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IN MEMORIAM
FLOR1AN CAJORI
Ia&w
ELEMENTARY ALGEBRA
BY
FREDERICK H. SOMERVILLE, B.S.
THE WILLIAM PENN CHARTER SCHOOL, PHILADELPHIA
NEW YORK .^CINCINNATI-:. CHICAGO
AMERICAN BOOK COMPANY
Copyright, 1908, by
FREDERICK H. SOMERVILLE.
Entered at Stationers' Hall, London.
el. alo.
W. P. I
\JMf
PREFACE
The plan of the early pages of this text is to offer a gradual
introduction to the subject without plunging the young student
too deeply into new difficulties. The treatment of negative
numbers as a natural extension of the familiar arithmetical
numbers, and a generous amount of detail in the explanations
of the early chapters, serve to clarify the beginnings of a sub-
ject so often troublesome. New processes are accomplished by
the use of the simplest of symbols, and the undivided atten-
tion of the student is centered upon the new elements of the
operation at hand. Definitions are introduced only as rapidly
as new processes call for them, and the young mind is not
confused and discouraged at the outset by the attempt to learn
the meaning of a bewildering mass of strange terms.
In arrangement the book does not differ widely from the
general scheme of the standard texts, but in some details there
are changes that have been found to be of genuine value in the
classroom. For example, in Factoring, the simple and the
difficult types are separated ; an elementary course gives a
treatment free from complex forms, while a supplementary
section provides for the preparation of college requirements.
The Lowest Common Multiple and its immediate application
to Addition of Fractions form a single chapter, and the plan
suggested by this order has proved to be natural, practical, and
sound. The classification of Simple Simultaneous Equations
and of the Theory of Exponents is both new and teachable,
and the logical arrangement of Affected Quadratic Equations
provides a chapter that, while omitting no essential, gives a
brief and clear general treatment of this important subject.
Throughout the early chapters exercises for oral drill are
frequent, and their introduction confines the simplest types
3
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4 PREFACE
of the problems to the oral discussion of the classroom. Such
discussions are of great value in a live class, and if supple-
mented by a practice of reading by members of the class each
step of every new illustrative solution, a most gratifying
progress results. The written exercises consist of new prob-
lems carefully graded, and the frequent reviews are constructed
on the lines of recent entrance questions of the leading col-
leges and universities. The treatment of Graphs is full and
complete, but is free from those elements of the advanced
discussions that so frequently confuse the young student. A
comprehensive introduction of the common Physical Formulas
familiarizes the student with a practical branch of applied
algebra; and this is accomplished without assuming that the
teacher is an expert in the laboratory practice of that science.
In those subjects where several methods of procedure are
possible, the text offers, as a rule, but one. To select arbi-
trarily and to suggest one method as the best of many is a
matter of personal choice and opinion, and the only claim for
the single methods chosen in the following chapters is that
they are uniformly practical. The text is planned on the
theory that one practical method thoroughly mastered is
sufficient for the needs of the young student, and that the
elementary classroom has neither time nor need for those
comparative discussions that interest only the mature mind.
The author gratefully acknowledges his indebtedness to
those friends whose suggestions and encouragement have
been of material aid in the preparation of this text.
FREDERICK H. SOMERVILLE.
The William Penn Charter School,
Philadelphia .
CONTENTS
CHAPTER PAGE
I. Introduction, Symbols, Negative Numbers ... 9
Symbols of Operation 11
Symbols of Quality 13
Negative Numbers 14
Algebraic Expressions 20
II. Addition, Parentheses 22
Addition of Monomials 22
Addition of Polynomials 24
Parentheses 26
III. Subtraction, Review 32
Subtraction of Polynomials . . .... 34
General Review 35
IV. Multiplication . . . . 38
Multiplication of a Monomial by a Monomial ... 42
Multiplication of a Polynomial by a Monomial . . 43
Multiplication of a Polynomial by a Polynomial . . 45
Multiplication of Miscellaneous Types .... 47
V. Division, Review 52
Division of a Monomial by a Monomial .... 54
Division of a Polynomial by a Monomial .... 55
Division of a Polynomial by a Polynomial . . . ,56
General Review^ 62
VI. The Linear Equation, The Problem . . .64
General Solution of the Linear Equation .... 68
The Solution of Problems 70
VII. Substitution 84
The Use of Formulas 87
VIII. Special Cases in Multiplication and Division ... 89
Multiplication 89
Division 95
5
CONTENTS
IX. Factoring
Expressions having the Same Monomial Factor
Trinomial Expressions
Binomial Expressions
Expressions of Four or More Terms factored by Grouping
Supplementary Factoring
X. Highest Common Factor
Highest Common Factor of Monomials .
Highest Common Factor of Polynomials by Factoring
XI. Fractions, Transformations .
The Signs of a Fraction .
Transformations of Fractions
XII. Fractions (Continued), Lowest Common Multiple, Lowest
Common Denominator, Addition of Fractions
Lowest Common Multiple
Lowest Common Multiple of Monomials .
Lowest Common Multiple of Polynomials by Factoring
Lowest Common Denominator . .
Addition and Subtraction of Fractions .
XIII. Fractions (Continued), Multiplication, Division, The Com
plex Form
Multiplication of Fractions .....
Division of Fractions
The Complex Form
XIV. Fractional and Literal Linear Equations, Problems
Special Forms . . . . * .
Problems leading to Fractional Linear Equations
XV. Applications of General Symbols, Review
The General Statement of a Problem
Use of Physical Formulas
Transformation of Formulas .
General Review
XVI. Simultaneous Linear Equations, Problems
Elimination by Substitution
CONTENTS
CHAPTER PAGE
Elimination by Comparison 176
Elimination by Addition or Subtraction .... 176
Systems involving Three or More Unknown Quantities . 178
Fractional Forms 181
Literal Forms . 185
Problems producing Simultaneous Linear Equations . 187
The Discussion of a Problem 193
XVII. Graphical Representation of Linear Equations . . .197
Graph of a Point 197
Graph of a Linear Equation in Two Unknown Numbers . 200
Graphs of Simultaneous Linear Equations . . . 202
XVIII. Involution and Evolution 206
Involution 205
Evolution 209
Square Root of Polynomials 213
Square Root of Arithmetical Numbers . . . .217
XIX. Theory of Exponents 221
The Zero Exponent 221
The Negative Exponent . . . . . . 222
The Fractional Form of the Exponent .... 223
Applications 227
XX. Radicals, Imaginary Numbers, Review .... 236
Transformation of Radicals 237
Operations with Radicals 242
Equations involving Radical Expressions . • . . . 252
Imaginary and Complex Numbers 254
Operations with Imaginary Numbers .... 255
General Review . . 261
XXI. Quadratic Equations 266
Pure Quadratic Equations 266
Affected Quadratic Equations 268
Discussion of Affected Quadratic Equations . . .277
Graphs of Affected Quadratic Equations .... 282
XXII. The Quadratic Form,. Higher Equations, Irrational Equa-
tions 286
The Quadratic Form ' . .286
CONTENTS
CHAPTER PAGE
Higher Equations solved by Quadratic Methods . . 286
Irrational Equations 292
Physical Formulas involving Quadratic Equations . 293
XXIII. Simultaneous Quadratic Equations, Problems . 294
Solution by Substitution 294
Solution by Comparison and Factoring . . . 295
Solution of Symmetrical Types . . . . 296
Solution of Miscellaneous Types 298
Graphs of Quadratic Equations in Two Variables . 302
Problems producing Quadratic Equations . . . 310
XXIV. Ratio, Proportion, Variation 315
Katio 315
Proportion . 317
Variation 330
XXV. The Progressions 338
Arithmetical Progression 338
Geometrical Progression 348
XXVI. The Binomial Theorem — Positive Integral Exponent . 358
Proof of the Binomial Formula 359
Applications 361
XXVII. Logarithms 364
The Parts of a Logarithm 365
The Use of the Four-place Table 367
The Properties of Logarithms 374
The Cologarithm 376
Use of Logarithms in Computations .... 377
Miscellaneous Applications of Logarithms . . . 379
XXVIII. Supplementary Topics, Review . . . . .385
The Remainder Theorem 385
The Factor Theorem 385
The Theory of Divisors of Binomials . . . . 387
H. C. F. of Expressions not readily Factorable . . 389
L. C. M. of Expressions not readily Factorable . . 392
Cube Root of Polynomials 394
Cube Root of Arithmetical Numbers .... 396
General Review 398
INDEX . 405
ELEMENTARY ALGEBRA
CHAPTER I
INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
1. The definite number symbols of arithmetic, 1, 2, 3, 4, 5, etc.,
are symbols that express in each case a number with one
definite value.
Thus, 3 units, 5 units, 7 units, etc., represent particular groups, the
symbols, 3, 5, and 7 having each a particular name and value in the
number system that we have learned to use.
2. The general number symbols of algebra are symbols that
may represent not one alone but many values, and for these
general symbols the letters of the alphabet are in common use.
THE ADVANTAGE OF THE GENERAL NUMBER SYMBOL
3. Many of the familiar principles of arithmetic may be
stated much more briefly, and usually with greater clearness,
if general or literal number symbols are employed. To
illustrate :
(a) The area of a rectangle is equal to the product of its height, or
altitude, by its length, or base. Or, arith-
metically,
Area = altitude x base.
** Using only the first letters of each word,
we may write,
Area = a x 6,
b
and this latter expression, while equally clear
in meaning, serves as a general expression for obtaining the area of any
rectangle with any values of a and b.
9
10 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
(6) The familiar problem of simple interest gives arithmetically,
Interest = principal x rate x time. .
A much simpler form of expressing the same principle is obtained
here as above, using only the first letters of each element for the general
expression.
Thus : I = p x r x t.
(c) Two principles relating to the operation of division in arithmetic
may be clearly and briefly expressed by the use of literal symbols.
Thus : Dividend -f- divisor = quotient. That is, — = q.
d
divisor x quotient = Dividend. Or, d x q = D.
4. In a more extended manner the literal symbol permits
a breadth and power of expression not hitherto possible with
the number symbols of arithmetic. Many problems involving
unknown quantities are readily stated and solved by means of
general symbols, and clearness of expression in such problems
is invariably gained by their use. To illustrate a common use
of literal symbols, consider the following problem :
Two brothers, John and William, possess together 200
books, and John has 20 books more than William. Write
expressions that clearly state these conditions. (Compare
carefully the two methods of expression.)
The Arithmetical Expression
(a) The number of John's books + the number of William's books =200
(6) The number of John's books — the number of William's books = 20
The Algebraic Expression
Let us assume that x = the number of books that John has,
and that y = the number of books that William has.
Then from the conditions given in the problem :
(a) x + y = 200.
(6) x-y = 20.
THE SYMBOLS OF OPERATION 11
By a simple process the values of x and y are readily determined.
From this parallel between arithmetical and algebraic forms of ex-
pression the brevity and the advantage of the literal or general symbol,
for number is clearly manifest. The later processes of algebra will con-
stantly furnish the means wherewith we may broaden our power of
expression, and the meaning of algebra will be interpreted as merely
an extension of our processes with number.
THE SYMBOLS OF OPERATION
5. The principal signs for operations in algebra are identical
with those of the corresponding operations in arithmetic.
6. Addition is indicated by the " pins " sign, + .
Thus, a + b is the indicated sum of the quantity a and the quantity b.
The expression is read " a plus 6."
7. Subtraction is indicated by the " minus " sign, — .
Thus, a — b is the indicated difference between the quantity b and the
quantity a. The expression is read " a minus 6."
8. Multiplication is usually indicated by an absence of sign
between the quantities to be multiplied.
Thus, ab is the indicated product of the quantities a and b.
abx is the indicated product of the quantities a, 6, and x.
Sometimes a dot is used to indicate a multiplication.
Thus, a • b is the product of a and b.
The ordinary symbol, " x ," is occasionally used in algebraic
expression.
An indicated product may be read by the use of the word
" times " or by reading the literal symbols only.
Thus, ab may be read "a times 6," or simply " a&."
9. Division is indicated by the sign " -j-," or by writing in
the fractional form.
12 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
Thus, a -*- b is the indicated quotient of the quantity a divided by the
quantity b.
- is the fractional form for the same indicated quotient.
b
Both forms are read u a divided by 6."
10. Indicated operations are of constant occurrence in alge-
braic processes, for the literal symbols do not permit the
combining of two or more into a single symbol as in the case
of numerals. Thus:
Arithmetically, 5 + 3 + 7 may be written " 15," for the symbol 15 is
the symbol for the group made up of the three groups, 5, 3, and 7.
Algebraically, a + b + c cannot be rewritten unless particular values
are assigned to the symbols a, &, and c. The sum is an indicated result.
Algebraic expression, therefore, confines us to a constant use
of indicated operations, and we must clearly understand the
meaning of :
I. An Indicated Addition, a +- b.
II. An Indicated Subtraction, a — b.
III. An Indicated Multiplication, ab.
IV. An Indicated Division, -•
b
11. Equality of quantities or expressions in algebra is indi-
cated by the sign of equality, =, read "equals," or "is equal to."
Thus, a + 6 = c + d
is an indicated equality between two quantities, a + b and c + d.
12. If two or more numbers are multiplied together, each of
them, or the product of two or more of them, is a factor of the
product ; and any factor of a product may be considered the
coefficient of the product of the other factors. Thus :
In 5 a, 5 is the coefficient of a. In acmx, a is the coefficient of cmx,
In ax, a is the coefficient of x, or ac may be the coefficient of
or x is the coefficient of a. mx, etc.
THE SYMBOLS OF QUALITY 13
Coefficients are the direct results of additions, for
5 a is merely an abbreviation of a + a + a + a + a.
4 xy is an abbreviation of xy + xy + xy + xy.
If the coefficient of a quantity is " unity " or " 1," it is not
usually written or read.
Thus, a is the same as 1 a. xy is the same as 1 xy.
13. The parenthesis is used to indicate that two or more
quantities are to be treated as a single quantity. The ordinary
form, ( ), is most common. For clearness in the discussion of
elementary principles, the parenthesis will frequently be made
use of to inclose single quantities.
14. An axiom is a statement of a truth so simple as to be
accepted without proof. Two of the axioms necessary in
early discussions are :
Axiom 1. If equals are added to equals, the sums are equal.
Axiom 2. If equals are subtracted from equals, the re-
mainders are equal.
THE SYMBOLS OF QUALITY
15. In scientific and in many everyday discussions greater
clearness and convenience have resulted from a definite method
of indicating opposition of quality. For example :
Temperature above and below the zero point,
Latitude north and south of the equator,
Assets, or possessions, and liabilities, or debts, in business, etc.,
represent cases in which direct opposites of quality or kind
are under discussion; hence, a need exists for a form of ex-
pression that shall indicate kind as well as amount, quality as
well as quantity.
To supply this need the plus and minus signs are in general
use, their direct opposition making them useful as signs of
14 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
quality as well as of operation; and we will now consider
their use as
THE + AND - SIGNS OP QUALITY
16. A few selected examples of common occurrence clearly
illustrate the application of quality signs to opposites in kind.
In the laboratory : %
Temperature above 0° is considered as +.
Temperature below 0° is considered as — .
In navigation :
Latitude north of the equator is considered as +.
Latitude south of the equator is considered as — .
In business administration :
Assets, or possessions, are considered as + .
Liabilities, or debts, are considered as — .
The following illustration emphasizes the advantage of in-
dicating opposites in kind by the use of the -J- and — signs
of quality.
A thermometer registers 10° above at 8 a.m.,
15° above at 11 a.m., 5° below at 4 p.m., and ' •■ ' , M
1am -4- 1 o
10° below at 10 p.m. At the right we have tabu-
lated the conditions in a concise form made possible .,_ , _
i ^ u .1. £ ,. 4 • 10 p.m. — 10°
only through the use of quality signs.
By applying this idea of opposition to arithmetical numbers
we may establish
NEGATIVE NUMBERS
17. It is first necessary to show that there exists a need for
a definite method of indicating opposition of quality in number.
Consider the subtractions :
(1) 5-4= (2) 5-5= (3) 5-6 =
There are three definite cases included.
In (1) a subtrahend less than the minuend.
In (2) a subtrahend equal to the minuend.
In (3) a subtrahend greater than the minuend.
NEGATIVE NUMBERS 15
The first two cases are familiar in arithmetical processes, but
the third raises a new question, and we ask,
5 — 6 = what number ?
18. On any convenient straight line denote the middle
point by " 0," and mark off equal points of division both to
the right and to the left of 0.
The two directions from are dearly defined cases of opposition,
and this / opposition may be indicated by marking the suc-
cessive division points at the right of with numerals having
plus signs,
-5 -4 -3 -2 -1 O +1 +2 +3 +4 +5
1
and, similarly, the division points left of with numerals hav-
ing minus signs.
The result is a series of positive and negative numbers
established from a given point which we may call " zero."
With this extended number system we may at once obtain a
clear and logical answer for the question raised above. The
minuend remaining the same in each case, the result for each
subtraction is established by merely counting off the subtrahend
from the minuend, the direction of counting being toward zero.
Therefore,
For (1) The subtraction of a positive number from a greater
positive number gives a positive result.
Or, 5-4 = 1.
For (2) The subtraction of a positive number from an equal
positive number gives a zero result.
Or, 5-5 = 0.
For (3) The subtraction of a positive number from a less posi-
tive number gives a negative result.
Or, 5 - 6 = - 1.
16 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
We may conclude, therefore, that :
19. The need of a negative number system is that subtraction
may be always possible. On the principle of opposition the idea
of negative number is as firmly established as that of positive
number, and the definition of algebra as an extension of number
is even further warranted.
20. The extension of arithmetical number to include nega-
tive as well as positive number establishes algebraic number.
21. It is often necessary to refer to the magnitude of a num-
ber regardless of its quality, the number of units in the group
being the only consideration.
Thus, -f 5 and — 5 have the same magnitude, or absolute value, for
each stands for the same group idea in the numeral classification. They
differ in their qualities, however, being exact opposites. In like manner,
-f a and — a, + xy and — xy, are of the same absolute value in each
respective case, no matter what values are represented by a or by xy.
POSITIVE AND NEGATIVE NUMBERS COMBINED
The simplest combination of algebraic numbers is addition,
and since two groups of opposite kinds result from the positive
and negative qualities, we must consider an elementary dis-
cussion of algebraic addition under three heads.
I. Positive Units + Positive Units.
If a rise in temperature is 16° -f 15°
and a further rise of 10° occurs, 4- 10°
we have a final reading of . . . . . + 25°
II. Negative Units + Negative Units
If a fall in temperature is 12° — 12°
and a further fall of 16° occurs, — 16°
we have a final reading of — 28 c
POSITIVE AND NEGATIVE NUMBERS COMBINED 17
III. Positive Units + Negative Units.
If a rise in temperature is 20° and -f 20°
an immediate fall of 15° occurs, — 15°
we have finally + 5°
If a fall in temperature is 30° and _ 30°
an immediate rise of 24° occurs, + 24°
we have finally — 6°
The student should verify these illustrations by making a
sketch of a thermometer and applying each case given above.
In general, addition of algebraic numbers results as follows :
(1) If a and b are positive numbers :
(+«) + (+&) = +« + &. i . . .
Numerical Illustration : V * 5 , p " f " 3 J
(+5) + (+3) = + 5 + 3=+8.
(2) If a and b are negative numbers :
(-«) + (- 6) = -a- b , ^ , , , 1 , , O
Numerical Illustration : V^ 3 Q 5 J
(_5) + (-3) = -5-3= -8.
(3) If a is positive and b negative :
(+«) + (-&)= +a-b. +5. . .
Numerical Illustration : ^-+2
(+5) + (-3)=+5-3=+2.
(4) If a is negative and b positive :
(-a) + (+6) = -a + 6. -5 Q
Numerical Illustration: r +3 ' — 2— '"
(_5) + (+3)=-5 + 3 = -2.
From the four cases we may state the general principles for
combining by addition any given groups of positive quantities,
negative quantities, or positive and negative quantities :
SOM. EL. ALG. — 2
18 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
22. The sum of two groups of plus, or positive, units is a posi-
tive quantity.
23. TJie sum of two groups of minus, or negative, units is a
negative quantity.
24. The sum of two groups of units of opposite quality is posi-
tive if the number of units in the positive group is the greater, but
negative if the number of units in the negative group is the
greater. *
JTrom Art. 24 we derive the following important principle :
25. The sum of two units of the same absolute value but oppo-
site in sign is 0.
In general: (+«) + (-«) = + a- a = 0.
Numerical Illustration : (+5) + (-5) = +5-5 = 0.
Exercise 1
1. If distances to the right are to be considered as posi-
tive in a discussion, what shall we consider distances to the
left?
2. If the year 20 a.d. is considered as the year " -f 20,"
how shall we express the year 20 b.c. ?
3. Draw a sketch of a thermometer, and indicate upon it
the temperature points + 35°, — 18°, + 12°, and — 12°.
4. On your sketch determine the number of degrees passed
through if the temperature rises from 8° to 35° ; * from to
15°; from - 15° to 10°.
5. On your sketch determine the number of degrees passed
through in a fall of temperature from 35° to 10° ; from 20° to
-10°.
* Note that the quality of a unit is considered plus when no sign of
quality is given.
POSITIVE AND NEGATIVE NUMBERS COMBINED 19
6. Show by your sketch that a rise of 25° followed by a
fall of 15° results in an actual change of 10° from the starting
point.
7. In example 6 is the conditi6n true for the single case
when you begin at 0°, or is the result the same no matter at
what point you begin ? Illustrate.
8. Show that a fall of 18° from the zero point succeeded by
a rise of 30° results in a final reading of + 12°.
9. Show that a rise of 10° from the zero point succeeded by
a further rise of 18°, and later by a fall of 40°, results in a
final reading of 12° below zero. Express " 12° below zero " in
a simpler form.
10. From your sketch determine the final reading when,
after a rise of 40° from 0, there occurs a fall of 15°, a succeed-
ing rise of 7°, and another fall of 35°. Express these changes
with proper signs.
11. C is a point on the line AC B
AB. A traveler starts at C, goes +10 ^
10 miles toward B, turns back 7 < ~+9
miles toward A, and returns 9
miles toward B. Determine his final distance from G, and
also his position at either the right or the left of C. (Assume
distances to the right of C as +.) Would the information
given be sufficient to determine the result without using the
sketch ?
12. Determine the result of a journey 8 miles from C toward
B, returning 6 miles toward A. Give the total distance trav-
eled and the final position.
13. Determine the result of a journey 9 miles from C
toward B, 16 miles back toward A, and then 8 miles toward B.
Make a drawing similar to that illustrating example 11, and
prove your answer by reference to the drawing.
20 INTRODUCTION. SYMBOLS. NEGATIVE NUMBERS
14. Determine the result of a journey 17 miles toward B,
10 miles back toward A, 3 miles more toward A, back 19 miles
toward B, the starting point being at G. Make a drawing
illustrating the entire journey.
15. With the same distances and directions as in problem 14,
determine the result if the starting point had been at A, giv-
ing for the answer the final distance of the traveler from G as
well as from A. (Assume that from G to A is 30 miles.)
ALGEBRAIC EXPRESSIONS
26. An algebraic expression is an algebraic symbol, or group
of algebraic symbols, representing some quantity. An ex-
pression is numerical when made up wholly of numerical
symbols, and is literal when made up wholly or in part of
literal symbols.
20 + 10 — 13 x 2 is a numerical expression.
ab + mn — xy and 5 a — 6 ex — 12 are literal expressions.
27. The parts of an expression connected by the + or -
signs are the terms.
In the expression
ab + ( C - d) _ mn + « + » _ g .4 - « » -P
b -y 2(d + k)
the terms are
ab, + (c — d), — mn, + a ~ x , and
b - y 2(d + k)
A parenthesis, or a sign of the same significance, may inclose a group
as one term. A fraction as an indicated quotient is also a single term.
The sign between two terms is the sign of the term following.
The sign of the first term of an expression is not usually written if it
is +. In the term ab of the given expression both the sign, +, and the
coefficient, 1, are understood.
ALGEBRAIC EXPRESSIONS 21
28. Terms not differing excepting as to their numerical
coefficients are like or similar terms.
5 ax and — Sax are similar terms. 3 ab and 14 mn are dissimilar
terms.
29. The common expressions of algebra are frequently-
named in accordance with the number of terms composing
them. The following names are generally used :
A Monomial. An algebraic expression of one term.
4 a, 5 win, — Sxy, and 17 xyz are monomials.
A Binomial. An algebraic expression of two terms.
a + b, 3 m — x, 10 — 7 ay, — 4 mnx + 11 z are binomials.
A Trinomial. An algebraic expression of three terms.
Sx — 7 m + 8, 4 ab — 11 ac — 10 mny are trinomials.
A Polynomial. Any expression having two or more terms.
While the binomial and the trinomial both come under this
head they occur so frequently that common practice gives each
a distinct name. Expressions having four or more terms are
ordinarily named polynomials.
Oral Drill
Kead the following algebraic expressions :
1. a + 3x — 4:mn-\-cdn — 3xy.
2. 2 mx — 3 acd -f- (a -f- x) — (m -f- w).
3. (a — x) — (c — d) + (m — y + z).
4. 5ayz — 2(2 m — n) -f- a(a — x) -f 3 a(2 a — 3y).
5. — mnx + 3 a(c — 2 d + 1) — (a — m + n)sc -f- 12(4 — sc).
6. (a - X )( c + y)-(x + 2)0/ - 3) - ■(«+ l)(a> + 2)(* + 3).
& 2/ 3(c-d) "c-2/
CHAPTER II
ADDITION. PARENTHESES
30. Addition in algebra, as in arithmetic, is the process of
combining two or more expressions into an equivalent ex-
pression or sum. The given expressions to be added are the
addends.
THE NUMBER PRINCIPLES OF ADDITION
31. The Law of Order. Algebraic numbers may be added in
any order.
In general : ' a + 6 = b + a.
Numerical Illustration : 5 + 3 = 3 + 5.
32. The Law of Grouping. The sum of three or more alge-,
braic numbers is the same in whatever manner the numbers are
grouped.
In general: a+b+c=a+(b+c) =(a+6) + c=(a+c) + 6.
Numerical Illustration: 2+3 +4 =2 +(3 +4) = (2 +3) +4= (2 +4) +3.
A rigid proof of these laws is not necessary at this point ;
but may be reserved for later work in elementary algebra.
The law of order is frequently called the commutative law,
and the law of grouping is called the associative law.
ADDITION OF MONOMIALS
The principles underlying the addition of the simplest forms
of algebraic expressions have already been developed, and they
are readily applied in the more difficult forms of later work.
22
ADDITION OF MONOMIALS 23
(1) The sum of like quantities having the same sign, all -f or
all - .
By Articles 22 and 23 :
+ 7
+ 12a
-7
-12a
+ 3
+ 5a
-3
— 5a
+ 10
+ 17 a
-10
-17 a
In general :
33. The coefficient of the sum of similar terms having like
sigiis is the sum of the coefficients of the given terms with the
common sign.
(2) The sum of like quantities having different signs, some +
and some — .
By Article 24 : +7 + 12 a - 7 - 12 a
-3 - 5q + 3 + 5q
+ 4 + la -4 - la
In general :
34. The coefficient of the sum of similar terms having unlike
signs is the arithmetical difference between the sum of the +- co-
efficients and the sum of the — coefficients, with the sign of the
greater.
Oral Drill
Add
orally :
1.
5a
Sa
2.
6a
9a
3.
4a;
7x
4.
9mn
Smn
5.
Sbcd
11 bed
6.
7 amx
19 amx
7.
17 cmy
19 cmy
8.
-4a6
-3a6
9.
— 8 en
—5 en
10.
— Sxz
—xz
11.
— 15cz
-11 cz
12.
—Sexy
— 14: cxy
13.
-12 abx
—19 abx
14.
— 5 dny
— dny
15.
10 ac
—6ac
16.
— 19 mx
9mx
17.
-Scy
15 cy
18.
12 ax
— Sax
19.
— 15 cz
15 cz
20.
27 axy
— 16 axy
21.
— Icdmn
45 cdmn
24 ADDITION. PARENTHESES
If the sum of three or more terms is required, we apply the
law of grouping (Art. 32), and separately add the + and
the — terms.
Thus, 5-8 + 11 -16 + 3 = 5+ 11 +3-8-16
= 19-24
= — 5. Result.
Let the student apply this principle in the following :
Oral Drill
Add orally :
1. 2-7 + 6-9. 7. — 8x + 9x-6x + 5x.
2. -8 + 3 — 12 + 7. 8. -llac + 14ac-ac-2ac.
3. —9 + 8 — 15 — 11. 9. 3 mnx — 8 mnx — 9 mnx + 13 mnx.
4. 7-14 — 3 + 10. 10. -1 cy-§cy + 3ey-llcy.
5. 13-18 + 7-21. 11. 3a-5a + 8a-lla-3a + a.
6. 3 a — 4a + 6 a — 3a. 12. — 4cxy-\-3xy—7xy+xy—xy-{-8xy.
13. 5 am — 8 am — 24 am + 13 am — am + 6 am — 11 am.
14. 6 m — 7 — 4 m + 11 — 5 m — 17 — m + 5 ra + 13.
15. — 4 ex + 8 ex — 3 ex + 2 ex — 3 ex + c# — 15 ex — 14 'ex.
ADDITION OF POLYNOMIALS
The principles established for the addition of monomials
apply directly to the addition of polynomials.
Illustrations :
1. Find the sum of 5a + 76 — 2c, 2a — 36 + 8 c, and
-3a + 26-10c.
In the customary form : 5a + 76— 2c
2a-36+ 8c
-3a + 26-10c
4a + 6&- 4c Result.
For the sum : the coefficient ofa= 5 + 2— 3 =
the coefficient of&= 7-3+2 =
the coefficient ofc=-2+8-10 =
4,
6,
-4.
ADDITION OF POLYNOMIALS 25
It frequently happens that not all of the terms considered are found in
each of the given expressions, in which case we arrange the work so that
space will be given to such terms as an examination shows need for.
2. Add 4a + 36 + 3m, 2b + 3 c — d, 2a + 3d + 2m — x,
and 5b — 5m — 3 x.
4a-f 36 + 3m
+ 26 + 3c- d
2a +3d+2m- x
+ 5 6 — 5 m — 3%
6a + 10 6 + 3c + 2(Z -4 a; Result.
The coefficient of the wi-term in the sum being 0, that term disappears.
In general, to add polynomials :
35. Write the given expressions so that similar terms shall stand
in the same columns. Add separately, in each column, the positive
coefficients and the negative coefficients, and to the arithmetical
difference of their sums prefix the proper sign.
Collecting terms is another expression for adding two or more
given expressions.
36. Checking results. The accuracy of a result may be
checked by substituting convenient numerical values for each
of the given letters. The substitution is made both in the
given expressions and in the sum obtained, and the work may
be considered accurate if both results agree.
Illustration : To check the sum of 5a + 76 + 2c and 2a — 36 — 5c,
let a = 1, 6 = 2, and c = 1.
Then 5a + 76 + 2c = 5 + 14 + 2=21
and 2a-36-5c = 2- 6-5 = -9
Whence, 7 a + 4 6 -
-3c = 7+ 8-
-3
= 12.
•
Exercise 2
Find the sum of :
1. 2.
3.
4.
3a + 76 _7a_f9
4 n — 9 cz
— acm-f- 3xyz
5a-36 12^-1
— n -f- 9 cz
2 acm—YI xyz
26 ADDITION. PARENTHESES
5.
6.
7.
5ab-
- 7 ac — ac?
- 2cx-16
m — nx —z
11 ab
+ 2 ad
ay + 20 c#
3nx — 5my + z
8. 4a+3 6, 5a — 2 6, — 7a+4 6, and 4 a —11 6.
9. 3a — 2c — jb, 4a — c + 7 x, and — a — 5c+9sc.
10. 4 x + 3 2/ — 11, — 5 a — 2 y — 8, and x + 19.
11. 4 m — 2 n — sc, — 3 n — 4 a?, 2 m + 3 n, and 7 % — 9 x.
12. 3a — 6 — 4c + 7, a — 4 6 — c+6, and —2a— 6 + 6c— 14.
13. — 3a+76— 2c — 7, c— 3a— 6+8, and 7—4 6 — a + 3 c.
14. a — 3y + m-2x + 7, 36-2?/ + 2a-4, 7x — 2a — y,
and66 — 5a-2a-12.
Collect similar terms in :
15. 5a6 + 66c — 3 am — 8a; + 4a6 — ac-f 3cd — am
+ 3ab—2cd—x.
16. 4m — 9 + d — 2/ + 3 — 2d — 6 — 5m — 3y-\-2m
-y-d+16.
17. 4a6c— 8 6cd — 5cd*a; — 18 + 13 6cd* — 18a6c— 5cdx
-7 + 12a6c.
18. — 9 dx — mn — 3 ay + 4 6w — 19 — 14 ay + 5 bn — 15 mn
+ 8 da; — 10 bn + 14 mn + 16 ay.
PARENTHESES
37. The parenthesis is used in algebra to indicate that two
or more quantities are to be treated as a single quantity.
Thus, a + (b + c) is the indicated sum of a and the quantity, b + c.
a — (6 + c) is the indicated difference between a and the quantity, b + c.
38. A sign + or — before a parenthesis indicates an opera-
tion to be performed.
a + (6 4- c) is an indicated addition,
a — (6 + c) is an indicated subtraction.
PARENTHESES 27
The Parenthesis preceded by the Plus Sign
Consider the expression 20 + (10 + 5).
By first adding 10 and 5, 20 + (10 + 5) = 20 + 15 = 35.
Adding separately, 20 + (10 + 5) = 20 + 10 + 5= 35.
The result is clearly the same whether the parenthesis is removed
before or after adding.
Again, consider the expression 20 + (10 — 5).
By first subtracting 5 from 10, 20 + (10 - 5) = 20 + 5 = 25.
Or, subtracting separately, 20 -f (10 - 5) = 20 + 10 - 5 = 25.
And, again, the same result from each process.
In general symbols :
For the + parenthesis : a + (b + c) = a + b + c.
a + (& — c) = a + b — c.
In general :
39. If a + ( ) is removed, the signs of its terms are not changed.
The Parenthesis preceded by the Minus Sign
Consider the expression 20 — (10 + 5).
By first adding 10 and 5, 20 - (10 + 5) = 20 - 15 = 5.
Or, subtracting separately, 20 — (10 + 5) = 20 — 10 — 5 = 5.
The result is the same from each process.
Again, consider the expression 20 — (10 — 5).
In this expression we are not to take all of 10 from 20, only the differ-
ence between 10 and 5, i.e. (10 — 5), being actually subtracted. There-
fore if all of 10 is subtracted, we must add 5 to the result.
By first subtracting 5 from 10, 20 - (10 - 5) = 20 - 5 = 15.
Or, separately, 20 - (10 - 5) = 20 - 10 + 5 = 15.
And, again, the same result from each process.
In general symbols :
For the — parenthesis : a — (b + c) = a — b — c.
a — (b — c) = a — b + c.
In general :
40. Ifa—() is removed, the sign of every term in it must be
changed.
28
ADDITION. PARENTHESES
Two important principles must be kept constantly in mind :
(1) The sign before a parenthesis indicates an operation of
either addition or subtraction.
(2) The sign of a parenthesis disappears when the parenthesis
is removed.
For example :
15 - (7 - 4 + 2) = 15 - 7 + 4 - 2.
The — sign of 7 in the result is not the original — sign of the paren-
thesis. The original sign of 7 was + , and this sign was changed to — by
the law of Art. 40.
The sign of a parenthesis shows the operation, and disappears
when we perform it.
Oral Drill
Give orally the results on the following :
1. (+5) + (+2). io. (_7a)-(+5a).
2. (+5)-(+2). 11. _(_5a)-(-7a).
3. (+5) + (-2). 12. -(+6m) + (-6m).
4. (+5) -(-2). 13. (-9 a?) -(9 a;).
5- (-7) + (+6). 14. (-9x)-(-9x).
6. (+5)-(-9). 15. -(-19ran)-(-20ran).
7. (-10) -(-19). 16. -(Jxyz)-(-Sxyz).
8. (+16) + (-17). 17. '(-16 6c)-(-116c).
9. (-14) -(-13). 18. -(±2xy)-(-15xy).
19. (_5)-(-2)-(-3) + (-4).
20. a-(4a) + (-4a) + (-a).
21. -5x + (-2x)-(-x)-(3x) + (-x).
22. _(-l)-(l)_(_l)_(l) + (-l)-(-l).
23. 2-(+3)-(-4) + (-6)-(-6).
24. -(-2a:)-(+3a>)-(2as)-(-3a!) + (-2a>).
25. — (— mn) + (— 3 raw) — (— 4 ran) + (— raw).
26. 2 a?-(- 3 «)-(+2o;) + (- 3 *)-( + ») + (-«).
PAKENTHESES 29
Parentheses within Parentheses
It is frequently necessary to inclose in parentheses parts of
expressions already inclosed in other parentheses, and to avoid
confusion different forms of the parenthesis are used. These
forms are: (a) the bracket [ ], (6) the brace { \, (c) the vin-
culum — . Each has the same significance as the paren-
thesis.
That is, (a + &) = [« + 6] = {a + b] = a + b.
In removing more than one of these signs of aggregation
from an expression :
41. Remove the parentheses one at a time, beginning either
with the innermost or with the outermost. If a parenthesis pre-
ceded by a minus sign is removed, the signs of its terms are
changed. Collect the terms of the result.
Illustration :
Simplify 4m- [3 m-(n+2)-4]-Sm+3n—(7i-l)j+n.
4 m - [3 m - (n + 2) - 4] - {m + 3 n - (n - 1)} + n =
4m- [3 m - n-2 - 4] - {m + 3 n — n + 1} + n =
4m- 3m + n + 2 +4 -{m + Sn - ra + 1} +w =
4m— 3 m + n + 2 +4 — m — 3 n + n — 1 + n = 6. Result.
Exercise 3
Simplify :
1. 4a+(3a-8). 7. 9 a + [3 a; -(a -2)].
2. 5x-(3x + 7). 8. 9z-[3a+-(a;+-2)].
3. 2c+(c-8)-6. 9. ±a-\a-(-a + 7)\.
4. 3m — (2 m — ?i+-3). 10. 2ac — (ac — 7 + 3ac).
5. 4z-(8-3z)-ll. 11. 5n+[9-{3n-n + 2\].
6. — 5m— (— m— p-6). 12. — 6 — (— xy+[ — 2 xy— (xy+3)~\).
13. -3a-[36-f2a-(6&-a)n.
14. m — [m + \m— (m — m — 1)}].
30 ADDITION. PARENTHESES
15. -(2m + l)-(-[-m + m-3]).
16. 7x-[2+(-3x-)x-(-6x-x-7)-x\)].
17. _(_ a _i)_2a-(a + l)-3a-5-8.
18. l_(_l) + )_l-lH-a-(-l)i,
19. 2-(-l)+j-l-l + a-(-l)-(-a)J.
20. -H-(_l)_{l + l-(a + l)+(-l)t.
21. 3m- [2m— \m — n — 3m— 9— 2 m— 7+3 nj — (— m— 5)].
22. 2a-(36 + c)-2a + (c-36)
-Jc-(2a-36)-[a-(6-c)]
23. l_(l+a)-[-j-[(-l-a-l-3a)-a]-lS-a].
24. ax— (- aa;)— | — (—ax)\ — [ — (—\ — ( — ax)\)].
25. (m- 1)— j— (— J — (— m + 1) — nj — m)— raj.
Inclosing Terms in Parentheses
This operation is the opposite of the removal of parentheses,
hence we may invert the principles of Arts. 39 and 40 and
obtain :
42. Any number of terms may be inclosed in a + ( ) without
changing the signs of the terms inclosed.
43. Any number of terms may be inclosed in a — ( ) provided
the sign of each term inclosed be changed.
Illustrations :
(Art. 42. ) 1. a + b + c-d + e = a + b + (c-d+e).
(Art. 42.) 2. a + 6-c-d + e = a + & + (-c-d + e).
(Art. 43.) 3. a + 6-c + d-e = a + &-(c-d + e).
(Art. 43.) 4. a-b + c-d + e = a-b-(-c + d-e).
PARENTHESES 31
Exercise 4
Inclose the last two terms of each of the following in a
parenthesis preceded by the plus sign :
1. a + 3 6-|-9 c. 4 12 mn + 4 mx — 5 my -f- 16.
2. 4mH-3n-f-2j9 — 8. 5. 5c — 5d — x — mny.
3. 4ae + 6ad + 9ae-ll. 6. -9c + lld~6 + a».
Inclose the last three terms in a parenthesis preceded by
the minus sign :
7. 4a-56 + 3c-9. 11. 3x + ±y + 5z + 2.
8. 2a — 66 — c + x. 12. 4 ac — 5 6c + 3 bd + 10 ad.
9. 5a; — 8y + 3« — 11. 13. 12 — w + ra — 2p + x.
10. — 7 — 5ra + 3w — 2 p. 14. m + n—p + x — y + z.
15. — 2 — x -f a» — xy — yz.
Without changing the order of the terms, write each of the
following expressions first in binomials and then in trinomials.
Before each parenthesis use as its sign the sign of the term
which is to be' first in that parenthesis :
16. a-f-6 — c-fd — x — y.
17. a — c + d — m — n + z.
18. c — ra — x + y — z — 1.
19. a — d — e — m — n + x.
20. m — n—p — x — y — z.
21. ab — ac + cd — mn — np + mp.
22. 2a + 3c-f 4a , + 5a — 3y-2z.
23. 7a6 — 3ac + 4 6d' — 5 6c + 3 ad — 4ca\
24. 5ay — 3az + 8z — 4:xy — 3xz + 2yz.
25. ama; — cny — bz — cnx — amy — dz.
CHAPTER III
SUBTRACTION. REVIEW
44. Subtraction is the inverse of addition. In subtraction
we are given the algebraic sum and one of two numbers, and
the other of the two numbers is required. The given sum is
the minuend, the given number the subtrahend, and the required
number the difference or remainder.
45. We may base the process of subtraction on the principle
of Art. 40, for,
The quantity to be subtracted may be considered as inclosed in
a parenthesis preceded by a minus sign. Consider the example,
From 10a + 36 + 7c take 6 a + b - c.
By definition, the first expression is the niinuend, the second
the subtrahend. Inclosing both expressions in parentheses,
and replacing the word " take " by the sign of operation for
subtraction, we have
(10 a + 3 b + 7 c) - (6 a + 6 - c).
Removing parentheses, 10a + 36 + 7c — 6a — b + c.
Collecting terms, 4 a + 2 b + 8 c. Result.
We have, therefore, changed the signs of the terms of the sub-
trahend and added the resulting expression to the minuend. In
practice the usual form would be
Minuend : 10 a + 3 b + 7 c.
Subtrahend : 6 a + b — c.
Difference : 4a + 2& + 8c. Result.
The change of signs should be made mentally. Under no cir-
cumstances should the given signs of the subtrahend be actually
altered.
32
SUBTRACTION 33
From the foregoing we make the general statement for sub-
traction of one algebraic expression from another.
46. Place similar terms in vertical columns. Consider the sign
of each term of the subtrahend to be changed, and proceed as in
addition.
Two important principles result from the process of sub-
traction.
47. Subtracting a positive quantity is the same in effect as
adding a negative quantity. In general :
By Art. 40: (+ a) -(+&) = «-&. By Art. 39 : («j-a) + (- 6)=a-&.
Numerical Illustrations :
(+ 5) - (+ 3) = 5 - 3 = 2. (+ 5) + (- 3) = 5 - 3 = 2.
48. Subtracting a negative quantity is the same in effect as
adding a positive quantity. In general :
By Art. 40: (+«)-(- &) = « + b. By Art.39: (+ a) + (+ b)=a + b.
Numerical Illustrations :
(+ 6) - (- 3) = 5 + 3 = 8. (+ 5) + (+ 3) = 5 + 3 = 8.
Oral DriU
From
1.
13
2.
12 a
3.
-9a
4.
-3a
5.
13 ab
6.
15 a
Take
_4
5 a
-2a
— 6 a
-1 ab
-3x
From
7.
9m
8.
-7c
9.
1 xy
10.
— Sac
11.
— 19 mn
12.
— 7 xz
Take
13 m
8c -
-5xy
— Sac
— 11 mn
23 xz
From
13.
— lac
14.
+ 7ac
15.
16.
17.
abc
18.
— 4 acx
Take
-f-7ac
— 7ac
9x
-9a
— 9abc
acx
SOM. EL. ALG.
34 SUBTRACTION. REVIEW
SUBTRACTION OF POLYNOMIALS
49. By application of the principle of Art. 46, we subtract
one polynomial from another. Results in subtraction may be
tested as shown in Art. 36.
Exercise 5
1.
From 5 a + 7
Take 3a-4
2.
8m-19
3 m- 11
3.
- 7 c + 14
-9c + 14
4.
-17ac-8
- 9ac + 9
5.
From 3a + 46*-7
Take 3 6-9
6.
6x-$y
x + 3y- z
7.
-3m
2 a — 5 m
+ 7
8.
a + b — c
9. From 7 a + 3 m — 9 take 2 a — 5 m 4- 8.
10. From 16a-|-6a;— 3y take a — 7 x — 15 y.
11. From 4a + 9a"-18 take -3a + 9d'-15.
12. From 5x—7y-\-3z take 3y — 1 z.
13. From — 7a + 6c-r-3cZ + 5 take 2 a — 5 c + 3.
14. From 11 a + 3 m — a; take 2 a + 7 # — m.
15. From 26 + 5c — 3n take — 2 a — 6 + 4 c — 3 n.
16. Subtract 15 aa? — 3 ay — 19 from 17 ax — hay — 11.
17. Subtract 10 — 3x — 1 y — z from z + ll — 2x-3y.
18. Subtract 5c + 6d — 4m + n from 3c — m.
19. Subtract 5a — 11 6 — 2c from 3a — 8 c + 2 ra.
20. From the sum of 3m — n + 2p and 3m-4n-5|) take
the sum of m -f- 3 n —p and 3 m — 7 n + 6 p.
21. From the sum of 5x— 3y + 2 z -f 3 and 3 x + 7 subtract
the sum of 3 a; — y 4- 2 and 7 — 2x — y-\-llz.
22. Subtract the sum of 4 a — 11 c + d and 3 b — c + 10 from
the sum of36 + d — 8c and 4 a — 4 c + 9.
23. Take the sum of 3 m— n+2y, 2n— 3 m— 4^ and m+p— z
from the sum of m — 3 3, 3 n +p, and p-f-m — 2y + n — z.
GENERAL REVIEW 35
50. Addition and Subtraction with Dissimilar Coefficients.
When the coefficients of similar terms are themselves dis-
similar, the processes of addition and subtraction result in
expressions with compound coefficients, i.e. coefficients having
two or more terms. The principle is easily understood from
the following illustrations :
Addition Subtraction
ax cm am cdz
bx_ — acm cm — 9 amdz
(a + b)x (1 — a) cm (a — c)m (c + 9 am) dz
Exercise 6
Find the sum of :
1. am + dm. 4. — 17 mxy + 2pxy. 7. bcdx + 12 cdx.
2. cdx + 5 nx. 5. cdn — 7 dn. 8. 5 cdx — 12 bdx.
3. 3 bed — 2 mcd. 6. acx — 19 bx. 9. — mxy — 11 cxy.
10. 6a; + ay and ma; + ny. 13. 5 ab — 2 cd and 3 6 + 7 a\
11. a# + nz and 6aj+pz. 14. 6 az -f- 5 by and raz — 3 y.
12. acm + 6n and dm + cfri. 15. 11 mxy + 5 ccz and ascy — acz.
16. 2 ftm + 3 a; + ?/ and 5m +cx — m?/.
17. 3 a6 + 7 ac + 11, m& — lie — 7, and 5 6 — nc + 3.
18. (3 + c)x + (2 c - 9)a> + (6 - 3 c)x + (a + c + 1)*.
(Examples 1-12 inclusive will serve as an exercise for subtraction ; the
first expression in each being the given minuend, the second expression,
the subtrahend.)
GENERAL REVIEW
Exercise 7
1. Find the sum of 3 a + 2 6 — 6 m, 2 b — 5 c + x, — 4 a— 76
-f c — m, and a — 3 c + x.
2. Simplify 1-J1- [1- (1-1 + m)]-mj -m.
3. Add 3a + 26-c, 2a-36 + 3c, 4a + 36-7c, and
-8a-46 + 5c.
36 SUBTRACTION. REVIEW
4. Subtract 4 x -f- 3 y — 2 from 5 as, and add x — 3 y + 2 to
the result.
5. Subtract 2# — 3y from 4aj+"2y, subtract this result
from 7 x — Sy, and add a; — ?/ to the final remainder.
6. What expression must be added to 4a — Sm-\-x to pro-
duce la — 5 ra + 3 a; ?
7. What is the value of (-3) - (+ 2) - (-3) + (- 5)
-(-2)?
8. Collect — 3a — & + 4 c, 2c — 3 a* -f #, a — 4a— d, 5 c + 2 a
+ 3 as, and 4 d + 6 - 7 c.
9. Simplify x — 1 — J# — 1 — [a — 1 — a; — 1 — ay } .
10. Subtract 4 a? — 3 y + 11 from unity, and add 5 a; — 3 y-h 12
to the result.
11. Combine the m-terms, the n-terms, and the avterms in the
following, inclosing the resulting coefficients in parentheses :
am -{■ 3 bn -\- x + m + n — ax + 2 n — ex.
12. A given minuend is 7 x + 12 y — 7, and the corresponding
difference, 4 x — y + 2. Find the subtrahend.
13. To what expression must you add 2a — 3c + m to pro-
duce 5 a -f 7 c — 9m?
14. Subtract 2a—7x+3 from the sum of 3 x + 2 a, 4 a
— 10 #, 5 a — 7, and — 4 a; — 11 a.
15. From (a + c) y + (m -f- n) z take (a — c)i/ — (m — n)z.
16. Subtract a; + 17 y — 2 from 12 x -f- 3 z and add the result
to3x+(y + llz)-4y.
17. Add the sum of 4 a; + 7 c and 2 a? + 3 c to the remainder
that results when x -f 4 c is subtracted from 5 c — 11 x.
18. From (a + 4) x + (a -f 3) y subtract (a + 1) x -f (a + 2) ?/.
19. Add 3 a? + 2 m — 1 and 2x — 3m-\-7, and subtract the
sum from 6 a; — m + 6.
GENERAL REVIEW 37
20. Simplify a — 1 — a + 1 + [a — 1 — a — a — 1 — (a — 1) —
21. From 7ra + 3<c — 12 take the sum of 3 a; + 7 and
2 m -3y-3.
22. What expression must be added to a + b + c to pro-
duce ?
23. What expression must be subtracted from to give
a + 6 + c?
24. From what expression must 4 x — 7 m + 10 be sub-
tracted to give a remainder of 3 # -f 6 ra — 4 n -f 2 ?
25. Inclose the last four terms of a — 86-f-4m — 5n + 7 x
in a parenthesis preceded by a minus sign.
26. Simplify and collect a + [ — 2m — 4 a + <c —
(— 2 <c — m -f- a) — 3 x\
27. If x + 7y — 9 is subtracted from 0, what expression
results ?
28. Isthesumof [-7 + (-2)-(-3)] + [-!-2 + (-l)i]
positive or negative ?
29. Prove that [3- (-2) -(-!)] + [(- 2) + (- 1)_(_3)]
+ [ _2_(_l) + (_5)]=0.
30. Simplify and collect
10_[9-8-(7-6-{5-4-3^2S-l)].
31. Show that
2 - ( - 3 + o^l) - 3 + ( - 2 - (Tfl ) = _ 2 a.
32. Collect the coefficients of x, y, and z in
abx — acy + abz — mnx + mpy — npz + x — y — z.
33. Simplify and collect
l-[-S-(-l + r^)-lf-l]-a.
34. What expression must be subtracted from a — x + 3 to
give -(a + [-«-(-2a -x + 1)- 3]) ?
CHAPTER IV
MULTIPLICATION
51. Multiplication is an abbreviated form of addition.
Thus, 3x4 = 4 + 4 + 4. 5a = a + a + a + a + a.
For the purpose of arithmetic multiplication has been
denned as the process of taking one quantity (the multiplicand)
as many times as there are units in another quantity (the
multiplier). This definition will not hold true when the mul-
tiplier is negative or fractional. Hence, the need for the
following definition :
52. Multiplication is the process of performing on one
factor (the multiplicand) the same operation that was per-
formed upon unity to produce the other factor (the multiplier).
The result of a multiplication is a product.
Illustration :
The Integral Multiplier.
3x5 means that 5 is taken three times in a sum. Or, 3x5 = 5 + 5+5.
By the same process the multiplier was obtained from unity, for
3 = 1 + 1 + 1.
The Fractional Multiplier.
Obtained from unity a multiplier, 2f = l + l + £ + £ + £. For unity
was taken twice as an addend and \ of unity taken three times as an
addend.
In like manner, 2fx5 = 5 + 5 + f + f + | = 10 + -^ = 13|.
THE NUMBER PRINCIPLES OF MULTIPLICATION
53. The Law of Order. Algebraic numbers may be multiplied
in any order.
In general : ab = ba.
Numerical Illustration : 3x5 = 5x3.
38
SIGNS IN MULTIPLICATION 39
54. The Law of Grouping. The product of three or more
algebraic numbers is the same in whatever manner the numbers
are grouped.
In general : abc = a(bc) = {ab)c = (ac)b.
Numerical Illustration : 2 . 3 . 5 = 2(3 . 5) = (2 : 3)5 = (2 . 5)3.
55. The Law of Distribution. The product of a polynomial
by a monomial equals the sum of the products obtained by multi-
plying each term of the polynomial by the monomial.
In general : a(x + y + z)= ax + ay + as.
Numerical Illustration : 2(3 + 4 + 5) = 6 + 8 + 10.
As in the case of addition, no rigid proof of these laws is
ordinarily required until the later practice of elementary-
algebra.
Here, as in # addition, the law of order is frequently called
the commutative law, and the law of grouping is called the
associative law.
SIGNS IN MULTIPLICATION
Upon the definition of multiplication we may establish the
results for all possible cases in which the multiplicand or
multiplier, or both, are negative numbers.
(1) A positive multiplier indicates a product to be added.
