution varies
as the cube of its distance from the sun. The
distances of the Earth and Mercury from the sun
being 91 and 35 millions of miles, find in days the
time of Mercury's revolution,
15. A spherical iron shell 1 foot in diameter weighs ^^
of what it would weigh if solid. Find the thickness
of the metal, it being known that the volume of a
sphere varies as the cube of its diameter.
16. The volume of a sphere varies as the cube of its diam-
eter. Compare the volume of a sphere 6 inches in
diameter with the sum of the volumes of 3 spheres
whose diameters are 3, 4, 5 inches respectively.
17. Two circular gold plates, each an inch thick, the diam-
eters of which are 6 inches and 8 inches respectively,
are melted and formed into a single circular plate
1 inch thick. Find its diameter, having given that
the area of a circle varies as the square of its
diameter.
18. The volume of a pyramid varies jointly as the area of
its base and its altitude. A pyramid, the base of
which is 9 feet square, and the height of which is
10 feet, is found to contain 10 cubic yards. What
must be the height of a pyramid upon a base 3 feet
square, in order that it may contain 2 cubic yards ?
CHAPTER XXL
Series.
366. A SUCCESSION of numbers which proceed according
to some fixed law is called a series; and the successive
numbers are called the terms of the series.
Thus, by executing the indicated division of :; , the series 1 +x
1 ■"" X
+ x2 4- x^ + is obtained, a series that has an unlimited number
of terms.
366. A series that is continued indefinitely is called an
infinite series ; and a series that comes to an end at some
particular term is called a finite series.
367. When x is less than 1, the more terms we take of
the infinite series 1-\-x-{-qi^-\-x^-\-.,..,, obtained by dividing
1 by 1 — X, the more nearly does their sum approach to the
value of :;
1 — x
1 13
Thus, if X = i, then _ = ^ = - , and the series becomes
l + i+ i + sV"^ » * ^^^ which cannot become equal to f how-
ever great the number of terms taken, but which may be made to
differ from f by as little as we please by increasing indefinitely the
number of terms.
368. But when x is greater than 1, the more terms we
take of the series l+« + ^^ + ^^ + the more does the
sum of the series diverge from the value of ;
the ser
1 + 3 + 9 + 27 + , a sum which dioerges more and more from — \y
Thus, if X = 3, then = :; ^ = *" o » *Dd the series becomes
1 — X 1 — o 2
300 ALGEBBA.
the more terms we take, and which may be made to increajse indefi-
nitely by increasing indefinitely the number of terms taken.
369. A series whose sum as the number of its terms is
indefinitely increased approaches some fixed finite value as
a limit is called a oonverg^ing series; and a series whose
sum increases indefinitely as the number of its terms is
increased, is called a diverg^ing series.
870. When x = 1, the division of 1 by 1 — a?, that is, of
1 by 0, has no meaning, according to the definition of divis-
ion; and any attempt to divide by a divisor that is equal
to zero leads to absurd results.
Thus, 8 + 4 = 8 + 4;
by transposing, 8 — 8 = 4 — 4;
or, dividing by 4 — 4, 2=1; a manifest absurdity.
871. When a: = l very nearly, then the value of ~
J. """ 35
will be very great, and the sum of the series 1 -\- x •\- a^ -\'
«'+ will become greater and greater the more terms we
take. Hence, by making the denominator 1 — x approach
indefinitely to zero, the value of the fraction may be
made to increase at pleasure.
372. If the symbol o be used to denote a number that
is less than any assignable number, and that may be con-
sidered to decrease without limit, not, however, becoming
0, and the symbol oo be used to denote a number that is
greater than any assignable number, and that may be
considered to increase "v^ithout limit, not, however, becom-
ing 00, then ^
= 00.
o —
a
In the same sense — = po^ where a represents any value
that may be assigned.
SERIES. 301
1 ^
373. If X in the fraction t^^ — be equal to 1, the mimer-
ator and denominator will each become 0, and the fraction
will assume the form — •
374. If, however, x in this fraction approach to 1 as its
limit, then the denominator 1 — x, inasmuch as it has some
value, even though less than any assignable value, may be
used as a divisor, and the result is 1 + x -{- a^ -\- x^ '\' x\
Hence, it is evident that though both terms of the fraction
become smaller and smaller as 1 — x approaches to 0, still
the numerator becomes more and more nearly five times
the denominator.
It may be remarked that when the symbol J is obtained for the
value of the unknown number in a problem, the meaning is that the
problem has no dqfinUe solution, but that its conditions are satisfied
if any value whatever be taken for the required number ; and if the
symbol J, in which a denotes any assigned value, is obtained for the
value of the unknown number, the meaning is that the conditions of
the problem are impossible.
376. The number of different series is unlimited, but the
only kind of series that will be considered at this stage of
the work are Arithmetical, Geometrical, and Harmonical
Series.
Arithmetical Series.
376. A series in which the difference between any two
adjacent terms is equal to the difference between any other
two adjacent terms, is called an Arithmetical Series or an
Arithmetical Progression.
377. The general representative of such a series will be
a, a-^-dy a'\'2d, a-^-Sd,,.,,,
in which a is the first term and d the common difference j
S02 ALGEBRA.
and the series will be increasing or decreasing according as
d is positive or negative.
878. Since each succeeding term of the series is obtained
by adding d to the preceding term, the coefficient of d will
always be 1 less than the number of the term, so that the
nth term = a-\-(n — l)d.
If the nth term is denoted by Z, this equation becomes
l = a + (n — l)d. (1)
879. The arithmetical mean between two numbers is the
number which stands between them, and makes with them
an arithmetical series.
380. If a and h denote two numbers, and A their arith-
metical mean, then, by the definition of an arithmetical
series,
A — a = ft — A.
381. Sometimes it is required to insert several arith-
metical means between two numbers.
If m = the number of means, then tti + 2 = n, the whole
number of terms ; and \im-\-2 is substituted for n in the
equation
l=a-\'(ii — V)dy
the result is / = a + (^ + 1) ^•
By transposing a, I — a = (7m + 1) d,
I — a
.\—-r-T = d. (3)
Thus, if it is required to insert six means between 3 and 17, the
Yj g
value of d is found to be = 2 ; and the series will be 3, 5, 7, 9,
11, 13, 16, 17.
SERIES. 303
382. If I denotes the last term, a the first term, n the
number of terms, d the common difference, and s the sum
of the terms, it is evident that
5= a +(a+d) + (a+2d)+ + (l-d) + Z, or
s= I j^li-d) + (l'^2d)+ + (a+d)+ a
.-. 2s=(a+l)+(a+l) +(a+l) + + (a+l) +(a+l)
= n (a 4" 0
.•.« = |(a + 0- (4)
383. From the two equations,
l=za + (n — l)d, (1)
« = |(« + 0^ (2)
any one of the numbers a, d, I, n, s may be found when
three are given.
Ex. Find n when d, I, s are given.
From (1), a =
From (2), a =
From (1), a=l—{n—l)d.
2 s — In
n
rwrn - , , , . , 2 8 — Zn
Therefore, l—(n — l)d =
.-. ln — dn'^ + dn = 2s — In.
,',dn^-(2l+,d)n= -2«.
.•.4(22n2- 0 + (2Z + d)2 = (2Z + (i)2- 8(fo,
.•.2dn- (2Z + d) = ± V(2 Z + d)2 - 8 (fo.
_.2^ + d±V(2Z + d)2-"8(to
•'•^ 2d
Note. The table on the following page contains the results of the
general solution of all possible problems in arithmetical series. The
student is advised to work these out, both for the results obtained
and for the praptice gained in solvmg literal equations in which the
unknown numbers are represented by other letters than x, y, z.
304
ALGEBBA.
No.
Given.
Required.
Results.
1
2
3
4
a d n
a d 8
a n 8
d n 8
I
l=a+(n-l)d.
Z=-id±V2(is+(a-id)2.
, 2a
1=1 a,
8 (n-l)d
*~n^ 2
6
6
7
8
a d n
a d I
a n I
d n I
8
s = in[2a+(n-l)(q.
l + a , l^-a^
*= 2 + 2d •
8=(l + a)l'
« = in[2Z-(n-l)d].
9
10
11
12
d n I
d n 8
d I 8
n I 8
a
a=Z-{n-l)d.
8 (n - 1) d
n 2
a = id±V(i + id)2-2d8.
28 ,
a = 1.
n
13
14
16
16
a n I
a n 8
a I 8
n I 8
d
n — 1
2(8-an).
n(n-l)
l^ - a2
^-28-/-a'
d-2(nZ-8).
n(n-l) .
17
18
19
20
a d I
ads
a I s
d I 8
n
n = —J- + 1.
a
d-2a±V(2a-d)2 + 8ds.
2d
28
n = z—. — •
l + a
2Z + d±V(2^ + d)2-8ds.
2d _J
SERIES. 305
Exercise 116.
1. Find the thirteenth term of 5, 9, 13..
ninth term of — 3, —1, 1
tenth term of — 2, — 5, —8
eighth term of a, a-\-3b, a + 6ft
fifteenth term of 1, f , {f
thirteenth term of —48, — 44, — 40..
2. The first term of an arithmetical series is 3, the thir-
teenth term is 55, Find the common difference.
3. Find the arithmetical mean between : (a.) 3 and 12 ;
(b,) — 5 and 17 ; (c.) a^ + ab-- b^ and a^ — ab + b\
4. Insert three arithmetical means between 1 and 19 ;
and four arithmetical means between — 4 and 17.
5. The first term of a series is 2, and the common differ-
ence -J. What term will be 10 ?
6. The seventh term of a series, whose common difference
is 3, is 11. Find the first term.
7. Find the sum of
5 4" 8 + 11 + to ten terms.
— 4 — 1 + 2 + to seven terms.
a 4" 4:a + 7a + to ^ terms.
§ + T^y + T^ + to twenty-one terms.
1 + 2^ + 4^ + to twenty terms.
8. The sum of six numbers of an arithmetical series is
27, and the first term is 1. Determine the series.
9. How many terms of the series — 5 — 2 + 1 + must
be taken so that their sum may be 63 ?
10. The first term is 12, and the sum of ten terms is 10.
Find the last term.
806 ALOEB&A.
11. The arithmetical mean between two numbers is 10,
and the mean between the double of the first and
the triple of the second is 27. Find the numbers.
12. Find the middle term of eleven terms whose sum is 66.
13^ The first term of an arithmetical series is 2, the com-
mon difference is 7, and the last term 79. Find the
number of terms.
14. The sum of 15 terms of an arithmetical series is 600,
and the common.difference is 5. Find the first term.
15. Insert ten arithmetical means between —7 and 114.
16. The sum of three numbers in arithmetical progression
is 15, and the sum of their squares is 83. Find the
numbers.
Let x — y,x,x+y represent the numbers.
17. Arithmetical means are inserted between 5 and 23, so
that the sum of the first two is to the sum of the last
two as 2 is to 5. How many means are inserted ?
18. Find three numbers of an arithmetical series whose
sum shall be 21, and the sum of the first and second
shall be J of the sum of the second and third.
19. Find three numbers whose common difference is 1,
such that the product of the second and third ex-
ceeds the product of the first and second by i,
20. How many terms of the series 1, 4, 7 must be taken,
in order that the sum of the first half may bear to
the sum of the second half the ratio 10 : 31 ?
21. A travels uniformly 20 miles a day ; B starts three
days later, and travels 8 miles the first day, 12 the
second, and so on, in arithmetical progression. In
how many days will B overtake A ?
22. A number consists of three digits which are in arith-
metical progression; and this number divided by
the sum of its digits is equal to 26 j but if 198 be
added to it, the digits in the units' and hundreds'
places will be interchanged. Find the number.
23. The sum of the squares of the extremes of four num-
bers in arithmetical progression is 200, and the sum
of the squares of the means is 136. What are the
numbers ?
24. Show that if any even number of terms of the series
1, 3, 5 is taken, the sum of the first half is to
the sum of the second half in the ratio 1 : 3.
26. A and B set out at the same time to meet each other
from two places 343 miles apart. Their daily jour-
neys were in arithmetical progression, A's increase
being 2 miles each day, and B's decrease being 6
miles each day. On the day at the end of which
they met, each travelled exactly 20 miles. Find
the duration of the journey.
26. Suppose that a body falls through a space of 16^^ feet
in the first second of its fall, and in each succeeding
second 32 J^ more than in the next . preceding one.
How far will a bodj^ fall in 20 seconds ?
27. The sum of five numbers in arithmetical progression
is 45, and the product of the first and fifth is | of
the product of the second and fourth. Find the
numbers.
28. If a full car descending an incline draw up an empty
one at the rate of 1^ feet the first second, 4^ feet
the next second, 7^ feet the third, and so on, how
long will it take to descend an incline 150 feet in
length? What part of the distance will the car
have descended in the first half of the time ?
308 ALGEBRA.
Geometrical Series.
384. A series is called a Geometrical Series or a Geomet-
rical Progression when each succeeding term is obtained by
multiplying the preceding term by a constant multiplier.
386. The general representative of such a series will be
a, ar, ar", ar», a7^.....,
in which a is the first term and r the constant multiplier
or ratio.
386. Since the exponent of r increases by 1 for every
term, the exponent will always be 1 less than the number
of the term ; so that the
nth term = ar*~*.
387. If the nth term is denoted by I, this equation
becomes
l=za7^\ (1)
388. The geometrical mean between two numbers is the
number which stands between them, and makes with them
a geometrical series.
389. If a and h denote two numbers, and G their geo-
metrical mean, then, by definition of a geometrical series,
^ = A
a g'_
.-. G = ^ab. (2)
890. Sometimes it is required to insert several geomet-
rical means between two numbers.
SERIES. 309
If w = the number of means, then m + 2 = », the whole
number of terms ; and if w + 2 is substituted for n in the
equation ;_-^,^i^
the result is Z = ar^+^
a ^ '
Thus, if it is required to insert three geometrical means between
3 and 48, the value of r is found to be
.•.r=2,
and the series will be 3, 6, 12, 24, 48.
391. If I denotes the last term, a the first term, n the
number of terms, r the common ratio, and s the sum of the
n terms, then
s = a + ar + ar* + ar^ -\- + ar'*~\
Multiply by r, rs == ar + ar* + ar* + + ai^~^ + ai^.
Therefore, by subtracting the first equation from the
second, y5_5=:^^_a,
or (r — 1) s = a (?•" — 1).
a(r" — 1)
392. When r is less than 1, this formula will be more
convenient if written
a(l — r")
1 — r
393. Since l=a7^-^,
then rl = ar",
and (4) may be written s= __ •
In working out the following results, the student will make use of
a (r» — 1)
the two equations, I = ar^-^ and 8 = — r-ITi '
310
AIiG13KA.
No.
T
2
3
4
Given.
Required.
Results.
a r n
a r B
a n B
r n 8
r
1(8-1)^-^ -a(8-a)*-i = 0.
a r n
a r I
a n I
r n I
rl-a
8= -•
10
11
12
r n I
r n B
r I 8
n I 8
r»— 1
a=W-(r-l)8.
a (s — a)«-i — Z (« — 0«-i = 0.
13
14
15
16
a n I
a n 8
a I 8
n I 8
n-lll
r" r H = 0.
a a
' 8-1
r« - - — r»-i + -^ = 0.
S—l 8—1
17
18
19
a r I
a r 8
a I 8
r I 8
logr
^^log[a + (r-l)3]-loga.
logr
log I — log a
- + 1.
log(a-a) — log(8 — Z)
^_logZ-logrZr-(r-l)gl_|.i
logr
SEBIE8. Sll
Exercise 117.
1. Find the seventli term of 2, 6, 18
sixth term of 3, 6, 12
ninth term of 6, 3, 1^
eighth term of 1, — 2, 4
twelfth term of a^, a*, a*
fifth term of 4 a, — 6via\ 9mV..
2. Find the geometrical mean between lSx*y and SOxy^z,
3. Find the ratio when the first and third terms are 5
and 80 respectively.
4. Insert two geometrical means between 8 and 125 ; and
three between 14 and 224.
5. If a = 2 and r = 3, which term will be equal to 162 ?
6. The fifth term of a geometrical series is 48, and the
ratio 2. Find the first and seventh terms.
7. Find the sum of
3 + 6 + 12 + to eight terms.
1 — 3 + 9 — to seven terms.
8 + 4 + 2 + to ten terms.
0.1 + 0.5 + 2.5 + to seven terms.
m — — + T^ — to five terms.
4 lb
8. The population of a city increases in four years from
10,000 to 14,641. What is the rate of increase ?
9. The sum of four numbers in geometrical progression
is 200, and the first term is 5. Find the ratio.
10. Find the sum of eight terms of a series whose last
term is 1, and fifth term -J.
312 ALGEBRA.
11. In an odd number of terms, show that the product of
the first term and last term will be equal to the
square of the middle term.
12. The product of four terms of a geometrical series is 4,
and the fourth term is 4. Determine the series.
13. If from a line one-third be cut off, then one-third of
the remainder, and so on, what fraction of the whole
will remain when this has been done five times ?
14. Of three numbers in geometrical progression, the sum
of the first and second exceeds the third by 3, and
the sum of the first and third exceeds the second
by 21. What are the numbers ?
15. Find two numbers whose sum is 3^ and geometrical
mean 1^.
16. A glass of wine is taken from a decanter that holds ten
glasses, and a glass of water poured in. After this
is done five times, what part of the contents is wine ?
17. There are four numbers such that the sum of the first
and the last is 11, and the sum of the others is 10.
The first three of these four numbers are in arith-
metical progression, and the last three are in geo-
metrical progression. Find the numbers.
18. Find three numbers in geometrical progression such
that their sum is 13 and the sum of their squares
is 91.
19. The difference between two numbers is 48, and the
arithmetical mean exceeds the geometrical mean by
18. Find the numbers.
20. There are four numbers in geometrical progression,
the second of which is less than the fourth by 24,
and the sum of the extremes is to the sum of the
means as 7 to 3. Find the numbers.
SEBIES. 313
21. A number consists of three digits in geometrical pro-
gression. The sum of the digits is 13 ; and if 792
be added to the number, the digits in the units' and
hundreds' places will be interchanged. Find the
number.
394. When r is less than 1, a geometrical series has its
terms continually decreasing; and by increasing n, the
value of the nth term, ar"~^, may be made as small as we
please, though not absolutely zero.
395. The formula for the sum of n terms,
a(l— r*)
l~r
a ai^
may be written
1-r
By increasing n indefinitely, the value of becomes
indefinitely small, so that the sum of n terms approaches
indefinitely to as its limit.
Ex. Find the limit of l — ^ + i — i+
Here a = 1, and r = — i,
and therefore the limit = :; ; — rr = 7-7-7 = «' -4^«-
1 — r 1 — (— I) 1 + 1 3
22. Eind the limits of the sums of the following infinite
series :
. 4 + 2 + 1+ 2-li+f-
* + i + f + 0.1 + 0.01 + 0.001 +
i-i^f + A- 0.868686
1-* + !^- 0.54444
i + A + iV+ 0.83636
314 ALGEBRA.
♦Harmonical Series.
396. A series is called a Harmonical Series, or a Har-
monioal Progression, when the reciprocals of its terms form
an arithmetical series.
Hence, the general representative of such a series will be
111 1
_, — : — 1?
a a + c? a + 2rf a-\-(n — V)d
397. Questions relating to harmonical series should be
solved by writing the reciprocals of its terms so as to form
an arithmetical series.
398. If a and b denote two numbers, and H their har-
monical mean, then, by the definition of a harmonical series,
H a b H
. 2 ^1 l_a + ^.
a + b
399. Sometimes it is required to insert several harmon-
ical means between two numbers.
Ex. Let it be required to insert three harmonical means
between 3 and 18.
Find the three arithmetical means between \ and ^y.
These are found to be if, f }, /^ ; therefore, the harmonical means
areHiHȴ; or3H,5}, 8.
* A harmonical series is so called because musical strings of uniform
thickness and tension produce harmony when their lengths are repre-
sented by the reciprocals of the natural series of numbers ; that is, by
the series, 1, J, }, J, J, etc.
SERIES. 315
Exercise 118.
1. Insert four harmonical means between 2 and 12.
2. Find two numbers whose difference is 8 and the har-
monical mean between them 1|.
3. Find the seventh term of the harmonical series 3, 3^,
4
4. Continue to two terms each way the harmonical series
two consecutive terms of which are 15, 16.
5. The first two terms of a harmonical series are 5 and 6.
Which term will equal 30 ?
6. The fifth and ninth terms of a harmonical series are
8 and 12. Find the first four terms.
7. The difference between the arithmetical and harmon-
ical means between two numbers is If, and one of
the numbers is four times the other. Find the
numbers.
8. Find the arithmetical, geometrical, and harmonical
means between two numbers a and h\ and show
that the geometrical mean is a mean proportional
between the arithmetical and harmonical means.
Also, arrange these means in order of magnitude.
9. The arithmetical mean between two numbers exceeds
the geometrical by 13, and the geometrical exceeds
the harmonical by 12. What are the numbers ?
10. The sum of three terms of a harmonical series is
11, and the sum of their squares is 49. Find the
numbers.
11. When a, b, c are in harmonical progression, show that
a\c: :a — b :b — c.
CHAPTER XXII.
♦Choice. Binomial Theorem.
400. If three paths, A, B, and C, lead to the top of a
mountain, there is obviously a choice of three different
ways of ascending the mountain ; and when the top of the
mountain is reached, there is again a choice of three differ-
ent ways of descending.
How many different ways are there of doing both ?
If a traveller ascend by A, he may descend by A, B, or
C. This makes three ways of doing both. If he ascend
by By he may descend by A, J5, or C; and again, if he
ascend by C, he may descend by A, B, or C.
Therefore, there are 3X3 = 9 ways in all of doing both.
These ways may be indicated as follows :
1.
A and A.
4.
BandA.
7.
C and A.
2.
A and B.
5.
B and B.
8.
C and B.
3.
A and C.
6.
Bajid C.
9.
C and C.
401. Suppose the traveller does not wish to ascend and
descend by the same path, then what choice has he?
He has a choice of three different ways in ascending.
But he has a choice of only two ways in descending.
If each of the three ways of ascending be joined to the
two eligible ways of descending, the result is 3X2 = 6
ways of doing both.
* This chapter is based upon Whitworth's Choice and Chance^ and
many of the examples have been taken from that elegant work.
CHOICE.
These ways are :
1. A and B,
3. BB.ndA,
6. (7 and A
2. A and C.
4. B and C.
6. (7and^.
317
402. If a box contains five capital letters, Ay By C, 2>, JF,
and three small letters, x, y, «, in how many different ways
may a capital letter and a small letter be selected ?
A capital letter may be selected in 5 ways.
With each capital letter selected, a small letter can be
joined in 3 ways. So that the number of different ways
in which the selection can be made is 3 X 5 = 15. These
ways are :
Ax Bx Cx Dx Ex
Ay By Cy Dy Ey
Az Bz Cz Dz Ez
403. Hence the fundamental principle of choice :
I. If one thing can he done in a different ways, and (when
it ha^ been done in any one of these ways) another thing can
he done in b different ways, then both can he done tn a X b
different ways.
For, corresponding to each of the a ways of doing the first thing,
there are h ways of doing the second thing. Therefore, altogether,
there are a X 6 ways of doing both things.
(1) On a shelf are 7 English and 5 French books. In how
many ways can one of each be chosen ?
7 X 5 = 36. Arts,
(2) On a shelf are 7 English, 5 French, and 9 German
books. In how many ways can two books be chosen
so that they shall be in different languages ?
An English book and a French book can be chosen in 7 X 5
= 36 ways. A French book and a German book in 5 x 9 = 45
ways. An English book and a German book in 7 x 9 = 63 ways.
Hence, there is a choice of 36 + 45 + 63 = 143 ways. Ans,
318 ALGEBRA.
(3) Out of 8 different pairs of gloves, in how many differ-
ent ways can a right-hand and a left-hand glove be
chosen which shall not form a pair ?
A right-hand glove can be chosen in 8 ways ; and when it
is chosen there are 7 left-hand gloves, any one of which may
be pnt with it without making a pair. Hence, the choice is in
8 X 7 = 56 ways.
(4) Out of the ten digits, ^, 1, 2, 3, 4, 5, 6, 7, 8, 9,0how
many numbers each consisting of two figures can be
formed ?
Since 0 has no value in the left-hand place, the left-hand place
can be filled in 9 ways.
The right-hand place can be filled in 10 ways, since repetitions
of the digits are allowed (as 22, 33, etc.).
Hence, the whole number is 9 X 10 = 90.
(5) How many odd numbers consisting of two figures can
be formed with the ten digits ?
The left-hand place can be filled in 9 ways ; the right-hand
place in only 5 ways, since it must be either 1, 3, 5, 7, or 9.
Hence, the number is, 9 x 5 = 45. Ans.
404. By a simple extension of Rule I. it is evident that,
II. If one thing can he done in a ways^ and then a second
thing can he done in b ways, then a third i?i c ways, then a
fourth in d ways, etc., the numher of ways of doing all the
things will fte a X b X 0 X d, etc,
(1) In how many ways can four Christmas presents be
given to four boys, one to each boy ?
The first present may be given to any one of the boys ; hence
there are 4 ways of disposing of it.
The second present may be given to any one of the other three
boys ; hence there are 3 ways of disposing of it.
The third present may be given to either of the other two
boys ; hence there are 2 ways of disposing of it.
CHOICE. 319
The fourth present mtlst be given to the last boy ; hence there
is only one way of disposing of it.
There are, then, 4x3x2X1 = 24 ways.
(2) In how many ways can five presents be given to two
children ?
Each present may be disposed of in 2 ways ; for it may be
given to either child. Hence, the whole number of ways of
giving the presents is 2X2X2X2X2 = 32.
(3) In how many ways can five presents be divided between
two children ?
This question differs from the last only in the fact that a
division of the gifts excludes the two ways in which either child
receives all the gifts.
Hence, there are 32 — 2 = 30 ways.
(4) In how many ways can x things be given to n persons ?
n*. Ans.
(5) In how many ways can a vowel and a consonant be
chosen out of the alphabet ?
Since there are in the alphabet 6 vowels and 20 consonants,
a vowel can be chosen in 6 ways and a consonant in 20 ways,
and both (Eule I.) in 6 X 20 = 120 ways.
(6) In how many ways can a two-lettered word be made,
containing one vowel and one consonant ?
The vowel can be chosen in 6 ways and the consonant in 20
ways ; and then each combination of a vowel and a consonant
can be written in 2 ways ; as ac, ca.
Hence, the whole number of ways is 6 X 20 X 2 = 240.
406. The last two examples show the difference between
a selection or combination of different things, and an
arrangement or permntation of the same things.
320 ALGEBRA.
Thus, ac forms a combination of a Vow^l and a consonant, and clc
and ca form two different arrangements of this combination.
From (5) it is seen that 120 different combinations can be made
with a vowel and a consonant ; and from (6) it is seen that 240 differ-
ent permutations can be made with the same.
Again, a, 6, c is a selection of three letters from the alphabet.
This selection then admits of 6 different arrangements or permuta-
tions, as follows :
abc bca cab
acb hoc cba
406. A selection or combination of any number of ele-
ments or things means a group of that number of elements
or things put together without regard to their order of
sequence. An arrangement or permutation of any number
of elements or things means a group of that number of
elements or things put together with reference to their
order of sequence.
Arrangements or Permutations.
407. In how many ways can the letters of the word
Cambridge be arranged, taken all at a time ?
There are nine letters. In making any arrangement any one of
the letters may be put in the first place. Hence, the first place can
be filled in 9 ways. Then the second place can be filled with any one
of the remaining eight letters ; that is, in 8 ways.
In like manner, the third place in 7 ways, the fourth place in 6
ways, and so on ; and, lastly the ninth place in 1 way.
If the nine places are indicated by Roman numerals, the result
(Rule II.) is as follows:
I. II. III. IV. V. VI. VII. VIII. IX.
9X8X7X6X6X4X3 X 2X 1 = 362,880 ways.
408. Hence, it will be seen that,
III. The number of arrangements or permutations of n
different elements or things taken qlljita time is
n(n — l)(n- 2) (n — 3) 3 X 2 X 1.
CHOICE. 321
For, the first place may be filled in n ways, then the second place
inn — 1 ways, then the third place in n — 2 ways, and so on to the
last place, which can be filled in only 1 way.
Hence (Rule 11.) the whole number of arrangements is the con-
tinued product of all these numbers,
n(n — l)(n-2)(n — 3) 3X 2X1.
409. For the sake of brevity this product is written [w,
and is read factorial n.
410. It will also be evident that,
IV. The number of arrangements of n different elements^
taken T at a time, is
n(n — l)(n—' 2) to r factors,
that isy n(n — 1) (/^ — 2) [n — (r — 1)],
or n(n — 1) (^ — 2) (n — r+1).
For the first place can be filled in n ways, the second in n — 1
ways, the third place in n — 2 ways, and the rth place in n — (r — 1)
ways.
(1) A shelf contains 4 English books, 5 French books, and
6 German books ; in how many ways can these books
be arranged ?
[16 = 1,307,674,368,000 ways.
(2) In how many ways can these books be arranged, if the
books of each language are kept together ?
The English books can be arranged (Rule III.) in [4 ways, the
French books in [5 ways, and the German books in [6 ways.
Also, the three sets can be arranged in [3 different orders. Hence,
the number of ways in which the whole can be done is (Rule II.)
|4 X [5 X IJS X |3 = 12,441,600.
822 ALOEBBA.
(3) Of the arrangements possible with the letters of the
word Cambridge:
(i.) How many will begin with a vowel ?
(ii.) How many will both begin and end with a
vowel ?
In filling the nine places of any arrangement in case (L), the
first place can he filled in only 3 ways, the other places in [8
ways.
In case (ii.) the first place can be filled in 3 ways, the last place
in 2 ways (one vowel having been used), and the remaining seven
places in [7 ways.
Hence, the answer to (i.) is 3 x [8 = 120,960 ;
to (ii.) is 3 X 2 X 7 = 30,240.
(4) With the letters of the word Cambridge, how many
arrangements can be made :
(i.) Each beginning with the word Cam?
(ii.) Keeping the letters Cam always together ?
For case (i.) the answer is evidently [6 ; since our only choice
lies in arranging the remaining six letters of the word.
Case (ii.) may be resolved into arranging Cam and the last six
letters, regarded as seven distinct elements, and then arranging
the letters Cam.
The first can be done in [7 ways, and the second in [3 ways.
Hence (Rule II.) both can be done in \]_x\^- 30,240 ways.
(6) In how many ways can the letters of the word Cam-
bridge be written :
(i.) Without changing the pUtce of any vowel?
(ii.) Without changing the order of any vowel ?
(iii.) Without changing the relative order of vowels
and consonants ?
In (i.) the second, sixth, and ninth places can be filled each in
only 1 way ; the other places in [6 ways.
Therefore, the whole number of ways is [6 = 720.
In (ii.) the vowels in. the different arrangements are always
kept in the order a, i, e. One of the six consonants can be placed
in 4 ways : before a, bettveen a and i, between i and e, and after e.
CHOICE!. 323
Then a second consonant can be placed in 5 ways, a third
consonant in 6 ways, a fourth consonant in 7 ways, a fifth con-
sonant in 8 ways, and the last consonant in 9 ways. Hence
(Rule II.) the whole number of ways is4x6X6x7x8x9
= 60,480.
In (iii.) the vowels can be arranged in |3 ways, and the con-
sonants in [6 ways. Hence (Rule II.) the number of ways is
|3 X |6 = 4320
(6) In how many ways can 4 persons, Ay J5, C, D, sit at a
round table ?
If the four places are not regarded as relative to each other,
then the whole number of ways is [4 = 24. But if the four
places are regarded as relative to each other, then by placing
one as ^, in one position, and by arranging the others in the
other three positions, the whole number of ways is [3 = 0.
(7) In how many ways can 6 persons form a ring?
Here relative position is required. Hence, the whole number
of ways is [5 = 120.
(8) How many three-lettered, words can be made from the
alphabet, no letter being repeated in the same word ?
26 X 25 X 24 = 15,600. Ans.
(9) How many four-lettered words ?
26 X 25 X 24 X 23 = 358,800. Ans.
(10) How many different arrangements can be made of the
letters in the word ei/e ?
Distinguish the e's thus, ei, 62, and arrange as follows :
ei y €2 ei 62 y y ei 62
62 y ei 62 Ci y y 62 e\
These six arrangements become three when the e's are not
distinguished. That is, each pair of arrangements produced by
permuting the e's is reduced to a single arrangement.
Hence, the number of different arrangements is found by
dividing [3 by [2 ; that is, by dividing the number of arrange-
ments possible when the letters are all different by the number
of ways in which the two 6's can be permuted.
324 ALGEBRA.
411. In how many different orders can the letters a, a, a,
X, yhe written ?
If the letters were all different, the answer (Rule III.) would be
[6 ; but the three a's may be permuted in [3 = 6 ways.
Hence, the [5 arrangements may be divided into six groups, each
group constituting but a single arrangement of the given letters.
Hence, the whole number of given orders is = — 20.
By the same course of reasoning, the whole number of
different orders of a, a, a, x, x is found to be
412. In like manner for any other numbers. Hence,
V. The number of arrangements of n elements, of which
p are alike, q others are alike, and r others are alike , is
\n
\p\q\r.
(1) In how many ways can the letters of the word Missis-
sippi be arranged ?
[11
"— = 34,660. Ans.
(2) In how many different orders can a row of 4 white
balls and 3 black balls be arranged ?
(3) In how many ways can 4 white balls and 3 black balls
be placed in a row, if the balls are all different in
size ?
[7 = 5040. An8.
CHOICE. 325
413. In case the n elements to be arranged are all differ-
ent, but repetitions of them are allowed, then in making
any arrangement, every place to be filled can be filled in n
ways, since we may always repeat an element already used.
Hence, corresponding to Eules III. and IV., the following
rules apply to cases in which repetitions are allowed :
VI. The number of arrangements of n different elements,
taken all at a time, when repetitions are allowed, is n"*.
VII. The number of arrangements of n different elements,
taken r at a time, when repetitions are allowed, is n'.
(1) How many three-lettered words can be made from the
alphabet, when repetitions are allowed ?
26«= 17,676. Ans,
(2) How many three-lettered words can be made from the
6 vowels when repetitions are allowed ?
68=216. Ans.
(3) A railway signal has 3 arms, and each arm may take
four different positions, including the position of
rest. How many signals in all can be made ?
48-1 = 63. Ana.
(4) In the common system of notation, how many numbers
can be formed consisting of not more than 5 digits ?
All the possible numbers may be regarded as consisting of
each 6 digits, by prefixing zeros to the numbers consisting of
less than 5 digits. Thus, 247 may be written 00247.
Hence, every possible arrangement of 6 digits out of the 10
digits will give one of the required numbers except 00000 ; and
the answer is 10* — 1 = 99999 ; that is, all the numbers between
0 and 100.000.
326 ALGEBRA.
(6) With the digits 0, 1, 2, 3, 4, 5, how many numbers
between 1000 and 4000 can be formed ?
Here we have to fill, in each possible case, four places. The
first place can be filled with 1, 2, or 3, that is, in 3 ways ; the
second, third, and fourth places each in 6 ways. Therefore,
the answer is 3 x 6« = 648.
(6) With the same digits, 0, 1, 2, 3, 4, 5, and between the
same limits, 1000 and 4000 :
(i.) How many even numbers can be formed ?