(2) A negative multiplier indicates a product to be sub-
tracted.
(1) The Positive Multiplier. Expressed with all signs (mul-
tiplier = + 3) :
(+ 3) x (+ 5) = + 5 + 5 + 5 = + 15.
(+ 3) x (- 5) = - 5 - 5 - 5 = - 15.
(2) The Negative Multiplier. Expressed with all signs (mul-
tiplier = — 3) :
(- 3) x (+ 5) = - (+ 5 + 6 + 5) = - (+ 15) = - 15.
(_ 3) x (- 5) = - (- 5 - 5 - 5) = - (- 15) = + 15.
+
5
- 5
—
3
- 3
—
16
+ 16
(-
a) ( + 6) =
— ab.
(-
<*)(-
-b) =
+ ab.
40 MULTIPLICATION
Comparing the four cases in the ordinary form of multiplica-
tion, we have
+ 6 - 6
± 3 ± 3
+ 15 -15
In general :
(+«)(+ &)= + «&.
(+a)(-b)= - ab.
56. Like signs in multiplication give a positive result.
57. Unlike signs in multiplication give a negative result.
Oral Drill
Give the products in the following, each with its proper
sign :
1. (3) (-6). 7. (-6) (-7). 13. (2)(-5)(0).
2. (4)(-5). 8. (-9)(0). 14. (_5)(3)(-8).
3. (-3) (6). 9. (-3) (-11). 15. -(2) (3) (-4).
4. (-4) (5). 10. (-4) (-3) (2). 16. - (3) (-5) (-2).
5. (-5) (-4). 11. (2) (-5) (3). 17. - (- 4)(- l)(-2).'
6. (5)(0). 12. (_2)(5)(-3). 18. -(-3) (-5) (-8).
COEFFICIENTS IN MULTIPLICATION
58. The coefficient of a term in a product of two algebraic
expressions is the product of the coefficients in the given multiplier
and multiplicand.
The principle is established by means of the Law of Grouping.
For the coefficient of mn in the product of am times en:
By Art. 64 : am x en = (a x c)(m x n)
= (ac) (mn)
= acmn.
And the required coefficient is ac.
EXPONENTS IN MULTIPLICATION 41
EXPONENTS IN MULTIPLICATION
59. An exponent is a symbol, numerical or literal, written
above and to the right of a given quantity, to indicate how
many times that quantity occurs as a factor.
Thus, if three a's occur as factors of a number, we write a 3 , and
avoid the otherwise cumbersome form of a x a x a. In like manner,
axaxaxbxb= a s b 2 , and is read " a cube, b square."
60. The product of two or more equal factors is a power.
Any one of the equal factors of a power is a root.
In common practice, literal and other factors having ex-
ponents greater than 3 are read as powers.
Thus, a 6 is read " a sixth power," or merely " a sixth."
a 3 y 7 z 2 is read " a cube, y seventh, z square."
The exponent "1" is neither written nor read. That is, a
is the same as a 1 . The difference between coefficients and
exponents must be clearly understood. A numerical illustra-
tion emphasizes that difference. Thus :
If 5 is a coefficient, 5x2 = 24-2 + 2 + 2 + 2= 10.
If 5 is an exponent, 2 5 =2x2x2x2x2 = 32.
The General Law for Exponents in Multiplication
By definition, a s = a x a x a,
a* = axaxaxa.
Therefore, a z xa* = axaxaxaxaxaxa
= a\
Similarly, a b x at = a 5+4 = a 9 ,
m?x m 8 xm = m 2+8+1 = m 6 .
In general, therefore, we have the following :
If m and n are any positive integers :
a m = ax ax a -"torn factors,
a n = ax ax a-"to n factors.
Hence, a m x a n = (a x a x a ••• to m factors) (a x a x a ••• to n factors)
= (a x a x a ••• to m + n factors)
— a m+n .
In the same manner, a m x a n x aP = a m+n+ P, and so on, indefinitely.
42 MULTIPLICATION
This principle establishes the first index law, m and n be-
ing positive and integral.
The general statement of this important law follows :
61. The product of two or more powers of a given factor is a
power whose exponent is the sum of the given exponents of that
factor.
Oral Drill
Give orally the products of the following :
1. a 2 X a 3 . 5. x* xx 9 . 9. c 7 x c 3 X c 5 .
2. m 5 x m 3 . 6. x 6 x x 23 . 10. y 9 xy 5 xy 3 .
3. d 7 x d 3 . 7. m X m 2 x m 3 . 11. a 2 x a 3 x a 5 X a 9 .
4. z 9 X z 8 . 8. a 3 X a 2 X a. 12. » 2 Xw 4 X n 6 X n 8 .
MULTIPLICATION OF A MONOMIAL BY A MONOMIAL
By application of the law of order and the principles for
signs and exponents, we obtain a process for the multiplication
of a monomial by a monomial.
Illustrations :
1. Multiply 3 a 2 b 3 by 12 a A b 2 .
By the Law of Order, 3 a 2 5 3 x 12 a 4 6 2 = 3 x 12 x a 2 x a 4 x 6 8 x b 2 .
By the Law of Grouping, =(3 x 12) (a 2 x a*)(6 3 x 6 2 ).
By Arts. 58 and 61, = 36 a 6 & 5 . Result.
2. Multiply -7 a 2 b 5 x 3 z by 5 a 7 b 2 y.
-7 a 2 b 5 x 8 z x 5 aWy = -7 x 5 x a 2 x a 7 x b 5 x b 2 x x s x z x y (53)
= ( - 7 x 5) (a 2 x a?) (6 5 x b 2 ) (x 3 ) («) (y) (54)
= -35a 9 & 7 z 3 «/z. Result. (57) (61)
Therefore, to multiply a monomial by a monomial :
62. Observing the law of signs, obtain the product of the
numerical coefficients. The exponent of each literal factor in the
product is the sum of the exponents of that factor in the multipli-
cand and multiplier.
MULTIPLICATION OF POLYNOMIAL BY MONOMIAL 43
t
Oral Drill
Give orally the products of the following :
1. 2. 3.
4. 5. 6.
7.
5a Sx — 4=a
5m — 5x —3x
- %y
— 3 — 7x —5a
■8m — 11 x 16 x
-13 2/
8. 9. 10.
11. 12.
13.
-Sab 2 4 m 3 n 5 x*y
— 3 xy 3 — 6 m 3 ny
— 3 mna;
2ab -3mn 3 -1 tftf
— IxPy — 11 mn 3 y
10 m 3 x
14. 4 abc by 3 acd.
20. - c 2 d 3 m 2 by -
5 m 3 n.
15. 4 axy by — 7 xyz.
21. c 3 dxy by — 11
cWy 3 .
16. aW by a 3 b 2 d.
22. — 10 tfyh by xPy 3 .
17. 4 x?yz by — a^/z 2 .
23. — 11 cmn 3 y by
— 5 m 2 n 2 .
18. 3 a 7 by — 4 amn.
24. 13 (Wa 8 by -
2c 2 d?.
19. — 12 a 3 m Q by — 2 m 3 w 2 z. 25. 15 an 2 a?z by — 8 ?i 3 ra2/.
MULTIPLICATION OF A POLYNOMIAL BY A MONOMIAL
The process of multiplying a polynomial by a monomial
results directly from the number principle for multiplication
assumed in Art. 55. That is :
a(x + y + z) = ax + ay-\-az.
In common practice the multiplicand and multiplier are
written as in arithmetic, excepting that the multiplier is
usually written at the extreme left.
Illustration :
Multiply 3 m 3 — 5 m 2 n -\- 7 raw 2 — 2 n s by — 2 mn.
3 w 8 - 5 m 2 n + 7 mn 2 -2n 8 Each term f the prod-
~~ 2mn uct is obtained by the
-6 m% + 10 m% 2 - 14 w 2 n 8 + 4 mri 4 Result, principles of Art. 62, for
the operation is made up of successive multiplications of a monomial by a
monomial.
44 MULTIPLICATION
Hence, to multiply a polynomial by a monomial :
63. Multiply separately each term of the multiplicand by the
multiplier, and connect the terms of the resulting polynomial by
the proper signs.
Exercise 8
Multiply :
1. 2. 3. 4.
3a+7x 2a 2 — 10a llax-15ay a^-lOx-ll
2 a 3 a 3 — axy x 2
5. 6. 7.
2 ra 2 — 10 ran + 15 n 2 —x* — x 2 y + xy 2 a 4 — 3 a 3 x + 3 ax 3 — x*
3n —x* ax
8. 10 a (7 ab - 8 ac + 11 be). 11. 3 a (a 3 - a 2 b-\-ab 2 ).
9. —3x(x 2 -x + ll). 12. -±bc(bx-bm + 3bn).
10. -2ra 2 (ra 3 n-ra 2 n 2 + ran 8 ). 13. 11 a 2 »(^ + 9»- 15).
64. The degree of a term is determined by the number of
literal factors in that term.
7 a 8 x 2 is a term of the 5th degree, for 3 + 2 = 5.
65. The degree of an algebraic expression is determined by
the term of highest degree in that expression.
5 m 2 n + mn + n 2 is an expression of the 3d degree.
66. An algebraic expression is arranged in order when its
terms are written in accordance with the powers of some letter
in the expression.
If the powers of the selected letter increase from left to right, the
expression is arranged in ascending order.
Thus, x - 2 z 2 + 5 x* - 7 a 4 + 10 x 5 .
If the powers of the selected letter decrease from left to right, the
expression is arranged in descending order.
Thus, 4 x 9 - 5 x 7 + 3 x*> - 2 z s - 3 x.
MULTIPLICATION OF POLYNOMIAL BY POLYNOMIAL 45
67. The degree of a product is equal to the sum of the degrees
of its factors.
68. A polynomial is called homogeneous when its terms are
all of the same degree.
Thus, sc* — 4 x s y + 6 x 2 y 2 — 4 xy z + y 4 is a homogeneous polynomial.
MULTIPLICATION OF A POLYNOMIAL BY A POLYNOMIAL
A further application of the law of distribution for multipli-
cation (Art. 55) establishes the principle for multiplying a
polynomial by a polynomial.
By Art. 55 : (a + b) (as + y) = a (x + y) + b (x + y)
= ax + ay + bx + by.
The polynomial multiplicand, (x +- y), is multiplied by each
separate term of the polynomial, (a + 6), and the resulting
products are added. The process will be clearly understood
from the following comparison :
Numerical Illustration : Algebraic Illustration :
a + 5 Multiplicand.
a + 7 Multiplier.
a 2 + ha
+ 7 q + 35
a 2 + 12 a + 35 Product.
Explanation :
a (a + 5) = a 2 + 5 a
7 (a + 5) = 7 a + 35
a 2 + (5 a + 7 a) + 35 = a 2 + 12 a + 35
We have, therefore, the following general process for multi-
plying a polynomial by a polynomial :
69. Arrange the terms of each polynomial according to the
ascending order or the descending order of the same letter.
Multiply all the terms of the midtiplicand by each term of the
multiplier. Add the partial products thus formed.
12 10+2
13 10+ 3
36 100 + 20
12 30 + 6
156 = 100 + 50 + 6
Explanation :
(10 + 2) = 100 + 20
3 (10 + 2) = 30 + 6
00 + (20 + 30) + 6 = 156
46 MULTIPLICATION
Illustrations :
1. Multiply 2 a +- 7 by 3a- 8.
2(1 + 7 T, ,
3 _ 8 Explanation :
6 a 2 + 21 a 3 a (2 a + 7) = 6 a 2 + 21 a
- 16 a - 56 - 8 (2 a + 7) = - 16 a - 56
6 a 2 + s a _56 Result. 6a 2 + (21a-16a) -56 = 6a 2 + 5a-56.
2. Multiply a 3 -2a 2 + 3a-2 by a 2 + 3a-2.
a 3_ 2a 2 + 3 a _2 Explanation :
a 2 +3q-2 a 2 (a 3 -2 a 2 +3 a-2)=a 6 -2 a 4 +3 a 8 -2a 2
a 5_2a 4 +3a 8 - 2 a 2 3 a (a 3 -2 a 2 +3 a-2)=3 a 4 -6 a 8 +9a 2 -6a
+3a 4 -6a 8 + 9 a 2 - 6 a -2 (a 3 -2 a 2 +3 a-2) = -2 a 3 +4a 2 -6a+4.
—2 a + 4a — 6 a+4 Adding the partial products, we have
a 6 + a 4 -5a 8 +lla 2 -12a+4 a 5 + a* - 5a 8 + 11 a 2 - 12a + 4.
Expressions given with their terms not arranged should
both be arranged in the same order before multiplication.
3. Multiply l — 7x* + a? + 5x by -4z- 1 + 23J 2 .
Arrange both multiplicand and multiplier in the descending order.
x 8 — 7se 2 + 5se-}-l (Let the student complete this multiplication,
2x — 4a: — 1 writing out a complete explanation in the same
form as those accompanying examples 1 and 2.)
70. Checking. A convenient check for work in multipli-
cation can usually be made by the substitution of a small num-
ber as shown in addition. Thus, in Ex. 2, if a = 2 :
Multiplicand a 3 -2a 2 +3a-2= 8-8 + 6-2=4
Multiplier a 2 + 3a -2= 4+6-2=_8
Product a 5 + a 4 - 5 a 3 + 11 a 2 - 12 a + 4 = 32 + 16 - 40 + 44 -24 +4=32
It is well to remember that this check will not always serve
as a test for both coefficients and exponents. If the value, 1,
had been used above, only the coefficients would have been
tested, for any power of 1 is 1.
MULTIPLICATION OF MISCELLANEOUS TYPES 47
Exercise 9
Multiply :
1. 4 a + 7 by 3 a + 5. 7. 7 ac- 3 by 5 ac 4-1.
2. 3a + 4 by 7<c-3. 8. llafc-3 by 5a6c + 2.
3. 5ra-9 by 4m + 7. 9. 4^-7 by 3a^ + l.
4. 2 c — 5 y by 3 c + 11 2/- 10. 7 mw — a? 3 by 3 raw + 2 x 3 .
5. 4 ra — 11 by 4 ra + 11. 11. 16 x 3 - 10 xy by 5 ar* + 11 a^.
6. 5cd-7 by 6cd + 5. 12. <lab 2 -b 3 xy by ab 2 -3b 3 xy.
13. c 3 -3^ + 3 c-1 by c 2 - 2c + 1. . .
14. ra 2 -2ra+l by ra 2 -2ra + l.
15. a 3 + 2a 2 -3a + 4 by a 2 -3a-l.
16. 3d 2 -5d + 2 by d 2 -d-l.
17. 4/-7?/ 2 -3y + 2 by ?/ 2 -52/-4.
18. x 3 — x?y + xy 2 — y 3 by a 2 -f cci/ + 2/ 2 -
19. 27 - 18 ra + 12 ra 2 - 8 ra 3 by 3 + 2 ra.
20. 1 - 2 ac + 4 aV - 8 a 3 c 3 4- 16 a 4 c 4 by 1 4- 2 ac.
2i. a;-7-3ic 2 4-» 3 by a-3 + 2aA
22. ra 2 -2ra 4 + 7-2ra 3 -ra by 2 ra 2 - 9 -3 ra 3 -2 ra.
23. 12-7x + 5x 3 -2x 2 by -Saf + x-Sx^x 3 .
24. -11 a- 7 a 3 4-17 + a 4 -3a 2 by 3a- 10- 7a 2 4- 2 a 3 .
MULTIPLICATION OF MISCELLANEOUS TYPES
71. Illustrations :
1. Multiply a 4- & 4- 2 by a 4-^—2.
a + 6 + 2
a + 6-2
a 2 + a6 + 2 a
+ ab + 6 2 + 2 6
-2a -26-4
a 2 + 2 a6 + 6 2 - 4 Result.
48 MULTIPLICATION
2. Multiply a 2 + 6 2 + c 2 + 2 ab — ac- be by a + b+c.
Arranging in the descending powers of a :
a? + 2ab - ac+ b 2 - bc + c 2
a + b + c
a 8 + 2 a 2 b - a 2 c + ab' 2 - abc + ac 2
+ a 2 & +2 a& 2 - a&c + 6 8 - b 2 c + 6c 2
+ a?c +2 abc- ac 2 + b 2 c - 6c 2 + c 8
a 8 + 3 a 2 6 +3 a& 2 + 6 8 + c 8 Result
3. Multiply (a - 2) 3 .
(a - 2) 8 = (a - 2) 2 (a - 2)
= (a 2 -4a + 4)(a-2)
= a 8 - 6 a 2 + 12 a - 8. Result.
4. (a + 5)(a + 6)(a-3).
(a + 6) (a + 6) (a - 3) = [(a +6) (a + 6)] (a - 3)
3= (a 2 +lla + 30)(a-3)
5.
(a + x)(a + y).
a + se
a 2 + a£
+ ay + xy
Or,
a 2 + ax + ay + xy
a 2 + (a; + y)a + icy. Result.
a 8 + 8 a 2 - 3 a - 90. Result.
6. (a-x)(a-y).
a — x
a-y
a 2 — ax
-ay + xy
a 2 — ax — ay + xy
Or, a 2 — (as + y)a + Xf. Result.
Exercise 10
Perform the following indicated operations :
1. ( c + x + 3)(c + x-3).
2. (a + m + y)(a + m — y).
3. (a + c + m + x)(a + c — m — x).
4. (ra 2 4- 2mn-\-n 2 + y)(m? + 2mw + w 2 - ?/).
5. (c 2 + aj 2 -r-z 2 -r-2ca; — cz— #z)(c + # + z).
6. (m + a; + w + 2/)(m — w — # — y).
MULTIPLICATION OF MISCELLANEOUS TYPES 49
7. (a + 3 bx) 2 . 9. (3ra-2?i 2 ) 3 . 11. (5 a 2 -2 ay)*.
8. (3 ran- 2 n?/) 2 . 10. (4 a 2 -7 or 5 ) 2 . 12. (3 cd 2 - 5 cdxf.
13. ( C 2_ C _1)( C 2 + C _ 1 )( C 2_ 1 ) >
14. 2 -n-f l)(n 2 + n + l)(n 4 -» 2 + l).
15. (x + 6)(x-7)(x-3).
16. (2x-5)(3x + l)(2x + 5)(3x-l).
17. (cd-3)(cc* + 7)(2cd-l)(3cd + 2).
18. (a 2 -l)(a 2 -5)(a 2 + l)(a 2 + 5).
19. (9x 2 -3x + l)(±x 2 + 2x + T)(3x + l)(2x-l).
20. (a + &)(a + m). 24. (a + c)(& + d).
21. (3a + x)(2a + y). 25. (3a + 2&)(2c-5d).
22. (am — x)(am — y). 26. (m 3 + 2)(m 2 — m).
23. (3cd-ra)(2cd + n). 27. (m + w+l)(m-y).
Perform the indicated operations and simplify :
28. 5x 2 -(x + l)(2x-3)-3x(x-l).
29. ( a -2) 2 +(a + 3) 2 -2(a 2 + a + 4)-l.
30. (2m-l)(m + 3)-(4m + l)(2m-5)-(l-3m)(l+2m).
31. (2a-3) 2 -3a(a-2)-(3-a) 2 .
32. cd(cd + 1) + cd(cd + l)(od + 2) - <?d\cd + 4).
33. b 2 (b 2 + 6 - 1) - 6(6 2 - 6 + 1) - & 2 (& 2 - 1).
34. mn(mn — 1) — [(ran — l) 2 — (1 — ran)].
35. (m-2n-3) 2 + ra(2n + 3-ra) + 2n(ra-2n-3).
36. (l-x)(l-y)+x(l-y)+y.
37. a(b — ra) + b(m — a) + ra(a — 6).
38. (x + a)(x — a) + (a 4- 2) (a — z) + (x + z)(z — a).
39. a\b -x) + b\x - a) + x*(a - b ) + (b - x) (x - a)(a - b).
40. (a + & + c + d) 2 — (a — b — c — d) 2 .
41. (a + x + l) 2 + 2(a+x + l)(a + x — l) + (a + x — l) 2 .
42. (a + lf_3(a + l) 2 (a-l)+3(a + l)(a-l) 2 -(a-l) 3 .
SOM. EL. ALG. 4
50 MULTIPLICATION
72. Multiplication with Literal Exponents. The literal expo-
nent is constantly used in the later discussions of algebra,
and familiarity with this form is readily attained in the pro-
cesses of multiplication and division.
Illustrations :
1. Multiply a 3w 4-a 2w -2a m + 3by a^ + cr — 1.
a Bm + a 2tn _ 2 a m + 3
a 2m + a m _ 1
a bm _j_ a im _ 2 a Sm + 3 a 2m
4 a 4m 4 a 3m - 2 a 2m + 3 a m
- a 3n » - a 2m 4 2a m -S
a bm + 2 a 4m - 2 a 3 ™ +5 a m - 3 Result.
2. Multiply a n+1 - 2 of + 3 a*" 1 - x n ~ 2 by a;" - 3 a;"" 1 .
»»+ 1 -2se n + 3 x n - 1 - x n ~ 2
x n — 3 x"- 1
£2n+l _ 2 X 2n + 3 X 2n_1 — x 2n ~ 2
-3x 2n + 6 a: 2 "- 1 - 9 x 2n ~ 2 4 3 a 2 "- 8
X 2n+i _ 5 x 2n + 9 x 2 "- 1 - 10 x 2 "- 2 4 3 x 2 "- 8 Result.
Exercise 11
Multiply :
1. a w + 7by a w + 3. 4. 2 a m+1 - 7 by 3 a w+1 - 4.
2. a m — 4 by a w — 9. 5.3 a"" 1 — 11 by 5 a w_1 — 8.
3. a n +7by x n -S. 6. 4 a w+2 -7 a by 3 a m+2 + 2 a.
7. a; n + 2a; n - 1 - 3 a;"- 2 — x w ~ 3 by x + 1.
8. a m+s — a m+2 — a m+1 + a m by a m+1 4- a" 1 .
9. a 4 " — ar* w + a; 2n — x n + 1 by af + 1.
10. c TO — c m ~ l 4- c m ~ 2 -f- c w ~ 3 by c 2 — c.
11. a 8m - 3 a 6w -f 3 a 4m -a 2m by a 2w + l.
12. a n+1 — 2 a w — 3 a; 71 - 1 + 4 of 1 - 2 by a* — 3 x"' 1 .
13. a 8 " 1 — a 2m 6 n 4- a m b 2n — b 3n by a 2m + a m b n + & 2w .
14. 2 a; m + 3n + 1 + a; 2m+2n + 3 iC 3w »+ w - 1 4. aj 4m_2 + x 5m ~ n ~ s
by aj w+n -1 -f- cc 2 ** -2 4- ar 5 " 1 -"- 3 .
MULTIPLICATION OF MISCELLANEOUS TYPES 51
73. Multiplication with Detached Coefficients. In many cases
the labor of multiplication and division is lessened by the use
of detached coefficients.
By Ordinary Multiplication : By Detached Coefficients :
2x 3 -3x 2 +4x-5 2-3+4-5
3x 2 + 2x - 1 3 + 2-1
6 x b - 9 x* + 12 x 3 - 15 x 2 6-9 + 12-15
+ 4x 4 - 6x 3 + 8x 2 -10x _l_4_6+8-10
- 2x 3 + 3x 2 - 4x + 5 _ 2+ 3- 4 + 5
6^-5 x 4 + 4x 3 - 4x 2 -14x + 5 6-5+ 4- 4-14 + 5
*
Since the product of two expressions is an expression whose degree is
the sum of the degrees of the given expressions (Art. 67), we supply the
necessary x-factors for the coefficients
6 -5 +4 -4 -14 +5,
obtaining 6 x 5 — 5 x 4 + 4 x 3 — 4 x 2 — 14 x + 5, the result.
Missing Powers. If any power of a literal factor is miss-
ing, its coefficient is 0, and the term must be provided for
in the sequence of powers by an inserted 0.
Thus, (x 3 -2 x 2 + 3) (x 8 + x-1) = (x 3 - 2 x 2 +0 x + 3)(x* + x 2 + x-1).
Multiplying with the coefficients detached,
1-2+0+3
1+0+1-1
1 _2+0+3
+1 _2+0+3
-1+2-0-3
1-2+1+0+2+3-3
Supplying the x-factors, x 6 -2x 5 + x 4 +2x 2 + 3x-3. Result.
For practice in multiplication with detached coefficients, use
Ex. 9, 13 to 24 inc.
At the discretion of the teacher the foregoing paragraph
and the corresponding operation suggested under Division may
be omitted on the first reading of the text. For review topics,
however, the methods have a distinct value.
CHAPTER V
DIVISION. REVIEW
74. Division is the process of finding one of two factors when
their product and one of the factors are given. The dividend
is the given product ; the divisor, the given factor ; arid the
quotient, the factor to be found.
THE NUMBER PRINCIPLE OF DIVISION
75. The Law of Distribution. The quotient of a polynomial by
a monomial equals the sum of the quotients obtained by dividing
each term of the polynomial by the monomial.
T i ax + ay + az ax , ay . az
In general : — y — = — + — + — = x + y + z.
a a a a
Numerical Illustration :
The significance of this law is that the divisor is distributed
as a divisor of each term of the dividend:
The process of division being the inverse of the process of
multiplication, the laws governing signs, coefficients, and ex-
ponents in multiplication form, when inverted, the correspond-
ing laws for division.
SIGNS IN DIVISION
By Art. 56: (+a)( + 6)=+a& Hence, +a&^+a=+& (1)
(— a)(— b)=+ab inversely, -\-ab-. — a=— b (2)
ByArt.57: ( + a)(-6)=-a& by - a b+4-a=-b (3)
(— a)(-h&) = — ab division: — ab-. — a=+b (4)
52
COEFFICIENTS AND EXPONENTS IN DIVISION 53
Therefore, from (1) and (4), and from (2) and (3) :
76. Like signs in division give a positive result.
77. Unlike signs in division give a negative result.
COEFFICIENTS IN DIVISION
78. The coefficient of a quotient of two algebraic expressions
is obtained by dividing the coefficient of the dividend by the
coefficient of the divisor.
That is, 12 a -=- 3= (12 -- 3) a = 4 a.
24 mx-=- 8 = (24 w -=- 8) x = 3 mx.
EXPONENTS IN DIVISION
By definition, a 5 =axaxaxaxa,
a 3 = a x a x a.
Therefore, * = axaxaxaxa =a*.
a 3 a x a x a
That is, a 5 + a 3 = a 5 " 3 = a 2 .
In general, therefore, we have the following :
If m and n are any positive integers and m is greater than n :
a m = a x a x a ••• to m factors,
fl n = axflXfl-ton factors.
a m a x a x a ••• to m factors
Therefore,
a n a x a x a ••• to n factors
— a x a x a,'" to m — n factors
Hence, the statement of the second index law is made as
follows :
79. The quotient of two powers of a given factor is a power
whose exponent is the exponent of the dividend minus the exponent
of the divisor.
54 DIVISION. REVIEW
80. Any quantity with a zero exponent equals 1.
By Art. 79, — = a m ~ m = a . But, — = 1. Therefore, a = 1.
a m a m ■
This principle inconstantly, in use in division. For example,
a% 8
We would obtain the same result by the old method of saying "x B in
X s once" ; and the quotient " 1 " obtained in this way is a factor of the
complete quotient.
DIVISION OF A MONOMIAL BY A MONOMIAL
By application of the laws established for signs, coefficients,
and exponents, we obtain a process for the division of a mono-
mial by a monomial.
Illustration :
1. Divide - 35 aVy 4 by 7 aWy 2 .
- 35 aHy = _ 5 ^-8358-8^4-2 - _ 5 a 2 y2> Result.
7 a?x s y 2
Hence, to divide a monomial by a monomial:
81. Divide the coefficient of the dividend by the coefficient of
the divisor, annexing to the result the literal factors, each with an
exponent equal to its exponent in the dividend minus its exponent
in the divisor.
Oral Drill
Give orally the quotients of :
1. 2. 3. 4. 5.
a 3 )a[_ a *)- a* -x 2 )^ -m ^-m 7 -a b)-a 2 b 3
6. 7. 8. 9.
4q )-12a 8 -Sx ^Wx'y 11 x y)-U&y* - 7 a 8 b ) - 28 a s b 2
10. 11. 12. 13. 14.
- 36 my - 39 c 3 dx 42 afyV - 108 ah 1 - 84 a s bc 2 d
9 m 2 y' 2 -13c 2 x -7xfz* - IS a 2 z - 12 a 2 bcd
DIVISION OF POLYNOMIAL BY MONOMIAL 55
x DIVISION OF A POLYNOMIAL BY A MONOMIAL
Applying the Law of Distribution (Art. 75), and the prin-
ciples governing the division of monomials (Art. 81), we obtain
a process of division when the dividend is a polynomial and
the divisor a monomial.
By Art. 75: ax + & ± az = ^ + & + ^ = x + y+ z.
a a a a
In practice the usual form of expression is a ) ax + a V + az
x + y + z Quotient.
Illustration :
1. Divide — 6 m A n -f 10 m 3 n 2 — 14 m 2 n 3 + 4 ran 4 by — 2 mn.
- 2 mn ) - 6 m 4 n + 10 w 3 w 2 - 14 m 2 n 3 + 4 wn 4
3 w 3 - 5 m% + 7 mn 2 -2n 3 Kesult.
Each term of the quotient is obtained by applying the principle of
Art. 81, for in the division of each term by the monomial divisor we have
a division of a monomial by a monomial.
Hence, to divide a polynomial by a monomial :
82. Divide separately each term of the polynomial by the
monomial and connect the terms of the resulting polynomial with
the proper signs.
Exercise 12
Obtain the quotient of :
1. 2. 3.
x )x-7xy 4 m )12m 2 -16m 8 - 2 q )8 a 3 b - 10 ac
4. 5.
- 9 a c)18ac-27a 3 c 3 a V-4a 3 + 2a 2
6. 7.
- 4 m) 8 m 4 - 1 2 m 3 + 8 m 2 - 4 m - 3 a 2 xz ) - 9 a 3 x V - 6 a W
8. 14 x 3 y — 21xy 3 by 7 xy.
9. 4c 3 d-2cd + 6cd 3 by -2cd
56 DIVISION. REVIEW
10. - 25 a?x 2 + 20 a 2 x 2 - 15 aV by - 5 a¥.
11. 5 a 3 6 - 15 a 2 b 2 + 30 ab 3 by 5 ao.
12. -36m 4 -48m 5 H-60m 6 -72m 7 + 84m 9 by -12m 3 .
DIVISION OF A POLYNOMIAL BY A POLYNOMIAL
By Art. 69 : (a + 6) (z + y) = ax + ay + 6x + 6y.
Then, by definition, ax + ay + 6x + fry may be considered a gri'vera divi-
dend, and x + y a given divisor. It remains to determine the correspond-
ing quotient. Both dividend and divisor are first arranged in order, a
step imperative in all algebraic divisions with polynomials. The x-term
is the selected letter for the arrangement. In algebraic divisions the
divisor is written at the right of the dividend for convenience in the later
steps of the process.
Dividend Divisor
ax -f- x = a ax + ay -+ bx + by (% + y
a(x-\-y)= ax-\-ay+ a + b Quotient, or Result.
bx -=- x = b +bx + by
b(x + y)= +bx+by
The steps of the operation are as follows :
(1) The expressions are arranged in the same order.
(2) The first term of the dividend, ax, is divided by the
first term of the divisor, x ; and the quotient, a, is written as
the first term of the quotient.
(3) The divisor, x + y, is multiplied by the first term of the
quotient, a, and the product, ax + ay, is subtracted from the
given dividend.
(4) The remainder, bx + by, is a new dividend, and the
process is repeated with this new dividend, the divisor always
being the first term of the given divisor.
The process applies the law of distribution, for
ax + ay + bx + by _ ax + ay . bx + by
x + y x + y x + y
= a(x + y) b(x + y)
x+y + x+y
DIVISION OF POLYNOMIAL BY POLYNOxMIAL 57
A simple comparison with a numerical process is frequently
an aid to beginners.
Numerical Illustration:
Algebraic Illustration :
12)156(13 100 4- 50 + 6(10 + 2
12 100 + 20 10 + 3
36 30 + 6
36 30 + 6
a 2 + 5 a + 6{a + 2
a 2 + 2 a a + 3 Quotient
3a + 6
3a + 6
Explanation : Explanation :
100 -r- 10 = 10, 1st term of quo. a 2 -r- a = a, 1st term of quo.
10(10 + 2) = 100 + 20. (Subtract.) a (a + 2) = a 2 + 2 a. (Subtract.)
100 + 50 + 6 - (100 + 20) = a 2 + oa + 6-(a 2 + 2c) =
30 + 6, the new dividend. 3 a + 6, the new dividend.
30 -=- 10 = 3, 2d term of quo. 3 a + a = 3, 2d term of quo.
3(10 + 2) = 30 + 6. (Subtract.) 3(a + 2) = 3 a + 6. (Subtract.)
30 + 6 - (30 + 6) = 0. 3 a + 6 - (3 a + 6) = 0.
Hence, the quotient is 10 + 3. Hence, the quotient is a + 3.
We have, from these illustrations and from the general
principle, the following process for the division of a polyno-
mial by a polynomial :
83. Arrange both dividend and divisor in the same order of
some common letter. Divide the first term of the dividend by the
first term of the divisor, and write the result as the first term of
the quotient.
Multiply the whole divisor by the first term of the quotient just
obtained, and subtract the product from the dividend.
Regard the remainder as a new dividend and proceed in the
same manner as before.
Illustration : »
1. Divide a 4 + a 3 -5 a 2 + 13 a- 6 by a 2 -2a + 3.
Dividend
a*+ a 3
- 5 a 2 + 13 a -
-6 (a 2
-2a
+ 3
Divisor
a 4 - 2 a 3
+ 3«2
a 2
+ 3a
-2
Quotient
+ 3 a 3
- 8 a 2 + 13 a
+ 3a 3
-6a 2 + 9 a
- 2 a 2 + 4a-
-2«2 + 4 a _
-6
-6
58 . DIVISION. REVIEW
Explanation :
(1) a 4 -j- a 2 — a 2 , the first term of the quotient.
(2) Multiplying (a 2 - 2 a + 3) by a 2 , we obtain a* - 2 a 8 + 3 a 2 , which
product is subtracted from the given dividend. The remainder, which
must have three terms, is 3 a z — 8 a 2 + 13 a, and this remainder is the
new dividend.
(3) 3 a 8 ■+■ a 2 = 3 a, the second term of the quotient.
(4) Multiplying (a 2 - 2 a + 3) by 3 a, we obtain 3 a 8 - 6 a 2 + 9 a,
which product is subtracted from the last dividend. The remainder,
with the last term of the given dividend, is — 2a 2 + 4a — 6, the new
dividend.
(5) — 2 a 2 -*- a 2 = — 2, the third term of the quotient.
(6) Multiplying (a 2 -2a + 3) by - 2, we obtain - 2 a 2 -f 4 a - 6,
which, subtracted from the last dividend, gives a remainder of 0, com-
pleting the division.
2. Divide 7x 2 -x A -x s -\-2x 5 + 2-9xhj3x-2x 2 -2 + x s .
Arranging both dividend and divisor in descending powers
of x:
Dividend
2x 5 - a*-
X* +
7s 2 -
-9s + 2(a; 8 -
- 2 a 2 + 3 a - 2
Divisor
2x 5 -4x i + 6x 8 -
4x 2
2 a 5
+ 3a; — 1
Quotient
+ 3**-
-7z 3 + llx 2 -
-9x
+ 3z*-
-6z 8 +
9x 2 -
-6x
■ SC 3 +
2x 2 -
-3x + 2
. cc 8 +
2z 2 -
-3x + 2
In certain expressions the intermediate powers of the terms
of the dividend appear during the process. In such cases order
must be observed in subtractions.
3. Divide x 3 — y 3 by x — y. 4. Divide a 4 — 16 by a + 2.
X * _ yS (X-y
a* -16(a + 2
x*-x 2 y* x 2 + xy + y 2
a* + 2 a 3 a 8 - 2 a 2 + 4 a -
-8
+ x 2 y*
-2 a 8
+ x 2 y - xy 2
- 2 a 8 - 4 a 2
+ xy 2 - y*
+ 4 a 2
+ xy 2 - y*
+ 4 a 2 + 8 a
- 8 a - 16
-8a -16
DIVISION OF POLYNOMIAL BY POLYNOMIAL 59
5. Divide a 3 -3 a&c+-6 3 + c 3 by a + 6 + c.
a 3 - 3 abc + 6 3 + c 3 ( q + &+c '
a 3 + d 2 b + a 2 c (a 2 - ab - ac + 6 2 - be + c 2
- a 2 6 - a 2 c - 3 a5c + 6 3 + c 3
- <z 2 5 - ab 2 - abc
- a 2 c + aft 2 - 2 abc + 6 3 + c 8
— a 2 c — abc — ac 2
+ ab' 2 - abc + ac 2 + b 3 + c*
+ ab 2 + 6 3 + b 2 c
— abc + ac 2 — b 2 c + c 8
- abc - b 2 c- be 2
+ ac 2 + be 2 + c 8
+ ac 2 + 6c 2 + c 8
Note that in all new dividends the letter a is selected for arrangement.
84. Checking. The work in division may sometimes be
checked as shown in multiplication. Thus, in Ex. 1, if a = 2 :
Dividend a 4 + a 3 - 5 a 2 + 13 a - 6 _ 16 + 8-20+26-6 _ 24 _ g
Divisor a 2 - 2 a + 3 4-4 + 3 3
Quotient a 2 + 3a-2 = 4 + 6-2 = 8.
It is well to remember that this method of testing is not
always reliable.
Exercise 13
Divide : »
1. 2o 2 + 7a>+-6 by x + 2.
2. 2c 2 + 5c-12by c + 4.
3. 3m 2 +-llm-4by 3m-l.
4. 3z 2 + 14z+-15by* + 3.
5. 16 - 8 a +- a 2 by 4 - a.
6. 72 + m-m 2 by 8 + m.
7. 7a 2 6 2 + 123a6-54by7a6-3.
8. 20a 2 -47a6 + 216 2 by 4a-76.
9. 21 c 4 d 4 -f 36 c 2 dV + 15 a 4 by 3 c 2 cP + 3 jb 2 .
10. 10 a 6 6 4 + 23 a 3 6 2 cd? - 21 c 2 ^ by 10 a?b 2 - 7 cd 2 .
60 DIVISION. REVIEW
11. a 3 +3a 2 + 3a + lby a + 1.
12. c 3 ^-\-c 2 x 2 -3cx-6 by cx-2.
13. 15 — 8 mn + 6 m 2 n 2 — m 3 7i 3 by 5 — mn.
14. a 4 + 2 a 3 6 4- 2 a 2 6 2 + ab 3 - 6 6 4 by a? + ab- 2 b\
15. 2 ra 4 n 4 — m 3 n 3 x — 3 m 2 n¥ + 5 mna? 5 — 2 # 4
by 2 m 2 7i 2 — 3 mnx + 2x*.
16. 8 aW - 8 aVd - 4aW + 11 aW - 10 acd 4 + 3 d 5
by 2a 2 c 2 -3aca , + a*.
17. 9 a 2 - a 4 - 16 a- a 3 + 6 a 5 + 3 by 5 a -2 a 2 - 1 + 3 a 3 .
18. 16a; 6 + l-4a 2 -4a; 4 by 2 a 2 - 1 + 4 ^-2 a.
19. 5m 2 -6-llm 4 + 6m 6 -5m 8 + 2ra 10 by 2 + 2m 4 -m 2 .
20.5 mV - 13 m 6 ?i + 6 m 3 n 4 + 3 m¥ + 6mn 6 + 6m 7
4- 8 ti 7 — 15 ra 4 7i 3 by 3 m 4 — mti 3 — 2 w 3 ti -2n 4 + m 2 n 2 .
21. m 2 — w 2 by m - w. 24. ra 4 — ti 4 by ra + ti.
22. m 2 -7i 2 by ra + 7i. 25. 8 a 3 -27 by 2 a-3.
23. m 3 — n 3 by m — n. 26. m 6 — 32 n 5 by m — 2 w.
27. 81 -w 4 by 3 +%.
28. 27 + a 6 by 3 + x 2 .
29. a^-27^by a;-32/ 2 .
30. 27 a,- 3 + 64 y 5 by 3a + 4?/ 2 .
31. 16ra 4 n 4 -81aV 6 by 2m7i + 3cey.
32. m 2 + 2 mn + ti 2 — x 2 by m + n + x.
33. c 2 — 4c + 4 — d 3 by c-2 + d
34. m 2 + 6m + 9— 25 a 4 by m + S-Bx 2 .
35. 16 m 2 n 2 + 40 m7iy + 25 2/ 2 - 81 by 4ra7i + 5y + 9.
36. a 2 + m 2 + a? 2 + 2 am + 2 asc + 2 m# by a + m + aj.
37. 9m 2 n 2 + 4 : x 2 z 2 + 25-12mnxz-30mn + 20xz
by 3 m» — 2 o& — 5.
38. m 3 + 7i 8 +p 3 — 3m7ip by m + n+p.
39. a 6 6 6 -2a 3 6 3 + lby a 2 6 2 -2a6 + l.
LITERAL EXPONENTS AND DETACHED COEFFICIENTS 61
85. Division with Literal Exponents. (See Art. 72.)
Illustration :
Divide 8 of 1 * 4 - 18 af n+3 - 13 x m+2 + 9 a^ 1 +- 2 x™
by 4 x m + x" 1 - 1 — 2 z m - 2 .
8 x m+i - 18 sc m + 3 — 13 x m+ ' 2 + 9 x w+1 + 2x m (4 z OT + x m-1 - 2 x m ~ 2
8 x m+4_|_ 2x n> + 3 - 4x m + 2 2 x 4 - 5 x s - x 2 Result.
— 20x w + 3 — 9x m + 2 + 9x m+1
-20x" t + 3 - 5x m + 2 +l0x m + 1
— 4x m + 2 — x m+1 + 2x m
— 4 a? n + 2 — x m+1 4- 2 x m
The exponent of cc in the first term of the quotient is : (m + 4) — m =
m + 4 — ra = 4.
Exercise 14
Divide :
1. ^ + 3^-18 by x m -S.
2. ar ?n -4ar Jn -20a; n +-3by x n + 3.
3. Sx 4m -Sx Zm -10x 2m -x m + lbj 3x^-^30^ — 1.
4. x^-S by » n -2.
5. x^ — y*" by x m + 2/ n .
6. 16 x m+1 - 46 a; m+2 + 39 x m+3 - 9 af +4 by 8 a; M+1 - 3 a w+2 .
7. # n+4 +- a n+3 - 4 z n + 2 + 5 x n+l - 3 # n by x n+2 + 2 a n+1 - 3 af .
8. 6a^+ 6 -a;^+ 6 +-4a^ +6 -5a^ w+6 -a; m+6 — 15 a; 6
by 2 a^ m — x m + 3.
86. Division with Detached Coefficients. The same process
described in Art. 73 is of advantage in division, as two simple
cases will illustrate.
1. Divide x*- 3 ar 5 - 36 x 2 - 71 x - 21 by x 2 - 8 x- 3.
Detaching the coefficients and dividing, we obtain :
1 _ 3 _ 36 _ 71 - 21 ( 1-8-3
1-8-3 1 + 5 + 7 Coefficients.
+ 5_33_71
+ 5-40-15
+ 7-56-21 z 2 + 5 Z + 7. Result.
+ 7-56-21
62 DIVISION. REVIEW
If occurs as a coefficient either in the given expressions or
in the process, the same provision is made as in multiplication.
2. Divide # 5 + 2 a^+^a 3 - 31 a+-9 ^+-15 by 3-2 x-x 2 .
Arranging the expressions in ascending powers of x and
detaching coefficients, we have :
15
-31 + 9 + 4+2 + 1 (3-
-2-1
15
-10-5 5 -
-7 + 0-
5-7*
1 Coefficients.
-21 + 14 + 4
-21 + 14 + 7
+ 0-3+2+1
-3+2+1
— X s . Result.
For practice in division with detached coefficients use Ex. 13,
11 to 20, inc.
GENERAL REVIEW
Exercise 15
1. Show that (x-l) 2 -(x-2)=l + (x-l)(x-2).
2. Simplify(a+l) 2 -(a+-l)(a-l)-[a(2-a)-(2a-l)].
3. Prove that (a + m)(a — m) +- (m -f- 1) (m — 1)
+ (l-a)(l+a) = 0.
4. Divide4a) 6 -2^-a?-2^-2a;-lby2^-£ 2 -a;-l.
5. Add the quotient of (a 3 — 1) -f- (x — 1) to that of
(tf-2x + l) + (x-l).
6. What is the coefficient of ac in the simplified form of
(ac+-3) 2 -3ac(ac-l)?
7. Simplify (m + l)(m + 3)(m + 5) -(m-l)(m-3)(ra-5).
8. Show that (x +- y +- z — 1) (x +- y — z +- 1) — 4 (xy +- z)
+ (a-2/+-z + l)(l-a; + 2/ + z)=:0.
9. Divide 2 x 2 — 3 xy—5 xz — 2 y 2 — 5yz—Sz 2 by 2x + y + z.
10. A certain product is 6 a 4 +- 4 a 3 # — 9 a 2 y 2 — 3 ay s +- 2 ?/ 4 ,
and the multiplier 2 a 2 -{-2 ay — y 2 . Find the multiplicand.
11. Simplify 3[ap — 2 joj — 3(2 a? — 3 aj + 7){].
GENERAL REVIEW 63
12. Show that (1 - 3 x + x 2 ) 2 + x (1 - x) (2 - x) (3 - x) - 1 = .
13. Simplify 2 x 2 -3-(3 x -}-3 x 2 )-x(x 2 -3)-(x + 1)(2 - x")
and subtract the result from 5 — 2 x.
14. Simplify (a + 4) (a + 3) (a + 2) - (a + 3) (a + 2) (a + 1)
_( a +2)(a+l)-(a + l).
15. Prove that (1 + c 2 ) (1 4- a) 2 - 2 (1 - be) (a - c)
= (l + c) 2 (l+a 2 ).
16. Subtract a + 3 from the square of a + 2, and multiply
the result by the quotient obtained when a 5 — 1 is divided by
a A + a z +a 2 + a + l.
17. Simplify a — x— [3 a— (x— a)] + [(2 x — 3 a) — (a— 2 a)].
18. Divide c?-3cd+cP + l by c 2 - cd -c + d 2 -a' + l.
19. Find the continued product of a 2 — ab -f 6 2 , a 2 + a& -f- 6 2 ,
anda 4 -a 2 6 2 +6 4 .
20. Simplify (a¥ — a¥ -f aa — 1) (ax + 1)
- (aV + 1) (a» + 1) (ax - 1).
21. Multiply 8 a 3 — 27 by a + 2 and divide the product by
2a-3.
22. Divide c + 36c 5 -18c 2 -73c 3 + 12 by _5c + 4-6c 8 .
23. Simplify [(x 2 + 3x + 2)(x 2 -9)]-*- [(* + 3)(^ -«- 6)].
24. Show that a^ + ^+l — 3a#— (1 — ajy)— ?/(# 2 - «) —
25. Divide 82 m 4 n 4 + 40 - 67 mV + 18 mV - 45 mV by
3 m 4 n 4 — 4 m 2 n 2 + 5.
26/ What must be the value of m in order that x 2 + 18 x
+ m may be exactly divisible by x -f- 4 ?
27. Show that 2(4 + o^ + a 2 -ax-2a-2x) = (2-a) 2 +
(a-2) 2 + (x-a)\
28. What must be the value of m + nm order that x* + 3 X s
+ 2 x 2 + Tnx + n may be exactly divisible by x 2 + 2 x + 1?
29. By how much does (aV -f- 3 ax + 2) 2 exceed 2(3 a 3 x 3 +
2a*s* + 6aa>)?
CHAPTER VI
THE LINEAR EQUATION. THE PROBLEM
87. An equation is a statement that two numbers or two
expressions are equal.
Thus, 3x + 5 = x + 7.
88. The expression at the left of the sign of equality is the
left member (or first member), and the expression at the right,
the riglft member (or second member) of the equation.
89. An equation is a conditional equation if its members are
equal for particular values of the unknown quantity.
Thus, 3x + 5 = x + 7isan equation only when the value of x is 1.
90. An equation is an identical equation when i*s members
are equal for any and all values of the unknown quantity.
Thus, x 2 — l = (x+l)(x— 1) is an equation for any value of x whatsoever.
A conditional equation is usually referred to as an equation ; an iden-
tical equation, as an identity.
91. To solve an equation is to obtain the value of the un-
known number that will, .when substituted for that unknown
number, make the members of the equation equal.
92. The value found to make the members of an equation
equal, or to satisfy the equation, is a root of the equation. A
root of an equation when substituted for the unknown quan-
tity reduces the original equation of an identity.
64
AXIOMS 65
93. A linear or simple equation is an equation which, when
reduced to its simplest form, has no. power of the unknown
quantity higher than the first power. Thus :
5 x = 15 is a linear or simple equation in x.
7 y — 35 is a linear or simple equation in y.
3aj + 2 = 2aj + 7isa linear equation in x, but is not reduced in form.
(x + 5) 2 =# 2 + 7a; + 6isa linear equation in x, for, when simplified,
the resulting equation will have only the first power of x.
While the final letters, x, y, and z, are most commonly used
for representing unknown quantities in equations, any other
letters may and will be used in later practice.
94. The solution of equations is based upon the truths
known as
»
AXIOMS
1. If equals are added to equals, the sums are equal.
2. If equals are subtracted from equals, the remainders are
equal.
3. If equals are multiplied by equals, the products are equal.
4. If equals are divided by equals, the quotients are equal.
5. If two quantities are equal to the same quantity, they are
equal to each other.
In general, these axioms may be illustrated as follows.
Given the equation A = B.
By Axiom 1 A = B By Axiom 2 A = B
Add C, C= C Subtract C, C= C
A+ C = B+ C A- C = B- C
Or, briefly :
95. The same number may be added to, or subtracted from,
both members of an equation.
By Axiom 3 A = B By Axiom 4 A-B
Multiply by C, C= G Divide by C, C= C
AC=BC 4=B
C C
SOM. EL. ALG. — 5
66 THE LINEAR EQUATION. THE PROBLEM
Or, briefly :
96. Both members of an equation may be multiplied, or
divided, by the same number.
By Axiom 5 If A = B and B *= D ; we have, A — D.
THE TRANSPOSITION OF TERMS
97. Most equations are given in such a form that the known
and the unknown terms occur together in both members.
Transposition is the process of changing the form of an equa-
tion so that the unknown terms shall all be in one member,
usually the left, and the known terms all in the other. The
process is based on Art. 95.
Given the general equation, ax — c = bx-\- d
By Axiom 1, adding c = c
(1)
(2)
(3)
ax =bx + c + d
By Axiom 2, subtracting bx = bx
ax — bx — c + d
■
Compare carefully (1) and (3).
In (3) we find c in the right member with its sign changed
from — to +.
In (3) we find bx in the left member with its sign changed
from + to — .
In general:
98. Any term in an equation may be transposed from one
member to the other member if its sign is changed.
As a direct consequence of the use of the axioms, we have :
(1) The same term with the same sign in both members of an
equation may be discarded.
Given the equation, Sx + a — n = 2x-\-a + m.
Whence, Sx — n = 2 x + m.
THE TRANSPOSITION OF TERMS 67
(2) The sign y of every term in an equation may be changed
without destroying the equality.
Given the equation, — 5x + m = b — a.
Multiplying by — 1, 5x — m = a — b.
The sign of a root in a solution depends upon the law of
\ signs for division. Thus :
5 a: = 20 5x=-20 -5a = 20 -5 a; =-20
x = 4: $e=— 4 x= — 4 se = 4
In general :
99. When both members of an equation are reduced to simplest
form, lilce signs in both members give a positive root, unlike signs
a negative root. #
If the coefficient of the unknown quantity in a simplified
equation is not exactly contained in the known quantity, the
root is a fraction ; and if, in a simplified equation, the member
containing the known quantities reduces to zero, the root is
zero.
That is: If 3 a = 5 - 4 x = 7 And if 5x = -9s =
x = f x= -£ x = x =
Oral Drill
Give orally the roots of the following :
1. 5x = S0. 6. 6z = -36. 11. 8aj = 7.
2. 7z = 42. 7. -7x = 21. 12. 7y = 13.
3. 4a = 28. 8. -8x = -56. 13. -5a=16.
4. 3y = -lS. 9. -3a = -39. 14. 7#=0.
5. -3y=l$. 10. 52=3. 15. -3a = 0.
THE VERIFICATION OF LINEAR EQUATIONS
100. To substitute a *root in an equation is to replace the
unknown literal factor in each term by the value of the root
obtained.
68 THE LINEAR EQUATION. THE PROBLEM
101. To verify a root is to show that, by the substitution of
this value, the given equation reduces to an identity. The
verification of a root, as illustrated in the solutions following,
should always be made in the original equation.
THE GENERAL SOLUTION OF THE LINEAR EQUATION
Illustrations :
1. Solve 5x-4:= 3a + 12.
5x-4 = 3x + 12.
Transposing 3 x to the left member and — 4 to the right member,
5<c-3x = 4 + 12.
Uniting terms, 2 x = 16.
Dividing both members by the coefficient of x,
x — 8, the root.
Verification :
In the original equation, 5x-4 = 3e + 12.
Substitute 8 for as, 5 (8) - 4 = 3 (8) + 12.
40 - 4 = 24 + 12.
36 = 36.
Therefore, 8 is the correct value of the root, for, by substituting 8 for
x in the original equation, we obtain an identity.
2. Solve 5«-[3-(ss-2a>-l)J = - 10.
5 x - [3 - (x - 2 x ~ 1)] = - 10.
Removing parentheses, 5 x — 3 + x — 2 x + 1 = - 10.
Transposing, &x + x — 2x = + S — 1 — 10.
Uniting, 4 x = — 8.
Dividing both members by 4, x = — 2, the root.
Verification :
In the original equation, 5 x - [3 -- (x — 2 x - 1)] = - 10.
Substitute - 2 for x, 5( -2) - [3 - ( - 2 - 2(-2)-l)] = - 10.
Simplifying, * - 10 - 3 - 2 + 4 + 1 = - 10.
- 10 = - 10.
GENERAL SOLUTION OE THE LINEAR EQUATION 69
3. Solve (x + 3)(2x-5) = 2(x-2) 2 -2(x + l).
(x + 3) (2 x - 5) = 2 (x - 2) 2 - 2 (as + 1).
Multiplying, 2 x 2 + X - 15 = 2 (x 2 - 4 x + 4) - (2 x + 2).
Removing parentheses, 2x 2 + x — 15 = 2 x 2 — 8x + 8 — 2x — 2.
Discarding x 2 -terms and transposing,
x + 8x + 2x = 15 + 8-2.
Uniting, 11 x = 21.
x = ff , the root.
Verification :
Substituting \ \ for x in the original equation,
(H + 3)(£i-5)=2(fi-2) 2 -2(fi + l).
(ff)(-i!)=2(-T 1 T) 2 -2(!f).
-m = tIt - Hf
4.' Solve .3 25 - (x - 3) 2 = 3.25 - (x - 3.5) (a + 2.1).
.3 x - (x - 3) 2 = 3.25 - (x - 3.5) (x + 2.1).
Multiplying, .3 x - (x 2 - 6 x + 9) = 3.25 - (x 2 - 1.4 x - 7.35).
Removing ( ), .3 x - x 2 + 6 x - 9 = 3.25 - x 2 + 1.4 x + 7.35.
Transposing, .3 x + 6 x - 1.4 x = 9 + 3.25 + 7.35.
Uniting, 4.9 x = 19.6.
Dividing by 4.9, x =s 4, the root.
Verification :
Substituting 4 for x in the original equation,
.3 (4) - (4 - 3) 2 = 3.25 - (4 - 3.5)(4 + 2.1).
1.2 - 1 = 3.25 - (.5) (6.1).
.2 = 3.25 - 3.05.
.2 = .2.
From the foregoing we may state the general method for
the solution of linear equations:
102. Perform all indicated multiplications and remove all
parentheses.
Transpose the terms containing the unknown quantity to one
member y and all known terms to the other member of the equation.
Collect the terms in each member.
Divide both members by the coefficient of the unknown quantity.
70 THE LINEAR EQUATION. THE PROBLEM
Exercise 16
Find and verify the roots of the following :
1. 4a; + 5 = 3aj + 9. 6. 7 -2x = 9 x — 92.
2. 6 x -j- 7 = 5 x— 11. 7. 5a — l = 3x + l.
3. 7x-2-5x-12 = 0. 8. 3# + 7-a=-3.
4. 8a-12 = 3a; + 8. 9. - 7 a- 5 = - 5 + 2x.
5. 4« + 5 = x- 28. 10. 12a>-7 — 14a; = 12a>.
11. 5a>- (2 a + 3) = 12.
12. (2a-l)-15 = 4-(5-7a).
13. 13-(2a> + ll) = 6-(« + l).
14. 3«-2-(l-3a>) = 0.
15. 12-2(a>-5) + [3aj-(2-aj)] = l.
16. ll-(2«-3)--5=-3»-(5-7a>).
17. (x+l)(x + 3) = (x-2)(x-B).
18. (2a-3)(x-7) = (a?-l)(a + 4) + a; 2 .
19. (aj-3) 2 + 2(a-4) 2 -3(a-5)(a; + 5) = 7.
20. 5(a-l) 2 -3(a-2) 2 ==(2a-l)(3 + a;)-6.
21. 4[3«-2(aj 8 + l)] = 7-4s(2a>-16).
22. - [2 (x - 3)(a> - 5) - (a + 7)(3 -»)] = - 3 (a 2 + 3).
23. (a,- + 2) 3 -(x-l) 3 -(3a; + l)(3a;-4)=0.
24. 2.7 x - (11 - 1.3 x) - 6.7 x = .62 + .4 a - 11.
25. 0.007 x - 2 (.0035 g + .07) = .017- (.14 - .85 x).
THE SOLUTION OP PROBLEMS
103. A problem is a question to be solved.
In general, a problem is a statement of conditions involving
an unknown number or numbers. We seek the value of that
unknown number, and by assuming a literal symbol for the
unknown we are able to state the given conditions in terms of
that unknown.
LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 71
104. The solution of a problem is (1) a translation of the
language into the symbol-expression of algebra by means of
an assumed value for the unknown ; (2) the translation of the
given conditions into equations ; and (3) the finding of the
root of the derived equation. The result of a solution should
always be verified in the given conditions.
LITERAL SYMBOLS FOR UNKNOWN QUANTITIES
105. A few simple statements will illustrate the ease with
which the conditions of a problem are translated, or expressed,
in algebraic symbols. It will be seen that in each case we
first assume a value for an unknown quantity upon which the
statements seem to depend, and then write expressions for the
different statements.
Illustrations :
1. If a denotes a certain number, write an expression for 10
more than x.
Since we may assume x = the given number,
By adding 10, x + 10 = the required number.
And, fulfilling the given condition, we have written a number greater
than the given number by 10 ; the words "more than" being translated
by the symbol of addition, + .
2. In a certain exercise John solved twice as many examples
as William. Write an expression for the total number of ex-
amples both together solved.
We assume that x = the number that William solved.
Then 2 x = the number that John solved.
By addition, 3 x = the number both together solved.
Here we assumed a literal symbol for that particular number upon
which the problem seemed to depend ; that is, the number of examples
solved by William. A simple application of multiplication and addition
completes the translation of the conditions.
72 THE LINEAR EQUATION. THE PROBLEM
3. A man is y years of age. (a) How old was he 10 years
ago ? (b) How old will he be in z years ?
(a) Since y = the number of years in his present age,
By subtraction, y — 10 = his age (in years) 10 years ago.
And "years ago," a decrease, is translated by the sign — .
(b) Since y = the number of years in his present age,
By addition, y + z — his age (in years) z years from now.
And " in " referring to the future is an increase translated by the
sign +.
4. m yards of cloth cost $ 3 per yard, and n yards of silk,
$ 5 per yard. Write an expression for the total cost of both
in dollars.
At 1 3 per yard m yards of cloth cost (3 x m) dollars = 3m dollars.
At $ 5 per yard n yards of silk cost (5 x n) dollars =5w dollars.
Therefore, adding, (3 ra + 5 ii) dollars equals the total cost of both.
5. Write three consecutive numbers, the least of the three
being x.
The difference between any two consecutive numbers is 1.
Therefore, when x = the first and least number,
x + 1 = the second number,
x + 2 = the third number.
6. Write three consecutive even numbers, the least being m.
The difference between consecutive even numbers is 2.
Therefore, as above,
m, m + 2, and m + 4, are the required numbers.
7. Write three consecutive odd numbers, the greatest being
a ; and write an expression for their sum.
i The difference between consecutive odd numbers is 2.
Therefore, a = the first and greatest number,
a — 2 = the second number,
a — 4 = the third number.
By addition, 3 a — = the sum of the three numbers,
LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 73
8. A boy has x dimes and y nickels and spends 10 cents.
Write an expression for the amount he has remaining.
Problems involving value must be so stated that different denominations
are all expressed in one and the same denomination.
Hence, the value of x dimes = (10 • x) cents = 10 a; cents.
Hence, the value of y nickels = (5 • y) cents = 5 y cents.
Therefore, he had at first, (10 x -f 5 y) cents.
Subtracting the 10 cents he spent, we have for an expression of his
final amount, (10 x + 5 y) — 10 cents. Result.
Oral Drill .
1. If x denotes a certain number, give an expression for 15
less than x.
2. If y denotes a certain number, give an expression for 12
less than the double of y.
.3. If x denotes the number of square inches in a certain
surface, how many square inches are there in m similar surfaces ?
4. John has x marbles, William y marbles, and Charles z
marbles. What is the expression for the total number all
together have ?
5. A boy had m marbles and lost 1 of them. How many
had he left?
6. A boy earned x cents, found three times as many, and
spent c cents. How many cents had he finally ?
7. A boy solved n examples and his sister solved 5 more
than twice as many. How many examples did the sister
solve ?
8. John caught m trout and his brother caught 3 less than
three times as many. How many did both together catch ?
9. If William solves x examples, how many examples must
John solve so that both together shall solve y examples ?
74 THE LINEAR EQUATION. THE PROBLEM
10. Three men together buy a field. B pays twice as much
as A, C four times as much as B. If A pays d dollars, what
is the cost of the field ?
11. A horse cost y dollars, a harness x dollars, and a wagon
as much as the combined cost of a horse and two harnesses.
What did all three together cost ?
12. If a line 10 inches in length is increased by n inches,
what is the length of the new line ?
13. How much remains of a line m inches long if n inches
are cut from one end andp inches are cut from the other end ?
14. A rectangle is x inches long and y inches wide, and a
strip z inches wide is cut from one end. What is the area of
the part cut off and also of the part remaining ?
15. The sum of two numbers is 10 and the smaller number
is n. What is the larger number ?
16. The sum of three numbers is 50; one is x, and another
35. What is the expression for the third number ?
17. Name three consecutive numbers, the least of the three
being n.
18. Name three consecutive even numbers, the least number
being n.
19. Name three consecutive odd numbers, the least of the
three being m.
20. Name five consecutive numbers, the middle one being x.
21. If x is an even number, what is the next odd number
above x ?
22. If a; is an odd number, what is the next even number
above x ?
23. Name the three consecutive odd numbers below n, n
being an odd number. ,
24. Name the three consecutive odd numbers below n, n
being an even number.
LITERAL SYMBOLS FOR UNKNOWN QUANTITIES 75
25. What is the sum of the three consecutive even numbers
below x + 1, x + 1 being odd ?
26. If x is the middle one of three consecutive odd numbers,
what is the expression for. their indicated product?
27. If a man is x years old, how old will he be in a years ?
28. If a man is x years old now, how many years will pass
before he is a years old ?
29. If a man is y years old now, how old was he z years
ago?
30. What is the sum of the ages of a father and son if the
son is x years old arid the father m times as old as the son ?
31. A man is m times as old as his son and n times as old as
his daughter. If the daughter is x years old, what is the sum
of the ages of all three ?
32. In a period of c year's a man earns d dollars. If he
spends n dollars each one of the c years, what is the expression
for his final saving ?
33. How many cents are represented by x dimes ?
34. Express the condition that x dimes shall equal y nickels.
35. One dollar is lost from a purse that had contained x
dollars and y quarters. Write the expression for the amount
remaining in cents.
36. A man travels a miles an hour for h hours. What is the
total distance he travels ?
37. A boy rides x miles in a train, then y miles by boat,
and, finally, by automobile twice as far as he has already
traveled. Write the expression for the total number of miles
in his journey.
38. What is the expression for the statement that the
square of the difference of two numbers, a and b, is 2 less than
the sum of the squares of two other numbers, m and n ?
76 THE LINEAR EQUATION. THE PROBLEM
39. A room is x yards long and y yards wide. What is the
expression for its area in square yards ? in square feet ?
40. If a man is now x + 1 years old, how many years ago
was he 30 years old? How many years must pass before he
will be 50 years old ? When was he x — 1 years old ? When
will he be 2 a? years old ?
PROBLEMS LEADING TO LINEAR EQUATIONS
106. From the oral work j ust covered the student will see that
no general directions can be given for the statement^of problems.
The following suggestions will be of assistance in the state-
ment of all problems, and will serve as a general outline of the
method by which we attack the different types.
107. In stating a problem :
1. Study the problem to find that number whose value is re-
quired.
2. Represent this unknown number or quantity by any conven-
ient literal symbol.
3. The problem will state certain existing conditions or
relations. Express those conditions in terms of your literal
symbol.
4. Some statement in the problem will furnish a verbal equation.
Translate this verbal equation into algebraic expression by means
of your stated conditions. The following illustrations will
show the ease with which certain common words and phrases
may be translated into the common operations of algebra.
Illustrations :
1. The greater of two numbers is 3 more than the less, and
four times the less number exceeds twice the greater number
by 8. Find the two numbers.
PROBLEMS LEADING TO LINEAR EQUATIONS 77
Let x = the smaller number.
Then x + 3 = the larger number.
Now the word "exceeds," in this case, may be translated by the sign
" — ," and the word " by " may be translated by the sign " =."
Hence, we select the conditional sentence :
" four times the less exceeds twice the greater by 8,"
and translate :4 a; — 2(x + 3) =8.
Solving the equation 4x — 2(x + 3) = 8,
x = 7, the smaller number.
Consequently, adding 3, x + 3 = 10, the larger number.
The results verify in the original condition, hence in the equation.
2. Find the number which, multiplied by 4, exceeds 40 by as
much as the number itself is less than 40.
Let x = the number.
Then 4 x = four times the number.
Translating the conditional sentence :
" number . . . multiplied by 4 exceeds 40 by as much as 40 exceeds the number,"
x X 4-40 = 40 - x.
Solving the equation, 4 x — 40 = 40 — as,
x = 16, the number.
Verification :
4(16) -
-40:
= 40-
-16.
64-
-40
= 40-
-16.
24:
= 24.
3. A father is four times as old as his son, but in 24 years
the father will be only twice as old as the son. What is the
present age of each ?
Let x = the number of years in the son's present age.
Then 4 x = the number of years in the father's present age.
Therefore, x + 24 = the son's age after 24 years.
4 x -f 24 = the father's age after 24 years.
Now, "... in 24 years the father will be only twice as old as the son."
Or, -24 + Ax = 2(« + 24).
Solving the equation, 24 + 4 x = 2(x + 24).
x = 12, the son's age now.
4 x = 48, the father's age now.
Verification :
In 24 years the father will be (48 + 24) = 72 years of age.
In 24 years the son will be (12 + 24) = 36 years of age.
78 THE LINEAR EQUATION. THE PROBLEM
The problems in the following exercise are at first classi-
fied in four groups involving only the simplest of commonly
occurring conditions.
Exercise 17
(I) Problems involving One Number.
1. Four times a certain number is 36. Find the number.
Let x = the required number.
From the problem 4 x = four times that number.
But the problem states that
36 = four times that number.
Hence, from our assumed condition and from the given condition, we
have two expressions, 4 x and 36, representing the same quantity.
Therefore, 4 x = 36.
x = 9, the required number.
Verification : 4(9) = 36.
36 = 36.
2. What is that number which, when decreased by 5, gives a
remainder of 19 ?
3. Three times a certain number is diminished by 7 and
the remainder is 11. What is the number ?
4. William has three times as many books as John, and
both together have 32 volumes. How many has each ?
5. If four times a certain number is added to five times the
same number, the sum is 36. What is the number?
6. Four times a certain number is subtracted from eleven
times the same number, and the remainder is 42. Find the
number.
7. I double a certain number and subtract 7 from the result,
and my remainder is 1 more than the original number.
What was the number ?
8. If a certain number is increased by 5, the sum is 8 less
than twice the original number. Find the number.
PROBLEMS LEADING TO LINEAR EQUATIONS 79
9. Twelve times a certain number is decreased by 5, and
the remainder is 15 more than seven times the original num-
ber. What was the number ?
10. Find that number which, if doubled, exceeds 60 by as
much as the number itself is less than 75.
11. What number is that which, if doubled and subtracted
from 50, gives a remainder 5 less than three times the origi-
nal number ?
(II) Problems involving Two or More Numbers.
12. The sum of two numbers is 24, and the greater number
is 3 more than twice the smaller number. Find the numbers.
Let x = the smaller number.
Then 24 — x = the larger number.
Now "the greater number is 3 more than twice the smaller,"
Hence,
24 - x = 3 + 2 x.
Solving,
x = 7, the smaller number.
Whence,
24 — 7 = 17, the larger number.
Verification :
24 - 7 = 3 + 2(7).
17 = 3 + 14.
17 = 17.
13. One number exceeds another number by 5, and their
sum is 49. Find the numbers.
14. One number is four times as large as a second number,
and their sum is 21 more than twice the smaller number.
Find the numbers.
15. The sum of three numbers is 108. The second number
is twice the first number, and the third is equal to twice the
sum of the first and second. Find the three numbers.
16. Find the three consecutive numbers whose sum is 54.
17. Find the three consecutive odd numbers whose sum is 39.
18. Find the five consecutive odd numbers whose sum shall
be equal to nine times the smallest of the numbers.
80 THE LINEAR EQUATION. THE PROBLEM
19. Find four consecutive odd numbers such that twice the
sum of the three smallest shall be 15 more than three times
the greatest one.
20. Divide 17 into two parts such that the smaller part plus
four times the larger part shall be 50.
(Hint : Let x = the smaller part • 17 — x = the larger part.)
21. Divide 64 into two parts such that three times the
smaller part added to twice the larger part shall be 158.
22. Divide 100 into two parts such that twice the larger
part shall be 50 more than three times the smaller part.
23. Divide 75 into two parts such that three times the
larger part decreased by 6 shall equal four times the smaller
part increased by 9.
(Ill) Problems involving the Element of Time.
24. A man is twice as old as his brother, but 5 years ago
he was three times as old. Find the present age of each.
Let x = the number of years in the brother's present age.
Then 2 x = the number of years in the present age of the man.
Now x — 5 = the brother's age 5 years ago.
And 2x — 5 = the man's age 5 years ago.
From the statement in the problem :
2x-5 = 3(x-5).
Solving, x = 10, the brother's age now.
2 x = 20, the man's age now.
25. A boy is 5 years older than his sister, and in 4 years
the sum of their ages will be 29 years. Find the present age
of each.
26. A man is twice as old as his son, but 10 years hence the
sum of their ages will be 83 years. What is the present age
of each ? i
27. Five years ago the sum of the ages of A and B was 50
years, but at the present time B is four times as old as A.
How old is each now ?
PROBLEMS LEADING TO ^LINEAR EQUATIONS 81
28. In 7 years the sum of the ages of A and B will be 26
years less than three times A's present age. If A is now three
times as old as B, find the age of each after 7 years.
29. In 7 years a 'boy will be twice as old as his brother,
and at the present time the sum of their ages is 13 years.
Find the present age of each.
30. A boy is three times as old as his sister, but in 4
years he will be only twice as old. What is the present age
of each ?
31. A young man is 23 years of age and his brother is 11
years old. How many years ago was the older brother three
times as old as the younger ?
32. A man 50 years old has a boy of 9 years. In how many
years will the father be three times as old as the son ?
33. The sum of the present ages of a man and his son is 60
years, and in 2 years the man will be three times as old as
the son. What will be the age of each when the sum of their
ages is 100 years ?
(IV) Problems involving the Element of Value.
34. A man pays a bill of $49 with five-dollar and two-dol-
lar bills, using the same number of each *kind. How many
bills of each kind are used ?
Let x = the number of bills of each kind.
Then 5 x = the value of the fives in dollars,
and 2x = the value of the twos in dollars.
Therefore, 7 x = 49.
From which x = 7, the number of bills of each kind.
35. Divide $100 among A, B, and C, so that B shall receive
three times as much as A, and C $20 more than A and B
together.
36. A has $ 16 less than B, and C has as many dollars as A
and B together. All three have $ 60. How many dollars has
each?
SOM. EL. ALG. — 6
82 THE LINEAR EQUATION. THE PROBLEM
37. A number of yards of cloth cost $ 3 per yard, and the
same number of yards of silk, $ 7 per yard. The cost of both
pieces was $ 100. How many yards were there in each piece ?
38. $41 was paid to 16 men for a day's work, a part of the
men receiving $2 per day and the other part $3 per day.
How many men worked at each rate ?
39. A boy has $42 in two-dollar bills and half dollars, and
there are three times as many coins as bills. How many has
he of each kind ?
40. $2.10 was paid for 8 dozen oranges, part costing 20 cents
a dozen and part costing 30 cents a dozen. How many dozen
were there in each lot ?
41. A merchant bought 50 postage-stamps, the lot being
made up of the five-cent and the two-cent denominations.
Twice the cost of the two-cent stamps was 48 cents more than
three times the cost of the five-cent stamps. How many of
each kind were bought, and what was the total amount paid
for them ?
(V) Miscellaneous Problems.
42. Find the two numbers whose sum is 70 and whose dif-
ference is 6.
43. A and B together have $90, and A has $12 more than
B. How many dollars has each?
44. One number exceeds another by 3, and the difference
between their squares is 51. What are the numbers ?
45. The difference between the ages of a father and son is
36 years, and the father is three times as old as the son. Find
the age of each.
46. Divide 70 into two parts such that ten times the smaller
part shall equal eight times the larger part.
47. A man divided $1500 among four sons, each receiving
$ 50 more than the next younger. How much did each receive ?
PROBLEMS LEADING TO LINEAR EQUATIONS 83
48. Divide 31 into two parts such that 1 less than eight
times the smaller part shall equal five times the greater part.
49. Ten times a certain number is as much above 77 as 43
is above five times the number. What is the number ?
50. The difference between the squares of two consecutive
even numbers is 52. Find the numbers.
51. The sum of two numbers is 16 and the difference of their
squares is 32. What are the numbers ?
52. One number is three times another, and the remainder
when the smaller is subtracted from 19 is the same as the re-
mainder when the larger is subtracted from 43. Find the two
numbers.
53. A man has three hours at his disposal and walks out
into the country at a rate of 4 miles an hour. How many miles
can he walk so that, by returning on a trolley car at the rate
of 12 miles an hour, he will return within just 3 hours ?
54. A walks over a certain road at a rate of 3 miles an hour.
Two hours after he leaves, B starts after him at a rate of 4
miles an hour. How many miles will A have gone when B
overtakes him ?
55. A and B are 60 miles apart and start at the same time
to travel towards each other. A travels 4 miles an hour and
B 5 miles an hour. In how many hours will they meet and
how far will each have traveled ?
56. A man walks 5 miles on a journey, rides a certain dis-
tance, and then takes an automobile for a distance four times
as great as he has already traveled. In all he travels 75 miles.
How far does he go in the automobile ?
57. How can you pay a bill of $5.95 with the same number
of coins of each kind, using only dimes and quarters?
58. The sum of the ages of a father and son is 96 years ;
but if the son's age is trebled, it will be 8 years greater than
the father's age. How old is each ?
CHAPTER VII
SUBSTITUTION
108. Substitution is the process of replacing literal factors
in algebraic terms by numerical or by other literal values.
Illustrations :
1. If a = 5 and 6 = 7:
(« + &) =
(5 + 7)
12. Result.
2.
If a
= 4 and b= -2:
(2a-a 2 6) =
. [2-4 -42 (-2)]
8 -16 (-2)
8 + 32
40. Result.
3.
If a;
= 4 m, y — 3 m, and
Z= — 5m:
(jc + y - z) =
(4 m + 3 m + 5 m)
12 m. Result.
4. With the same values for x, y, and % as in (3) :
(aj + 2/2)(2a;-^-2«2) = [4m+(3m)2][2(4m)-3m-2(-5m)2]
= (4m + 9w 2 )[8ra-3m-2(25m 2 )]
= (4 m + 9 m 2 ') (5 m - 50 m 2 )
= (20 m 2 - 155 m 8 - 450 wi 4 ). Result.
From these illustrations we state the general method for
substitution :
109. Replace the literal factors of the terms of the given ex-
pression by their respective given values. Perform all indicated
operations and simplify the result.
84
SUBSTITUTION OF NUMERICAL VALUES 85
I. SUBSTITUTION OF NUMERICAL VALUES
Exercise 18
Find the numerical values of the following when a = 4, 6 = 3,
c = 4, and d = 1.
1. a+b + c. 7. ab — 3 be + 5 ad.
2. a-26 + 5c. 8. 2ab-3cd + 2bd.
3. 46-3c + 2a\ 9. 5ad , -36a , + 8cd'.
4. 7a— a" + 9c — 6. 10. ab—(bc—cd).
5. 10a + 6-(3c-d). 11. 8a6c-5 6ca' + 2aca\
6. 46 — [3 c — (2a + a")]. 12. acd* — (abc — bed — acd).
13. acd-2 6cd-3(a6-c(f).
14. 2a6- (3cd + 2a — a6 + 6c — bd).
15. 6c-[-6d-(a6-cd") + (a6 — 6d*)].
16. Simplify a(a+ c) —c(c — a) when a = 4 and c = 2.
17. Simplify (a + 6) 2 - (a + 6) (a - b) - (a - 6) 2 when a = 5
and 6 = 4.
18. Simplify 10 (a -f l) 2 - (3 a - 2) 2 when a = -2.
19. Simplify 4 (2 a; -f- y) 2 — S(x +y) (x — y) when x = — 2 and
2/ = 0.
20. Simplify 3(ra -2x) 2 - 2(ra + 2 »)(m + 3 x) -f 4 x when
m = — 5 and a; = 2.
21. Simplify d — 2(c + d)(c - 2 d) - 3(e - d) when c =2 and
a" =-2.
22. Simplify a(a + 6) - b(a - 6) 2 - (a + &) 3 when a = - 3
and 6 = — 2.
23. Simplify (3 a + m)(a- 5 m) +[9 a 2 - |ra-a(2m-9a) j]
when a = and m = 10.
86 SUBSTITUTION
24. Simplify 7 a(a + 1) - (2 + a) (3 a - 5) - 3 a(a - 1) when
a = 0.
25. Simplify (3 m- 2) 2 +(w + n-l)»-(2w- 1) (m + 1)
when m = n= — 3.
II. SUBSTITUTION OF LITERAL VALUES
When literal values are substituted, the results will be in
terms of those literal values.
Exercise 19
Simplify :
1. (a — x) 2 -\-(a — 2x) 2 when a = 2x.
2. (a — 3 x — x 2 ) when x = 2a.
3. (a + x)(a — 2x) 2 — (a — x) 2 when a= — x.
4. (a? + n)(x + ft — 1) when x = m and n = — m.
5. (2 a — 3 c -\- x)(x — 2 a) — 2 acx when or = x and c = 0.
6. 5 ac — (a + c)(a — 4 c) -f- (a — c) 2 when c = 0.
7. 2(ra — a) — 2 (m + 3 x)(m — 2 x) — 3 mx when ra = a and
x = — 2 a.
8. (2a>+l)*-3a?-4a>(a!-5) whena= -2a.
9. (3x-l) 2 -(3x+l)(2x-5) whenx = ab.
10. (3a 2 -2a + l)-(3a;-2) 2 when »= — mnl
11. (m + a?)(m — #) — (m — a;) 2 + 2 m when m= —x.
12. (a-26 + 8) 2 -x(a; + 2a)-(a-2&) 3 -l when a = 2&
and x = 0.
13. (6 ar> - 11 o?y - 10 y 2 ) - 6(a^ — 2xy) + 11 2/ 2 when aj = m
and y——m.
14. (a? — a)(aj + a — y) — (x — y) (x -\» a) a — (x — a) 2 when
x= — m and a = 0.
15. c(c + c7 + j) — (c — d-f-l)(c+d— 1) — cd— 1 whenc=d=0.
THE USE OF FORMULAS 87
HI. THE USE OF FORMULAS
110. A formula is an algebraic expression for a general
principle. For example : If the altitude
of a triangle is represented by a, the base
by b, and the area by S, we have the
general expression for its area in the
formula
o ab
Given the values of a and b for a par-
ticular triangle, we obtain its area by substituting those values
in this general formula and simplifying.
A few common formulas will illustrate the value of this brief
method of expression.
(1) The Formula for Simple Interest.
By arithmetic : The interest (I) on a given principal at a
given rate for a given time equals principal x rate x time. Ex-
pressed as a formula:
If p = the principal expressed in dollars,
r = the rate of interest expressed decimally,
t = the time expressed in years,
Then / = prt is the general formula for simple interest.
(2) The Formula for Compound Interest. (Interest com-
pounded annually.)
If p = the principal expressed in dollars,
r = the rate of interest expressed decimally,
n = the number of years in the interest period,
Then /=p(l +r) n — p is the general formula for interest com-
pounded annually.
(3) The Formula for the Transformation of Temperatures.
Both the standard thermometers, the Fahrenheit and the Centi-
grade, are in everyday use in physical investigations, and the
formula given below is used to change Fahrenheit readings to
the Centigrade scale.
88 SUBSTITUTION
If F = the given reading from a Fahrenheit scale,
C = the required equivalent reading on the Centigrade scale,
Then C sa $ (F — 32) is the formula for temperature transformation.
Exercise 20
1. What is the area of a triangle having a base of 14 inches
and an altitude of 9 inches ?
2. Find the simple interest on $ 1500 for 12 years at 5 per
cent.
3. What is the interest on $6200 for 3 years 7 months
10 days at 4.5 per cent ?
4. Find the interest, compounded annually, on $ 1200 for
4 years at 5 per cent.
5. On a Fahrenheit thermometer a reading of 50° is taken.
What is the equivalent reading on a Centigrade thermometer?
6. Find the circumference (C) of a circle, the radius (R)
being 5 feet, and the constant (it) in the formula 0=2 ttR
being 3.1416 + .
7. Find the area (S) of a circle whose radius (R) is 4 feet,
the constant (71-) in the formula S = irR 2 being 3.1416 +.
8. With the formula of Example 7, find the area of a cir-
cular pond whose diameter (D) equals 100 feet. (D — 2 R.)
9. If the diameter (D) of a sphere is 2 feet, what is the
volume (V) of the sphere from the formula, V— ? (tt =
3.1416 +.) 6
10. Find the last term (Z) in a series of numerical terms of
which the first term (a) is 3, the number of terms (n) is 8,
and the difference between the terms (d) is 2, the formula be-
ing l = a+ (n — l)d.
11. If a = 3, r = 2, and n m 5, what is the value of I in the
expression I = ar n_1 ?
12. If r = 5, s = 4, and n = 6, find the value of a in the ex-
pression a = (r — l)s -f- r n_1 .
CHAPTER VIII
SPECIAL CASES IN MULTIPLICATION AND DIVISION
MULTIPLICATION
The product of simple forms of binomials may often be
written without actual multiplication, the result being obtained
by observing certain laws seen to exist in the process of actual
multiplication. Three common cases are :
The square of the
sum of two quan-
tities,
a + b
a + b
a 2 + ab
+ ab + b 2
II
The square of the
difference of two
quantities,
a — b
a — b
a 2 - ab
- ab + b 2
III
The product of the
sum and difference
of two quantities.
a + b
a — b
a 2 + ab
-ab-b 2
a 2
F
a 2 + 2 ab + b 2 a 2 - 2 ab + b 2
Therefore, we may state, from (I), (II), and (III),
respectively :
111. The square of the sum of two quantities equals the square
of the first, jplus twice the product of the first by the second, plus
the square of the second.
112. The square of the difference of two quantities equals
the square of the first, minus twice the product of the first by the
second, plus the square of the second.
113. The product of the sum and difference of two quantities
equals the difference of their squares.
Ability to apply these principles rapidly is essential in all
later practice.
89
90 SPECIAL CASES IN MULTIPLICATION AND DIVISION
Oral Drill
Give orally the products of :
1. (a + m) 2 . 4. (a^+4) 2 . 7. (3a + 2) 2 .
2. (x + z) 2 . 5. (a + 3m) 2 . 8. (4a + 5) 2 .
3. (a; + 3) 2 . 6. (c+5d) 2 . 9. (5c+7) 2 .
10. (c-z) 2 . 13. (c-8) 2 . 16. (cd-4) 2 .
11. (&-4) 2 . 14. (3a-5) 2 . 17. (a 2 6 2 -6) 2 .
12. (m-5) 2 . 15. (7a-3) 2 . 18. (cd-Scx) 2 .
19. (a + a;)(a-aj). 27. (3 mn + 5 ma) 2 .
20. (m + y)(m-y). 28. (4c 2 d 2 -5) 2 .
21. (a + 4) (a- 4). 29. (2m 3 + 9)(2ra 3 -9).
22. (2a + l)(2a-l). 30. (5 xyz + 7 y 2 z 2 ) 2 .
23. (3a + 5)(3a-5). 31. (3a 4 -13)(3a; 4 + 13).
24. (7a + 10)(7a — 10). 32. {am 2 + xyz)(am 2 -xyz).
25. (5a-8 2/)(5a; + 8 2/). 33. (3c 5 + ll) 2 .
26. (3 a 2 + 5) (3 a 2 -5). 34. (7 m 6 -9 a 7 ) 2 .
IV. THE DIFFERENCE OF TWO SQUARES OBTAINED FROM TRINOMIALS
114. Many products of two trinomials may be so written
as to come under the binomial case of Art. 113. In such
multiplications we group two of the three terms in each quantity
so as to produce the same binomial expressions in each, the pa-
renthesis being treated as a single term. Three different cases
may occur :
(1) (2) (3)
(a + 6 + c)(a + b-c) (a + b + c)(a- b + c) (a + 6-c)(a -b + c)
The terms inclosed in parentheses must form the same binomial
in each expression.
THE DIFFERENCE OF TWO SQUARES 91
From 1. (a + 6 + c)(a + b - c) = [(a + b) + e][(a + 6) - c]
=s (a + &) 2 - c 2
= a 2 + 2 a& + 6 2 - c 2 . Result.
From 2. (a + b + c) (a - b + c) = [(a + c) + 6] [(a + c) - 6]
= (a + c) 2 - 6 2
= a 2 + 2 ac + c 2 - 6 2 . Result.
From 3. In this case only one term has the same sign in each ex-
pression, the a-term. Hence the last two terms of each expression ara
inclosed in parentheses, the parenthesis in one case being preceded by
the minus sign.
(a + b - c)(a - b + c) = [a + (6 - c)][a - (6 - c)]
= a 2 - (6 - c) 2
= a 2 - (6 2 - 2 6c + c 2 )
= a 2 - 6 2 + 2 be - c 2 . Result.
Exercise 21
Write by inspection the following products :
1. [(a + 0) + 4][(a+«)— 4]. 8. (ra + a + 2)(ra + a; — 2)
2. [(ra + a;) + 2][(m + a)— 2]. 9. (m — w + c)(ra — n — c).
3. [(c-2) + m][(c-2)-m]. 10. ( c 2 + c + l)(c 2 + c-l).
4. [(a 2 + l) + a][(a 2 + l)-a]. 11. (c-d + 2)(c-d-2).
5. [a -r-(d + a>)] [a — (d + a;)]. 12. (m + w — 4)(m — n + 4),
6. [d+(y-«)][<i-(y-«)3. 13. (^-2-^(^-2 + ^),
7. ( a+a . + y)(a + a;-^). 14> (x*j tX -2){x i -x + 2).
15. (aj 4 + ^ + l)(» 4 -^+l).
16. (m 4 -2m 2 + l)(m 4 + 2m 2 + l).
17. (x 4 -x 2 -6)(x A + x 2 -6).
18. [( c + d) + (m + l)][(c + d)-(m + l)].
19. [0 + x) - (y - z)] [(« + <*) + &- *)J
20. (t^-^-c-lX^ + ^ + c-l).
92 SPECIAL CASES IN MULTIPLICATION AND DIVISION
V. THE PRODUCT OF TWO BINOMIALS HAVING A COMMON
LITERAL TERM
By actual multiplication : a + 7
a + 5
a 2 + 7 a
+ 5a + 35
a 2 + 12 a + 35
Hence, in the product :
a? = a x a, the product of the given first terms.
+ 12a=(+7 + 5)a the product of the common literal term by the
sum of given last terms.
+ 35 = (+ 6) (+ 7) the product of the given last terms.
In like manner : (1 ) (x - 4) (x - 9) = z 2 - 13 x + 36.
x* = x-x. -13z=(-4-9)z. + 36=(-4)(-9).
(2) (e + 9)(z-3)=z 2 + 6z-27.
x 2 = x-x. +6x=(+9-3)x. -27=(+9)(-3).
(3) (w-15)(m+7)=m 2 -8m-105.
m a = OT-m. -8w=(-15 + 7)to. -105=(- 15)(+7)
In general : (x + a) (x + b) = « 2 + (a + b)x + ab.
Therefore, in the product of two binomials having a com-
mon literal term:
115. The first term is the product of the given first terms.
The second term is the product of the common literal term by the
algebraic sum of the given second terms. The third term is the
product of the given second terms.
Oral Drill
Give orally the following products :
1. (a> + 3)(a> + ±). 6. (c + l2)(c + l).
2. (« + 4)(o; + 5). 7. (tt + ll)(n + 12).
3. (a> + 5)(» + 7). 8. (m + 3)(m + 20).
4. (m + 6)(m + 4). 9. (d + 12)(d + 16).
5. (y + 9)0/ + 4). 10. (c + 15)(c + 16).
THE PRODUCT OF ANY TWO BINOMIALS 93
11. (6-4)(6-7). 15. (xz-9)(xz-l).
12. (n-ll)(n-10). 16. (c 2 - 3) (c 2 - 4).
13. (ax-3)(ax — 5). 17. (x 2 - xy)(x> - 2 xy).
14. (cd— T)(cd — 3). 18. (mn- 3n)(mw- 7 w).
19. («-3)(sc + 5). 24. (a-13)(a+10).
20. (c-3)(c + 8). 25. (a 2 + 5)(a 2 -3).
21. (x-9)(x + 7). 26. (m 3 4-8)(m 3 -3).
22. (a?4-9)(a?-10). 27. (a& - 9)(ao + 12).
23. (m + H)(m-12). 28. (cx 2 -7)(cx 2 + S).
VI. THE PRODUCT OF ANY TWO BINOMIALS
By actual multiplication
2a +5
3a +4
6 a 2 + 15 a
+ 8 a -f 20
6 a 2 + 23 a + 20
The two multiplications resulting in the terms + 15 a and
+ 8 a are cross products. It will greatly assist the beginner
to imagine that the terms entering the cross product have this
connection :
I r=i 1
(2a+5)(3a+4)
And, by inspection, the middle term results from
(+2a)(+4)+ (+3a)(+5) = + 8 a + 15 a = + 23 a.
Considering other possible cases :
(1) In (4 a — 7) (3 a — 5) the middle term in the product will be
(+4a)(-5) + (-f3a)(-7) = _ 20 a -21 a = - 41 a.
(2) In (2 a + 5) (3 a — 4) the middle term in the product will be
(4- 2 a)(- 4) + (+ 3 «)(+ 5) = - 8 a + 15 a = + 7 a.
(3) In (3 a — 2) (9 a + 4) the middle term in the product will be
(+3a)(-4) + (+9a)(-2) = + 12 a - 18 a = - 6 a.
94 SPECIAL CASES IN MULTIPLICATION AND DIVISION
In general : (ax + b) (ex + d) = acx 2 + (acd + bc)x + bd.
Therefore, in the product of any two binomials :
116. The first term is the product of the given first terms.
TJie second term is the algebraic sum of the cross products of the
given terms. The third term is the product of the given second
terms.
Exercise 22
Write by inspection the following products :
1. (3» + 2)(aJ + l). 13. (7aj-2)(3o?-7).
2. (2 a -fl) (3 a + 2). 14. (10 c- 11) (3 c- 7).
3. (3m+2)(2m + 3). 15. (7 n -9) (8 n- 5).
4. (4a? + JL)(a + 3). 16. (5mw — 1)(3 mn-4).
5. (5c + 4)(2c + 3). 17. (4a -f 7m)(3a - 2 m)
6. (66 + 5)(26 + 3). 18. (7 c - Sd)(Sc + 3d).
7. (5z + 7)(3z + 4). 19. (2 ac - 3 a>)(ac + 11 x)
8. (7m + 2)(2m + 9). 20. (14m — 5 no;) (2m + nx).
9. (3a-7)(2a-5). 21. (11 x + 9y)(9x - 2 y).
10. (3m-l)(2m-3). 22. (3 c - 13 mn) (4 c + 5 mn).
11. (5c-2)(3c-l). 23. (mna-ll)(2mna; + 3).
12. (6y-l)(2y-3). 24. (13 ac - 5 x)(3ac +2 x).
VII. THE SQUARE OF ANY POLYNOMIAL
By actual multiplication :
(a + b + c) 2 = a 2 + b 2 + c 2 + 2 ab + 2 ac + 2 6c.
( a + & _ c _ <Z)2 _ a 2 + &2 + c 2 + ^2 + 2a& _ 2ac - 2aa* - 26c - 26a" +2ca\
It will be seen that, in each product,
(1) The square terms are all positive in sign.
(2) The other terms are positive or negative according to
the signs of their factors.
THE DIFFERENCE OF TWO SQUARES 95
(3) The coefficient of each product of dissimilar terms is 2.
Applying the principle to a representative example we have :
(a-2& + 3c) 2 = (a) 2 + (-2 6)2+(3c) 2 + 2(a)(-2&) + 2(a)(+3c) +
2(_2&)(+3c) = a 2 + 46 2 + 9c 2 -4a& + 6 ac- 12 be. Result.
In general :
117. The square of any polynomial is the sum of the squares
of the several terms together with twice the product of each term
by each of the terms that follow it.
Exercise 23
Write by inspection the following products :
1. (a -\- m + rif. 7. (a + c + m + x) 2 .
2. (c + d + x) 2 . 8. (m-2w + 3x- 2) 2 .
3. (a + c-m) 2 . 9. (3a-26-c-l) 2 .
4. ( a + 26 + 3c) 2 . 10. ^(l-c-c 2 -^) 2 .
5. (2m-3rc + 4) 2 . 11. *(m 3 - m 2 + m - l) 2 .
6. (3m-4n-5) 2 . 12. *(d 3 - 3d 2 + 4d - 2) 2 .
DIVISION
In certain cases where both dividends and divisors are bi-
nomials we are able to write the quotients without actual divi-
sion. Such divisions are limited to the cases where the terms
of the dividend are like powers ; that is, both terms must be
squares, both cubes, both fourth powers, etc. The powers of
the binomial divisors are also like.
I. THE DIFFERENCE OF TWO SQUARES
By Art. 113, (a + b)(a - 6) = a 2 - 6 2 .
Therefore, by division :
^^! = a _ 6 , and «^? = a + 6 .
a + b a — b
* After squaring, the terms should be collected.
96 SPECIAL CASES IN MULTIPLICATION AND DIVISION
Hence the general principle may be stated as follows :
118. The difference of the squares of two quantities may be
divided by either the sum or the difference of the quantities. If
the divisor is the sum of the quantities, the quotient will be the
difference of the quantities ; and if the divisor is the difference
of the quantities, the quotient will be the sum.
- Oral Drill
Give orally the quotients of :
1. (a 2 -ra 2 )-=-(a-m). 8. (16-9m*)-s-(4-3m).
2. (m* -«■)-*-(«* -a). 9. (25 x 2 - 49 y*) + (5 x- 1 y).
3. (a 2 -4)+ (a + 2). 10. (49 c 2 cZ 2 - 9) -- (7 cd + 3).
4. (aj»_9)-*-(a> + 3). "■■ (100 -8 la 4 ) -(10 -9 a 2 ).
5. (c 2 -36)^-(c-6). 12. (rfy 2 - 121) + (xy + 11).
6. (4m 2 -l)-f-(2m-l). 13. (81mW-169c 4 )-s-(9mw-13c*).
7. (9d 2 -25)-j-(3d+5). 14. (196c 6 -81d 4 )-(14c 3 + 9d 2 ).
II. THE DIFFERENCE OF TWO CUBES
By actual division : qB ~ &B = a 2 + ab + & 2 .
a—b
From the form of the quotient and its relation to the divisor
we state:
119. The difference of the cubes of two quantities may be
divided by the difference of the quantities. The quotient is the
square of the first quantity, plus the product of the two quantities,
plus the square of the second quantity.
Oral Drill
Give orally the quotients of:
1. (a 3 -c 3 )-5-(a-c). 3. (or* - z 3 ) -*- (x - *).
2. (m 8 -a?)-*-(m-a;). 4. (x*-l)-*-(x-l).
THE SUM OF TWO CUBES 97
5. (c8_8) + (c-2). 9. (c 3 d 3 - 125) + (cd- 5).
6 . (^_27)_^(d_3). io. (8 ar>- 27 m 3 ) + (2 x -3 m).
7. (64 -a?) -s- (4 -x). 11. (512c 3 d 3 -729)-=-(8cd-9).
8. (ary-l)-r-(a;?/-l). 12. (216 afy 8 - 1000 z 3 )^ (6ay -10 2).
m. THE SUM OF TWO CUBES
By actual division : q3 + b * = a?-ab + 6 2 .
a-tb
From the form of the quotient and its relation to the divisor,
we state:
120. The sum of the cubes of two quantities may be divided by
the sum of the quantities. The quotient is the square of the first
quantity, minus the product of the two quantities, plus the square
of the second quantity.
Oral Drill
Give orally the quotients of :
1. (a* +c *)+<a+c). 7. (125 + cP) + (5 + d).
2. (m* + a?)-*-(m + x). 8. (27 + d?) -+- (3 + d).
3. (c 3 + ^)-(c + 2/). 9. (mY + l) + (my + l).
4. (a 3 + l)-f-(a + l). 10. (m 3 x 3 + 125)^-(maj + 5).
5. (aj» + 8)-*-(» + 2). 11. (8c» + 27a?)-*-(2c + 3*). ■
6. (n 3 + 27)-r-(7i. + 3). 12. (125 6 3 + 64) -s-(5 b + 4).
13. (729 m 3 n 6 + 1000 a; 9 )- (9 mn 2 + 10 a 3 ).
IV. THE SUM OR DIFFERENCE OF AWT TWO LIKE POWERS
(a) The Difference
Even Powers Odd Powers
a 2 - 62 a s _ & 8 ,
a 4 — 6 4 may each be divided by a 5 — 6 6 may each be divided
a 6 - 6 6 (a + 6) or (a - 6). a 7 - 6 7 by (a - 6) only.
a 8 - ft 8 a 9 _ 59
etc. etc.
SOM. EL. ALG. 7
98 SPECIAL CASES IN MULTIPLICATION AND DIVISION
(b) The Sum
Even Powers
Odd Powers
a 2 + & 2
a 8 + 6 8
a* + 6*
are not divisible by
a 5 + 6 5
may each be divided
a 6 + 6 6
either (a + b) or (a — 6).
a 7 + 6 7
by (a + 6) only.
a 8 + 6 8
a 9 +6 9
etc.
etc.
A general proof for these cases is considered in later practice.
Illustrations :
(1) a4 ~ &4 = o 8 + a 2 5 + a& 2 + 6 8 - (2) q * ~ ~ = a 8 - a 2 6 + a& 2 - 6 8 .
(3) «L±A 7 = a 6 - a 5 6 + a 4 6 2 - a 3 6 3 4- a 2 & 4 - a& 6 + 6«.
a + b
From the form of the quotients and their relation to their
divisors we state :
121. The number of terms in the quotient is the same as the
exponent of the powers in the dividend.
The exponent of "a" in the first term of the quotient is the
difference between the given exponents of " a " in the dividend and
divisor, and this exponent decreases by 1 in each successive term.
The exponent of " b " is 1 in the second term of the quotient,
and this exponent increases by 1 in each successive term until
it is equal to the difference of the given exponents of " b " in the
dividend and divisor.
If the sign of "b" in the divisor is -{-, the signs of the quotient
are alternately + and — , and if the sign of'b" in the divisor is
— , the signs of the quotient are all +•
Exercise 24
Write by inspection the following indicated quotients :
% a A -b\ f gt+y 5 g c 5 + 1 ? m^
a — b a + y c-fl m 2 — n
i 5 -b 5 4 c 4 -d* 6 xY-z\ g fl* + 32
a—b c + d xy — z x+ 2
CHAPTER IX
FACTORING
122. An algebraic expression is rational with respect to any
letter if, when simplified, the expression contains no indicated
root of that letter. Thus :
x 2 — xy + y/y is rational with respect to x, but irrational with respect to y.
The symbol -y/ is used to indicate a required square root.
123. An algebraic expression is integral with respect to any
letter if that letter does not occur in any denominator. Thus :
x + — + m is integral with respect to m, but fractional with respect to x.
x
124. An algebraic expression is rational and integral if it is
rational and integral with respect to all the letters occurring
in it. Thus :
4 m 2 — 5 mn + se 4 is both integral and rational.
125. The factors of a rational and integral algebraic ex-
pression are the rational and integral algebraic expressions
that, multiplied together, produce it. Thus :
atbZm = axaxbxbxbxm,
and a, a, 6, ft, b, and m are the factors of a% z m.
One of two equal factors of a number is a square root of the
number ; one of three equal factors, a cube root ; of four equal
factors, a fourth root ; etc. Thus :
x is the square root of x 2 ; y is the cube root of y s ; etc.
100 FACTORING
126. An algebraic expression is a prime expression when it
has no factors excepting itself and unity. In the process of
factoring we seek prime factors, and we extend our work until
the resulting expressions have no rational factors.
Unless otherwise stated, the use of the term "algebraic expression"
in elementary algebra is understood to refer to rational and integral ex-
pressions only.
127. An expression is factored when written in the form of a
product.
128. Since factorable expressions result from some com-
pleted multiplication, we may repeatedly refer to multiplication
as the origin of particular type forms for which we have definite
methods of factoring. The forms are classified by
1. The number of terms in the expression.
2. The powers and coefficients of the terms.
3. The signs of the terms.
WHEW EACH TERM OF AN EXPRESSION HAS THE SAME MONOMIAL FACTOR
Type Form • • • ax + ay -\-az.
The Origin : If a polynomial (x + y + z) is multiplied by a
monomial (a), we obtain (Art. 55) :
a(x + y + z) =zaz + ay + az.
Therefore, ax + ay + o,z = a(x + y + z).
That is, the factors of ax + ay + az are " a " and " x + y + z."
In general, to factor an expression in which each term has
the same monomial factor :
129. By inspection find the monomial common to the terms of
the given expression.
This monomial is one factor, and the expression obtained by
dividing the given expression by it is the other factor.
TRINOMIAL EXPRESSIONS 101
Illustrations :
1. x 8 - x 2 + 3x = x(x 2 - x + 3).
2. 3 m 4 + 9 ro 8 - 12 ro 2 + 15 ro = 3 ro (ro 8 + 3 ro 2 - 4 ro + 5).
3. 3 a 2 6 2 - 6 a 8 *) 2 + 6 a 8 6 4 - 9 a 4 6 5 = 3 a 2 b 2 (1 - 2 a + 2 a& 2 - 3 a 2 6«).
Exercise 25
Factor orally :
1. 5a- 106 + 15c. 7. 5x — 20xy + 35xz.
2. 8m — 16n-f-24a;. 8. a& + 3ac+7aa\
3. 12 a + 18 y + 24 2. 9. 5 m — 20 ran — 40 raz.
4. 5 c + 10 a" + 15 k. 10. a 4 + a 3 -a 2 + a.
5. 15x + 30y + 4:5z. n. ic 8 — a^ — a; 4 — ic 2 .
6. 14ra-21n + 14. 12. x 2 - x 5 + x* - x 9 .
Write the factors of:
13. 5 <*b-20 cW + 35 car 5 .
14. 17 a 4 »+ 51 a 3 ar 2 -34aV-85aa! 4 .
15. 27 ra 3 n + 18 ra 2 n — 9 ran 2 — 36 ran 3 .
16. cPa?y — a 2 xy + aa^ 2 — aary.
17. a s m 3 n 3 — a 3 m 4 n 2 — cPmhi 3 — a 3 m 4 n 5 .
18. 14 a 4 6 + 21 a 3 6 2 - 35 a 2 6 3 - 42 a& 4 -
19. 3a?z-6<x& + 9a&-12a& + 15x*.
20. a 3 ™, — 3 a 2 cm — a6 2 ra — ara 3 + 2 ara 2 4- ara.
TRINOMIAL EXPRESSIONS
(a) The Perfect Trinomial Square
Type Form ... x 2 ± 2 xy +/.
The Origin: If a binomial (x±y) is multiplied by itself, or
squared, we have :
by Art. Ill : (x + y) 2 = x 2 + 2 xy + t/ 2 .
or by Art. 112 : (x - y) 2 = x 2 - 2 xy + y 2 .
102 FACTORING
In each product :
The first term is a perfect square ; ]
The third term is a perfect square. J Both are + *
The second term is twice the product of 1 Its sign may be
the square roots of the square terms, j either + or — .
These are the conditions of a perfect trinomial square. In like manner :
c 2 - 18c + 81, 4a2_i2a& + 962, 16 xY - 40 x 2 y 2 z + 25 z*
are all perfect trinomial squares, for each trinomial has two positive
square terms, with a second term that is twice the product of the square
roots of the square terms.
Hence, to factor a perfect trinomial square :
130. If necessary ; arrange the terms of the given expression in
order.
Take the square roots of the first and last terms, and connect
them with the sign of the second term.
The resulting binomial is one of the two required equal factors.
Illustrations :
1. x 2 + 16 x + 64 = (x + 8) (x + 8) = (as + 8) 2 .
2. 9 m 2 - 12 mn + 4 n 2 = (3 m - 2 n) (3 m - 2 n) = (3 m - 2 w) 2 .
3. x 6 - 26 « 8 + 169 = (x 3 - 13) (x 3 - 13) = (x 3 - 13) 2 .
Exercise 26
Factor orally :
1. ^ + 6^ + 9. 5. m 2 -r-14m + 49. 9. raV '-22 mn + 121.
2. x 2 + 8^+16. 6. c 2 - 18c + 81. 10. 64 + 16z + x 2 .
3. ^-6^ + 9. 7. w 2 + 20n + 100. 11. 81+m 2 -18m.
4. 7? - 8 x + 16. 8. a 2 b 2 - 12 ab + 36. 12. 144 - 24 xy + x 2 y 2 .
Write the factors of :
13. 9c 2 -6cd + d 2 . 16. 4(P+12dn + 9n a .
14. 4P-4fcm + m 2 . 17. 16 c 2 + 24 ex + 9 ar 2 .
15. 8a2/ + 2/ 2 + 16n 2 . 18. 25 p 2 - 30# + 9.
TRINOMIAL EXPRESSIONS 103
19. 36^ + 25m 2 -60cm. * 22. 64 x 4 - 80 x*y 2 + 25 y 4 .
20. 72 cc? + 16^ + 81 d 2 . 23. 49 a 2 b 2 + 140 a&c f 100 c 2 .
21. 36^ + 84^ + 49. 24. 81aV-198a 2 a*/ + 121^.
(6) The Trinomial whose Highest Power has the Coeffi-
cient Unity
Type Form ••• x 2 + (c + </)* + cfl ^
The Origin : If two binomials, (x + c) and (x + d), are multi-
plied, we have (Art. 115) :
(x + c)(x -{• d)= x(x + d) + c (x + d)
= x 2 + dx + cx + cd
= x 2 + (c + d) x + cd.
In the resulting product :
The first term x 2 = the product of the given first terms.
The second term (c+d)x = the product of the given first term by the alge-
braic sum of the given second terms.