(ii.) How many odd numbers can be formed ?
(iii.) How many numbers divisible by 5 ?
Evidently (i.) and (ii.) are like Ex. (5), except that the last
place can be filled in only 3 ways. In (i.) the last place must
be filled by 0, 2, or 4; in (ii.) the last place must be filled by
1, 3, or 6.
Hence, the answer in each of these cases is3x6x6x3 = 324.
In (iii.) the last digit must be either 0 or 5 ; and the answer
for this case is3x6x6x2 = 216.
Combinations.
414. In how many ways can 3 vowels be selected from
the 5 vowels a, e, i, o, u?
The number of ways in which we can arrange 3 vowels out of 5 is
(Rule IV.) 5 X 4 X 3 = 60.
These 60 arrangements might be obtained -by first forming all the
possible selections of the 3 vowels out of 5, and then arranging the 3
vowels in each selection in as many ways as possible.
The 3 vowels of each selection may be arranged in |3 = 6 ways.
Hence (Rule II.),
The number of selections X 6 = the number of arrangements = 60.
Therefore, the number of selections = ^^ = 10.
415. In general,
VIII. Oict of n different elements, the number of selections
of r elements is equal to the number of arrangements of n
elements divided by [r.
/
CHOICE. 327
For, let 8 = the number of ways of selecting r elements out of n
elements. Then the r elements thus selected may be arranged
(Rule III.) in [r different ways. Therefore (Rule I.) « X [r= the
number of arrangements of n elements taken r at a time.
_ the number of arrangements
\r
(By Rule IV.) The numerator of this fraction is equal to
n(n-l)(n- 2) [n-(r- 1)].
_ n (n - 1) (n - 2) ..... jn-r+l)
\r
If both terms of this fraction are multiplied by |n — r,
In
the result is s = , — r^
If n — r = p, then n = p-\- r^ and this formula may be written
Ip + r
[p\r
416. The value of this fraction is not altered if p and r
are interchanged. Hence,
IX. Out o/ p + r different elements, the number of ways
in which p elements can he selected is the same as the number
of ways in which r elements can be selected.
Thus, out of 8 elements, 3 elements can be selected in the same
number of ways as 5 elements ; namely,
__ = _ = 56ways.
(1) Out of 20 consonants, in how many ways can 18 be
selected ?
The 18 can be selected in the same number of ways as 2 ; and
the number of ways in which 2 can be selected (Rule VIII.) is
20 X 19 ,^^ .
— = 190. Ans.
2
328 ALGEBRA.
(2) In how many ways can the same choice be made so as
always to include the letter B ?
Taking B first we must then select 17 out of the remaining 19
consonants. This can be done in
19 X 18
2
= 171 ways.
(3) In how many ways can the same choice be made so as
to include B and not to include C?
Taking B first, we have then to choose 17 out of 18, C being
excluded. This can be done in 18 ways.
(4) A society consists of 50 members, 10 of whom are
physicians. In how many ways can a committee of
6 members be selected so as to include 1 physician ?
The 6 members not physicians can be selected in
[40
\b\S6 "^^y®'
and the 1 physician in 10 ways. Hence, the 6 can be selected in
|40
10 X iTT^ different ways.
I o \oo
(5) In how many ways can a committee of 6 members be
selected so as to include at least 1 physician ?
Six members can be selected from the wfwle society in
[50
Six members can be selected from the whole society, so as to
include no physician, by choosing them all from the 40 members
who are not physicians, and this can be done in
[40
[50 [40
^^^°^* 1 6~|44 ~ 16734 ~ *^® number of ways of select-
ing the committee so as to include at least one physician.
CHOICE. 829
(6) Out of 20 Eepublicans and 6 Democrats, what choice
is there of appointing a committee consisting of 3
Eepublicans and 2 Democrats ?
20 X 19 X 18
The Republicans can be selected in — — = 1140 ways ;
6X5 lX2Xd
and the Democrats in 7——^ =15 ways. Hence, the whole com-
1 X ^
mittee can be appointed in 1140 x 15 = 17,100 ways.
(7) From 20 Republicans and 6 Democrats, in how many
ways may 5 different offices be filled, three of which
must be filled by Republicans, and the other two by
Democrats ?
The first three offices can be assigned to 3 Republicans in
20 X 19 X 18 = 6840 ways (Rule II.) ; and the other two offices
can be assigned to 2 Democrats in 6 x 5 = 30 ways.
There is, then, a choice of 6840 X 30 = 205,200 different
ways.
(8) Out of 20 consonants and 6 vowels, in how many ways
can we make a word consisting of 3 different conso-
nants and 2 different vowels ?
on. * 1. , , , . 20X 19X 18 ,,.^
Three consonants can be selected m . ^ ^ ^ o = ^^^^
6x6 I X z X 6
ways, and 2 vowels in rrT^o ~ ^^ ways. Hence (Rule I.) the 5
1 X ^
letters can be selected in 1140 x 15 = 17,100 ways.
When they have been so selected, they can be arranged (Rule
ni.) in [5 = 120 different orders. Hence, there are 17,100 X
120 = 2,062,000 different ways of making the word.
(9) How many words of 2 consonants and 1 vowel can be
formed from 6 consonants and 3 vowels, the vowel
being the middle letter of each word ?
The two consonants can be selected in 16 ways ; the vowel in
3 ways. Each combination of the two consonants and 1 vowel
can be arranged in 2x1x1=2 ways. Hence, the number of
words that can be formed isl5x 3x 2 = 90.
330 ALGEBRA.
(10) How many words of 3 consonants and 3 vowels can
be formed from the alphabet, if one of the vowels
is to be always a ?
nru * X. ^ ^ A • 20 X 19 X 18 ,, .-
The consonants can be selected m » = 1140 ways,
5X4 IXzXo
and the vowels in -r-—^ = 10 ways.
1 X ^
Then the six letters of each combination can be arranged in
|6 = 720 ways. Hence the number of words that can be formed
is 1140 X 10 X 720 = 8,208,000.
417. To find for what value of r the number of selections
of n things, taken r at a time, is greatest.
^ ^ , n(n-l){n-2) (n-r+1)
Theformula s= /^ , ^ 8 x .....r "^
may be written s = - x —— x — - — •
•^ 12 3 r
The numerators of the factors on the right side of this equation
begin with n, and form a descending series with the common differ-
ence 1 ; and the denominators begin with 1, and form an ascending
series with the common difference 1. Therefore, from some point in
the series, these factors become less than 1. Hence, the maximum
product is reached when that product includes all the factors greater
than 1.
When n is an odd number, the numerator and the denominator
of each factor will be alternately both odd and both even ; so that
the factor greater than 1, but nearest to 1, will be the factor whose
numerator exceeds the denominator by 2. Hence, in this case, r must
have such a value that
)i — r+l = r + 2, or r= — ^— •
When n is an even number, the numerator of the first factor will
be even and the denominator odd ; the numerator of the second factor
will be odd and the denominator even ; and so on, alternately ; so
that the factor greater than 1, but nearest to 1, will be the factor
whose numerator exceeds the denominator by 1. Hence, in this case,
r must have such a value that
n — r+l = r+l, or r = -'
CHOICE. 331
(1) What value of r will give the greatest number of selec-
tions out of 7 things ?
Here n is odd, and r = — r — = — r— = 3.
7X6X5 „, .
••••=nr2-x3=^^- ^"'•
TT ^ *!. 7 x 6X 6X 4 „_
Ifr=4,then , = -_^-^_^ = 36.
So that, when the number of things is odd, there will be
two equal numbers of selections ; namely, when the number
of things taken together is just under and just over one-half
of the whole number.
(2) What value of r will give the greatest number of
selections out of 8 things ?
Here n is even, and r = - = - = 4.
8X7X6X5 ^^ ^
•••*=Ix2X3X-4='^- ^^-
So that, when the number of things is even, the number
of selections will be greatest when one-half of the whole
are taken together,
418. It may be shown that,
X. The number of ways in which x + y different elements
can be divided into two classes, so that one shall contain x
and the other y elements, is equal to the number of ways in
which either x elements or j elements may be selected from
the X -|- y elements ; or
\^ + y
For each division ofx + y elements into two classes, one consisting
of X elements, the other of y elements, is evidently effected by making
a selection of x elements from x + y elements, and leaving y elements
not selected.
332 ALGEBRA.
In like manner the number of ways in which x-^y-^-z
different elements can be divided into 3 classes, containing
X, y, and z elements respectively, is
\x + y + z
; and so on.
(1) In how many ways can 18 men be divided into 2 classes
of 6 and 12 ?
(2) In how many ways can 18 men be divided into 2 groups
of 9 each ?
According to the rule, the answer would be
The two groups, considered as groups, have no distinction ;
therefore, permuting them gives no new arrangement, and the
true result is obtained .by dividing the preceding by [2, and is
[18
[2 [9^
If any condition is added that shall make the two groups
differeid^ as, if one group wear red badges and the other blue,
then the answer will be
[9(9
419. Whenever the groups are indifferent, Kule IX. gives
each arrangement repeated as many times as the groups can
be permuted one with another ; that is, [2 when there are
2 groups, [3 when there are 3 groups, and so on.
Hence, the result found by the rule must be divided by
|2, [3, etc., in order to obtain the true result.
CHOICE. 333
(1) In how many ways can the 52 cards in a pack be
divided among 4 players, each to have 13 ?
Here the assignment of each group to a different player makes
the groups different ; and the answer is
|_62
[13 [13 [13 [13*
(2) In how many ways can the 52 cards of a pack be
divided into 4 piles ?
Here the groups are ind\ff'erent, and the answer is
|52
[4[13|^[J3|J3*
(3) A boat's crew consists of 8 men, of whom 2 can row
only on the stroke side of the boat, and 3 can row
only on the bow side. In how many ways can the
crew be arranged ?
There are left 3 men who can row on either side ; 2 of these
must row on the stroke side, and one on the bow side.
The number of ways in which these three can be divided is
^ = 3 ways.
When the stroke side is completed, the 4 men can be arranged
in IJt ways ; likewise, the 4 men of the bow side can be arranged
in [4 ways. Hence (Rule II.) the arrangement can be made in
3x[4[4=1728way8.
(4) In how many ways can 10 copies of Homer, 6 of
Virgil, and 4 of Horace be given to 20 boys, so
that each boy may receive a book ?
The boys have to form themselves into a group of 10 for
Homer, of 6 for Virgil, of 4 for Horace. This can be done in
[20
[10|6[4
different ways.
334 ALGEBBA.
(6) In how many ways can 3 copies of one book, 2 of
another, and 1 of a third be given to a class of 12
boys, so that no boy shall receive more than 1 book ?
In every possible way of assigning the books, 6 boys receive
them. These 6 may be selected from the whole 12 (Rule VIII.)
in [12
[6|6 ^*y«-
When thus selected, the books may be assigned to them
(Rule V.) in [6
^i2^^"y^-
Hence, the whole number of ways of giving the books is
|12 \6
— X -
\6\6 [3[2[l
(6) In how many ways can the same books be given to
the 12 boys, so that no boy shall receive more than
1 copy of any book ?
In allotting the three copies of the first book, the boys are to
be separated into two groups of 3 and 9 ; and the group of 3
will in each case receive the books.
This can be done in -r—^ ways.
Likewise the two copies of the second book may be given in
[12
The single copy of the third book can be given in 12 ways.
Hence (Rule II.) the books may be given in
[12 1 12
1319 ^ iTTo ^ ^^ different ways.
(7) In how many ways can 2 letters be selected from a, b,
c, d, e, f, if the letters may be repeated in making
the selection ?
Without repetitions, two letters can be selected from 6 in 15 ways.
With repetitions, as aa, etc. , 6 selections can be made. Hence, there
are 15 4- 6 = 21 different selections.
CHOICE. 336
(8) In how many ways can selections of 3 letters be made
from a, 5, c, d, e, /, if letters may be repeated ?
These selections will be of 3 kinds :
(i.) All three letters different,
(ii.) Two letters alike, the third different,
(iii.) All three letters alike.
(By Rule VIII.) (i.) gives 20 ways.
(ii.) gives 6 X 5 = 30 ; for we can choose 2 alike in 6 ways,
and then join a different letter to each pair in 5 ways,
(iii.) gives evidently 6 ways.
Hence, there are in all 20 + 30 + 6 = 56 different selections.
420. This may also be shown as follows :
By adding the 6 letters, a, h, c, d, e, f, to each selection of
the kind required, they will become selections of 6 + 3 = 9
letters out of 6, in which selections each letter occurs at
least once, and in which, therefore, there must be exactly
3 repetitions. These repetitions may be all of the same
letter, or divided among the different letters.
Any one of these selections may be made by writing 9
places, and then filling them in alphabetical order, taking
care to make exactly three repetitions in passing from one
end of the row to the other.
Thus:
1
2
3
4
5
6
7
8
a
a
a
a
h
c
d
e
/
a
b
c
c
c
d
e
e
/
a
b
c
c
d
d
e
e
/
By striking out the 6 letters, a, b, c, d, e, /, each once,
there is left in the first row aaa, in the second row cce, in
the third row cde ; that is, three of the required selections.
Now, in filling the 9 places for each row, 9 — 1 or 8 steps
must be made, of which exactly 3 are repetitions, and each
of the other 5 is a change to a different letter.
336 ALGEBRA.
The 3 repetitions may be chosen from the 8 steps
(Kule Vlll.) in
8X7X6 ^^
This, then, is the number of ways in which 3 letters can
be selected out of 6 letters, when repetitions are allowed.
In other words, 3 elements can be selected from 6
elements, when repetitions are allowed, in as many ways
as 3 elements can be selected from 6 + 3 — 1, or 8 elements
without repetitions,
421. Hence the general rule :
XI. The number of ways of selecting r elements from, n
different elements^ when repetitions are allowed, is the same
as the numher of ways of selecting r elements from, n + r — 1
elements without repetitions.
And this number of ways is (Rule VIII.).
1^ + ^—^ ___ n{n + l){n + 2). »^^n + r— 1) ^
I r I n — 1 [r
(1) In how many ways can 4 elements be selected from n
elements, when repetitions are allowed ?
n(n+l)(n + 2)(n + 3) ^^
li
(2) How many dominoes are there in a set mumbered from
double blank to double nine ?
Each domino is made by selecting two numbers out of the ten
digits, and repetitions are included ; that is, the two numbers on
a domino may be the same.
Hence, the number of dominoes is equal to the number of
selections of 2 from 10 + 2 — 1, or 11, without repetitions.
IPX 11 ^^ .
CHOICE. 337
(3) In how many ways can 4 glasses be filled with 5 kinds
of wine, without mixing ?
The number is equal to the number of ways in which 4 things
can be selected from 6 + 4 — 1, or 8 things, without repetitions.
5X0X7X8
1X2X3X4
= 70. Ana.
(4) In how many ways can six rugs be selected at a shop
where 2 kinds of rugs are sold ?
The number is equal to the number of ways of selecting 6
from 2 + 6—1 = 7, without repetitions.
2 X 3 X 4 X 5 X 6^XJ7 _
1X2X3X4X6X6"
If a and b represent the two kinds of rugs, the 7 ways are as
follows :
aa aaa a aa abbb
a a a a ab aabb bb
aaa a bb abb b b b
bbbbbb
422. It may be shown that,
XII. The number of ways in which a selection (of some,
or all) can be made from n different things is 2" — 1.
For each thing can be either taken or left, that is, can be disposed
of in two ways.
There are n things ; hence (Rule II.) they can all be disposed of in
2" ways. But, among these ways is included the case in which all are
rejected ; and this case is inadmissable.
Hence, the number of ways of making the selection is 2» — 1.
(1) In a shop window 20 different articles are exposed for
sale. What choice has a purchaser ?
220-1 = 1,048,676. Ans,
(2) How many different amounts can be weighed with
1 lb., 2 lb., 4 lb., 8 lb., and 16 lb. weights?
26-1 = 31. Ans.
(Let the student write out the 31 weights.)
338 ALGEBRA.
423. It may be shown that,
XIII. The whole number of ways in which a selection can
he made from p + q + r things , of which p are alike, q are
alike, r are alike, etc., is (p + 1) (q + I) (r + I) — 1.
For the set of p things may be disposed of in jp + 1 ways, since
none of them may be taken, or 1, 2, 3, , or p, may be taken.
In like manner, the q things may be disposed of in ^ + 1 ways ;
the r things in r + 1 ; and so on.
Hence (Rule II.), all the things may be disposed of in (p+1)
(q + 1) (r + 1) ways.
But the case in which all the things are rejected is inadmissable ;
hence, the whole number of ways is (p+ 1) (q+ 1) (r+ 1) — 1.
(1) 111 how many ways can 2 boys divide between them
10 oranges all alike, 15 apples all alike, and 20
peaches all alike ?
Here, the case in which the first boy takes none, and the case
in which the second boy takes none, must be rejected.
Therefore, the answer is 1 less than the result, according to
Rule XIII, 11 X 1(5 X 21 — 2 = 3694. Ans.
(2) If there are vi kinds of things, and n things of each
kind, in how many ways can a selection be made ?
In this case p, q^ r, etc., are all equal, and each is equal to n.
Hence the result is (n + i)'" — 1.
(3) If there are m kinds of things, and 1 thing of the first
kind, 2 of the second, 3 of the third, and so on, in
how many ways can a selection be made ?
|m+ 1 — 1. Ans.
Exercise 119.
1. How many different permutations can be made of the
letters in the word Ecclesiastical, taken all together?
2. Of all the numbers that can be formed with four of the
digits 5, 6, 7, 8, 9, how many will begin with 56 ?
CHOICE. 33d
3. If the number of permutations of n things, taken 4
together, is equal to 12 times the permutations of
n things, taken 2 together, find n.
4. With 3 consonants and 2 vowels, how many words of 3
letters can be formed, beginning and ending with a
consonant, and having a vowel for the middle letter ?
5. Out of 20 men, in how many different ways can four
be chosen to be on guard ? In how many of these
would one particular man be taken, and from how
many would he be left out ?
6. Of 12 books of the same size, a shelf will hold 5.
How many different arrangements on the shelf may
be made ?
7. Of 8 men forming a boat's crew, one is selected as
stroke. How many arrangements of tlie rest are
possible ? When the 4 who row on each side are de-
cided on, how many arrangements are still possible ?
8. How many signals may be made with 6 flags of differ-
ent colors, which can be hoisted either singly, or
any number at a time ?
9. How many signals may be made witli 8 flags of differ-
ent colors, which can be hoisted either singly, or
any number at a time one above another ?
10. How many different signals can be made with 10 flags,
of which 3 are white, 2 red, and the rest blue,
always hoisted all together and one above another ?
11. How many signals can be made with 7 flags, of which
2 are red, 1 white, 3 blue, and 1 yellow, always dis-
played all together and one above another ?
12. In how many different ways may the 8 men serving a
field-gun be arranged, so that the same man may
always lay the gun ?
340 ALGEBRA.
13. Find the number of signals which can be made with
4 lights of different colors when displayed any
number at a time, arranged above one another, side
by side, or diagonally.
14. From 10 soldiers and 8 sailors, how many different
parties of 3 soldiers and 3 sailors can be formed ?
15. How many signals can be made with 3 blue and 2
white flags, which can be displayed either singly,
or any number at a time one above another ?
16. In how many ways can a party of 6 take their places
at a round table ?
17. Out of 12 Democrats and 16 Republicans, how many
different committees can be formed, each consisting
of 3 Democrats and 4 Republicans ?
18. From 12 soldiers and 8 sailors, how many different
parties of 3 soldiers and 2 sailors can be formed ?
19. Find the number of combinations of 100 things, 97
together.
20. With 20 consonants and 5 vowels, how many different
words can be formed consisting of 3 different conso-
nants and 2 different vowels, any arrangement of
letters being considered a word ?
21. Of 30 things, how many must be taken together in
order that having that number for selection, there
may be the greatest possible variety of choice ?
22. There are m things of one kind and n of another ; how
many different sets can be made containing r of the
first and s of the second ?
23. In how many ways may 10 persons be seated at a
round table, so that in no two of the arrangements
may every one have the same neighbors ?
CHOICE. 341
24. The number of combinations of n things, taken r to-
gether, is 3 times the number taken r — 1 together,
and half the number taken r-\-l together. Find n
and r.
25. In how many ways may 12 things be divided into 3
sets of 4 ?
26. How many words of 6 letters may be formed of 3
vowels and 3 consonants, the vowels always having
the even places?
27. From a company of 90 men, 20 are detached for mount-
ing guard each day. How long will it be before the
same 20 men are on guard together, supposing the
men to be changed as much as possible ; and how
many times will each man have been on guard ?
28. Supposing that a man can place himself in 3 distinct
attitudes, how many signals can be made by 4 men
placed side by side ?
29. How many different arrangements may be made of 11
cricketers, supposing the same two always to bowl ?
30. Five flags of different colors can be hoisted either
singly, or any number at a time one above another.
How many different signals can be made with them ?
31. How many signals can be made with 5 lights of differ-
ent colors, which can be displayed either singly, or
any number at a time side by side, or one above
another ?
32. The number of permutations of n things, 3 at a time,
is 6 times the number of combinations, 4 at a time.
Find n.
33. At a game of cards, 3 being dealt to each person, any
one can have 425 times as many hands as there are
cards in the pack. How many cards are there ?
342 ALttEBKA.
Binomial Theorem.
424. By performing the indicated multiplication
{x + a)(x + b) = T' + (a + b)x + aI>,
Multiply each side hy x-\-c, and the result is
(x-\-a)(x+b)(x + c)=x^+(a+b)x^ +abx
-\- cx^ + (ac-\-b€)x-\-(ibc
=x^-\-(a-\-b^-c)x^-\-(ab-{-ac-\-bc)x-j-abc
From this result certain laws are to be observed :
I. The number of terms is one more than the number of factors on
the left side.
II. The exponent of x in the first term is the same as the number
of factors, and decreases by 1 in each succeeding term.
III. The coefficient of x in the first term is unity ;
in the second term, the sum of a, &, c ;
in the third term, the sum of the products two and two, ab, ac,bc;
and the fourth term is the product abc.
Do the same laws hold, whatever be the number of factors ?
Suppose that these laws hold for r factors, so that
(X + a)(x + b) (X + m)=x^ + piX'-i+ PaX'-a + psP^*^ + +Pr
where pi stands for a 4- & + + m, the sum of the second terms ;
P2 stands for ab + ac+ , the sum of the products, two and
two;
Ps stands for abc + dbd + , the sum of the products, three
and three ;
Pr stands for obcd , the product of the r letters.
Multiply by another factor (x + n), and the product of the r + 1
factors is
a.r+14. p^x"- 4- P2X''-'^ + psx"—^ + + PrX
4- nx'' 4- pinx'^^ -f pjux''-^ + + p,.^iwx +Prn
=x''+i+(piH-n)x''4-(i)24-pin)x'-H(p84-i)2n)x'-24- +(pripr-in)x-\-prn
Here the laws I. and II. evidently hold ; and as to the coefficients
Pi4-n=a4-64- 4- m 4- n, the sum of r 4- 1 letters ;
BINOMIAL THEOREM. 343
1>2 + Pin = {ab-\- ac+ ) + (an + bn + + mn), the sum of the
products, two and two ;
- 1>3 + P2» = (ctbc + abd + ) + (abn + acn + ), the sum of the
products, three and three.
PrU = abc mn, the product of the r + 1 letters. Hence,
The laws hold for r + 1 factors if they hold for r factors.
But they have been shown to hold for three factors, therefore they
hold tor four factors, and therefore for five factors ; and so on for any
number of factors.*
426. When the factors in the preceding proof are all
equal, so that by c, d, n, are each equal to a, the left
side of the equation becomes
(x + a) (x -\-a) taken n times ; that is, (x + a)».
On the right side.
Pi = a+ a-V = a taken n times = na ;
P2 = art + «rt + = a"^ taken as many times as there is a choice
of 2 letters from n letters,
that is, P2 = ^i?-^a2; §415.
1 X ^
Pz = acta + aaa + = a' taken as many times as there is a choice
of 3 letters from n letters,
XV- X • n(n — l)(n — 2) , ..-.
that IS, P8 = \y^2\o^ «^- 5 *1^-
p^n = a X aX a = a taken n times as a factor = a".
.-. (X + a)» = x« + nax»-^ + ^fx2^ "^ ^""^
426. The expression on the right side is called the
expansion of (x + cr)".
If a and x are interchanged, the expansion will proceed
by ascending powers of Xj as follows :
(a -h x)» = a" + na**-! x + ^ i^>72^^ a«-2x2 + + nax»-i + x».
* A proof of this kind is called mathematical induction.
344 ALGEBRA.
If a = 1, then,
(1 + x)» = 1 + rw + ^l\~^^x^ + + nx— 1 + x»
1 X ifi
If X ia negative, the odd powers of x will be negative and the even
powers positive.
(a- x)» = a«-no»-ix + ^-^^^a"-%B2
__ n(n-l)(n-2)
1X2X3
427. It will be observed that the last factor in the de-
nominator of the coefficient is one less than the number of
the term, and is the same as the exponent of the second letter;
also, that the last factor of the numerator of the coefficient
is found by subtracting the last factor in the denominator
from n-\-l, and that the exponent of the first letter is found
by subtracting the exponent of the second letter from n.
So that,
The rth (or general) term in the expansion of (a + x)" is
Mn-1) (n-^r+2) ^^^
1 X 2 (r-1) " -^ •
Thus, the third terra of (a 4- x)"^ is
20 X 19
1X2
ai8x2=190a28x2.
428. The coefficient of the rth term from the beginning
is equal to the coefficient of the rth term from the end.
For, the coefficient of the rth term from the beginning is
n(n— 1) (71 — r + 2) n(n — 1) (n — r+ 2)
1 X2.....(r-1) ' ^^ |r-l '
and this becomes, when both terms are multiplied by jn — r + 1 ,
\^
|r-l \n-r+ \
BINOMIAL THEOREM. 345
The coefficient of the rth term from the end, which is the (n— r-f 2)tA
term from the beginning, is
n(n — 1) r
|n-r+l '
and this also becomes, when both terms are multiplied by |r — 1 ,
__Jn__
|r-l |n-r+l'
429. It will be evident from § 417, that in the expansion
of (a + ^)"> ^^^ middle term will have the greatest coeffi-
cient when n is even ; and when n is odd the two middle
terms will have equal coefficients, and these will be the
greatest.
Exercise 120.
Expand :
1. {i+2xy. y^. (2^-3.^)^ 5. (i-x)'"
2. (x-sy. 4. (2-xy. 6. (i-^)'-
y;f-^. Find the fourth term of (2x — 5 ?/)'^
/x y\^^
"v^ 8. Find the seventh term of ( - + ^ ) *
^J^ 9. Find the twelfth term of (a^ — axy^.
/ 4 10. Find the eighth term of {5a^y — 2xf)\
'^ 11. Find the middle term
12. Find the middle term of
13. Find the two middle terms
\y «/
\y «/
y
346 ALGEBRA.
14. Findther^Aterm of (2a + ir)».
15. Find the rth term from the end of (2a + a;)*.
16. Find the {r-\-^)th term of (a + a;)*.
17. Find the middle term of (a + ^)*'-
18. Expand (2a + a;)^, and find the sum of the terms if
a = l, a; = — 2.
When the Exponent is Fractional or Negative.
430. When ti is a whole number, §§ 424-426.
P
Suppose n to be a positive fraction -• We may take
(l-\-xyi^l-\-Ax-\-
By raising both sides to the qth power, we have
(i^xy={i^Ax-\- )s
or (l-\-px-\- ) = l + 5'^ic4- •
P
Hence, fjA=2>, and therefore A= -'
That is, the numerical coefficient of x is the same as the
exponent of the binomial, when the exponent is a positive
fraction.
Suppose n is negative = — m.
Then (1 + xy = (1 + x)-" ^
(1+xy
1
- , . , - §§ 424-426.
1 + mx +
= 1 — 7JIX + , by division.
I That is, the numerical coefficient of x is the same as the
exponent of the binomial when the exponent is negative.
BINOMIAL THEOREM. 347
Therefore, whether n is positive or negative, integral
or fractional, in the expansion of (14-^)" the numerical
coefficient of x will be n.
To obtain the coefficients of x^, x', etc., in the expansion
of (1 -f- ^)"> we separate x into two parts, y and z, and
write tlie expression 1 + ^ in two ways : first, retaining the
parts of X in connection ; and secondly, separating them.
We then compare the two results.
First form :
{l+xy = {l + y + zY
=.l^n{y + z)^B{u + zy+C{y + zy+
Ll + ny-^Bf+Cif^
+ nz^2Bzy + 'dCxy^ +
+ +
Second form :
In this last equality it is assumed that the power of a
product is the product of the powers of the factors, what-
ever be the exponent.
a+-)-(i+rf;)"
= (1 + ;.)" + «(l+«)'-'y+ 5 (l + «)«-y+
=l+«y -\-B,/+
-\-nz + n{n-l)zy+B{n-2)zf-\-
+ +
348 ALGEBRA.
Since these two expansions are the expansions of the
same thing, in the same form, the coefficients of «, zy, zi^^
etc., must be the same in both.
In the first expansion : In the second expansion :
The coefficient of « is n. The coefficient of z is n.
The coefficient of zy is 2 B. The coefficient of zy is n (n — 1).
The coefficient of 22^ is 3 C. The coefficient of zy^ is B(n — 2).
Therefore, n = ?i; 2B = n(n — l)', 3(7=^(n — 2).
Hence, ^=-A^; ^ = 3^ = ""T^rfW^ ' ''''
Accordingly, whether n is positive or negative, integral
or fractional,
431. In the expansion of (1+^)"? if ti is a positive
integer, the numerator of the last factor of the coefficient
of the (r-^l)th term, n — r-f-1, will be equal to 0 when
7* = n + 1 ; this term, therefore, and all following terms
(for they will also have this factor) will vanish. Hence,
the series will end with rth, or (n + l)th, term. But if n is
fractional or negative, no value of r will make n — r 4-1=0,
and the series will be infinite. Hence, the sign = in these
cases will mean " is equal to the limit of the sei^ies^
432. The sum of the numerical coefficients of the expan-
sion of (l-\-xY is equal to 2"; for the sum is equal to
(1 + 1)", as appears by writing 1 for x in (1 +a;)*.
433. The sum of the positive coefficients is equal to the
sum of the negative coefficients in the expansion of (1 — a;)";
as appears by writing 1 for x in (1 —a;)".
Note. The proof in 5 430 is taken from the article on Algebra in
EncyclopsBdia Britannica.
BINOMIAL THEOREM. 349
Expand to four terms :
1
(2) T
Va^ — 2ax
1 _/.o n-^x-l_.-4(. 2x)-l
-^^
■2ax
= (a2-2aic)-* = a-* 1- —
a
a4( ^2a^ 1x2 \aj 1X2X3 \a/j
a4( 2a
5x^ 15x8
8a2'^16a8
A root may often be extracted by means of an expansion.
(3) Extract the cube root of 344 to six decimal places.
344 = 343(l + ^i,) = 73(l + ^i^).
.•.-^3ii=7(l + ^h)i,
= 7 (1 + 0.000971815 - 0.000000944),
= 7.006796.
(4) Extract the fifth root of 3128 to six decimal places.
3128 = 55 + 3 = 55(l + ^y
...^3128 = 5(l + |)\
=>(1 + 0.000192 - 0.000000073728 +
= 6.000959.
360 ALQEBBA.
Qd
1.
Exercise 121.
to four terms :
(l + x)i 7. (2x^Sy)-l.
2.
o
(l+x)l 8. ^l-5x.
(n 4- fW Q
^ ' V(4a='-3aa;)'
4
(\-r\-* 10 -^ 1
6.
6.
(a»-a;»)«. 11. (l + a; + a^l.
(«» + a;y)-l. 12. (1 - a; + a;")!.
13. Find the rth term of (a + ic)4.
14. Find the rth term of (a — x)~^.
15. Find ^^b to five decimal places.
z,
16. Find VI g^^ to five decimal places.
17. Find vl29 to six decimal places.
18. Expand (1 — 2a; + Zx^yi to four terms.
('14-2icV
19. Find the coefficient of x^ in the expansion of ) J_ (g-
(1 +ojr)
20. By means of the expansion of (1 + ar)i show that the
limit of the series
1 + 1 j_^_J^X3 1X3X5 , . rr
^2 2X2^ "^2X3X2^ 2X3X4X2^"*" '^"^
CHAPTER XXIIL
Chance.
434. If an event can happen in a ways and fail in h
ways, and if all these ways are equally likely to happen ;
if, also, only one can happen, and one must happen, then
the mathematical probability or chance of the event hap-
pening is expressed by the fraction
a + h
I. The probability of an event happening is expressed by
the fraction whose numerator is the number of favorable
ways, and denominator the whole number of ways.
Thus, if 1 ball is drawn from a bag containing 3 white balls and
9 black balls, the chance of drawing a white ball is ^^ ; or, as it is
expressed, one chance in four.
II. The probability of an event not happening is expressed
by the fraction whose numerator is the number of unfavor-
able ways, and denominator the whole number of ways.
Thus if a denote the number of favorable ways, and h the number
of unfavorable ways, then the fraction will express the proba-
bility of the event not happening. If, for example, 1 ball is drawn
from a bag containing 3 white and 9 black balls, the chance that it
will not be a white ball is y\.
436. Since — tt H n = ^y
it is evident that the chance of an event happening, added
to the chance of its not happening, is equal to 1; and,
352 ALGEBRA.
since an event is certain to happen or not happen, it follows
that in the theory of chances
III. Certainty is expressed by unity.
436. Since — tt = 1 rr? it follows that,
a-\-b a-\-b
IV. The chance of an event not happening is found by
subtracting from unity the chance that it does Jiappen,
437. If the number of favorable ways is equal to the
number of unfavorable ways, then
The chance of the event = — t-z = ;7- = i-
a-\-b 2a
This is expressed by saying " the event is as likely to
happen as not," or " there is an even chance for the event,"
or " the odds are even " for and against the event.
Thus, in tossing a cent, there is an even chance that it will fall
with the head up.
438. li a>b, the chance of the event happening is > ^.
This is expressed by saying " the event is probable," or
" the odds are as a to ^ in favor of the event."
If a < 6, the chance of the event happening is < ^.
This is expressed by saying " the event is improbable,"
or "the odds are as ^ to a against the event."
Thus, the odds are as 6 to 3 in favor of drawing a white ball at
the jyrst tried from a bag containing 5 white and 3 black balls.
Again, since a die has 6 faces, on one of which is an ace, the chance
for an ace the first throw is \ ; and the odds are 6 to 1 against an ace.
439. It may be shown that,
V. If there are several events of tvhich only one can hap-
pen, the chance that some of them will happen is the sum,
of their respective chances of happening.
CHANCE. 353
For, let a,b,c denote the number of ways favorable to the first,
second, third event respectively ; and let p denote the whole num-
ber of ways, all equally probable, and of which one, and only one, must .
happen. Then the chances of the first, second, third, event are
-, - -, respectively.
P P P
Since there are a-{-h + c+ ways favorable to some one or
other of the events happening, the chance in favor of some one or
other of the events is
a + 6+ c+ a , h , c ,
or - + - + -+
P P P P
If, for example, a bag contains 3 white, 4 black, 5 red, and 6 green
balls, the chance of drawing at the first trial a white or a black ball is
/y + j*g = r^g ; the chance of drawing a white or a black or a red ball
is 3^5 + y*y + y\ = jl ; the chance of drawing a white or a black or a
red or a green ball is ^\ + y*y + j\ + ^s^ = i.| z= i j that is, certainty,
(1) When two dice are thrown, what is the chance of
throwing double aces ?