The third term cd = the product of the given second terms.
Hence, the coefficient of the second term of the product is the
sum of the given second terms.
Illustrations :
1. x 2 + 8 a? + 15 = (x + ?) (x + ?).
We require the two factors of + 15 whose sum is + 8 : +3 and + 5.
Therefore, x 2 + 8 x + 15 = (x + 3) (x + 5). Result.
2. x 2 - 8 x + 15 = (x - ?) (x - ?).
We require the two factors of + 15 whose sum is — 8 : — 3 and — 5.
Therefore, x 2 - 8 x + 15 = (x - 3) (x - 5). Result.
3. x 2 + 2a? - 15 = (as + ?) (x - ?).
In this expression the sign of 15 is — , hence the signs of its factors
are unlike. The sign of 2 is +, hence the greater factor of — 15 is +.
We require the two factors of — 1 5 whose sum is + 2 : +5 and — 3.
Therefore, x 2 + 2 x - 15 = (x + 6) (x - 3). Result.
104 FACTORING
4. x 2 -2x~ 15 = (x-?) 0+?).
In this expression the sign of 15 is — , hence the signs of its factors are
unlike. The sign of 2 is — , hence the greater factor of — 15 is — .
Therefore, x 2 - 2 x - 15 = (x - 5) (x + 3) . Result.
From the illustrations we make the following important
conclusions :
131. (1) The first aid to factoring such expressions is the sign
of the third term.
(2) If that sigyi is +, the signs of the second terms in the
required factors are like ; but if that sign is — , the signs of the
second terms of the factors are unlike.
(3) The sign of the given second term is the same as that of the
greater factor of the given third term.
Exercise 27
Give orally the factors of the three groups, a, b, and c.
(a) The third term + . The second term -f- . The sign of
the last term of each factor +•
Illustration : x 2 + 10 x + 24 = (x + 4) (x + 6).
1. a? + 7x + 12. 5. d 2 + 20d + 36.
2. m 2 + 8m-}-l5. 6. f + 19 y + 48.
3. c 2 + l2c + 20. 7. c 2 + 31c + 58.
4. a 2 +12a + 32. 8. p 2 + 37p+70.
(b) The third term +. The second term — . The sign of
the last term of each factor — .
Illustration : x 2 - 10 x + 24 = (x - 4) (x - 6).
9. a 2 _8a + 12. 13. x 2 -18x + 32.
10. ^-9a; + 18. 14. c 2 -16c + 39.
11. c 2 -9c + 14. 15. ?i 2 -ll7i + 24.
12. m 2 -10m + 21. 16. ?/ 2 - 14 */ + 24.
TRINOMIAL EXPRESSIONS 105
(c) The third term — . The second term either + or — .
The signs of the last terms of the factors unlike, the greater
last term having the same sign as the second term of the given
trinomial.
Illustrations : x 2 + 2 x - 24 = (as + 6) (as - 4).
x 2 - 2 x - 24 = (x - 6) (x + 4).
17. tf + x-20. 21. c 2 -15c-34
18. ra 2 -3ra-10. 22. a 2 -9a-70.
19. C 2_5 C _14. 23. z 2 + 13z-48.
20. a^ + 8a;-20. 24. a?-6x-72.
(d) Write the factors of :
25. m 2 - 15 m -54. 34. ra 2 n 2 - 8 ran - 84.
26. ^-c-m 35. x 2 y 2 -72-xy.
27. ?/ 2 - 11 2/ -26. 36. 28 - 16 xz + arV.
28. 110-53C-C 2 . 37. ah 2 -15 az- 76.
29. afy 2 -25a;2/ + 46. 38. 96 + 28 xyz + afyV.
30. n 2 -9n-112. 39. c* + 33c?-70.
31. 2 a; -120 -far 2 . 40. 63 a 2 6 2 +a 4 6 4 -130.
32. & 2 + 13 6-140." 41. ra 2 n 2 - ran - 210.
33. d 2 -6d -135. 42. aW - 7 a 2 6 2 c - 144.
43. a 2 c?x 2 + 5acx-36.
44. 45 + 4 ran 2 — ra 2 n 4 .
45. 33+8cti-cftft
46. 54-15ar 5 -a*
47. ac 2 d-aW + 12.
48. 2 mux — mWx 2 + 143.
49. 380-cc?V-<fW,
106 FACTORING
(c) The Trinomial whose Highest Power has a Coefficient
Greater than Unity
Type Form • • • acx 2 4- (ad 4- be) x 4- bd.
The Origin: If two binomials, (ax + b) and (cx + d), are
multiplied, we have (Art. 116) :
(ax + b) (ex 4- a") = ax (ex 4- a") -f 6 (ex + d)
= acx 2 4- adx + 6cx + bd
= acx 2 + (ad 4- 6c) x + bd.
Now the coefficient of x in the result (ad -{-be) is made
up of the coefficients that, multiplied, would produce abed.
Therefore,
132. We require the two factors of abed that, added, will
produce ad + be.
The application of the principle in practice will be readily
understood from the following illustrations :
1. Factor 6x* + 25 x + 14.
6 x 14 = 84.
Required the factors of 84 that, added, equal 25.
By trial we find them to be 4 and 21.
Therefore, 6 x 2 + 25x + 14 = 6 x 2 + 4 x + 21 x + 14
= (6x 2 4-4x) + (21x + 14)
= 2x(3x + 2)+7(3x+2)
= (2 x + 7) (3 x + 2) . Result.
2. Factor 14 a 2 + 31 a -10.
14 x -10= -140.
Required the factors of —140 that, added, equal 31.
By trial they are found to be 4-35 and —4.
Therefore, 14 a 2 4- 31 a - 10 = 14 a 2 + 35 a - 4 a - 10
= (14a 2 4-35a)-(4a4-10)
= 7a(2a4-5)-2(2a4-5)
= (7 a - 2) (2 a + 5) . Result.
Several excellent methods for factoring a trinomial of this type form
might be given, but to understand and apply accurately one method is a
better plan for the beginner.
BINOMIAL EXPRESSIONS 107
Exercise 28
Write the factors of :
1. 4m 2 + 8m + 3. 14. 9m 2 + 23m + 10.
2. 2c 2 + 3c + l. . 15. 3z 2 + 52-22.
3. 6z 2 -7z + 2. 16. 6a 2 + 7a&-5& 2 .
4. 6d 2 -lld + 4. 17. 12m 2 -23m7i-f-10w 2 .
5. 9?i 2r -9n + 2. 18. 10c?-3cd-18d 2 .
6. 6 2/ 2 — 13 2/ -h 6. 19. 6 z 2 - 31 zz + 35 z 2 .
7. 6c 2 + 17c + 12. 20. 16c 2 + 18cd-9d 2 .
8. 8a 2 -2a-3. 21. 10 a 2 + ax - 24= a 2 .
9. 9m 2 + 21m + 10. 22. 9 v 2 - 15 vx - 50 x 2 .
10. 6z 2 -7z-5. 23. 7c 2 -50cz + 7z 2 .
11. lOrf-Ux + S. 24. 20a 2 -27am-14m 2 .
12. 10 a 2 + <e -3. 25. 10 z 2 -zm- 21m 2 .
13. 6z 2 -z-12. 26. Sn 2 -10nx-25x 2 .
27. 20a 2 -37az-18zV
BINOMIAL EXPRESSIONS
(a) The Difference of Two Squares
Type Form ••• x 2 —/*.
The Origin: If two binomials, (x + y) and (x—y), are mul-
tiplied, we have (Art. 113):
(x + y)(x-y)=x*-y*.
Therefore, to factor the difference of two squares :
133. Extract the square roots of the given square terms.
One factor is the sum of these square roots ; the other factor,
their difference.
108 FACTORING
Illustrations :
1. 4a 2 -25a 2 = (2a + 5z)(2a
2. 25a 4 - 9s 6 = (5a 2 + 3s 3 ) (5
3. 16 a 4 - 81 = (4 a 2 + 9) (4 a 2
= (4a 2 + 9) (2a
-5x).
a 2 -3s»).
-9)
+ 3) (2a -3).
Note that the second factor of (3) can be refactored into
two other factors.
Exercise 29
Factor orally :
1. tf-y 2 . 6. d 2 -16. ll. 9^-25.
16. 4tx*-25y 2 .
2. a?-z 2 . 7. a 2 -25. 12. 16 m 2 -9.
17. 16ra 2 -49n 2 .
3. c 2 -1. 8. m 2 -49. 13. 4w 2 -25.
18. 36 a 2 - 25 z 2 .
4. a 2 -4. 9. w 2 -81. 14. 9a 2 -64.
19. 49 n 2 - 100 # 2 .
5. a 2 -9. 10. 4c 2 -9. 15. a?-9f.
20. 121a 2 6 2 -144c 2 .
Write the factors of :
21. a A — x A . 25. a 8 — a 8 .
29. ra 4 -92 6 .
22. c 4 -16d 4 . 26. c 10 -16.
30. c 12 -64m 4 .
23. 64 x 4 - if, 27. c 8 -256.
31. 36aV-25y«.
24. 81m 4 -16w 4 . 28. <c 16 -l.
32. 49 m 10 -4n 1 V 4 .
(6) The Difference of Two Cubes
Type Form ••• x*—f.
The Origin : Since the product of a divisor by a quotient
equals the corresponding dividend, we have, from Art. 119, an
expression (x* — if) equal to the product of the expressions
(x - y) and (x 2 + xy + y 2 ).
Therefore, for the factors of the difference of two cubes :
134. One factor is the difference of the cube roots of the
quantities. The other factor is the sum of the squares of the cube
roots of the quantities plus their product.
BINOMIAL EXPRESSIONS 109
Illustrations :
1. c 3 -8 = (c-2)(c 2 + 2c-f 4).
2. 27 ra 3 - 64 = (3 w - 4) (9 ro 2 + 12 m + 16).
3. 64 s 6 - 125 y% 3 = (4 x 2 - 5 yz) (16 a* + 20 afys + 25 y 2 * 2 ).
Factor orally :
Exercise 30
1. c 3 -d 3 .
5.
m 3 -8.
9.
27 -s 3 .
2. a 3 — m 3 .
6.
a 3 -27.
10.
64-a 3 .
3. w 3 -^.
7.
f-64:.
11.
cW-64.
4. c 3 -!.
8.
c?-125.
12.
a 3 m 3 - 125
Write the factors of :
13. 8^-27. 17. 8 m 3 -343. 21. 125m 9 -l.
14. 27 -125 m 3 . 18. 8 c 6 -27. 22. 27 m 6 n 3 - 64 x 9 .
15. 125 x 3 - 64 f. 19. 512 ra 3 - a 6 . 23. 125 x« - 729 z 12 .
16. 216 a 3 -ay. 20. 125 ay - 64 z 6 . 24. 729 m 12 - 1000 ny.
(c) The Sum of Two Cubes
» Type Form ••• **+/.
The Origin : As in the preceding case, we refer to the
principle of division by which (aP + y 3 ) is shown to contain
I(x + y), the quotient being (zP — xy + y 2 ) (Art. 120).
Therefore, for the factors of the sum of two cubes :
135. One factor is the sum of the cube roots of the quantities.
The other factor is the sum of the squares of the cube roots of the
quantities minus their product.
Illustrations :
1. c 3 + 27 = (c+3)(c 2 -3c+9).
2. 8+125c 3 = (2 + 5c)(4-10c + 25c 2 ).
3. 27 z 6 + 64 w 9 = (3 a 2 + 4 n 3 ) (9 x* - 12 se 2 w 3 + 16 n«).
110 FACTORING
Exercise 31
Factor orally :
1. a 8 +C 8 . 4. n 8 + l. 7. m 3 + 125. 10. 64 + afy 3 .
2. rrt + tf. 5. a^ + 8. 8. 2/ 3 + 216. 11. m 3 n 3 + 343.
3. d s +tf. 6. z 3 + 27. 9. 27 + a 3 6 3 . 12. 2/V + 512.
Write the factors of :
13. 8 a 3 + 27& 3 . 17. 125 a 6 + 1. 21. 512 + 125 aP.
14. 64 a 3 +125. 18. 64 a 6 + 27^. 22. 729 + 64 aft/ 3 .
15. 27ra 3 + 64n 3 . 19. xY + 125z\ 23. 1000 v? + 27 y l \
16. 125 tt 3 + 216 ar 5 . 20. ay + 216z 3 . 24. 1728 a 15 + 1331 y u .
EXPRESSIONS OF FOUR OR MORE TERMS FACTORED BY GROUPING
(a) The Grouping of Terms to show a Common Polynomial
Factor
Type Form • • • ax + ay + bx -f by.
The Origin: If any binomial, (% + y), is multiplied by a
binomial having dissimilar terms, (a + 6), we have (Art. 55) :
(a + b) (x + y) = a(x + y) + b (x + y)
= ax + ay + bx + &y.
In the resulting product note that a is common to the first two terms,
and that b is common to the last two terms. Note, also, that dividing
the first two terms by a, and the last two terms by 6, gives the same
quotient, (x+y).
Therefore : (x +y) is a common polynomial factor.
And from a (x 4- y) + b (x + y) we obtain by adding the coefficients
of the common factor : (a + 6) (x + y) , the factors.
Illustrations :
1. ac + bc+ ad+bd = (ac + 6c) + (ad + bd)
= c(a + b) + d(a + b)
= (c + d)(a + b).
EXPRESSIONS OF FOUR OR MORE TERMS 111
2. 2 c 2 - 6 c + cd - 8 d = (2 c 2 - 6 c) + (cd - 3 d)
= 2c(c-3) + d(c-3)
= (2c + d)(c-3).
3. a 8 6 - 14 - 7 a 2 + 2 a& = a 3 5 + 2 a& - 7 a 2 - 14
= (a 3 6 + 2 aft) - (7 a 2 + 14)
= a&(a 2 +2)-7(a 2 + 2)
= (a6-7)(a 2 + 2).
136. Note that the proper arrangement of an expression is
I first necessary. We are assisted in grouping the terms by
noting that terms bearing to each other the same relation are
grouped together.
With expressions of more than four terms the principle is
unchanged.
Exercise 32
Write the factors of :
1. am + an +mx + nx. 10. abxy — cxy—cz + abz.
2. am + an — cm — en. 11. x 2 + bx + ax -f- ab.
3. ac — ad — be + bd. 12. y 2 — my + 2y — 2m.
4. ax + ay -f- x + y. 13. .x 3 + ax 2 + x -f a.
5. ax — az — x + z. 14. z 3 + 2 2 — 3 z 2 — 6.
6. aai-f 3 a + 2 a; + 6. 15> a 6 + 5 a 3 + 10 + 2 a 2 .
7. a2/-4a + 52/-20. 16 x 4 + i4 1 _2x-7 3?.
8. afoc — 2 a& -f ca> — 2 c. 17. — arfcc -f 3 d — 3 ex -f- aca^.
9. 2rana;-5a;-6mw+l5. 18. 3 d- 10 d 2 -15 + 2 d*.
(b) The Grouping of Terms to form the Difference of
Two Squares
Type Form ... x 2 + 2 xy +/ - z 2 .
The Origin : By multiplication,
(x + y + z)(x + y-z) = [(a + y) + *][(*+» - z]
= ( X + t/) 2 - z 2
= a; 2 + 2 a;y + y 2 - s 2 .
112 FACTORING
The product is the difference of a trinomial square and a
monomial square. Therefore, an expression of this type is
factored by grouping three of the terms that will together
form a perfect trinomial square ; the fourth term being a per-
fect monomial square. The result is a difference of two
squares.
Illustrations :
1. a 2 + 2 ab + b 2 - c 2 = (a 2 + 2 ab + 6 2 ) - c 2
st (a + by - c 2
= (a + b + c)(a + b-c).
2. x 2 - y 2 - 2 2/0 - z 2 = x 2 _ (y2 + 2 ^ + s*)
= x 2 - (y + zY
«[•+ ft + ! «)K^- &■+"#)]
= (» + y + «)(&-?-«)•
3. a 2 - 6 ax + 9 at 2 - 4 m 2 - 12 m - 9
= (a 2 - 6 ax + 9 x 2 ) - (4 m* + 12 m + 9)
= (a-3x) 2 - (2m + 3) 2
= [(a - 3 x) + (2 m + 8>] [(a -3 x) - (2 m -f 3)]
= (a - 3 x 4- 2 m-\- 3) (a - 3 x - 2 m - 3).
The process consists mainly in finding three terms that, when
grouped, form a perfect trinomial square. The key to the group-
ing is the given term that is not a perfect monomial square. Or,
considering the given square terms, we may state :
137. (1) When only one given square term is plus, it is written
first, and the other three terms are inclosed in a negative paren-
thesis.
(2) When only one given square term' is minus, it is written
last, and the other three terms are written first in a positive
parenthesis.
Exercise 33
Write the factors of :
1. a 2 + 2ax + a?-m 2 . 3. c 2 + d 2 -y 2 -2 cd.
2. tf-2yz + z*-4:. 4. l-c 2 -2cd-tf.
REPEATED FACTORING 113
5. v?-m 2 -l + 2m. 10. c 2 - 10 ex + 25 x 2 - 49 m 2 .
6 . 4 c 2_<f_j_ 1 _4 c> X1> a ^-4a;2/-9» 2 2/ 2 4-4?/ 2 .
7. 9-m 2 -2m2/-2/ 2 . 12. 2/ 2 + 16z 2 - 16 + 8 yz.
8. 4c 2 + c 4 -4c 3 -4. 13. ary-a 2 & 2 + 16a&-64.
9. 12x + ±x 2 -9y 2 + 9. 14. 490^ + 10 c 2 ^ 2 -.^ 4 - 25.
15. 81 -100 x*y* + 60 x 4 y 2 z -9 z 2 .
16. 4iB 2 -4» + l-9m 2 4-6wi?i-7* 2 .
17. c? — a? + x 2 — y 2 — 2cx — 2 ay.
18. c 2 -d 2 -x 2 + m 2 — 2cm + 2dx.
19. 36c 4 4-l-49ic 6 -92/ 2 -12c 2 -42of l y.
REPEATED FACTORING
138. Any process in factoring may result in factors that
| may still be resolved into other factors. The review examples
following frequently combine two or more types already con-
\ sidered. The following hints on factoring in general will
i assist.
1. Remove monomial factors common to all terms.
2. What is the number of terms in the expression to be fac-
tored ?
(a) If two terms, which of the three types ?
(b) If three terms, which of the three types ?
(c) If four terms, how shall they be grouped ?
3. Continue the processes until the resulting factors are prime.
Illustrations :
1. Sa 7 -3ax* = 3a(a?-x 6 )
= 3 a (a 3 + x 8 ) (a 8 - x 3 )
= 3 a (a + x) (a 2 - ax + x 2 ) (a - x) (a 2 + ax + x 2 ).
2. 12x5-75x 8 + 108x=3x(4x 4 -25x 2 + 36)
= 3 x (4 x 2 - 9) (x 2 - 4)
= 3x(2x + 3)(2x-3)(x + 2)(x-2).
SOM. EL. ALG. 8
J
114 FACTORING
3. 8 a 8 x - 2 ax + 8 a 5 x - 2 a 4 x = 8 a 8 x + 8 a 6 x - 2 a 4 x - 2 as
= 2 ax (4 a 7 + 4 a 4 - a 8 - 1) •
= 2 ax [(4 a 7 + 4 a 4 ) - (a 8 + 1)]
= 2 ax [4 a 4 (a 3 + 1) - (a 8 + 1)]
= 2ax(4a 4 -l)(a 3 +l)
= 2ax(2a 2 +l)(2a 2 -l)(a+l)(a 2 -a + l).
MISCELLANEOUS FACTORING
Exercise 34
Write the factors of :
1. 5a? + 25x + 20. 21 75 a 3 -90 a 2 -{-27 a.
2. 5 0^-45 a 3 . 22. 6a 2 x+9abx-8a 2 y-12aby.
3. 8 a 2 - 24 a + 18. 23. 5 m 3 + 135.
4. a 4 -8a. 24. 15 c 3 +33 c 2 + 6c.
5. 2am + 2ma;-4a-4aj. 25. 242 ar*- 98 a.
6. 2-2c 2 -4cd-2d 2 . 26. 12 m 4 + 10 m 3 -8 m 2 .
7. 3 m 4 + 81 m. 27. 160 a,- 5 + 20 x 2 .
8. 3a 2 -9a-84. 28. 3 ar 5 - 15 a 3 + 12 a;.
9. 2c 5 - 128c 2 . 29. 98 ar> + 18 a*/ 2 - 84 afy.
10. 16 m V + 8 mnx + x 2 . 30. m 4 - 2 m 3 + 8m- 16.
11. 15 - 60 xy + 60 x*y 2 . 31. 15 a 3 - 25 a 2 a - 10 ax 2 .
12. 7 aV - 175 ax. 32. 81 afy 4 - 3 a#.
13. c 4 +c 2 -3c 3 -3c. 33. 4cd + 2c 2 d 2 + 2-2ar J .
14. Ux*-S2x 2 -12x. 34. 250c 4 -16c.
15. 49m 3 -84m 2 + 36m. 35. 8mV+4m¥-112mw.
16. 4o 3 -36ar J + 56a;. 36. 5 c 5 - 10 c 3 - 315 c.
17. 7 a 7 - 7x. 37. 21 ar> + 77 a 2 -140 a;.
18. 4 a 3 + 16 a 2 + 15 a. 38. 512 - 32 m 4 n\
19. 8a 2 -18c 2 + 24a& + 186 2 . 39. 27^+215^-8.
20. 24^-55^-24^ 40. 3 x» -51 a^ + 48 x.
SUPPLEMENTARY FACTORING 115
41. 686 x* - 2 xtf. 45. 242 a 4 - 748 a 2 + 578.
42. 8m 5 + 8m 2 -18m 3 -18. 46. 2x 5 +64y 2 -$x s tf-16x 2 .
43. 8a 3 + 2a 5 -8a 4 -2a. 47. 64a 6 + 729a 2 .
44. 32 a 4 - 2ajy. *»• 8 a* - 10 a 8 - 432 a.
SUPPLEMENTARY FACTORING
(a) Compound Expressions in Binomial and in Trinomial Forms
Comparative illustrations with corresponding processes :
1. z 2 + 12z+35 = (x + 5)(z + 7).
Similarly,
(a _ &)2 + 12 (a - 6) + 35 = (a - 6 + 5)(a -6 + 7).
2. 3£ 2 + 10zy + 32/2 = (3s + y)(z + 3y).
Similarly,
3 (a - b) 2 + 10 (a - 6) (c - d) + 3 (c - d) 2
= [8(a-&) + (c-<I)][(a-&)+3(c-tf)]
= (3a-3 6 + c-d)(a-6 + 3c-3d).
3. a 2 -9a 2 = (a + 3z)(a-3x).
Similarly,
(« + 2) 2 -9(x-l)2 = [( a;+ 2)+3(x-l)][(x + 2)-3( a ;-l)]
= (z + 2 + 3o;-3)(a; + 2-3a; + 3)
= (4x-l)(-2x-f 5)
= _(4z-l)(2x-5).
4. a 8 + 8 = (a + 2)(a 2 -2a + 4).
Similarly,
(a - 6)8 4- 8 = [(a - 6) + 2] [(a - 6) 2 - 2 (a - 6) + 4]
s (a-6 + 2)(a 2 -2a6 + 6 2 -2a + 26 + 4).
139. If similar monomial terms occur in the different compound
terms of an expression, the factors can usually be simplified by
collecting like terms.
116 ' FACTORING
Exercise 35
Find and simplify the factors of :
1. 16(a+l) 2 -9a^. 6. (x 2 - 1) 2 - 9 (x + 2) 2 .
2. 49(c-l) 2 -4. 7. 27(a-l) 3 + a 3 .
3. (a + %) 2 + 10 (a + x) + 24. 8. 2 (x + l) 2 + 11(*+1) + 12.
4. (m + 2) 2 - 15 (m+2) + 56. 9. (2 m- l) 4 - 16 (2 m + 1) 4 .
5. ( 2/ _3)2_7(^_3)_30. io. 8(a-2) 3 -27(a + l) 3 .
11. 6a 2 -12aaj + 6^-5a + 5a;4-l.
12. 3(s+ l) 2 + 7(> 2 -l)-6(a-l) 2 . .
(6) The Difference of Two Squares obtained by Addition
and Subtraction of a Monomial Perfect Square
140. No general statement of this process can be given in a
simple form.
Illustration :
Factor 9^+6^ + 49.
The first and third terms are positive perfect squares.
Hence, if the middle term were twice the product of their square
roots, the expression would be a perfect trinomial square.
For a perfect trinomial square the middle term should
be 2 (3 2^(7) = 42 x 2 . Adding + 36 x 2 to the expression, we
obtain the perfect trinomial square required, and subtracting
the same square, + 36 x 2 , the expression is unchanged in value
but is now the difference of two perfect squares. Therefore :
9z 4 + 6z 2 + 49 = 9a 4 + 6 x 2 +49
+ 36 x 2 - 36 x 2
= 9 & + 42 x 2 + 49 - 36 x 2
= (3x 2 + 7) 2 -36s 2
= (Sx 2 + 7+6x)(3x 2 + 7-6x).
It is important to note that a positive perfect square only can
be added.
SUPPLEMENTARY FACTORING 117
Exercise 36
Write the factors of:
1. a^ + ^ + l. 7. 25^-51^ + 25.
2. a 4 + 3a 2 + 4. 8. 81 m 4 + 45 ra 2 + 49.
3. n 4 -7n» + l. 9. 36c 4 -61c 2 m 2 + 25m 4 .
4. (j* -28 c 2 + 16. 10. 64 a 4 + 79^ + 100 a 4 .
5. 9c 4 + 5cV + 25a; 4 . 11. 64 m 4 + 76 m 2 n 2 + 49 n 4 .
6. 9 d 4 - 55 d?x 2 + 25 a; 4 . 12. 81 x 4 - 169 zV + 64 z\
(c) The Sum and the Difference of Equal Odd Powers
By actual division :
<z 5 + x 5
a + x
a 6 — x 5
a* — a s x + ah? — ax 3 + x 4 .
= a 4 + a 3 x + a >2 x 2 + ax 3 + x 4 .
a — x
Hence, for the factors :
a 5 + x 5 = (a + x) (a 4 - a 3 x + a 2 x 2 - ax 3 + x 4 ).
a 5 - x 6 = (a - x) (a 4 + a 3 x + a 2 x 2 + ax 3 + x 4 ).
By Art. 121, the factors of any similar cases may be found.
In general, therefore :
141. (1) One of the factors of the sum of equal odd powers is
the sum of the quantities. The other factor is the quotient obtained
by using the binomial as a divisor.
(2) One of the factors of the difference of equal odd powers is
the difference of the quantities. The other factor is the quotient
obtained by using the binomial as a divisor.
Write the factors of :
Exercise 37
-
1. ra 5 -l.
5.
ra 5 + n 5 .
9.
32(^+243.
2. tf + x*.
6.
32 - n 5 .
10.
a¥-32.
3. m 7 — n 7 .
7.
128 +c 7 .
11.
m 11 + w 11 .
4. a 7 + x 7 . ■
8.
a?Y- 343.
12.
a io c iD_ a jy.
118 FACTORING
MISCELLANEOUS FACTORING
Exercise 38
1. 3a*- 54^ + 243a. 14. 72 m 8 + 94 m 4 + 128.
2. ( c 3 + 8) + 2c(c + 2). 15 . 2a 2 (a+2)-5a 2 -8a+4.
3. 4a 2 + 9ar>-*-(12aa; + 16). 16. (c 2 + a 2 ) 2 - 4 c'x 2 .
4. 5(c? + l)-15c-15. 17. (4ar 5 +32)-5a 2 -17a-14.
5. (m 2 -12) 2 -m 2 . 18. 98 a 5 - 16ajV + 8aw*.
6. (a + 2) 2 -7(a + 2) + 12. 19. m 6 -3m 4 + 3m 2 - 1.
7. 4 + 2 a? — (am + 05) am. 20. (a^ + a- l) 2 -(a 2 -a-l) 2 .
8. 3(a + l) 2 -19(a + l)+6. 21. 2(ar 3 -l) + 7(a 2 -l).
9. (m 2 -12) 2 -(m 2 -6) 2 . 22. m 4 -9a 4 + m 2 + 3a 2 .
10. 2 a 6 + 38 a 3 -432. 23. 12m 2 -m(w-l)-(n-l) 2 .
11. (a 2 + 4) 3 -125a 3 . 24. 2 d 10 - 1024 a\
12. (m+fl?) 2 — 7— 3(m+a;+l). 25. cx+my—mx—cz—cy-\-mz.
13. (c 2 + 4) 3 -16c 4 -64c 2 . 26. 4 mV - (m 2 + n 2 - 1) 2 .
27. (c + d)(m 2 -l)-(m + l)(c 2 -c? 2 ).
28. 126 2 -18ma + 12a6-3a 2 + 3a 2 -27m 2 .
29. 7m 2 — 7x 2 + 7n 2 — 14(az — mri) — 7 z 2 .
30. (m — l)(a 2 — asc) + (l — m)(ax — x 2 ).
31. (a-3) 3 -2(a-3) 2 -15(a;-3).
32. (a 2 + 4a + 3) 2 -23(a 2 + 4a + 3) + 120.
33. (x-2)(x-3)(x-4:)-(x-2) + (x-2)(x-3).
34. a?b + b 2 c + ac 2 - a?c - a& 2 - 6c 2 .
35. (3a + 2)(9a 2 + 2a; + 12)~(27aj 3 + 8).
36. 8(a-a) 2 + 5a 2 -5a 2 -3(a + a) 2 .
37. (c-l)(c 2 -9)-(c + 5)(c- 3)- 3(^ + 9 0.
38. (x + 2y) 2 -3(x + 2y + l)-W.
39. m 2 (x — 1) — 2 mx — m + x 2 (m — 1) — x.
40. ^(a-5) 2 + 2a(a 2 -a;-20) + (a; + 4) 2 .
CHAPTER X
HIGHEST COMMON FACTOR
142. A common factor of two or more algebraic expressions
is an expression that divides each of them without a re-
mainder.
Thus : a s b s , a 3 5 4 , and a?b 5 may each be divided by ab.
Therefore, ab is a common factor of a 3 6 3 , a 3 6 4 , and a 2 b b .
In this definition the algebraic expressions are understood
to include only rational and integral expressions.
143. Expressions having no common factor except 1 are said
to be prime to each other.
Thus : 3 abx and 7 cmn are prime to each other.
144. The highest common factor of two or more algebraic ex-
pressions is the expression of highest degree that divides each
of them without a remainder.
Thus,
{a 3 6 3
3 4 a 2 6 3 is the expression of highest degree that will
... divide each of the three without a remainder.
a 2 b 5
That is, a 2 b 3 is the highest common factor of a 3 & 3 , a 3 6 4 , and a?b & .
145. The highest common factor of two or more expressions is
the product of the lowest powers of the factors common to the
given expressions.
The abbreviation " H. C. F." is commonly used in practice,
119
120 HIGHEST COMMON FACTOR
THE H. C. F. OF MONOMIALS
146. The H. C. F. of monomials is readily found by inspection.
Oral Drill
Give orally the H. C. F. of :
1. a 3 6 4 anda 2 6 5 . 8. 15 mn z and 20 m 2 x.
2. m 2 nx 2 and mn 2 x. 9. 12 m s n 7 and 15 m 4 n 9 .
3. 2m¥and4m¥. 10. 8 c z d z m and 12 <j*#n.
4. 3 c 3 c? 5 and 6 c 2 a?. 11. 35 arfy 2 z and 42 m 2 ?/ 2 .
5. 3 atmy 3 and 9 am 2 ?/ 2 . 12. 51 a s y 2 z 3 and 17 a?/z 2 .
6. 10 a 2 m 2 n 2 and 15 m 2 n. 13. 12 arfyz, 16 afy s z 2 , and 20 aj?/ 2 2.
7. 16 x*y and 24 ?/ 2 2. 14. 18 cdm, 24 cmn, 30 cdn, and 36 dmn.
15. 5 c%, 10 cd 3 2/, 15 cdy 3 , and 20 c 2 dfy
16. 33 mnx, 44 mw?/, 55 ma;?/, and 66 mnxy.
THE H.C.F. OF POLYNOMIALS BY FACTORING
147. The H. C. F. of factorable polynomials is readily found
by inspection of the factors.
Illustrations :
1. Find the H. C. F. of
a 3 - 3a 2 - 10 a, a 3 - 8 x 2 + 15 a, and a 3 - 25a.
Factoring, a 8 - 3 a 2 - 10 a = a(a - 5) (a + 2)
a 3 - 8 a 2 + 15 a = a(a - 3) (a - 5)
a 3 -25a = a(a + 5)(a- 5)
Therefore, H. C. F. = a{a - 5) Result.
2. Find the H. C. F. of m 3 - 27, 9 - m 2 , and m 3 - 6 m 2 + 9 m.
Factoring, wi 8 - 27 = O - 3) (m 2 + 3 m + 9)
9- m 2 = -(m + 3)(m- 3)
ro 8 - 6 m 2 + 9 m = m(m - 3) 2
Therefore, H. C. F. = (m - 3) Result.
The student will recall that
(9-m 2 ) = (3 + m)(3-ra) = (wi + 3)(-m + 3)= -(ra + 3)(m-3).
H.C.F. OF POLYNOMIALS BY FACTORING 121
Exercise 39
By factoring obtain the H. C. F. of :
1. am + m,ax + x. 8. a 2 — 1, (1 — a) 2 .
2. am + m, mx + m. 9. a 3 4- 1, (1 -f a) 2 .
3. cx — dx,cy — dy. 10. c 3 — 1, 2^-f 2c + 2.
4. mx + m,mx — m. 11. c 3 — cd 2 , c 3 — 2 c 2 ^ + cd?.
5. c* + c, c 2 — c. 12. 4ar J — 4 a, aar*— 6a#4-9a.
6. m 2 + m, ra 2 -l. 13. c 3 -4c 2 -12c, 2 (^ + 4 c 2 .
7. 0^-9,^ + 33;. 14. 27m-m 4 , m 3 +3m 2 +9m.
15. ^ + 30 + 2, c»-l, c* + c.
16. §-x 2 y i ,2x i y 2 -&xy,x 2 tf-xy-§.
17. 7c?-Ucd + 7d 2 ,Uc i -Ud?.
18. c 6 -l, c3_ c , c 4 -l, 1-c 3 .
19. a 4 -« 4 , a 4 + 5aV + 4a; 4 , a 4 + aV.
20. m 2 + ma; — m — #, m* — m, m 4 — m 3 .
21. ,c 2 -3c + 2, c«_c-2, 6-c-c 2 .
22. 2 m 3 + 4 m 2 - 30 m, 4 m 3 - 20 m 2 + 24 m, 6 m 3 - 12 m 2 -18 m.
23. 8 a 3 + 64 b 3 , 12 a 3 + 48 a 2 b 4- 48 a& 2 , 96 a 2 b 2 - 24 a 4 .
24. m 4 — n 4 , m 3 4- n 3 , m 5 4- n 5 , m 3 4- 2 m 2 n 4- mn 2 .
25. (^-(m + l) 2 , (m4-c) 2 -l,m 2 -(c4-l) 2 .
26. 4ar>-14z + 12, 8^-32 a + 30, 18 a 2 - 69 x 4- 63.
27. a 2 + ax — 2 a — 2 #, am 4- a 4- m# 4- as, a 3 — ax 2 .
28. 6a 2 x + 6a 2 -6x 2 -6x,4:a 2 m-4:a 2 -±mx+4:X,2a 2 x-2x 2
-2a 2 -\-2x.
29. 3a 2 + 6a2/4-3aa;4-6iB2/, 6a 2 + 6ai/ + 6aa; + 6a^,9a 2 -9a2/
4- 9 ax — 9 #?/.
130. am + an + a + mx + nx + x, m 2 — n 2 + m — n, m 2 + m 4- w
4- 2 mn 4- n 2 .
CHAPTER XI
FRACTIONS. TRANSFORMATIONS
148. An algebraic fraction is an indicated quotient of two
algebraic expressions.
Thus : 2, ^, a2 + a6 + 62 are algebraic fractions.
b x a-2b
149. The numerator of a fraction is the dividend; the denomi-
nator, the divisor.
The numerator and the denominator are the terms of a
fraction :
The following principle is of importance in processes with
fractions :
Since, by definition, a fraction is an indicated quotient, we
may let the quotient of a divided by b be represented by x.
Then
b
The dividend being equal to the product of the divisor by the quotient,
a = bx.
Multiplying by m, am = bmx. (Ax. 3)
Considering am a dividend, bm a divisor, and x a corresponding
quotient,
a ^ = x.
bm
Hence, ^m = a (Ax. 4)
bm b
150. That is : The value of a fraction is unchanged when both
numerator and denominator are multiplied or divided by the same
quantity.
122
THE SIGNS OP A FRACTION 123
THE SIGNS OF A FRACTION
151. Three signs are considered in determining the quality
of a fraction. From the law of signs,
a _ — a _ _ — a _ _ a
b~ -b~ b -j> '
Since each fraction has the same value, -,
o
From the second fraction :
If the signs of both numerator and denominator of a fraction
are changed, the value of the fraction is not changed.
From the third and fourth fractions :
If the sign of either numerator or denominator and the sign of
the fraction are changed, the value of the fraction is not changed.
152. Consider the signs of the factors of the terms of a fraction :
■ (+<0 =| (-a) ■ (-a) ■ (+o) m
(+&X+C) (+&)(- <0 (-&)(+e) (-&)(_<>) " KJ
Note that in each fraction two signs are changed and that the sign of the
fraction is not changed.
+ (+«) = _ (-«) - _ (+«) _. _ (-«) (2 \
(+*)(+•) C+^X+c) (-6)(+c) (-ft)C-c)' W
Note that in these fractions one sign is changed or three signs are
changed, and that the sign of the fraction is changed.
(1) We may change the signs of an even number of factors in
either numerator or denominator of a fraction without changing
the sign before the fraction.
(2) We may change the signs of an odd number of factors in
either numerator or denominator of a fraction if we change the
sign before the fraction.
The following is a common application of these principles :
, etc.
(w — n) (x — y) (in — ri) (x — y) (n — m) (x — y)
124 FRACTIONS. TRANSFORMATIONS
TRANSFORMATIONS OF FRACTIONS
To reduce a Fraction to its Lowest Terms.
153. A fraction is in its lowest terms when- the numerator
and denominator have no common factor.
Illustrations :
! 45 aWc _ 3 x 3 x 5 aWc = 3 ab 2 Regult
60 a 2 b 2 c 4x3x5 a 2 b 2 c 4
2 a* — ax* _ a(a - x) (a 2 + ax + x 2 ) _ a 2 + ax + x 2 R esu i t
a 8 — ax 2 a(a — x) (a + x) a + x
3 ab + bm — am — m 2 _ (a + m) (b — m) _ — (a + m) (m — b)
m 2 - b 2 ~ (m + b) (m - b) ~ (m + b) (m - b)
a + m
Result.
m + b
For (b - m)= ( - m + 6) = - (w - 6) (Art. 152).
In general, therefore, to reduce a fraction to an equivalent
fraction in its lowest terms :
154. Factor both numerator and denominator.
Cancel the factors common to both.
To cancel is to divide both numerator and denominator by a factor
common to both. The expression "cancel" cannot be applied to any
other operation in algebra. The terms of an expression in a numerator
cannot be canceled with like terms in a denominator, for such an operation
is not division.
Oral Drill
Eeduce orally to lowest terms :
8 m 2 35afy AZrfyz 48 c 3 fe 2
' 12m* ' 2Sxtf ' 28xyz' ' 72 c 2 d 4 x
2 15 a& 39 m 2 n 2 19 a 2 fz A . lUa s x 2 z
20ac* ' 65m 3 n* ' 76a 3 y 2 z' ' 95a 3 x'z'
_ 14m 2 yi 24 c 3 d 2 q 69 m 2 ny 12 81 a 2 mn 3 x
21 mn ' 36 c A d' ' 46 m 2 n 2 y 2 ' ' 135 m 2 naT
TRANSFORMATIONS OF FRACTIONS 125
Exercise 40
Eeduce to lowest terms :
4a 2 -8a 10 m 2 -(x + l)\
4a 2 + 20a* ' (m-xf-1
3m 2 -3n 2 ■ (^-(c + l) 2
6m 3 -6r/ ' (c + l) 2 -^
1.
3.
2a*4-2a 2 + 2a 9-(a-2) 2
a 2 + a + l ' ' (3-a) 2 -4*
4 8c 3 -8 13 ar i + (a + c)a; + ac
'4c2_4 as 2 -|- (-J7L -f- c)x + cm
2a 3 -18a 4 s»-4a? + 4-<?
* 2(a-3) 2 ' ' ic2-4 + 4c-c 2 *
a* -$a* + 7x _ 4^ + 4aa; + 12a + 12a;
a?-x ' ' 6x 2 + 6ax-9x-9a '
2c 3 -16c 2 + 30c a 3 a;-3a 2 -aa + 3
2c 3 -llc 2 + 5c* ' 9a?-aW-9 + »""
4 a 3 + 4 a 2_24a 16m 4 -2m
4a 3 -36a ' ' 32m 5 + 8m 3 + 2ra*
4 c 3 + 5 c 2_21c a 2_2a5 + 6 2 -a; 2
3c 4 + 81c ' ' « 2 -a 2 + & 2 + 25a;'
To transform a Fraction to an Integral or a Mixed Expression.
155. A mixed expression is an expression having both frac-
tional and integral terms.
Thus, a + -, x — 1 H — , are mixed expressions.
a * + 1
The principle by which a fraction ivhose numerator is of
higher degree than its denominator is transformed to an integral
or mixed expression depends upon Art. 75, by which
ab + r _ab . r _ ,r
b ~ 6 6 b
126 FRACTIONS. TRANSFORMATIONS
Illustration :
£C 3 _|_'K 2 o
Change — -^ — - — to an integral or a mixed expression,
ar + 1
x 8 4- £ 2 - 2( a 2 + 1
s 3 + x (« + 1
+ x 2 - £C - 2
-fa 2 +1
-x~3
Hence,
x s 4 x* - 2 = (a*+ !)(« + !)- a; -3 = (a; 2 + 1) (x 4 1) , -s-3
« 2 + l x 2 + l x 2 + l x* + l
By changing the sign of the numerator the form of the result becomes
g + l + -»-g = a . 4 a + -(* + 8 ): = s + l -■*+£. Result.
T x 2 + l x 2 + l a 2 4 1
Therefore, to change a fraction to an integral or a mixed
expression :
156. Divide the numerator of the fraction by the denominator.
Write the remainder over the denominator, and annex the result-
ing fraction to the integral quotient obtained. If the sigyi of the
first term of the remainder is negative, change the signs of the
entire remainder, and the sign of the fraction annexed will be
negative.
Exercise 41
Change to integral or to mixed expressions :
c 4 4-l
1. C -JLL±. 5.
c-fl
m 2 + m + 1 : 6
m — 1
a 2 -Sa-2
7.
a — 5
h4a-
a 2 + a — 1 m
4 a 4 -f- 4 a + 1 .
c 3 -c-3
c 2 +2
2n 8 -3n 2 + 2
n 2 - n ■ + 3
a 3 + a 2 + a-7
a 2 +2
3 m 4 — m 2 — ra
+ 1
CHAPTER XII
FRACTIONS (Continued) — LOWEST COMMON MULTIPLE.
LOWEST COMMON DENOMINATOR. ADDITION
THE LOWEST COMMON MULTIPLE
157. A common multiple of two or more algebraic expres-
sions is an expression that may be divided by each of them
without a remainder.
Thus, a 6 & 6 will contain a 3 6 8 , a 8 6 4 , and a 2 6 5 .
Therefore, a 6 6 6 is a common multiple of a 8 & 3 , a 3 6 4 , and a 2 6 6 .
In this definition the algebraic expressions are understood
to include only rational and integral expressions.
158. The lowest common multiple of two or more algebraic
expressions is the expression of lowest degree and least numeri-
cal coefficient that will contain each of them without a re-
mainder. Thus :
(3 a s b s 30 a 8 6 5 is the expression of lowest degree and least
2 a 3 6 4 numerical coefficient that will contain each of them
5 a?b 6 without a remainder.
That is, 30 a 3 6 5 is the lowest common multiple of 3 a 8 6 3 , 2 a 8 & 4 , and
5 a 2 6 5 .
159. The lowest common multiple of two or more expressions is
the product of the highest poivers of all the factors that occur in
the given expressions.
The abbreviation "L. CM." is commonly used in practice.
127
128 FRACTIONS
THE L.C.M. OF MONOMIALS
160. The L. C. M. of monomials is readily found by in-
spection.
Oral Drill
Give orally the L. C. M. of :
1. <?m and cm 2 . 7. 16 ah and 24 a 2 c.
2. ra 2 ?/ 3 and m 3 y 2 . 8. 9 x 2 \f and 18 a^i/V.
3. 2 m 2 n and 3 m s n. 9. 10 c 2 dx and 8 cd 2 y.
4. 3 xy 2 and 6 ax. 10. 16 mn s y and 12 m 2 ^ 2 ?/ 3 .
5. 8 my and 12 wy. 11. 15 ah and 20 cV.
6. 10 ra 2 n 2 a; and 6 mtfx 2 . 12. 25 aW and 20 aWc 5 .
THE L.C.M. OF POLYNOMIALS BY FACTORING
161. The L. CM. of factorable polynomials is readily
found by inspection of the factors.
Illustrations :
1. Find the L. C. M. of a 2 - ab and ab - b 2 .
Factoring, a 2 — ab — a (a — 6)
ab-b' 2 = b(a-b)
Therefore, L. C. M. = ab (a- b) Result.
2. Find the L.C.M. of 0?+ a - 2, a 2 - a- 6, and a?-4« + 3.
Factoring, x 2 + x - 2 = (x + 2) (a - 1 ) ,
SC*-x-6 = (x-3)(x + 2),
x 2 - 4 x + 3 = (x - 3) (x - 1),
Therefore, L. C. M. = (x + 2) (x - 1) (x - 3). Result.
3. Find the L.C.M. of 2 a 3 + 8 a 2 + 8 a, ±x-a 2 x, and
5a 2 -20ci + 20.
Factoring, 2 a 8 + 8 « 2 + 8 a = 2 a (a + 2) (a + 2)
4 x - a 2 x = - x (a + 2)(a - 2)
5 a 2 - 20 a + 20 = 5 (a - 2) (a - 2)
Therefore, L. C. M. = 10 ax (a + 2) 2 (a - 2) 2 . Result.
THE LOWEST COMMON DENOMINATOR 129
Exercise 42
By factoring obtain the L. C. M. of :
1. x 2 + x, xy -f y. , 5. 2 a(a + 1), 3a 2 + 3 a.
2. am — a,m 2 — m. 6. mV-9,m¥-6mn + 9.
3. c* + c,c?-c. 7. (W-l, (W-l.
4. 2/ 2 — 1, (y — l) 2 . 8. a 3 -4a 2 + 4a, a 4 -8a.
9. ^-ec 5 ^^, (^-81 c.
10. m 3 -f 3ra 2 + 2m, m 3 + 4m 2 + 3ra.
11. 3 a; (a? + a; + 1), 4a 4 + 4a.
12. 7a 3 -175a, 3 a 3 -30 a 2 + 75a.
• 13. (x + y) 2 , (y- x y,x>-tf.
14. tf(y-z\y(tf-#),y + z.
15. 3(aj» + ajy), 8(^- t/ 2 ), 12 (x 2 -i/ 2 ).
16. 2c 3 -c 2 -c, 2c 3 -3c 2 -2c.
17. 5« 4 -5^, 4^-8^ + 40?.
18. 7a 3 -7a, 4a(a-l) 2 , 3a 3 + 6a 2 + 3a.
19. m 3 + l, 3(2 + 3ra + m 2 ), 4ra-4m 2 + 4m 8 .
20. 18-2^,3^-81, 4a; 4 -324.
21. am-f-a+ m + 1, am 2 + am — m 2 — m.
22 . 5 c (c - 2) 2 , 4 ca> + 12 c - 24 - 8 x, 24 c 2 - 3 c 5 .
23. (a + xf-1, a 2 -(x-l) 2 , (a-lf-x 2 .
24. 4a 2 -14a; + l2, 8 »* - 32 a + 30, 18 as" - 69 » + 63.
25. 6 a 2 A5 + 6 a 2 — 6 x 2 — 6 a, 4 a 2 m — 4 a 2 — 4 m# + 4 as,
2a 2 <c-2z 2 -2a 2 '+2a.
THE LOWEST COMMON DENOMINATOR
162. The lowest common denominator of two or more algebraic
fractions is the lowest common multiple of their denominators.
The abbreviation " L. C. D." is commonly used in practice.
SOM. EL. ALG. — 9
130 FRACTIONS
Illustrations :
1. Change — , — , and - to equivalent fractions having a
o 4 o
common denominator.
The L. C. M. of the denominators 3, 4, and 6 is 12. L. C. D. s 12.
Dividing each denominator into the common denominator, we have :
12 = 4, « = 8 ,!? = 2.
3 4 '6
Multiplying both numerator and denominator of each fraction by the
respective quotients from the division, we have :
5^ = 5^ x 4 = 20^ H
3 3 4 12
3z_3z x 3 = 9z. Regult
4 4 3 12
x_ x 2_2_a
6 6 2 12
x = x x 2 = 2_x i Regult>
2. Change — and x „ to equivalent fractions having
af + x xr — x
a common denominator.
The L. G. M. of x 2 + x and x 2 - x is a: (jc + 1) (x — 1). Hence
L. C. D. = x (x + 1) (x - 1).
Dividing each denominator into the L. C. D., we have :
x(x + l)(x~l) _ x 1 . a (g + 1) (a - 1) = g | x
X 2 + X x 2 - x
Multiplying both numerator and denominator of each fraction by the
corresponding quotients obtained :
£rl x *=± = I^zDL. Result.
X 2 + X X-l x(x' 2 -l)
x + l x x+l = (x+l) 2 . Regult
x 2 -x x + 1 x(x 2 -l)
In general, to change two or more fractions to equivalent
fractions having a common denominator:
THE LOWEST COMMON DENOMINATOR 131
163. If necessary, reduce the fractions to their lowest terms.
Find the lowest common multiple of the given denominators,
for the common denominator.
Divide each given denominator into the common denominator.
Multiply both numerator and denominator of each given fraction
by the respective quotients obtained.
Exercise 43
Change to equivalent fractions having a common denominator :
3m 2m 5ra 2 3
l " T' 3 ' 2 ' 7 '
5a 3a a
2 - T' T' S' 8 *
AAA
2a 4a'3a
9.
4. A, A, 1. 10
ax 1 bic 9 ab
5 ± 2- ± 11
' a°b' aV aV
6. -i-. -i-, JU. 12
c+1' c-
m
1
m
m + n n
c
— m
c
<?-!' (c
2
-i) 8
5
a
(y-1) 2
a
3a + 4'
a;
9 a 2 -16
a;
m 2 n2/' mny ' mny 2 ' oj 8 — 1 ' a^ + a^ + o?
3 2
13.
14.
m 2_ m _e ? m 2 + 2m-15
1 2
2c 2 -c-l' 6c 2 -c-2
15
16.
17.
a 2 + 4a + 3' a 2 -a-12' a
1 2 3
3+1' (a + 1) 2 ' (a; + l) 3 *
c c 3c
3(c + 1)' 4(c - 1)' 2(c 2 + 2 c + 1)
132 FRACTIONS
18.
19.
20.
m m m
m 2 + m m 2 — m -j- 1' m 4 -f- m
3a; 4a; 5 a;
2a7^2' 3a; + 3' 4a^-4*
1 1 1
3^-15' 52/ 3 -125y'72/ + 35'
ADDITION AND SUBTRACTION OF FRACTIONS
__ . __ a , b . c a + b + c
By Art. 75, -H h-= —
J a; a; a? a;
That is, two or more fractions having the same denominator
may be added by adding their numerators, and writing the sum
over that common denominator.
By Art. 163, any two or more fractions may be changed to
equivalent fractions having the same or a common denominator.
From these two statements it is clear that any given frac-
tions may be added.
The term " simplify " is used to include both operations of
addition and subtraction.
Illustrations :
x x
1. Simplify
a; — 1 # + 1
The L. C. D. is x* - 1.
Dividing each given denominator into the L. C. D. and multiplying the
corresponding numerators by the respective quotients, we obtain :
x _ (x + l)x — (x — l)x
1 x + 1 x*-l
_ x* + x-x* + x
x*-\
2x _ u
= -2 — r Result.
x 3 — 1
ADDITION AND SUBTRACTION OF FRACTIONS 133
o _. ... 2m 3 9 ra + 1
2. Simplify — 3 — r — 2 ' .
Every integer may be considered as having a denominator 1. There-
fore, in adding fractions and integers the integers are multiplied by the
L. C. D.
The L. CD. = to 3 - 1.
Dividing each denominator into the L. C. D., and multiplying as be-
fore, we have :
2 to 3 2 to + 1 _ 2 to 3 -2(to 3 - 1) - (to + 1) (to- 1)
to 3 — 1 1 to 2 + to + 1 to 3 — 1
2 to 3 - (2 to 3 - 2) - (m 2 - 1)
TO 3 — 1
=l=J#. Result.
TO 3 — 1
In general, to add algebraic fractions :
164. Reduce the given fractions to lowest terms.
Divide each denominator into the lowest common denominator,
multiply the corresponding numerators by the quotients obtained,
and write the sum of the resulting products for the numerator of
the result.
The sign of each fraction becomes the sign of its numerator in
the addition.
Oral Drill
Simplify orally :
i. M.
X X
6.
M-+2.
XXX
11.
ax ay xy
2. ---•
m m
7.
a ,b c
m m m
12.
a b , c
x y z
3 -4-^.
3 ' 3 + 4
8.
3_2_a
XXX
13.
2+1 + - 6 .
b a
4- 2+f.
a Z a
9.
Sx 2x
14.
m 1 n
n m
2 a 3 a
x 2x
10.
J>_ _ 3_
8 a 7 x
15.
i-^7
134 FRACTIONS
Exercise 44
Simplify :
i. ~+ x 4 1 - i2. -^-2 • a
a + 1 a — 1
2 a + x ~ 3 _l 1 ~ a + x , 13 3
4 a 2 + a a 2 — a
3 £±I_1_^+A 14 3c 4c
2 6 ' 2c-2 3c + 3*
4 2m-l m-1 7m+l 12 5
5 3 10 " 7^-7 3a; 2 + 3*
5 c ~ - 2 c ~ ^ - 16 m + 3 m + 1
12 6 ' m + 4 m + 2*
a 2 4- 2 a — 1 a + l < m + 9 m — 9
4 a 2 ""3a 2a* * m-9 m+9'
a—b.b—c.c—a ■ _ 5 a + 7 5 # — 7
7. — r~ + "T 1 18 - i s*— •= =•
a& 6c ca 5<e — 7 5sc+7
x x + y c 2 — c + lc 2 +c + l
9. #- + -^« 20. *«-* 4 2C + 1
2m m + 1 (2c + 1) 2 (2c - 1)
in 5 3 01 m — 5» . 10m#
io. _ _ _. 4i, ___ + —
8xy 5xy—l m + 5x m 2 — 25 a 2
11. X !£-. 22 . *" 2 *+ 2
ax ax 2 -! x 2 -2x + ± a? + 2x + 4:
1
23.
c^-f 7c 2 + 12c c 8 + 8c 2 +15c
24. JL-h*-. 3
ax ax 2CUC 2
a -J- <c a — a; a* — x 2
26. g3j| 3 * + l| (3*+7)tt
x + 1 *— 1 ar 1 — 1
27.
28.
29.
ADDITION AND SUBTRACTION OF FRACTIONS 135
a* + ra 4 a 4 m 4
"ay aY + tf* a 4 + ay*
a-1 a + 1 a 2 4- 11
2a + 2 3a-3 6a 2 -6
3a; + 2 13a?-38 2a;-3
a; — 5 a 2 — 4# — 5 x + 1
30 4c 2 2c 8(^ + 20 + 1
* 4c 2 +2c + l" r 2c-l 8c 3 -!
31 a — 1 a+1 | a-i
32.
a 2 -5a + 6 a 2 -3o + 2 a 2 -4a + 3
1 1 3a
a^ + 2a; + 4 ^-2^ + 4 a 4 + 4^ + 16
165. The process of addition of certain forms of fractions
is simplified by application of the principles governing the
signs of fractions (Art. 152).
3 2 5a
1. Simplify
a -f- 1 1 — a
Changing the form of the second fraction, in order that the terms of
the factors in all the denominations may be in the same order, we have
(Art. 152) :
2 2 2 2
1 — a — a + 1 — (a — 1) a — 1
Therefore, 3 2 6a _8(a- 1) + *« + lV-fi«
a + la-1 «2_i a 2_!
— ~~ ' ? Rpfmlt
a 2 _ 1 1 _ a 2'
2 Simplify * * *
1 " (a-l)(a-c) (l-a)(l-c) (c-a)(c-
-1)
Changing the signs of both factors of the denominator of the second
fraction, we have changed the signs of an even number of factors, and
the sign of the fraction is not changed. Changing the sign of one factor
in the denominator of the third fraction changes the sign of the fraction.
136 FRACTIONS
Therefore,
1 1
(a-l)(a-c) (l-a)(l-c) (c-a)(c-l)
1 1
(a -1 ) (a - c) (a - 1) (c - 1) (a - c) (c - 1)'
L.C.D. =(a-l)(a-c)(c-l)
= (c - l) - (g - C ) + (a - 1)
(a-l)(a-c)(c-l)
m 2c-2
(« - 1) (« - c) (c - 1)
= ? . Result.
(a-l)(a-c)
Exercise 45
3c 2 2
Simplify
3.
8.
1_C2 c _l c + l
1 1_ 10a?
5x + l 5x-l 1-25 a^
m 2 + 3m + 9 m 2 -3m + 9
m + 3 3 — m
4 c — 2d , 2c — d . 3(cy — ay)
.■_ m + ft . a + ft
o. — ■
(m — a)(c — m) (a — m)(c — a)
i + i + i
(a, -l)( x -2) (x- 2) (3 - a;) T (1 - a?) (a? - 3)
a; + a 2(c — a) x + c
(a? — a) (# — 6) (c — a) (a; — a) (b — x) (c — a;)
1.1 1
(b — c)(b — a) (c — a) (c — b) (a — 6) (c — a)
(1 — a — a;) a?(a; — 1 — a) a(a — 1 — x)
(x-l)(l-a) (x-a)(x-T) (a-l)(a-»)"
CHAPTER XIII
FRACTIONS (Continued) — MULTIPLICATION. DIVISION.
THE COMPLEX FORM
MULTIPLICATION OF FRACTIONS
(a) A Fraction multiplied by a Fraction
The product of two fractions is obtained as follows :
Given two fractions, - and -, and let
(3)
(Ax. 3)
(Ax. 3)
X
y
m :
= 2 (l) andw=^(2).
x y
Multiplying (1) by (2),
mn = - . — •
x y
Multiplying (1) by x,
Multiplying (2) by y,
Therefore, multiplying,
mx = a.
ny = b.
mnxy = ab.
Dividing by xy,
ab
mn = —
xy
But from (3),
mn = - . — •
x y
Therefore,
In general :
a b_ab m
x ' y xy
(Ax. 4)
(Ax. 5)
166. The product of tivo fractions is a fraction whose
numerator is the product of the given numerators, and whose
denominator is the product of the given denominators.
(h) A Fraction multiplied by an Integer
Since any integral expression, b, has a denominator, 1 :
From Art. 166, «x& = «.£ = ^.
X X 1 X
137
138 FRACTIONS
That is :
167. A fraction is multiplied by an integral expression if its
numerator is multiplied by that expression.
The process of multiplication of fractions is simplified if
factors common to numerators and denominators of the given
fractions are canceled before multiplication.
Illustrations :
8c 8 c 2 x 25 m 2 s 4 = 8 ■ 25 a*c*m*& = 5a 2 c Regult
15 am s x 8 16 ex 15-16 acm z x* 6 m
The cancellations are : 8 in 16, 5 in 25 and 15, a in a 8 , c in c 2 , ro 2 in
m 8 , and x* in x 4 .
2. Multiply 2+1* ,("*-!)' X 5^i.
m — 1 m 2 — 3m + 2 m 2 — 1
Writing each fraction with numerators and denominators factored,
m + 1 x (m - l) 2 x m 2 -4 _ m + 1 x (m-l)Q-l) x (m+2)Q-2)
m-1 m 2 -3m + 2 m 2 -l m-1 (m-l)(m-2) (m+l)(m-l)
Canceling common factors = m + 2< t jjgsult.
In applying cancellation select factors for divisors from the
numerator only. Begin at the left of the numerator and seek
a possible cancellation for each new factor considered.
Multiply orally:
Oral Drill
1. aa %/ x 6m .
3 m ax
5.
17 c 2 2n 8
6n 4 51c 4 '
2 Say y 10c 2
5 c 9y
6.
5aa^ 39^2
13 xh 10 afy
3 2 mn 9 c?
3 cd 7 cm
7.
42^ 2 x 10 z
15 xy ' 28 afyz
4 5m 2 x 21c 2 d #
3 cd 10 am
8.
32 x 2 y K 27 c 4 m 2
9 <?x 16 my
MULTIPLICATION OF FRACTIONS 139
cd
Exercise 46
Simplify :
o + l ^ (c + 1? (c-1)'
*' a -2 a 2 -l c c (c 2 -!) 5
2 . ^ X / + 1 ■ 5. ?^l.(?*-2x + ±).
05 + 1 a^-^ + a; ar* + 8 v y
m 2 — mn ^ m 2 — 4mn+3 w 2 fl a^4-2a; + 4 a^ + 8
(m-rc) 2 m 2 -n 2 a?-4 ^-8
7 3(c + l) 2 7(c-l) 3 4(c + l)
' 2(c-l) 2 6(c + l) 2 7 c
8.
^-9a;4-18 ^-9
^-27 ^-3aj-18
• c 2 -7c+12 N/ c?-c-2
c2_3 c _4 ^-50+6
10 3-4a+o 2 x a 3 -16a
5a + 4 a 3 -7a 2 + 12a
n . ^=9 x(a 2 + 2a-15)x a2 - 5a + 25
a 3 + 125 v y a + 3
m 2 + 2 mn + n 2 — x* m 2 — 2 mx + x* — n 2
12.
13.
(m -f- n -f- a;) 2 m 2 — 2 mn + n 2 — x*
c + 3 c 4 -81 c*-9 (3-c) 2
(c-3) 2 '(c + 3) 2 'c 2 + 9 (c-3) 2 '
14 ^—9 am + 2 a am — 2m — 2a + 4:
m 2 -± bx-3b ax + 3a-2x-6
15 s 3 -8 y S'-2ft + 4 g + 2
* a? + 8 aj» + 2a? + 4 a>-2
16.
3tf2_a;_2' 3a 4 -3a; 10-10<e
3*»+2* 5a^-10a; + 5 6^ + 6^ + 60;
140 FRACTIONS
DIVISION OF FRACTIONS
(a) A Fraction divided by a Fraction
The quotient of a fraction divided by a fraction is obtained
as follows:
We will assume that - +- - = m. (1)
x y w
Now a dividend equals the product of the corresponding divisor and
quotient.
Therefore, - = - x m.
x y
Multiplying by £ , 2 x %*:$ x m x *..
6 x b y b
Whence, -x y =m. (2)
x b , v
Hence, from (1) and (2) , 5 - 5 = 5 x y - •
w v " x y x b
In general :
168. The quotient of a fraction divided by a fraction is the
product of the dividend by the inverted divisor.
169. The reciprocal of a quantity is the quotient obtained by
dividing 1 by that quantity. Thus :
- is the reciprocal of a.
a
1 -f- - = -, the reciprocal of -.
3 2 3
(b) A Fraction divided by an Integer
From Art. 168, £ -- y = 2 + ^
a; x 1
a 1
= - x -
x y
xy
DIVISION OF FRACTIONS 141
In general :
170. The quotient of a fraction by an integer is the product of
the given fraction by the reciprocal of the integer.
The first step in a division of any fraction is, therefore, the
inversion of the divisor; whence the process becomes a multi-
plication of fractions.
Illustrations :
1. Divide JJ& by M.
S2cy J 40 cy
7 a 2 x . 21 q 2 s _ 7 q 2 x 40 cy _ 5 x Result
32 cy " 40 cy 32 cy 21 a 2 z 12 z
2 . Divide q2 - a " 2 by a2 ~ 3a ~ 4 .
a 2_5 a+ 6 y a 2 -8a + 15
q 2_ g _2 | a 2-3a-4 = (a-2)(g + l) . (o-4)(g + l)
a 2 - 5 a + 6 ' a" - 8 a + 15 (a - 2) (a - 3) ' (a - 3) (a - 5)
= (q+l) x (a-3)(a-5)
(a -3) (a-4)(a+l)
a -
a-4
Result.
3. Divide - ^ + 27 by ^-3^+9.
*» + 27 ^, (x2 _ 3a;+9)= x3+27 1
x» + 6x 2 + 9x x 3 + 6x 2 + 9x a j2_3 iB + 9
= (s + 3)(a«-3g+9) x 1
a; (x + 3) (a + 3) x 2 - 3 x + 9
= Result.
x (x + 3)
Exercise 47
Simplify :
1 a 2 -^ . a-x (m 2 -!) 2
a 2 + ar> ' 3a*+3aj»" m 3
2 c 2 - 9 c-3 (ft + 2) 2 . a? -4
c 2 + 2c ' c 2 + 3c + 2* ' a?-5 * as 8 -125'
4- (m 2 - 1).
142 FRACTIONS
k c 2 — ccc— 6ar* . c+3x x 2 — 4a; + 3 t x — 1
(P-9CX 2 ' c+2x 3-x ' 2 — x
n a? + 6x-7 . x 2 -3x+2 aj"-l. 1-x
x*-±x-2l x* + x-6 1-x 3 ar' + l+a;
^_8aj + 15 lx-10-x 2
9.
10.
6 — 50? + ^ ' x 2 — 4a;+4
(a.-l)2_ a 2 ^_(i_ a )2
(a._ a )2_i ' ^_( a _i)2'
X1 a 2 -3a x <i 2 + 3a + 9 . a 3 -27
12
13
a 2 + 3a a+3 (a + 3) 2
^-^-12 g^-2a;-3 . a^+a;
a^-9 a?2-2a;-8 ' x-2
a? — ax — a-\-x_a? — 8 8 — a 3
2-3a + a 2 a 3 -^ a? + ax + x*
14 a 2 + l x a 2 + (a + l) 2 (a-l) 2 . a g + l
' a 2 -l a 2 (a 2 +1)+1 ' a 3 -l
171. When addition and subtraction with multiplication and
division occur in the same fractional algebraic expression the
additions and subtractions are first performed.
Illustration :
Simplify f^+l-^U^.-G + ^-Y
* J \x-2 x + 2J V x + 2j
(x + 1 x-l\.( 3x 6 12 \ r ( x + l)( x + 2)-(x-l)(x-2) -l
\x-2 x + 2/ ' V £ + 2/ L x 2 -4 J
. r (3s-6)(g + 2) + 12 l
L x+2 J
6 a; 3 x 2
a; 2 - 4 x + 2
6s a + 2
x 2 -4 ' 3a: 2
: - Result.
*(»-2)
DIVISION OF FRACTIONS
143
Simplify :
Exercise 48
1.
fx a\ f aV y
\a xj \x -f- a J
V3 m]\m 2
3m
6m + 9
6a;
>
' \x J\m 2 -2mx-3x 2 J\3 J
\a + z J\a — z J ar — z 2
(2. + « + D(8.-7 + ?) + (.-g.
+2 y\
( c + 2 y-(P / 2d Y
(c + d + 2f \ ^c + 2-d)
10. / r 2a? + 5 + ?V3a-7 + -
11. (c + 2y-(c-2)' ; 8c '
3m + 3
14. r * 4 i^-iu^+i+^j.
LC^-ox 2 ) 2 ]^ 6 ) W a 1 )
Vm+i^ J\m-i A™- 1 A^+i /
[2Va + 2 a-2y/ 2a 2 +4aJ V ;
144 FRACTIONS
[_m J n mn J |__ftMr mn m'nj
" [(•^)("8(- + ;)> ' w+ * w ' +1 '
ti4 !
a-\-xj c(acx + a + x)
/'/7_1^2 ,v7_/*_9/7_i_9
\ cy \acx -f-
c 4_2c 3 -c + 2 (d-1) 2 . cd-c-2d + 2
±-d 2 ' c 2 -c ' d + 1
FRACTIONS IN THE COMPLEX FORM
172. A. complex fraction is a fraction whose numerator or
denominator, or both, are fractions.
The order of processes used in simplifying complex fractions
varies with different types, and no general statement will cover
all possible cases that arise. The student will easily under-
stand the following
Illustrations :
1. Simplify f ■ x *
1_1 c 3 + d 3
d c
c 2 + cP c 2 + d? - cd
d y c*-cP _ d x c 2 -d*
1_1 c 8 + # c-d c 3 + d 3
d c cd
tf-cd+d* cd ' (c + d)(c-d) _ c
d c-d (c + d)(c 2 -cd + d 2 )
x
2. Simplify * + 2 x ~ 2 ,
x-2 x+2
The L. C. D. of both numerator and denominator is (x + 2)(x
Multiplying both numerator and denominator by (x + 2) (a; — 2),
FRACTIONS IN THE COMPLEX FORM 145
x . x
x + 2 x-2 _ x(x-2) + x(x + 2) _ x 2 -2x + x 2 + 2x _2x 2 _x
~lc ~ x(x + 2) -x(x-2) x 2 + 2x-x 2 + 2x 4x 2*
x - 2 x + 2 Result.
3. Simplify 1 + ^
(The work begins by first sim-
■j . 2 plifying the lowest fraction, etc.)
1 — x
i+ 1 \ = i+ t-V= i+ 7to =1+ ^
t , 2 3-s 3-x 3-x
1-x 1-x
= 1 +f_=^ = _i_. Result.
1+x l+x
Exercise 49
Simplify :
6 . ^-1- "- 1 .
1.
•*■■■ M . ■' s
x — o
. 3 \
x — o
x + 3 2
2 x + 3
1 1
2 z + 3
c-1
2o- c -±|
c + 1
x*-tf-l ^^
2a
<C + 1
7.
x x+1 , x— 1
2 a-l + a; + l
5-6s 2 + a 4
^Zd + l £=2 + 1
^+4_^^+2_
a^-4 * „ x-2
9 c 2 + 4 c 2
ar* + 4 x + 2
4 c 2 + 4
a*-ar* + l c-2 c + 2
2a c+2 c-2
1-1 ,-1 2 + 1 x- 1 -
m 2 -9 3 m 2 y j
m 3 + 27 m 2 + 9 1 ,1 1
-—Z m 2-- 2/ + - y--
3 a? 2 a;
SOM. EL. ALG. — 10
146
FRACTIONS
11.
12.
m
m
+i
i+»
l+m + m !
i + 2
■i-(H-m) 2 .
c —
2 +
13.
14.
a + 1
— 1
a-1
+ 1
a-1
2c + <b
Hi
a + 1
1
-1
a-1
a + 1
2c + x
2 ex
2c + x
U-x 2
15.
16.
17.
18.
a + 2 2 1 a(2-a)
l-2a l+2a
l + 2 (" + 2 ) a + !^L
T l-2a ^l + 2a
1 +
a — 1 a+1 a— 1
a + 1
a-1
a—1 a+1 a—1 a+1
a+1 . a—1
-1 +
a+1 a—1 a+1
1 + A+I
9 3a a 2
[ ~a 3 + 27 . a 2 + 9~ | . f (a-3) 2 + 3a . ft IV]
[a 8 -27 ■ a 2 -9j ' [_(a+3) 2 -3a ' \ 9 a 7J
U + 2 JL (* + 2) 8 J
(i rnL\ ( ^ + i + ^_A
V (x + 2yj V(* + 2) 2 ^x + 2j
CHAPTER XIV
FRACTIONAL AND LITERAL LINEAR EQUATIONS
PROBLEMS
173. To clear an equation of fractions is to change its form
so that the fractions shall disappear. This change is ac-
complished by the use of the L. C. D. of the given fractions.
Given; «±« = *±«>.
4 o
Multiplying both members by 12,
12 (x + 6) 12 (x + 10) t
4 6
Reducing, 3 (* + 6) = 2 (x + 10).
And the original equation is merely changed in form and is
free from fractions.
In general, to clear an equation of fractions :
174. Multiply both members of the equation by the L. C. D. of
the fractions, remembering that the sign of each fraction becomes
the sign of its numerator. Solve the resulting integral equation.
Illustrations : #
I. Solve^±l_^z2 = 6- £ .
3 ' 5 2
The L. C. D. = 30. Multiplying both members by 30,
10(2x + l)-6(3x-2)=15(6-x).
20 x + 10 - 18 x + 12 = 90 - 15x.
20x - 18x + 15x = - 10 - 12 + 90.
17x = 68.
x = 4. Result.
147
148 FRACTIONAL AND LITERAL LINEAR EQUATIONS
x-1 x + 2 „ 2x?
2. Solve
2-
x + l x — 1 x 2 — l
Multiplying both members by the L. C. D., (% + l)(x - 1),
!)•-(* +!)(* + 2) = 2(x 2 - 1)
x 2 -2a; + l
From which
x 2 -Sz-2 = 2x 2
1
2 a*.
-2x2.
x = — Result.
5
Solve :
x 4-1 ,a5 — l
3
*-l
1.
3.
4.
4-
4
* + l
2
a; + 1
3
x-1
= 4.
3.
Exercise 50
' 2x-l A-x 2x+l
o.
6.
3
2*4-1
5
3*4-1
14
15'
7.
= 4:' 8.
3
5 10
4
»-l
Q 2*4-1
3
J_ 5
3
x + 2
3*4-1 2*4-1
4
3 2
3
a; 4-2
*-6 1-2*
= 0.
4-1.
9. i.(*_l)_f(*4-l)4-2=0.
10. f(2*-l)- T V = -f(5 + 3*).
11. (*4-|)(^-i) = * 2 .
12. (x-l)(x + i) = (x-l)(x-i).
13. i(*_2)-|( a; - r -l)-i(*-2) = 0.
14. 2f
*-fl
2x-
15.
(*4-l) 2 ' (*4-2) 2 ^ 14-* !
rA
16.
17.
18.
3
x — 6
#4-5 *+3
2*-3 4*-5
3* + 4
3*-2
6*-7
2*-5
6* — 1 4*4-1
= 0.
19.
20.
21.
4*-l 3*-l
a? 4-1 " x-1
4 * *
1.
* — 5 x
5*-l
3 = 0.
*4-3
4-2
7*-3
*4-l
= 0.
SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS 149
00 2x Sx x . 2x 5
Z6. — -J- — — — -+-
23.
» + l a? + l 2a; + 2 3a? + 3 6
a^ + 2a? + 4 2 a; ^-2^ + 4
a + 2 4-a 2 x-2
•24. _6 2 _ 6.
25.
26.
27.
2a; + 3 3-2a 4a 2 -9
3a;-4 18g 2 + a? = 3a? + 4
3a> + 4 9^-16 4-3z*
9a^ + 3a? + l 6 a;== 9a;2 - 3a? + 1
3a; + l l-3a;
2 3 4
<e 2 + 3» + 2 ^ + 405 + 3 x* + 5x + 6
28. * + 1 + £-1 = * + * ,
2^-a;-l 2x 2 -3x-2 ^-3^ + 2
SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS
(a) The Independent Monomial Denominator
Illustration :
175. Solve ^±i_A^l_^-l = 7.
3 2a + l 2 6
The L. CD. of the monomial denominators, 3, 2, and 6, is 6. Multi-
plying both members of the equation by 6, we have,
2(3x + 4) _ 6 ( a: - 1 ) -3(2a;-l) = 7.
From which 6 x + 8 - 6 ( z ~ *) - 6 x + 3 = 7 •
2x+ 1
At this point note that all x-terms outside of the fraction disappear.
In any simple equation of this type the unknown term
similarly disappears when the equation is cleared of monomial
denominators. If the unknown term does not disappear except-
ing from the fraction having the binomial denominator, error
has been made in the work.
150 FRACTIONAL AND LITERAL LINEAR EQUATIONS
Transposing and collecting,
Dividing by — 2,
Clearing of fractions,
6(s-l) = j
2x + l
2x + l
3(x-l) =2(2x + l).
3x-3 = 4x4-2.
x = — 5. Result.
Solve
1.
2.
4.
5.
6.
7.
8.
9.
Exercise 51
x 4-1 a; — 1 _ x— 1
3 «4-l"~ 3
3a?-l a? + l == 2a? + l t
3 a; 4- 2 2
4a?+5 x + l = 2x — l
4 »— 1~ 2
2fl + l a;— 1 5fl? + l = Q
2 z + 10 5
2x — l ^z2l. — a lZll —0
4 2^4-4 2 ~~ '
3a; — 4 9a; — 4 a; 4- 11 x
3 12 12a; -27 4
2a?4-7 , x Sx — 1 5a;4-6
= 0.
12
3a;-3
10.
3a?4-5 4a?-l 2a? + l = 75
2a;-5 6 3 6'
8a;-5 2j 16 a; + 3 = x-H
8 2 16 a?-5*
2 a;- 5 3a? + l = 3a? + £ , 3a? — 7
4 8 a?— $■ 24
SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS 151
(b) Forms having Four or More Dissimilar Denominators
176. Illustration :
, x— 3 x—2 x—7 x — 6
Solve - = -•
x—2 x—1 x—b x—o
To avoid the use of a common denominator having four binomial
factors, each member of the equation is first simplified independently of
the other.
Combining separately,
(x - 1) (x - 3) - (x - 2) (x - 2) = (x - 7)(x - 5) - (x - 6) (x - 6)
(x - 1) {x - 2) (x - 6)(x - 5)
imp ymg, _______ = ______ .
The fractions being equal and having the same numerators, it follows
that the denominators must be equal. Therefore,
(x - 1) (x - 2) = (x - 6) (x - 5).
x 2 - 3 x + 2 = x 2 - 11 x + 30.
8 x = 28.
x=\. Result.
Solve
3.
Exercise 52
1
1
1
1
„-5
x — 4
x — 3
x-2
1
1
1
1
x + 1
x + 2
x + 3
« + 4
2
1
x — 4
1
x-1
2
2a-3
2x-
7
11
11
7
7
6.
x + 3 x — 4 a-j-2 x-9
x + 2 a;+3 _ a; + 4 a; + 5 <
a;+5 x+6~~x + 7 x + S'
x — S x— 7 _ x— 5 a;— 4
#-10 a;-9~a;-7 #-6
152 FRACTIONAL AND LITERAL LINEAR EQUATIONS
(c) Literal Fractional Equations
177. In literal equations either a portion or all of the assumed
known quantities are represented by letters, the first letters of
the alphabet being ordinarily chosen for these known quanti-
ties.
Illustration :
Solve £±^ + i=l^.
x — a 2x — a
Multiplying both members of the equation by L. C. D. (x — a) (2 x — a),
we have,
(x + a) (2 x — a) + (x — a) (2 x - a) = (4 x - a) {x - a).
2x* + ax - a 2 +2x 2 - 3 ax + a 2 = 4x 2 - 5 ax + a 2 .
2x 2 + 2s 2 - 4& 2 + ax - 3 ax + 6 ax = a 2 - a 2 + a 2 .
3 ax = a 2 .
Result.
Exercise 53
Solve :
1. 3 a — 4: ex = 7 a — 6 ex.
2. (c + d)»+ (c-d)a = 4c 2 .
3. m(sc + n) -J- n(sc + m) = 1 + 2 mn.
4. (a + a)(a + l) = (a; + a + l) 2 .
5". (a-3a-c) 2 = (a; + a + 3c) 2 .
6. (m — no?) (71 — m#) = ran(sc* — 1).
m sc a; n
~ cd , 1 . ,
8. l. a = - -f- dm.
x x
xxx
9. =ra— n-}-».
mn mp np
10. — 7— = — f—
x + n ' x + m
SPECIAL FORMS OF FRACTIONAL LINEAR EQUATIONS 153
2a + 5c _ 5a; + 4c
2x — 5c 5 a? — 4 c*
i2. r a^n = 2- Sx - p .
n-\-x x+p
j£±% 1 _2_
10 26 Sx 26 3x
13.
14.
15.
16.
17.
18.
19.
20.
1
36
3
2x
36 2x
X
c
X
a + 1
1-
-a a 2 -l"
x + c
_x +
c + 2
x — c
X —
c-2'
2x —
x — c
i i
x+c
x — a
+ 2_
-2
x — a-\- 1
x— c
X— c + 1"
c
d
__c — d
x—c
X —
d x
ex
-<*+*+i§i
c-\-x
1 n.
x y
\ f X
21.
22.
23.
24.
d — 2 ' x+l d+2 '
x-
x — 1 2(1 — n) _ x — n
x—n n+l—x x—X
x — m x + 2m _ 5mx + 9m 2
x — 3m a — 4ra x 2 — 7mx+12m*
1 1 = 1 1_ .
x — 2c x — 6c x — 4:C x — Sc
154 FRACTIONAL AND LITERAL LINEAR EQUATIONS
PROBLEMS LEADING TO FRACTIONAL LINEAR EQUATIONS
Exercise 54
1. The difference between the fourth and the ninth parts
of a certain number is 2 more than one twelfth of the number.
Find the number.
Let x = the required number.
Then ?_^ = the difference between the
fourth and ninth parts.
-*+2:
12
= one twelfth the
number in-
Hence, from the conditions,
creased by 2.
X x m
4 9'
= — + 2.
12
Clearing,
9
X — 4x :
2X:
= 3a + 72.
= 72
Verification
: 4
9"
_36
" 4 "
36
" 9
X :
= 9-4:
= 36, the required number.
= 5.
X
12
+ 2:
_ 36
12
+ 2
= 3 + 2:
= 5.
2. Find that number the difference of whose fifth and sixth
parts is 3 less than the difference of its third and fourth parts.
3. When the sum of the third, eighth, and twelfth parts of
a number is divided by 2 the quotient is 1 more than one fourth
the number. Find the number.
4. Find three consecutive numbers such that the first
divided by 6, the second by 5, and the third by 2, give quo-
tients whose sum is 4 less than the greatest number.
5. The first digit of a number of three figures is two thirds
the second digit and 4 less than the third. If the sum of the
digits is 18, what is the number?
6. The sum of two numbers is 38, and if the greater number
is divided by the less increased by 2, the quotient is 3 and the
remainder 4. Find the number.
PROBLEMS WITH FRACTIONAL LINEAR EQUATIONS 155
Let * = the smaller number.
Then 38 — x = the greater number.
Now Dividend - Remainder = Quotient<
Divisor
Hence, <«-«)-* = 8.
' x + 2
Or, 38 - x - 4 = 3 x + 6.
From which x = 7, the smaller number;
and 38 — x = 31, the larger number.
Verifying: (38 ^ ^ - 4 = 38 - 7 - 4 = 27 = 3
3 * x+2 7+2 9
7. The sum of two numbers is 57, and if the greater num-
ber is divided by the less, the quotient is 3 and the remainder 9.
Find the numbers.
8. The difference between two numbers is 23, and if the
greater is divided by 4 less than twice the smaller, the quotient
is 3. Find the numbers.
9. The sum of the third, fifth, and sixth parts of a number
is divided by one half the number, and the quotient is 1 and
the remainder 6. Find the numbers.
10. If the sum of three consecutive numbers is divided by
the smallest number increased by 7, the quotient is 2 and the
remainder is 6. Find the three numbers.
11. In 3 years a certain man will be half as old as his
brother, and the sum of their present ages is 69 years. What
is the present age of each ?
Let x = the man's age in years at the present time.
Then 69 — x = the brother's age in years at the present time.
Hence, x + 3 = the man's age after 3 years,
and 72 — x = the brother's age after 3 years.
Then
3 + 3:
72- a;
2
and
69-z
x-
= 69 - 22 =
= 22, the man's present age.
= 47, the brother's present age.
Verification :
2(22 + 3) =
60 =
= 47 + 3.
= 50.
156 FRACTIONAL AND LITERAL LINEAR EQUATIONS
12. A is one third as old as B, but in 8 years he will be
only one half as old. Find the present age of each.
13. A's age is one fourth that of B, but in 5 years A will be
one third as old as B. Find the present age of each.
14. A child is 1| times as old as his brother, but 2 years
ago he was 1 \ times as old. How old is each now ?
15. The sum of the ages of a father and son is 72 years, and
if the son were one year older and the father one year younger,
the son's age would be one third that of the father. Find the
present age of each.
16. If the length and the width of a certain rectangular field
were each increased by 10 feet, the area of the field would be
increased by 800 square feet. If the length is now 10 feet
more than the width, what are the dimensions of the field ?
Let x = the width of the field in feet.
Then x + 10 = the length of the field in feet.
Since the area of the field equals the product of the length by the width,
x (x + 10) = x 2 + 10 x = the present area of the field in
square feet.
In like manner,
(x + 10) (x + 20) = x 2 + 30 x + 200 = the area of the field in square
feet if length and width are
increased.
Hence, (x 2 + 30 x + 200) - (x 2 + 10 x) = 800.
x 2 + 30 x + 200 - x 2 - 10 x = 800.
20 x = 600.
x = 30, the present width of the field ;
and x + 10 = 40, the present length of the field.
Verification : (x + 10) (x + 20) - x (z + 10) = (40) (50) - 30 (40)
= 2000 - 1200
= 800.
17. The length of a certain rectangle is 10 feet greater, and
the width 5 feet less, than a side of an equivalent square. Find
the dimensions of the rectangle.
PROBLEMS WITH FRACTIONAL LINEAR EQUATIONS 157
18. A square has the same area as a certain rectangle whose
length is 20 feet more, and whose width is 12 feet less, than
the side of this particular square. Find the dimensions of the
rectangle.
19. The length of a certain rectangle is 12 feet greater than
the width. If each dimension is increased by 3 feet, the area
of the rectangle will be greater by 225 square feet. Find the
dimensions of the rectangle at present.
20. A square is cut from a certain rectangular field, the length
and width of the field being respectively 50 feet and 30 feet
greater than the side of the square cut out. If 9500 square
feet remain in the field, what was the original length? the
original width ? the original area ?
21. A can do a piece of work in 5 days, B the same work in
6 days, and C the same in 7 days. How many days will be re-
quired for the work if all three work together ?
Let x = the number of days all three together require.
\ — the portion done by A in 1 day.
\ = the portion done by B alone in 1 day.
\ = the portion done by C alone in 1 day.
Hence, \ + \ + \ = the amount all together can do in 1 day.
Therefore, 1 + 1 + 1 = 1.
5 6 7 a;
42s + 35z + 30x = 210.
107 x = 210.
x s Ifljf days. Result.
22. A can do a piece of work in 4 days, B in 5 days, and C
in 8 days. How many days will be required if all three work
together ?
23. A and B can together build a wall in 7 days, and C
alone can build the wall in 15 days. How many days will be
required if all three work together ?
158 FRACTIONAL AND LITERAL LINEAR EQUATIONS
24. A and B can paint a house in 8 days, A and C to-
gether in 9 days, and A alone in 12 days. In how many days
can B and C together do the work ?
25. Two pipes enter a tank, the first of which can fill it in
7 hours while the second pipe requires 9 hours to fill. How
many hours will be required to fill the tank if each runs alone
1 hour, and then both run together until filled ?
26. At what time between 4 and 5 o'clock are the hands of
a clock together ?
At 4 o'clock the hour-hand is 20 minute-spaces ahead of the minute-
hand. Hence,
Let x = the number of spaces the minute-hand passes over,
and x — 20 = the number of spaces the hour-hand passes over.
Now the minute-hand moves 12 times as fast as the hour-hand.
Therefore,
12 (x — 20) = the number of spaces the minute-hand passes over.
Hence, 12 (x - 20) = x.
12 x - 240 = x.
\\x= 240.
x = 21 A-
That is, the hands of the clock will be together at 21 T 9 T minutes after
4 o'clock.
27. At what time between 3 and 4 o'clock will the hands
of a clock be together ?
28. At what time between 7 and 8 o'clock will the hands of
a clock be together ?
29. At what time between 10 and 11 o'clock will the hands
of a clock be at right angles to each other? (Hint: In this
case the position of the minute-hand will be 15 minute-spaces
behind the hour-hand.)
30. Find the time between 4 and 5 o'clock when the hands
of a clock are at right angles to each other. (Hint : Two pos-
sible positions may be found in this case, for the minute-hand
may be 15 minute-spaces ahead or behind the hour-hand.)
PROBLEMS WITH FRACTIONAL LINEAR EQUATIONS 159
31. The denominator of a certain fraction is greater by 2
than the numerator. If 1 is added to both the numerator
and denominator, the fraction becomes -|. Find the original
fraction.
32. Out of a certain sum a man paid a bill of $30, loaned
I of the remainder, and finally had left $ 56. How much had
he at first ?
33. The largest of three consecutive odd numbers is divided
into the sum of the other two, the quotient being 1 and the
remainder 9. Find the numbers.
34. The sum of the ages of a father and son is 80 years,
but if each were 2 years older the son's age would be f the
father's age. How old is each ?
35. A certain number is decreased by 12 and the remainder
is divided by 4. If the resulting quotient is increased by 7,
the sum is the same as if the original number had been in-
creased by 7 and then divided by 3. Find the number.
36. A man gave -| of a certain sum to relatives, y 1 ^ to each
of two churches, ^ to a library, and the remainder, $ 60.00, to
a hospital. What was the total bequeathed ?
37. In a certain baseball game a total of 13 runs was made
by both teams. If the winning team had made 2 more runs,
and the losing team 3 less, the quotient obtained by dividing
the winning runs by the losing runs would have been 5. How
many runs did each team make ?
38. The distance around a rectangular field is 96 rods, and
the length of the field is f the width. Find the length and
the width of the rectangle, and the number of square feet it
contains.
39. In eight games a certain fielder made 2 less runs than
hits. If 5 times the number of hits he made is divided by the
number of runs increased by 3, the quotient is 4. How
many hits and how many runs did he make?
160 FRACTIONAL LITERAL AND LINEAR EQUATIONS
40. Find three consecutive even numbers such that 3 less
than one half the first, plus 2 less than one half the second,
plus 1 less than one half the third equals 15.
41. In going a certain distance an automobile moving 20
miles an hour required 3 hours less time than a second auto-
mobile making 16 miles an hour. What was the distance in
miles ?
42. A number is composed of two digits whose difference
is 4. If the digits are reversed, the resulting number is ^ the I
original number. Find the number.
43. A can run 10 yards in 1 second, B 8 yards in 1 second. .
If A gives B a start of 3 seconds, in how many seconds will
A overtake B ?
44. The length of a rectangle is 9 rods more than its width.
If the length is increased by 6 rods and the width decreased
by 3 rods, the area is unchanged. Find the length and breadth
of the rectangle.
45. An automobile going 25 miles an hour is 40 minutes
ahead of one going 30 miles an hour. In what time will the
second automobile overtake the first ?
46. At what time between 8 and 9 o'clock do the hands of
a clock point in opposite directions ?
47. A freight train goes from A to B at 15 miles per hour.
After it has been gone 4 hours an express train leaves A for B,
going at a rate of 45 miles per hour, and the express reaches
B ^ hour ahead of the freight. How many miles is it from
AtoB?
48. In traveling a certain distance a train going 45 miles
an hour requires 5-J- hours less time than an automobile going
the same distance at 27- miles per hour. What is the distance
between the two points ?
CHAPTER XV
APPLICATIONS OF GENERAL SYMBOLS. REVIEW
STATEMENTS. PHYSICAL FORMULAS. DERIVED
EXPRESSIONS
THE GENERAL STATEMENT OF A PROBLEM
178. From the following illustrations it will be seen that
when the given numbers of a problem are literal quantities,
the statement and the solution result in a formula or general
expression for that particular kind of problem.
Illustrations :
1. If A can mow a field in m days, and B can mow the
same field in p days, in how many days can both together
mow the field ?
Let x = the number of days both working together require.
Then - = the portion of the work both together can do in 1 day.
x
Also — = the portion of the work that A alone can do in 1 day.
m
— = the portion of the work that B alone can do in 1 day.
P
Hence, — h — = the portion of the work both together can do in 1 day ;
m p
and — |- - = - is the required general equation for the condition.
m p x
Solving, mx +px = mp.
(m +p)se = mp.
X= _™P_. Result.
m +p
This expression is, therefore, a formula for finding the time
in which two men whose individual ability is known, can, work-
SOM. EL. ALG. — 11 161
162 PRACTICAL APPLICATIONS OE GENERAL SYMBOLS
ing together, accomplish a given task. By substituting in this
formula, any problem involving the same condition can be
solved. For example :
A requires 4 days to do a certain task, and B requires 5 days
for the same work. In how many days can both working to-
gether complete the work?
Here we have A's time alone (or m) = 4 ; B's time alone
(or p) = 5. In the formula x = -^£- = ijA = ^ = 2f days, ,
v i} m+p 4 + 5 9 9
the time in which both together can do it.
2. Divide the number a into two parts such that m times
the smaller part shall be contained q times in the larger part.
Let x = the smaller part.
Then a — x = the larger part.
From the conditions,
«^ = g,
mx
and a — x = mqx.
mqx + x = a.
(mq + l)x as a.
x s= — - — , the smaller part required.
mq+ 1
Also, a-x = a 2_ = amg + a-a = «m<? ? the larger part
m«+l «g + l «« + l required.
To use this formula : Suppose we are required to divide 60
into two parts such that twice the smaller part shall be con-
tained 7 times in the larger part. We have
a = 60, x = — — — = 4, the smaller part required.
w» = 2. 2-7 + 1 '
q = 7. f>A o <t
a _ x _ w - * > t _ 56 the i ar g er part required.
2-7 + 1
And, verifying, 56 +■ 2(4) = 7.
GENERAL STATEMENT OF A PROBLEM 163
Exercise 55
1. Divide the number c into two parts such that m times
the larger part shall equal n times the smaller part.
2. Divide a into two parts such that the sum of -th of the
n
larger part and -th of the smaller part shall be q.
r
3. The sum of two numbers is s, and if the greater number,
g, is divided by the less number, the quotient is q and the
remainder r. Find the numbers.
4. If A and B can together mow a field in t days and B
alone can mow the same field in b days, find the number of
days that A working alone will require to do the work.
5. Show that the difference of the squares of any two con-
secutive numbers is 1 more than double the smaller number.
6. m times a certain number is as much above k as d is
above c times the same number. Find the number.
7. A and B are m miles apart, and start to travel toward
each other. If they start at the same time and A goes at a
rate of & miles an hour while B goes at the rate of s miles, how
far will each have gone when they meet ?
8. When a certain number is divided by a, the quotient is
c and the remainder m. Find the number.
9. The front wheel of a wagon is m feet in circumference,
and the rear wheel n feet in circumference. How far has the
wagon gone when the rear wheel has made r revolutions less
than the front wheel ?
10. A and B can together build a barn in r days, B and C
the same barn in s days, and A and C the same in t days. In
how many days can each alone build it ?
164 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS
THE USE OF THE LABORATORY FORMS OF PHYSICS
179. Density is defined as mass per unit of volume, and may
M
be calculated by the formula D—^, in which D is the density
of the body under examination, M its mass, and Fits volume.
Volume is another name for cubical contents. The mass of a
body may be found by weighing it, and the formula for density
W
may be written D = — , in which W is the weight of the body
to be considered.
Exercise 56
1. A block of iron 3 feet 6 inches long, 2 feet wide, and 6
inches thick weighs 1706.25 pounds. Calculate its density in
pounds per cubic foot.
2. The density of water is approximately 62.5 pounds per
cubic foot. What is the volume of a ton of water ?
3. A cubical block of wood 9 centimeters on an edge has a
mass of 371.79 grams. Calculate its density in grams per
cubic centimeter.
4. A block of iron having an irregular cavity weighs 3.265
kilograms. When the cavity is filled with mercury, the whole
weighs 3997 grams. The density of mercury being 13.6 grams
per cubic centimeter, calculate the volume of the cavity.
5. The density of a certain substance being a grams per
cubic centimeter, calculate the mass of this substance necessary
to fill a vessel of 1000 cubic centimeters' capacity.
180. To change a reading on a Centigrade thermometer to a
corresponding reading on a Fahrenheit thermometer use is
made of the formula
_F=.f 0+ 32,
In which F is the reading in degrees on the Fahrenheit scale,
and C the reading in degrees on the Centigrade scale. To
LABORATORY FORMULAS OF PHYSICS 165
change Fahrenheit readings to Centigrade readings we use the
formula
C = f(F- 32).
Exercise 57
1. Change the following Centigrade readings to correspond-
ing Fahrenheit readings : 20°; 40°; 0°; -15°; -273°; -180°.
2. Change the following Fahrenheit readings to correspond-
ing Centigrade readings : 212°; 32°; 70°; -10°.
3. At what temperature would the reading on a Centigrade
thermometer be the same as the reading on a Fahrenheit
thermometer?
4. The Centigrade scale is marked 0° at the freezing point
of water, and 100° at the boiling point of water. The Reaumur
thermometer is marked 0° at the freezing point of water and 80°
at the boiling point. Prepare (1) a formula for changing Reau-
mur readings to Centigrade readings, and (2) a formula for
changing Centigrade readings to Reaumur readings.
181. In the figure, we have a straight bar whose weight is
__ ,m m to be neglected. The bar is sup-
——d- \~jf~ ! ported at the point C arid is
1 acted upon by the several forces,
,«//
f C
t» /? /'? /"> etc v * n tne directions
indicated by the arrows. These
forces tend to cause rotation of
the bar about the axis C. The tendency of a force to produce
rotation is called its moment, and the moment is calculated by
multiplying the magnitude of the force by the distance of its
point of application from the axis. Moments tending to rotate
a body clockwise are given a positive sign ; those tending to
produce rotation in the opposite direction are given a negative
sign. In order that a body under the influence of moments
may be in equilibrium, i.e. stationary, the algebraic sum of the
166 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS
moments acting on it must be 0. Thus in the figure, if the
bar is in equilibrium,
f'd" + f'"d'" -fd -f'd' -f""d"" = 0,
or, better :
(f'd" + /'"<*'") - (fd +fd' +f""d"") = 0.
(/'»/"»/"'» etc., are read "/prime," "/second," "/third" etc.)
Exercise 58
1. Two weights of 4 and 12 pounds respectively are bal-
anced on a bar, the distance from the support to the further
weight being 6 feet. What is the distance from the support
to the nearer weight ?
2. A bar 11 feet long is in equilibrium when weights of
15 and 18 pounds are hung at its ends. Find the distance of
the point of support from each end.
3. If weights are distrib- ,
— v: — ,o--
ED EI
uted upon a bar as in the
figure, where must a weight
of 40 pounds be placed to keep
the bar in equilibrium ?
4. A 10-pound weight hangs at one end of *a 12-foot bar,
and a 15-pound weight hangs at the same side of the support-
ing point, but 2 feet nearer it. If a 40-pound weight at the
other end keeps the bar in equilibrium, at what distances from
the ends is the point of support located ?
THE VALUE OF ANY ONE ELEMENT OF A FORMULA IN TERMS OF THE
OTHER ELEMENTS
182. The statement of a mathematical law by means of a
formula always gives an expression for the value of the par-
ticular element to which the law refers. By transpositions
and divisions we are able to derive from any formula another
expression or formula for any oue of the other elements.
TRANSFORMATION OF FORMULAS 167
Illustrations :
1. Given the formula, R = & s . Derive a formula for s.
9 + s
Clearing of fractions, (g + s) B = gs.
Multiplying, Bg+ Bs = gs.
Transposing, Bs — gs = — Bg.
Dividing by — 1, gs — Bs = Bg.
Collecting coefficients, (g — B)s = Bg.
Dividing by (g — B) , s = — 2-, the required formula for s.
g — B
2. Given the formula, I = a + (n — 1) d. Find an expression
for n. *
l = a+ (n-l)d.
Multiplying, I = a + nd — d.
Transposing, I — a + d = nd.
Dividing by d, n= ~ a + , the required formula for n.
d
THE TRANSFORMATION OF FORMULAS
Exercise 59
PHYSICAL FORMULAS
1. Given v = at, find the value of t in terms of a and v.
2. Given S = % g (2 1 — 1), find a formula for t in terms of s
and g.
3. Given C = , derive a formula for S in terms of E,
f*9
b, P, C, and R.
4. Giv
p, and p'
4. Given -=--{-—, find an expression for each element, /,
168 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS
5. Given F = f C + 32, derive a formula for C in terms of F.
6. Given F = — , v = gt, and w = .Fs, find a value for to
in terms of m, g, and s.
7. Given ^1 = ^M, derive expressions for 2\ and v 2 -
MISCELLANEOUS FORMULAS
8. Given 0=2 irR, obtain a formula for M.
9. Given / = a -f (n — 1) d, derive an expression for a in
terms of I, n, and d.
10. Given 8 = ^ (a ■+■ Q> find the value of a in terms of Z, n,
and &
11. Given T=nR(R + L), find £ in terms of the elements
involved.
12. Given S = r ~ a , find each element in terms of the
r-
others.
13. Given d=-±- - 1 , find a formula for S.
n(n — l)
14. Given a=p-\-prt, find each element in terms of the
others.
GENERAL REVIEW
Exercise 60
1. Simplify 2a- [3-2{a-4(a-a + l)}].
2. Solve (a>-l)(« + 3)-2(«-l)(3»+l) = (3-«)(2+5a).
3. Showthatfl-f-4^— ^-Y-^- i_+lVl=0.
\ a + 1 a-f-3/\a — 3 a — 1 /
GENERAL REVIEW
169
8
or a
r l 1
a 2
M
la 2)
a a z J
4. Simplify
5. Factor a 5 -S(a 2 -x 2 )-a 3 x i .
6. What number added to the numerators of the fractions
- and -, respectively, will make the results equal? Is there
b d
an impossible case ?
7. Factor 6 x 3 + 6 a 2 x - 37 ax 2 .
9. Solve
+
l-2a; l-4a; l-3z
10. Factor (a 2 - 9) (a + 2) - a - (4 + a) (3+ a) - 3.
11. Solve ^±^ S_ = ^-m
3 a + ra 3
12. By three different methods factor (sc 2 — 6) 2 — a£
5
13. Solve JtlJfl-lL
2,-1 4^,-1 37
6(1-2/)
14. For what value of aiscc 4 — 3a^ + 2a^ + 21a; + 3a divis-
ible by x 2 + x- 3 ?
15. The sum of the numerator and the denominator of a
certain fraction is 39. If 3 is subtracted from both numerator
and denominator, the result is \. Find the original fraction
16. Factor 225 - 4 x 2 (9 + x) (9 + x).
17. Find the H.C.F. and the L. CM. of x 4 - ax* - 2 a 2 x>,
2ar 5 -2a 2 a;, and 3 ar> + 12 aa 2 + 3 a 2 x.
170 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS
18. Simplify ( C -±±- C -=i) + ( c ±A + c ^±). ■
19. Solve and verify 2 -^=± + ^ = *JL=1 .
3 7 2s + 2 28
20. Prove that -i- 1 — = cd-
1+i 1-1 od-I
cd cd cd
21. Simplify (a-l) 2 -4^r_3 21 (a-3}
p J (a+2) 2 -l [a + 3 a-3j a-U
22. Solve
2
15
(M-.l a + 2 a + 3
23. A boy has a dollar, and his sister has 28 cents. He
spends three times as much as she spends, but has left four
times as much as she has left. How much did each spend ?
24. What is the value of m if 3m + 5n = 1 when n = - 1.
m — n
■2 S_ x _7
25. Solve £_ + -* -— 4— J-
f(a^l) 1(1+,) 1-1
26. Prove that the sum of any five consecutive numbers
equals 5 times the middle one.
27. Factor (a 2 -{-5 a- 10) 2 + 2(a 2 + 5 a- 10)a - 8 a 2 .
28. Find three consecutive numbers such that if the second
and third are taken in order as the digits of a number, this
number will be 7 more than four times the sum of the 1st and
3d given numbers.
29. If a = 2, b = 3, and c = — 4, find the value of
(a 2 + c*)(3 a - c)V(7a + c)(4 6-a).
30. Simplify -
x
GENERAL REVIEW
1
171
1 Sx2 -*
i 5 1 +
3 3a + l I
J — 2x
2x
2x
+3
31. At what time between 8 and 9 o'clock are the hands
of a watch 5 minutes apart ?
32. Find the H. C. F. and the L. C. M. of
aj»-4, 8-a? 8 , -4+4a-a 2 , (2-oj)*, and (a?-2)(l-a?).
/>»2 ^yj ~,
33. Solve and verify -^ (- -
4^ — m 4 2a? + m
34.
Simplify g + i+^)+j
a?+l . x(x-l)+l ~\ . a 2
cc 2 » 3 — 1 a^— 1
35. What value of x will make (4 x + 3) (3 x — 1) equal to
(6o> + 5)(2»-l)?
36. Simplify (6-c)(c-a)(a-5) 5-c c-a g-6,
J (&4-c)(c + a)(a + &) & + c c + a a +6
37. By what must x 4 -}-2x s -\-Sx 2 -\-4:X-{-5 be divided to
give a quotient of ar* + 4 a? -}- 14 and a remainder of 44 a? + 47 ?
8^-1
38. Simplify _.
1
1-
l+»
39. Show that if a = - 1,
i(l-"H 1('-)K 1+ -)
SFFi " JF+3R
40. Factor mV - mV - 27 »V + 27.
41. Solve fg- > + c ) 2 - CT = a- 3ca; .
n a
42. Factor 6 a 2 — 6np — (4n — 9p)a.
172 PRACTICAL APPLICATIONS OF GENERAL SYMBOLS
43. The head of a certain fish weighs h pounds, the tail
weighs as much as the head and -J the body, and the body as
much as the head and tail. What is the weight of the fish
in terms of h ?
3a 2 -3al
44. Simplify
I 6a?-6
4a
a + 1
48^~
45. In how many years will s dollars amount to a dollars
at r per cent, simple interest ?•
46. Solve for m : (2m + 2)(3 b — c)+2ac = 2c(a — m).
47. If a certain number of wagons is sold at $80 each,
the same amount is received as when 10 less are sold at $ 100
each. How many are sold in each case ?
48. Find the value of
\a? a J a? cr + 1 a 2 — 1 m
49. Factor a 2 + a 2 -2(1 -ax)— (x + a).
50. Two bills were paid with a 10-dollar bank note. One
bill was 25 % more than the other, and the change received
was -i-f the smaller bill. Find the amount of each bill.
51. Solve for s:
(s - a) (s + b) - (s + a)(s - b) - 2(a - b) = 0.
52. Factor aJ 6 - X s — 2 (3 a? + 4).
53. Solve for n, I = (n — 1) (a — I) + a.
54. An orderly is dispatched with an order, and 3 hours
after he leaves a second orderly is sent after him with instruc-
tions to overtake the first in 6 hours. To do this he must
travel 4 miles an hour faster than the first traveled. How
many miles an hour does each travel ?
55. Factor 50 x + 2 x 5 - 38 X s .
CHAPTER XVI
SIMULTANEOUS LINEAR EQUATIONS. PROBLEMS
183. If x and y are two unknown quantities and their sum
equals 7, we may write
x + y = 7.
Clearly, an unlimited number of values of x and y will satisfy
this equation. For example :
If x = 2,
y = 5.
If* = l,
y = 6.
If x = 0,
y = 7. •
11*= -
1,
2/ = 8.
Iix= -
2,
y = 9, etc.
184. Such an equation in two unknown quantities, satisfied
by an unlimited number of values for the unknown quantities,
is an indeterminate equation.
185. If, however, we have with this equation a second equa-
tion stating a different relation between x and y } as
x - y = 3,
then the pair of equations,
x + y = 7,
x - y = 3,
is such that each is satisfied only when x = 5 and y = 2.
For x + y = 5 + 2 = 7,
And x-y = 5-2 = 3.
173
174 SIMULTANEOUS LINEAR EQUATIONS
No other values of x and y will satisfy this pair of equations.
Hence,
186. Simultaneous equations are equations in which the same
unknown quantity has the same value.
187. A group of two or more simultaneous equations is a
system of equations.