Each die may fall in any one of 6 ways ; therefore both dice
in 6 X 6 = 36 ways (§ 403). Of these ways only one will give
double aces. Hence, the chance of double aces = ^^g. Ana.
(2) What is the chance of throwing doublets in a single
throw with two dice ?
The dice may fall in 36 ways. Of these, 6 will be doublets.
Hence, the chance of throwing doublets = /^ = J. An^,
(3) What is the chance of throwing a six and a five by a
single throw of two dice ? ""
The dice may fall in 36 ways. Of these ways the first die may
turn up a six and the second a five, or the first may turn up a
five and the second a six. Hence, the chance is ^ = xV- -^^•
(4) With two dice, wliat is the chance of making a throw
so that one and only one die may turn up a five ?
In 6 of the 36 possible ways one die will turn up a five, and
the other also will turn up a five in 6 ways. Two of these 12
ways will be double fives; so that there are 10 ways in which one
die, and only one^ will turn up a. five, and the chance is Jf . Ans,
>r
354 ALGEBRA.
(6) What is the chance of making a throw that will
amount to five ?
Of the 3G possible ways, 1 and 4, 4 and 1, 2 and 3, 3 and 2
amount to Jive, Hence, the chance is y^ = J. Ana.
(6) In a single throw with two dice, if the player may
count the number on one of the dice, or the sum of
the numbers on the two dice, what is the chance of
throwing five ?
The chance is Jg + J + ^'^ (double fives), that is, J|. Ans.
(7) If A's chance of winning a prize is J, and B's J, what
is the chance that neither will obtain a prize ?
The chance that one will win is J + } = /t* Hence, the
chance that neither will win is 1 — /:f = JJ. Ans.
(8) If 4 cards are drawn from a pack of 52 cards, what is
the chance that there will be one of each suit ?
Four cards can be selected (§415) from the pack in
■ 62 X 51 X 50 X 40 __- ___ V
-T^rT>r3i^4- = 2^^'^2^^^y^-
But 4 cards can be selected so as to be one of each suit in
13* = 28,561 ways. § 404.
Hence, the chance is 2V0W5 ~ tu» nearly.
(9) If 4 cards are drawn from a pack, what is the chance
that they will be the 4 aces ?
There are 270,725 ways of drawing four cards, and only one
way of drawing the four aces. Hence the chance is J7-5V13' °^
1 chance in 270,725.
(10) Three balls are to be drawn from an urn containing 5
black, 3 red, and 2 white balls. What is the chance
of drawing one red and 2 black balls ?
10x0x8
Three balls can be selected from the whole 10 in r— -^
1x2X0
= 120 ways. Also, 2 black balls can be selected from the 6 black
V 11 . 5 X 4
balls m — — = 10 ways, and 1 red ball from the 3 red balls in 3
CHANCE. 355
ways. Hence, 1 red and 2 black balls can be drawn in 3 x 10
= 30 ways. That is, there are 120 different ways of drawing
3 balls, and 30 of these ways give 1 red and 2 black balls.
The chance, then, of 1 red and 2 black balls is ^^^ = J. Ana.
(11) If 2 tickets are drawn from a package of 30 tickets,
marked 1, 2, 3, , what is the chance that both
will be marked with odd numbers ?
30 X 29
Two tickets can be drawn from 30 tickets in -7-— „ ways ;
1 X ^
and 2 odd numbers can be drawn from the 15 odd numbers in
^ ^ 2 ^*y®' Hence, the chance is ^^-^^ = /g- ^ns.
(12) In a bag are 5 white and 4 black balls. If they are
drawn out one by one, what is the chance that the
first will be white, the second black, and so on,
alternately ?
The 9 balls can be arranged in [9 ways. The 5 white balls
can be arranged in the odd places, and the 4 black balls in the
even places, in |5 x [4 ways. Hence the chance of alternate
order is ^5 ^ [4
— T9— = ih- ^w«.
(13) -From a bag containing 10 balls, 4 are drawn and re-
placed, then 6 are drawn. Find the chance that
the 4 first drawn are among the 6 last drawn.
The second drawing could be made in
110
j^ = 210 ways.
But the drawing can be made so as to include the 4 first drawn
^ [6
^=15 ways.
since the only choice consists in selecting 2 balls from the 6 not
previously drawn. Hence, the chance is jV^j = it* -^^s.
356 ALGEBRA.
(14) The chance of an event is f . What are the odds in
favor of the event ?
4 to 3. An8.
(15) The odds against an event are 3 to 1. What is the
chance of the event ?
i. Ana.
(16) The odds against an event are m to n. What is the
chance of the event ?
Ans,
m-\- n
(17) If 4 coppers are tossed, what is the chance that ex-
actly 2 will turn up heads ?
Since each coin may fall in two ways, the 4 coins may fall
in 2* = 16 ways. The 2 coins to turn up heads can be selected
4x3
from the 4 coins in = 6 ways. Hence, the chance is
1 X iS
T ff = I ; ^^^ ^^6 odds are 5 to 3 against it.
(18) A has 3 tickets in a lottery where there are 3 prizes
and 6 blanks. Find his chance of winning one
prize, two prizes, three prizes, respectively.
9x8x7
Three tickets can be selected from 9 tickets in
1 X iS X o
= 84 ways. A prize ticket can be selected from the 3 prize
tickets in 3 ways, and 2 blanks can be selected from the 6
6X6
blanks in =15 ways ; therefore, 1 prize and 2 blank
1 X ^
tickets can be selected in 3 X 15 = 45 ways. Hence, the chance
of drawing one prize is f |.
Again, 1 blank and 2 prize tickets can be selected in 6 X
3X2
r-— = 18 ^ays. Hence the chance of two prizes is J|.
Also, the 3 prize tickets can be selected in only 1 way.
Hence, the chance of drawing three prizes is y^j.
CHANCE. 367
(19) What is the chance that A in Ex. (18) wins at least
one prize?
The chance is jj + if +• ^V = ft = }f . For, he will have at
least one prize In any one of the three cases given in (18).
Or, the chance may be found in this way : A gets a prize
unless his three tickets all turn out blanks. Three tickets can
be selected from the whole number in 84 ways, and from the 6
blanks in = 20 ways. Hence, the chance that they
1 X z X o
will all be blanks is f { = /y > ^^^ ^he chance against this result
is 1 - A = if •
440. If a person is to receive a prize in case a particular
event happens, the sum of money for which he may equit-
ably sell his chance for the prize is called his expectation
from the event.
VI. The expectation from an uncertain event is the product
of the chance that the event will happen by the sum to be
realized in ca^e the event happens.
Thus, if there is a lottery with 40 tickets, and 1 prize worth .flOO,
a person might equitably pay $100 for the whole 40 tickets, since one
of them is sure to draw the $100. Now, all of the tickets are of equal
value before the drawing ; hence the value of each ticket is ^\^ of $100.
The value of 6 tickets is ^ of $100, or $12.60 ; that is, the product of
the chance which the holder of 6 tickets has of winning the prize and
the value of the prize ; that is, ^ of $100.
(20) If a lottery has 1 prize of $50, 2 prizes of $5 each,
4 prizes of f 1 each, and 13 blanks, what is the
expectation of the holder of 1 ticket ?
The chance of drawing the prize of $60 is ^Vj a-^d the expecta-
tion is -^ of $60 = $2.60. The chance of drawing a prize of
$6 is 5%, and the expectation is -^^ of $6 = $0.60. The chance
of drawing a prize of $1 is /^, and the expectation is ^V o^ $1
= $0.20. Hence, the whole expectation is $2.60 +$0.50 +
$0.20 = $3.20. Atis.
368 ALGEBRA.
Exercise 122.
1. If I throw a single die, what is the chance that it will
turn up :
(i.) An ace ?
(ii.) An ace or a two ?
(iii.) Neither an ace nor a two ?
2. The chance of a plan succeeding is J. What is the
chance that it fails ?
3. If the odds are 10 to 1 against an event, what is the
probability of its happening ?
4. If the odds are 5 to 2 in favor of the success of an
experiment, what are the respective chances of
success or failure ?
5. The chance of an event is f. Find the odds for or
against the event.
6. What is the chance of a year, not a leap-year, having
63 Sundays ?
7. Two numbers are chosen at random. Find the chance
that their sum is even.
8. If 4 cards are drawn from a pack, what is the chance
that they will all be hearts ?
9. If 10 persons stand in a line, what is the chance that
2 assigned persons will stand together ?
10. If 10 persons form a ring, what is the chance that 2
assigned persons will stand together ?
11. Show that, if n persons sit down at a round table, the
odds against 2 particular persons sitting next to
each other are n — 3 to 2.
CHANCE. 359
12. If 2 letters are selected at random out of the alphabet,
what is the chance that both will be vowels ?
13. Five men, A, B, C, D, E, speak at a meeting, and it is
known that A speaks before B. What is the chance
that A speaks immediately before B ?
14. A, B, C have equal claims for a prize. A says to B,
"You and I will draw lots, and the winner shall
draw lots with C for the prize." Is this fair ?
15. A person is allowed to draw 2 tickets from a bag
containing 40 blank tickets, and 10 tickets each
entitling the holder to a prize of $100. What is
his expectation ?
16. One of two events must happen. If the chance of one
is §■ of that of the other, find the odds on the first.
17. There are 3 events. A, B, C, one of which must happen.
The odds are 3 to 8 on A, and 2 to 5 on B. Find
the odds on C.
18. In a bag are 7 white and 5 red balls. Find the chance
that if one is drawn it will be (i.) white or (ii.) red ;
or, if two are drawn, that they will be (i.) both white,
(ii.) both red, or (iii.) one white and the other red.
19. If 3 cards are drawn from a pack, what is the chance
that they will be king, queen, and knave of the
same suit?
20. A general orders 2 men by lot out of 100 mutineers to
be shot ; the real leaders of the mutiny being 10 in
number. Find the chance (i.) that one only, (ii.)
that two, of the leaders will be shot.
21. Show that the odds are 8 to 1 against throwing 9 in a
single throw with 2 dice.
360 ALGEBRA.
22. Show that in a throw with 3 dice the chance of either
a triplet or a doublet is |.
23. In a bag are 5 white and 4 black balls. If drawn out,
one by one, what is the chance that the first will be
white, the second black, and so on, alternately ?
24. A bag contains 2 white balls, 3 black balls, and 5 red
balls. If 4 balls are drawn, find the chance that
there shall be among them :
(i.) Both the white balls,
(ii.) Two only of the black balls,
(iii.) Two at least of the red balls.
441. A series of events, such that only one of them can
happen, may be called a series of exclusive, or dependent,
events.
Two or more events, such that both or all may happen,
are called non-exclusive, or independent^ events.
Thus, if a copper be thrown twice in succession it may fall head
up both times ; and, if it be thrown ten times, it is possible for it to
fall head up each time.
442. If there are two or more independent events, the
occurrence of all of them simultaneously or in succession
may be regarded as a single compound event.'
Thus, in tossing a copper twice, the event of its falling with head
up at both trials may be regarded as an event compounded of two
simple events ; namely, with head up at the first trial, and with head
up at the second trial.
(1) In tossing a copper twice, what is the chance of its
falling head up both times ?
The chance of a head at each trial is i. If these separate
chances were added (according to Rule V.), the result would be
1 ; that is, certainty ; a result obviously false. Rule V. applies
only to dependent or exclusive events. In this case, however,
the events are independent, or non-ezclusive.
CHANCE. 361
Now, each time the copper is thrown, it can fall in two ways.
Hence, the double fall can occur in 2 x 2 = 4 ways : $ 403.
1. Both times a head.
2. First time a head, second time a tail.
3. First time a tail, second time a head.
4. Both times a tail.
Only one of these four ways gives heads both times. Hence,
the chance of heads both times is J = i X | ; that is, the product
of the separate chances of a head at each trial.
In general,
VII. The chance that two independent events both happen
is the product of their separate chances of happening.
For the product of the denominators of the separate chances is the
whole number of ways in which the compound event can happen;
and the product of the numerators is the number of ways favorable
to its happening.
(2) A bag contains 3 balls, two of which are white;
another contains 6 balls, five of which are white.
If a person draws 1 ball from each bag, what is
the chance that both balls drawn will be white ?
The first ball can be drawn in 3 ways and the second in 6
ways. Hence, both can be drawn in 3 X 6 = 18 ways. Also,
the first ball can be a white ball in 2 ways, and the second in 5
ways. Hence, they can be both white in 2 x 5 = 10 ways. The
chance of both white therefore is if = f x ^ ; that is, the product
of the separate chances of a white ball at each trial.
443. In like manner, whatever the number of simple
events that unite to produce a compound event, it may be
shown that :
VIII. The chance of a compound event is the product
of the separate chances of the simple events that unite to
produce it.
Note. It is important not to confound exclusive events with non-
exclusive, and not to apply Rule V. to problems to which Rule VII.
applies.
362 ALGEBRA.
(3) The chance that A can solve a given problem is f , and
the chance that B can solve it is ^^^. If both try,
what are the chances (i.) that both solve it; (ii.)
that A solves it and B fails ; (iii.) that A fails and
B solves it ; (iv.) that both fail ?
A's chance of success is f , A's chance of failure is i.
B's chance of success is j\, B's chance of failure is ^j.
Therefore, the chance of (i.) is f x j\ = |§ ;
the chance of (ii.) is |^ X ^^ = || ;
the chance of (iii.) is ^ X y\ = /^ j
the chance of (iv.) is i X y^^ = -^^.
The sum of these four chances is i? + H + W + ^V — 1» *s ^^
ought to be, since 1 of the 4 results is certain to happen.
(4) In Ex. (3) what is the chance that the problem will be
solved ?
The chance that both fail is j^^. Hence, the chance that both
do not fail<, or that the problem will be solved, is 1 — /^ = |§.
(5) There are 3 bags, the first containing 1 white, and 1
black ball ; the second, 1 red and 2 white balls ; the
third, 3 white and 2 green balls. If a person draw
a ball from each bag, what is the chance that all
three balls drawn will be white ?
i X f X I = J. Ans,
(6) Under the conditions of the last problem, what is the
chance that no one of the balls drawn will be white ?
The chances of failing to draw a white ball at the three trials
are h h h respectively. Therefore, the chance of failing alto-
er Is i X ^ X I = yij. Ans.
(7) What is the chance in Ex. (5) of drawing at least one
white ball?
One white ball will be drawn unless all three trials fail. The
chance that all three fail is ^^. Therefore, the chance of draw-
ing at least one white ball is 1 — y^ = ii- ^»i«-
CHANGE. 363
(8) What is the chance in Ex. (5) that oTie, and only one,
white ball should be drawn in the three trfals ?
The chance of a white ball from the first bag and not
from the others is i X i X i = t a
The chance of a white ball from the second bag and
not from the others is ixfx§ = 3^
The chance of a white ball from the third bag and not
from the others is i^^xj = j\
Therefore, the sum of these chances is ...... . /^
(9) When 6 coins are tossed, what is the chance that at
least one will fall with the head up ?
The chance that all will fall heads down isixix^xixl
X i = ?V- Hence the chance that this will not happen is
(10) When 6 coins are tossed, what is the chance that one,
and only one, will fall with the head up ?
The chance that the first alone falls with heads up is 4^ x |
X|X|X|X|=^^j; the chance that the second alone falls
with the head up is -^j, and so on.
Hence, the chance that some one, and only one, falls head
upis^V+^'T + ^V + ^'^ + 7i^T+<jV = /i = iV ^^' ■
(11) When 4 dice are thrown, what is the chance that two,
and only tivo, turn up aces ?
The chance that any particular two of the 4 dice turn up
aces, and the other two something else, isjxjx^Xg = y||^.
Now the number of ways in which this can happen is the
. number of ways in which two dice can be selected from 4 dice,
or 6 ways. Hence, the chance that two, and only two, turn up
aces is 6 X -,-|5^ = ^2^^^. Ans.
(12) When 4 dice are thrown, what is the chance that
they will all turn up alike ?
The chance that the first and second turn up alike is J.
The chance that the third turns up like the first and second
to*.
The chance that the fourth turns up like the others is J.
Hence, the chance that the four turn up alike is ^J-^. An9,
364 ALGEBRA.
(13) When 4 dice are thrown, what is the chance that two,
and only two of them, should turn up alike ?
The chance that any two should be alike is ^, the chance
that the third should be different is f , and the chance that the
fourth should be different from all the rest is J. Hence, the
chance that a pair should agree while the others should differ
from the pair and from each other is J x J X J = ^\Q^. The
4x3
pair to agree may be selected in = 6 ways. Hence, the
1 X i2
total chance is 6 X ^x\ = f • -^^•
(14) When 4 dice are thrown, what is the chance that
they should all fall different?
The chance that the second should differ from the first is f ,
that the third should differ from both the first and the second
is i, and that the fourth should differ from all the others is f .
Hence, the required chance is J X | X | = 5^^. Ans.
(15) A single die is thrown until it turns up an ace.
What is the chance that it must be thrown at least
10 times? What is the chance that it must be
thrown exactly 10 times ?
The chance of failing the first 9 times is (Rules IV. and VH.)
(D®. This, then, is the chance that at least 10 trials must be
made. Since (1)^ is the chance of failing the first 9 trials, and
J the chance of success the next trial ; therefore (Rule VII) (f )*
X ^ is the chance that exactly 10 throws must be made.
(16) What is the chance that a person with 2 dice 'will
throw double aces exactly 3 times in 5 trials ?
The chance of throwing double aces at any particular trial is
1 X J = yi^, and of failing is ||. Hence, the chance of succeeding
at 3 assigned trials, and failing at the other 2 trials, is (3^)^ X (f 1)^.
Now double aces will be thrown exactly three times if thrown in
any set of 3 trials that may be assigned out of the 5 trials, and
fail in the other 2 trials. 3 trials can be assigned out of- 5 trials
6x4x3
in ^ ^ ^ ^ g = 10 ways. Hence, the chance is {^\)^ X (f|)2x 10,
CHANCE. 365
(17) A and B throw with a single die alternately, A throw-
ing first ; and the one who throws an ace first is to
receive a prize of |10. What are their respective
expectations ?
The chance for the prize at the first throw is } ; at the second,
4 X J J at the third, (f )2 X i ; at the fourth, (^)s x ^ ; and so on.
As A has the first, third, etc., and B the second, fourth, etc.,
throws,
A^s chance
1^ ^+(5)«of} + =6.Bothat
^ tofi + m8ofi+ -f>«othat,
B's chance f of J + (|)8
A's expectation is ^j of $10 = |6 j\, and
B's expectation is ^y of |10 = ^x\.
(18) A and B play at a game that cannot be a drawn game,
and on an avej'age A wins 3 games out of 5 games.
Out of 5 games what is the chance that A wins at
least three ?
The chance that A wins 3 assigned games out of 6 games is
(1)^ ^ (i)^~ ITSTJ- Th® 3 games may be assigned in 10 ways.
Hence, A's chance for 3 games is 10 X ^^/s — MfS-
The chance that A wins 4 games, and B the other game, is
ttV? X 5 = ^^^y. The chance that A wins all the games is
For A to win at least 3 games, he must win 3, 4, or 5 games.
Hence, A*s chance for at least 3 games is JJJJ + -g\^^^ + -^^-^
= f IB.
(19) A's skill at a game, which cannot be a drawn game,
is to B's skill as 3 to 4. If they play 3 games, what
is the chance that A will win more games than B ?
Their respective chances of winning a particular game are
f and ^. For A to win more games than B, he must win all 3
games or 2 games. The chance that A wins all three is (f )' =
^. The chance that A wins any assigned set of 2 games out
of the 3 games, and that B wins the other, is (f )2 x ^. As there
are 3 ways of assigning a set of 2 games out of 3, the chance
that A wins 2 games, and B the other, is (f )« x ^ X 3 = JJf .
Hence, the chance that A wins more than B is ^^ + Jf f = ^f J.
366 ALGEBBA.
(20) In the last example, find B's chance of winning more
games than A.
B's chance of winning all three games is (^)' = yVy. His
chance of winning 2 games, and A the other game, is (f )^ X ^
X 3 = JJf . Hence, his chance of winning more games than A
Notice that A's chance added to B's chance, Jff + |Jf = 1.
Why should this be so ?
(21) A plays a set of games (drawn games excluded) with
B, his chance of winning a single game being to
B's as 3 : 2. What is the probability :
(i.) That A will win 4 games at least out of 7 ?
(ii.) That he will win 4 games before B wins 3 ?
(i.) A's chance of winning a single game is |, and B^s chance
is f . The chance that A wins at least 4 games out of 7 is the
sum of the chances that he wins 4, 5, 6, or 7 games out of 7,
in any possible order.
The chance that he wins all 7 games = (f )'^.
The chance that he wins 6 games = 7 (1)^ X §.
The chance that he wins 5 games = 21 (f)^ X (f )2.
The chance that he wins 4 games = 35 (§)* X (§)».
The sum of these values = — tt— . Ans.
5*
(ii.) Here the chance required is that A shall win at least 4
games ; that is, 4, 6, or 6 games out of 6,
The chance that he wins all 6 games = (|)®.
The chance that he wins 5 games = 6 X (3)5 x |.
The chance that he wins 4 games = 15 X (3)* x (f)2.
The sum of these values = — r^. Ans.
(22) In a certain locality it is found that, on the average
for 10 years, out of 100 persons 40 years old at the
beginning of the decade, 20 die ; out of 100 per-
sons 50 years old, 30 die ; and out of 100 persons
60 years old, 40 die.- What is the odds against a
person 40 years old living 30 years longer ?
CHAi^cfi. Set
The chance that he dies between 40 and 60 is ^ ; that he lives
till 60, and dies between 50 and 60, is J X ^^^ = ^^ ; that he lives
till 60, and dies between 60 and 70, is ^ X ^^ X j*f^ = ^2^8.. Hence,
the chance that he dies between 40 and 70 is J + /§ + ^ =
^^. Therefore, the odds against his living for 30 years are 83
to 42, or about 2 to 1.
(23) A is 40 years old and B 50 years old. What is the
probability that at least one of them will be alive
10 years hence ?
The chance that A dies is ^, and the chance that B dies is ^.
Hence, the chance that both die is J x j^^ = ^\ ; and the chance
that one at least will be alive is 1 — ^^ = |J.
444. Cases often occur where the simple events which
iinite to form the compound event are so related that the
happening of one of them alters the chances of the others.
(1) What is the chance of drawing in succession 2 vowels
from the alphabet ?
The chance of drawing a vowel the first time is /^ ; but, if
one vowel is drawn, the chance of drawing another is /^. Hence,
(Rule VII.) the required chance is ^% X /^ = ■^^.
(2) A bag contains 5 white and 6 black balls. What is
the chance of drawing 5 times in succession a white
ball, the balls drawn not being replaced ?
■A X A X 5X f X } = 1^^' ^ns.
(3) What would have been the chance in the last example
if after each drawing the ball had been replaced ?
(A)6. Ans.
(4) If the chance of an event at first is as a to b, and if
whenever it happens, the number of favorable ways,
as well as the whole number of ways, is diminished
by unity ; find the chance that the event will occur
n times in succession.
a(a-l)(a-2) (a-n + 1). .
6(6-l)(6-2) (6-n+l)
368 ALGEBRA.
(6) A bag contains 5 white and 6 black balls. If 5 balls
are drawn in succession, and no one of them replaced,
what is the probability that the first three will be
white, and the fourth and fifth black ?
The separate chances for the 6 simple events are respectively
T*T» A» l» s» ?• Hence, the chance for the compound event is
(6) Find the probability in the last example that the 5
balls drawn will be 3 white and 2 black balls.
Here the chance required is that 3 white and 2 black should
be drawn not in any assigned order^ as in the last case, but in
any possible order. Now 5 things, of which 3 are alike and the
other 2 alike, may be arranged (§ 411) in
-^=10 ways.
Hence, the probability is 10 X r|f = 8f.
(7) Find the respective probabilities in Examples (5) and
(6) if after each drawing the ball is replaced.
In Ex. (6), (,*r)8 X (,\)a. Ans,
In Ex. (0), 10 X (i\)8 X (j\)^ Ans,
(8) A purse contains 9 silver dollars and 1 gold eagle, and
another contains 10 silver dollars. If 9 coins are
taken out of the first purse and put into the second,
and then 9 coins are taken out of the second and
put into the first purse, which purse now is the
more likely to contain the gold coin ?
The gold eagle will not be in the second purse unless it (i.)
was among the 9 coins taken out of the first and put into the
second purse, (ii.) and not among the 9 coins taken out of the
second and put into the first purse. The chance of (i.) is ^^^ and
when (i.) has happened the chance of (ii.) is }§. Hence, the
chance of both happening is /^^ X J§ = j\. Therefore, the chance
that the eagle is in the second purse is /g, and the chance that
it is hi the first purse is 1 — /g = {^. Since fj is greater than
x®s, therefore the gold coin is more likely to be in the first purse.
CHANCE. 369
(9) In a bag are 2 red and 3 white balls. A is to draw a
ball, then B, and so on alternately ; and whichever
draws a white ball first is to receive $10. Find
their expectations.
A's chance of drawing a white ball at the first trial is |. B*s
chance of having a trial is equal to A^s chance of drawing a red
ball = |. In case A drew a red ball there would be 1 red and 3
white balls left in the bag, and B's chanpe of drawing a white
ball would be f. Hence, B's chance of having the trial and
drawing a white ball is f X f = ^^^ ; and B's chance of drawing
a red ball is | X ^ = j\.
A's chance of having a second trial is equal to B's chance of
drawing a red ball = ^J^. In case B drew a red ball there would
be 3 white balls left, and A's chance of drawing a white ball
would be certainty i or 1.
A's chance, therefore, is f + jV = A ; and B's chance is ^^j.
A's expectation, then, is |7, and B's $3.
446. In general, when it is required to find which of
two doubtful events is more likely to happen, it is neces-
sary to find their respective chances, and then to compare
the results obtained.
(1) In one throw with two dice which sum is more likely
to be thrown, 9 or 12 ?
Out of the 36 possible ways of falling, four give the sum 0
(namely, 6 + 3, 3 + 6, 5 + 4, 4 + 5), and only one way gives 12
(namely, 6 -h 6). Hence, the chance of throwing 9 is /our times
as good as that of throwing 12.
(2) With three dice what are the relative chances of
throwing a doublet and a triplet?
The chance of throwing a doublet is
The chance of throwing a triplet is
6 X 1 X 1 _ J
68 ^®'
Hence, the chance of a doublet is 15 times that of a triplet.
370 ALGEBRA.
(3) A bag contains 1 black and 4 white balls, and another
bag contains 7 black and 3 white balls. If a person
draws a ball from one of the bags, (i.) what is the
chance that it be a white ball ? (ii.) what is the
ratio of the chance of its being drawn from the first
bag to that of its being drawn from the second bag ?
The person (so far as we know) is as likely to choose one bag
as the other. Hence, the chance of his choosing the first bag is
i ; and the chance of his drawing a white ball from the first bag
is jj. Therefore, the chance of drawing a white ball from the
first bag is i X I = J. In the same way, the chance of drawing
a white ball from the second bag is found to be | x ^\ = ^^.
Therefore, the chance of drawing a white ball is § + ^% = J J ;
and the ratio of the separate chances is 8 : 3.
(4) Suppose in the last example that at the first trial a
white ball is actually drawn. What are now the
chances that it came from the first bag, and from
the second, respectively ?
Let X and y represent the chances required.
Then, by Ex. (3), ^ = |.
Also, X + y = 1,
since the ball must have come from one or the other bag.
The solution of these equations gives
x=r\, and y=^\,
446. From Examples (3) and (4) it will be seen that,
IX. If a doubtful event may happen in some one of several
ways, the actual happeniufj of the event changes its proha-
bllify, and the separate probabilities of the several ways of
hapjteninfj, in the same ratio; and this ratio is the reciprocal
of the fraction that expresses the chance of the event before
it actually happens.
Thus, in Ex. (3), the chance of the event before it happened, and
the chance of the two separate ways of happening, were found to be
CHANGE. 371
ih /{ji f'd 't ^° ^^- (^) ^^ ^^ shown that q/Ker tAe event happened these
chances became 1, j^j, ^^^.^ respectively. The values I, y\, j?j are ob-
tained by multiplying JJ, /^y, ^^^ by f f ; that is, by the reciprocal of J^.
Evidently, the happening of the event must change to unity the
chance of the event ; and must therefore increase to unity the sum of
the separate chances.
So long, however, as the only additional knowledge about the event
is the fact that it has happened, the relative probabilities of the sepa-
rate ways of happening remain unchanged. Therefore, the several
fractions which before expressed the probabilities of the separate ways
of happening must now be multiplied by the same factor, and that
factor is the reciprocal of the fraction that expressed the probability
of the event before it happened.
Exercise 123.
1. The chance that A can solve a certain problem is J, and
the chance that B can solve it is f . What is the
chance that the problem will be solved if both try ?
2. What is the chance of throwing at least one ace in 2
throws with one die ?
3. If n coins are tossed up, what is the chance that one,
and only one, will turn up head ?
4. What is the chance of throwing double sixes at least
once in 3 throws with 2 dice ?
5. A copper is tossed 3 times. Find the odds that it will
fall:
(i.) One head and two tails without regard to order,
(ii.) Head, tail, head.
6. If a copper is tossed 4 times, find the odds that it will
fall 2 heads and 2 tails sooner than 4 heads.
7. If from a lottery of 30 tickets, marked 1, 2, 3, , four
tickets are drawn, what is the chance that 1 and 2
will be among them ?
372 ALGEBRA.
8. If 2 coppers are tossed 3 times, find the odds that they
will fall 2 heads and 4 tails.
9. There are 10 tickets, five of which are numbered 1, 2,
3, 4, 5, and the other five are blank. Find the
chance that the sum of the numbers on the tickets
drawn in 3 trials will be 10, one ticket being drawn
and then replaced at each trial ?
10. Find the chance in Ex. 9 if the tickets are not replaced.
11. A bag contains 4 white and 6 red balls. A, B and C
draw each a ball, in order, replacing. Find the
chance that they have drawn :
(i.) Each a white ball.
(ii.) A and B white, C red.
(iii.) Two white and one red.
12. Find the answer to Ex. 11 if the balls are not replaced.
13. A draws 4 times from a bag containing 2 white and 8
black balls, replacing. Find the chance that he
has drawn :
(i.) Two white, two black.
(ii.) Not less than two white,
(iii.) Not more than two white,
(iv.) One white, three black.
14. Find the odds against throwing one of the two numbers
7 or 11 in a single throw with 2 dice.
15. If a copper is tossed 5 times, what is the chance that
it will fall heads either 2 times or else 3 times ?
16. Find the same chance if the copper is tossed 6 times.
17. In one bag are 10 balls and in another 6 ; and in each
bag the balls are marked 1, 2, 3, etc. What is the
chance that on drawing one ball from each bag the
two balls will have the same number ?
CHANGS. 373
18. A bag contains n balls. A person takes out one ball
and then replaces it. He does this n times. What
is the chance that he has had in his hand every ball
in the bag ?
19. K on an average 9 ships out of 10 return safe, to port,
what is the chance that out of 5 ships expected at
least 3 will return ?
20. What is the chance of throwing double sixes at least
once in 3 throws with a pair of dice ?
21. What is the chance of throwing 15 in one throw with
3 dice ?
22. In 5 throws with a single die what is the chance of
throwing an ace :
(i.) Three times exactly ?
(ii.) Not less than three times ?
(iii.) Not more than three times?
23. In a bag are 3 white, 5 red, and 7 black balls, and a
person draws three times, replacing. Find the
chance that he has drawn:
(i.) A ball of each color.
(ii.) Two white, one red.
(iii.) Three red.
(iv.) Two red, one black.
24. A and B play at chess, and A wins on an average 2
games out of 3. Find the chance of A's winning
exactly 4 games out of the first 6, drawn games
being disregarded.
26. A and B engage in a game in which A's skill is to B's
as 2 : 3. Find the chance of A's winning at least 2
games out of the first 5, drawn games not being
counted.
874 AL6EBBA.
26. The skill of A is double that of B. Find the odds
against A's winning 4 games before B wins 2.
27. If B's skill in a certain game is equal to three-fifths of
A's, find A's chance of winning 5 games out of 8.
28. A bag contains 4 red balls and 2 others, each of which
is equally likely to be red or white. Three times
in succession a ball is drawn and replaced. Find
the chance that all the drawn balls are red.
29. A man has left his lunbrella in one of 3 shops which
he visited in succession. He is in the habit of
leaving it, on an average, once every 4 times that
he goes to a shop. Find the chance that he left it
in the first, second, and third shops, respectively.
30. A bets B $10 to $1 that he will throw heads at least
once in 3 trials. What is B's expectation ? What
would have been a fair bet ?
31. A draws 5 times (replacing) from a bag containing 3
white and 7 black balls; every time he draws a
white ball he is to receive $1, and every time he
draws a black ball he is to pay 50 cents. What is
his expectation ?
32. From a bag containing 2 eagles, 3 dollars, and 3
quarter-dollars, A is to draw one coin and then B
three coins ; and A, B, and C are to divide equally
the value of the remainder. What are their expec-
tations ?
33. A, B, and C, staking each $5, draw from a bag in which
are 4 white and 6 black balls, each drawing in order,
and the whole sum is to be received by him who first
draws a white ball. What are their expectations ?
(i.) Replacing the balls,
(ii.) Not replacing the balls.
CHAPTER XXIV.
Formulas.
Simple Interest.
447. If in Interest
The principal is represented by P,
interest on $1 for one year by r,
amount of $1 for one year by R,
number of years by n,
amount of P for n years by A,
Then E = l-\-r,
Simple interest on P for a year = Pr,
Amount of P for a year = PR,
Simple interest on P for n years = Pnrj
Amount of P for n years = P (1 + nr).
That is A = P{l + nry
■448. When any three of the quantities A, P, n, r are
given, the fourth may be found.
Ex. Required the rate when $500 in 4 years at simple
interest amounts to $610.
Here r is requii^d, A, P, n are given.
A = P{l-\-nr),
or A = P-\- Pnr.
.-. Pnr =A — P.
A — P 610 — 600 ^ ^.-
. ». ;::; := r= 0.066.
Pn 2000
bi per cent. Arts,
376 ALGEBRA.
449. Since P will in n years amount to A, it is evident
that P at the present time may be considered equivalent in
value to A due at the end of n years ; so that P may be
regarded as the present worth of a given future sum A,
Ex. Find the present worth of $600, due in 2 years, the
rate of interest being 6 per cent.
A = P (1 + nr),
' P = ^ = ^^^ = «635 71
••^ 1 + nr 1 + 0.12 '^^-^l-
Compound Interest.
460. When compound interest is reckoned payable annur
ally.
The amount of P dollars in
1 year is P(l + r) = PR,
2 years is PB (1 + r) = PB',
n years = PR\
That is, A = PR\
Hence, also, -P = ^;*
When compound interest is payable semi-annually.
The amount of P dollars in
iyear =p(n-Q»
lyear =p(l + g',
^ "'"2/
That is, ^ = p(l+^)*"'
When the interest is payable quarterly,
FORMULAS. 377
When the interest is payable monthly ^
When interest is payable q times a year,
Ex. Find the present worth of $500, due in 4 years^ at
5 per cent compound interest.