188. Two possible cases of equations that the beginner may
confuse with simultaneous equations must be carefully noted.
(a) Inconsistent Equations.
Given : x + y = 9,
x + y = 2.
It is manifestly impossible to find a set of values for x and y that
shall satisfy both given equations. The equations are inconsistent.
(b) Equivalent Equations.
Given : x + y — 4,
3a + 3^ = 12.
If the first equation is multiplied by 3, it becomes the same as the
second equation, and every set of values that satisfies the second satis-
fies the first as well. The equations are equivalent.
189. For a definite solution of a pair of simultaneous equa-
tions we must have a different relation between the unknown
quantities expressed by the given equations.
Equations that express different relations are independent
equations.
•
190. Simultaneous equations are solved by obtaining from
the given equations a single equation with but one unknown
quantity. This process is elimination. Each of the three
methods of elimination in common use should be thoroughly
mastered.
ELIMINATION BY SUBSTITUTION 175
ELIMINATION BY SUBSTITUTION
Illustration :
(5x + 2y = ll, (1)
Solve the equations : { ^ . '
H {3x + ±y = l. (2)
l-4y
3
From (2),
Substituting in ( 1 ) , 5 1 1-4y > \ + 2 y = 11.
From which 5-20j + 2 y = n.
o
Clearing, 5 - 20 y + 6 y = 33.
- 14 y = 28.
y=-2.
Substituting in (1), 5x + 2(- 2) = 11.
5x-4 = ll.
x= 3.
l Result.
y = - 2. J
Check :
Substituting in (1), 5 (3) + 2 (- 2) = 15 - 4 = 11.
Substituting in (2), 3 (3) + 4 (- 2) = 9 - 8 = 1.
From the illustration we have the general process for elimi-
nation by substitution :
191. From one of the given equations obtain a value for one
of the unknown quantities in terms of the other unknown quan-
tity. Substitute this value in the other equation and solve.
The method of substitution is of decided advantage in the
solution of those systems in which the coefficients of one equa-
tion are small numbers. In later algebra a knowledge of this
method is indispensable.
In applying this process of elimination care should be taken
that the expression for substitution is obtained from the equa-
tion whose coefficients are smallest. The resulting derived
equation will usually be free from large numbers.
176 SIMULTANEOUS LINEAR EQUATIONS
ELIMINATION BY COMPARISON
Illustration :
5x + 2y = 9, (1)
Solve the equations : ,
'2a5 + 32/ = 8. (2)
From(l), x= 9 ~ 2y ' From (2), x-*^^-
2
By Ax. 5, 9-2 1 = 8-3_ y<
6 2
Clearing, 2 (9 - 2 y) = 5 (8 - 3 y).
18 - 4 y = 40 - 15 y.
11 y = 22.
2/ = 2.
Substituting in (1), 5 x + 2 (2) = 9.
Hence, *, _
X = 1,1
2/ = 2. J
Check :
Substituting in (1), 5 (1) + 2 (2) = 5 + 4 = 9.
Substituting in (2), 2 (1) + 3 (2) = 2 + 9 = 8.
In general, to eliminate by comparison :
192. From each equation obtain the value of the same unknown
quantity in terms of the other unknown quantity. Place these
values equal to each other and solve.
The method of comparison is particularly adapted to those
systems of simultaneous equations in which the coefficients are
literal quantities.
ELIMINATION BY ADDITION OR SUBTRACTION
Illustrations :
1. Solve the equations : \ Sx -±y= 5 >
H l5a> + 3y = 18. (2)
Choosing the terms containing y for elimination, we seek to make the
coefficients of y in both equations equal ; for, if these coefficients were
ELIMINATION BY ADDITION OR SUBTRACTION 177
equal, adding the equations would cause y to disappear. The L. C. M.
of the coefficients of y, 3 and 4, is 12. Dividing each coefficient into 12,
we obtain the multipliers for the respective equations that will make the
coefficients of y the same in both.
Multiplying (1) by 3, 9 x - 12 y = 15 (3)
Multiplying (2) by 4, 20 x + 12 y = 72 (4)
Adding (3) and (4), 29 x = 87
x =3.
Substituting in (1), 3 (3) - 4 y = 5. T
9-4, = 5. Hence,* = 3,l Regult
, = 1. y = 1 *i
It is to be noted that the signs of y, the eliminated letter, being unlike,
the process of addition causes the y-term to disappear.
2. Solve the equations
I2x-3y = l, (1)
\3x + 7y = 13. (2)
Multiplying (1) by 3,
Multiplying (2) by 2,
Subtracting (4) from (3),
6 x - 9 y = 3. (3)
6 x + 14 y = 26. (4)
- 23 y = - 23.
2/ = l.
2s-3(l)=l. tt
2a- 3 = 1. Hence '* = M Result.
x=2, y =1 -i
Substituting in (1),
In this example since the signs of the sc-term are like, the equations
are subtracted to eliminate the sc-term having the same coefficients.
In general to eliminate by addition or subtraction :
193. Multiply one or both given equations by the smallest num-
bers that will make the coefficients of one unknown quantity equal.
If the signs of the coefficients of the term to be eliminated are
unlike, add the equations ; if like, subtract them. The necessary
multipliers for the elimination of a letter may be found by dividing
each coefficient of that letter into the lowest common multiple of the
given coefficients of that letter.
The method of addition or subtraction is of advantage when
it is desirable to avoid fractions, and is used in the solutions
of systems involving more than two unknown quantities.
SOM. EL. ALG. 12
178 SIMULTANEOUS LINEAR EQUATIONS
Exercise 61
Solve:
1. x + y = 3, 10. 3s + 7y = -8,
3x + 2y = 7. s + y = 0.
2. a;-2/ = 4, 11. 5x + 3y = 10,
2x + 3y = 13. Sx-5y = 16.
3. 4 x + 3 y = 7, 12. 7# — 3 = 4y,
3<c-2/ = 2. o; + 2/-2 = 0.
4. 6cc-5?/ = 7, 13. 5# + 8?/-2 = 0,
a; + 92/ = ll. 7y + 72-3x = 0.
5. 3o; + 4z = 2, 14. 10m + llw = 32,
4 a;- 2 = 9. 15 m + 31 = 23™.
6. 3s-5* = 13, 16. 3y-±x = 77,
2s + 7t = -12. 6y + x = l.
7. 3x + y = -10, 16. 5s + 3* = 112,
2a?-5y = -l. 4*- 5 s -49 = 0.
8. 4v-3w = 0, 17. 5^4-13^ = 4,
2 v + 5 w = 26. 7 n + 19 v = 4.
9. 5« + 7?/ = 24. 18. 12p-7Z-9 = 0,
x-y=0. 15 t -7 p =-300.
SIMULTANEOUS LINEAR EQUATIONS CONTAINING THREE OR MORE
UNKNOWN QUANTITIES
194. A system of three independent equations involving
three unknown numbers is solved by a repeated application
of the process of elimination. The method of addition or sub-
traction is most commonly used with systems having three or
more unknown quantities.
THREE OR MORE UNKNOWN QUANTITIES
179
Illustrations :
-3a;-22/ + 3z = ll,
(1)
1. Solve the equations :
2x — 3y + 2z = 9,
(2)
.3# + 52/ + 4z = 6.
(3)
Multiply (1) by 8,
9x-6y+9z = S3
(4)
Multiply (2) by 2,
4x-6y + 4z = 18
(•5)
Subtract (5) from (4),
5x +5^ = 15
(6)
Divide by 5,
x + z = 3.
(7)
Multiply (2) by 5,
10 x - 15 y + 10 z = 45
(8)
Multiply (3) by 3,
9x + 15^ + 12 = 18
(9)
Adding (8) and (9),
19 x + 22 z = 63
(10)
Multiply (7) by 19,
19 a; + 19 z = 57
Subtracting,
30 = 6
= 2.
Substituting in (7),
x + (2) = 3, x = 1. x- 1, ]
Substituting in (1),
3(1) -2^ + 3(2) = 11. y=-l,l
y=-l. 2=2. J
Result.
r*+2y=l,
(i)
2. Solve the equations : \3 x — z = — 8,
(2)
[y + 2z=0.
(3)
Multiply (1) by 3,
3x + 6y =3
(*)
Subtracting (2),
Multiply (5) by 2,
Subtract (3) from (5),
6y + =
12 y + 2 =
= 11
= 22
y +20 =
=
Substituting in (3),
Substituting in (1),
11 y =22
y = 2.
(2) +20 = 0, 0=-l.
x + 2(2)=l, x=-3.
53
(5)
Result.
195. It is important to note that if three unknown quanti-
ties are involved, three independent equations must be given.
Similarly, with four unknown quantities, four independent
equations are necessary, etc.
180
SIMULTANEOUS LINEAR EQUATIONS
From the illustrations we may state the general process :
196. Eliminate one unknown quantity from any convenient pair
of equations, and the same unknown quantity from a different pah-
of equations. Solve the resulting equations by any of the methods
already given.
Exercise 62
Solve :
1. x + y + z = 9,
x-y + z = 3,
x — y — z = l.
2. x — y -fz = 5,
y-x + z = -l,
z + y + x = 19.
3. x + y-z = 4,
y — x — z — — 10,
z + y — x = 0.
4. u + 2v + w = 4:,
u — v + 2w = 2,
2u-\-v — w — 2.
5. r-f 2s + 3£ = 14,
2 r - s + 1 = 3,
3r-2s-* = -4.
6. 3x — 2y— z=— 8,
5x — 3y + z = l,
2x + 7y + 3z=38.
7.. 4 a; — 3 y 4- 5 z = — 15,
4z-3z + 2?/ = 3,
4 2/ 4- 7 a; — 3z = 4.
8. 10m + 3n-2jt> = 22,
3m — 5n4-7p = — 1,
8m-9n-5jp = 21.
9. a-hz=10,
2/-z = 2,
«-2/ = 2.
10. 2 # + 2/ =3,
3z-x = -3,
±y-3z = -12.
11. a?4-2/ + z = 0,
5x-3y = 50,
2z + y = -20.
12. y = 5 x + 17,
3^-22/4-37 = 0.
71 = - 7 * 4- 2 x.
13. a?4-2/ + z=24,
x + y + u = 25,
x + z + u = 26,
y + z+u = 27.
14. a?4-y=18,
2/ 4- * = 14,
z + w = 10,
w 4- w = 6,
a 4- u = 12.
FRACTIONAL EQUATIONS 181
FRACTIONAL FORMS OF SIMULTANEOUS LINEAR EQUATIONS
(a) When the Unknown Quantities occur in the Numera-
tors of the Fractions
197. Systems of simultaneous linear equations in which the
fractions have unknown quantities in the numerators only,
are solved by first clearing of fractions and then eliminating
by either of the three methods.
Illustration: [ V - * 1_ ^~ 4 m
~2~ 3 ' W
Solve the equations :
3 4 w
From (1), By -Bx-6 = 2x-S.
Simplifying, 5 x - 3 y = 2. (3)
From (2), 4 (2y + 1) - 3 (a + 3) = 0.
Simplifying, 3 x — 8 y = — 5. (4)
From (3), x = ?JL±2. From (4), x = ^^ .
o 3
Therefore, Sy±2 = Sy-5 t (5)
5 3
Solving (5), y = l,l Regult>
Substituting in (1), x = 1> J
198. Extraneous Roots. In fractional simultaneous equa-
tions we may reject any solution that does not satisfy both
given equations.
Exercise 63
Solve:
1.
* , y -
3^4
6,
4^~2
7.
2.
X
5~
y _
3
•■ —
X
4
.y -
7
:2.
3,
3x
2
T 3 "
= 17,
2x
3
■3
= 13.
4<c
3
5
= 12,
Sx
4
T 2 '
= 34.
182 SIMULTANEOUS LINEAR EQUATIONS
x+3 y+l+ q x+1 2-y_ 5
x-2 y-2_ i 3a?-l y-S_ 1
8.
2 3 4
6> 2^fl + 3^1 = 6> 10 ?_4 = |+6,
3a; — 1 2y — l _g a; + y a; — 2 y = 35 a;
2 + 5 ' 10 *" 8 2 6*
5y-2 a;-9 = 5 3s + l 4y + l = 1
3 2 2' 3 2 2'
3a?~l y + 4 1 =0 2a?-l 4y-l ^l
2 3 6' 2 32*
2 3 ~4' iL_?.4-5 5-5
a2 J -l_y+l == _5. ' 3 + 4 ~8 ?
3 2 6* 5 (x +2/) = a; — y.
2y-a;-3 y-2x-3 _,
13. j - -4,
4y-3a;-3 2y-4a; + 9 _
14. * + 2>+* = 10, 15. ^-^ + ^ = 0,
2 T 3 4 3 2 6'
» y_z_ 8 4a; . 2y 3z_l
3 2 8~ ' 3 3 2 2'
a:_y 2_o 5a; 5y 4z_7
6 5 + 4~ '
16. | + | = 2, 17.
2 3
2>+*=4.
3 2
4
6
• y__ *
2 3 12'
18
a; z_ 1
3 4~24'
2/ *_ * .
2 3 12
a;
3"
y_ 11
2 72
2/
4
2 1
"3 48'
a;
2
z _ 5
"5 24*
FRACTIONAL EQUATIONS
183
(b) When the Unknown Quantities occur in the Denom-
inators of the Fractions, giving Simultaneous Linear
Equations in - and —
x y
199. A solution of this type of simultaneous equations is
obtained by considering the unknown quantities to be - and -,
x y
and the process of elimination is carried through without clear-
ing of fractions. Much difficulty is avoided by this method.
Illustrations :
f3 2 == _31
x y 40 '
5_10 = 11 #
x y 8
1. Solve the equations :
I
Multiplying (1) by 5,
Adding (2),
Dividing by 20,
Clearing of fractions,
Substituting in (2) , - - - — = —
16 ,
10
31
V
—
— —
- —
X
y
8
5_
10
11
X
y
8
20
20
X
8
1
1
X
Y
X
=-
-8.
11
10
16
(1)
(2)
(3)
8' - so = 16 y< y=- 5 -
Hence, x
y
::;:)
Result.
200. If the coefficients of x or y are greater than 1, the
equations may be changed in form by multiplying each equa-
tion by the L. C. M. of those coefficients. The resulting equa-
tions will be of the same form as the equations just solved.
2. Solve the equations : .
2a; 3^
4a; 6y
(i)
(2)
Multiplying (1) by 6
i£ + l? = _12.
-
Multiplying (2) by 12, ?! + — = 12.
x y
From the system (3)
and (4),
jc = - and y = —
1
2 *
3'
Exercise 64
Solve:
, 1 , 3 o
3 5_7
5 2 _11
1. - + - = 2,
x y
6.
x y 6'
11.
2a; Sy 6'
?-§ = 2.
5_6 = 7
9 2/ — 8 x = xy.
x y
a; 2/ 2
12.
1,1 17
1 ~ TrT J
A 2,3 .,
1 1 _ 5
a: y z 12
2. - + - = 1,
x y
7.
2a; Sy 12'
1 1,1 5
1 — -i r>>
4_3_1
1 1 ___ 1.
a; 2/ z **
a 2/ 2
3a; 2y 12*
x y z 12
„ 3.5 o
1 1.7
3. - + - = 2,
a ' 2/
8.
3a; 2y 18'
13.
?-?+!=?
3_10 = 1
x y
1 \- 1 - 1 .
a? y z 5
2a; 32/ 2*
2,3 2_17
x y z 30
4. ?_9 = _ 2 ,
2 3 35
4 4 3_83 (
9.
— — SB •
a; y
3a; 2y 6
a; 2/ z 60
1-1 = 0.
•Li-ik
3 3 _3
a; y
2a; 32/
14.
^""^""i'
5. i- 3 -=-2,
10.
^_ _5_ = _4
2 3 __1
x y
3a; 22/ 3*
3a; 2y 6'
5 2 13
6 3 _19 #
2 3 19
i i"'8 f
5a; 2y 5*
3y 2% 36*
(3)
(4)
LITERAL SIMULTANEOUS LINEAR EQUATIONS 185
LITERAL SIMULTANEOUS LINEAR EQUATIONS
Illustration :
Solve the equations : \ r ~ ' )J
{mx + ny=d. (2)
From (1), x = l^M. .From (2), x = <L=M.
a rr
Therefore, c-by == d_ =lMt
a m
Clearing, cm — bmy = ad — any.
Transposing, any — bmy = ad — cm.
Therefore, (an — bm) y = ad — cm,
ad — cm
m
and y =
an — bm
The labor of substituting a root found for one unknown and reducing
the resulting expression for the value of the other unknown, is frequently
as great as that of making a new solution for the value still undetermined.
Therefore, in practice it will be well to make a separate elimination for
each unknown.
From(l), 2/ = ^-=^- From (2), y = d ~™ x .
Therefore, c_^ax = d JZ mx.
b n
Clearing, en — anx = bd — bmx.
Transposing, bmx — anx = bd — en.
Therefore, (6m — an)x = bd — en,
and «=*|izJHL
bm— an
201. As in other fractional forms of simultaneous linear
equations, solutions that do not satisfy both given equations
are rejected.
Exercise 65
1.
x + y = m,
x — y = n.
2.
2x+3y=
c,
3x+2y=
d.
3. mx + ny =1,
nx —my = l.
4. mv + nw = 3,
sv -f- tw = 3.
186
SIMULTANEOUS LINEAR EQUATIONS
5. x + y = a + l, 7. m(x + y) = 5,
x — y = a — l. n(x — y)=10.
6. x + y = 2a + b, 8. ax + cy = a + 2c,
x-y = a + 2b. cx+ay = a-2c.
9. (a + l)x — (a — 1)?/ = 4 a,
(a + l)a + (a-l)2/ = 2(a 2 + l).
10. (m + 2)a?=(m — 2)y,
x — n = y.
11. (a + m)»-(a-m)y = 4am,
(a — m) & — (a — ra) ?/ = 0.
12. c» + m?/ = c(c + ra) 2 ,
mx+ cy =m(c + m) 2 .
13.
m n
14.
35 1 y c
c + 1 ' c-l -0 '
c
15.
m .n 1
cc 2/
16.
2 , a o .
— + — = 2 + a,
ax 2 y
- + - = 4 + a 2 -
a; y
17.
x y _ 2
a + 1 a - 1 a 2 - 1'
x > y - 2
18.
19.
x — y+ n
y+x-m =3
y — x— n
ax _ ..
my
(a -f- m)x ..
(ra ~ a)y
by
ax
2,
by + cz ^ 2
a
cz + ax
21.
22.
2.
ay , bx _ cy dx
6 6 ~9 9
— a
52
3 '
a-1 a+1
a — c
x-1
1-c
^=i,
+
a-1
y + i
a
PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 187
PROBLEMS PRODUCING SIMULTANEOUS LINEAR EQUATIONS WITH TWO OR
MORE UNKNOWN QUANTITIES
202. A problem may be readily solved by means of a state-
ment involving two or more unknown quantities provided that
(1) there are as many given conditions as there are required
unknown numbers, and
(2) there are as many equations as there are required un-
known numbers.
Exercise 66
1. If a certain number increased by 3 is multiplied by
another number decreased by 2, the product is 9 more than
that obtained when the first number is multiplied by 1 less
than the second number. The sum of the first number and
twice the second number is 30. Find the numbers.
Let x = the first number,
and y = the second number.
Then (a + 3) (y-2) = the product of (the 1st no. +3) by (the 2d no.- 2).
Also, x{y — 1) = the product of (the 1st no.) by (the 2d no. — 1).
Therefore, from the condition,
(x + S)(y-2)-9 = x(y-l). (1)
Also we have, x + 2 y = 30. (2)
From (1), -x + 3y = 15. (3)
From (2) and (3) , y = 9, the second number.
x = 12, the first number.
Verifying in (1) :
(12 + 3) (9 - 2) - 9 = 12(9 - 1).
(15) (7) -9 = 12(8).
105 - 9 = 96.
96 = 96.
2. If the larger of two numbers is divided by the smaller
increased by 5, the remainder is 1 and the quotient 3; but if
their product decreased by 43 is divided by 1 less than the
larger number, the quotient is 1 more than the smaller number.
Find the numbers.
188 SIMULTANEOUS LINEAR EQUATIONS
3. If A gives B $ 30, each will have the same amount; but if
B gives A $ 30, the quotient obtained by dividing the number
of dollars A has by the number of dollars B has will be -J.
Find the amount each has.
4. If 1 is added to both the numerator and the denominator
of a certain fraction, the result is f ; but if 2 is subtracted
from the numerator, and 2 is added to the denominator, the
fraction becomes £. Find the fraction.
Let x = the numerator of the fraction,
y = the denominator of the fraction.
Then,
By the first condition,
By the second condition,
Solving (1) and (2),
The required fraction is, therefore, -.
- = the required fraction.
y
x + 1 8
y + i 9
(1)
x-2 1
«/ + 2 2*
(2)
x = 7, and y = 8.
5. A certain fraction becomes -§- when 3 is subtracted from
its numerator and 4 is added to its denominator. The same
fraction is increased by T ^- if -J- is added to its numerator only.
Find the fraction.
6. If a certain fraction is divided by 3, and the result is
increased by 2, a new fraction, |-£, is obtained ; but if the
original fraction is multiplied by 2, and then both numerator
and denominator are decreased by 2, the result is f . Find the
fraction.
7. A certain number is made up of three digits. The
hundreds' digit equals the sum of the units' digit and the tens'
digit; the units' digit is 3 more than the tens' digit, and the sum
of the three digits is 14. Find the number.
PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 189
Let x = the digit in the hundreds' place,
y = the digit in the tens' place,
z = the digit in the units' place.
Then 100 x + 10 y + z = the number.
From the 1st condition, x =y + z. (1)
From the 2d condition, z — y = 3. (2)
From the 3d condition, x + y + z = 14. (3)
Solving the system (1), (2), and (3), x = 7, y=2, z =6.
Therefore, the required number is 725.
8. The sum of the two digits of a certain number is 13, and
if 45 were added to the number, the digits would be reversed.
Find the number.
9. If a certain number is divided by the difference of its two
digits, the remainder is 1 and the quotient 18 ; but if the
digits are interchanged and the new number is divided by the
sum of the digits, the quotient is 3 and the remainder is 7.
Find the number.
10. A certain sum of money placed at simple interest
amounted to $ 1400 in 3 years, and to $ 1500 in 5 years. What
was the sum at interest and what was the rate of interest ?
Let x = the number of dollars in the principal,
y ss the rate of interest.
The interest for 1 year = -^-ths of the principal, = ^- dollars.
100 100
Therefore, ^-^ = the interest for 3 years,
100
— ^ = the interest for 5 years.
100 J
Hence, 3 4.^= 1400. (1)
100
x+^=1500. (2)
100 # Vf
From (1) , 100 x + 3 xy = 140000.
190 SIMULTANEOUS LINEAR EQUATIONS
From (2), 100 x + 5xy = 150000.
500 x + 15 xy = 700000. (Multiplying (1) by 5.)
300 x + 15 xy = 450000. (Multiplying (2) by 3.)
200 x = 250000.
x = 1250.
Substituting in (1), y = 4. Principal = f 1250 1 R esu it
Rate =4% J
11. A sum of money at simple interest amounted to $ 336.96
in 8 months, and to $348.30 in 1 year and 3 months. Find
the sum at interest and the rate of interest.
12. A banker loaned $15000, receiving 5% interest on a
portion of the amount, and 4 % on the remainder. The income
from the 5 % loan was $ 60 a year less than that from the 4 %
loan. What was the sum in each of the loans ?
13. Two loans aggregating $ 6000 pay 3 % and 4 % respec-
tively. If the first paid 4 % and the second paid 3 °/ the total
income from the loans would be $ 12 more each year. What
is the amount of each loan ?
14. There are two numbers whose sum is 10, and if their
difference is divided by their sum the quotient is $•. Find the
numbers.
15. The two digits of a certain number are reversed, and the
quotient of the new number divided by the original number is
If, The tens' digit of the original number is 4 less than the
units' digit. Find the number.
16. If the greater of two numbers is divided by the less,
the quotient is 2 and the remainder, 1. If the smaller number
is increased by 20 and then divided by the larger number de-
creased by 3, the quotient is 2. Find the numbers.
17. Find two numbers such that the' first shall exceed the
second by m ; and the quotient jof the greater by the sum of the
two shall be s.
PROBLEMS — SIMULTANEOUS LINEAR EQUATIONS 191
18. A certain number of two digits is 9 more than four times
the sum of the digits. If the digits are reversed, the resulting
number exceeds the original number by 18. Find the number.
19. A rectangular field has the same area as another field 6
rods longer and 3 rods less in width, and also has the same
area as a third field that is 3 rods shorter and 2 rods wider.
Find the length and width of the first field.
20. Two automobilists travel toward each other over a dis-
tance of 60 miles. A leaves at 8 a.m., 1 hour before B starts
to meet him, and they meet at 11 a.m. If each had started at
8.30, they would have met at 11 also. Find the rate at which
each traveled.
21. Three pipes enter a tank, the first and second together
being able to fill the tank in 3 hours, the second and third
together in 4 hours, and the first and third together in 5 hours.
How long would it require for all three running together to fill
the tank?
22. The numerator of a certain fraction is the number com-
posed by reversing the digits of the denominator. If the
denominator is divided by the numerator, the quotient is 2 and
the remainder, 5. If the numerator is increased by 18, the
value of the fraction becomes 1. Find the fraction.
23. A man seeks to purchase two different grades of sheep,
the whole to cost $210. If he buys 12 of the first grade and
13 of the second grade, he lacks $ 3 of the amount necessary to
buy them. If he buys 13 of the first grade and 12 of the sec-
ond grade, he still lacks $ 2 of the necessary amount. What is
the cost of each grade per head ?
24. A tailor bought a quantity of cloth. If he had bought
5 yards more for the same money, the cloth would have cost f 1
less per yard. If he had bought 3 yards less for the same
money, the cost would have been $1 more per yard. How
many yards did he buy and at what price per yard ?
192 SIMULTANEOUS LINEAR EQUATIONS
25. An automobile travels over a certain distance in 5 hours
If it had run 5 miles an hour faster, the run would have been
completed in 1 hour less time. How far did it run and at
what rate ?
1
1
26. If the length of a certain field were increased by 6 rods,
and the breadth by 2 rods, the area would be increased by 102
square rods. If the length were decreased by 7 rods, and the
breadth increased by 4 rods, the area would be unchanged.
Find the length and breadth of the field.
27. A company of men rent a boat, each paying the same
amount toward the rental. If there had been 3 more men, each
would have paid $1 less; and if there had been 2 less men,
each would have paid $1 more. How many men rented it,
and how much did each pay ?
28. A boat runs 12 miles an hour along a river with the cur-
rent. It takes three times as long to go a certain distance
against the current as it does with it. Find the rate of the
current and the rate of the boat in still water.
29. 144 voters attend a meeting and a certain measure is
passed. If the number voting for it had been ^ as large, and
the number voting against it had been f as large, the vote
would have been a tie. How many voted for and how many
against the measure ?
30. The total weight of three men is 510 pounds. The first
and third together weigh 340 pounds, and the second and third
together 350 pounds, and the first and
second 330 pounds. Find the weight
of each of the three.
31. The sum of the three angles of
a triangle is 180°. The angle at A is
£ of the angle at B\ and the angle
at C is f the sum of the angles at A and B. Find the number
of degrees in each angle.
DISCUSSION OF A PROBLEM 193
THE DISCUSSION OF A PROBLEM
203. Since a problem is solved by means of equations based
upon given conditions, it follows that a true solution can result
from possible conditions only. Briefly,
(a) Impossible conditions result in no solution ; or,
(&) An impossible answer indicates impossible conditions.
Some important cases are considered under the following
heads :
(a) A Negative Result may indicate an Impossible Problem
1. If a dealer doubles the number of horses he owns and
also buys 9 additional head, he will then have -J- the number
he might have possessed by adding 20 head to 5 times the
original number. Find the number he originally possessed.
Let x = the original number of horses.
2 x + 9 = the number under the first condition named.
x + 20 = the number under the second condition.
Then 2x + 9 = 6 -*±™.
O
6x + 27 = 5z + 20.
6x- 5z = 20-27.
x=-7.
And the negative result indicates an impossible problem, due to a fault
in the statement of the conditions.
If the statement had been "and then sells 9 head" instead of "and
also buys 9 additional head " the problem would have been possible.
For 2x-9 = 6x + 20 ,
3
6x- 27 = 5z + 20.
x = 47. Result.
(b) A Statement in a Problem may be Reversed
2. A certain man 42 years of age has a son 18 years old.
How many years ago was the father twice as old as the son ?
SOM. EL. ALG. 13
194 SIMULTANEOUS LINEAR EQUATIONS
Let x = the number of years since the father was twice as !
old as the son.
Then 18 — x = the son's age x years ago.
42 — x = the father's age x years ago.
From the condition,
42 -x = 2(18 -x).
42 - x = 36 - 2 x.
2sc-x = 36-42.
x — — 6. Result.
The result indicates that 6 years have still to elapse before the condi-
tion named will hold. (That is, in 6 years the father will be 48 and the
son 24 years of age.) If the problem were changed to read " In how
many years will the father be twice as old as the son ? " the solution
would be possible.
(c) A Fractional Result may indicate an Impossible Problem
3. If the number of boys in a certain schoolroom is de-
creased by 5, there will be left 2 more than one third the
original number. How many boys were there at first ?
Let x = the number of boys at first.
From the given condition,
X
-5
= |+2.
Sx-
-15
= a + 6,
Sx
— X
= 6 + 15.
2x
= 21.
X
= 10*.
Clearly, the fractional result indicates an impossible condition.
(d) Possible Discussions of Given Values and their Relations
to each Other
A general problem admits of discussion by assuming differ-
ent relations between given values.
DISCUSSION OF A PROBLEM 195
4. Two trains, an express and a mail, pass along the same
railroad in the same direction, the express train traveling m
miles per hour, and the mail n miles per hour. At 12 o'clock
the mail is k miles ahead of the express. In how many hours
will the two trains be together ?
We may assume that they are together x hours after 12 o'clock, the
express having traveled mx miles, and the mail nx miles. But, by the con-
ditions, the express has traveled k miles more than the mail at 12 o'clock ;
hence,
mx — nx = k.
k
m — n
Discussion :
1. Suppose m is greater than n.
The value of x will be positive.
The express will overtake the mail after 12 o'clock.
2. Suppose m less than n.
The value of x will be negative.
(For, if n is greater than w, the rate of the mail was the faster rate.)
The trains were together before 12 o'clock.
This assumption, therefore, is impossible, for we are going contrary to
the condition that the trains are to be together after 12 o'clock.
3. Suppose m equal to n.
The value of x will become _.
If the rate m equals the rate n, the trains have not been together, are
moving at the same constant distance apart, and will never be together.
In this case, therefore, the supposition has led to the impossible con-
k
dition denoted by the symbol -.
This symbol is usually denoted by oo, and is read " infinity."
4. Suppose m equal to w, and k equal to 0.
If k equals 0, the mail and the express started together ; and since m
equals n, the trains have been, and will continue to be, together.
Expressing this final condition in symbols, we have,
x = — - — = - = any finite number.
m — n •
196 SIMULTANEOUS LINEAR EQUATIONS
That is, there is an infinitely great number of points at which the two
trains are together. In this case, therefore, - is the symbol of inde-
terminate value.
In general, therefore :
From (2), A negative result indicates an error in statement.
From (3), A result x = — = co indicates no possible solution.
From (4), A result x = — indicates an infinitely great number of solu-
tions.
Exercise 67
Discuss the following problems and interpret the solution
for each :
1. A is 12 years old, and B is 17 years old. In how many
years will B be twice as old as A ?
2. The total number of boys and girls in a certain school
is 32, and four times the number of boys plus twice the num-
ber of girls equals 95. How many boys and how many girls
are there in the school ?
3. A and B can together paint a sign in 8 hours, and B
alone can paint the same sign in 5 hours. How many hours
will A require if working alone ?
4. A boy has 45 coins, the value being in all 78 cents.
A portion of the number consists of nickels, and the remain-
der of cents. How many are there of each kind ?
5. A group of b boys bought a boat, agreeing to pay d dollars
each, but / of the boys failed to pay their shares, and each
remaining boy had to pay e dollars more than he had agreed.
Find the cost of the boat in terms of b, e, and /
Ans. ( 6 ~/> 6e .
/
CHAPTER XVII
THE GRAPHICAL REPRESENTATION OF LINEAR
EQUATIONS
THE GRAPH OF A POINT
In the figure two straight lines, XX' and YY', intersect at
right angles at the point 0, the origin. These lines, XX' and
YY'j are axes of reference.
From O measure on OX
the distance, OA = a. From
measure on OF the dis-
tance, OB — b. Through A
draw a line parallel to Y,
and through B draw a line
parallel to OX. These lines
intersect at P. And P is
the graph of a point plotted
by means of the measure-
ments on OX and OY.
204. OA and OB, or
their equals, a and 6, are
the rectangular coordinates of point P.
205. A coordinate parallel to the XX' axis is called an
abscissa.
A coordinate parallel to the YY f axis is called an or-
dinate.
206. Abscissas measured to the right of are positive, to
the left of 0, negative.
197
Y
B
P
/>
x
a
X
jA
Y
198
GRAPHS OF LINEAR EQUATIONS
Y
P
M
X
X
O
T
N
Y
Ordinates measured upward from are positive, downward
from 0, negative. In the accompanying diagram it will be
seen that
The abscissa of P is + 3 ; the
ordidate, + 4.
The abscissa of M is — 4 ; the
ordinate, + 3.
The abscissa of N is — 4 ; the
ordinate, — 5.
The abscissa of T is + 2 ; the
ordinate, — 3.
207. It will be seen that
the axes of reference di-
vide the plane of the axes
into four parts or quad-
rants, and these quadrants are named as indicated, I, II, III
and IV. From the principle governing the signs (Art. 206), ,
we observe that in
Quadrant I : abscissa + ; ordi-
nate + •
Quadrant II : abscissa — ; ordi-
nate + .
Quadrant III : abscissa — ; ordi-
nate — .
Quadrant IV : abscissa + ;
ordinate — .
In naming the coordi-
nates of a point, the ab-
scissa is always named first,
and the ordinate last.
In the illustrations of this chapter the scale of the graphs is
so chosen that one unit of division on the axes corresponds to
one numerical unit in a solution. In later chapters the student
will easily apply other scales as occasion requires.
Y
It
X" _ x
— -g
II IE
7^
THE GRAPH OF A POINT
199
In the diagram let the
student name the location
of each of the several points,
making a table in which
the position of each is re-
corded.
4= (2, 4),
**<
),
B= (-5,1),
0=(
)i
C=(-3,-5),
R={
),
#=(7, -7),
K={
),
fc=( ),
£ = (
)•
Y
H
A
K
B
X
X
G
O
L
F
E
C
D
Y
Exercise 68
On properly ruled paper plot the following points :
I 1. (2,5). 6. (-3, 4). 11. (4,0).
2. (3,4). 7. (5, -2). 12. (0,4).
3. (7,1). 8. (3,3). 13. (-5,0).
4. (1,7). 9. (-2,-1). 14. (0,-5).
15. (0, 0).
5)?
5. (-2,5). 10. (-4,-7).
16. In what quadrant does (- 3, 5) lie ? (2, -4)? (-3,
(4, 7)?
17. In what quadrant does (5.5,. 7.5) lie ? (- 4.8, - 9.75) ?
(- 5.4, 8.7) ? (- 14.75, - 6.1) ?
18. Using two divisions on your paper for one given unit of
measure, plot the points ; (4,3), (2 f 5) f (-2,3), (3,-2),
(0, 4), and (7, 0).
19. Using one division on your paper for two given units of
measure, plot the points; (4, 6), (10, 18), (-8, 14), (20,-16),
(-18,-18), (0,-24), (10,15), (-17,-21), (-3,25),
(25, -3), and (-19, -12).
200
GRAPHS OF LINEAR EQUATIONS
THE GRAPH OF A LINEAR EQUATION IN TWO UNKNOWN NUMBERS
208. A variable is a number that, during the same discus-
sion, may have an indefi-
X
A
J>r-
\B
s
n
5s
Sn
S? _
Ss -
X"~
m SEL
- x
5 — ~S C
s
V
S^
^
_ B
7^
nitely great
values.
number of
If 2/=0, *=4, P=(4,0).
y=l, x=3, P 1= (3, 1).
y=2, a>=2, P 2 =(2,2).
2/=3, a=l, P 3 =(l,3).
209. A constant is a
number that, during the
same discussion, has one
and only one value.
If, in the given linear
equation, jc-f y = 4, we as-
sume a succession of differ-
ent values for y, the corre-
sponding values of x are as
follows:
2/=4, 0=0, P 4 =(0,4).
y=5, a=-l,P 5 =(-l,5).
y=6, x=-2,P 6 =(-2,6).
etc., indefinitely.
Kegarding the respective pairs of values as the coordinates
of points which we may designate by P, P lt P 2 , etc., we plot
these points and draw through them the line AB. Therefore,
the line AB in the figure is the graph of the linear equation,
x + y = 4.
An indefinite number of points might have been obtained by
assuming fractional values for y, and finding the corresponding
values of x.
Exercise 69
Plot the graphs of the linear equations :
1. x + 5y = 7. 3. x + 3y = ±. 5. x+y = 0.
2. 3x + y = 2. 4. 2x + 7y = 3. 6. #-4?/ = 0.
LINEAR EQUATIONS IN TWO UNKNOWNS
201
A Shorter Method for Obtaining the Graph op a
Linear Equation in Two Variables
210. It can be proved that the graph of every linear equa-
tion in two variables is a straight line. (This fact justifies the
use of the word "linear" in naming so-called equations of
the first degree.) Now a straight line is determined by two
points, hence the graph of a linear equation should be deter-
mined by the location of any two points that lie in its graph.
The two points most easily determined are those where the
graph cuts the axes. Therefore,
(1) Find the point where the graph cuts OX by placing y = 0,
and calculating x.
(2) Find the point where the graph cuts OFby -placing x — 0,
and calculating y.
Illustration:
Plot the graph of
Sx-
If t/ = 0, x = 2.
Plot Pi (2,0).
If x = 0, y = - 3.
PlotP 2 (0-3).
Join PiP 2 .
AB is the required graph of
3cc-2y=6.
2y = 6.
Y
Q-/k ™~"
mmmm
b ZZZ—Z
Exceptions to the Shorter Method
(a) A Linear Equation whose Graph passes through the Origin.
211. If a given linear equation has the form of ax = by, it is
evident that when x = 0, y = also. That is, a graph of such
an equation passes through the origin. Therefore, to plot the
graph of an equation in this form, at least one point not on
either axis of reference must be determined.
202
GRAPHS OF LINEAR EQUATIONS
(b) A Linear Equation whose Graph is Parallel to Either Axis.
212. If a given linear equation is in form of #=7, it is evident
that the value of a? is a constant. That is, the abscissa of every
point in the graph is 7. Therefore, the graph of this equation
is a straight line parallel to the axis, YY' and 7 units to the
right of it.
Exercise 70
Plot the graphs of the following :
1. x+2y = 5. 5. a? = 4. 9. Sx — 5y = — 3.
2. y = 3x + 4:. 6. y + 5 = 0. 10. 5x + 3y = 0.
3. x = 2y + 3. 7. 3x-2y = -10. 11. 12 a; -17 2/ =15.
4. x + ±y = 5. 8. y = 7 x. 12. 4?/ = 16 — 18 a.
THE GRAPHS OF SIMULTANEOUS LINEAR EQUATIONS IN TWO UN-
KNOWN NUMBERS
(a) Independent Equations
In the figure, the line AB
is the graph of the linear
equation, x + y = 5. The
line CD is the graph of the
linear equation, 4 a?— 3y =6.
It will be seen that the
graphs intersect at the
point P, (3, 2). That is,
the point, (3, 2), is common
to both graphs. Solving the
given linear equations,
x + y = 5 and 3 x — 4 y = 1,
we obtain as a result, x = 3,
- s i
*
^
y ss 2. And, clearly, the values obtained for a; and 2/ are the
coordinates of the intersection of the graphs.
213. The coordinates of the point of intersection of the graphs
of two simultaneous linear equations form a solution of the two
equations represented by the graphs.
SIMULTANEOUS LINEAR EQUATIONS
203
Exercise 71
Solve the following simultaneous linear equations and verify
the principle of Art. 213 by plotting their graphs :
1. 5x — 3y = l,
Sx + 5y = 21.
2. 5a -3?/ = 36,
7x— 5y = 56.
3. 8a + 3^ = 12,
12 a + 5y = 16.
4. 4x + 6y = -3,
2</ + a = 0.
(ft) Inconsistent Equations
Given two equations
Multiply (1) by 2,
2a + 3i/ = 8,
4* + 6y= -25.
4se + 6y = 16.
(1)
(2)
The equations are therefore inconsistent, for it is impossible
to find any values of x and y that will satisfy both equations.
Plotting the graph of
2x + Sy = 8 (^4jB), and the
graph of £x + 6y = — 25
(CD), we obtain two parallel
lines, and as parallel lines
never meet, we find no point
common to the graphs.
Thus the graphical repre-
sentation of the linear
equations in this case shows
with great clearness the
real meaning of inconsist-
ent equations, and serves
to emphasize more than
ever the need of independent equations if a solution is to be
possible.
1
A
s s
^- C ^ ""
5Cjv 5*-J- .X
*S"~ o" • S~
X ^B
^^
s
^^
D
Y^
204
GRAPHS OF LINEAR EQUATIONS
(c) Indeterminate Equations
Y
B
X
1
X
A
Y
simultaneous equations by means of the graph
Given two equations
2x — 3y = 10, (1)
4a -6 ?/ = 20. (2)
Multiply (1) by 2,
4 x - 6 y = 20.
That is, the first equa-
tion is put in the form
of the second, and their
graphs are found to coin-
cide. Here, also, we fail
to find independent equa-
tions; and we are again
assisted to a clearer con-
ception of indeterminate
Exercise 72
Apply the principle of graphical representation to the follow-
ing, and determine the pairs of independent, inconsistent, and
indeterminate equations.
1. hx + 2y = l,
5a + 2 y = -S.
2. 3x-±y = 7,
9x — 12y = 0.
3. 3 x + ly = — 4,
4a-3y-7 = 0.
4. 4 x -f- y — 9,
3#-22/ = 4.
5. 2x — 5y =3,
6x = l& + 15y.
6. 4 x — 5 y = 3,
8a>-10y = 6.
7. 2a> — 3y = l,
16 oj - 24 y = 8.
8. a?+% = 19,
lly-3a> = 19.
9. 4#-32/ = 8,
16-8a; = -6y.
10. 3a; — 5*s%,
9z = 15 + 18y.
CHAPTER XVIII
INVOLUTION AND EVOLUTION
INVOLUTION
214. Involution is the operation of raising a given expression
to any required power. All cases of involution are multiplica-
tions, the factors in each case being equal. In all elementary
work the exponents of the powers are positive and integral.
THE GENERAL PRINCIPLES FOR INVOLUTION
(a) The Power of a Power
When m and n are both positive integers :
By Art. 61, a m = a x a x a ... to m factors.
Therefore, (a m ) n = (a x a x a ... to m factors) (a x a x a ... to
m factors) ... to n groups of factors,
= a x a x a ... to mn factors,
as a mn . The Third Index Law.
Hence, to obtain any required integral power of a given
integral power :
215. Multiply the exponent of the given power by the exponent
of the required power.
(b) The Power of a Product
When n is a positive integer :
(a&) n = ab x ab x ab ••• to n factors,
= (a x a x a ••• to n factors) (b x b x b ••• to n factors),
= (a«)(6»),
= a n b n . The Fourth Index Law.
205
206 INVOLUTION AND EVOLUTION
Hence, to obtain any required power of a product :
216. Multiply the factors of the required product, first raising
each factor to the required power.
(c) The Power of a Fraction
When n is a positive integer :
(a\ n a /. a „ a - .
[ - =7 x T x 7-ton factors,
\bj b b o
— ( a x a x a '" to n ^ actors )
(b x b X b ••• to n factors)
~b»
Hence, to obtain any required power of a fraction :
217. Divide the required power of the numerator by the same I
required power of the denominator.
These laws are general for unlimited repetitions of the
powers or of the factors.
Thus : C(o n ) m ] p = a mnp , etc., (abc ••• w) m = a m b m c m ••• n m , etc.
THE SIGNS OF POWERS
218. All even powers of any quantity, positive or negative, are
positive.
Thus, (+a) 2 = + a 2 . (+ a) 4 = + a 4 . (-a) 2 =+a 2 . (- a) 4 = + a 4 .
219. j£B odd powers of any quantity, positive or negative, have
the same sign as the given quantity.
Thus, (+ a) 8 = + a 3 . (+a) 6 =+a 5 . (-a) 8 =-a 8 . (-a) 5 =~a 5 ,
(a) The Involution of Monomials
Illustration :
1. I -1*LY '..(«*^'.-«^ ■ Result.
V 3m»y<y (3«V)' 27WV 2
INVOLUTION OF BINOMIALS 207
In general, to raise a monomial to a required power :
220. Determine the sign of the power. Raise each factor of
the given quantity to the required power, the numerical factors by
actual multiplication, and the literal factors by multiplying each
exponent by the exponent of the required power.
Oral Drill
Give orally the value of:
\3 mnj \ xy )'
2. (3xy. 8< (a*x\\ ( 4aV y
W 14 - rwj-
3 - ( - 2 ^- 9 -(fSJ- n jlm
ii ( 2x ^w
5. (2mrc 2 ) 5 . " lg TO «J
2^2/V*
16. _(^_5^Y
V 3cdJ
17. -(|m 3 nY) 6 .
6. (3aV) 4 . _ V^M 18 - -(-faWrc) 4 .
- -&>'
(b) The Involution of Binomials
The process of raising a binomial to a required power, or, the
expansion of a binomial, is best shown by comparative illustra-
tions.
By actual multiplication :
(a + 6) 2 = a 2 + 2 ab + 6 2 .
(a + 6)3 = a 3 + 3 a 2 6 + 3 a62 + 6 3.
(a + 6)4 = a 4 + 4 a t 6 + 6 a i b 2 + 4 a?> 3 + 6 4.
(a + 6) 5 = a 5 + 5 a 4 6 + 10 efib* + 10 a 2 ^ + 5 ab* + & 5 .
208 INVOLUTION AND EVOLUTION
*
221. From a study of the form of each product we note :
1. The number of terms in the expansion exceeds by 1 the ex-
ponent of the binomial.
2. The exponent of a in the first term is the same as the expo-
nent of the power to which the binomial is raised, and it decreases
by 1 in each succeeding term.
3. b first appears in the second term with an exponent 1, and
this exponent increases by 1 in each succeeding term until it is the
same as the exponent of the binomial.
4. TJie coefficient of the first term is 1, and of the second term
the same as the exponent of the binomial.
5. The coefficient of each succeeding term is obtained from the
term preceding it, by multiplying the coefficient of that term by
the exponent of a, and dividing the product by the exponent of b
increased by 1.
6. If the second term of the given binomial is positive, the sign
of every term of the expansion will be positive; and if the sign of |
the second term of the given binomial is negative, the signs of the
terms of the expansion will be alternately positive and negative.
Illustrations :
1. Expand (a — #) 4 .
For the a-factor : a 4 a 3 a 2 a
x-f actor: x x 2 X s x*
coefficients: 4 6 4
signs : - + - +
Combining, a 4 — 4 a 8 x + 6 a 2 x 2 — 4 ax* + x*. Result.
2. Expand (3 a -|Y.
By observing the formation of the expansion of (a — &) 6 , we write :
(a - h)'° = a 6 - 5 a 4 6 + 10 a 8 6 2 - 10 a 2 6 3 + 5 a& 4 - b & .
(3a-^-(3a)5_5(3a) 4 ^+10(3a)3^ 2 _10(3a) 2 (|y
EVOLUTION 209
Simplifying,
2 4 16 32
A polynomial is raised to a power by the same process, if
the terms are so grouped as to express the polynomial in the
form of a binomial.
3. Expand (1-a + ar 2 ) 3 .
(1 - X + Z 2 ) 3 =[(1 - x) + *2]3
= (1 - X) 3 + 3(1 - X )*X* + 3(1 _ s)(z2)2 +(>2)8
= (1 _ Sx + 3x 2 - x* + 3x 2 - 6x 8 + 3x* + 3x 4 - 3x6 + 36)
= (1- 3x + 6x 2 - 7x 8 + 6x* - 3x 6 + x 6 ). Result.
Exercise 73
Expand :
1. (m + w) 3 . 8. (3a-l) 4 .
13.
(*-i)r-
2. (m + ny. 9. (ax-2yy.
3. (m-*) 6 - i . (c 2 -^) 5 . 14 - (^ + ^ + l) 2 -
4. (c + 3) 4 . / 3.N3 15. (m 2 -2m + 3) 2 .
5. (a-2) 5 . n ' ( C + 2y* 16- C^ + c + l) 3 .
6. (3a-2y)\* / _2V 17 « (rf-x + lf-
7. (2c 2 -3) 3 . I 37 18. (2a J -aj 2 + 2) 3 .
EVOLUTION
222. Evolution is the process of finding a required root of a
given expression.
223. The symbol for a required or expressed root is the
radical sign, ^/.
Thus : Va = the square root of a ; y/a = the cube root of a. ; etc.
224. The number written in the radical sign and indicating
the root required is the index of the radical.
SOM. EL. ALG. — 14
210 INVOLUTION AND EVOLUTION
Thus : 3 is the index of the indicated cube root above. It will be noted
that in the case of a square root the index is not usually written, the
absence of an index being an " understood " square root.
The vinculum is commonly used to inclose the expression affected by a
radical.
' 225. From the definition of root (Art. 60) it is clear that
y/a means " Required : One of the two equal factors of a."
v/& means " Required : One of the three equal factors of &."
y/c means " Required : One of the four equal factors of c."
{Vx means " Required : One of the n equal factors of x," etc.
THE GENERAL PRINCIPLES OF EVOLUTION
In the discussion of these principles both m and n are posi-
tive and integral numbers.
(a) The Root of a Power
By Art. 215, (a m ) n = a mn .
Therefore, by definition (Art. 60),
a m is the nth root of a mn .
For a m is one of the n equal factors of (a m ) n .
That is, a m = {Var™. The Fifth Index Law.
The conclusion is a direct result of a division of the exponent of the
given quantity by the index of the required root. Or,
mn
y/a^n- — a/" = a m .
Hence :
226. Any required root of a power is obtained by dividing the
exponent of the power by the index of the required root.
(5) The Root of a Root
By Art. 226, ( m KVa) mn = a.
Extracting the nth. root, ("\/a) m = Va.
Extracting the with root, ( w -v/a) = Zjtya.
GENERAL PRINCIPLES OF EVOLUTION 211
Hence :
227. The mnth root of an expression is equal to the mth root
of the nth root of the expression.
(c) The Root of a Product
By Art. 226,
({Vab) n = ab.
Therefore,
(Va x {Vb) n = (Va) n x ({Vb) n .
Or,
(y/aby=(Vax y/iy.
Hence,
{Vab={Vax Vb.
Therefore :
228. Any required root of a product of two or more factors is
equal to the product of the like roots of the factors.
w
The Root of a Fraction
By Art. 217,
/a\ n _a n
\b) b»
Therefore,
>&» b
Hence :
229. Any required root of a fraction is obtained by finding the
like roots of its numerator and denominator.
THE SIGNS OF ROOTS
(a) Positive Even Powers
By Art. 218, (+ a)(+ a) = + a 2 and (- a)(- a) = + a 2
Therefore, V+a 2 = + a or — a.
Hence :
230. Every positive number has two square roots whose abso-
lute value is the same, but whose signs are opposite in kind.
The double sign, ± , is used to indicate two roots. Thus, Va 2 = ±a.
The sign ± is read, " plus or minus."
212 INVOLUTION AND EVOLUTION
(b) Negative Even Powers
Since — a 2 = (+ a)(— a), we have a product of unequal factors.
Hence :
231. It is impossible to obtain an even root of a negative
number.
(c) Positive and Negative Odd Powers
ByArt.219, (+a)( + a)(+a) = -f a 3 . Also, (_a)(_ a )(-a) = -a 8 ,
etc.
Therefore, V+ a 3 = -f a, V— a 8 = — a, etc.
Hence :
232. !Z7ie odd roots of a positive quantity are positive, and the
odd roots of a negative quantity are negative; or, briefly, the odd I
roots of a quantity bear the same sign as the given quantity.
THE EVOLUTION OF MONOMIALS
233. Illustrations :
1. Eequired the cube root of 8 a% 3 c 9 .
The root is odd ; the given quantity, positive ; the sign of the result, +.
Dividing each exponent by the index of the root,
\/S a*b*c* = \/2*cfib*<P = 2 a?bc 8 . Eesult.
2. Eequired the fifth root of - 243 x l0 y 15 .
The index is odd ; the given quantity, negative ; the sign of the result, — .
Dividing each exponent by the index of the root,
V-243x 1( y 6 = V - 3 6 £c 10 2/ 15 = - 3 x 2 y*. Result.
3. Eequired the fourth root of 16 aWc 12 .
The root is even; the given quantity, positive; the result bears the
double sign.
Hence : ^16 a*b*c 12 = \/2*a 8 &*c 12 = ± 2 a 2 6c 3 . Result.
SQUARE ROOT OF POLYNOMIALS 213
4. Required the square root of 1587600.
A root of a large number may frequently be obtained from its prime
factors.
Hence, V 1587600 = V2* • 3* . 5 2 . 7 2 = ± (2 2 . 3 2 . 5 . 7) = ± (4 . 9 . 5 • 7)
= ± 1260. Result.
In the consideration of numerical quantity we shall consider only the
positive roots in our results.
Oral Drill
Find the value of :
1. V121 a 4 c 6 . 10. -v / -243m 10 z 30 . 18 . 4 /625^
2. V64m 4 7iy. 11. vSSW. 16zl6
3. VWtftf. 12. ^-1024m 15 n 20 . 19. Jj/EZ.
4. ^-27 aft/ 3 . 13. -\/64a% 18 . 64<f
5. ^-343 (W. 14. ^729 aW°. 20 ' <\F
6. \/16m & /i M . i5. ^/i28c¥" 4 .
243 m 15
21 3/125^"
7. ^ 81 <Wy. 16 . ^ 2 56m<V«. ' V"27V *
9. -^-64aW. N 64w 6 \1024ic 30
23. V1296. 24. ^3375.
THE SQUARE ROOT OF POLYNOMIALS
234. If a binomial, (a + 6), is squared, we obtain (a 2 + 2 ab
+ ft 2 ). We have in the following process a method for extract-
ing the square root, (a + 6), of the given square, (a 2 + 2 a& + 6 2 ).
a 2 + 2 a& + 6 2 | a + 6 square root. The first term of the root,
a? a, is the square root of the
«6 + 6
b
:
given expression, the remainder, (2 ab + 6 2 ), results. Dividing the first
term of this remainder, 2 ab, by twice the part of the root already found,
+ 2 ab + 6 2 first term of the given ex-
-f 2 a& + 6 2 pression, a 2 .
Subtracting a 2 from the
214
INVOLUTION AND EVOLUTION
2 a, the quotient is b, the second term of the root. This second term, b, is
added to the trial divisor, 2 a ; and the sum, 2a-\-b, is multiplied by b.
The result, (2a+&)6, = 2«H b' 2 , and completes the process. It will be
seen that much of the work depends directly upon the trial divisor, 2 a.
The reason for this prominence of the trial divisor will be seen from the
following
a*
ab
ab
V
Graphical Representation of a Square Root
235. In the accompanying figure we have a graphical repre-
sentation of a square constructed upon a given line, a + b. If
the square whose area is a 2 is subtracted
from the whole area, there remain the areas
ab + ab+b 2 =2ab + b 2 .
(Note that this corresponds with the sub-
traction above.) Now the length of the
side of the square removed being a, we are
to provide for a remaining area built upon
two sides of that square, or a + a = 2 a i
(the trial divisor). Now
ajb
ab
&
Area
Length
= Width. Hence,
2ab
2a
b.
That is, b is the width of the remaining
area whose length is known. Now, because
of its position in the original square, there
still remains unprovided for the square 6 2 ,
whose side is b in length. Hence the length of the total area
necessary to complete the square is 2 a -\-b. (This explains
the addition of the second term of the root to the trial divisor
above.) Multiplying our known length by the width ascer-
tained by division, we have, as above,
(2a + b)b = 2ab + b 2 ,
which area completes the square required.
SQUARE ROOT OF POLYNOMIALS
215
236. By the principle of Art. 234, we obtain the square root
of any polynomial, the trial divisor at any point being in every
case twice the part of the root already found.
Illustrations :
1. Extract the square root of x A -f- 6 X s -f- 19 x 2 -f- 30 x + 25.
A polynomial must be arranged in order if its root is to be found
without difficulty.
x*46x 3 + 19x 2 + 30x+25| x 2 +3x+5
x 4 Result.
First Trial Divisor, 2 (x 2 ) = 2 x 2 + 6 x 3 + 1 9 x 2
First Completion, (2 x 2 + 3 x) (3 x) — \ +6 x 3 + 9 x 2
Second Trial Divisor, 2 (x 2 + 3 x) = 2 x' 2 + 6 x
Second Completion, (2x 2 +6 x+5)( + 5) =
+ 10x 2 +30x+25
+ 10x 2 +30x+25
2. Extract the square root of 1 — 2 a to 4 terms.
The approximate square root of expressions not in themselves perfect
squares is obtained by the principles of Art. 191.
2a
+
»(1) = 2
(2-q)(-q)
2a
2« + a 2
2(1 -a) =2-2 a
t-"-s(-a
-a 2
a 2 +a 8 + «L
('-'-*-f)(-f)-
4
2 4
Result.
In extracting the square root of fractional expressions care must be
iken that the expression is properly arranged. The descending powers
of a letter occurring in both numerator and denominator of a fraction are
written thus :
o8 + a + + i + I + I + 1 + I .
3. Extract the square root of — + 11 -\ h - ■ H — -
x 2 x a z a
216 INVOLUTION AND EVOLUTION
In descending powers of a :
* + y + u + £* + *«. + B4 .«. R e SU i t .
x 2 x a a 2 \x a
x 2
2^= —
\xj X
(¥ +3 ) (+3) =
a;
a;
2 («+?\ ^ + 6
\X ) X
\ x a/ \ a/
+ 2+ — + -!
a a 2
+ 2+ — 4~
a a 2
Exercise 74
Extract the square root of :
1. a 4 + 4<e 3 + 6£c 2 +4a;+l.
2. a 4 -4a 3 4-10ff 2 -12a;4-9.
3. a 4 -6a 3 + 12a + 4 + 5a 2 .
4. a 6 + 4^-2a; 4 -l6a? + ^ + 12a; + 4.
5. 13m 2 -30m + 4ra 4 +20ra 3 +9.
6. 4 a; 4 + 4 y?y + 9 ic 2 ?/ 2 + 4 ai/ 3 -f- 4 y 4 .
7. 9a 4 -12a 3 & + 34a 2 & 2 -20a& 3 + 25& 4 .
8. 25c 4 -20c 3 d-4cd 3 +14c 2 d 2 + d 4 .
9. 4-12» + 17aJ»-32a? + 34a! 4 -20aJ , + 25a*
10. 16a^-16afy-28a^ + 30a^-20a;y+29afy 4 +
11. ^ + 3a; + 9.
4
12. a; 4 + 2jK 3 +2^4-a;4-i.
25 V 6 .
13.
13 a 2
2 + 4
3a +9.
SQUARE BOOT OF ARITHMETICAL NUMBERS 217
14. !l 2 + 2a + 3+2_ c + c°.
& c a ar
4 a; 4 4 a? 7 a? x 1
'9 9 9 3 4*
9 a; 4 6 a? .13 a? x 1
' 25 5 + 10 2 + 16*
4 a 2 2 a a? 19 g
' 9ar> 3a; a 2 12 a'
9 9 ■'■ T 4 T 9 ^16^ 4
Find three terms of the square root of :
19. 1+9 x. 21. ar 2 -}-^. 23. 9^ — 1.
20. a?-S. 22. 4 0^-5 0;. 24. 36 -12a;.
THE SQUARE ROOT OF ARITHMETICAL NUMBERS
237. Since an arithmetical square integer is the result of
the multiplication of some integer by itself, we are assisted in
obtaining arithmetical square roots by noting a certain relation
that exists between such numbers and their squares.
I 2 = 11 A number of one place has not more than two
9 2 = 81 J places in its square.
10 2 = 100 1 A number of two places has not more than four
99 2 = 9801 J places in its square.
100 2 = 10000 1 A number of three places has not more than six
999 2 = 998001 J places in its square.
Conversely, therefore :
If an integral square number has two figures, its square root has one
figure.
If an integral square number has four figures, its square root has two
figures.
If an integral square number has six figures, its square root has three
figures.
a
+ 2 ab + b' 2 \a + b 1
2 302 =
a + b
1296 180 + 6 Result.
900
396
396
__
+ 2 ab + 6 2 2 (30) = 60
+ 2ab + b 2 (60 + 6) (6) =
218 INVOLUTION AND EVOLUTION
Hence :
238. Separate any integral square number into groups of two
figures each, and the number of groups obtained is the same as
the number of figures in its square root.
Illustrations :
1. Find the square root of 1296.
Beginning at the decimal point,
separate into periods of two figures
Parallel Algebraic Process each.
(*)* =
2(a) = 2 a
(2 a + 6) (6)
In the square root of 1296 : The greatest square in 1296 is 900.
The square root of 900 is 30.
The trial divisor is 2 (30) = 60.
The second term of the root is (396 -f- 60 =6).
For the completion, (60 + 6) (6) = 396.
The process is repeated in the same order if the given integer is of a
higher order.
2. Eind the square root of 541,696.
The following process is given in a form commonly used in practice.
Separating into periods of two figures each : , • • •
-n, i ,. 541696 /So
Explanation :
The greatest square contained in the first period
(54) is 49. The square root of 49 is 7. 7 is, there-
fore, the first figure of the root. Subtracting 49
from 54, and bringing down the two figures of the
next period, we have 516, the remainder. Annex-
ing a to the first figure of the root, 7, our trial divisor is 2 (70) = 140.
Dividing 516 by 140, we obtain 3, the second figure of the root.
(140 + 3) 3 = 429, which product is subtracted from 516. With the
remainder (87), we bring down the last two figures (96), and the new
remainder is 8796. Annexing a to the figures of the root already found,
our trial divisor is 2 (730) = 1460. Dividing 8796 by 1460, we obtain 6,
541696
140 + 3
49
1516
3
[429
1460 + 6
8796
6
8796
SQUARE ROOT OF ARITHMETICAL NUMBERS 219
the third figure of the root. (1460 + 6)6=8796, which product, sub-
tracted from 8796, gives a remainder of 0, and the square is completed.
The addition of the to the figures of the root already obtained gives
a trial divisor of the same order as the remainder, or of the next lower
order. The process gives fewer figures and less likelihood of error.
If a given square number has decimal places, we point off by
beginning at the decimal point, first separating the whole
number as before, finally separating the decimal into periods
from left to right. Ciphers may be annexed, if necessary, to
complete any period.
If a given number is not a perfect square, its approximate
square root can be found to any desired number of places.
The square root of a common fraction is best found by chang-
ing the fraction to a decimal and extracting the square root of
the decimal to the required number of places.
Illustrations :
1. Find the square root of 2. Find, to three decimal
19920.4996. places, Vl2f.
20 + 4
4
280 + 1
1
2820 + 1
1
28220 + 4
4
19920.4996|
1
99
96
320
281
3949
2821
141.14 f = .375. 12| = 12.375.
Result. The required three decimal places
necessitate six decimal figures in the
square. Hence, with three ciphers
annexed, we have to obtain the
square root of
12.375000 13.517+.-.
9 Result.
112896
112896
The decimal point in the result
is located easily by noting between
which periods of the given example
the given decimal point lies. In the
example above there are two periods
to the right of the decimal point
given ; therefore, there will be two
decimal places in the root obtained.
60 + 5
5
337
325
700 + 1
1
1250
701
7020 + 7
7
54900
49189
220 INVOLUTION AND EVOLUTION
Exercise 75
Find the square root of :
1. 9216. 5. 186624. 9. .717409. 13. .00002209.
2. 67081. 6. 4202500. 10. 9617.7249. 14. .0001752976.
3. 32761. 7. 49.434961. 11. 44994.8944. 15. .009409.
4. 182329. 8. 9486.76. 12. .00119716. 16. .0000879844.
Find, to three decimal places, the square root of :
17. 3. 19. 7. 21. .5. 23. f 25. .037.
18. 5. 20. 10. 22. .05. 24. f. 26. .0037.
Find, to three decimal places, the value of :
27. 5 + 3 V'2~. 29. 2V3-V5T 31. Ve + VK
28. V7+-VI0. 30. 3V7-7V3. 32. ^VS-Vf.
33. How many rods in the side of a square field whose area
is 2,722,500 square feet?
34. Simplify and extract the square root of
^(10^ + 13)-2(2a 4 + 3 + 7^)a; + (ar J + l) (a 4 - a 2 + 1).
35. Simplify and extract the square root of
(x 2 - 2x - 3) (x 2 - x - 6) (x 2 + Sx + 2).
36. Show that the required square root in the preceding
example can be readily obtained by factoring and inspection.
37. If a, b, and c are the sides of a right triangle, and a lies
opposite the right angle, we may prove by geometry that
a 2 = b 2 + c 2 . Find the length of a in a right triangle in which
b and c are 210 feet and 350 respectively.
38. Applying the principle given in the preceding example,
find, to three decimal places, the distance from the " home 1
plate to the second base ; the four base lines of a regulation
baseball diamond forming a square 90 feet on each side.
CHAPTER XIX
THEORY OF EXPONENTS
239. The Index Laws for Positive Integral Values of m and
n have been established :
(1) For Multiplication: a m x a n = a m + n (61)
(2) For Division (m greater than n) : a m ■* a n = a m ~ n (79)
(3) For a Power of a Power : (a m ) n = a mn (215)
(4) For a Power of a Product : (ab) n = a n b n (216)
(5) For a Root of a Power : Va"*™ = am ( 2 ~ 6 )
240. The extension of the practice of algebra requires that
these laws be also extended to include values of m and n other
than positive integral values only; and the purpose of this
chapter is to so extend those laws. We shall, therefore :
1. Assume that the first index law (a m x a n = a m+n ) is true for
all values of m and n.
2. Define the meaning of the new forms that result under this
assumption, m or n, or both, being negative or fractional.
3. Show that the laws already established still hold true with
our new and broader values for m and n.
THE ZERO EXPONENT
If m and n may have any values, let n = 0.
Then, a m x a° = a TO +° (61)
= a m
d m
Dividing by a m , a = —
That is, a° = 1.
Hence, we define a as equal to 1. Or :
221
222 THEORY OF EXPONENTS
241. Any quantity with the exponent equals 1.
Illustrations :
1. s° = 1. 2. (wm)° = 1. 3. 3 m° = 3-1
4. 4aP + (2y)° = 4.1 + l=5.
Oral Drill
Simplify orally :
1. a°x. 4. 4a°6°c. 7. 5 a? -5. 10. 3a + 3 + a°6 .
2. 2aa°. 5. 3a° + l. 8. 2a°+3a°. 11. 2a° + (a°-l).
3. 3aV- 6- 4a°-3. 9. 4a; - 52/°. 12. 4?x + (a + 3 a?) .
THE NEGATIVE EXPONENT
If m and w may have any values, let n be less than and
equal to — m.
(61)
(241)
Then,
a m x
a -7 " = a mr ~ m
lence,
a m x
a- m = 1.
Dividing by a~ m ,
1
Dividing by a m ,
1
a _m = — •
a™
Hence, we define ar™ as 1 divided by a m .
From this definition we obtain an important principle of con-
stant use in practice :
242. Any factor of the numerator of a fraction may be trans-
ferred to the denominator, or any factor of the denominator may
be transferred to the numerator, if the sign of the exponent of the
transferred factor is changed.
Illustrations :
1 <,&-*-« 2 ar^s-i-JL. <* 2- l mx-* _3mn
FRACTIONAL FORM OF THE EXPONENT 228
Oral Drill
Transfer to denominators all factors having negative
exponents :
1. ax~ 2 . 4. 2a 2 ~b~ 2 . 7. crtc" 3 . 10. 2 ar^ar 1 .
2. crf-V" 3 . 5. 2- 1 ar 2 b- s . 8. x~ l y-h-\ 11. 5 ary 3 * 4 -
3. 2ab~ 2 . 6. Stf-Vr 1 - 9 - S-'a-^c- 1 . 12. iHar^Mr^.
Read the following without denominators :
a 2 cd Sa c 2 1
13. -• 15. 17. ~- 19. -^r- . 21.
b 2 mn a?y 2 ' ar 2 b~ 2 2~ 2 x~ 2 f
xyz ax m 5 x AA - 2 a ab
14. - 2 -. 16. — • 18. — • 20. -rt. 22. 53 2
m z J a l m 6 n t —3 "era
Read the following with all exponents positive :
23. 5a -1 &c. 27. = ; , - 31.
2ar>" 1 2- 1 m 2 w- 1 2
24. 12 a- 1 ^. 28. 5 5-5-5* 32. r — rr— j.
25. -5«*V. 29. — 53. 33. 3^.
26. 3a-Vd 30. |S^C 3 4.
3-V^z - 3 m- 1 ^-^ y- 1 ^ 2
THE FRACTIONAL FORM OF THE EXPONENT
■
If the expression a 2 can be shown to conform' to the first
index law, we may find a definition for the fractional form of
exponents. By the first index law,
(a*) 2 = at x J = a? + % = a.
Hence the meaning of a% is established, and the exponent in
this form still agrees with the fundamental index law.
That is:
224 THEORY OF EXPONENTS
a? = Va is one of the two equal factors of a.
Similarly, (a^) 3 = eft x eft X eft = cft + * + $ = eft = a 2 .
2
That is, a 3 is one of the three equal factors of a 2 .
In like manner, a? = Va = one of the three equal factors of a.
3 4 .
a* = va 3 = three of the four equal factors of a.
Therefore :
243. In the fractional form of an exponent we may define the
denominator as indicating a required root, and the numerator
as indicating a required power.
Illustration :
Vs 2 = 8» = 2 2 = 4. And 4 contains two of the three equal factors of 8.
\/8P = 81^=3 8 =27. And 27 contains three of the four equal factors of 81.
In general :
m m m m m m
a n • a n • a n ••• to n factors — a n n n ••• to n terms
mn
= a m .
m
That is, a n is an expression whose nth power is a m .
m
And a n is the nth root of a m -
m
Or, as above, a n is one of the equal factors of a m . %
It is understood that while a* must equal either + Va or — Va, we
i r -
consider the positive value only ; and a 2 is denned as -f V a, the principal
square root of a.
In future operations we may apply the definition of Art. 243
to expressions given in radical forms, observing that
244. The index of a radical may be made the denominator of
an exponent in the fractional form, the given exponent of the
power of the quantity becoming the numerator of the fractional
form.
FRACTIONAL FORM OF THE EXPONENT 225
Illustrations :
1. Vrn* = wA 2. Vtf = x\ 3. y/b^ = b^.
And in the converse operation of the principle of Art. 244 :
4. **=v^. 5. a$=Va?. 6. c^=«v^
In the reduction of numerical forms :
7. v / l66 = (\/l6)5=(2)5=32. 8. v^8* = (\/=8)6 = (-2*) = -32.
It is to be noted that the root is extracted first.
Oral Drill
Express with radical signs :
1. eft. 4. cfix?. 7. y*z\ 10. x$y%z$.
2. x\ 5. m^/*. 8. cW. 11. rftp^q*.
3. c*. 6. «&*. 9. e^mi 12. ir&rfix?.
Express with exponents in fractional form :
13. Va?. 16. Vz\ 19. y/a • Vc 3 . 22. V4~a* • V^" 8 .
14. Vm 5 . 17. -v^S*. 20. V^-^c 3 . 23. ^^-V367 4 .
15. Va 7 . 18. <^. 21. ^/m^.-^/n 3 . 24. VlScSJ* • \/8?.
Give the numerical value of:
25. 4* 28. 64*. 31. 4*. 34. 81*. 37. 32^.
26. 9i 29. 4*. 32. 8*. 35. 25*. 38. 64*.
27. 27*. 30. 8*. 33. 27*. 36. 49*. 39. 128*
Having established a meaning for the new forms of ex-
m
ponents, a , a -1 , and an, we must show that the index laws hold
true for these new forms; thus fulfilling the third and final
clause of our agreement in Art. 239.
SOM. EL. ALG. 15
226 THEORY OF EXPONENTS
Proof of the Index Laws for Negative, Fractional, and
Negative and Fractional Values of m and n
245. In the following proofs, m and n are rational integers
or rational fractions.
The Law a m xa n = a m+n .
1. When m and n are negative and integral.
a~ m x a~ n — — x — = = a-*-*.
a m <x n a w + n
2. When m and n are positive and fractional.
Let m = ■£ and n = -, », q, r, and s being positive and integral.
q s
P r ps qr ps+qr
a« x a* = a«* x a«* = VaP* x Va« r = Va** x a* r - tyap+Hr = a « 8
= a« •
3. When m and n are negative and fractional.
Let wi = — £ and n — — I, p, g, r, and s being positive and integral.
- p - -- i i i -M-9-
p r p + r
a? a* a? »
Let the student discuss this law when m is a positive and n a negative
fraction.
TheLaw(a w ) n = a mn .
1. When m and n are negative and integral.
Let m = — p and w = — q, p and a being positive and integral.
(a m ) n = (a~p)~ 9 =(—) = (a p ) q = a** — a(-*X-«) = aw. >
2. When w is jpo^iYwe and fractional.
Let w =^, p and g being positive and integral.
Q
P mp m p
(a m ) n = (a wl ) < * = V(a m )p = Va m » = a « — a '«,
APPLICATIONS OF PKINCIPLES OF EXPONENTS 227
3. When n is negative and fractional.
Let n = — ■?, p and q being positive and integral.
0.
= — =a « =a v « y
(a m ) n =(a m ) * =
(a m ) * «
The Law (a6) w = a n b n .
1. When n is negative and it
Let n — — p, p being positive and integral.
(ab) m = (db)-p = — — = — ?— = a-vb-P.
(aby aPbP
2. When n is positive and fractional.
Let n = -, p and <? being positive and integral.
p p p.
(ab) m = (aft) « = ty(ab)p = -^a*>&p = i'a*' • VbP = a* 6«
3. When w is negative and fractional.
Let n = — -, p and q being positive and integral.
-E i i _£ _£.
(«&)"»=(«&) « = ; = "T7 = a ?6 q
(ab)i aib*
Let the student discuss this law (1) when m and n are both positive
and fractional, and (2) when m and n are both negative and fractional.
APPLICATIONS OF THE PRINCIPLES OF EXPONENTS
(a) Simple Forms involving Integral Exponents
246. In processes with exponents no particular order of
method can be said to apply generally. Experience with dif-
ferent types will familiarize the student with those steps that
ordinarily produce the clearest and best solutions. As a rule,
results are considered in their simplest form when written
with positive exponents.
228 THEORY OF EXPONENTS
247. Illustrations :
1. Simplify x~ A • x 5 • x~ 7 .
x-* • sc 5 - x- 7 = z-4+6-7 = a;-«= — . Result.
x 6
2. Simplify m 3 n~ 2 • m~ 2 n~ 5
m 8 n- 2 • m _2 n -6 = wi 3_
3. Simplify ^^ - ^Vl
ar 1 (ma; -1 ) ~ J
m 8 n- 2 • m- 2 n- 5 = wi 3 -%- 2 " 5 = mn~ 7 = S . Result.
ML 7
^m-gft 4 (a" 2 ) 3 _ a 2 m~ 3 y>cg- 6
X" 1
(jnix- 1 )- 2 x-im-W
= a 2-6 m~
5+2^4+1-2
= a^w-ix 3
-
-_^!_. Result.
a 4 m
Simplify :
Exercise 76
1. a 5 x a -3 .
5. m 2 • m~ 3 • m 5 .
9. x- 3 + x- 2 .
2. c~ 3 x c 7 .
6. x* • ic° • or 4 .
10. m~ 4 -r- m~ 7 .
3. ra" 2 X m 3 .
7. a-V- 1 ^- 8 ^
11. a~ s m~ 2 -r- a 2 m~ 3 .
4. a; -4 • x 3 .
8. c^d-cM" 4 .
12. ar 1 ^" 5 + x~ 5 y~ l .
13.
C"
2 dV 3
'd-'z 5
x®y~ l m 3 n~ 9
m 5 n~ 7 x°y- 8
14.
a
a
~ 2 b-
~ 2 b
" 2 ^°. 16.
a?m~hr 3 x~ 5
a~ 5 m~ 3 n 5 x 7
17. (a" 3 ) 2 .
22. (»-^l|)" 1 .
27. (a° + ay)- 3 .
18. (c- 3 )- 2 .
19. (a)" 5 .
23. (a-V 2 ) 2 .
24. (a- 1 ^- 2 ) 3 .
28. (mV)°x(3a; ) 2 .
20. (a" 1 ) 5 .
21. (a- l b)\
25. (a 2 a°a- 3 )- 2 .
26. (c- 1 ^ • c ^- 1 )-
, >°-m'-
APPLICATIONS OF PRINCIPLES OF EXPONENTS 229
(6) Types involving the Fractional Form
248. In the following illustrations attention is called to
each important feature of the process, and the order of the
principles that is emphasized in each is such as will, under
similar conditions, produce the best form of solution.
Illustrations :
1. Simplify (a^V 1 )" 2 .
(asrtr 1 )" 2 = <r 4 &^ 2 = h ^~- • Result.
Note (1) that the first step is the application of the law (a m ) n = a mw ,
and (2) that the result is given with all exponents positive.
2. Simplify (c 2 V0 3 .
(c 2 \/<Fi) 8 = (c 2 • c~*) 8 = (cfy = A Result.
Note that the law a m x a n = a m+n is first applied so as to unite c-f actors.
3. Simplify {^(V^)-|}~*
Note that the reduction is accomplished outward.
Result.
L Simplify (
9Vm-
s\-f
>25Vm/
/ 9 v^F2 \-f __ p^ -f _ / 9 \"i = / 25 mM = 25? ro* _ 125 m^
V25Vm/ V 26m i/ \ 25m V V 9 / " 9* 27
Result.
Note that in the third step inverting the fraction changes the sign of
the exponent of the fraction. In general, ( -\ ~* = 225 = £ = (k\ x .
230 THEORY OF EXPONENTS
V m f Vn~i _ j m Vn" 4 _ [" w^n""* _ / mn~ 2 T
= [mJrT* -T- Vm%- 8 ] 6
b= (wi*n~^ -h m« _i ) 6
= (m^VM)*
= (m^) 6
= m 2 . Result.
Exercise 77
Simplify :
1. (8a-2a" 1 )i
6. (^Txyt+i-l/Slx 7 )- 1 .
if
3/ =- 7. JaY(*X\Z
2. \(aV- 3 Vaary. Af * V2T7 2/
3. [m^m-^m-?) 2 ] 6 . g ^ c~ V § J m* y/tfc _
r a~Wm m*a 16 Vc
4. [V a -i c i(c-W)^]" 8 .
r 3/ \ 1-2 9. \ r- X =
5. L^ ax -Vax-U ' c*d- 2 cy/d
10
A c~Wx 4/ I6 n Jc^n*
n As a ~s/x 2 + 27 xi Va-<aQ-*
(9a?*-+-4at)
12.
[yV < Vy 3
L^ 2 ^ y^.
13. |Jm 3 a;- 1 ^/ m 2 a ,-2 % /^v
APPLICATIONS OF PRINCIPLES OF EXPONENTS 231
(c) Types involving Numerical Quantities Only
249. The exponents of a numerical expression must be
made positive before reduction is attempted.
Illustration :
Simplify 9* + 27"$ + 5° - 8"* + 1.
9 t + 27~8 + 50_ 8 -f + i = 33 + JL +1 __L +1
27* 8$
= 27 + | + 1 - | + 1
= 28§£. Result.
Exercise 78
Simplify :
l- (f)- 3 + (f)- 2 -(t) 3 - 3. 4 a?+ (4 *)•+!*•.
2. (J)-'- 8* + (!)-«. 4. -^s-s^+c^)- 1 .
6. 16^ + 16« -16°-(^-)- 2 .
•[ 8 -^-' ! ][© ,+ <^>(r + (r}
(rf) Types involving Literal Exponents
250. Illustration :
m m _i 2m— 1 n*
Simplify [a» -a" -^a^ 7 "] .
p m m 2m— -i«x r- m+n m— n 2m l-.nx
\_an + • a»~ ■+■ a «~ J = [_a « • a~^*~-r- a - ** - _
Cm+n m— n_2m— 1-ji
a~n~ « «~J
a*. Result.
232 THEORY OF EXPONENTS
Exercise 79
Simplify :
1. a 2m+n • a m+n -!-a? m+n . 7 f *+i . 1 \^
2. (m*) x+1 + m x+2 . K ^ J
3. («*)•*■ + (<**•). 8. ^_ 2n+2 x8r
4. (m* - ")' (m y ~*y -*- m _(ae+2)y .
16"
5. (03& 2C6)ax.
9 - [^'(^t^)^}
a x+2y a y+z a x+3a 4~ T~
(e) Miscellaneous Processes involving the Principles op
Exponents
251. Illustrations :
1. Multiply a - 3 J - 2 a$ by 2 - a* + 3 a~i
g — 3 gf — 2 g^
2-
- a "i + 3 g~$
2a-
- 6 g$ - 4 g$
- gt + 3gi + 2
+ 3 a \ _ 9 _
-6a
-1
2 a - 7 of + 2 rf - 7 - 6 g~£ Result.
2. Divide 9 a* - 3 a* + 1 + 7 <T* - 6 a~* by 3 + cf* - 2 cf*.
9g* - 3 a* + 1 + 7 g"? - 6 a - * ( 3-f g-i-2g~£
9 gi + 3 g£ - 6 3 g£ - 2 g£ + 3. Result.
- 6 gi + 7 + 7 o~i
-6gi-2+4g"?
+ 9 + 3g _ ?-6g"i
+ 9 + 3 g~i - 6 g"i
APPLICATIONS OF PRINCIPLES OF EXPONENTS 233
3. Simplify qq + q^ + q-^i-q).
a (1 + a)- 1 + <r l (l ~ «) _ j+g
a . 1 — a
a(l + a)_ 1
I - a 2a 2 -l 2 a 2 - 1
Result.
1 -t-a a
a (1 + a)
4. Expand (a 2 -2 a" 2 ) 4 .
(a 2 -2a- 2 ) 4 = (a 2 ) 4 -4(a 2 ) 3 (2a-2) + 6(a2)2( 2a -2)2_ 4 ( a 2) ( ;2a-2)3 + (2a- 2 >*
= a 8 - 8 a 4 + 24 - 32 a~* + 16 <r 8 . Result.
5. Extract the square root of
--^-2^ + ^ + 17 + 12^5 + 4^
tf*-2aT*-3-2a*. Result.
- 4 art - 2 af £
- 4 aT? + 4 a;"^
2afz--2ari
-2x~i
2 x~i - 4 x~ X i - 3
-3
2 af* - 4 af * - 6 - 2 a4
-2a*
6aT* + 8 a;"? 4- 17
6aT* + 12 aT4' + 9
- 4ari+ 8 + 12x£ + 4a^
- 4af* + 8 + 12xT + 4a^
6. Find the value of x in the equation x * = 8.
(Give to both members the exponent necessary to make the exponent
of x equal 1.)
aft = 8, (art ) - f = (8)~t, » = 8 - $, x = — , a; = J> Result.
7. If ar 3 = 2/ -2 , and 2/ -1 = —8, what is the value of x ?
We first require the value of y~ 2 from the second equation.
Hence, if y~ 1 = -S, Tnen > x" 3 = (-8) 2 .
(y-i) 2 zz(-8) 2 , * = [(-8) 2 r*
(-8) 2 .
= -8-1
= J. Result.
-2m
234 THEORY OF EXPONENTS
Exercise 80
Multiply :
1. a;" 2 + 3 a;" 1 -2 by a;- 2 -2 a;- 1 -4.
2 . a* + 2a±&* + 6*by a*-2a*&* + &i
3. 4ar 3m + 6ar 2m — 5ar OT -3 by 3 x~ 2m + 2 ar TO - 1.
4. a*-9a* + 27a~*-27a"Hy a^-6 + 9a"i
5. va^ ^ + 2v^-2V^by -^ + Va + -^.
W Va 3 Va
Divide :
6 . c -4_ c -3_ 8c -2 + llc -i_ 31:?v c -2 + 2c -i_3.
7. 2a-a* + 4a* + 4a*-3by a*-a* + 3.
8. 35+4a- w -16a- 2m +19a- 3w -6a- 4m by 7+5or w -3cr
9. a 2 + 2a;*-7ar*-8af£ + 12ar 3 by a? - 3 oT* + 2 af*.
10 . a-M? + ^- 8 K + H 7 V-a-^+±.
■y/c Vc Vc 3 c Vc Vc
Simplify :
11. [( x + 3)(x-3)- 1 -(x-3)(x+3)- 1 ] + [l-(x i +9)(x+3)- 2 ].
1 2 . [(o^-SXc-^- ^^^^] x [Sc^-O)-}- 1
Expand :
/ 2V 17, I — = F/
14. (V^+2^)\ 16 " ( 3ViF -aJ' W ° 2V ^
Extract the square root of :
18. 4 ar 4 + 12 ar 3 + ar 2 - 12 a;- 1 + 4.
19. 9a- 2 " l -6a- w -ll + 4a m + 4a 2w .
APPLICATIONS OF PRINCIPLES OF EXPONENTS 235
20
. x~% — 4 x ^ + 8 x %y* — 8 x~*yl + 4 y^.
* -y/x 3 \y \x
Find the value of x in :
22. a£==3. 25. y~* = 4. 28. #"*=£.
23. a;* = 2. 26. 2"* = 9. 29. x» =2.
24. a>* = 25. '27. af* = — 8. 30. x m = 3 n .
Multiply by inspection : Divide by inspection :
31. (3 a" 3 - 2 a 3 ) 2 . 35. (a" 3 - 9) by (a* + 3).
32. (2 or 1 + 3) (3 ar 1 + 7). 36. (a - 81) by "(a* + 3).
33. (3a- 1 -2a)(5a" 1 -3a). 37. (a- 3 -86" 8 )by (a- 1 -26~ 1 ).
34. (2a- 2 + a~ 1 H-3) 2 . 38. (27a"*+125)by(3a~ J +5).
Find the value of a; in each of the following :
39. x' 1 = y, and y 2 = 9. 42. a£ = y _1 , and y = 3.
40. sb = 2/ -1 > and y* = 3. 43. a£ = 2/^, and y» = 9.
41. x~ 3 = ?/, and y~ 2 = 2. 44. aT* = IT 3 , and y~* = 9.
Simplify :
45. (a" 1 - or 1 ) 2 - (a" 1 + ar 1 ) (a" 1 - x- 1 ).
46. (a- ffl + 3a m ) 2 -(a- m -3a w ) 2 .
47. (a- 3 + l)a- 3 -(a- 3 -l) 2 + (l-a- 3 ).
V d»-c*A c^ + dU J___L
49 f ^ + 13 ajy* a^ + 4yH . |" 5^ + 2^ "|
L^4-2a-3^ — 3^ a* + 3^J L*t— 9yt.J
CHAPTER XX
RADICALS. IMAGINARY NUMBERS. REVIEW
252. A radical expression is an indicated root of a number j
or expression.
Thus : V2, fyl, VlO, and Vx + 1 are radical expressions.
253. Any expression in the form -\/# is a radical expression,
or radical. The number indicating the required root is the
index of the radical, and the quantity under the radical is the
radicand.
In \/7, the index is 3, and the radicand, 7.
254. A surd is an indicated root that cannot be exactly ob-
tained.
255. A radical is rational if its root can be exactly obtained,
irrational if its root cannot be exactly obtained.
Thus : V25 is a rational expression ; VlO is an irrational expression.
256. A mixed surd is an indicated product of a rational
factor and a surd factor.
Thus : 3 V6, 4V7x, abVa + b are mixed surds.
257. In a mixed surd the rational factor is the coefficient of
the surd.
Thus : In 4VEx, 4 is the coefficient of the surd.
258. A surd having no rational factor greater than 1 is an
entire surd.
Thus : Vbac is an entire surd.
236
TRANSFORMATION OF RADICALS 237
259. The order of a surd is denoted by the index of the
required root.
Thus : . a/5 is a surd of the second order, or a quadratic surd.
\/7 is a surd of the third order, or a cubic surd.
260. The principal root.
Since (+ a) 2 = -f- a 2 and (— a) 2 = -f a 2 , we have V+a 2 = ± a.
That is, any positive perfect square has two roots, one + and
the other — , but in elementary algebra only the -f value, or
principal root, is considered in even roots.
THE TRANSFORMATION OF RADICALS
TO REDUCE A RADICAL TO ITS SIMPLEST FORM
261. A surd is considered to be in its simplest form when
the radicand is an integral expression having no factor whose
; power is the same as the given index. There are three com-
mon cases of reduction of surds.
(a) WJien a given radicand is a power whose exponent has a
factor in common with the given index.
By Art. 244, tfp = a* = a% - Va.
Hence, to reduce a radical to a radical of simpler index :
262. Divide the exponents of the factors of the radicand by the
index of the radical, and write the result with the radical sign.
Illustration :
V%ahfi = y/WahP = 2 y a M a Wcfix- -Result.
Exercise 81
Simplify :
1. -v/aV. 5. VW&. 9. a/243.
2. -Vote*. 6. VWm*. 10. J/WaW.
3. vW. 7. ^36. 11. \/ffiS.
4. tyrftf*. 8. -y/lSS. 12. ^216 aV.
238 RADICALS. IMAGINARY NUMBERS. REVIEW
(b) When a given radicand has a factor that is a perfect power
whose exponent is of the same degree as the index.
By Art. 244, Vcfib = (a 2 &)* = aM = a$ = aVb.
Hence, to remove from a radicand a factor of trie same power
as the given index :
263. Separate the radicand into two factors, one factor the
product of powers whose highest exponents are multiples of the
given index. Extract the required root of the first factor and write
the result as the coefficient of the indicated root of the second factor.
Illustrations :
1. vW = V4 a 2 x 3 a = 2 aVSa. Result.
2. 2V72 aWy 1 - 2 V36 a 2 x*y« -2 ay = 2(6 ax 2 y s ) y/2 ay
= 12 ax 2 y* y/2ay. Result.
Exercise 82
Simplify :
1. V28". 8. VI§2. 14. |V*p;
2. vs: 9 2 * /mu «• 1^375-
3 - ^* io. 3VI62: 16 - *W
4. V98T 17. V9 a 3 .
11. 3^80.
5. </W. XX ' ovo ^_ 18. Vl6aY.
e. VIM 12 - AV720. i9. ivTe^:
7. a/54. 13. 2-J/243. 20. </54aV.
21. -i-Vl62^y. 23. |^-V54 mV
3 y 2 9 m 2
22. V675mV. 24. ■y/4S<*<P 1 aP.
TRANSFORMATION OF RADICALS 239
»■ & - m- ■• f#§
6 3 jl25^fz 16& 3/ ^686"
' * \ 216 m 3 ' 30, 14 \125z 6 '
29 . J2B0 (« + «)« 31 . ((B + 1) 'I
i*
a + 1) 4
(c) TT^en ^e given radicand is a fraction whose denominator
is not a perfect power of the same degree as the radical.
If the denominator of a fractional radicand can be made a
perfect power of the same degree as the index of the radical,
the fractional factor resulting may be removed from the radi-
cand as in the previous case. By multiplying the denominator
by a particular factor we produce the desired perfect power.
Multiplying both numerator and denominator by this particular
factor introduces 1 under the radical, and the value of the
radicand is unchanged.
Ingeneral: ^.^-^^g^ily^
Illustrations :
Hence, to reduce any fractional radicand to an integral
radicand :
. Multiply both numerator and denominator of the radicand
the smallest number that will make the denominator a perfect
power of the same degree as the radical. By the method of the
preceding case remove the fractional power thus formed.
240 RADICALS. IMAGINARY NUMBERS. REVIEW
Exercise 83
2
Simplify :
,4
>-4
W!
10. M
,4
11. 4—-
*a§
12. J*.
-4
13. mM
>-4
14 c 4 /^\
SC Ac 3
-4
K 2c s/T
15. — a •
3 V'2c
•■#
- 6 *VS-
17.
m
3/54 m 4
18. J-1%
*# + 1
19. (l_a)^f-—
20 . bi^SS
a + 1 *# — 1
21 1 3/ (m-iy(m + 2) 6
m + 2\ (m + 1) 2
22 m * p ^-12a; + 38
^ — ft \
a? — o * fir
23. (a-2) x f^±
a — 1 w
J^+2^+1
TO CHANGE A MIXED SURD TO AN ENTIRE SURD
The process is the reverse of that of Article 261, Section (a).
In general : ay/x = a • x% = a% 2 = Va 2 #.
Hence, to change a mixed surd to. an entire surd :
265. liaise the coefficient of the surd to the same power as the
degree of the radical, and multiply the radicand by the result.
TJie indicated root of the product is the required entire surd.
Illustrations :
1. 3 V5 = VPT5 = \/9^5 = V45. Result.
2. 2 VI = \f¥Ti = %%7i = ^32- Result.
TRANSFORMATION OF RADICALS 241
Exercise 84
Change the following to entire surds :
1. 2V3.
2. 5V2.
3. 3^/4. u 3™ /~8^
»X 2
4.2^2. 2a;V27TO
5. 3^4. 12. i^Vg.
a *ar
6. 2xV2x. ,
7. fcrfffii 13 ' ( a + 1 )\(^i)^-
8. So 2 ^*;.
14. g±Sj ^- 1
» — 1 ^ar J + 4a;H
9. 3a-\/3a; 2 . ' a-1 \ar 2 + 4a; + 4*
TO CHAISE RADICALS OF DIFFERENT INDICES TO EQUIVALENT RADICALS
HAVING THE SAME INDEX
We have Va = a?
and -\/a = a 5 .
Expressing the exponents as equivalent fractions having a
common denominator, a%=a? = -\/a 3 ,
• a* = a* = -v'a*.
Hence, to change radicals of different indices to equivalent
radicals having the same index :
266. Express the given radicals with exponents in fractional
form and change these fractions to equivalent fractions having a
common denominator. Rewrite the results in radical form.
Illustrations :
1. Change V5 and VlO to radicals having the same index.
V5 = 5^ = 5* = #5* = #125.
#10 = 10^ = 10^ = #10 2 = #100.
SOM. EL. ALG. 16
Result.
242 RADICALS. IMAGINARY NUMBERS. REVIEW
2. Arrange ^11, V5, and V90 in order of magnitude.
vTl = 113 = ni = ^n-2 = ^121.
V5 = 5* = 5^ = \/p = vT25.
^90 = 90^ = #90.
The order of magnitude is, therefore : V5, vTI, #90. Result.
Exercise 85
Change to equivalent radicals of the same index :
1. V2, VS. 4. -y/n, -v/6. 7. V2, #3, ^5.
2. V3, -Vl. 5. ^6, ^/8. 8. V5 </9, ^15.
3. </2, -VS. 6. >^^5. 9. y/l,</7, VS.
Arrange in order of magnitude :
10. V5, -v^IO. 12. \/iJ, VK 14. V3, -v/8, -v^5.
11. -v/4, V3. 13. -\/5, </4. 15. V2, -#6, ty^3.
OPERATIONS WITH RADICALS
267. Kadicals that, when reduced to simplest form, differ
only in their coefficients are similar radicals.
ADDITION AND SUBTRACTION OF RADICALS
268. Similar radicals may be added or subtracted by add-
ing or subtracting their coefficients.
Illustration :
Find the sum of 2Vl2 + 2 V27 - 9 V^~+ 5 V3.
2vT2= 4V3
o-v/97 — fiVs " tne &* ven radicals are dissimilar, such as
~~ are similar may be added as illustrated, theex-
~~ 9v ^ * = — 12V3 pression for the sum being indicated.
6V3 = 5V3
Sum = 3V3 Result.
OPERATIONS WITH RADICALS 243
Exercise 86
Simplify :
1. VI8 + V8-V32. 5. ^2-^/16 + ^250.
2. V75-V48+V27. 6 . ^ _ -</512 + Vl62.
3. 2V3-2V27 + 2V108. 7. V| + V6- V|-iV6.
4. 3V98-2V75-3V32. 8 . V|- Vf +iVf-iVff.
9. V45~S- V20o¥- V5~a¥- V80o*a; + V180 a 2 s.
10. V8aa + V32aa — f Vl8 ax — V16 ax.
11. 2V7- V80 + V63-Vil2 + Vi5.
12. 4\/24 -2^/81 + 11^3-3^192.
13. 10Vi-5VJ| + 24Vi-7V2 + 16VS.
14. 3^^-4^ + -\/l28-2^|- + 6^/J-9^/|.
15. 8V| + 14V|-V75-iV5i2 + 5V||-6V||.
MULTIPLICATION OF RADICALS
269. Any two radicals may be multiplied.
Illustrations :
1. Multiply 3V6 by 2V15.
(3>/6)(2 Vl5) = (3 • 2)( V6 . Vl5) = 6V90 = 18\/l0. Result.
2. Multiply V28 X V42 X Vl5.
Expressing each radical in prime factors,
V28)(V42)(Vl5) = V(7.22)(7-2.3)(5.3) = V(7 2 • 22 . 3 2 ) (5 . 2)
= (7-2.3) VTO = 42 VlO. Result.
3. Multiply 2-v/4 by 3</8.
Changing to radicals of the same index,
2^4 = 2 • 4* = 2 . 4 T * = 2\/4*.
3\/8 = 3 . 8* = 3 • 8T2 = 3y/&.
(2 y/i*) (3 v/p) =(2-3) ( \/5*7p) ==6v / 2^^=6\/2^=6v / 2i 2 26". =12^32.
244 RADICALS. IMAGINARY NUMBERS. REVIEW
Hence, to multiply two or more monomial radicals :
270. Multiply the product of the given coefficients by the prod-
uct of the given radicands, first changing the radicals to radicals
having a common index.
The principle is in no way changed in application to frac-
tional radicands. It is usually best to multiply fractions in the
form given, reducing the product to its simplest form.
Exercise 87
Multiply :
1. V3by V6. 5. |V6by^Vl5. 9. fV^- by |V28f
2. Vl2 by V8. 6. -\/4 by ^2. 10. V35ac by VUax.
3. 2V7by|V21. 7. V6 by ^9. 11. ^Im^byVBlm^.
4. 3-\/4 by -v/12. 8. V 5 ? by VJf. 12. J/2nm by Vl2msc.
13. (2V3 + 3V6-2Vl2)(2V3).
14. (4^/2 -3 ^4 + 2 a/12) (^12).
15. (V| + iVf-3V|)(V|).
16. (2 V6) (2 V3 - 3 V6 + i V15 + i VlO).
MULTIPLICATION OF COMPOUND RADICALS
271. Illustration:
Multiply 3 V2m-2 V6m by 2V2m + 3 V6m.
3 V2ro - 2 V6m
2 V2m + 3 V6m
6 • 2 m - 4 Vl2 m 2
+ 9 V12 m 2 - 6 • 6 m
12 m + 5 V12 to 2 - 36 m = 5 Vl2 ra 2 - 24 m - 10 m V3 - 24 m. Result.
OPERATIONS WITH RADICALS 245
Exercise 88
Simplify:
1. (2+V2)(3 + V2). 5 . (4V3 + 2)(2V3-2).
2. (3+V5)(2 + V5). 6. (3^3-2-v/9)(2-v'3 + 3-v/9).
3. (4 + 2V3)(3 + 2V3). 7. (4 V5 - 2 V3) (5 V5 + 3 V3).
4. (5-2V2)(2-V2). 8. (2a/4 + V2)(3^'4-V2).
9. (Va+ V^)(Va + V»).
10. (2Vm-3Vw)(3Vm-2Vn).
11. (4ajV2-?/V3)(2a;V2 + 2/V3).
12. (3V2^-2V5^)(2V2^ + 4V5y).
13. (3V3-2V2-4V5)(2V3 + V2-2V5).
14. (3V6 + V2-V3)(4V2+V6 + V3).
15. (V2 + V3)(V3 + V6)(V3-V2).
16. (V6-V3)(V2-V6)(V3-V2).
17. (V^TT-Va^l)(V^+l-2V^^l).
18. (2VaT^-3Va^)(3VaT^ + 2V^T).
19. (3Va^ + a + l-2Va; + l)(2V^ + a; + l + 2Va> + l).
DIVISION OF RADICALS
272. As in multiplication of radicals we may divide any
two radical expressions of the same index.
Illustrations :
1. Divide -^96 by y/2.
M = ^.= m = 2Vz. Result.
2. Divide y/l by V6.
n = w = yii = ^i^ Result
V6 W >216 *27 3
246 RADICALS. IMAGINARY NUMBERS. REVIEW
3. Divide yfiff by >^f.
Exercise 89
Divide :
1. V8by Vl2. 5. 6V5by2Vl5. . 9. ^36 by -y/U.
2. V6by V18. 6. 4-^4 by 3^12. 10. </\ by VJ.
3. V27by V12. 7. JV72 by fVJ. 11. Vffby^f.
4. 4Vl5by2V5. 8. -?^ by ^50. 12. V^ by ^£.
DIVISION BY RATIONALIZATION
273. If a given divisor involves quadratic surds only, a
division is really accomplished by the process of rationalization ;
or, by multiplying both dividend and divisor by the expression
that will free the divisor from surds.
(a) When the Divisor is a Monomial.
Illustrations :
1. Divide 8 by 2 V3.
_^ = _8_ X V3 = 8V3 = 8^ = 4 V 3 > Result> .
2V3 2V3 V3 2.3 6 3
2. Divide 2 -tys by -y/L
2jft = 2V8xtt = 2( l Vy^ ) = 2WW) = 2 i^ Regult
The process of rationalization of the divisor simplifies
numerical calculations with radical expressions.
3. Find, to the three decimal places, the value of " Y, '
Vl2
3V6 3V6xV3 3V18 9V2 3 _ S„„ A N mmm £
—=z = —= 7= = — =■ = -7— = - v5 = o (1.414+-..) = 2.121+....
Vl2 Vl2 x V3 V36 6 2 2 ^ >
OPERATIONS WITH RADICALS 247
Exercise 90
Eationalize the denomi- Find, to three decimal places,
nators of : the value of :
1. — -• . 4. -• 7. 10.
V2 . SV2 V5 V75
& 5. A. 8 . ju ii. b*
V6 V2 3V2 3V2
-*-. 6 . _L. 9. ?V5. 12 A.
2V3 a/9 3V3 ^i
(5) When Either or Both of the Terms of a Binomial Divisor
are Quadratic Surds.
By Art. 104, (« + 6) (a - b) = a 2 - 6*.
Similarly, ( Va + V&)( Va - Vb) = a - b.
274. Two binomial quadratic surds differing only in sign
are called conjugate surds.
( Va + Vb) and ( Va — Vb) are conjugate surds.
From the multiplication above we may conclude :
275. The product of two conjugate surds is rational.
Illustrations :
2V3-3V2
1. Rationalize the denominator of
2 V3 - V2
2V3-3v^ _ 2V3 -3\/2 x 2\/3+ V2
2V3-V2 2V3-V2 2V3 + V2
_ 6-4V6 _, 2(3-2V6) ^ 3-.2V6 Result
10 10 5
Two successive multiplications will rationalize a trinomial
denominator in which two quadratic surds are involved.
248 RADICALS. IMAGINARY NUMBERS. REVIEW
2. Eationalize the denominator of — ^t— •
V3+ V2-1
V3_ V2+ 1 = V3-(V2-1) V3-(V2-1)
V3+V2-1 V3+(V2-1) V3-(V2-1)
= 6-2V6 + 2a/3-2V2*
2V2
3 - V6 4- V3 - V2
V2
3_V6+V3-V2 x V2_ 3\^-2V3+ V6 - 2
V2 V2 2
Exercise 91
Eationalize the denominators of:
4 V2+ V3 7> 2V2-4
2 + V2 V2-V3 3V2+2
3 5 5-V2 g 4V3-3V2
V2-2* ' 2 + V2* ' 2V3 + 3V 2 '
5_ 3V2 + 1 9 5V6-2V3
V7-V2 V2-1 3V2 + 3
10 V^+V^ - 15. Vs — 2 — Va^
Va— V# V# — 2 + Va;
„ 2Vm+Vn . -„ Va + 1 + V2 a — 1
11« = r* ■*■*>. — rzz= z'
3 Vm — Vw Va + 1 — V2 a — 1
2Va + V2a , 17. V3 + V2-1 ,
3Va-V2^ V3-V2 + 1
13 qVft + cVa? 3- V2 + V3
aVaj — cVa 3 + V2 — V3
V^fl+2 4 - V 3 + V5
14. — ■ ly- — -=•
Va + 1-2 4_VB-V5
12.
INVOLUTION AND EVOLUTION OF RADICALS 249
INVOLUTION AND EVOLUTION OF RADICALS
276. By the aid of the principles governing exponents, we
may obtain any power or any root of a radical expression.
Illustrations :
1. Find the value of (2 Va 5 ) 2 .
(2^)2= (2 afy = W = 4 a? = 4 Vol Result.
2. Find the cube root of VaV.
VVaW sk VaW = a?x* = Vatf = xVax. Result.
Exercise 92
3/ — —
\
Find the value of :
1. (Va) 6 . • 5. y/^/S. 9. V^a + fc) 2 *.
2. (-^y) 3 . 6. </^ io. i/yfajfy.
3. (-^)« 7. Vj^L 11. ^vS
4. (-V1T0) 6 . 8. V^f^. 12 . e ^i^
PROPERTIES OF QUADRATIC SURDS
277. A quadratic surd cannot equal the sum of a rational ex-
pression and a quadratic surd.
That is, Va cannot equal b + Vc.
If b is a rational quantity and Va and Vc are quadratic surds,
If Va = & + Vc,
we may square, and a = b 2 + 2 6 Vc + c.
From which, 2 & Vc = a — & 2 — c.
w
Or, Vc «-& 2 -c
c
26
That is, we have a quadratic surd equal to a rational expression, an
impossible condition because of the definition of a surd. Therefore,
Va cannot equal b + Vc.
250 RADICALS. IMAGINARY NUMBERS. REVIEW
278. Given, a -f- V& = c + Vd, V& and -Vd 6emgr surds,
2%eft a = c and b = d.
If a is not equal to c, we have, by transposition,
a = c+ Vd — V&,
which, by Art. 277, is impossible. Therefore a must equal c. Hence it
follows that Vb = Vd.
279. Given, V a + V& = Vx + Vy.
7%ew, V«_ V6 = V5 — Vy,
to^en a, 6, a;, anc? y are rational expressions.
Va + V& = Va- + Vy .
Squaring, a + V& = x + 2 Vsey + y.
By Art. 278, a = x + y. (1)
And Vb = 2 Vsy. (2)
Subtracting, a — V& = x — 2 Vxy +y.
Extracting sq. rt., V a — Vb = Vx — Vy.
THE SQUARE ROOT OF A BINOMIAL SURD
The square root of an expression consisting of a rational
number and a quadratic surd is obtained by application of
the principles of Arts. 277, 278, and 279.
(a) The General Method
280. Illustration:
Find the square root of 11 + V96.
We may assume that, Vll + V96 = Vx + Vy. (1)
Then (Art. 279), Vll - V96 = Vx - Vy. (2)
Multiplying (1) by (2), Vl21 - 96 = x - y.
Or, x — y = 5. (3)
SQUARE ROOT OE A BINOMIAL SURD 251
(4)
Squaring (1),
11 + V96 = x + 2 y/xy + y.
Hence (Art. 278),
11 = x + y.
From (3) and (4),
x-y = b.
3 + ^ = 11.
Whence,
x = 8, y = 3.
Therefore,
V» + y/y = V8 + V3 = 2 V2 + V3.
That is,
Vll + V96 = 2V2+ V3. Resu
(b) The Method of Inspection
281. If a binomial quadratic surd can be written in the form
a + 2 V&, its square root may be obtained by inspection.