••• ^ = (TT^ = (S = »*"•««• ^-
Sinking Funds.
461. If the sum set apart at the end of each year to be
put at compound interest is represented by aS^^ then.
The sum at the end of the
first year = S,
second year = /SH- SH,
third year =S+SB+SE^,
nth year =S+SE+SE^+ + SB^\
That is, the amount A = S+SE+SE^+ + SE^\
.\AE=SE+SE'+SE'+ + a9JB«.
,\AE-A = SE^ — S,
' : "^ E-i '
r
(1) If $10,000 be set apart annually, and put at 6 per cent
compound interest for 10 years, what will be the
amount?
_ ^ (1ft - 1) _ 810,000 (i.oe^Q - 1)
r 0.00 '
By logarithms the amount is found to be |131,740 (nearly).
378 ALGEBRA.
(2) A county owes $60,000. What sum must be set apart
annually, as a sinking fund, to cancel the debt in 10
years, provided money is worth 6 per cent?
„ Ar J>60,000 X 0.06 ^.^^^ ,
^ = ^r=ri= 1.0610-1 =^55MneaWy).
Note. The amount of tax required yearly is '$3600 for the iTvterest
and 14555 for the sinking fund ; that is, |8155.
Annuities.
452. A sum of money that is payable yearly, or in parts
at fixed periods in the year, is called an annuity.
I. To find the amount of an unpaid annuity when the
interest, time, and rate per cent are given.
The sum due at the end of the
first year = S,
second year = /5+ SB,
third year ^S+SR+SIP,
nth year ^ S -\' SR + SIP + + SBr-\
That is, A = ^ ^« $ 451.
r
Ex. An annuity of $1200 was unpaid for 6 years. What
was the amount due if interest is reckoned at 6 per
cent?
_^(/?n-l)_ {11200(1.066-1) _
A- ^ - ^^ - $8370.
II. To find the present worth of an annuity when the time
it is to continue and the rate per cent are given.
Let P denote the present worth. Then the amount of P for n
years will be equal to A^ the amount of the annuity for n years.
But the amount of P for n years
= P(l>r)»=PJ?",
and ^ = ^^lZlD. 5 461.
FORMULAS. 379
This equation may be written
^ R-1^ R-
If the annuity is perpetual, the fraction
approaches to unity as its limit.
.-. P = limit of — — - X — - — = — — - = —
ie — 1 R» R—l r
(1) Find the present worth of an annual pension of $105,
for 5 years, at 4 per cent interest.
(2) Find the present worth of a perpetual scholarship that
pays $300 annually, at 6 per cent interest.
S_$300_
III. To find the present worth of an annuity that begins
in a given number of years, when the time it is to continue
and the rate per cent are given.
Let p denote the number of years before the annuity begins, and
q the number of years the annuity is to continue.
Then the present worth of the annuity to the time it terminates is
S Rp+1 - 1
Rp+<i R-l '
and the present worth of the annuity to the time it begins is
Rp E—1
Hence, P = (^^, x -^z-t) " ( K^ ^ ■¥^}'
• • iJP+« B-l
880 ALOEBKA.
If the annuity is to begin at the end of p yean, and to be perpetual,
tbefomraU
JtP+t B-l
R9— 1
And since the limit of — - — is unity,
P = the limit of rr-T^ — 77 X ^ " ^ - ^
RP{R-1) B9 BP{R-l)
(1) Find the present worth of an annuity of $5000, to
begin in 6 years, and to continue 12 years, at 6 per
cent interest.
Rp+9 R-1
.$6000 1.06"- 1
1.06M 0.06
= 129,550.
(2) Find the present worth of a perpetual annuity of
$1000, to begin in 3 years, at 4 per cent interest.
8 _ 11000 _^.»..,
^ ~ UP (iJ - 1) - 1.04» X 0.04 ~ ^^'^^^'
IV. To find ths annuity when the present worthy the time,
and the rate per cent are given.
S(g— 1)
If{R—l)
J2" — 1 iJ» — 1
(1) What annuity for 5 years will $4675 give when inter-
est is reckoned at 4 per cent ?
Rh 1 04,6
S=Prx ^IT— [ = W«75 X 0.04 X ^^^^i-j. = |1060.
formulas. 381
Life Insurance.
453. In order that a certain sum may be secured; to be
payable at the death of a person^ he pays yearly a fixed
pretntum.
If P denotes the premium to be paid for n years to insure an
amount ^, to be paid immediately after the last premium, then
A = ^f!^- J451.
ii 1
...p=^<^-l) =
Ar
If ^ is to be paid a year after the last premium, then
j.^A{R-l)^ Ar .
i2(i2»-l) R{R^-1)
Note. In the calculation of life insurance it is necessary to
employ tables which show for any age the probable duration of life.
Bonds.
454. If F denote the price of a bond that has n years to
run, and bears r per cent interest, *Si the face of the bond,
and q the current rate of interest, what interest on his
investment will a purchaser of such a bond receive ?
Let X denote the rate of interest on the investment.
Then P (1 + »)* is the value of the purchase money at the end of
n years.
8r (1 + 9)«-i + Sr (1 + q)^^ + + -8r + -8 is the amount of the
money received on the bond if the interest received from the bond is
put immediately at compound interest at q per cent.
But5r(H-g)*-i + 5r(l + g)'^2+ +Sr-\-8=8-\- M(^+g)*-^]-
Pq
_/Sq-^8r(l + q)^-Sr\^
'{'
S82 ALGEBRA.
(1) What interest does a person receive on his investment
if he buys a 4 per cent bond, at 114, that has 26
years to run, and if money is worth S^ per cent ?
^ . ^ _ /3.6 + 4 (1.035)26 -4\ A
^"^^"v 3.99 ; *
By logarithms, 1 + x = 1.033.
That is, the purchaser will receive 3^ per cent for his money.
(2) At what price must 7 per cent bonds be bought, run-
ning 12 years, with the interest payable semi-annu-
ally, in order that the purchaser may receive on his
investment 5 per cent, interest semi-annual, which is
the current rate of interest ?
pa +x)n= Sq+Sr{l + q)n^Sr^
. p- Sq+Sr(l + q)^-Sr
?{l + x)»
In this case S = 100 ; and, as the interest is semi-annual,
q = 0.025, r = 0.035, n = 24, z = 0.025.
Hence P - 2.5 + 3.5(1.025)^-3.5,
Hence, r- 0.026(1.025)2*
By logarithms, P = 1 18.
Exercise 124.
1. In how many years will $100 amount to $1050, at 5
per cent compound interest ?
2. In how many years will $A amount to $B (i.) at sim-
ple interest, (ii.) at compound interest, r and B being
used in their usual sense ?
3. Eind the difference (to five places of decimals) between
the amount of $1 in 2 years, at 6 per cent compound
interest, according as the interest is due yearly or
monthly.
4. At 5 per cent, find the amount of an annuity A which
has been unpaid for 4 years.
FOBMULAS. 383
6. Find the present value of an annuity of ^100 for 5
years, reckoning interest at 4 per cent.
6. A perpetual annuity of $1000 is to be purchased, to
begin at the end of 10 years. If interest is reckoned
at 3^ per cent, what should be paid for it ?
7. A debt of $1850 is discharged by two payments of
$1000 each, at the end of one and two years. Find
the rate of interest paid.
8. Reckoning interest at 4 per cent, what annual premium
should be paid for 30 years, in order to secure $2000
to be paid at the end of that time, the premium
being due at the beginning of each year ?
9. An annual premium of $150 is paid to a life-insurance
company for insuring $5000. If money is worth 4
per cent, for how many years must the premium be
paid in order tliat the company may sustain no loss ?
10. What may be paid for bonds due in 10 years, and
bearing semi-annual coupons of 4 per cent each, in
order to realize 3 per cent semi-annually, if money
is worth 3 per cent semi-annually ?
11. When money is worth 2 per cent semi-annually, if
bonds having 12 years to run and bearing semi^
annual coupons of 3^ per cent each, are bought at
114 J, what per cent is realized on the investment?
12. If $126 is paid for bonds due in 12 years, and yielding
3^ per cent semi-annually, what per cent is realized
on the investment, provided money is worth 2 per
cent semi-annually ?
13. A person borrows $600.25. How much must he pay
annually that the whole debt may be discharged in
35 years, allowing simple interest at 4 per cent ?
384 ALGEBRA.
14. A perpetual annuity of $100 a year is sold for 1^2500.
At what rate is the interest reckoned ?
16. A perpetual annuity of $320, to begin 10 years hence,
is to be purchased. If interest is reckoned at 3^
per cent, what should be paid for it ?
16. A sum of $10,000 is loaned at 4 per cent. At the end
of the first j^ear a payment of $400 is made ; and at
the end of each following year a payment is made
greater by 30 per cent than the preceding payment.
Find in how many years the debt will be paid.
17. A man with a capital of $100,000 spends every year
$9000. If the current rate of interest is 6 per cent,
in how many years will he be ruined ?
18. Find the amount of $365 at compound interest for 20
years at 5 per cent.
19. In how many years will $20 amount to $160, at 4 per
cent compound interest ?
20. At what rate per cent, compound interest, will $2500
amount to $3450 in 7 years ?
21. If the population of a state increases in 10 years from
2,009,000 to 2,487,000, find the yearly rate of in-
crease.
22. The population of a State now is 1,918,600, and the
yearly rate of increase is 2.38 per cent. Determine
its population 10 years hence.
23. A banker borrows a sum of money at 3^ per cent,
interest payable annually, and loans the same at 5
per cent, interest payable quarterly. If his annual
gain is $441, determine the sum borrowed.
CHAPTER XXV.
Continued Fractions.
456. A FKACTioN in the form of
a
'' + 7+ etc.
is called a Crontinned Fraction, though the term is commonly
restricted to a continued fraction that has 1 for each of its
numerators^ as \
^^r+etc.
456. Any proper fraction in its lowest terms may he con-
verted into a terminated contintced fraction.
Let - be such a fraction ;
a '
then " ^
a a , c
(if p is the quotient and c the remainder of o -r 6) ;
1 1 1
and
1 <^ j^ 1^ I
q + -
c ^ c
(if g IB the quotient and.cZ the remainder of & -r c),
__ 1
386 ALGEBRA.
The soccesslTe steps of the process axe the same as the steps for
finding the G.C.M. of a and h ; and since a and b are prime to each
other, a remainder, 1, will at length be reached, and the fraction
terminates.
457. The fractions formed by taking one, two, three,
of the quotienti( p, q, r, , are
11 1
— > -}
P+- P+-j_
which simplified are ^ r
1 q qr + 1
i?' pq + l' (lJq + l)r+p'
and are called the first, second, and third conyergents,
respectively.
468. The successive convergents are alternately greater
and less than the true valu^ of the given fraction.
1
Let X be the true valae of
p+^
then, since j), g, r, are positive iidegerSy
P<P + J
«"^r + etc.
•'• ~ > 7 J t-l^at is, - > a;.
^ P+-4 ^
Again, 9<7 +
r + etc.
« + r + eto.
• ^< t; that is, 7<x;
P + - P+—J P + \
f + etc., and so on.
CONTINUED FRACTIONS. 387
459. Tf —y —y ~ are any three consecutive convergents,
and if tslx, tsl^, nis are the quotients that produce them, then
lis III8TI2 + Hi
For, if the first three quotients are p, q, r, the first three conver-
gents are
1 1 1 ,. ,
that is, i, -L, , Zt\ ' ("•)
p pq+1 (pq-\-l)r + p ^ '
From (i.) it is seen that the second convergent is formed from the
first by writing in it p + - f or p ; and the third from the second by
writing q+ - for q. In this way, any convergent may be formed
r
from the preceding convergent.
t/fl Us 1
Therefore, — will be formed from — by writing 1112 H for mz.
Vs t2 wig
In (ii.) it is seen that the third convergent has its numerator
= r X (second numerator) 4- (first numerator) ; and its denominator
= r X (second denominator) + (first denominator).
Assume that this law holds good for the third of the three consecu-
tive convergents
uo til M2 *v * ^ W*2Wl + t*o
— » — I — » so that, — = ; — ;
1?0 ^1 V2 1>2 WI2U1 "T Vq
then, since, by (ii.), — is formed from — by using ni2 H for m2,
U9 __ V msJ ms (mgUi + up) + Ui.
Substitute U2 and «2 for their values mzUi + uq and m2Vi + Vq;
then «.^m,u, + ui.
Us Tflz'02 -T Vi
Therefore, the law still holds good ; and as it has been shown to be
true for the third convergent, the law is general.
388 ALGEBRA.
480. The difference between two eaneectUive eonvergents
Hi . Ha . 1
— and — IS
The difference between the first two conyergents
1 2_= 1
P 1>7+1 P{pq+l)
Let the sign ^ mean *Hhe difference between/* and assome the
propoeition true for
— and — 80 that — "^ — — — ;
i>o 'i ©0 t?i ©ot^i •o'^i
tt« Ml _ U20i^Mi02
tnen T^^~ ~ »
__ (matti + Mq) Pi ^ ui (mgDi + Vp)
(by BubBtituting for Us and Oa their values, miui + uo and msVi + no)*
= — (by the assumption).
Hence, if the proposition be true for one pair of consecutive eon-
vergents, it will be true for the next pair ; but it has been shown to be
true for the^r^ pair, therefore it is true for every pair.
Since the real value of x lies between two consecutive eonver-
gents, — and — » -^ will differ from x by a quantity less than -^^^— ;
•i t2 ©1 Vi v%
1 Ui
that is, by a quantity < — ; so that the error in taking — f or x is
ViVi Vi
< — , and therefore < —i'
Any convergent, — , is in its lowest terms ; for, if Ui and Vi had any
common factor, it would also be a factor of mvi ^^^ Mjti (5 146) ; that
is, a factor of 1.
CONTINUED FRACTIONS. 389
461. The sticcessive convergents approach more and more
nearly to the true value of the continued fraction.
Ua Ui Us
Let — » — » — be consecutive conversents.
uo ©1 «a
Now — differs from a;, the true value of the fraction, only by using
©2
m% instead of m2 + \ •
ws + etc.
Let this complete quotient, which is always greater than unity, be
represented by M,
^, . U2 wiatti + uo Mu\ + 1*0
Then, smce — = ; ' x = .. , •
Mi __ MU\ + Uq U\ _ UqVi — U\Vq __ 1
^ Vi~ Mvi + i?o '^ »i "" Ui (Mvi + Vo) ~ Vi(Mvi +Vo)
And ^o,-^j; = ^Q,^-^"' + Uq_M{uoV]->-U]Vo) _ M
Now 1 < Jf and ti > i?o, and for both these reasons :
a; 'w — < — ^^ a;.
ui ro
That is, — is nearer to x than — is.
«i - i?o
462. Any convergent — is nearer the true value j: than
any other fraction with smaller denominator.
Let r he a fraction in which 6<ui.
If 7- is one of the convergents, x ^ -<C. — '^x, % 461.
If r is not one of the convergents, and is nearer to x than — is,
0 V\
then, since x lies between — and — » r must be nearer to — than —
is ; that is,
O U2 ^U\ W2 'C2a'^U2h ^ 1
O ^2 ''l ^2 ^2^ ^i'H
and since & < »i» this would require t>2a ^ uzb to be < 1 ; but wi ^
U2& cannot be less than 1, for a, &, 1/21 V2 are all integers.
390 ALGEBRA.
Applications.
(1) Find the continued fraction equal to ^^, and also the
successive convergents.
By following the process of finding the G.C.M. of 31, 75, the
successive quotients are found to be 2, 2, 2, 1, 1, 2. Hence the
continued fraction is
!+-^
2 + -1
2 + -L
l + -i
1 + i
To find the successive convergents :
Write the successive quotients in line, ^ under the first
quotient, i under the second quotient^ and then multiply
each term by the quotient above it and add the term to the
left to obtain the corresponding term to the right. Thus,
Quotients = 2, 2, 2, 1, 1, 2.
Convergents = ?, i, §, y\, ^j, i|.
Note. It is convenient to begin to reckon with f , but the next
convergent, in this case i, is called the first convergent.
463. A quadratic surd may be expressed in the form of
a norirterminating continued fraction.
(2) To express Vs in the form of a continued fraction.
Suppose , Vs = 1 4- - (for 1 is the greatest integer in VS),
then 1 = V3 — 1.
X
CONTINUED FBACTION8. 391
Suppose — - — = 1 + - (^ for 1 is the greatest integer in — - — ji
,, 1 V3 4-1 , V3-1
then - = — 1 =
y 2 2
2 V34-1
..,_ _ {by§278).
V3-1 1
Vs + i 1/ V3 + l\
Suppose — :; = 2 + -las2is the greatest integer in ; ) J
then l=.V3_f.l_2^^_j
z 1
.-. z = —= (the same as x above).
V3-1
.-. VS = 1 + — r , of which will be a continu-
^ + 2 + etc. ^ + 2
ally-repeating series, and the whole expression may b6 written,
1 + i i.
^1 + 2
The convergents will be 1, 2, |, J, |f , f f, Jj, etc.
464. A continued fraction in which the denominators
recur is called a periodic continued fraction.
465. The value of a periodic continued fraction can be
expressed as the root of a quadratic equation.
(3) Find the surd value of - -•
1 -f-z
then
Let X be the value ;
1
1 + :^'
2 + x
that is, X = r-r— »
whence x = — 1 + Vs.
392 ALGEBRA.
466. An exponential equation may be solved by con-
tinued fractions.
(4) Solve 10* = 2, by continued fractions.
Suppose X = 0 + - ,
then 10^ = 2 ;
or 10 = 2*^.
.-. y = 3 + - (as 10 lies between 2^ and 2^).
Then 10 = 2* X 2^ ;
or 2^=Y.=:j,
and 2 = {|)*.
.•.2 = 3+--
Then 2 = {J)« X (|f ;
or ^ (J)^ = Hi,
and i={HI)^
The greatest integer in v will be found to be 9.
1
Hence, x = 0 +
34~i
^■^94-etc.
The convergents will be ^, y\j, jf , etc.
.-. X = If = 0.3010, nearly.
Exercise 125.
1. Find continued fractions for ^fj; Yt^-; V5 ; Vll ;
4 VG ; and find the fifth convergent to each.
2. Find continued fractions for ^j ; JjJ ; joy . ^/^i. .
and find the third convergent to each.
3. Find continued fractions for V2i ; V22; V33; V55.
4. Obtain convergents, with only two figures in the denom-
inator, that approach nearest to the values of VlO ;
Vi5; Vl7; Vl8; V20.
CONTINUED FRACTIONS. 393
6. Find the proper fraction which, if converted into a con-
tinued fraction, will have quotients 1, 7, 6, 2.
6. Find the next convergent when the two preceding con-
vergents are ^^ and Jf , and the next quotient is 6.
7. If the pound troy is the weight of 22.8167 inches of
water, and the pound avoirdupois of 27.7274 inches,
find a fraction with denominator < 100 which shall
differ from their ratio by < 0.0001.
8. The ratio of the diagonal to a side of a square being
V2, find a fraction with denominator < 100 which
shall differ from their ratio by < 0.0001.
9. The ratio of the circumference of a circle to its diame-
ter being 3.14159265, find the first three convergents,
and determine to how many decimal places each may
be depended upon as agreeing with the true value.
10. Two scales whose zero points coincide have the dis-
tances between consecutive divisions of the one to
those of the other as 1 : 1.06577. Find what divi-
sion-points most nearly coincide.
11. Find the surd values of
3.1 i. i i i. 1 + 1 i i.
^1 + 6' 3 + 1 + 6' ^2 + 3 + 4
12. Show that the ratio of the diagonal of a cube to its edge
may be nearly expressed by 97 : 56. Find the limit
of the error made in taking this ratio for the true
ratio.
13. Find a series of fractions converging to the ratio of
5 hours 48 minutes 51 seconds to 24 hours.
14. Find a series of fractions converging to the ratio of a
cubic yard to a cubic meter, if 1 cubic yard = 0.76463
of a cubic meter.
CHAPTER XXVI.
Theory of Limits.
467. When a quantity is regarded as having a fixed
value, it is called a Constant; but when it is regarded,
under the conditions imposed upon it, as having an indefi-
nite number of different values, it is called a Variable.
468. When it can be shown that the value of a variable,
measured at a series of definite intervals, can, by indefinite
continuation of the series, be made to differ from a given
constant by less than any assigned quantity however small,
but cannot be made absolutely equal to it, the constant is
called the Limit of the variable ; and the variable is said
to approach indefinitely to its limit
469. In order, then, for a fixed value to be the limit of
a variable value, it is necessary and sufficient that there be
some difference between the variable and the fixed value,
but that this difference may be made as small as we please.
Consider the series
^> h i» i> T^ffj iJ^>
The general formula for the sum of n terms of this series
is (§ 395), ]
a ar"
1 — r 1 — r
When n = 3, the sum equals 7—7 =2 — 7*
it 4
When n = 4, the sum equals 7 — ""^ = 2 — q*
THEOBY OF LIMITS. 395
When n = 5, the sum equals T "" ^ = 2 — — •
When n = Q, the sum equals 7 — ^ = 2 — — •
As n increases, the sum approaches 2 as a limit ; for, however great
n becomes, there will continue to be some difiference between the
variable and 2, and this difference, by increasing n, may be indefi-
nitely diminished.
• ■
470. In the above series the variable sum is increasing
towards its limit, and each successive value is less than its
limit.
A variable may decrease towards its limit, and each suc-
cessive value be greater than its limit.
A variable, in approaching its limit, may be sometimes
greater and sometimes less than its limit.
Thus, in the series
li — ii +i> — ii + tV "¥21 + '
for which a of the general formula, § 395, is 1, and r is — i.
When n = 3, the sum equals - — -^ = o + TS
I 1 2 1
When n = 4, the sum equals t — ^ ~ q ~" I^
I 1 2 1
When n = 5, the sum equals - — — ^ = o + 7^
i I o 48
1 ' 2 1
When n= 6^ the sum equals r — ^ ~ 5 "~ i^
2 1 o vo
As n increases, the sum evidently approaches f as a limit ; and for
every odd number of terms is greater ^ for every even number of terms
is less, than the limit. Each successive value of the variable differs
from f less than the preceding value ; and though the successive
values of the variable are alternately greater and less than the limit,
the difference between the variable and the limit diminishes indefi-
nitely by increasing n.
396
ALOEBKA.
471. As a geometrical representation of the indefinite
approach of a variable to its limit,
suppose a square ABCD inscribed
in a circle ABCD,
Connect the extremities of each
side of the square with points of
the circumference equally distant
from them, namely,, the points Ey
F, H, K.
The length of the perimeter rep-
resented by the dotted lines is greater than that of the
square, since two sides replace each side of the square and
form with it a triangle ; and two sides of a triangle are
together greater than the third side.
By continually repeating the process of doubling the
number of sides of each resulting inscribed figure, the
length of the perimeter will increase with the increase of
the number of sides ; but it cannot become equal to the
length of the circumference, for the perimeter will continue
to be made up of straight lines, each one of which is less
than the part of the circumference between its extremities.
The length of the circumference is therefore the limit
of the length of the perimeter as the number of sides of
the inscribed figure is indefinitely increased.
Theorems of Limits.
472. If two variables are equal and are so related that a
change in one produces such a change in the other that they
continue equal, and each approaches a limits their limits are
equal.
Let X and y represent two equal variables which increase towards
their respective limits a and 6, and which continue equal in approach-
ing their limits ; then o = 6.
THEORY OP LIMITS. 397
If a and b are not equal, one, as a, is the greater.
Let d = a — b.
Since x approaches a indefinitely, the difference between a and x
may be made less than d.
Then a — x<d;
that is, a — x^Ca — b.
Therefore, x>6.
Since b is the limit of y, and y increasing,
y<b.
But x>6.
Therefore, y<ic.
But this is contrary to the supposition that x and y are equal.
Therefore, a and b cannot be unequal. Hence, a = 6.
When the variables are decreasing towards their limits, if a and b
are unequal, one, as 6, is the greater.
Let d = b — a.
Since x approaches a indefinitely, the difference between x and a
may be made less than d.
Then x — a<d;
that is x — a<Cb — a,
Therefore, x<6.
Since b is the limit of y, and y decreasing,
y>b.
But x<b.
Therefore, 2/ > x.
But this is contrary to the supposition that x and y are equal.
Therefore, a and b cannot be unequal. Hence, a = 6.
473. If two variables hatie a fixed ratio and are so related
that a change in one produces such a change in the other that
they continue to have this ratio, and each approaches a limit,
tlieir limits are in the same ratio.
Let X and y represent the two variables, r their ratio, a and b their
respective limits.
Then x : y = r : 1 ; that is, x = r X 2^.
By $ 472, the limit of x = r x (the limit oty)\ _
thatiSy a = r X 6; hence, a :6 = r:l.
398 ALGEBRA.
474. T?ie limit of the sum of two or m/yre variables is
equal to the sum of their respective limits.
Let X, 2^, z, represent the variables.
a, 6, c, represent their respective limits,
and d, d\ d'% represent the differences between the variables
and their respective limits.
If the variables are increasing,
z = a— d,
y = h-d%
z = c — d"; and so on.
.•.x + y + «+ = a+6 + c+ -(d + d'+d"+ ).
It is required to show that d + d' + d" + can be made less than
any assigned quantity.
Let q represent any assigned quantity ; let d, d', and d", be n
in number, and d the greatest of them. Then
d + d' + d''+ <nd.
Now, since d, the difference between x and its limit a, may be
diminished at pleasure, it may be made less than -, so that,
d < - , and therefore nd<^q,
n
But d + d'+d''^ <nd.
Therefore, d + d'-{-d''+ < g ;
that is the difference between the sum of the variables x+ y-\- z+
and the sum of their respective limits a + 6 + c + ...... can be made
less than any assigned quantity.
Therefore, the limit of x + y + z + = a+ b^ c+
475. The limit of the product of two variaJ)les is equal
to the product of their limits.
Let X and y represent two variables, a and h their respective limits,
d and' d' the differences between the variables and their limits.
If the variables are increasing,
x — a — d^
y=h-d\
.; xy = ab — db — d'a + dd'
= a6- (dbi-d'a-dd").
THEORY OF LIMITS. 399
Since d and d^ may be made as small as we please, and a and b are
finite values, the value of the expression db+ d'a — dd^ may be made
as small as we please; that is, the difference between the product
of the two variables and the product of their limits can be made less
than any assigned quantity.
Therefore, the limit of xy = ab.
If there are three or more variables, the proposition may-
be proved for two variables, then for this product and a
third variable, and so on.
476. By considering the variables x, y, z, all equal,
and the limits a, b, c, equal, it follows that,
The limit of any power of a variable is equal to that
power of its limit,
477. The limit of the quotient of two variables is equal to
the quotient of their limits.
Let X and y represent the variables,
a and b represent their respective limits,
d and d' represent the differences between the variables and
theb limite.
If the variables are increasing,
x = a — d and y = b — d'.
Q — d
Therefore, , _ is the quotient of the variables,
also - is the quotient of the limits,
and T — - _ -^ is the difference of the quotients.
But
a a — d ad' "bd
b b-d' b{b-d')
and since d and d' may be decreased at pleasure, the dividend ad' — bd
may be made less than any assigned value.
Therefore, if b is not zero, this quotient can be made less than any
assigned value.
Hence, the limit of , _ = r ; that is, the limit o^ ~ = I*
400 ALGEBRA.
478. To find the limit of the quotient — — — , when x
approaches a as a limit.
By division, ^^^* = x^i + ox^* + a2x»-« + o^i.
As a; approaches a, «»-^ approaches a"-*,
ax»-* approaches a*^*,
a*sc"-* approaches a"-* ; and so on.
Since there are n terms in the right member, the limit of
— — — = ?ia»-i as X approaches a.
This result may be shown to be true whether n is inte-
gral or fractional, positive or negative.
CONVEEGENCY OF INFINITE SeRIES.
479. A convergent series is a series whose sum, as the
number of its terms is indefinitely increased, approaches
some fixed finite value as a limit. § 369.
480. The sum of n terms of the geometrical progression
a, ar, a/^, has been shown to be
-m
391.
When r > 1, r" increases with the increase of », and the
sum of the series approaches no fixed limit. Hence, in
this case the series is not convergent.
When r = ly the sum equals §, which represents no defi-
nite value. Hence, in this case the series is not convergent.
When r =: — 1, the sum equals 0 or a, according as n is
an even or an odd number. Hence, in this case the sum
of the series does not indefinitely approach to a fixed limit
by increasing n, but fluctuates between 0 and a, and the
series is not convergent.
THEORY OF LIMITS. 401
When r < 1, r", with the increase of », approaches 0
as a limit, and the sum approaches the fixed value
Hence, in this case the series is convergent.
a
1-r
481. It is necessary and suifficient for the convergency
of an infinite series that the sum of the remaining terms
after the nth should approach 0 as a limit, as n increases
indefinitely.
482. If, in a series of positive terms, the sum of the
remaining terms after the nth approaches 0 as a limit, it is
evident that each separate term after the nth must approach
0 as a limit.
483. If each term after the nth approaches 0 as a limit,
it does not necessarily follow that the sum of all the terms
after the nth approaches 0 as a limit.
(1) To determine whether the sum of the Harmonical
series is convergent :
11111 1
1,
2 3 4 n n + 1 n + 2
Each tf^rm after the nih approaches 0 as n increases.
The sum of n terms after the nth term is
1 .+. 1 + 1 +..,.+ 1.
n+ 1 n+ 2 n + 3 * 2n
which is greater than h ^;— + to n terms; and therefore
greater than n x -— ; that is, greater than -•
Now, the first term is 1, the second term is i, the sum of the next
two terms \& greater than i, the sum of the succeeding four terms is
greater than \ ; and so on. So that, by increasing n, the sum will
approach a number greater than any finite multiple of \,
Therefore, the series is divergent.
402 ALGEBRA.
(2) To determine whether the following series is con-
vergent :
1*1X2'1X2X3' ' 1 71-1 ' \n ' \n + l
1
The nth term of this series is evidently
Llril
The sum of the remaining terms is , — h ; — r— + ; — r-^ +
{n |n+ 1 |yi + 2
which
ichis r(H 17T + : — TTTT — Tl^ + V
[n\ n + 1 (n + 1) (n + 2) /
and this is <r('l + -+-„ + )
therefore, < —( j , for, expanded = 1 + - + -^ + •
n n
therefore, <! — i ; I '
n\n—l/
therefore, < — ( r ) »
|n\n— 1/
therefore, <
(ii-l)[nj^l
But T—j approaches 0 as n increases.
(n — 1) 1 71 — 1
Therefore, the sum of the series is convergent.
Tests of the Convergency of Series.
484. If the terms of a series are all positive, and the
limit of the nth term is 0, theii if the limit of the ratio of
the (n'\-l)th term to the nth term is less than 1, the series
is convergent.
In the infinite series
ai, oa, as, On, On+i, a„+2, ,
let g, after a certain value of n, represent the limit of the ratio ^^
and be less than 1. ^
THEORV OF tIMITS. 403
Ako, let k be some fixed valae between q and 1. Then
^^<k, ^^<K -'•-^-'<ik,
On Am+I Ofi+S
.-. On+i < A:a«, a„+2 < *:^a«, On+s < ^^^n.
.-. a„+i + a„+2 + a„+3 <«« (fc + A^-^ + A:^ +
Since A: is less than 1, = fc + A:^ + ^ +
1 ~~ ic
k
.-. a„+i + a„+2 + ai»+3+ <art X ^ _ ^*
A:
Since a« has a certain finite value, a„ X _ has a certain finite
value.
Hence the infinite series beginning with a„+i is convergent; and
by adding to this the fixed finite sum ai + 02 4- as + •\- Ont the
whole series is convergent.
Thus, in the series
11X21X2X3 1X2X3 (n - 1)
a„ n
Hence, the ratio — ^^ approaches zero with the increase of n, and
the series is convergent ; which agrees with what is shown in example
(2), $ 483.
If Q'>1, there must be in the series some term from which the
succeeding term is greater than the next preceding term ; so that the
remaining terms will form an increasing series, and therefore the series
is not convergent.
If q' = 1, this value gives no explanation as to whether the series is
convergent or not ; and in such cases other tests must be applied.
If q approaches 1 indefinitely as a limit, then no fixed value k can
be found which will always lie between q and 1, and other tests of
convergency must be applied.
Thus, in the infinite series
1,1.1. ■ 1 , 1 ,
1^ "^ 2'- "^3'- "^ n'- "^ (n + l)** "^
g, the ratio of the (n + 1)^^ term to the nth term, equals
( - j ~ ( ^ ZT ) ' ^^^^^ approaches 1 as n increases.
404 ALOEBRA.
SappoBe r positiye and greater than 1 ; then the fiist term of the
2
series is 1. The snm of the next two terms is leas than — * The sum
4 ^"^
of the next four terms is leas than — * The sum of the next eight
8 *
terms is less than — ; and so on.
Hence, the sum of the series is less than
i_i_2.4.8. ^,. 1, 1 , 1.
which is evidently convergent when r is positive and greater than 1.
If r is positive and equal to 1, the given series becomes
. l + i+i + i+ ,
which is the Harmonical series, and is shown in % 483 to be divergent.
If r is negative, or less than 1, each term of the series \& then
greater than the corresponding term in the Harmonical series, and
hence the series \& divergent.
486. If the terms of a series are altematelt/ positive and
negative, then vjhen the terms continually decrease, and the
limit of the nth term is zero, the series is convergent.
Consider the infinite series,
«i — «2 + «3 — «4 + =F «« ± ««+i =F a,+2 ±
The sum of the terms after the nth term is
i [fl||+l "■ («i«+2 — Oii+s) — {dn+A — On+s) — ]»
which may be written
± [dn+i — a»+2 + (a„+3 — 011+4) + (Oii+s — Oii+e) + ].
Since the terms are continually diminishing, the series of groups
in either form of expression are positive, and therefore the absolute
value of the required sum is seen, from the first form of expression,
to be less than an^i ; and from the second form of expression, to be
greater than On+i — an+2. But both 0^+1 and 0^+2 approach zero
with the increase of n ^ therefore the sum of the series after the rUk
term approaches zero, and the series is convergent.
It will be seen that if, in finding the sum of an infinite decreasing
series of which the terms are alternately positive and negative, we
stop at any term, the error will be less than the next succeeding
term.
CHAPTER XXVIL
Indeterminate Coefficients.
486. A SERIES, ax-\- bx^-\- cx^ + dx* -\- , in which the
coefficients, a, b, c, d are finite, may, by taking x suffi-
ciently small, be made less than any assigned quantity.
For if q be any assigned quantity, and k the greatest of the coeffi-
cients a, 6, c, , then
ax + 6x2 + ex' + <to-hfcc2 + A:aj«+
kr
But kx-\-kx'^+kx'^+ = -^^^»
1 — a;
(as is evident by dividing fee by 1 — x).
kz
.'. ox + 6x2 + cx3 -h < , if X be taken less than 1.
1 — X
kx
Hence, if ;; be taken less than o.
1 —X ^
that is, if X «< — 7-r »
q-h k
then ox + 6x2 4- cx8 + ^ill be less than q.
Theorem of Indeterminate Coefficients.
487. If the two series,
A+Bx -\- Cx^ +
and A' + B'x+C'x^+
are equal for all finite values of x, then
A = A', B=B', C=C\ and so on.
406 ALGEBKA.
For, since A + Bx-hCx^-^ = ^' + B'x + C«a + ,
by transposition, A - A' = (R — B)x + (C - C)z^ -^
Now, by taking x sufficiently small, the right side of this equation
can be made less than any assigned value whatever (f 486) ; and there-
fore less than A — A\ it A — A' have any value whatever. Hence,
A — A' cannot have any value.
Therefore, A — A' = 0 or A = A',
Hence, & + Ca;2 + 2Xc» 4- = B'x + C'x^ +l>'x» +
or (B-B0x=(C~C)x2+(2y-D)x8+ ;
by dividing by x,
B-ir=(C-C)x+(2y~2))x«+ ;
and, by the same proof as for -4 — A\
B-R=0 or B=Jr.
In like manner, C = C, D = ly, and so on.
Hence, the equation
A + Bx+Cx^+ = A' + B'x+Crx'+ »
if true for all finite values of x, is an identical equation;
that is, the coefficients of like powers of x are the same.
Applications of the Theorem.
(1) Expand t-j^ — x~2 ^^ ascending powers of x.
Assume .V^V^ . = ^ + & + Cx^ + Dx« + t
1 + X + x-*
then, by clearing of fractions,
2 + 3x = A + J5x + Cx2 + Dx8 +
+ Ax+i?x2+ Cx8+
-hAx^-h Bx^-h ;
.-. 2+3x=A + (B+A)x + (C4-^+A)x«4-(Z) + C+B)x« +
Therefore, by the Theorem,
A = 2, J5 4-A=3, C4-B + A = 0, 2)+(7 + B=0;
whence, B=l, C=— 3, D=2, and so on.
2 + 3x
•••n:^^+^ = 2 + x-3x2 + 2x3-
INDETERMINATE COEFFICIENTS. 407
In employing the method of Indeterminate Coefficients, the form
of the given expression must determine what powers of the variable
X must be assumed. It is necessary and sufficient that the assumed
equation, when simplified, shall have in the right member all the
powers of x that are found in the left member.
If any powers of x occur in the right member that are not in the
left member, the coefficients of these powers will be found to be 0,
and they will vanish, so that in this case the process will not be
vitiated ; but if any powers of x occur in the left that are not in the
right member, then the coefficients of these powers of x must be put
equal to 0 in equating the coefficients of like powers of x ; and this
leads to absurd results. Thus, if it were assumed in problem (1) that
2 + 3x
1 + X + x2
= ^x + Bx^ +0x8+'
there would be in the equation simplified no term on the right cor-
responding to 2 on the left j so that, in equating the coefficients of like
powers of x, 2, which is 2x'^, would have to be put equal to Ox<^ ; that
is, 2 = 0, a manifest absurdity.
(2) Expand (a — x)l
Assume (a — x)* = ^ + ^x + Cx^ + Dx^ +
Squarea-x = -42+2^Bx+(2-4C+-B2)x2+(2^D+2Ba)x8+
Therefore, by the Theorem,
A^ = a, 2AB = - 1, 2^C + B2 = 0, 2 AD+2BC = 0, etc.,
and A = a^, B= -—., C = - —^, D= ^:-
2 a* 8 a* 16 a*
TT / %i A a; x2 x*
Hence, la — x)* = a* 1 — — r —
2 a* 8 a* 16a*
(3) Find the fraction in the form of — r— ^ o which will
p-\-qx-\-rx^
produce, by executing the indicated division, the
series l — 5x + 6x^ + Sx^'-^Ox^—
Assume z—r-7: — . ^ » to be the fraction required.
1 + Cx + Dx^ ^
408 ALGEBRA.
and, by clearing of fractions,
l + BiB=l-6x + 6ara +8x8 _
+ Cx-5Cx2 + 6Cx84-
-^Dx^ -&l>x»+ (i.)
.•.fi=-5+C, 6-5C + i>=0, 8 + 6C-6D = 0;
whence D=4, C = 2, and B=— 3.
l-3x
, the fraction is
14-2x + 4x2
Note. It will be seen in the product (1.) that the column which
contains x>^ may be obtained by multiplying the third term of the
given aeries by 1, the second term by Cx, and the first term by Dx^;
also, that the column which contains x^ may be obtained by multiply-
ing the fourth term of the given series by 1, the third term by Cx, and
the second term by 2>x^ ; and that each column is equal to 0.
A series, any given number of whose consecutive terms are thus
related, is called a reeorring series; and the expression, 1 + Cx + Dx*,
is called its scale of relation.
(4) li y=^ ax + hx^ -\-ca?-\- find x in terms of y.
Assume x=-4y -{• By^ + Cy* + ;
But since y=ax-\-bz^-hcx^-h , by substituting this value of
y in the equation x = Ay + By- + Cy* +
the result is x = Aax + Abx'^ + Acx^ +
+ Ba2x2 4.2J5a6x3+
+ Ca^^ 4-
r.Aa=l, Ab-^Ba^ = 0, Ac + 2Babi- Ca^ = 0;
u >. 1 x> ^ r, 2 62-ac
whence, A = -^ B= :» C= 1
. J-- y ^^1 (26^-ac)y«
a a^ a*
Note. Ex. (4) is an instance of reversion of series. If x in the
given series is equal to 0, ^ will be equal to 0, and therefore no term
in the required series will be clear of y. If, however, the given series
is in the form of y = a + 6x + cx^ + , it is necessary to substitute
2 for y — a, so that z = bx + cx^-h , and to express x in terms of z ;
then to put y — a in place of z.
If the given series ox + ftx^ + cx^ + is an infinite series, then x
must be less than 1, or the series will not be convergent ; also, y, the
sum of the given series, must be less than 1, or the assumed series
Ay + Bj/^ + Cy« + will not be convergent.
INDRTKRMINATE COEFFICIENTS. 409
(6) Resolve 7 xr-7 rr: into partial fractions.
^ ^ (x — 2)(x — 3)
The denominators will be a; — 2 and a; — 3.
3x-7 A , B
^^'^^ (x-2)(x-3) = i^+^^'
then 3x--7= ^ (x- 3) + B(x-2).
.-. ^ 4- B = 3 and 3^ + 2 fi = 7 ;
whence, A = 1 and B = 2.
3X--7 1 . 2
Therefore,
(x-2)(x — 3) x-2 x-3
3
(6) Resolve 3 into partial fractions.
Since x* + 1 = (x 4- 1) (x^ — x + 1), the denominators will be
x+l andx2 — x4- 1.
3 A , Bx+C
Assume , , .. = — , — r +
x«+l x+ 1 x'^-x+ 1
then 3 = A (x2- X + 1) + (Bx + C) (x + 1)
= (A + B)x^+{B+C — A)x-{'(A-^C);
whence, 3=A + C, B-hC-A^O, A-h B = 0;
and A=\, B= — \y (7=2.
3 1 x-2
Therefore,
x8 + 1 X + 1 x'^-x+1
42j3 ^2 3^ 2
(7) Resolve ^ _li\2 ^"^^ partial fractions.
X (x -|- 1)
The denominators may be x, x^, x + 1, (x + l)^.
4x3-x2-3x-2 A , B , C , D
Assume T7~_rT^> ="~ + "5+ ~i~^ +
x'i {x+iy^ X ^ x2 ^ X + 1 (X + 1)-^
.-. 4x3-x2-3x-2=/lx(x+l)2 + «(x+l)2 + Cx2(x+l)+i)x2
= (A + C)x^-h{2A+ B+ C + D)x^ + {A-\-2B)xi-B
whence, A + C = 4,
2A + 2^4-0+ D= -1,
A + 2«= -3,
/^=-2;
or J5=-2, A = l, (7=3, 2)= -4.
^ . 4x8-x2-3x-2 1 2 , 3 4
Therefore, r-; — riTTj = z^^ rr "" : — rrrs*
' x2 (x + 1)2 X x2 X + I (x + 1)2
410 ALGEBBA.
In decomposing a given fraction into its simplest partial
fractions, it is important to determine what form the
assumed fractions must have.
Since the given fraction is the sum of the required par-
tial fractions, each assumed denominator must be a factor
of the given denominator ; moreover, all the factors of the
given denominator must be taken as denominators of the
assumed fractions.
Thus, if the given denominator can be expressed by
a" (X ± ay (x2 + 6)H (x^ ± ax + 6)»,
the denominators of the assumed fractions must be
35, x^ ;
(x±a), (x±a)2 ;
(x2+6). (x2 4-6)2 ;
(X ± ox + 6), (x--^ ± ax + 6)2,
Since the required partial fractions are to be in their
simplest form incapable of further decomposition, the
numerator of each required fraction must be assumed with
reference to this condition.
Thus, if the denominator is x* or (x ± a)", the assumed fraction
A A A.Z 4* B
must be of the form — or ■; — ; — — ; for, if it had the form
Ax + B ^" (* =t «)" *"
or ■ ^^, it could be decomposed into two fractions, and the partial
fractions would not be. in the simplest form possible.
When all the monomial factors, and all the binomial
factors, of the form x±:a, have been removed from the
denominator of the given expression, and there remains as
a factor (x^ztax+hy or (x^+ft)", this is a quadratic which
cannot be further resolved ; and the numerator, therefore,
may contain the first power of a, so that the assumed frac-
tion must have the form
Ax + B Ax + B
: Qj. : •
a^ztax-\-b x^-\-b
indeterminate coefficients. 411
Exercise 126.
Expand to four terms in ascending powers of x :
1 1+x 3 — 2aj 1 — x
2~3a: 2 + 3a; 4 — 3a; ' 1-x + x^
1 5 — 2x ^ 4:x — 6x'
6. T-T— ; ;• 7.
l~2a: + 3a;2 l + 3a; — a^ l-2a; + 3x2
Eevert the series :
8. y=x+x^+x^+ 10. y=x—ix^+^a^—\x'^+'
x^
9. y=x-2x'+3x'- 11. y=x+T— r-F
X
.3
1X2 ' 1X2X3 '
12. Find the fractions in the form — ; -. — ^ whose ex-
/? "T~ ffX ~T" 7*31/
pansions produce the series :
l + 3x + 2x^'-x^- ;
3 + 2aj + 3x2 + 7x« + ;
Resolve into partial fractions :
13.
14.
15.
7a: + l
16.
x — 2
««— 3a;— 10
19.
3a;»-4
(x+4)(x-5)
x'ix + S)
6
17.
3
20.
l3? — X
(x + 3)(x + 4)
{x-l)\x+2)
5a;-l
18.
x'-x-3
«(«='— 4)
21.
2a^-7x+l
(2=^-l)(*-5)
a^ + 1
CHAPTER XXVIII.
The Exponential Theorem.
488. To expand a* in a series of ascending powers of
a'=|l + (a-l)|-
therefore, by the Binomial Theorem,
»'=l + i(o-l) + S^^ («-!)■
+ ^^^f^ («-»•+■■■•
By performing the indicated operations, this series will consist of
powers of x, and the whole series can be rearranged in ascending
powers of x.
The coeflScient of x will be found to be
(a-l)-i(a-l)2 + i(a-l)3-i(a-l)*+ »
which may be represented by A.
And if the coefficients of x'^^ x'^, x*, be represented by J?, C, D<,
the series may be written
a^=l + Ax + Bx^-{'Cx^ + I)x* +
Now, jB, C, D, may be found in terms of A ; for this series is
true whatever be the value of x, since A, B, C, D, are functions
of a, and therefore wholly independent of the value of x. Hence,
aa:+y = 1 + ^ (X + y) + 5 (X + 2^)2 + C (X + y)8 + D (X + y)* + •-
But a^ + y = a* X ay = ay (1 + ^x + Bx^ + Cx^ + Dx* + ).
These two series are identically equal, and therefore, by § 487, the
coefficient of x in the first series is equal to the coefficient of x in the
second series ; that is,
THE EXPONENTIAL THEOREM. 413
A+2By + 3Cy2 +42)y» + = AcUf
= A + A-^y + ABy'^ + ACy^+
And since these expressions are identically equal for all values of
y, the coefficients of y, y^, y*, in the one are equal to the coeffi-
cients of y, y2^ y8^ respectively, in the other.
whence, 2B=A^; SC=AB] 4D=AC;
_ ^2 ^ AB r. ^C*
^^ ^ = T> ^="3"' ^="4">
Sothat a^=l + ^.+ -- + ^^^- +
in which ^ = (a - 1) - i (a - 1)^ + ^ (a - 1)8 -
489. If X is taken such that Ax = l ] then ^ = -1^ and
the last series becomes
1 1 ^
1X2 ' 1X2X3 '
The value of tliis series can easily be computed to any degree of
approximation, and is 2.7182818
This constant is denoted, for shortness, by the letter e, and is the
number whose logarithm is 1 in the Napierian system of logarithms.
In this system no base is assumed, but a logarithm is defined to be
such that the increment of the number shall be the product of the
number by the increment of the logarithm.
490. Since a'* = e, a = e'*, and A = log^a ; so that
«-=i+(i»^).+<'^-)''-+a^+
491. The preceding proof has proceeded on the assump-
tion that a is not greater than 2 ; for the sum of the series
(a — 1) — ^ (a — iy-\-i(a — iy — is not convergent if a
is greater than 2.
To extend the proof to cover all values of a, put b' for a
in the theorem
^^ = ^ + (1^^^^)^+ 1>^ + 1>^3 +
414 ALOEBJiA.
Now, by taking y small enough, 6 may be made as great as we
please, while a is not greater than 2. Then
^.=,^„^.,,„^<te=^MW5^^
and by patting z for yx,
6^ = l + (loge6). + e^^-^^ + fi^^^^+
1-r ViogeC'^zT j^2 ^1X2X3^
Hence the series is true for all values of o.
492. In the theorem
put e for a, and observe that log^e = 1 ; § 294.
then6' = l + x + -^+ "^ '
1X2 ' 1X2X3 ' '
a result true for all values of x.
Computation of Logarithms.
493. To expand log^ (1 + x) in a series of ascending pow-
ers of X.
Since, by % 491 , log^a = ^ = (a - 1) - i (a - 1)« + i (a - 1)» - ,
by putting 1 + x for a, and therefore x for a — 1,
loge(l + x) = x-ix2+ Jx«-}x*+
494. This series is called the Logarithmic Series, but is
not of a form for practical use, as may be seen by substi-
tuting 10 for X ; which gives
I ^^ in 102 , W 10*.
logell=10-— + — - — +
in which the values of the successive terms are increasing, and no
number of them can be taken that will give an approximate value of
logell.
It is necessary, therefore, to obtain from it other formulas,
as follows :
THE EXPONENTIAL THEOREM. 415
10g,(l + X) = X-ix2 + ix8-iX*+
loge (1 — x) = — X — ix2 — Jx8 — Jx* — (by putting — x for x).
/.10ge(l + X) - loge (1 -X) = 2x + 2 (ix^) + 2 (ix^) +
= 2(x + ix8 + Jx6+ ).
But loge (1 + X) - loge (!-«) = log« ( J^) * « 308.
•••iog«(r^)=2(x+iaj»+ix6+ ).
In this series put — ; — for x, and therefore — for ;
^ m+n n 1 — x
flien, iog.5» = 2 j !!Lp + 1 (^^L^)' + 1 (IlpV + j .
' ^ n (m-^n 3 \m + n/ 5\m + n/ )
In this last series put n + 1 for m, and therefore - — pr: for — 7-- •
^ . ' 2n+l m + n
n+ 1
loge — ^ = loge (n + 1) — logen
^^ 1 2^rfi'^ 3(2^1^1) "*■ S'
.Moge(n+l) = logen + 2J^ + l(^J+ j;
by which the logarithm of any number can be obtained from that of
the preceding number.
495. To compute logarithms to the base e for 1, 2, 3, 4,
etc.,
loge 1 =0. Hence, by the last series,
log,(l+l) = log.2 = 0 + 2 11 + 3313-3+^.+ -^}^, + j
= 0.69314718 (by computation);
l0ge(2+l) = l0ge3 = l0ge2 + 2J|+33^3-h^-,+ j
= 1.09861228
loge 4 =2 loge 2 5 306.
= 1.38629436
logeS =iog,4 + 2 jJ+3^+^+ j
= 1.60943791
loge 10 =l0ge5 + l0ge2
= 2.30258609
416 ALGEBRA.
496. If iV= any number,
The multiplier rri by means of which every natural logarithm
may be changed to the corresponding decimal logarithm, is called the
ModvliiB of the common system, and is represented by M. Hence,
^ = 2.30258609 = ^■'^^'^-
497. Common logarithms may be obtained from natural
logarithms, as explained in § 496, or they may be computed
directly by adapting to the common system the series
log«(«+l) = log.n + 2{2;ip^ + -j(^y+ }•
Thus, logio(n+l)
= 'og>on + 2 3fj^ + l(^)%|(.^-)% j;
and by means of this series a logarithm in the common system may be
computed from that of the preceding number.
498. In finding the logarithm of a number which consists
of more digits than are given in the tables, it is assumed
that when the diflFerence of two numbers is small in com-
parison with either of them, the diflFerence of their logarithms
is proportional to the diflFerence of the numbers. The truth
of this assumption may be shown as follows :
logio (n + d) — logio n = logio ( ^^-^ J = logio ( 1 + ")
When d is very small in comparison with n, the terms of the right
side, except the first, will be so small that they may be neglected ;
Md
so that logio (n + d) — logio n = — »
n
and as 3f is constant and n a given number,
{logio (n -f- d) - logio n} « d.
CHAPTEE XXIX.
The Differential Method.
499. If we have a series whose terms proceed according
to some law, but are not immediately equidifferent, we
may find the difference of every two consecutive terms, and
thus form a series of differences called the first order of
differences. If the differences of the terms of this new
series be similarly taken, another new series will be formed
called the second order of differences, and so on.
Let a, 6, C| d, he the terms of the series.
Then h — a^ c — h^d-- c^ , which may he denoted hy ai, 61, ci,
, will be the first order of differences.
Again, 61 — ai, Ci — 61, d\ — c^ » which may be denoted by a^t
^2) C2, , will be the second order of differences.
This process may be continued as long as there are any differences.
The given series and the successive orders of differences arranged
in lines will be :
a h c d e f
ai7 &i ci d\ ei
02 ^2 ^2 d2
as 63 C3
500. Let it be required to express the (n + l)th term of
the series a, b, c, d, in terms of a, %, ag?
As b — a = ai, .*. & = a + cti ;
as 61 — ai = 02, .'. 61 = «! + a2 ;
as 62 "- <H = ct3» ••• &2 = ^2 + as ; and so on.
418 ALGEBRA.
In like manner, c =b +hi = a + 2ai 4- Oj,
Ci = 6i + 6s = ai + 202 4- a«i
ca = 6s + 63 = 02 + 2a8 + 04, and so.on.
Likewise, d = c + Ci = a + Soi + 802 + a«, and so on.
The coefficients, therefore, of o, Oi, 02, Os? » in the expressions
for 6, c, d, , are the same as the coefficients obtained from the
expansion of the expression (o + 6)» ; and since by the Binomial
Theorem the coefficients of (o + 6)« are
n(n-l) n(n-l)(n-2)
' "' 1X2 ' 1X2X3'
the (n + 1)^^ term of the series o, 6, c, d, will be
, . n(n-l) . n(n-l)(n-2) ,
« + "^^'+ 1^^+1x2x3 «»+
501. Let it be required to find the sum of n terms of the
series a, h^ c, d, in terms of a, ai, aj?
The sum of two terms = a + 6,
of three terms = a + 6 + c,
of four terms =o-h6 + c + d,
and so on ; so that, if another series be formed,
0, a, 0+ 6, 0 + 6+ c, ,
the (n + l)th term of this series will be the sum of n terms of the
series a, b^ c^ d^
But, by the preceding proof, the (n + l)th term of the series
0, o, a + by o + 6 + c,
n. I I w(n — 1) , n(n— l)(n — 2) ,
Hence, the sum of n terms of the series o, 6, c, d, is
, n (n — 1) , n (n — 1) (n — 2) ,
-,, (.j.^-1 , (n-l)(n-2) , )
THE DIFFERENTIAL METHOD. 419
(1) Find the sum of the squares of the first n natural
numbers, 1^ 2^, 3^ 4«, n\
1 4 9 16 n2 = given series.
3 6 7 9 = first order of differences.
2 2 2 = second order of differences.
0 0 = third order of differences.
Therefore, a = 1, ai = 3, Os = 2, aj = 0.
These values substituted in the general formula give
= ^{6 + 9n-9+2n!'-6n + 4}
o
= g{2n^-h3n+l} = Mn-HM2n+lK
n n = 12, the sum equals 2 X 13 X 26 = 660.
(2) Find the sum of the cubes of the first n natural num-
bers, 1«, 2«, 3», 4«, 5^ n\
1 8 27 64 125 = given series.
7 19 37 61 = first order of differences.
12 18 24 = second order of differences.
6 6 = third order of differences.
0 = fourth order of differences.
Hence, a = 1, ai = 7, 02= 12, as = 6, a^ — 0.
These values substituted in the general formula give for the
sum
(. ,(n-l)7 (n-l)(n-2)12 (n- l)(n~2)(n- 3)6 )
"1^ 2 ^ 2X3 ^ 2X3X4 )
= j{4+14n-14 + 8n2-24n+16 + n8-6n2-f-lln-6}
»r «j_o 1A. X w' (n + 1)2 ( n (n -f 1) ) 2
= -{n« + 2n2+n}= ^^ / = | -^ — ^j .
Now, -^^ — ^ = the sum of the first n numbers, § 382; hence.
The sum of the cubes of the first n natural numbers is equal to
the square of the sum of the numbers.
420 ALGEBRA.
(3) Find the twelfth term and the sum of 12 terms of the
series 300, 270, 242, 216,
300 270 242 216 = given series.
— 30 — 28 — 20 = first order of differences.
2 2 = second order of differences.
0 = third order of differences.
Hence, a = 300, ai = — 30, Oa = 2, Os = 0, and n = 12.
These values substituted in the general formula give
Twelfth term = 300 - 11 X 30 + ^\^^ X 2
= 300 - 330 + 110
= 80.
Sum = 12(300-»^^x30 + y-J]-»x2)
= 12 (300 - 166 + 30f )
= 2060.
Exercise 127.
1. Find the fiftieth term of 1, 3, 8, 20, 43,
2. Find the sum of the series 4, 12, 29, 55, to 20 terms.
3. Find the twelfth term of 4, 11, 28, 55, 92,
4. Find the sum of the series 43, 27, 14, 4, —3, to 12
terms.
5. Find the seventh term of 1, 1.235, 1.471, 1.708,
6. Find the sum of the series 70, 66, 62.3, 58.9, to 15
terms.
7. Find the eleventh term of 343, 337, 326, 310,
8. Find the sum of the series 7 X 13, 6 X 11, 5 X 9, to
9 terms.
9. Find the sum of n terms of the series 3 X 8, 6 X 11,
9 X 14, 12 X 17,
10. Find the sum of n terms of the series 1, 6, 15, 28, 45,
the differential method. 421
Piles of Spherical Shot.
502. When the pile is in the form of a triangular pyra-
mid, the summit consists of a single shot resting on three
below ; and these three rest on a course of six ; and these
six on a course of ten, and so on, so that the courses will
form the series,
1, 1 + 2, 1+2 + 3, 1 + 2 + 3 + 4, ,1 + 2 + + «.
1 3 6 10 16 = given series.
2 3 4 5 = first order of differences.
1 1 1 = second order of differences.
0 0 = third order of differences.
Hence, a = 1, ai = 2, Og = 1, as = 0.
These values substituted in the general formula give
o I 1 . w-1 X. .-> . (n-l)(n-2) )
Sum = n|l + -^-X2+^ 2X3 )
= nJH-n— IH
= ^{(n + l)(n + 2)}
^n(n-f l)(n + 2)
1X2X3
in which n is the number of balls in the side of the bottom course.
503. When the pile is in the form of a pyramid with a
square base, the summit consists of one shot, the next
course consists of four balls, the next of nine, and so on.
The number of shot, therefore, is the sum of the series,
V, 2\ 32, 42, , n\
Tin.- 1. V « AiAi • w(n+ 1) (2n + 1)
Which, by § 501, is ^ ^ ^ '^^ ^ ^'
in which n is the number of balls in the side of the bottom course.
422 AlGEB&A.
504. . When the pile has a base which is rectangular, but
not square, the pile will terminate with a single row. Sup-
pose jp the number of shot in this row ; then the second
course will consist of 2 (/? + !) shot; the third course of
3 (7? + 2); and the nth course of n{p-\-n — 1). Hence
the series will be
i?,2i? + 2,3i? + 6, n{p + n-l).
p 2p + 2 3p + 6 4p+12 = given series.
p-l-2 p + 4 jJ-hC = first order of differences.
2 2 = second order of differences.
0 = third order of differences.
Hence, a = p, ai = p -h 2, aj = 2, as = 0.
These values substituted in the general formula give
Sum = n{p+^\p4-2)-h(^'^^)Jy^)x2}
= f {6p + 3(n-l)(p + 2) + 2(n-l)(n-2)}
= ^(6p + 3np-3p + 6n-6 + 2n2-6n+4)
= ^(3np + 3p + 2/12-2)
= ^(n+l)(3p + 2n-2).
If n' denotes the number in the longest row, then n" = p + n — 1 ,
and therefore p = n' — n + 1 ; and the formula may be written
^(n+l)(3n'-n+l).
in which n denotes the number in the width, and nf in the length, of
the bottom course.
506. When the pile is incomplete, compute the number
in the pile as if complete, then the number in that part of
the pile that is lacking, and take the difference of the
results.
the differential method. 423
Exercise 128.
1. Determine the number of shot in the side of the base
of a triangular pile which contains 286 shot.
2. The number of shot in the upper course of a square
pile is 169, and in the lowest course 1089. How
many shot are there in the pile ?
3. Find the number of shot in a rectangular pile having
17 shot in one side of the base and 42 in the other.
4. Find the number of shot in five courses of an incomplete
triangular pile which has 15 in one side of th^ base.
5. The number of shot in a triangular pile is to the num-
ber in a square pile, of the same number of courses,
as 22 : 41. Find the number of sliot in each pile.
6. Find the number of shot required to complete a rec-
tangular pile having 15 and 6 shot, respectively, in
the sides of its upper course.
7. How many shot must there be in the lowest course of
a triangular pile so that 10 courses of tlie pile, begin-
ning at the base, may contain 37,020 shot ?
8. Find the number of shot in a complete rectangular pile
of 15 courses which has 20 shot in the longest side
of its base.
9. Find the number of shot in the bottom row of a square
pile which contains 2600 more shot than a triangular
pile of the same number of courses.
10. Find the number of shot in a complete square pile in
which the number of shot in the base and the num-
ber in the fifth course above differ by 225.
11. Find the number of shot in a rectangular pile which
has 600 in the lowest course and 11 in the top row.
424 ALGEBRA.
Series of Separable Terms.
506. It is evident from the appearance of certain series
that they are the sums or the differences of two other
series.
(1) Find the sum of the series
111 1
1X2 2X3 3X4 n(n+l)
Each term of this series may evidently be expressed in two
parts:
1111 1 1
1 2' 2 3' 'n n+l'
BO that the sum will be
(f-i)+(.i-5)+a-i)* Hj-d^>
in which the second part of each term, except the last, is can-
celled by the first part of the next succeeding term.
Hence, the sum equals 1 — r*
n -h 1
If n increases without limit, -— : approaches 0 as a limit, and
the sum equals 1.
(2) Find the sum of the series
111 1
Sum
3X5 4X6 5X7 n(7i + 2)
Each term may be written,
2V3 5/' 2\4 ar '2Vn n + 2/
= 1/1 + 1 + 1 + 1+ + 1_1_1_ _1 1 1_\
2\3^4 5 0 n 5 6 n n+l n + 2j
= 1(1 + 1 I ^V
2\3^4 n+l n + 2/
Hence, the sum equals ^ - ^^^^ - -^-
7
If n increases without limit, the sum equals — •
a4
THE DIFFEKENTIAL METHOD. 426
(3) Find the sura of the series
111
3X86X12 9X16 3n(4n + 4)
By multiplying each term by 12 the series becomes
1 1 1 1
1x2' 2x3' 3x4' ' n(n+l)
which is the same as the series in Ex. (1).
Hence, when n increases without limit, the sum equals — •
(4) Find the sum of the series
4 5 6 71-1-3
1X2X3 2X3X4 3X4X5 n(n + l)(n + 2)
To determine whether the terms of this series can each be
separated into two parts, assume
w + 3 ^ A B
n (n + 1) (M + 2) n(n+l) (n + l){n+2)'
Reduce the right member to a common denominator ; then
n-f 3 {A + B)n + 2A
n (n + 1) (n -H 2) ~ n (n + 1) (m + 2)*
.'.2A = S a,ndA-\- B=l. § 487.
3 1
Whence ^ = o ^.nd B= —-•
Hence, each term may be separated into two terms, and the
series will then become
?(^-L- + _i-+_i_ + + _J_^
2VIX 2 2X33X4 n(M+l)/
2 \2 X 3 3 X 4 n{n+l) (n-f l)(n + 2)J
•+r-^+ ' -f + ^ ^ i
L V2 X 3 3 X 4 ?i (?i -f 1) / 2 (n + 1) (/I + 2)
^3
4
"^4 ' JV2 3; ■ V3 4; ' ' Vn n + lj\ 2(n+l)(n+2)
-ia-i)Mi-i)- Hi-M-
4 2 n+l 2(n+ l)(n4-2)
.6 3n + 5
'4 2(n+l)(n + 2)*
If n increases without limit, the sum equals 7*
4
426 ALGEBRA.
Exercise 129.
Sum to n terms, and to infinity, the following series :
111
1.
2.
3.
4.
5.
1X4' 2X5' 3X6'
1 1
1
1X3X6' 2X4XG'
1 1
3X5X7'
1
2X4X6' 4X6X8'
4 7
6X8X10'
10
2X3X4' 3X4X5'
1 1
4X5X6'
1
1X2X3 2X3X4 3X4X5
Interpolation of Series.
507. As the expansion of (a + />)" by the Binomial The-
orem has the same form for fractional as for integral values
of n, so tlie formula
n (n — 1)
a + nai + - ^-^^^ ^a +
may be extended to cases in which n is a fraction, and be
employed to insert or interpolate terms in a series at
required intervals between the given terms.
(1) The cube roots of 27, 28, 29, 30, are 3, 3.03659, 3.07232,
3.10723. Find the cube root of 27.9.
0.03659 0.03573 0.03491 = first order of differences.
— 0.00086 — 0.000.82 = second order of differences.
0.00004 = third order of differences.
These values substituted in the general formula give
^4.^m^qfli;Q^ ^t 1 \/0,00086\ 9 / 1 w ll\/0.00004\
3 + j^(0.03669)~-(--)(— ^-) + -(--)(--j(^— g— j
= 3 +0.032931 + 0.0000387 -h 0.00000066
= 3.03297.
THE DIFFERENTIAL METHOD. 427
(2) Given log 127 = 2.1038, log 128 = 2.1072, log 129 =
2.1106. Find log 127.37.
0.0034 0.0034 = first order of differences.
0 = second order of differences.
Therefore, the differences of the second order will vanish, and
the required logarithm will be
2.1038 -ftVV of 0.0034
= 2.1038 + 0.001368
= 2.1052.
(3) The latitude of the moon on a certain Monday at noon
was 1° 53' 18.9", at midnight 2° 27' 8.6"; on Tues-
day at noon 2° 58' 55.2", at midnight 3° 28' 5.8" ;
on Wednesday at noon 3° 54' 8.8". Find its lati-
tude at 9 P.M. on Monday.
The series expressed in seconds, and the differences, will be
6798.9 8828.6 10735.2 12485.8 14048.8
2029.7 1906.6 1750.6 1563
-123.1 -156 -187.6
-32.9 -31.6
1.3
As 9 hours = J of 12 hours, n = J.
Also, a = 6798.9, ai = 2029.7, a2= — 123.1, as = — 32.9
tti = 1.3.
These values substituted in the general formula
, , 7i(n— 1) , n (n— 1) (n — 2)
n(n-l)(n-2)(n-3)
^ 1X2X3X4 "*^
give 6798.0 + I (2029.7) -|(-l)(lfi)
-!(-l)(-!)(¥)-
= 6798.9 + 1522.27 -f 11.64 - 1.29
= 8331.4 = 2° 18' 61.4''.
CHAPTER XXX.
The Theory of Numbers.
systems of notation.
M)8. A System of Notation is a method of expressing
numbers by means of a series of powers of some fixed
number, called the Kadix, or Base, of the scale in which
the different numbers are expressed.
609. Integral numbers expressed in any system are
polynomials arranged according to the descending powers
of the base, and containing multiples of these powers
written in a condensed form by omitting the exponents
and indicating them by the place of the digits.
Thus, 2384 in the decimal system means
2 X 108 -f 3 X 102 -f 8 X 10 + 4 ;
and in the nonary system means
2X9« + 3x92-f8x9 + 4.
610. If r be any integer, any integral number N may he
expressed in the form
iV= ar« + &?-"-^ + +pi^ +qr-\-8,
in which the coefficients a, h, c, , are each less than r.
For, divide N by r", the highest power of r contained in N, and let
the quotient be a with the remainder Ni.
Then, N=ar^-h Ni.
In like manner, Ni = br"-^ + N2, N^ = cr**-^ + JVs ; and so on.
By continuing this process, a remainder a will at length be reached
which is less than r. So that.
THE THEORY OF NUMBERS. 429
Some of the coefficients s, q,p, may vanish, and each
will be less than r; that is, their values may range from
zero to r — 1. Hence, including zero, the scale of r will
contain r digits.
611. To express any integral number N in the scale of r.
It is required to express N in the form of
Qjfa 4- 5^1.-1 ^ ^. py2 4- qry 4- g^
and to show how the digits a, 6, may be found.
If N- ar^ -f 6r»-i + -f pr2 + gr -f «,
then - = ar^^ + bi^-^ -f + pr + g + - •
r r
That is, the remainder on dividing ^ by r is «, the last digit.
Let Ni — ar"*-^ + 6r»-2 4- + pr + g,
then ^ = ar«-2 4- ^r«-3 + + p -f ^•
r • r
That is, the remainder is q, the last bat one of the digits.
Hence, to express an integral number in a proposed scale.
Divide the number by the radixy theii the quotient by the
radixy and so on ; the successive remainders will be the suc-
cessive digits beginning from the units* place.
(1) Express 42897 (common scale) in the senary scale
(scale of six), and transform 37214 from the octo-
nary scale (scale of eight) to the nonary scale (scale
of nine).
(ii.) 9)37214
9)3363 1
9)305. ... 6
9)25 8
2 3
(i). 6)42897
6)7149. ... 3
6)1191. ... 3
6)198. ... 3
6)33. ... 0
5 3
Ans. 630333.
Ana. 23861.
In (ii.) the radix is 8 ; and hence the two digits on the left,
37, do not mean thirty-senen^ but 3 X 8 + 7, or thirty-one^ which
contains 9 three times, with the remainder 4.
The next partial dividend is 4 X 8 4- 2 = 34, which contains 9
three times, with the remainder 7 ; and so on.
430 ALGEBRA.
(2) What is the radix of the scale in which the number
140 (common scale) is expressed by 362 ?
Let r denote the required radix ;
then the value of 362 = 3 r2 + 5 r + 2.
Hence, 3r2-f5r+2= 140.
Whence, r = 6. Ana.
512. As in the decimal scale it is convenient to express
quantities less than the unit by means of decimal fractions,
so radix fractions may be employed in any other scale.
Thus, 0.2341 in the decimal scale means
A+A + A + J..
10 102 ^ 108 104
And in a scale whose radix is r it means
r r^ r^ r^
618. Computations are made with numbers in any scale,
by observing that one unit of any order is equal to the
radix-number of units of the next lower order ; and that the
radix-number of units of any order is equal to one unit of
the next higher oi'der,
614. The practice of many centuries has decided in
favor of the decimal system. For, when the radix is a
small number, as 2 or 3, large numbers are expressed in a
form so extended that they are not easily comprehended ;
and when the radix is a large number, the series of separate
symbols that must be used is inconveniently large.
616. In some cases different systems are employed for
expressing the same kind of quantity.
Thus, in measuring time, the smallest unit is the second; the next
higher unit is the miniUe, which is equal to 60 seconds ; the next higher
unit is the hour, which is equal to 60 minutes ; while the next higher
unit is the day, which is equal to 24 hours.
tHE THEORY O*' NUMBERS. iSl
Exercise 130.
1. If 6, 7, 8, 3, 2 are the digits of a number in the scale of
r, beginning from the right, write the algebraical
value of the number.
2. Find the product of 234 and 125 when r is the base of
the scale.
3. In what scale will the common number 756 be expressed
by 530 ?
4. In what scale will 540 be the square of 23 ?
5. Show that 1234321 will in any scale be a perfect square,
and find its square root.
6. In what scale will 212, 1101, 1220 be in arithmetical
progression ?
7. Multiply 31.24 by 0.31 in the scale of 5.
8. Find the least multiplier of 13168 which will make the
product a perfect cube.
The Common System of Notation.
616. A number which contains the factor a, since it is a
multiple of a, may be written ma,
617. An even number, since it contains the factor 2,
may be written 2m; and an odd number may be written
2m + l.
618. A number which ends in zero, since it is a multiple
of 10, may be written 10 m. A number which is a multi-
ple of 100 may be written 100 m ; and so on.
432 ALGEBRA.
619. If a prime number ^ is a factor of a product ab,
and is not a factor b,, it is a factor of b.
p) a(qi
PSi
~ri) P (?a
riq%
Tt etc.
In finding the 6. CM. of p and a, let q^ 92) denote the succes-
sive quotients, and ri, r2, the successive remainders.
Then a = pqi + ri ; p = q2ri + Vt ; etc.
By multiplying each of these equations by 6,
ab = hpqi + bri; bp = hqzTi + br^ ; etc.
Hence, 6ri = oft — bpqi ;
and since p is a factor of ab and of bpqi, it is a factor of their differ-
ence 6ri.
Also, since p is a factor of 6ri, it is a factor of hqzVi.
Again, hr2 = bp — bq^ri ; and since p is a factor of bp and of hqzTu
it is a factor of their difference 6r2.
In like manner it may be shown that p is a factor of bvs, br^, brg,
and so on.
Now, since p is a prime number, it has no common factor with a,
except unity ; and the last remainder, in finding the G.C.M. of p and
a, will be 1.
Hence, p will be a factor of 6 X 1 ; that is, a factor of 6.
520. From the above proposition it follows that,
A prime number which is not a factor of either of ttoo
other numbers is not a factor of their product.
In other words,
The product of two or more prime numbers loill contain no
prime factor except themselves.
Hence,
A composite number can be separated into only one set of
prime factors.
If two numbers are prime to one another, every power
of one will be prime to every power of the oth&r.
THE THEORY OP NUMBERS. 433
1^
521. A fraction - in its lowest terms cannot be equal to
c
another fraction - unless d is a multiple of b.
For, if - = -, then, by multiplying by d, — = c ; and since c is
integral, — is integral.
But b is prime to a, and must therefore be contained in d an inte-
gral number of times.
522. From the last proposition it follows that,
A common fraction in its lowest terms will not produce
a terminating decimal if its denominator contains any prime
factor except 2 and 5.
For a terminating decimal is equivalent to a fraction with a denom-
inator lO". Therefore, a fraction t in its lowest terms cannot be
0
equal to such a fraction, unless 10*» is a multiple of 6. But lO", that
is, 2'» X S", contains no factors besides 2 and 5, and hence cannot be
a multiple of 6, if 6 contains any factors except these.
523. If a square number is resolved into its prime factors,
the exponent of ea^h factor will be even.
For, if any number N = aP X I/i X c" ,
N^ = a^pxl^x c^r
524. Conversely : A number which has the exponents of
all its prime factors even will be a perfect square ; there-
fore, to change any number to a perfect square,
Resolve the number into its prime factors, select the factors
which have odd exponents, and mAiltiply the given number by
the product of these foMors,
Thus, to find the least number by which 260 must be multiplied to
make it a perfect square.
250 = 2 X 68, in which 2 and 6 are the factors which have odd
exponents.
Hence, the number required is 2 X 6 = 10.
434 ALGEBBA.
Divisibility op Numbers.
625. If two numbers JX and Ki, when divided by the same
number a, have tJie same remainder r, their difference is
divisible by a.
For, if N when divided by a has a quotient q and a remainder r,
<^^en N=qa-{'r.
And if Ni when divided by a has a quotient qx and a remainder r,
^'Jien iV^i = qia + r.
Therefore, ^ — Ni= (q — ^i) a.
626. iy* the difference of two numbers K" and Ki ts divisi-
ble by a, ^Aen K" and "Si when divided by a will have the
same remainder.
For, if
N-Nx=(q-qx)a,
then
N Ni
~ = q — qv
Therefore,
N Ni
That is,
JV'— aq = Ni — aqi.
627. If two numbers K" and Kj, tvhen divided by a given
number a, have remainders r and ii, ^Aew KBTi aTic? rii «£;/ie/i
divided by a t^^iVZ /ia<;e ^/ie? say/ie remainder.
For, if N=qa + r,
and ^1 = ^la 4- ri,
then NNi = qqia^ 4- gari -f g'lar + rri
= (qqia + g^i + gir) a + rri.
Therefore, § S26, iSTi^i and rri when divided by a will have the
same remainder.
As a particular case, 37 and 47 when divided by 7 have remainders
2 and 6, respectively.
Now 37 X 47 = 1739, and 2 X 5 = 10.
The remainder, when each of these two numbers is divided by 7,
will be 3.
*HE theorV of numbeks. 435
528. From the preceding propositions it follows that :
A number is divisible by 2, 4, 8, if the numbers de-
noted by its last digit, last two digits, last three digits,
are divisible respectively by 2, 4, 8,
A number is divisible by 5, 25, 125, if the numbers
denoted by its last digit, last two digits, last three digits,
are divisible respectively by 5, 25, 125,
If from a number the sum of its digits is subtracted, the
remainder will be divisible by 9.
For, if from a number expressed in the form of
a 4- 106+ 102c 4- Wd-f
a-h 6 4- c 4- d4- is subtracted,
the remainder will be (10 - 1)6 4- (102 - l)c 4- (10* - l)d 4- ,
and 10 — 1, 102 — 1, 108 — 1, will be a series of O's.
Therefore, the remainder is divisible by 9.
Hence a number N may be expressed in the form of
97i + 5 (if s denotes the sum of its digits);
and N will be divisible by 3 if s is divisible by 3 ; and
also by 9 if 5 is divisible by 9.
A number will be divisible by 11 if the difference
between the sum of its digits in the even places and the
sum of its digits in the odd places is 0 or a multiple of 11.
For, a number N expressed by digits (beginning from the right)
a, 6, c, d, may be put in the form of
iV=a4-106 4- 102c 4- 108^4-
,:N-a-hh-c+d- = (104-1)64-(102- I)c4-(1084-l)d4-
But 10 4- 1 is a factor of 10 4- 1, 102 - 1, W 4- i,
Therefore, iV--a4-6 — c4-d— is divisible by 10 4- 1 = 11.
Hence, the number N may be expressed in the form of
lln+(a + c+ )^(b + d+ ),
and will be a multiple of 11 if (a + c + ) — (b + d+ )
is 0 or a multiple of 11.
CHAPTER XXXI.
Imaginary Numbers.
629. Any real number, whether positive or negative,
integral or fractional, rational or surd, may be represented
in magnitude by a horizontal line extending from the zero-
point to the place it occupies.
X' I X
O
Thus, if the line X'X represent a horizontal line of unlimited
length, and 0 be the zero-point, OX the positive, and OX' the nega-
tive direction, then all positive real numbers will have their place in
OX, and all negative real numbers in 0X\
580.
If the straight line OA, which may be supposed
to represent h units of
length, turns around O
180°, it comes into the
negative direction 0^';
-f- h becomes — h. After a
further turn of 180°, it
X comes again in the posi-
tive direction OA, and — h
becomes -f- h.
The turn of 180° signi-
fies, then, the multiplicar
tion by the factor —1.
■^' The double turn of 180°
signifies the multiplication twice by the factor —1.
IMAGINARY NUMBERS. 437
Since — 1 is the product of V— 1 X V— 1, the turn of
one-half of ISC' ' signifies the multiplication by the factor
V=i.
Hence^ a definite geometrical meaning may be assigned
to such an expression as ftV—l.
Draw FF through 0, perpendicular to X^X; then
5V— 1 may be represented by b units counted on a line
perpendicular to the original straight line at the zero-point.
And the point B will be the place which this imaginary
number occupies on the perpendicular.
Multiplying again by the factor V — 1 brings the line
OB to the position OA', opposite OA, and changes + ^ V— 1
to —b. Multiplying a third time by the factor V~ 1 brings
the line into the position OB' and changes — ^ to — 5V— 1.
Multiplying by the factor V— 1 a fourth time brings the
line into its original position OA, and changes — 5V— 1 to
-\-b, which corresponds to the laws governing the powers
of V^. § 281.
Hence, the direction factors (as they may be called), + 1,
+ V— 1, — 1, — V— 1, indicate that b units are to be
counted from 0 in the directions OX, OY, OX, 0Y\ re-
spectively. To the positive real number + b the negative
real number —b is opposed; to the positive imaginary
number +^V— 1 the negative imaginary number — ftV— 1
is opposed.
631. The expression V— 1 is called the Imaginary Unit,
and every pure imaginary number can be e:^pressed in the
form of dzftV—l.
For convenience the letter i is \ised for V — 1. Hence,
i =V^
ia=(V-l}^-l,
i«=-iV-i=-V=^,
i* = *8xi2 = -lX-l= + l.
438 ALGEBRA.
Operations with Imaginary Numbers.
638. In order to add two imaginary numbers, we begin at
the place which the first number occupies and count in the
direction of the second number as many units as are equal
to the absolute value of the second number. Therefore,
ai + bi =(a-\-b') i.
Hence, the sum of two imaginary numbers is also imag-
inary.
It follows that
at -f- ^' = ftt -f- at.
688. In order to find the difference between two imag-
inary numbers, we begin at the place that the minuend
occupies, and count in the direction opposite to that of the
subtrahend as many units as are equal to the absolute
value of the subtrahend. Therefore,
ai — bi=(a — b)i.
Hence, the difference of two unequal imaginary numbers
is also imaginary.
534. To multiply an imaginary number by a real num-
ber means to form from the multiplicand a new number,
in the same way that the multiplier is formed from the
real unit.
Therefore, aiXb = abi.
For 6 is formed from the positive real unit by taking this unit b
times to be added ; ai must, therefore, be taken h times to be added.
That is, aix 6 = ai + at -f- at -f- taken h times = a6t.
Again, a X bi=^ abi.
For bi is formed from the positive real unit by forming the imag-
inary unit i from the positive real unit, and taking this imaginary
unit b times to be added.
IMAGINABY NUMBERS. 439
Therefore, from the multipUcand a the corresponding imaginary
number ai must be formed, and ai taken b times to be added.
Hence, axbi = aixb = dbi.
Also, ai X bi = — ab.
For aixbi = ai:2xb= — axb = — ab.
Hence, the product of an imaginary number and a real
number is imaginary ; the product of two imaginary num-
bers is real.
It follows, also, that
aiXb =b Xai,
and ai Xbi = biX ai.
635. If ab = d is substituted in these three equations,
then a = T> and z-iXb = di:
0 V
■--Xbi = di, rriXbi = — d, or — r *' X bi = d.
b b b
Whence it follows from the general notion of division,
di d , di d d d ,
J^'b^' bi^b' bi'^'^V'
Hence, when the dividend and divisor are the one real
and the other imaginary, the quotient is imaginary ; when
both are imaginary, the quotient is real.
From the second of these equations it is evident that a
fraction remains unchanged if both numerator and denomi-
nator be multiplied by i,
636. Since ^^ = — 1, i®= — t, i* = -f-l, it follows, in gen-
eral, when n signifies any positive whole number,
that is, (ai)" = ai X ai X ai n times = a" i'\
440
ALGEBRA.
So that the power of an imaginary number is real or
imaginary according as the corresponding power of i is real
or imaginary.
637. The sum of a real and an imaginary number, as
a + bi, is called a Complex Number, in distinction from a
pure imaginary number hi. Two complex numbers of the
form a + hi and a — hi are called Conjugate.
588. In order to represent the complex number a -{-hi,
let X'X be the line of
real numbers, O the zero-
point, and Y Y' perpen-
dicular to X'X at 0 be
the line of imaginary
numbers. Let A be the
place of the real num-
ber a, B the place of the
imaginary number hi.
Through A draw a line
parallel to YY', and through
B draw a line parallel to X'X. Then M, the point of
intersection, shows the place of the complex number a+hi;
and the straight line OM represents the absolute value of
this complex number.
The point M can also be reached by laying off a units
from 0 on the line OX, and erecting at A, the last division
point on this line, a perpendicular, and laying off h units
on this perpendicular.
In like manner, the complex numbers — a -[-hi, — a — hi,
a — hi, correspond respectively to the points M\ Jtf" ", M'".
589. If a and h have all values from — oo to + oo the
point M will have every position in the infinite plane
determined by XX and YY\
IMAGINARY NUMBEBS.
441
The expression a -{- hi is then the general expression for
all numbers. This expression includes zero when a = 0
and 5 = 0; includes all real numbers when 5 = 0 ; all imag-
inary numbers when a = 0; all complex numbers when
a and b both differ from 0.
Operations with Complex Numbers.
640. In order to add two complex numbers, a -{-hi and
c-^di, it is necessary to proceed from the place occupied
by the first number in the direction of the second number
c + di, as far as the absolute value of the second number.
The sum thus reached consists of the sum of the real and
the sum of the imaginary parts of the two numbers.
Find the sum of (2 +3i) + (— 4 + t).
2-f 3i=03fand —44- i = OJlf ' (§ 638). If now we proceed from
Jf, the extremity of OM, in
tiie direction of OM' as far
as the absolute value of 0M\
we reach the point M'\
Hence, OM'' = - 2 4- 4 i, the
sum of the two given com-
plex numbers.
The same result is
reached if we first find
the value of 2 + (—4)
= — 2. That is, if we count from 0 two real units to A",
and add to this sum 3i+i = 4t; that is, count four imag-
inary units from -4" on the perpendicular A"M^\
In general, the sum of two complex numbers is a com-
plex number.
The sum of two conjugate numbers is a real number.
(a + bt) + (a — bi) = 2a.
442 ALGEBRA.
641. To subtract e + di from a + W it is necessary to
proceed from the place occupied by the minuend in a
direction opposite to that of the subtrahend as far as the
absolute value of the subtrahend. The value thus reached
is the difference of the real and the difference of the imagi-
nary numbers.
Thus, (a + bi) — (c + di) = (a — c) + {h — d) l
In general, the difference of two complex numbers is a
complex number.
642. If a-\-bi = c-\- di, then must a = c and b = d] for
otherwise a — c=(d — b) i, or - — r = i ; that is, a real
number equal to an imaginary number, which is impos-
sible.
643. If a + ^* = 0, then must a = 0, and ft = 0 ; for,
otherwise — - = {, which is impossible.
644. In order to multiply a complex number a + ^ by a
real number c, the number a + bi must be taken c times, so
that the real and the imaginary parts must be multiplied
by c. That is,
(a -\- bi) X c= (a-\- bi) -\- (a -\- bi) -\- c times = ac -|- bci.
Thus, to multiply — 2 + i by 3 : Take 0A = —2 on 0X% and
erect at A the perpendicular AM= 1. Then 0M= — 2 + £; and,
by taking OM three times, the result
is OM' = — 6 H 3 i the product of
(-2 + 1) by 3.
646. In order to multiply a-\'bi
by c + di it is necessary to obtain
a number from a + bi in the same
way that c-\-di is obtained from
the positive real unit. It is necessary, therefore, to take
a -f- bi itself c times, then to take a-^-bi in the imaginary
IMAGINARY NUMBEBS.
443
direction d times, and join the two results in one sum.
The result corresponds to the algebraic product of two
factors.
Find the product of (3 — t) (2 + 3i).
Find OM = 3 — i. It is required to find a product from- 2 + 3i in
the same way that OM is ob-
tained from 3 — i. ^
With OM as a new positive
unU, lay ofE OM' = 2 OM. At
3r erect a perpendicular, and
lay off from M% on this perpen-
dicular, M'M'' = 3 OM. Then
(3- i) (24-3i) = 03f"= 9+ 7i.
If 2 + 3i is the multiplier,
the new positive unit will be
0M''\
The same result is ob-
tained if the product is
found by the algebraic
method of multiplication.
Thus,
(3--i) (2 + Si) = 6 — 2i + 9i + S = 9 + 7l
In general, the product of two complex numbers is a
complex number.
The product of two conjugate numbers is a real number.
Thus,
(a + bi) (a — U) = a^ + h\
646. Since (a + hi) X c = ac-\- bci, it follows that
ac + bci , , ,
c
That is, a complex number is divided by a real number
by dividing both the real and imaginary parts by this
number.
444 ALGEBBA.
^^"^'dia + bi^-d
That is, a quotient is unchanged if both the dividend
and divisor are multiplied by the same complex number.
647. In order to divide one complex number by another,
it is only necessary to multiply the dividend and divisor
by the conjugate of the divisor, and the result is a complex
number to be divided by a real number. Thus,
a-\'bi (a + bt^c — di) (ac -\- bd) -\- (he — ad) i
e + di" lc + di)(c'-d'C)~' c^+d^
ac-\-bd . (be — ad) i
The quotient of two complex numbers is a complex
number. Thus,
3 + i^ (34-0(2 — 5^') ^ll-13i
2 + 5i (2 + bi)(2-b%) 29
648. The power of a complex number is a complex
number. Thus,
(a + W)«=(a + ^t)(a + ^f) = (a« — ^ + 2aW,
(a + 6iy = (a + biy (a + bi)
=:(a^ — Sab^ + (Sa^b — b^i, and so on.
649. The meaning of an imaginary expmient is deter-
mined by subjecting it to the same operations as if it were
a real exponent.
It follows that such an expression as k"^^ where A; is a
real number and a + ^i a complex exponent, may be simpli-
fied by resolving it into two factors, one of which is a real
number, and the other an imaginary power of e (e being
used as in § 489).
By the ordinary rules for exponents,
IMAGINARY NUMBERS. 445
Put A^ = e", u denoting log [A**] in the natural system ;
then
The value of e«» is (§ 492)
,.==! + ,,+ _ + _ + _+__ +
This value (§§ 531 and 536) may be resolved into two
series, as follows :
-='-i+g-i+ +<«-i+i- )■
The series
1 _ ?^ j_ ^* _ ^* j_
is a function of u called Cosine of u, and is written cos u.
The series
"-il+^-
is a function of i^ called Sine of u, and is written sin u,
660. Hence, the value of e'" may be written
e«* = cos u-\-i sin w. (1)
If, in the development of e"', u be changed to — ii, cos w
will remain unchanged, because its terms contain only even
powers of u ; but sin u will become — sin u, because its
terms contain only odd powers of u. Therefore,
e""' = cos u — i sin u, (2)
The product of equations (1) and (2) is
e"* X e~"*' = (cos w)^ — i^ (sin w)^,
that is, 1 = (cos uy — P (sin w)*, (§ 266)
or 1 = (cos uy + (sin w)l (§ 531)
446 ALGEBBA.
It is customary to omit the parentheses^ and to write
this equation as follows :
cos*ttH-sin*tt = l. (3)
661. If nu is substituted for u, and if e^ is developed as
e** has been developed, the result will obviously be
6^ = cos nu-\-i sin nu.
But equation (1) raised to the nth power becomes
e«« = (cos u-\'i sin u)\
Therefore,
cos nu-^-i sin nu = (cos w + * sin u)\ (4)
If, in this equation n = 2, the resulting equation is
cos 2u-\-i sin 2 w = cos^ u — sin^ U'{'2i sin u cos u;
whence it follows, from § 542, that,
cos 2 M = cos* u — sin* u, (5)
sin 2u = 2 sin u cos u. (6)
By substituting x, y, x-\- y, successively, for u in equa-
tion (1),
e^ = cos x-\-i sin x, (7)
e^ = cos y-\-i sin y, (8)
g(x+y)i _ (jQg (3. _|_ y) _j. ^- sin (aj + 2/). (9)
The product of equations (7) and (8) is
+ i (sin aj cos 3/ + cos x sin y).
Therefore,
cos (x + y) + i sin (a; + y) = cos a cos y — sin x sin y
+ * (sin X cos 2/ + cos x sin y) ;
whence (§ 642),
cos (x -}- 2/) = cos X cos y — sin aj sin y, (10)
sin (a; + y) = sin a: cos y + cos x sin 3^. (11)
CHAPTER XXXII.
Loci of Equations.
552. It is possible to represent equations by diagrams,
some of whicb are regular geometrical figures.
For this purpose, the lines X^X and YY^ are drawn
perpendicular to each other, intersecting at the point 0.
These lines may be of any length in drawing the diagram
of an equation.
B
G
H
E
XX and YY' are called the Axes of Eeference; XX
is the Axis of Abscissas, and YY' the Axis of Ordinates.
In order to determine the position of the point A, with
reference to these axes, AE is drawn perpendicular to XXy
and is called the Ordinate of the point A, The line OE
from the point 0 to the foot of the ordinate is the Abscissa
of A. OE and AE are the Co-ordinates of A, What are
the co-ordinates of B, C, and D ?
448
ALGEBRA.
The abscissa of a point is usually designated by the let-
ter «, and its ordinate by y. For the point A, x=OE,
and y=^AE.
0 is the Origin of Co-ordinates, or simply the Origin.
663. In representing equations whose roots are arith-
metical numbers, numerical values are assigned to the
co-ordinates. All ordinates above X'X are considered posi-
tive, and all below, negative. Abscissas drawn to the right
of FF' are positive, and those to the left are negative.
T
B
-i-i-
H-+-
0--
d
Equal lengths of any convenient magnitude are measured
on the axes from 0, and each length represents a unit. To
locate the point whose abscissa is 4 and whose ordinate is
7, a distance of 4 units is measured on X'X from 0 to the
right, and at the point reached a perpendicular 7 units
long is erected. A, the extremity of this perpendicular, is
the required point.
The co-ordinates of B are x== — 6, and y = 9; of C,
x = — 3, andy = — 6; of D, x=:S, andy = — 8.
LOCI OF EQUATIONS.
449
664. The equation 2x-\-y = 3 may be satisfied by an
infinite number of corresponding values of x and y, By
changing the equation to the form y^3 — 2x, the follow-
ing table is readily computed :
If
x = l,
y= 1.
If
x = -l,
y= 5.
i(
x=2,
y=-\.
a
x = -2,
y= 7.
((
05 = 3,
y=-3.
a
x= — 3,
y= 9.
a
a; = 4,
y= — 5.
((
x = -A,
y=n.
«
X=:5,
y=-7.
<(
x = — 6,
y=13.
M
\
\
\
\ ::
-\
N
If
This table contains the co-ordinates of ten points,
the points are lo-
cated with refer-
ence to the axes,
and the line MN
is drawn through
them, the line MN
is the Locus of the
given equation.
If MN be pro-
longed, the value
of y which cor- y,
responds to any
given value of x
may be found by
laying off an ab-
scissa equal to the
given value of oj,
erecting at its ex-
tremity an ordinate
terminated by MN,
and measuring the length of the ord. ^ate.
If a: = 0, the diagram shows tha"; y = 3; and if a: = 1.5,
y = 0. If a = — 3.5, what is the v lue oi y?
450
ALGEBRA.
666. In plotting any equation containing x and y, we
assiune values of x, compute the corresponding values of y,
locate the points whose co-ordinates are thus found, and
draw the locus through them. When the equation con-
tains only the first powers of x and y, and is of the first
degree, the locus is always a straight line. In this case it
is necessary to locate only two points of the line ; the locus
may then be drawn through these points.
666. Let it be required to draw the locus of
x^+ 1^=72.25.
y = dbV72.25-x«.
From this equation, by assuming values of x and com-
puting the values of y, the following table may be formed :
/
/;
/!
/ :
-L4-
1
\:
;
If X is
y is
±1
±8.44 +
±2
±8.26 +
±3
±7.96 +
±4
±7.60
±5
±6.87 +
±6
±6.02 +
±7
±4.82 +
±8
±2.87 +
If a curve is drawn
through the extremi-
ties of the ordinates,
the locus is found to
be a circle whose CvV^tre is at the origin. The diagram
shows that x does n reach the value 9. If this value
be substituted in the e lation, y will be imaginary.
LOCI OP EQUATIONS.
451
657. Construct the locus of 9«* + 252/*=900.
y = dbj V900 — 9aj2.
From this equation the following table may be formed :
Ifzis
2/ is
±1
±6.96 +
±2
±5.87 +
±3
±5.72 +
±4
±6.49 +
±6
±5.19 +
±6
±4.80
±7
±4.28 +
±8
±3.60
±9
±2.61 +
±10
0.00
0
±6.
— r-
"T-i
>>.
j
\
\
1 1 I 1 •• 1 1
^ ! ' I 1 1 !
^ • 1 ! •
■
—
—
_J-
i^-^'
y^
y
1
The locus is an ellipse.
668. Construct the locus of ^ = 5x.
In this case y=± -s/dx.
Ifzis yia
Ifzia
yis
1 ±2.23 +
6
±6.47 +
2 ±3.16 +
7
±5.91 :-
3 ±3.87 +
8
±6.32 +
4 ±4.47 +
9
±6.70 +
6 ±6.00
10
±7.07 +
../ I
v; I
■f-i-
■Ki :
-■N ;
For negative values of x the
ordinates are imaginary. The
curve is a parabola; and as x
and y may have infinite values, the curve will extend to an
infinite distance.
..TlT
.^r
452
669.
lie locus of y=^3?-
-«» + «-
•6.
Ux\a y is
Ifzis
yis
0.6 -4.626
2.6
+ 6.875
1.0 -4.000
0.0
— 5.000
1.6 —2.376
-0.5
— 6.875
2.0 + 1.000
-1.6
— 12.125
Tf
This locus is not a regular geometri-
cal figure.
Computation shows that, if y = 0,
X = 1.88 + . Does this result agree with
the figure ? What is x when y is + 5 ?
When y is + 10 ? When y is — 15 ?
660. When two simultaneous equations containing x and
y are given, the values of x and y which satisfy both are
shown by constructing the Joci of both equations with ref-
erence to the same axes. In the equations
3a; + 7//
= — 7 and
y = -(2 + ^x), (1):
2x'-y __
5
and
y=2x — 19.
(2)
AB is the locus of equation (1), and CD is the locus of
^ ^ equation (2). The values
' of the co-ordinates of any
f point on AB satisfy (1), and
those of any point on CD
-^ki I I I I 1 I I i 1 I I i/i- satisfy (2). Since the point
E lies on both AB and CD,
its co-ordinates satisfy both
equations. By measurement,
^ they are found to be a; = 7,
'^ and y = — 5. The correct-
ness of these values is easily proved by substituting them
.^/
LOCI OF EQUATIONS.
453
\
\
\
IV-
V
in the given equations. Errors in drawing the diagram
will of course affect the accuracy of the results.
661. In the equations
a^ + 2/' = 66, (1)
x-y=ll, (2)
the circle ABCD is tlie
locus of equation (1), and
the line MN is the locus
of equation (2). The co-
ordinates of A and B sat-
isfy both equations, since
A and B lie on both loci.
The co-ordinates of A are
x = 7f ?/ = — 4 ; the co-ordinates of B are x = 4, y = — 7.
Since the square on the hypothenuse of a right triangle
equals the sum of the squares on the other two sides,
inspection of the diagram shows that (if r is the radius
of the circle),
r'^x^ + if.
Hence, the locus of any equation of the form x^ + y^ = 7^
is a circle whose radius is the square root of the right-hand
member. In the present example
^ = ± V65 = ±8.06
y^
Exercise 131.
Solve the following equations by constructing their loci :
1. 2a; + 3i/=8'i 3. 2x-'9ij=ll
3x + 7y--
2. Sx — 6f/=2
2x+7i/
= 2 .
= 22 J
x-'9ij=ll -)
x — 12y = 15)
3x — 12y--
4. 4x — 2y=20
6a; = 9y
454
ALGBBBA.
6. 2x'-3y = 4: ^
3x + 2y=^S2)
6. 2x + 3y = 7'\
7. 2a: — 9y = lll
3aj-4y = 7 J
8. 3a5 — 4y = — 61^
4x-5y = l J
9.
x-2y = 4.
2x — y = 5
10.
x y
^-5 = 6
X y J
11.
X y
« y J
12. a:* + 2/» = 104^
13. aj — y = 10 ^
a;« + 2/« = 178;
14. a:y — 12 = 0^
x-2y = 5 ;
15. x + y = 13^
a;y = 36 J
17.
18.
5y2_4a:2 = 12>»
16. 3y2 — 4a:* = 12^
2i
5 + y 12 + x
2a; + 5y = 35
4 ^
a 3y 27
1 ,1^11
Ja'^y'"72
662. When a single equation is given containing only-
one unknown number, all the terms must be transposed
to the first member. Thus,
x' + a: — 15.75 = 0;
or, if 2/=0, o^ + a;— 15.75=y.
Assuming values of «, we compute the corresponding
values of y, and construct the locus. Now, any value of x
which makes y = 0 satisfies the equation, and is a root ;
hence, any abscissa whose corresponding ordinate is zero
represents a root. The roots, therefore, may be found by
measuring the abscissas of the points where the locus meets
X'X, for at these points y = 0.
LOCI OF EQUATIONS.
455
From the given equation the following table may be
formed :
zv
Itxia
yis
Ifxis
Via
0
-16.75
-1
-16.76
1
-13.76
-2
-13.76
2
— 9.74
—3
— 9.76
3
-3.76
— 4
-3.76
4
+4.26
-5
+ 4.26
The table shows that one root
is between 3 and 4 (since y
changes from — to +, and there-
fore passes through zero); and,
for a like reason, the other is
between — 4 and — 5. The dia-
gram shows that the roots are
3.5 and— 4.5.
V
668. Construct the locus of x* — 5x — 3 = 0.
From this equation the following table may be formed :
Y
X
y
X
y
0
— 3
-1
+ 1
1
-7
-2
-1
2
-5
-3
-16
3
+ 9
/\ -I-
To plot this locus accurately, the
values of y must be computed for
aj = — 1.1, —1.2, —1.3, etc.
. The necessary portions of the lo-
cus are given in the diagram, and
the roots are found to be 2.5-, ^
— 0.6 + , and— 1.8+ ; nearer values are 2.49 + , — 0.65 + ,
and— 1.83 + .
456
ALGEBRA.
An equation of any degree may be thus plotted, and the
locus will be found to cross the axis X^X as many times as
the number of real roots in the equation.
664. When an equation has no real roots, the locus does
not meet XX.
In the equation
Y\
XU-
i
\J
-lilt
r'
/
X'.
.V
^ — I — I — I "»K I
Yf
a;*— 6a: + 13 = 0,
both of whose roots are imaginary,
the locus, at its nearest approach, is
4 units distant from XX.
666. The locus of an equation
whose roots are equal touches X'X,
but does not intersect it.
The equation
aj«+4aj + 4=0
has the roots — 2 and — 2, and
its locus is shown in the mar-
gin.
666. Loci are chiefly valued
as illustrations of the properties
of equations.
Exercise 132.
Construct the loci of the following equations :
1. a;2 + 3a; — 10 = 0. 3. a;* — 20a^ + 64 = 0.
2. cc' — 2x2 + 1 = 0. 4. x^ — 4a: + 10 = 0.
5. X* — 5x^+4 = 0.
CHAPTER XXXIII.
Equations in General.
667. Every higher equation of which we shall treat can
be reduced to the following typical form, which is called
the General Equation of the nth Degree :
^n _j. ^^«-l + ^^«-2 _^ C^n-^ ^ _^ ^=,0. (1)
In this the coefficients are real and rational, and the
exponents positive integers ; the coefficients may be inte-
gral or fractional, positive or negative. The equation is
arranged according to the descending powers of the un-
known number : K is called the Absolute Term, because it
does not contain x ; but it is classed among the coefficients,
since it may be regarded as the coefficient of x°. If any
one of the coefficients is zero, the corresponding term
vanishes.
Reduce to the typical form,
6x* - 2ar« -f 3o^ + 20 - 14a; = 37.
The coefiBcient of the highest power of x must be made unity.
Transposing, arranging, and dividing by —2, and supplying the
missing power of x we have,
a;6 - 3a;* + 0x8 - f aj2 + 7 X + V = q.
Here^= -3, 5 = 0, C=-f, D=7, JK:= V^ andn=6.
An Algebraic Function of x is an algebraic expression
whose value depends upon that of x, A function of x is
denoted by such symbols as F(x),f(x), and <l>(x). Thus,
7a;» + 4ar^ + 2a;-5
is a function of x.
458 ALGEBRA.
We shall designate equation (1) by F(x) = 0 ; and when
its coefficients as well as exponents are all integral, we shall
denote it ^y/(aj) =0.
668. A Cubic Equation is one of the third degree ; as^
A Biquadratic Equation is one of the fourth degree ; as^
a;*-3a;» + 4a;-15 = 0.
A Boot of an equation is an expression, which, when
substituted for the unknown number, will satisfy the
equation. If the equation is in the typical form, the first
member will become zero.
Imaginary roots are found in Higher equations as well
as in quadratics.
If an equation having literal coefficients is lower than
the fifth degree, it can be solved by methods discovered by
Ampere, Clausen, Descartes, Euler, and others. Abel has
proved that such an equation cannot be solved if its degree
is higher than the fourth.
Equations of any degree can be solved if the coefficients
are numerical.
669. J^(x) is exactly divisible by x — a, if a is a root of
F(x) = 0.
For the division can be continued until the remainder
does not contain x. Let Q represent the quotient, and It
the remainder. Since the dividend equals the product of
the divisor by the quotient, added to the remainder,
(x — a) Q + E = 0.
But, by supposition, x = a', hence, (x — a) Q = 0, and so
^ = 0. Therefore, the division is exact.
670. Conversely, if x — a is an exact divisor of F(x), a
is a root of F(x) = 0.
EQUATIONS IN GENERAL. 459
Let Q be the quotient, then
Q(x-a)=0.
Now, this equation is satisfied when a is substituted for
X ; hence, by definition, a is a root of the equation.
671. If a is a root of F (x) = 0, U is a factor of the abso-
lute term.
For, if a is a root, x—a is a divisor of F(x); and, by
the principles of division, — a must be a factor of the abso-
lute term; hence, +« is also a factor.
672. A ready method of determining whether a given
number is a root of a given equation is furnished by §§ 570,
571. Thus it may be shown that 4 is a root of
a^-llx' + 2Sx' + 2a^-16x + 32 = 0,
We see that 4 is a factor of the absolute term, and proceed as
follows :
aj6 - 11 aj* + 28aj8 + 2aj2 - 16aj + 32 |g~4
x^- 4x^ x4-7aj8 + 2x-8
-7aj4 + 28aj8
~7g*H-28a;8
+ 2rc2-16»
+ 2x2- 8g
-8X+32
-8xH-32
The work may be shortened by omitting the letters, and
using the numbers only. If any power of x is wanting, it
should be supplied, the coefficient being zero. The work
now appears as below :
1-11+28 + 2-16 + 3211-4
1- 4 1-7+2-8
-8 + 32
-8 + 32
-7 + 28
-7 + 28
+ 2-
16
+ 2-
_8
460 ALGEBRA.
But the operation may be still further abridged. As the
first term of the divisor is unity, the first term of each re-
mainder is the next term of the quotient ; also, the product
of the first term of each remainder by the second term of
the divisor gives the second term of the remainder. The
quotient will not be altered if we use 1 + 4 as a divisor,
and substitute addition for subtraction in the foregoing
process. Thus we have, omitting the first coefficient,
-11 + 28+2-16 + 32
4-28 + 0+ 8-32
-7 +0 + 2 -8 +0
The last horizontal line contains the coefficients of x in
the quotient, except that of the highest power, which is 1.
The number 0 is the remainder which comes from the
division.
If the coefficients are denoted by A^ B^ (7, etc., and the
trial root by r, we have the following rule :
Write A, B, C, , K in a horizontal line.
Under A write r, and write their sum under r.
Multiply the sum by r, and add the product to B ; multi-
ply this sum by r, and add the product to C ; and so on. If
the last sum is zero, the^ division is exact.
This method is called Synthetic Diyision.
When the coefficient of the first term of the dividend is
not unity, it must be multiplied by r, and the product
added to A ; this sum multiplied by r is then added to B ;
and so on.
Exercise 133.
Determine whether the number placed in a parenthesis
after each equation is a root of the equation :
1. aj* + 6a;* — 10a:« — 112a;2 — 207a; — 110 = 0. (—5.)
2. a:^'-Sx'+7x^ + x' — Sx + 2=:0, (1.)
EQUATIONS IN GENERAL. 461
3. x^ + 21x + 7a^ + U7 = 0. (-7.)
4. cx^ + 8x^—Tx' — 54:X + lQ = 0. (—8.)
6. a;* — 4aj» — 3aj2 — 2a; — 8 = 0. (2.)
6. x^ + Ua^ + 65x + 112 = 0, (-7.)
7. 2a;* — 4ir«-62a;2 + 114ir- 180 = 0. (6.)
8. a;* — 7a; — 2x2 — 15 = 0. (—5.)
9. x* + 2.3a;3 + 3.6a;2 + 4.9a; + 1.2 = 0. (—0.3.)
10. a;«-ia:2-^a;-^ = 0. (f.)
673. If F (x) = {x — a) (x — h)(x — c), etc., a, by c, etc.,
are the roots of F(x)=^0,
For, by hypothesis,
(X — a) (a — 6) (a — c){x — d) .... = 0.
This equation is satisfied if x = a, for this supposition will reduce
the first member to zero. The equation will be satisfied if x = a, or
if X = 6, or if X = c, or if x = d, etc. ; so, by definition, a, 6, c, d, etc.,
are roots of F (x) = 0.
The degree of the equation equals the number of bino-
mial factors. When there are n factors, the equation is of
the nth degree ; conversely, when the equation is of the
nth degree, and resolvable into factors of the form (x — a),
there are n of those factors.
674. When F(x)=0 is of the nth degree, it has n roots
and no more,*
The equation is, § 567,
«« + ^a;«-i + Bx^-^ + Cx"^ + + ^= 0. (1)
♦ Note. It is assumed that every equation has at least one root.
The demonstration of this fact, which is long and difficult, need not
be given here.
462 ALGEBRA.
Let a be a root of equation (1); then, by S 569, the equation is
divisible by x — a, and the quotient will be of the form
X— 1 4- ^'x— 2 4-^x»-3 + C'x"--* + 4- JT = 0. (2)
Now (2) must have at least one root, and let b be that root.
Dividing (2) by (x — 6), dividing the resulting equation by (x — c), c
being a root of that equation, and continuing the process until n — 1
divisions have been made, we reach a result of the form
x--/=0. (8)
It is plain, therefore, that the first member of equation
(1) is the product of n factors of the form (x — a), and so,
by § 573, it has n roots. If there were more than n roots
there would be more than n factors, and the degree of the
equation would be higher than the nth. These roots are
not necessarily all different. The roots of
x^'-'x' — Sx + 12 = 0
are 2, 2, and — 3.
This equation has two equal roots,
676. When the roots of an equation are given, the equa-
tion is readily found.
Find the equation whose roots are 3, 7, — 1, and — 2.
(x — 3) (x — 7) (x + 1) (a;+2)=0j
or, by multiplying,
X*— 7x^ — 7 x' + 4:3x + A2 = 0.
Exercise 134.
Find the equations whose roots are given below :
1. 2, 6, and —7. 3. 2, 3, —2,-3, and —6.
2. 1, 4, —1, and —3. 4. 0.2, J, and —0.4.
5. 5, 3+ V^, and 3 — V^.
EQUATIONS IN GENERAL. 463
676. The relation between the coefficients and the roots
of an equation is to be investigated.
According to § 675, a and h are the roots of
«« — (a + *)aj + a* = 0; (1)
a, hy and c are the roots of
aj« — (a-f5 + c)x» + (a^ + ac + ftc)a; — a^c = 0; (2)
a, 6, c, and d are the roots of
a* — (a + ft + c + (f)a:'+(a^ + ac + a^ + ftc + 5(f + ccf)a;'
— {abc + ohd + acd + hcd^ x -\- abed = 0 ; (3)
and so on.
From these the following conclusions may be drawn :
I. The coefficient of the second term, with its sign
changed, equals the sum of the roots; the coefficient of
the fourth term, with its sign changed, equals the sum
of the different products that can be formed by taking
the roots in groups of three ; and so on.
II. The coefficient of the third term equals the sum of
the different products that can be formed by taking the
roots in groups of two ; the coefficient of the fifth term
equals the sum of th'e different products that can be formed
by taking the roots in groups of four ; and so on.
III. The absolute term equals the continued product of
the roots with their signs changed.
Exercise 135.
By the above principles, form the equations whose roots
are given below :
1. 2, 4, and —3. 4. 6, 6, and 6.
2. 2, —1, and —7. 6. 2, 1, —2, and —1.
3. 2, 0, and —2. 6. 2, i, — 2, and —-J-.
464 ALGEBRA.
Solutions by Factoring.
677. J^\x) is sometimes factored by inspection, and the
roots of F(x) = 0 are thus found:
(1) Solves* — Tx — 18 = 0.
By factoring, we have {x — 9) (x + 2) = 0.
Therefore, by § 670, x = 9, or — 2.
(2) Solve aj« — 8 = 0.
(x-2)(x2 + 2x + 4)=0. (1)
Now, as equation (1) may be satisfied by putting either factor
equal to zero, we have
X - 2 = 0, (2)
and x2+2x + 4 = 0. (3)
From (2) we see that x = 2, and the solution of (3) gives
x=-l-\- V^, or - 1 - V^.
(3) Solve4a;* — a;2 + 2a; — 1=0.
4x4-(x2-2x+ 1) = 0,
(2x2)2-(x-l)2 = 0,
(2 x2 + X - 1) (2 x2 - X + 1) = 0.
2x2+ X- 1 = 0. (1)
2x2-x+l = 0. • (2)
From (1), X = I or — Ij
from (2) X = i (1 + V^) or i (1 - V^).
Exercise 136.
Find the roots of the following :
1. a:2 + lla; + 24 = 0.
2. 7x2+ 161ic + 714 = 0. (Divide by 7.)
3. a;* — 4aV + 3a* = 0.
4. a^ + 4x^ + Sx^ + 32 = 0. (Divide by a« + 8.)
6. 12a;2 — 5a: — 2 = 0.
EQUATIONS IN GENERAL. 465
6. 4:X^ — 9a^ + 6x — l = 0. 8. «« — 64 = 0.
7. 4:9x^ — 112bx + 6U^ = 0, 9. 3oi^ — x^ + Sx — l = 0,
10. « — 27aj* = 0.
678. Cubic equations in which at least one root is inte-
gral are easily factored. Let —a be an integral root; then
the equation may be written in the form
(x + a)(x^ + mx + 7i) = 0, (1)
or a:^ + (a-\'w)x^-\-(7i-\'ani)x + an = 0. (2)
a n
m am am -r- m = a.
By trial, a, m, and n are to be obtained. Resolve the absolute
term into two trial factors. If these factors are the correct values of
a and n, when a is subtracted from the coefficient of x^, and n from
the coefficient of x, as above, the second remainder divided by the
first remainder will give the first subtrahend, as shown above. In
this example, a is the first subtrahend, m the first remainder, and n
the second subtrahend ; — a is one root, and by solving x2+mx+n=0,
the other roots are determined.
Solve a:^ — 9x^ + 260: — 24=0.
x8- 9x2 + 20x- 24 = 0
-3+8
-6+18 18+ -6 = -3.
.•. X — 3 = 0, andx2 — 6x + 8 = 0; whence, x=3, 2, and 4.
Exercise 137.
Solve the following :
1. 0^3+3x2 — 25a; — 12 = 0. 6. x«—3a;2-54x — 104=0.
2. a:» — 4x* — 8x + 8 = 0. 7. x^ + 9o(^ + 2x'- ^S = 0.
3. a;' — 7x*+19a; — 21 = 0. 8. a:^ — 2a;2 — 25a: + 50 = 0.
4. ar»--8aj2 + 21aj — 18 = 0. 9. a;^ — 3a:2 — 61ar + 63 = 0.
5. x« — 26aj — 5 = 0. 10. a;' — 37ar — 84 = 0.
466 ALGEBRA.
679. Biquadratic equations can be solved if they can be
resolved into two factors of the second degree, even when
no root is integral.
(05* + vix + n) (x^ -\-px + ^)
= X* + (w+J») ^+ (^ + ^i^+ 2') a^+ (wp + wi^) « + M-
Solve x*+ 13x^ + 33 x2 + 31a: + 10 = 0.
The following are the values of the coefficients :
m + i)=13, (1)
n + rnp + ^ = 33, (2)
np+mq- 31, (3)
and nq = 10. (4)
Take trial-values of n and q from (4). From (2) find the value of
mp. Combine this with (1), and get by inspection the values of m
and p. If the values of m, n, p, and 9, satisfy (3), they are correct.
First Hypothesis.
Let n = — 2, and g = — 5.
Then, from (2), mp = 40,
and, using (1), m = 5, and j) = 8.
These values will not satisfy (3).
Second Hy'pothesis.
Let n = 1, and q = 10.
Then, from (2), mp = 22,
and, using (1), m = 2, and j) = 11.
These values satisfy (3). Since the two factors of the biquadratic
are (x^ + t/ix + n) and (x^ + px + g), «ach of which equals zero, we
have, in this example,
x2+ 2x4- 1 = 0,
and x2H-llx+10 = 0;
the solution of which gives the values of x.
The values of n and q can be interchanged without affect-
ing the values of m and p. If any hypothesis makes m
and p contain surds, it is to be rejected, unless n = q\ for,
if m and p contain surds, (rip-\'mq), the coefficient of aj,
will contain surds, unless n=^q) in which case the surds
may disappear.
EQUATIONS IK 6ENEHAL. 467
Exercise 138.
Solve the following :
1. a;* — 2x3— 13x» + 38aj — 24 = 0.
2. x' — 5x^ — 2a^ + 12x + S=^0.
3. «* — 4x« — 8a; + 32 = 0.
4. aj* — 12a;«+50x2 — 84a; + 49 = 0.
6. aj* — llic2 + 18a;~8 = a.
6. a;*— 10aj*-20x~16 = 0.
7. a;* — 7a;3 + 23a;2 — 47a; + 42 = 0.
8. a:* + 2a;3 — 9a;2 — 8a; + 20 = 0.
9. a;* — 4a;3 — 102aj2 — 188a; — 91 = 0.
10. a;* — lla;3 + 46a:2-117a; + 46 = 0.
Descartes' Eule of Signs.
680. A Complete Equation is one in which no power of
X is wanting.
If two successive terms have like signs, there is a perma-
nence of sign ; if they have unlike signs, there is a variation.
Thus, in the complete equation
a;' — aj« — 5a;* + 4a;* + 2aj3 — 3a:2 + 6a; + 7 = 0,
there are three permanences and four variations.
Descartes discovered that the signs of the roots are
related to the signs of the terms of an equation.
581. Descartes' Eule (modified) is as follows :
iVb complete equation has a greater number of positive
roots than of variations of sign, nor a greater number ofneg-
ative roots than of permanences of sign.
468 ALGEBRA.
Suppose that the successive terms of a certain complete
equation have the signs
+ - + + + -- +
Here are three permanences and four variations. To in-
troduce a new positive root into the equation, we multiply
it by a? — a. The resulting signs are shown in the follow-
ing scheme :
+ - + + + +
+ -
+ - + + + +
~ + + + -
+ ~ + ±±-±+~
The sign it indicates that there is doubt whether the
term is positive or negative. Inspection shows that where
there were permanences in the multiplicand, there are
ambiguities in the product. Therefore, the introduction of
the root + a has not increased the number of permanences.
But, as it has made the number of terms one greater, the
number of variations must have been increased by one, at
least.
As the introduction of a positive root increases the
number of variations, the whole number of positive roots
cannot exceed the number of variations. The root — a can
be introduced into an equation by multiplying it by a: + a ;
and, by reasoning similar to the preceding, it can be shown
that the number of negative roots cannot exceed the num-
ber of permanences.
682. When all the roots of a complete equation are real,
the number of positive roots equals the number of variations,
and the number of negative roots equals the numher of per-
manences.
EQUATIONS IN GENERAL. 469
Let jp be the number of positive roots, n the number of
negative roots, v' the number of variations, and p' the num-
ber of permanences.
Now the degree of the equation equals jp + n, and also
equals v'+p'. Therefore,
p + n = v'+p'',
or, by transposing, v' —p = n — p\ (1)
Now, by § 581, the first member of equation (1) must be
positive, or zero ; also the second member must be negative,
or zero. Since the first member equals the second, both
are zero. Hence, v'=py and n=p\
583. In an equation which lacks one power of x, the
presence of imaginary roots may sometimes be detected.
For illustration, take the equation
x^:h0x^ + 5x + T = 0.
We are at liberty to assume that the second term is
positive, or that it is negative.
Taking it as positive, there is no variation, and the
equation has, therefore, no positive root.
Assuming the second term to be negative, there is only
one permanence, and so there cannot be more than one
negative root.
As there are three roots, and as imaginary roots come in
pairs (§ 586), we conclude that one root is negative, and
two are imaginary.
684. A complete equation whose signs are all positive
can have no positive real root, for there are no variations
of sign. When the signs are alternately positive and nega-
tive, there are no negative real roots, for there are no
permanences of sign.
470 AtiGS^Ki.
Exercise 139.
All the roots of the equations given below are real;
determine their signs.
1. a* + 4ar' — 43a:* -68a: + 240 = 0.
2. x» -22 a:* + 155a: — 350 = 0.
3. a:* + 4ar»-35a:*-78a: + 360 = 0.
4. a:»- 12x2 -43 a: -30 = 0.
6. a:«-3a:*-5a;3+15x« + 4a:-12 = 0.
6. a:^ — 12a:2 + 47a: — 60 = 0.
7. a:*— 2a:» — 13a:» + 38a: — 24 = 0.
8. a:* — a;* — 187 a:' — 359 a;2+ 186a: + 360 = 0.
9. a:«— 10a:« + 19a:< + 110a:» — 536a:* + 800a:-384 = 0.
10. a:'-10i:«+22a:* + 32a;*— 131a:«+50a:*+108x— 72=0.
Fractional and Imaginary Eoots.
685. A rational fraction is never a root off(x) = 0.
Suppose that -, a simple fraction in its lowest terms, is a
root of
x^ + Ax'^' + Bx»-^+ Cx»-^+ + K=0. (1)
Substituting - for x, and multiplying by t'^\ we obtain
~ + Al»-' + Btl'^-' + Ct^l^-^ + + Kt^' = 0. (2)
But the sum of a fraction and a number of integers can-
ot equal zero. Therefore
not a root of equation (1).
not equal zero. Therefore, equation (2) is false, and - is
EQUATIONS IN GENERAL. 471
686. Imaginary roots enter equations in conjugate pairs.
Any imaginary root may be represented by the expres-
sion a ± bL (See § 538, and the last sentences of §§ 541
and 546.) When a is zero, the root is a pure imaginary.
There must be an even number of imaginary roots, or the
absolute term, which is the product of all the roots with
their signs changed, would be imaginary ; hence these roots
are in pairs.
The product of x—{a-{-hi) hj x-—(a — bi) is real. If
the roots could not be grouped in conjugate pairs, the prod-
uct {x — a) (x — b) (x — c) would involve imaginaries.
687. If F(x)=0 is of an odd degree, it has at least one
real root ; if of an even degree, all the roots may be im^ag^
inary.
An equation of an odd degree has an odd number of
roots; hence, all the roots cannot be imaginary. § 586.
When the equation is of an even degree, it can be resolved
into an even number of factors, as follows :
(x--a)(x--b){x'--c){x'-d)(x—e){x--f) =0. (1)
If (x — a)(x — b) = x^-^ px-\- q,
and (x — c) (x — d)=^x^ -\- i\x + q^, and so on,
equation (1) becomes
(aj2 + i?aJ + !?) (a^ + Pv^ + ?i) {^ + V^ + <1^ = 0.
Putting each of these trinomial factors equal to zero, and
solving the resulting equations,
if \'^^^ ^^^ 1l"<2'i> and so on,
all the roots will be imaginary.
472 ALGEBRA.
Equal Roots.
588. In finding the equal roots of an equation, expres-
sions called Derivatives are used, which play a very impor-
tant part in the Differential Calculus. •
Let a; be a variable (§ 467). If it is gradually increasing
or diminishing, Sx will increase or diminish three times as
fast; 7x, seven times as fast; and mx, m times as fast;
m is the ratio of the rate of change in mx to the rate
of change in x.
Thus, if x = 3, 7aj = 21. Suppose that x increases from
3.0 to 3.1, then F(x) will increase from. 21.0 to 21.7 in the
same time. That is,
the increase of F(x) _ 0.7 _ 7
the increase of a; 0.1 "~ i
But while x increases at a uniform rate, F(x) may not
increase at a uniform rate.
Thus, if F{x)=5x^—2^-\-4.x — l, when x = l, F(x) = 7',
when x = 2y F(x) = 43 ; when x = 3, F(x) = 137.
Hence, to find the ratio between the increments of F(x)
and of X at the instant x begins to change from any par-
ticular value, the increment of x must approach indefinitely
to zero. In general.
The derivative of F(x) is the limit of the ratio of the iruyre-
ment of F(x) to that of x a^ the increment of x approaches
indefinitely to zero.
An increment may be positive or negative.
The derivative of F{x) may be found as follows :
I. In F(x) substitute x-^-h for x, h being an increment
that approaches zero as a limit. Denote the result by
F{x + h), Subtract F(x) from F(x-\-h)i the remainder
is the increment of F(x).
EQUATIONS IN GENERAL. 473
II. Divide this remainder, [_F(x'\- h) — F(x)']yhj h: the
quotient is the ratio of the increment of F(x) to that of x.
III. Any term of this quotient that contains A as a
factor, will have zero for a limit, and therefore vanish.
589. Find the derivative of aic^+ ft.
ByL, F(x) = ax^ + b,
and Flx + h) = aa^ + 2 axh + ah^ + b.
.\F(x + h) — F(x) = 2 axh + ah\
By III., 2 ax is the derivative of ax^-^-b.
Find the derivative of a^.
(x + hy=a^ + 2xh + h^,
2x + h,
(x + hy—x^
h
By III., 2a; is the derivative of x*.
Find the derivative of x^,
h
By III., 3x2 jg ^YiQ derivative of x^.
In general, the derivative of x" is nx^~^.
For ic" = X taken n times as a factor ; and if x increases,
the increase of each x is multiplied by the continued product
of all the others ; that is, by ir"~^ But there are n of these
increasing quantities, and the derivative is therefore nx^^\
Hence, the derivative of a simple expression is found by
multiplying the expression by the exponent of the variable,
and diminishing the exponent by unity.
690. When F{x) contains a number of terms, involving
different powers of x, the derivative of the entire polynomial
is equal to the sum of the derivatives of the separate terms.
474 ALGEBRA.
For, if /', /", /"', 7"^, etc., represent the increments of
the successive terms of the polynomial (when a; + A is sub-
stituted for a;),
/'+/" + 7"' + 7'^
h "^
is the derivative of F{x) (when h approaches indefinitely
to zero). Likewise,
h h h h
are the derivatives of the different terms. It is plain that
7' + 7" + 7"' + 7'^ _J^,J^yI^_,lZjL.
h h h h h
Find the derivative of 4ar' + 35c* + 6a; + 7=0.
Let F(x) = 4x» + 3x2 + 5x + 7.
By substitution,
F(x + A) - F(x) = 12^2+ (12 A2+ 6ft)x + 4A8 + 3^2 + 6^
.-. ^^"^ ^ hy-n^) ^ 12x2 4- (12;^+ 6)x 4- 4/^2 + 3 j^4- 5.
By letting h approach zero as a limit, and denoting the derivative
l>y ^'(»)» 2r'(a;) = 12 x2 + 6 X + 5.
But 12x2, 6x, and 5, are respectively the derivatives of 4 x*, 3x2,
and 5 x. By considering the last term of F(x) as the coefficient of x°,
its derivative would be 7 X Ox-^, which equals 0.
The derivative of a given expression is called its first
derivative ; the derivative of the first derivative is called
its second derivative, etc. The second derivative is denoted
by F'\x)] the third by F''\x), etc.
From what has preceded, the following rule for finding
the first derivative of F{x) is obtained :
Multiply each term by the exponent of x in it, and dimin-
ish the exponent by unity.
EQUATIONS IN GENERAL. 475
591. Find the successive derivations of
F^ (a:) = 6a:* + 15aj*-8xS-15a;2 + 4a; + 7,
i^" (a) = 30a;^ + 60aj«~24a;*--30a; + 4,
ii^'" \x) = 120 a^ + 180 aj2 - 48 a - 30,
ii^'v (a;)=360x3 + 360a:-48,
1^ (x) = 720 a; + 360,
F^y^ (a;) = 720,
iP^"(aj) = 0.
Exercise 140.
Find the successive derivatives of the polynomials :
1. x^^2x^Z. 3. aj* + 2a;» — 5a;2 + 64.
2. a^-3a;2 + 7x + 25. 4. x* + a;^- 6a:3 + 3x2-4a; + 27.
5. «* — 3ax^ + 6&a;^ — 9<?a; + mn.
592. TAe derivative of F(x) is the coefficient of the first
power of h in F(x-\-h),
For, F(x 4- h) is composed of F(x) and terms each of which con-
tains some power of h.
F(x + h) — F(x) is composed of those terms of F(x + h) of which
each contains some power of h,
-^ ^ — is composed of the coefficients of the first powers
of h in F{x + h), and of terms containing h ; rejecting the terms con-
taining h (III., § 688), the derivative of F(x) is the coefficient of the
first power of h in F{x + h).
Find the derivative of F(x) = ax^ + bx^ -{-cx-^d.
F(x + h) = ax^ + Sax^h + Saxh^ + ah^
+ bx^ + 2bxh + bh^
-{- ex + ch
+ d.
F{x-\-k) - F(x) = (3ax2 + 2bx-\- c)h+ {Sax-^h)h^+ ah^,
476 ALOEBBA.
593. Apply these principles to the detection of the equal
roots of fIx) = 0.
This equation may be written in the form
(x-a)(x-6)(x-c)(x-d) = 0. (1)
Substituting x + hior x,
(X - a + h)(x -b + h)(x-c + h)(x -d+ h) = 0. (2)
If (2) be expanded, and powers of h higher than the first be
neglected, the h in the first parenthesis will be multiplied by
(x-6)(x-c)(x-d) ;
the h in the second parenthesis will be multiplied by
(x — a) (x — c) (x — d) ; and so on.
The total coefficient of h will be
(x-6)(x-c)(x-(i) -]
+ (x-a){x-c){x-d) I (3)
+ (x-a)(x-6)(x-d) J
Now (3) is the derivative of F{x), by § 592.
When (1) has two equal roots, let a = b. We see that
a; — a is, in this case, the H. C. F. of (1) and (3). When (1)
has three equal roots, let a = ft = c ; then ' (x — ay is the
H. C. F. of (1) and (3). In general, when F(x) =0 has w
roots equal to a, and p roots equal to b, the H. C. F. of F(x)
and F'(x) is (x—a)"^^ (a;— 5)^-^
694. Hence, in seeking the equal roots of F(x) = 0, if
<l> (x) represents the H. C. F., it will be seen that :
The roots of <^ (a;) =0 will each occur once more in F(x) =0
than in <^ (x) = 0.
When there is no H. C. F. there are no equal roots. In
some cases, <^(a;) = 0 is of so high a degree that it cannot
EQUATIONS IN GENERAL. 477
be solved conveniently. If, however, it has equal roots,
they may be found as above.
Solve x^ + 4x'' — 20x^ — 50x^ + 175x^
+ 118x2 _ 5833.^350 = 0. ^^j
F'(x) = 7afi + 24x5 - 100 x* - 200x8 + 625x2 + 236x - 688, (2)
0{x) = x8 — x2 — 8x+ 12,
and the equation to be solved is,
x8-x2-8x+12 = 0. (3)
Apply § 678 to this case, or else proceed to find the equal roots of
(3). The first derivative is
8x2 -2x- 8. (4)
The H. C. F. of (3) and (4) is x — 2. Since x — 2 = 0, x = 2, and
the root 2 occurs twice in (3). Dividing (3) by (x — 2)2 gives x + 3 = 0;
hence, — 3 is the other root of (3). As the root 2 occurs twice in (3),
and — 3 once, we know that (1) has as roots, 2, 2, 2, — 3, and — 3.
By dividing (1) by (x — 2)8 (x + 3)2, the quotient is
x2 4-4x — 5 = 0.
The solution of this equation gives x = 1 or — 6. The seven roots
of (1) are, therefore, 1, 2, 2, 2, — 3, — 3, and — 5.
Exercise 141.
Find all the roots of the following ;
1. a^'-Sx^ + 13x — 6 = 0. 3. x^ — 6x^-'Sx — S = 0.
2. a?'-'7x^ + 16x-12 = 0. 4. x^-2x'—15x + S6=0.
5. a;* — 7x3+9a:2 + 27x-54 = 0.
6. x* — 24 0^2 + 64^ — 48 = 0.
7. x* — 10x^ + 24.x^ + 10x — 25 = 0,'
8. a;«-ll^^ + 19a;«+115a;2-200aj-500 = 0.
9. a^ — 2x*+Sx^—7a^ + Sx — S = 0.
10. x^ + 6x^ + x^''24:X + 16=^0.
478 ALGEBBA.
Transformation of Equations.
595. The solution of equations is usually facilitated by
reducing them to the form /(a;)=0; and, since we treat
only of equations whose exponents are positive integers, we
have simply to make the coefficient of the first term unity,
and the succeeding coefficients integral. If the exponents
were negative or fractional, the equation could still be
reduced to the form f(x)=0.
The coefficient of the first term may be reduced to unity,
without altering the values of the roots, by the following
rule :
Divide the equation by the coefficient of the highest power
of X,
When this division makes any of the coefficients frac-
tional, the equation is to be transformed into another whose
jiGoefficients are integers.
Let m be the L. C. M. of the denominators of the frac-
tions.
Substitute — for x in the equation
m ^
jr« + Ax"-"^ + ^ic«-3 + Cx^-^ + -\-K= 0, (1)
and clear of fractions ; this gives
2^''+^my«-^4-^mV""2_|_^^^8y„-s_|_ -[.Km''=0. (2)
The coefficients of (2) must be integral ; because m is the
L. C. M. of the denominators of A, B, C, etc. The roots of
(2) divided by m are the roots of (1) ; for, by hypothesis.
y
m
Hence, to remove fractional coefficients.
Multiply the coefficient of x'*-^ by m, that of af^^ by m\
etc,, the absolute term being multiplied by m\
EQUATIONS IK GENERAL. 479
A little study will sometimes enable us to find a number
less than the L. C. M., whose successive powers will clear
the equation of fractions.
Reduce 3a5«— 4x* + ^a; — i = 0
to the form/(ic) = 0.
Divide by 3, 2^-ix^+ix-i=0,
The L. C. M. of the denominator is 12 ; and, by the last rule,
x8 - 16x2 - 24x - 432 = 0,
which is in the form required.
Exercise 142.
Put these equations in the form f(x) =0 :
1. 2a^+fa;2— x+J=0. 3. 5a;*— a;^— Jj^aj*-V-a^+l=0.
2. 3a;»+5«2— Jx— 8=0. 4. a^+ix*+ia^—i<t'+X'-S=0.
5. x^ — 2x^-\-ix — 14: = 0. (Supply the term Oa^,) . ^.
Hereafter we shall treat only of equations which are in the
formf(x) = 0.
596. The signs of the roots of f(x) = 0 are changed by
changing the signs of the alternate terms, beginning with the
second.
By § 576, I. and II., it is plain that changing the signs
of the roots changes the signs of the alternate terms, begin-
ning with the second, and does not change the signs of the
other terms. Conversely, if the signs of the alternate terms
of a given equation are changed, beginning with the second
term, an equation is obtained whose roots are the roots of
the given equation with their sighs changed. Missing
powers of x must be supplied with the coefficient 0.
480 ALGEBRA.
597. The roots of f(x) = 0 are rmdtiplied by m, by multu
plying the second term by m, the third by m^, etc.
In removing fractional coeflBicients, we multiply aj*~^ by
m, «*"* by m*, etc. ; and it has been shown, in § 596, that
the roots of the resulting equation are m times the roots
of the original equation.
598. To obtain an equation whose roots are the reciprocals
of those of f(x) = 0, "ivrite the coefficients in reverse order.
In the given equation
«« + ^x«-' + Bx"^^ + + 1^ +J^^ + K= 0, (1)
substitute - for aj, multiply by y", and change the order
of the terms ; the result is
Ky- + Jy-' + hf-' + ^Btf + Ay + 1=0. (2)
Comparison of (1) and (2) shows that the order of the
coefficients has been reversed.
By dividing by JT, (2) may be reduced to the form
/(*)=0.
Thus, the equation whose roots are the reciprocals of the roots of
{C* - 6a* - 7 x* + 6x8 + 13x2 - 9x 4- 25 = 0,
is 26x6-9x5+13x* + 6x8-7x2-5x+ 1 = 0.
599. In solving some higher numerical equations, it is
necessary to transform f{x) = 0 into an equation whose
roots are less by A.
In the equation
«« + ^a:»-i + ^a;«-* + + Jx'\- K= 0, (1)
by putting x = y-^h (in which case y=:x — h,SLS required),
(y+hy+A(y+hy-'-\-B(y+hy-'+ +J(y+h)+K=0, (2)
Expanded by the binomial theorem, equation (2) becomes
y^' + A
-{•nh
yr^ +B
+ (n — l)Ah
I +in(n'-l)h^
y^'+ +Jr=0. (3)
EQUATIONS IN GENERAL. 481
Let A + nh = A', B + (n — 1) Ah + in(n--l) h^ = B',
etc., and x — h=^y\ equation (3) then reduces to the form
+ B\x - hy-^ + + J\^ - A) + JT' = 0. (4)
Divide the first member of (4) by {x — h), and the re-
mainder is K' ; divide the quotient just found by (x — h),
and the remainder is J\ By continuing the divisions, all
the coefficients of (4) can be found.
The values of x are the same in (4) and (1) ; therefore,
the first member of (4), when developed, equals the first
member of (1). Hence, if the successive divisions were
performed upon (1), the remainders would be the same.
600. The rule, therefore, for transforming f(x) = 0 into
an equation whose roots are less by A, is as follows :
Divide f(x) by x — h, and the remainder will be the abso-
lute term of the transformed equatio?i. Divide the quotient
just found by X — h, and the remainder will be the coefficient
of the last term but one of the transformed equation. Con-
tinue the process until all the coefficients are determined.
By using »+ A as the divisor, f(x)=0 can be transformed
into an equation whose roots are greater by h.
The above rule will apply when the coefficient of x" is not
unity. It is best to employ Synthetic Division (§ 572).
Transform into an equation whose roots are less by 5,
x*'-Aa^ — 7x^ + 22x + 24: = 0.
-4- 7+ 22 + 24
+ 5+ 6- 10 4-60
+ 1 — 2 + 12 + 84 (84 = first remainder).
+ 6 + 30+140
+ 6 + 28 + 162 . . . (162 = second remainder).
+ 6 + 66
+ 11 + 83 (83 = third remainder).
+ 6
+ 16 (16 = fourth remainder).
482 ALGEBRA.
The required equation is, therefore,
y* + 16y» + 88ya + 152y + 84 = 0.
The roots of this equation are — 1, — 2, — 6, and — 7 ; the roots of
the original equation are 4, 3, — 1, and — 2.
Exercise 143.
Transform each equation below into another whose roots
are less by the number placed in the parenthesis directly
after the equation :
1. a;»~lla;2 4- 31a; -12 = 0. (1.)
2. x^'-6x^ + 4.a^ + lSx — o = 0, (2.)
3. x« + 10^2 + 13a; — 24 = 0. (—2.)
4. ar»- 9a:* + 22 a; — 12 = 0. (3.)
5. x' + a^ — 16x'-4x-\-^S = 0. (4.)
6. a;* + 2a;«-25a;*-26a; + 120 = 0. (0.7.)
7. a;^ + a;2-3x + 4 = 0. (0.3.)
8. a;« + a:* + 3a;*-2a; — 16 = 0. (0.5.)
9. a;*-3a;*-2a:» + 3a;*-7a; + 12 = 0. (-1.)
10. a;« — ar^ + 2a;*-3a;« + 4a;«-5a; + 6 = 0. (0.2.)
601. To transform /(x^^-O into an equation whose second
term is wanting, substitute y for x.
n
In (3), § 599, to make the second term vanish, nh must
equal — A, and so h must equal In that case, x =
A ^ n
y Hence, to make the second term vanish, sub-
"" A
stitute y for x,
n
CHAPTER XXXIV.
Higher Numerical Equations.
situation op the roots.
602. A commensurable root is rational, and is either
integral or fractional.
An incommensurable root is a real root which is not
commensurable.
In the solution of higher numerical equations, we first
determine, by § 572, the commensurable roots. These are
integral, when the equation is in the form /(a) = 0 (§ 585).
In finding the first root, which we shall denote by a, we
get the coefficients of the equation
^=0;
we then find a commensurable root, ft, of that equation, and
get the equation
m =0.
(x — a)(x — h)
Each division depresses the degree of the equation divided,
and when all the commensurable roots have thus been
divided out, if the resulting equation is of too high a degree
to be solved easily, we proceed to find the incommensura-
ble roots by Horner's Method. Imaginary roots may then
be sought ; the student, however, will not be required to
search for these.
484 ALGEBRA.
603. Find the commensurable roots of
The factors of 20 are zb 1, =b 2, db 4, db 5, and =b 10. We
try 1 at first, as follows :
+ 3-2-15-15+ 8 + 20
+ 1 + 4+ 2-13-28-20
+ 4 + 2-13-28-20 0
Hence, 1 is a root ; and the last line contains the coeffi-
cients (except the first) of the equation obtained by divid-
ing the original equation by x — 1. Next try — 1.
+ 4 + 2-13-28-20
-1-3+ 1+12 + 16
+ 3-1-12 — 16— 4
) fails. We then
Hence, -
- 1 is not a root. A trial
try -2.
+ 4 + 2-13 — 28-20
-2-4+ 4 + 18 + 20
+ 2-2- 9-10" 0
Hence, — 2 is a root ; and the last line contains the co-
efficients of the next depressed equation. A trial of 2 fails,
but — 2 succeeds, giving as the coefficients of the depressed
equation q 2—5
Trials of the factors of — 5 all fail. The roots so far
found are 1, — 2, and — 2.
The equal roots might have been found by § 594, but the
work would have been tedious. Three exact divisions have
been made, depressing the degree of the original equation
by three, and the last equation is, therefore,
ic' + Ox*- 2a; — 5 = 0,
whose roots are incommensurable.
higher numerical equations. 485
Exercise 144.
Find the commenstirable roots of each equation below.
The number of these roots is given in the parenthesis.
1. x* — 4:a^ — Sx + 32 = 0. (2.)
2. x^— 6x2 + 10x — 8 = 0. (1.)
3. x* + 2a^ — 7x^ — Sx + 12 = 0, (4.)
4. a^ + 3x^ — 30x + 36 = 0, (1.)
5. x^ — 12r^ + 32a;2 + 27a; — 18 = 0. (2.)
6. ir^ — 9a;^+ 17x2 + 270; — 60 = 0. (2.)
7. ic* — oa;^ + 3a;3 + 17a;2 — 28x + 12 = 0. (5.)
8. x'-10x^ + 3ox^--50x + 24: = 0. (4.)
9. x' — Sx^ + llx^ + 29x^ — 36x — 4.5 = 0, (3.)
10. x' — x^ — 6x^ + 9x^ + x — 4. = 0. (1.)
604. After the commensurable roots are removed, the
situation of the incommensurable roots is to be determined.
Many methods of doing this have been discovered, the most
noted of which is Sturm's Theorem ; but the application is
tedious, and it is generally best to proceed by trial.
605. If two trial values of x, substituted in f(x), give
results with unlike signs, at least one root lies between the
values. If the signs of the results are alike, no root, or some
even number of roots, lies between the values.
Let the trial-values of x be w and n, and the resulting
values oif{x) be JIf and N, which have unlike signs. If x
be supposed to vary continuously from m to n, f(x) will
pass from M to N, and must pass through zero, because M
and N have unlike signs.
The value of x which makes f(x) = 0 is a root ; hence, at
486 ALGEBRA.
least one root lies between m and n, f(x) may pass through
zero any odd number of times ; but if the number of times
is even, M and iV will have like signs.
Let M and N have like signs. When x varies from m
to 71, f(x) may not pass through zero at all. It cannot pass
through it an odd number of times, for M and iV" would
then have unlike signs. It may pass any even number of
times ; and therefore no root, or an even number of roots,
lies between m and n.
These theorems are beautifully illustrated by the loci of
equations. Thus, if any two abscissas are chosen (repre-
senting values of x), the signs of the corresponding ordi-
nates, which represent values of f(x), are unlike when the
number of intersections of the curve with the axis of X is
odd; the ordinates have like signs when the number of
intersections is even.
606. The value of f(x), when x = m,is the remainder left
after dividing f(x) by (x — m).
Let Q be the quotient, and B the remainder which does
not contain x. Then
f(x) = (x-m)Q+E.
Now, when m is substituted for x,
f(x) 01 f(m) = E.
Thus, when a; = 5, the value of
a^-4.x* + 7x^-2x + ^7
is computed by § 572 as follows :
-4 + 0+ 7- 2+ 47
+ 5 + 5 + 25 + 160 + 790
+ 1 + 5 + 32 + 158 + 837
Therefore, /(a;)= 837 when x = 5, and this value is
obtained more expeditiously than by direct substitution.
higheb numerical equations. 487
Exercise 145.
Compute the value of each expression below, the value
of X being enclosed in the parenthesis following each :
1. x* — 5a^ + 26x^ — 4.x + 7. (6.)
2. x^-4:a^ + &x-22. (—7.)
3. a^-2x' + Sx^ + x^ — 2S. (2.)
4. a^ + 7a;« — 2^2 — 49. (—3.)
6. x*- 14x3 + 473. (6>)
6. x'' — 2x' + 3x* + 2a^ + x^ — 7x'-96, (—2.)
7. x^ — x'-'2x*+a^ — 6x + U. (3.)
8. x' — 4x^ + 2x''-7x + 16. (10.)
9. x' — x^-2x'-3x* + 2x^+x^'-x + 4. (-2.)
10. x' — &a^ + 6x^ + 3x — l. (4.)
607. Let it be required to find the situation of the roots
of
x^ — 2j^—llx' + ex + 2 = 0.
They are real. Since there are two variations and two
permanences of sign, there are two positive and two nega-
tive roots. Inspection of the signs and coefficients shows
that these roots are small numbers.
The values of f(x) may be computed by § 606, assuming
values of x from — 4 onward, until the situations of all the
roots are discovered. These values are tabulated below :
Ifxis
/(*)i8
Ifxis
/(a;)i8
-4
+ 186
+ 1
-4
— 3
+ 20
+ 2
-30
-2
-22
+ 3
-62
— 1
-12
+ 4
— 22
0
+ 2
+ 6
+ 132
488 ALGEBRA.
The changes of sign in the columns headed f{x) show
that the first root lies between — 3 and — 2, the second
between —1 and 0, the third between 0 and + 1, and the
fourth between + 4 and + 5. When any root lies between
0 and ± 1, before using Horner's Method, it is necessary
to find the first significant figure.
Search for the root between 0 and + 1 ;
If a; = 0.5, /(«) = + 2.1.
« a: = 0.6, /(a:) = + 1.3.
« re = 0.7, /(ic)= + 0.4.
" a; = 0.8, /(a;):=-0.9.
The root therefore lies between 0.7 and 0.8.
The root between 0 and — 1 is next sought :
If a; = -0.2, /(a;) = + 0.4.
« a; = — 0.3, /(a;) = — 0.7.
This root therefore lies between — 0.2 and — 0.3.
The first significant figures of the other roots are evi-
dently — 2 and + 4.
608. In the preceding example, the changes of sign of
f{x) led to the easy discovery of the situation of the roots.
But two successive values of f{x), having the same sign,
may, by § 605, have two or some other even number of
roots between them.
Take the equation
whose roots are all real. The law of signs gives two posi-
tive and one negative root. It is easy to see that the
positive values of x do not exceed + 2, and that the nega-
tive value is about — 5. We therefore tabulate values of
HIGHER NUMEKICAL EQUATIONS. 489
Ifxia
/(*)i8
n«is
/(x)is
+ 2
+ 16
-3
+ 41
+ 1
+ 1
— 4
+ 46
0
+ 2
-6
+ 37
-1
+ 13
-6
+ 8
-2
+ 28
-7
-47
One root is — 6 + ; the values of f(x) indicate by their
rapid diminution and close approach to zero in the first
part of the table, that f(x) becomes zero when x is between
Oand+1.
If a: = 0.4, /(a;) = + 0.1.
« aj = 0.5, /(a;) = -0.1.
« aj = 0.6, /(a;) = — 0.2.
« a; = 0.7, /(a;) =-0.1.
" a; = 0.8, /(aj) = + 0.1.
The other roots are therefore 0.4+ and 0.7 + .
609, The figure sought may also be found by a method
which depends upon the principles of differences. Take
the example in § 607. In the table below, the column
headed di contains the differences of the first order ; that
headed d^ contains the differences of the second order ;
and so on. The differences which are of the same order as
the degree of the equation, are constant, and furnish a test
of the accuracy of the computation of the values oif{x),
X f(x) di dz c?8 d^
— 3 +20 _
-2 -22 T'n +52
— 1 -12
0+2 ^'-;^ _20
+ 10 J^^; _48
+ 14 +^ _-24 +24
+ 1 -4 -.26 -20 I +24
+ 2 -30 f +4 +24 ^
+ 3 -52 -"^"^
490 AL6EBBA.
In Example (3), § 507, » = i, and it is required to find
the number which added to 6798.9 will give the answer.
If the answer 8331.4 had been given and n required, the
following equation might have been written :
8331.4 =6798.9+71 (2029.7) + ^(^^^) (123.1) + y
or, in a general form,
n(w — 1) ,
A==a-f- nai -\ ^^-r — ^ a^ + j
the solution of which would give an approximate value
of n.
To find the first figure of the root which lies between
0 and + 1 in the present example,
A = f(x) = 0', a = + 2', ai = — 6; a2 = — 20.
The equation is
0=2-6«-20^^^^- ;
or, reducing,
lOn^ — 4n = 2;
whence n = 0.7, roughly ; and, adding this to 0 (the value
of X which stands opposite a), 0.7 is an approximate value
of one of the roots.
The next figure of the root which lies between — 3 and
— 2 may be obtained by means of the equation,
0 = 20-42^ + 52 ^?^^^f^;
or, reducing,
2671^ — 68n = — 20;
whence, 7i = 0.3, roughly; and, adding this to —3 (the
value of X which stands opposite a), — 2.7 is an approxi-
mate value of this root.
HIGHER NUMERICAL EQUATIONS. 491
610. By applying principles already learned, and by the
exercise of due discretion, the student will find little diffi-
culty with the following examples. Skill is acquired only
by practice.
Exercise 146.
Determine the first significant figure of each root in the
following equations :
1. x» — ifc^ — 2ic + l = 0. 6. x^ — 6x^ + Sx + 5 = 0.
2. x' — 5x — 3 = 0. 7. x^ + 9x^ + 24:x + 17 = 0.
3. a;»-5a;2 + 7 = 0. 8. x^-15x^ + 6Sx — 50 = 0.
4. a^—7x + 7=0. 9. x*—Sa^+Ua^+^x—S=d(}.
6. x»+2ic2-30a;+39=0. 10. ic*-12a;2-f 12a; — 3=0.
611. In difficult cases, where the roots are widely differ-
ent in value, Sturm's Theorem is sometimes useful ; and it
is given, without proof, simply for reference :
Divide out the equal roots of the equation, and let the
resulting equation be f(x) = 0. Let fi(x) be the first de-
rivative of f(x). Perform the operations for finding the
H. C F. of fi(x) and f(x)f changing, however, the sign of
each remainder before using it as a divisor. Denote these
successive remainders thus changed ^y f2{^)y fz{^)y y and
let r be the final remainder with its sign changed; this
remainder does not contain x.
f{x),f(x),fi{x), , are called Sturm's Functions.
Let a and b be trial-values of x. Assuming that a = a,
compute in order the value of Sturm's Functions, and note
the number of variations of sign in these values ; likewise,
note the number of variations when a; = ft. The difference
between the number of variations in the first case and the
number in the second equals the number of real roots
between a and ft.
492 ALGEBRA.
Hobneb's Method.
612. Having found one or more figures of the root, we
employ for the remainder of the work Homer's Method,
which was discovered in the early part of this century by
W. G. Horner, of Batli, England. This process will be
explained by the solution of an example.
Take the equation
x»-6a;« + 3ic + 5 = 0. (1)
Suppose that we have found, by one of the methods
already given, that a root of this equation lies between 1
and 2. By the method of § 600, we transform the equa-
tion into one whose roots are less by 1.
-6 +3 +5
+1 -5 -2
-5 -2 +3
+ 1 -4
-4 -6
±1
— 3
The transformed equation is, therefore,
y»-3y2-6y + 3 = 0; (2)
or, 6y = 3-3y2 + y^.
Neglecting for the moment the terms — 3^* + ^^, since
(as will be seen presently) their numerical value is not
large, we find y=0.5. With this assumed value of y, com-
puting the value of — 3^ + 2^> and substituting, we obtain
6y = 2.375;
whence y = 0.4, approximately. We next transform (2)
into an equation whose roots are 0.4 less.
HIGHEB NUMERICAL EQUATIONS. 493
—3
-6
+ 3
+ 0.4
— 1.04
— 2.816
-2.6
-7.04
+ 0.184
+0.4
—0.88
— 2.2
—7.92
+ 0.4
-1.8
The second transformed equation, whose roots are 0.4
less than those of (2), and 1.4 less than those of (1), is,
therefore,
«« — 1.8 «2 — 7.92^ + 0.184 = 0. (3)
Neglecting the first two terms, because « is a small deci-
mal, and «' — l,Sz^ is still smaller, we find « = 0.02 + .
Equation (3) is now to be transformed into an equation
whose roots are 0.02 less.
-1.8
— 7.92
+ 0.184
+ 0.02
-0.0356
-0.159112
-1.78
-7.9556
+ 0.024888
+ 0.02
-0.0362
— 1.76
-7.9908
+ 0.02
-1.74
The third transformed equation, whose roots are 1.42 less
than those of (1), is, therefore,
i;3-1.74t;2-7.99082; + 0.024888 = 0; (1)
whence t; = 0.003 + . As the root of (4) is 1.42 less than
the root of (1), the root of (1) must be 1.42 + 0.003 + , or
1.423 + . The preceding process can be continued until
the root of (1) is found to any required degree of accuracy.
Since the values of y, z, v, etc., are to be added to the
first rough value of x, they are all positive ; and hence the
494 AL6EBBA.
coeflScient of the first power of the unknown quantity in
each transformed equation is unlike in sign to the absolute
term. If, in (4), the signs of the last two terms were alike,
the value of v would be — 0.003 + . This would show that
the value assumed for z was too great, and we should
diminish the value of z and make the last transformation
again. In beginning an example, one is very likely to
assume too large a value for the next figure of the root ; in
solving (2), for instance, the first solution gave y = 0.5,
and, had that value been tried, it would have proved to be
too great.
613. It is not necessary to write out the transformed
equations in full, as in the last section. But when the co-
efficients have been computed, the next figure of the root
may be found by dividing the last coefficient (the absolute
term), with its sign changed, by the coefficient which pre-
cedes it. Thus, instead of writing out (4) § 612, we might
have obtained the value of v by simply dividing — 0.0248 +
by — 7.99 + . Indeed, the following will be f oimd to be a
good practical rule :
Whenever a true significant decimal figure of the root
has been found by trial, and the figure found by dividing
the last coefficient by the coefficient which precedes it
proves to be the same, the subsequent figures of the root
may also be successively found by dividing the last coeffi-
cients of the succeeding transformed equations by the
coefficients which precede them.
If n decimal places are required in the root, the num-
bers in the last column of the following scheme need not
be carried further than the nth decimal place. In the
other columns, a sufficient number of decimals must be
employed to insure the accuracy of the nth place of the
last column.
HIOHBB NUMEBICAL EQUATIONS.
49fi
-6 +3 +6
1.423 +
1 -6
-6 -2
-2
3
1
-4
-4
-6
-2.816
0.184
1
-3
-1.04
-7.04
-0.169112
0.4
-2.6
— 0.88
-7.92
- 0.023988
n nnnonn
0 4
-2.2
0.4
-1.8
- u.uooo
- 7.9656
- 0.0362
- 7.9908
0.02
-1.78
0.02
-1.76
- 0.0062
- 7.9960
- 0.0062
- 8.0012
0.02
-1.74
0.003
- 1.737
0.003
- 1.734
0.003
- 1.731
The coefficients of each transformed equation are readily
discerned by observing the lines which separate the differ-
ent parts of the work.
The next three figures of the root can now be found by
dividing — 0.000900 by — 8.0012, giving 0.000112 + . The
reason for this is shown by consideration of the last trans-
formed equation,
w;« — 1.731 w;» — 8.0012 «; + 0.000900 = 0.
As the value of w is about 0.0001, the value of i^;* —
1.731 i^;^ is only about —0.00000002; therefore we may
assume, without sensible error, that
8.0012 w; = 0.000900;
whence w = 0.000112, as above.
496 AL6EBBA.
614. One of the roots of the equation of § 612 is — 0.6+ ;
and to avoid, as far as possible, the inconvenience of com-
puting with negative numbers, it is customary to transform
the equation into one whose root is +0.6 + . By § 696,
the transformed equation is
aj« + 6x* + 3a; — 6 = 0,
a root of which is 0.66966+ ; hence, a root of the original
equation is — 0.66966 + .
616. Sometimes the coefficient of the first power of the
unknown number in one of the transformed equations is
zero. To find the next figure of the root in this case,
divide the absolute term by the coefficient of the second
power of the unknown number, and extract the square
root of the quotient.
For instance,
2/* -7.6y^ + 3.642/2 + 0^-0.0007829 = 0.
whence, 3.64 ^/^ = 0.0007829, approximately,
and 2^=0.01 + .
616. In rare cases two of the roots are so nearly equal
that Horner's Method, carried out as above, will not find
them both. Take, for example,
a;3 + lla:2-102a: + 181=0,
in which we have found that there are two roots between
3 and 4. Horner's Method gives for the first transformed
equation yi^20i/'-9j/ + l = 0.
Since the coefficient of y^ is large, it is best to neglect y^
only; and the solution gives y = 0,2.
The next transformed equation is
^ + 20.6 z^ — O.SSz + 0.008 = 0,
and — 0.008 -^ - 0.88 = 0.009 + .
HIGHER NUMERICAL EQUATIONS. 497
Perceiving that 20,6 z^, when transposed, will increase
the right-hand member numerically, we conclude to assume
0.01 as the next portion of the root. Proceeding in the
usual manner, we find one value of x to be 3.21312 + . The
first two decimals of the other value of x are found by trial.
If a; = 3.21, f(x) = + 0.001261.
" aj = 3.22, /(aj)=- 0.001352.
" x = 'S.23, /(a;) = + 0.000167.
The change of sign, and the numerical values of f(x),
show that the second value of x is nearly 3.229.
The application of the usual process gives 05 = 3.22952 + .
617. When the root sought is a large number, in finding
the successive figures of its integral portion we do not
divide the absolute term by the coefficient of the first
power of the unknown number, because the neglect of
the higher powers, which are in this case large numbers,
would lead to serious error.
Let it be required to find one root of
re* — 3 a;2 + 11 a: — 4842624131 = 0. (1)
By trial, we find that a root lies between 200 and 300.
Diminishing the roots of (1) by 200, we have
y^ + 800 2^ + 239997 if + 31998811 y - 3242741931 = 0. (2)
If 2/ = 60, f(y) = — 273064071.
" y = 70, /(?/) = + 471570139.
The signs of f(i/) show that a root lies between 60 and
70. Diminishing the roots of (2) by 60, we obtain
;5:*+1040«^+405597.^2_|_ 70302451;$:— 273064071=0. (3)
The root of this is found by trial to lie between 3 and 4.
Diminishing the roots by 3, we may find the remaining
figures of the root of the original, equation, which are
decimal, by the usual process.
498
ALGEBRA.
The scheme of the work is as follows :
0
-3
+ 11
- 4842624131
200 40000 7999400 1599882200 |
200 39997 7999411
- 3242741931 |
200 80000 23999400
2969677860
400 119997
31998811
- 273064071
200
600
i:iuuuu
239997
17496820
494946ai
214585887
200
800
oloUO
291597
70302451
68477522.9
(V\
f^f^onn r
1226178
860 346797
60 58800
71628629
1235592
920
406597 1
72764221
%
60
3129
332682.6
980
60
408726
3138
73096903.6
333356.9
1040
411864
75l/15in9.ftft ft
3 iJ14/
1043
416011
3
842.2
1046
416853.2
3
842.9
1049
416696.1
.
3
843.5
1052
417539.6
0.8
1052.8
0.8
1053.6
661.1
= 0.000009+.
73430260.5
0.8
1064.4
x = 263.800009 +.
0.8
1055.2
618, Any root of a number can be extracted by Horner's
Method. Find the fourth root of 473.
Here x* = 473,
or X* + 0x8 -f 0x2 + Ox - 473 = 0,
and x = 4.66353+.
If the number be a perfect power^ the root will be
obtained exactly.
HIGHBB KUHEBIGAL EQUATIONS. 499
619. From the preceding sections are derived the fol-
lowing general directions for solving a higher numerical
equation whose roots are real :
I. Eemove commensurable roots by § 603.
II. By inspection and Descartes' Eule find roughly the
situations of the incommensurable roots.
III. Find one or more figures of each incommensurable
root by §§ 607-609.
IV. Apply Horner's Method as set forth in §§ 612-618.
620. Compute one root of each of the following equa-
tions, carrying each result to six decimal places. The last
two or three places may be found as in the last part of
§ 613. The root to be sought lies between the two numbers
placed in the parenthesis following each equation, except
in examples 8-12.
Exercise 147.
1. a;« + 10»2 + 6ic — 120 = 0. (2,3.)
2. a? + x^ + X''100 = 0. (4, 5.)
3. x* — 2a^ + 21x--23 = x. (1,2.)
4. x^ — 5a? + Sx^ + S5x-70 = 0. (2,3.)
5. ic*-12a;2-f 12x-3 = 0. (-3,-4.)
6. x' + 2x* + 3x' + 4.a^ + 5x-54.321 = 0. (8,9.)
7. x^-59x^ + 840 = 0. (4, 5.)
8. a;8 — 35499 = 0. 11. aj« — 147008443 = 0.
9. 2^-242970624 = 0. 12. x«~ 551791 = 0.
10. x* — 707281=0. 13. aj«— 17a + 70.3=0. (7,8.)
14. a^ + 9x^ + 2^x + 17 = 0. (-4, -5.)
15. a:*— 8a:« + 14a:* + 4a; — 8 = 0. (0,-1.)
500 ALGEBRA.
621. Many methods have been found of abbreviating
the computation^ when it is desired to get the root to a
large number of decimal places. We give one simple way
which can sometimes be profitably used after three or four
decimal places have been computed.
Consider the example in § 613. If that computation had
not been abbreviated, the last transformed equation would
have been
v^ - 1.731 v^ - 8.0012131^; + 0.000899967 = 0. (1)
We have found that w is between 0.00011 and 0.00012.
Computing the value of v^ — 1.731 w;* with these values of
ti;, and substituting the results in (1), we have
8.001 213 1^ = 0.000899946 *, (2)
and 8.001213 w; = 0.000899942*. (3)
From (2), w;== 0.0001124762*. (4)
From (3), i/; = 0.0001124757*. (5)
Since w is about one-fourth the way from 0.00011 to
0.00012, take a value one-fourth the way from the one in
(4) to that in (5), and
,1; = 0.0001124761*.
If we wish it more accurately still, v? — \!lZ\v^ can be
recomputed with the value of w just found ; and substitu-
tion in (1) will give a value of w extending a few decimal
places further ; and so on.
i. When all but two roots of any equation have been
determined, § 576 enables us to get both these roots at
once.
Take the equation
a;»-6x2 + 3a:-f 5 = 0,
one root of which is 1.423112. Let this root be represented
by a, and the other two roots by h and c.
HIGHER NUMERICAL EQUATIONS. 601
By § 676, a + 6 + c = 6, and a6c = — 6.
.-.6+0 = 4.676888,
&c=- 3.613427,
62 + 2 6c + c2 = 20.94790,
4 6c =-14.05371,
62 - 2 6c + c2 = 36.00161,
6 - c = 6.916216,
26=10.493104,
2 c = — 1.339328,
6=5.246662,
and c = — 0.669664.
The above solution involves little labor when the com-
putation is made with the aid of logarithms.
Two roots of
a* — 16ic« + 79a;2 — 140a; + 58 = 0
are 0.58579 and 8.64575 ; find the others.
Solutions op Special Forms,
recurring equations.
A Becurring Equation, or Beciprocal Equation, is
one in which the coefficients of terms equally distant from
the extremes are numerically equal ; the signs of the cor-
responding terms are either all alike or all unlike.
Below are examples of recurring equations :
x^--7a^+15x^-7x + l=0,
x^ — 3a^ + 5x^ — 5x^ + Sx — l = 0.
The last equation shows that when the degree of the
equation is even, and the signs of the corresponding terms
unlike, the middle term is wanting. By definition, the
middle term of such an equation is both positive and nega-
tive, and therefore must be zero.
502 ALGEBBA.
624. If B, he a root of a recurring equation, - is also a
a
root For, in order to obtain an equation whose roots are
reciprocals of those of the given equation, we simply write
the coefficients in reverse order (§ 598) ; this, however, does
not alter a recurring equation (except to change all its
signs when in one form), and so, if a is a root of it, - is
also a root.
Eecurring equations are called Seciprocal Equations,
because of this property.
625. A recurring equation of an odd degree hxis for one
root — 1 or + 1, according a^ the signs of the corresponding
terms are alike or unlike.
a;« + ^a;«-^ + j5x''-*+ ±Bx^:^Axdtil = 0 (1)
is a typical recurring equation, and may be written in the
form
(a;«±l) + ^(a;«-^±a;) + j5(a;«-*±x^ + =0. (2)
When the signs of the corresponding terms are alike, we
consider the upper signs in (2), since the coefficient a" is
positive. As n is an odd number, the substitution of — 1
for X will satisfy the equation, because the expression in
each parenthesis becomes zero. Likewise, when the signs
of the corresponding terms are unlike, we must consider
the lower signs in (2). The value +1 substituted for x
will then satisfy the equation.
626. A recurring equation of an even degree has -\- 1 and
— 1 05 roots, when the corresponding terms have unlike signs.
Equation (2) of § 625 may be written, when the corre-
sponding terms have unlike signs, as follows :
(x« — 1) + Ax (a:«-2 — 1) + Bx^ (aj""* — 1) + = 0.
HIGHER NUMERICAL EQUATIONS. 503
In this case w is an even number, and every exponent of
X is even. Each parenthesis is therefore divisible by x*— 1,
and the whole equation is divisible by (x + 1) (a; — 1) ;
and consequently + 1 and — 1 are roots.
627. A recurring equation of an even degree is reducible
to an equation of half its degree, when tJie corresponding
terms have like signs.
The first and last terms are positive, but the interme-
diate terms need not be. For convenience, we shall repre-
sent them all as positive, and write the equation
T^^+Ax^-^+Bx^-^^ + ^x2+^a;+l=0. (1)
Dividing (1) by a", and rearranging it, we have
The middle term of (1), since it is of the form Eaf", will
appear as a known number {E) in (2).
Let (« + i) = l..
Then (a5«+jj)=(a! + |)''-2 = y«-2;
^x*-h^) = (a;2+iy-2 = y*-4y2+2; and so on.
In general,
( ^" "I" ^ ) "^ ^" " ^^'^'^
Hence, (2) may be expressed in terms of y, involving no
power higher than the nth, and being, therefore, of half the
degree of (1).
504 ALGEBRA.
628. Any recurring equation is reducible to one of half
its degree.
For the equations of § 625, being divided respectively
by a: + 1 and x — 1, are reduced to the form of § 627 ; the
equation of § 626, being divided by a? — 1, is reduced to
the form of § 627. These reduced equations can now be
treated by the method of § 627.
Solve a;*-llaj* + 17aj' + 17a*-llaj + l=0.
One root is — 1, and dividing the equation by x + 1, we have
X*- 12x8 + 29x2- 12x+ 1 = 0.
Divide by x^, rearrange the terms, and add 2 to both sides.
Then (x + ^ V - 12 (x + i) + 29 = 2 ;
whence Tx + - j = 9 or 3,
and X = 1 (9 ± V77) or 1(3 ± \fb),
1 (9 + V77) is the reciprocal of 1 (9 - V77), and 1 (3 + VH) is the
reciprocal of J (3 — Vs), the product in each case being unity.
Exercise 148.
Solve :
1. a:* + 7a^ — 7aj — 1 = 0.
2. aj* + a:3 + a;2 + a; + l=0.
3. aj«-3a:* + 5aj* — 5aj« + 3a:-l=0.
4. a;* — 5x« + 6a;* — 5a; + l=0.
5. 2aj* — 5a;' + 6a;* — 5x + 2 = 0.
6. a«-4a;* + x» + aj«-'4a: + l = 0.
7. ic*-10x« + 26a* — 10aj + l=0.
8. ic' + ^'wc' + ma; + l = 0.
9. a:« + l = 0.
10. So* — 2a;* + 5aj» — 5a;« + 2a: — 3 = 0.
HIGHER NUMERICAL EQUATIONS. 505
Exponential Equations.
629. An Exponential Equation is one in which an un-
known number appears as an exponent.
I. Solve a* = m. Taking the logarithms of both mem-
bers, we have x log a=log m; whence a; = 7-^ — •
log a
II. Solve x^ = m. As before, x log x = log m. Find by
trial, from a logarithm table, a value of x which satisfies
the equation. Thus, in the equation a:* = 7 :
By trial, 2.3 log 2.3 = 0.88197 ;
2.32 log 2.32 =0.84793;
2.316 log 2.316 = 0.84474. Hence, x = 2.316.
Exercise 149.
Solve :
1. 11^ = 346. 6. 4^=3.74. 8. 14.74^ = 8.64.
2. 3' = 10. 6. 146^ = 12984. 9. ar^ = 2.767.
3. 10^=745. .7. 0.2^ = 0.4. 10. aj' = 23.10.
4. 7^=324.
In Compound Interest, if P denotes the principal, A the
amount, r the rate, t the integral number of years (or other
periods of time), we have the formula,
^ = P(l+r)';
vhence (1 + »•)'= p*
and ^^log^-^l^gP
log (1 + r)
Find t in the following cases :
11. P = 760, ^ = 1797.42, r = 6%.
12. P = 780, ^ = 1559.22, r = 8%.
13. P = 5630.75, ^ = 21789.22, r = 7%.
14. P=300, ^ = 515.46, r = 7%.
15. P= 84.66, ^ = 289.47, r = 7i%.
606 ALGEBBA.
Cardan's Method for Cubic Equations.
630. The general form of a cubic equation is
oi^-^ma^ + nx + c = 0.
0)
Let iB=y--.
(2)
By { 601, equation (1) reduces to the form
lf' + PV+q = 0.
(3)
Let y = '-iz-
W
Sabstitate this value of y in (3) and reduce :
whence z«= -lg± -^^+ ^»
and ^=^^:^q±^J^+t,
Sz 3
(6)
From (2) and (4), x=z-^-^- (6)
When z has been determined from (5), x can be found
from (6). In solving these equations, it is best to get
from (5) the value of z most easily obtained. Then find
the value of x from (6), and designate this value by r ;
then divide (1) by a — - r, and solve the resulting quadratic.
Or the method of § 622 may be employed. If the original
equation does not contain the square of the unknown num-
ber, it will be in the form of (3), and the value of y can
be found from (5) and (4).
Solve a^-6a^~12a;+ 112 = 0.
By(2), x=y-^ = y + 2.
Substitution gives y* — 24 y + 72 = 0.
Here i) = — 24, and q = 72.
HIGHER NUMERICAL EQUATIONS. 507
^ /,cx JI QAO-^r 13824 .6184
.-. from (5), 2= \- 36 ± \- -^ + -j-
= -^-36 ±28 = - 2.
From (6), x = — 6.
Diyide the original equation by x + 4 ; the resulting equation is
x2-10x + 28 = 0;
whence x = 6 ± V— 3.
631. If p is negative, and p and q have such values
that ^ + 4; IS negative, equation (5) shows that Cardan's
Method is unsatisfactory, since z is the cube root of an
expression which is partly imaginary, and we have no
general rule for extracting such a root. This case, which
is called the irreducible case, can be solved by Trigonometry,
and may be illustrated by an example.
Solve 2/* — 15y — '* = ^-
Substitution in (6) gives
5 = \2±Y
- 3376
27 "'■^^ V2±llV=^.
It is possible to determine, by trial, that
(2±V^l)8 = 2±llV=l;
.•.z = 2±V^,
and, from (4), y = 4.
Divide the given equation by y — 4, and solve the resulting quad-
ratic; then
y=-2±V3.
gQlye . Exercise 150.
1. a;'+12aj« + 46a; + 50 = 0. 5. y« + 48y + 504 = 0.
2. x» — 21a:^+159a; — 490 = 0. 6. 2^-213/ — 344 = 0.
3. a;»-6a;« + 13x-10 = 0. 7. y» — 3y + 2 = 0.
4. a;» + 3x» + 9x — 13 = 0. 8. 2/^ — 60^ + 671 = 0.
608 ALGEBRA.
Trigonometric Solution of Cubic Equations.
632. Trigonometric formulas are used in solving quad-
ratic and cubic equations ; and we subjoin those for cubics,
so that the student may use them when he has acquired a
knowledge of Trigonometry.
In the formulas, p is essentially positive, but ^ may be
either positive or negative. We distinguish three cases :
I. a?'+2WJ + g' = 0;
11. a:»^^ + ? = 0, and^<|';
III. x»-i>x + y = 0, and ^>|'.
The formulas for Case I. are as follows :
X8= •%/^COt0+ C8C0>^V^-1.
Xs = a/? cot 0 — CSC 0 '^ V— 1.
The formulas for Case II. are as follows :
Xi=-2-^|csc0.
X2 = -l/^ CSC 0 + cot^'v^V— 1.
xs = -1/^ CSC 0 — cot 0 Vp V^.
HIGHER NUMERICAL EQUATIONS.
The formulas for Case III. are as follows :
Bine-
609
2p\p
Xi=2'J^Bmie.
X2 = 2^|sin(60°-i^).
XB=-2^|
8in(60« + i^).
If sin 6 or tan 6 is positive, 6 is in the first quadrant ; if
it is negative, 6 is in the fourth quadrant.
In the preceding formulas, x^ X2, x^ denote the three
roots in each case.
633. Take the difficult equation
which comes under Case III.
log 403 = 2.60631
^ = 68°32'0''
log 441 = 2.64444
i^= 22*^60' 40"
log 46=1.66276
log8lnitf= 9.68909 -10
log 147 = 2.16732
log 2 = 0.30103
log2-^| = 0.04291
log p = 9.96087 -
-10
logsin (60*^ - i^) = 9.78102 ♦- 10
log 3 = 0.47712
log q = 9.49544 -
log 3g= 9.97266-
colog 2i)= 9.73810-
-10
■10
-10
log sin (60° + ie) = 9.99660 - 10
logaJi= 9.63200 -10
loga:2= 9.82393 -10
log (-X8) = 0.03951
log-J-= 0.26812
xi = 0.42855
X2 = 0.66670
log8in^= 9.96878 -
•10
Xs = - 1.09626
510 ALGEBRA.
This is an unfavorable example, because sin 6 is large,
and 6 is therefore inaccurately determined ; yet the values
of the roots are near the true values, which are 0.42857,
0.66667, and —1.09524. If log Jl, which was 0.258125,
had been assumed to be 0.25813, the answers would have
been 0.42857, 0.66668, and —1.09525.
Exercise 151.
Solve :
1. x« + 3a; — 5 = 0. 3. a^ — 7x + ll = 0.
2. a^+7x + 3 = 0, 4. a;» — 4a; — 5 = 0.
5. a* — 5a; + 4 = 0.
Wddi
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