Illustration :
1. Find the square root of 23 + 6 VlO.
6 VlO may be written thus: 6 VlO = 2 (3 VlO) = 2 V90.
Then 23 + 6 VlO = 23 + 2 V90. (The required form of a + 2 Vb.)
The factors of 90 whose sum is 23 are 18 and 5.
Hence, the square root of (23 + 2 V90), or the square root of
(18 + 2 V90 + 6) = Vl8 + V5 = 3 V2 + V6. Result.
Exercise 93
Find the square root of :
1. 7 + 4 V3. 9. 146-56V6.
2. 11-6V2. 10. 124-30VTL.
3. 17 + 12V2. 11. 3x + 2x^2.
4. 28 + 6V3. 12. c 2 + m + 2cVm.
5. 33-8V2. 13. 2m + 2Vm 2 -l.
6. 30-12V6. 14. 2m 2 -l + 2mVm 2 -l,
7. 67 + 16V3. 15. c 2 + c + l-2cVc + l.
8. 138 + 30V21. 16. a 2 +4a+l + (2a+2)V2o:
252 RADICALS. IMAGINARY NUMBERS. REVIEW
EQUATIONS INVOLVING IRRATIONAL EXPRESSIONS
282. Equations in which the unknown number is involved
in radical expressions are called irrational equations.
283. An equation containing radicals is first rationalized by
involution. If there are two or more radicals in the given equa-
tion, it may be necessary to repeat the process of squaring
before the equation is free from radicals.
284. Roots of Irrational Equations. It can be shown that
(1) The process of squaring may introduce a root that is not
a root of the given equation.
(2) An irrational equation may have no root whatever.
Hence, we conclude :
285. Any solution of an irrational equation must be tested, and
a root that does not satisfy the original equation must be rejected.
Illustrations :
1. Solve a> + 3 = Vaj" + 15.
X + 3 = Vx 2 + 15.
Squaring, x 2 + 6 x + 9 = x 2 + 15.
Transposing, x 2 — x 2 + 6 x = — 9 + 15.
Uniting, 6 x = 6.
Whence, , x = l.
Substituting 1 in the original equation, and taking -the positive value of
the square root, we have,
1 + 3 = vT+To.
4=4.
Therefore, the solution, x = 1, is a correct solution.
2. Solve VaJ — 2 + Vx + 5 = 7.
y/x-2 + Vx+5 ss 7.
Transposing, Vx — 2 = 7 - Vx + 5.
Squaring, x-2 =49-14V» + 5 + X + 5.
Whence, 14 Vx + 5 = 56.
Dividing by 14, Vx + 5 = 4.
Squaring, x + 5 = 16.
EQUATIONS INVOLVING IRRATIONAL EXPRESSIONS 253
And, a; = 11.
In the original equation, Vll — 2 + Vl 1 + 5 = 7.
V9+ Vl6 = 7.
7 = 7.
Hence, x = 11, is a solution of the given equation.
3. Solve VaT+3 — 5 = Vx — 2.
Vx + 3 _ 5 = Vx - 2.
Squaring, x + 3 - 10 Vx + 3 + 25 = x - 2.
Transposing, x — x — 10 Vx + 3 = - 3 — 25 — 2.
Collecting, - 10 Vx + 3 = - 30.
Dividing by - 10, Vx + 3 = 3 .
Squaring, x + 3 = 9.
Whence, x = 6.
Substituting in the original equation,
V6 + 3 - 5 = V6 - 2.
V9 - 5 = VI.
3-5=2.
Since the original equation is not satisfied by the solution x = 6, this
solution must be rejected.
It will be found by trial that the solution, x =6, satisfies the equation
5 - Vx + 3 = Vx^2 .
Exercise 94
Solve and test the solutions of :
1. Va;+1 = 2. 8. V7+a;-7 + VaJ = 0.
2. v^i=3. 9 - Vt+s-vs+4
3. V ^T = ,-1. 10 - 2(V^5)(V^-5)=-35.
11. Vx — 1— Va = 2.
4. 2VS-3-V5+2. 12 V -_ [=V5 _ TI _ V ^
5. ^a?-l = V6. 13 3V^+V^ = 2= V4»-l.
6. 2VaJ+3=VaJ + 2. 14. V# + a + Va 4- 2 a = V4 a; — a.
7. V« 2 + 3 = a;-l. 15. yz-5-V4a;-2+- Va + 3 = 0.
16. V4 a + c = V25 x — 3 c — V9 <c — c.
254 RADICALS. IMAGINARY NUMBERS. REVIEW
17. V36a-1 = V4a-1 + Vl6a;
18. Va + 1 + V5 + 2 = V5 — 1 — Va; — 3.
19. VaJ-5 + Va + 8 = Va?-3 + Vaj + 2.
20. — —-VC^-Vi 2 2. V 6 + V5 + V^ = 3.
Va? — 7
2i. vss+i — -Jg-^vssL 23. Vfl; - 9+v ^=-i.
V2a+1 VaJ-9— Va:
24 . J*±|_J*E| = 0.
2 V^+ 3 4VaT+l A
25. 7= 7= = v.
3Va;-l 6Va;-2
26. V?a-V3»=V3a + 2Vo[6aj-(4-a)].
IMAGINARY AND COMPLEX NUMBERS
286. The factors of -fa 2 are either (+a) and (+a), or
(—a) and (—a). The factors of —a 2 are (4- a) and (—a).
Clearly, therefore, no even root of a negative number is pos-
sible. Hence :
287. An imaginary number is an indicated even root of a
negative number.
V— 2, V— 5, V— 10, are imaginary numbers.
288. The symbol, V^l, is the unit of imaginaries.
289. The rational and irrational numbers hitherto con-
sidered are known as real numbers in contradistinction to this
new idea of imaginary numbers.
290. Imaginary numbers occurring in the form of V— b,
where b is a real number, are pure imaginaries.
OPERATIONS WITH IMAGINARY NUMBERS 255
291. An imaginary number occurring in the form of
a-f-V— b, where a and b are real numbers, is known as a
complex number.
The Meaning of a Pure Imaginary.
By definition, Va is one of the two equal factors of a. Or,
(v^) 2 = VJaj* = («)t = a.
In like manner, ( V^l) 2 = V(- l) 2 = (- 1)1 = -1.
OPERATIONS WITH IMAGINARY NUMBERS
293. The fundamental operations with imaginary numbers
are based upon the principles governing the same operations
with radical expressions.
294. A pure imaginary number in which the real factor is a
perfect square may be changed to the form aV— 1, in which
form the elementary processes involving imaginary numbers
are greatly simplified.
Thus, (1) * V^"4 = V(4)(-l) = 2 V^l.
(2) 2\^l9 = 2V(49)(- 1)= 14\/^I.
(3) av^x* = aV(x*)(-l) = ax 2 V^T, etc.
ADDITION AND SUBTRACTION OF IMAGINARY NUMBERS
We assume that the principles underlying the addition and
subtraction of real numbers will still hold true in the addition
and subtraction of imaginary numbers.
Illustration :
Find the sum of 5V^9 + \f^2& - 3 V^IB.
5V^9 = 15V^T
v^26= 5v^T
- 3V^l6=-12\/^T
Hence, 8 V— 1 Result.
256
RADICALS. IMAGINARY NUMBERS. REVIEW
Exercise 95
Find the sum of :
i. V^ + v
2
9 + V-16.
V^i9 _ V- 25 + V- 36.
3. V-81 + V-64
100.
4 2V-25-3V-49-2V-64 + 3V-81.
5. 4V :r I-5V^16 + 3V^25 4-iV :r 64.
6. V^^ + V^Ta 2 - V- 16 a 2 - V- 25a 2 .
7. v^l^- V-(a + 2) 2 +V-4.
2V^"
8. 3V^i + 5V-2V-- v- T
-L / 9 9 -L / 9 79 -1-
+ 6V-
1
ST'
1
9. -y/ZTM-±^-c>d?-±^-a 2 <? + ±V-aW.
a d c d
2 > 3
10. cV-l-f- V-16c 2 + V-c 2
c
11. (2 + 5V^) + (7-2V^)-(8-2V^T).
12. (x + y^/^) + (x-z^~^) + (y-x->J^l).
MULTIPLICATION OF IMAGINARY NUMBERS
295. The positive integral powers of the imaginary unit,
By Art. 292, (V^T) 2 =- 1.
Therefore, (V^H) 3 = (\/^T) 2 (\/^T) =(- 1) (V^I) = - \T^\.
( V31 ) 4 =( V31 ) 2 ( V31)2=(_1)(-1) = +1^
(V^T)^(v r ^l)HV^l) = (+i)(V^I)-v r -i.
That is, the first power of V— 1 = V— 1.
the second power of V— 1 = — 1.
the third power of V— 1 = — V— 1.
the fourth power of V— 1 = 1.
And this succession repeats itself in order for the following higher
powers.
OPERATIONS WITH IMAGINARY NUMBERS
257
In the multiplication of imaginary numbers it is helpful to
remember that
296. The product of two minus signs under a radical is a
minus sign outside the radical.
Illustration :
1. Multiply V^3 by V^6.
VZ3 by V^6 = V(-3)(-6)= V(18)(-l) 2 = - Vl8= -3 V2. Result.
Exercise 96
Multiply :
1. V^ by V"
2. V^G by V-
3. V- 12 by V- 3.
4. V^36by -V^16.
5. _V-18by -V-54.
6. -yf^-2 by V^^T by V-
7. -yf^l by V^9 by V-
= »)(
28.
16.
(•
V- 25) (V-ioo).
9. (V3^)(_V-4c 2 )(- V-16 6 2 ).
10. (-V^^C-aV^H-aV^^.
11. (-V^l) (2V^1) (-3V=i).
12. (-2V" = l8)(-3V" :r 8)(5V^ : 32).
13. (V^aV) (- -V^^x) ( - V - c 2 x).
14. (a V" 11 ^) (a" 1 -/-!) ( - a-V^l).
15. (— Vm — ») ( vn - to).
16. (V-^-2x-l) (V^I) (VT=^).
MULTIPLICATION OF COMPLEX NUMBERS
297. We assume that the principles underlying the multipli-
cation of real numbers still hold true in the multiplication of
complex numbers.
SOM. EL. ALG. 17
258
RADICALS. IMAGINARY NUMBERS. REVIEW
Illustrations :
1. Multiply 3 + V^2 by 2 - V^2.
2 - Bv^-2
6 + 4V^2
- 15V^2-10(~2)
6-llV^2 + 20 = 26- 11V^2. Result.
2. Expand (3 - V^) 4 .
In processes involving frequent repetitions of the imaginary
unit, V— 1, it is convenient to substitute t in place of V— 1,
replacing the imaginary unit after the simplification is com
plete.
Let (3 - v/^T)* = (3 - iy. Then :
(3 - iy = 81 - 108 i + 54 & - 12 + &
= 81 - 108V^T + 54(-l) - 12C-V~T) + (1) (Art. 292)
= 81 - 108v^I - 54 + 12V^1 + l
= 28 - 96 V^l. Result.
Simplify
Exercise 97
3).
1. (2 + V- 1) (3 + V- 1).
2. (3-V^l)(4-V^l).
3. (2 + 2 V^T) (3 + 2V" =: T).
4. (V^+V^)(V^T+V-
5. (2V ::: l-3V :r 3)(2V" :r T-
6. (3 V2 + 2 V^5) (2 V2 - 3V^
7. (V— a — x) (V— a + a).
8. (-V^l + 2 V^2 + 3 V"^) 2 .
9. (2 + 3V^I) 8 . 11. (3V2 - 2V = T) 8 .
10. (3 + 2V-T)« 12. (aV^T- bV^iy
-2).
4V"
rg).
OPERATIONS WITH IMAGINARY NUMBERS
259
DIVISION OF IMAGINARY AND COMPLEX NUMBERS BY RATIONALIZATION
OF THE DIVISOR
298. Illustrations:
1. Divide V- 28 by V^8.
■ = ) . {} -= =± = \T = \T = 5 VIl. Result.
2. Divide V- 12 by V3.
yrrT2 = (V—2)xVs = V^ == 6V=T =2Vzr - v Regult
V3 V3xV3 3 3
3. Divide 2 + 2V :r 3 by 2 -2V :r 3.
2 + 2V^3 2 + 2 V^l? = (2 + 2\Z^3)« _ 4 + 8 V^3 - 12
2 + 2V-3 2 2 -(2V-3) 2
= 8V~3-8 _-
16
4 +
"3-1
Result.
Exercise 98
Simplify
5
2.
V^2
V18
V^3*
V-12
4.
13
V-12
2V^3*
2V^3-
5.
3V-
x +
3+V-
«-V-l
V-27
2V^~3
2V^~4
-SV^
-14
V^2
-4V20
V-5
15.
16.
10.
11.
12.
5-V-2
2+V-
3+V"
3-V-3
V2+V^T
V2-V^l*
V^2~+V^3
V-2-V-3
Vc + 3V-2c
Vc— V— c
2 m - 3 # V^^T
2m
-1
260
RADICALS. IMAGINARY NUMBERS. REVIEW
299. Conjugate Imaginaries. Two complex numbers differing
only in the sign of the term containing the imaginary are con-
jugate imaginary numbers.
Thus, a +bV— 1 and a — by/— 1 are conjugate imaginaries.
By addition, (a + b v^T) + (a - 6 V^l) = 2 a.
By multiplication, (a + b
l)(a
1) = a 2 + & 2 .
300. T7ie sum and the product of two conjugate complex num-
bers are real.
MISCELLANEOUS PROCESSES WITH IMAGINARY NUMBERS
Exercise 99
Simplify :
i. (V"^i) 5 + (V=T) 3 . 4. (_V^I) 5 + (-V^i) 4 .
2 . (_v^T) 3 -(-V"=T) 2 . 5. (_V^2) 2 -(-V^2) 4 .
3 . (_2V^2) 2 + 3(V^3) 3 . 6. (V^I-1) 2 -(1 + V^T) S
7. (V"=3-l) 4 .
8. (1 + V"^T) 3 + (l - V^T) 2 - (1 - V^l).
9. 3(V^) 3 -3(V"^)(V^-1) 2 .
10. (m + w V— l)(m — nV- 1).
11.
12.
13.
V-
-1+1
1 v:
-i-i
=1-1
1 _
ri+i
(V
:r 2) 3 +
(V-
^y.
14.
(2 + V-l) 2 + (2-
1£
3V-1
15 a + #V— 1 a — x
1
2V-2 + V-1
(V3i)4, ( V=2)4
yzri_V-2
a — xV—l a + x-V— 1
i6. ft^i-ff-v=i ;
17.
V-i
( V3l)6 _ ( V3T)5 + ( V3l)4
V^2
GENERAL REVIEW 261
GENERAL REVIEW
Exercise 100
1. Expand (xk - 2 -s/xf.
2. Find the square root of (x 2 + 3 x + 2)(x* — l)(a? + x — 2).
3. Simplify Vf + V^f + V| - VJ - V8.
5. Factor 27" V - 8" 1 .
6. Find the approximate numerical value of
2V2 + 1 . 3V2-1
V2-1 I V2 + 1 '
[ Simplify (^-y(^)« 1
8. So i ve f^-^ = w2 -^
[ my — %# = ra J — w .
9. Find the square root of a* + a* — 4 a 6 -f- 4 a + 2 a« — 4 al
10.
| si mp iif y (V^W^) 2
11. Simplify (- V^) 3 - (- V^T) 4 - (- V- l) 5 .
12. How may the square root of (x 2 — x— 2)(x 2 — 4)(ar J +3a;+2)
be found without multiplying ?
i(* + 2/)-z = -3,
13. Solve J 2x + (z-y) = 9,
_z + \(x + y + z) = 2.
14. Factor a; (x + 1)+ a (2 a? + 1) + a 2 .
15 - Simplify , ^C^%#£±gg •
1 J (x-l)- 1 -(x-2)- 1 -(x-3)- 1
262 RADICALS. IMAGINARY NUMBERS. REVIEW
1
16. Simplify <?*+*> • —
17. Simplify(V^-V^-V^3)(V^I+V^2-V^3).
18. Solve x — 2y = 3 f 3y + z = ll, 2z — 3x=l.
19. Simplify [^.^B 4 .
L ar 1 V a -1 a" 2 V ar 3 J
20. Draw the graphs of 2 a? + 3 y = 13 and 3 a? — y = 3, and
check by solving the system.
21. Solve 5x + 2y — z = 3x + 3y + z = 7x + 4:y — z = 3.
22. Factor ar 4 — 4ar 2 — 21.
23. What is the square root of x (x 3 - 4) + 1 - 2 x 2 (2 x - 3) ?
24. Find the cube of - 2 V^l + (V^I)" 1 .
rixl6«x2*l-i;
25. Simplify Li___J_^.
26. Simplify cc + [x + a? (as + ar 1 )" 1 ]- 1 .
27. Simplify (- 2 V^ - V"^) (3 V^J.+ V-2).
28. Expand (2 a 2 -3 a;) 5 .
«~ o i J? j ra — n + 1 2 „ , 3m+n- 1 r
29. Solve for m and n : ! — - = — and — =$
2m + n—l 13 m — ti + 2
30. Simplify g^+g^i;
31. Factor c(<* -f » + 2 c) + c(a? + 1) + (» + 1)*
««ai 3 4 1 4 . 2 06,8 1
32. Solve = 1, - + -=3, - -f - = 1.
x y x z z y
33. Explain the meaning of a~ n — — •
GENERAL REVIEW 263
34. Simplify — - 3 x° + 27"^ - 1 3 *.
8"~*
35. Simplify 16c4 + 8c3 - 2c - M" 2 -l .fl L^\
J l-9a 2 4c 2 + 2c ^ 1_2-^V #
36. Solve for s : V 4 s — 9 — Vs + 1= Vs — 4.
37. Find two numbers whose sum is m and whose differ-
ence is n.
38.
Simplify |^ + flz£ . (c-Q(l + c-i) T
39. Find to three decimal places the value f 2 ^+3V2 t
3V3-2V2
40. Solve 2x = 5y, y- X -2z = l, 12y-3 + z = 4:X.
41. Simplify - ^-l-f.
2+V-l 5 2-V-l
42. Find the square root of 1 ~^( 2 ~^) 1£
43. Is the expression a 4x + 2 a 3 * + a 2x + 2 a* -{- 2 + a" 2 * a per
feet square ?
44. Simplify 2m ( m ~ 1 ) + L_ .
(l_ m 2)f (l- m 2)i ^
45. Factor 1 + 2 ( 3 + *> + ?*±i.
(a 2 ^^)- 1 ^ a- 1 ^ 2- 1
af)
46. Solve -i + -i = m,—-i = n. •
ex ay dx cy
47. If a~% = c" 1 , and c* = f, find the numerical value of a.
48. Simplify
Y— L^-f ^Y— -i L^
V3+V-1 3-V-1A3+V-1 3-v"^iy
49. Collect 3Vl|-4V3-6V5| + 9V8j.
264 RADICALS. IMAGINARY NUMBERS. REVIEW
50. Under what condition is a solution of mx + ny = a and
kx + ly = b impossible ?
51. What must be the values of m and n in order that the
division of x 4 + x 8 + mx 2 + nx — 3bj x 2 — x + 1 may be exact ?
52. Extract the square root of
4 a 2n - 24 a~ n + 9 a~ 2n - 16 a n + 28.
53. Simplify V(a + l)(a>*- 1) - Va^-ic 2 - VaT^l.
54. A discount of 10 % was given on a bill of rubbers, and
one of 5 % on a bill of shoes. The amount of both bills was
$ 250, but, with discounts off, the merchant paid both with a
check for $230. What was the amount of each bill ?
55. What is the value of 2 a°[2°+(2 a )°] ?
56. Simplify
[ab-h- 1 + a-'bc- 1 + a^b^c) ' \ (a + b + c)~ 2 J '
57. Expand |> + x l — 2] 3 .
58. Draw the graphs of x-\-3y — 12 and 3x — 4?/ = 10.
Checlf by a solution.
59. Calculate to three decimal places the value of
(1 - V3) + (1 + V3).
60. Simplify °-°~ 1 a + a_1
a* — a » a* + a"J
61. Solve (« + l)(»-l)->- — — 1— -— = 8ar>.
62. Introduce the binomial coefficient under the radical
and simplify : (V3 - 2)V7 + 4V3.
'
GENERAL REVIEW 265
63. Simplify 1«- - 3 aP+ 3 (3 x)° - 1° + (1 + a°)(l - a ).
64 - tt * ^M-.^ +»-W
65. Find the value of
( a _ V6)(a-f V6)(a— V^l>)(a+ V^6) when a = 2 and 6=3.
66. Solve and verify \
I a; -f 2 y — m 2 + mw -f ™ 2 »
67. Simplify T^_^!l ( e \ .
68. Show that
(m 2 ^-?! 2 )?!" 1 — m m 2 — n 2 _ m 2 -mw
w _1 — »i _1 m 3 + w 3 7i — m
69. Solve and test the solution of
^/5~x~^a — V5 a; + . : = .
V5a; — a
70. Draw the graphs of 2sc + 3?/ = 7 and 4 <c + 6 y = 14,
and discuss the result.
71. Simplify $ V 1 - x + a + (» - a) (1 - x + a) ~*.
3 Vm + n + 2Vm-n
72. Rationalize the denominator of
3 Vm + n — V'
m — n
73. Solve V2aj + 5-f V8aj-3= V2o? + l, and test the
solution.
74. Simplify [a^afW^] • [^ar^??] +
75. Simplify
V2+V^l • V2-V := l ' V3 + V^2 • V3-V^2.
CHAPTER XXI
QUADRATIC EQUATIONS
301. A quadratic equation is an equation that, in its simplest
form, contains the second power, but no higher power, of the
unknown quantity.
A quadratic equation is an equation of the second degree
(Art. 64).
PURE QUADRATIC EQUATIONS
302. A pure quadratic equation is an equation containing
only the square of the unknown quantity. Thus :
ax* = a, ^ = 16, 2/2 = 20 a\
A pure quadratic equation is called also an incomplete
quadratic equation.
303. Every pure quadratic equation may be reduced to the
form
x* = c.
In this typical form we note :
(1) The coefficient of x 2 is unity.
(2) The constant term, c, may represent any number, posi-
tive or negative, integral or fractional.
Extracting the square root of both numbers of the equation
a? = c,
we obtain x = ± Vc7
and both values, + Vc and — Vc, satisfy the given equation.
PURE QUADRATIC EQUATIONS 267
The omission of the double sign before the square root of
ie left member has no effect on the result, no root being lost
by the omission. For:
(1) 4. x = +- Vc, (3) +aj = — Vc, and
(2) -z=-Vc, (4) -z= + Vc.
From (1) and (2), x — Vc, and from (3) and (4), x— — Vtf.
Clearly, nothing is gained or lost by the omission of the
double sign before the square root of the left member, and it
is not customary ,to write it.
From the foregoing we have the method for solving pure
quadratic equations :
304. Reduce the given equation to the form x 2 = c.
Extract the square root of both members, observing both posi-
tive and negative roots of the nght member.
Illustrations :
1. Solve 9 a 2 - 7 = 2 a 8 + 21.
9x 2 -7 = 2a;2 + 21.
Transposing,
9ie 2 -2z 2 =
= 7 + 21.
Uniting,
7x2 =
= 28.
Dividing by 7,
X2 =
= 4.
Extracting square root,
X =
: ± 2. Result.
Verifying,
9
(±2)2-7 =
: 2 (±2) 2 + 21.
9-4-7 =
= 2.4 + 21.
29 =
= 29.
2. Solve a + x _
a — x
x —2a
x + 2a
Clearing of fractions,
2a a
+ 3ax + x 2 =
-x* + Sax-2a*.
Transposing,
x 2 + x* =
-2a 2 -2a 2 .
Uniting,
2x* =
-4a a .
Dividing by 2,
« 2 =:
X =
-2a*.
Extracting square root,
± V-2a 2 .
Or,
X-
: ± a V-2. Result
268
QUADRATIC EQUATIONS
Solve:
•1. ^ = 49.
2. 3ar» = 75.
3. 5a 2 -80 = 0.
10. 5ar* + 3
11. 2^ + 7
Exercise 101
4. 4a 2 = 9.
5. 6a 2 -18=0.
6. 2a^==5.
2x 2 + 27.
5x> - 20.
7. 3^ + 1 = 0.
8. ±ax 2 -l6a 2 = 0.
9. 7/ = 35a 4 .
12. 3a 2 + 18 - <e (x + 1) + x = 0.
13. aj(3» + 2)-3 = (z + 1) 2 .
14. 4s 2 - 3* + 2 = (2* - 1)(« - 1).
15. (x + l) 3 -{x- l) 3 = 26.
16. (3 + x) (x 2 - 2) = 3 + x (x 2 - 2).
(x - 2) 3 - 6 (aj - 2) 2 = x (x - 4) (a>
a; 2 - 5 1 1
17.
18.
19.
20.
21.
9).
x 2 -\-x — 6 07+3
X + 1 X
0.
a — 3 a + 4
or* + a; — 1 x 2 — x —
x+2 ~x—2
a^ + a ?-2 = a3 2 + a?-l
<c2-a; + 2 tf-x+i
= 0.
AFFECTED QUADRATIC EQUATIONS
305. An affected quadratic equation is an equation con-
taining both the square and the first power of the unknown
quantity, but no higher power.
z 2 +12 a = 18. 3x 2 -2z-7 = 0. ax 2 + bx + c = 0.
An affected quadratic equation is called also a complete
quadratic equation.
AFFECTED QUADRATIC EQUATIONS 269
Completing the Square
306. The solution of an affected quadratic equation is based
upon the formation of a perfect trinomial square. In the per-
fect trinomial square,
x 2 -f- 2 ax + a 2 ,
we note (1) The coefficient of x 2 is unity.
(2) The third term, a 2 , is the square of
one half the coefficient ofx.
That is, a 2 =f~\\
Similarly in x 2 + 12 ax + 36, 36 = f~\\
lnx 2 -8ax + 16a 2 , 16a 2 = f- ^Y.
307. Given, therefore, an expression containing the first
power of x, and the second power of x with the coefficient unity,
we may form a perfect trinomial square by adding to the ex-
pression the square of one half the coefficient ofx. Thus, given,
x 2 + 6 x, x 2 + 6 x + (3) 2 , or x 2 - 6 x + 9, a perfect square.
x 2 - 18 x, x 2 - 18 x+ ( -9) 2 , or x 2 - 18 x + 81, a perfect square.
3 x 2 + 4 x, x 2 + (|) x + (f ) 2 , or x 2 - 1 x + f, a perfect square.
Oral Drill
With each expression in the proper form, give the term
necessary to make a perfect trinomial square.
1. a^ + 10a;. 5. y 2 -S2y. 9. 3^ + 4^.
2. tf + Ux. 6. ar> + 7a;. 10. 5a^-8a;.
3. ^-16x. • 7. tf-Sx. 11. 2a^-3aj.
4. ^ + 24^. 8 . 2/ 2 +ll2/. 12. 6^-17 x.
The principle of completing the square is applied to the solu-
tion of affected quadratic equations in the following :
270 QUADRATIC EQUATIONS
Illustrations :
1. Solve a 2 + 12 a =13.
Adding to both members the square of one half the coefficient of x,
x 2 + 12 x + (6) 2 = 13 + 36
= 49.
Extracting square root, x+ 6 = ± 7.
Whence, x = ± 7 - 6.
That is, x= + 7-6orx=-7-6.
From which, x = 1 or x = — 13. Result.
Both values of x are roots of and satisfy the given equation.
Verification :
Ifx = l: l 2 .+ 12(l) = 13, 1+12 = 13.
Ifx=-13: (-13) 2 + 12(-13)=13, 169-156 = 13.
2. Solve 2z 2 -3a;-20 = 0.
2x 2 -3x-20 = 0.
Transposing, 2 x 2 - 3 x = 20.
Dividing by 2, x 2 — | x = 10.
Completing the square, x 2 — f x + (f ) 2 = 10 + ^
Extracting square root, x — f = ± l ¥ 3 -.
Whence, x = \ 6 -, or - J^.
Or, x = 4, or — f . Result.
Verification :
If X = 4 : 2 (4) 2 - 3 (4) - 20 = 0. 32 - 12 - 20 = 0.
Ifx=-f: 2(-f) 2 -3(-|) -20=0. ^ + Y__ 2 0=0.
« o i x* — x-\-l q? + x — 1 .,
3. Solve J J — - — =1.
x—2 x+2
Clearing of fractions,
(x 2 - x + l)(x + 2) - (x 2 + x - l)(x - 2) = (x + 2)(x - 2).
x 8 + x 2 - x + 2 - x 3 + x 2 + 3 x - 2 = x 2 - 4.
Whence, x 2 + 2 x = - 4.
Completing the square, x 2 + 2 x + 1 = — 4 + 1
= -3.
Extracting square root, x + 1 = ± V— 3.
From which, x = — 1 ± V— 3. Result.
AFFECTED QUADRATIC EQUATIONS 271
Irrational roots result when, after completing the square, the right
member of an equation is irrational.
The original equation will be satisfied by either (— 1 + V— 3) or
(-l-v^3).
From these illustrations we may state the general process for
completing the square in the solution of affected quadratic
equations:
308. Simplify the given equation and reduce to the form
x 2 -j- bx = c.
Add to each member the square of one half the coefficient of x.
Extract the square root of each member, and solve the two
resulting simple equations.
Exercise 102
Solve and verify :
1. ^-4a?==12. 11. 15x* + l±x=-S.
2. a? + 2x=z35. 12. 14 a 2 - 5 a; = 24.
3. ar J -6a? = 27. 13. 20<c 2 + aj = l.
4. a^ + 3a; = 10. 14. 12ar* + 23 o: + 10 = 0.
5. x 2 -\-x = 6Q. 15. a 2 + 4z = 3.
6. 2x 2 — Sx = 2. 16. x 2 -6x=6.
7. 2>x*-lx = Q. 17. x 2 -\-5x-S = 0,
8. 3^ + 17 a;+20 = 0. 18. 2ar s -8a = l.
9. 4:X 2 -5x — 6 = 0. 19. x 2 -4:X=-9.
10. 6x 2 -x = 2. 20. 3a^-a; + l=0.
21. 2x 2 -{-Sx= -10.
22. (3a + l) 2 -(4« + l)(2a;-l) = ll.
23. (x 4 -l)-(x 2 + 2)(x 2 -3)-(x + 5)=0.
24. (2x + 3)(x-2)-(3x-iy = x(x-3)+l.
25. (2 z + 1) (3a -2)-(z + l) (2z -!) = (» + l)(3a?-l).
272 QUADRATIC EQUATIONS
26. x 2 -2(x-l) + x(x-T)-(x-l)(x-2) = 0.
27. x(x 2 -x-3) + (x + 3)(x + ll) = (x — 1) (x 2 + 1).
28 2 't 3 -1. 34. ^±±Z^«^±£±2.
' # + 1 x — 1 ^ + # + 2 a^ + x — 3
89 . «±! + i«4. 35. E±| + l = ^3.
oj + 3 a; 3 x — 3 # + 3
30. i±i + ^L-^l. 36. -A- =6+1-
05 + 1 a? — 4 a? + 4 x' 2 — 16
31 a? + 1 I Ez^ asl 37 - I X == a?2 + 5a; + 8 ,
' x — 1 a + 2 " ' a + 2 a + 5 a 2 + 7 a + 10*
32 g-1 , g-r2^3 a+3 g-2 = 2(s-5)
' x -2 ai-3 4* x 2 a 2 + l x 4 + a^ '
33 ?* + l ,3* + 2 = 39 2^ + 1 3a?-l = 5a?-2
2x-l 'Sx-2 ' ' a? + l x-1 " a+2
3z + 4 2 x 2 + x-l
40.
^ + 2^ + 4 2 — aj a^-8
The Quadratic Formula
309. Every affected quadratic equation may be reduced to
the form
ax 2 + bx + c — 0,
in which, form the coefficients, a, b, and c, represent numbers,
positive or negative, integral or fractional.
310. Solving this affected quadratic equation by completing
the square, we have,
ax 2 + bx = — c.
a a
a \2a] a 4a 2
= 6 2 - 4 ac
4 a 2
AFFECTED QUADRATIC EQUATIONS 273
x +
b
2a
X
Vb 2 - 4 ac
4 a 2
b V6 2 - 4 ac
2a 2a
_ -b±Vb 2 -,4ac
2a
This value of x from the general equation, ax?-{- 6a; + c = 0,
serves as a formula for the solution of affected quadratic equa-
tions. The formula is expressed in terms of the coefficients of
the given general equation, and by substitution of particular
values for a, b, and c from a given equation we obtain the roots
of that equation. The formula is the most practical of the
many methods of solution and it should be memorized.
To obtain the solution of an affected quadratic equation by
means of the general formula :
311. Transpose all terms of the given equation to the left
member, and reduce to the form ax 2 +- bx + c = 0.
In the formula substitute the coefficient of the given x 2 for a,
the coefficient of the given x for b, and the given constant for c.
Simplify the resulting expression.
Illustrations :
1. Solve by the formula, 2 x 2 + 5 x = 12.
Transposing, 2 x 2 + 5 x -
- 12 = 0.
For the formula :
Then in
a = 2,
6 = 5,
c = - 12.
x _ - 6 ± V6 2 - 4 ac
2a
-(5)±V(5) 2 -4(2)(-12)
2(2)
_ - 5 ± V25~+ 96
4
_ -5 ±11
4
= f , or — 4. Result.
SOM. EL. ALG. — 18
274 QUADRATIC EQUATIONS
2. Solve by the formula, x 2 — ±g. x = -§-.
Transposing and clearing of fractions,
3x 2 -10z-8 = 0.
For the formula, a = 3, b = — 10, c = — 8.
Substituting, ^ = -(-10)±V(-10) 2 -4(3)(-8).
6
x ='4, or — f . Result.
Exercise 103
Solve by the formula :
1. x? -11 x + 24 = 0.
2. o 2 -9# = 22.
3. ^ + 9 # + 14 = 0.
4. aj*-lla>=-28.
5. ^ + 9z = 52.
6. a^-a;-72 = 0.
7. o*-21 z = 46.
8. 6o 2 -a = 2.
9. 2ar>+7a; = l5.
10. 8x 2 + 2x = 3.
11. 15^ + 44^ = 20.
The Solution by Factoring
312. Many affected quadratic equations may be solved by
an application of factoring. This method is based upon the
principle that :
313. The product of two or more factors is zero when one of
the factors is equal to zero.
12.
21 a* + 29^-10 = 0.
13.
14^ + 53^ + 14 = 0.
14.
30 a 2 -11 a = 30.
15.
42 <& — m so = - 20.
16.
a; 2 --u- # + 2 = 0.
17.
^ + -V 3 ^ = f
18.
x 2 -- = 2.
6
19.
4 8.
20.
^_*? = o.
30 3
AFFECTED QUADRATIC EQUATIONS 275
To solve an affected quadratic equation by factoring :
314. Reduce the given equation to the general quadratic form,
ax 2 + bx -f- c = 0, and factor the resulting trinomial.
Assume that each factor in turn equals zero, and solve the other
factor for the value of the unknown quantity.
Illustrations :
1. Solve by factoring, x 2 + 6 x = 7.
Transposing, x 2 -f 6 x — 7=0.
Factoring, (x + 7) (x - 1) = 0.
From which, x + 7 = 0, or x — 1 = 0.
Solving, x = - 7 and x = 1. Result.
2. Solve by factoring, 2x 2 + 1 ^x=%.
Simplifying, 10 x 2 + 11 x - 6 = 0.
Factoring, (2 re + 3) (5 x - 2) = 0.
Whence, 2 as + 3 = and 5 x - 2 = 0.
And x = -f, or x = % . Result.
Exercise 104
Solve by factoring:
1. a^-5a; = 14. 10. 8a*-38a; = -35.
2. a?-8x + 15 = 0. 11# 15 x 2 - 77 a + 10 = 0.
3. « 2 -3a; = 4.
4. z 2 - 13 a + 12 = 0.
5. a^-12a;-13 = 0.
6. ar J -19<c = 42.
12. 12a? 2 -23a; + 10 = 0.
13. x*-lf-x=*.
14. ^ + ^+1=0.
7. 2z 2 -9z + 4 = 0. 1S iB 2_» == I.
8. 4^-5^=6. 4 2
9. 6^-17a; = -12. 16. f a> = 3»-ff.
276 QUADRATIC EQUATIONS
LITERAL AFFECTED QUADRATIC EQUATIONS
315. Any one of the three methods given applies readily to
literal affected quadratic equations. .
Illustrations :
1. Solvec?(x i -l)=x(x + 2c).
Clearing of parentheses, c 2 x 2 — c 2 = x 2 + 2 ex.
Transposing, c 2 x 2 — x' 1 — 2cx = c 2 .
Uniting coefficients of x 2 , (c 2 — 1) x 2 — 2 ex = c 2 .
Dividing by (c 2 - 1) , x 2 - (j^) * = j£j ■
c 2
Completing the square, x 2 - -^- + ( —^—r) = -^—- + —
c 2 — 1 \c 2 — 1/ c 2 — 1 (c
2 - l) 2
= c 2 (c 2 -l) + c 2
(C 2 - l) 2
(c 2 -!) 2 '
Extracting square root, x -—r— - = ±
c 2 — 1
± *
C 2 _ 1 - C 2 _ X
From which, x = — ^— orx= — • Result.
c- 1 c + 1
2. Solve *+»•-*£•
Transposing and clearing of fractions,
n 2 x 2 + mnx - 2 m 2 = 0.
For the formula : a = n 2 , b = ran, c = — 2 m 2 .
x _ -mn± V (m») 2 - 4 (n 2 ) (-2 m 2 )
2(w 2 )
_ — mn ± V9 m 2 n 2
In 2
_ — mn ± 3 mn
2n 2
= », or- 2 -^. Result.
DISCUSSION OF AFFECTED QUADRATIC EQUATION 277
Exercise 105
Solve:
1. x 2 -2ax = 3a\ 8. a?-2ax + a* = l.
2. x 2 = 5 ax — 6 a 2 . 9. cV + c(m — ri)x = mn.
3. x 2 + ax = 2a 2 . 10. mx 2 — (m + l)x + l = 0.
4. 3 aV + acx — 2 c 2 . 11. acx 2 + anx = cmx + mn.
5. IOcAb 2 - 21 c«a> + 9 = 0. 12. cV-c(a + l)« + a = ().
6. 35c 6 x 2 — c 3 x = 6. 13. (2*+a)(*— c)=(aj— a)(»+c).
7. cc 2 + 4a» = 4a + l. 14. (as + a -f- mf = (a — m) 2 .
15. « 2 -2aa;-26x = c 2 -a 2 -6(2a + 6).
1iS a;-i-a . a 5 -„ 1 1 _ 1 1
lb. 1 ; — — • 17. — •
a x + a 2 m — x m c—x c
ma x-2a . x + 2a 2x 2 + 3a 2
lo. 1 -— " • •
cc + 3 a x — 3 a 0^ + 9 a 2
DISCUSSION OF THE AFFECTED QUADRATIC EQUATION
Character of the Roots
316. If the two roots of the affected quadratic equation,
ax 2 + bx -+- c = 0, be denoted by t-j and r 2 , we have (Art. 310),
— b + V& 2 — 4 ac — 6 — V & 2 — 4 etc
2 a 2a
The character of the result in each root depends directly
upon the value of the radical expression V& 2 — 4ac, for the
radicand, b 2 — 4 ac, may be positive, zero, or negative.
(I) When b 2 — kac is positive.
(a) The roots are real, for the square root of a positive quan-
tity may be obtained exactly or approximately.
(6) The roots are unequal, for Vb 2 — 4 ac is -f in r x and —
in r 2 .
(c) The roots are rational if 6 2 — 4ac is a perfect square,
irrational if not. ■
278 QUADRATIC EQUATIONS
Illustrations :
(1) In2sc 2 + lla; + 12=0, a=2, 6 = 11, c=12. 6 2 -4ac=121-96=25.
Hence, the roots of 2<c 2 +ll x + 12=0 are real, unequal, and rational.
(2) In5z 2 + lla + 3 = 0, a = 5, 6 = 11, c = 3. 6 2 -4ac=121-60=61.
Hence, the roots of 5 x 2 + 11 x + 3 = are real, unequal, and irrational.
(II) When b' 2 — 4:ac equals 0.
(a) The roots are real, and ] -™ -. -, b
\( _. \ For each root reduces to .
(6) The roots are equal. J 2 a
Illustration :
In 4s 2 -20x+25=0, a=4, 6 = -20, c=25. 6 2 -4ac=400-400=0.
Hence, the roots of 4 x' 2 — 20 x + 25 = are real and equal.
(III) When b 2 — 4 ac is negative.
(a) The roots are imaginary, for the square root of a nega-
tive number is impossible.
Illustration :
In 3s 2 + 2s + 2=0, a = 3, 6=2, c = 2. 6 2 -4ac = 4 -24 = -20.
Hence, the roots of 3 x 2 + 2 x + 2 = are imaginary.
317. We may, therefore, by inspection of the discriminant,
b 2 — 4 ac, summarize the foregoing conclusions as follows :
I. If b 2 — 4 ac > 0, the roots are real and unequal.
II. If b 2 — 4 ac = 0, the roots are real and equal.
III. If b 2 — 4 ac < 0, the roots are imaginary.
Illustrations :
1. Determine the character of the roots of x* — 5 x = 6.
a a - 5z-6 = 0. a = l, 6 = -5, c = -6.
62 _ 4 ac = [(-5)2- 4 (1)(- 6)] = (25 + 24) = 49.
Therefore, the roots are real, unequal, and rational. (316, 1.)
2. Show that the roots ofar 2 — 4sc+5=0 are imaginary.
6 2 -4ac=[(-4) 2 -4(l)(5)] = (16-20) = -4.
Hence, the roots are imaginary.
DISCUSSION OF AFFECTED QUADRATIC EQUATION 279
3. Determine the value of m for which the roots of
4or J + 10a; + m=0 are equal.
By Art. 316, II, the discriminant must equal 0.
Hence, (10) 2 - 4 (4) (m) = 0, 100 - 16 m = 0, m = ^.
That is, the roots of 4 z 2 + 10 * + ^ = are equal.
4. For what value of m will the roots of
(m + l)a; 2 + (6m + 2)a; + 7m + 4 = 0be equal ?
The discriminant must equal 0.
a = (m + 1), 6 = (6m + 2),c = (7m + 4).
Then (6 m + 2) 2 - 4 (m + 1) (7 m + 4) = 8 m 2 - 20 m - 12.
Solving, 8 m 2 - 20 w - 12 = 0, m = 3 and - J. Result.
By substituting these values, 3 and — §, in the given equation, two dif-
ferent equations result, both of which have equal roots.
Exercise 106
By inspection of the discriminant, determine the character
of the roots of :
1. a? + 7x=&. 7. 2 x* + l a + 3 = 0.
2. a 2 -6a; = 40. 8. 3 ar 2 - 5 x= - 12.
3. tf 2 + 5a-84 = 0. 9. a? + 10« + l = 0.
4 . ^-3^ + 54 = 0. 10. 3a^-12a;=-17.
5. x* + 12x = -36. 11. 5ar> + 40a; = l.
6. 3a 2 + 8a; + 4 = 0. 12. 7 ar> + 11 a? + 12 = 0.
Determine the values of m for which the two roots of each
of the following equations are equal :
13. 4x 2 + 20a + ra = 0. 16. a 2 -}- (m -f 5) a; + 5m+l f=0.
14. 9x* + mx + 25 = 0. 17. (m-f-l)a5 2 -(m-2)a;-hl = 0.
15. 2m^- 30 # — 15 = 0. 18. 2ma^+7ma!=^+5a;-5m.
19. (m + l)a; 2 + ma;=-9(a; + l).
20. m — 7 + mo 2 — m# = — 2 a; — x 2 .
280 QUADRATIC EQUATIONS
Relation of the Roots and Coefficients
318. If the roots of
ax 2 + bx + c = Oorx 2 + -x+- =
a a
are obtained by Art. 310, and are denoted by r x and r 2 respec-
tively, we have
2a 2a
1*^ «aam^ „ _l v ~ b + V6 2 - 4 ac - b - V& 2 - 4 ac
By addition, n + r 2 = !
2a
Or, ri + r2= _&.
a
By multiplication, nn = (- » + V5' - 4«c) (- 5 - Vt° -4 ac)
4 a 2
_ ft 2 - ft 2 + 4 qc
4 a 2
Or, nr 2 =--
a
Hence, we may state :
319. In any affected quadratic equation of the form
a a
(a) The sum of the roots equals the coefficient of x with its sign
changed.
(b) The product of the roots is equal to the constant term.
Illustrations :
1. Find by inspection the sum and the product of the roots
of6x 2 + 5x = 6.
Changing to the required form, x 2 + f x — 1 = 0.
From Art. 319 :
Sum of the roots = — f .
Product of the roots = — 1. Result.
2. One root of 2 x 2 + 5 x = 12 is — -§. Find the other root.
Transposing and dividing, x 2 + f x — 6 =0.
Dividing the product of the roots, - 6, by the known root, - f, we
have, (-6)-(-f)=+(6x|)=4, Result.
DISCUSSION OF AFFECTED QUADRATIC EQUATION 281
Exercise 107
Find by inspection the sum and the product of the roots of :
1. x> + 9x + U = 0. 5. 3ar>-10a; + 3 = 0.
2. x 2 -21x = ±6. 6. 2ar> + 3a = 2.
3. 2 a^+7 »«.!«. 7. 7^ + 9^-10 = 0.
4. 6x 2 -x = 12. 8. 15^ + 14^ + 3 = 0.
9: One root of 5 x 2 — 26 x + 5 = is 5. Find the other root.
10. One root of 8 x 2 = 15 x + 2 is — -|-. Find the other root.
11. With the values of r, and r 2 from the equation
ax 2 + foe + c = 0, find the value of
rs—r
1 . 1
n + r 2
Also the value of — +
Formation of an Affected Quadratic Equation with
Given Roots
b c
Consider again the form x 2 + -x + - = 0. (1)
a a
If, as before, i\ and r 2 denote the roots of this equation, we
have (Art. 319) :
7*1 + 7*2=
a
Whence, - = - r s - r 2 (2). Also (319), -= r x r 2 (3)
Ct CL
Substituting in (1) the values found in (2) and (3),
x 2 + ( — r x — r 2 )x + r x r 2 = 0.
x 2 — r x x — r 2 a* + r^ = 0.
(x 2 — r&) — (i\x — rfo) = 0.
x(x — r 2 ) — r x (x — r 2 ) = 0.
(»-r 1 )(a;-r 2 )=0.
282 QUADRATIC EQUATIONS
Therefore, to form a quadratic equation that shall have any
given roots :
320. Subtract each root from x, and equate the product of the
resulting expressions to 0.
Illustrations :
1. Form the equation whose roots shall be 3 and 7.
By Art. 320, (* - 3) (x - 7) = 0.
Or, x 2 - 10 x + 21 = 0. Result.
2. Form the equation whose roots shall be f and — -|.
By Art. 320, (z_f)( x _(_|)) = o ;
(x-f)(x + §) = 0;
l5x 2 + x-6 = Q. Result.
Exercise 108
Form the equations whose roots shall be :
1.
2,5.
5. -2*
8.
a — 1, a -f- 1.
! 3
9.
a — 1, 2 a.
2.
3, -8.
6 5 7
6 * 2' ~8
10.
2±V=3.
3.
-4,-7.
11.
V2, -1.
4.
3,-L
7. -?,-3.
12.
Va + 1 Vtt— 1
' 2
3' 4
2 ' 2
GRAPH OF A QUADRATIC EQUATION IN ONE VARIABLE
321. The graph of a quadratic equation is obtained by
application of the principles governing the graphs of linear
equations. The given equation is written in the typical form,
ax 2 + bx + c — O, and the left member is equated to y. By as-
suming successive values for sc, the corresponding values of y
are obtained as before.
GRAPHS OF QUADRATIC EQUATIONS
GRAPHS OF QUADRATIC EQUATIONS HAVING UNEQUAL ROOTS
283
322. Plot the graph of
a? + 2x = 8,
Assume y = x 2 + 2 x — 8.
In the figure,
If x = 3, y = 7. P.
X= 2,y = 0. Pi.
x = 1, y = - 5. P 2 .
X = 0, y = - 8. P 8 .
x = - 1, y = - 9. P 4 .
x= -2,y= -8. P 6 .
x = - 3, y = - 5. P 6 .
x = - 4, y = 0. P 7 .
x = - 5, y = 7. P 8 .
etc.
Bsag
i
The curve representing the equation, ^ + 2^ — 8 = 0, might
be indefinitely extended by choosing further values of x. In-
termediate points on the curve may be obtained by assuming
fractional values of x and obtaining corresponding values of y.
323. The lowest point of the graph of a quadratic equation
in one variable may, in general, be obtained by completing the
square.
Thus, x 2 + 2x + 1=8+ 1; (x + l) 2 = 9; (x+l) 2 -9 = 0.
Now (x + l) 2 — 9 has its greatest negative value when x = — 1.
Hence, the coordinates of the lowest point of the curve are (— 1, — 9).
Plot the graph of 2 x 2 + 7 x - 4 = 0.
Let y = 2 x 2 + 7 x - 4. Then in the first figure on p. 284 :
(The student will note that, in the figure, the scale of the graph is so
chosen that one unit of division on the axis corresponds to two units from
the solutions.)
284
QUADRATIC EQUATIONS
If x = 2, y = 18.
x — 1,2/ = 5.
x = 0, y = - 4.
s= — 1, 2/ =-9.
P.
Pi.
P 2
P s
2
[T
P
f i
i .
1 .
-1
-it
f
X' (
)t x
in:
X
t-
T^
t.
ll|
fL
J
I
B
~Y"
a; =-2, y =-10.
x=-S,y=-7.
x=-4,y = 0.
x=-b, y = 11.
P 4 .
Ps-
P 6 .
P 7 .
On completing the square the lowest
point in the curve is (— |, - - 8 ^).
Exercise 109
Plot the graphs of:
1. x>-§x + 5z=0.
2. x 2 + 6x + 8 = 0.
3. a^ + a; — 12 = 0.
4. 4 x 2 — 5 x = 0.
5. 2^-5x = -3.
6. 8a 2 + 2z-3 = 0.
GRAPHS OF QUADRATIC EQUATIONS HAVING EQUAL ROOTS
324. Plot the graph of
x>-
-6x
+ 9
= 0.
Let y :
= x*-
-6s
+ 9.
the figure
Ifx =
0,2/
= 9.
P.
« =
1*9
= 4.
Pi.
x =
2,2/
= 1.
P 2 .
X =
3,2/
= 0.
P 3 .
X =
4, y
= 1.
P 4 .
X =
5, y
= 4.
Ps.
X =
6,2/
= 9.
P 6 .
In
«
PV
45
Now, ce 2 — 6se + 9 = 0, may-
be written (x — 3) 2 = ; and
x in this equation can never be 0. Hence, the curve cannot cut the XX 1
axis, but is tangent to it at (3, 0), the lowest point of the graph.
GRAPHS OF QUADRATIC EQUATIONS
285
GRAPHS OF QUADRATIC EQUATIONS HAVING IMAGINARY ROOTS
325. Plot the graph of x 2 - x + 2 = 0.
Let y = x' 2 — x + 2. In the figure :
x = 3, y = $.
x = —l, y = 4.
x = - 2, y = 8.
lix =
0, y = 2.
P.
x -
1, y = 3.
Pi.
X -
2, y = 4.
P 2 .
x =
in
Completing the square
x 2 -x + 2 = 0:
X 2
- x+2
= (z 2 -
= (x-
-i) 2 + |.
i + 2
p 3 .
p 4 .
p 5 .
- 3, y = 14. P 6 .
Pif 1
EEEEEEEEK2EEEEEEE
FTP
And this expression not
being or negative for
any value of x, it follows
that y cannot be or nega-
tive ; that is, the graph
cannot touch the XX' axis.
Important Conclusion.
The graph of any quadratic
equation in one variable
and in the form ax 2 + bx + c = 0, intersects the XX' axis at
two points whose abcissas are roots of the equation, provided
that the given equation has real and unequal roots ; has one
point in the XX' axis if the roots are equal ; and does not
cut the XX' axis if the roots are imaginary.
Exercise 110
Plot the graphs of :
1. a 2 -4# + 4 = 0.
2. 4ar>-|-4:r + l=0.
4. x 2 + 2x + S=0.
5. x? + x+6 = 0.
6. 2x 2 -6x + ll = 0.
CHAPTER XXII
THE QUADRATIC FORM. HIGHER EQUATIONS.
IRRATIONAL EQUATIONS
EQUATIONS IN THE QUADRATIC FORM
326. An equation in the quadratic form is an equation
having three terms, two of which contain the unknown num-
ber ; the exponent of the unknown number in one term being
twice the exponent of the unknown number in the other term.
Thus:
HIGHER EQUATIONS SOLVED BY QUADRATIC METHODS
327. It will be seen at once that many equations in the
quadratic form must be of a higher degree than the second.
The method of factoring permits the solution of many such
equations, and is generally employed in elementary algebra.
328. If quadratic factors result from the application of
factoring to higher forms of equations, they may ordinarily
be solved by completing the square or by the quadratic
formula. Such factors most frequently occur in connection
with binomial equations.
Illustrations :
1. Solve x 4 —
13 x 2 + 36 = 0.
Factoring,
And, + 2) (i
"Whence,
(z 2
c - 2) (x
- 4) («2 _ 9) =
+ 3) (x - 3) =
x =
286
= 0.
= 0.
-2,2,
3, and 3. Result.
EQUATIONS IN THE QUADRATIC FORM 287
2. Solve a? + 7a?-8 = 0.
Factoring, (x 3 + 8) (x 3 - 1) = 0.
And, (as + 2) (x 2 - 2x + 4) (x - 1) (x 2 + x + 1) = 0.
By inspection, x = — 2 and 1.
Solving, x 2 -2x + 4=0.
x = 1 ± V^3.
Solving, x 2 + x + 1 = 0.
Therefore, x = - 2, 1 ± V^3, 1, and ~ + ^ ~ ■ Result.
3. Solve ^? - 4a/^ = 12.
Expressed with fractional exponents and in the transposed form, we
have,
a;f _4»i_ 12 = 0.
Factoring, (x* - 6) (x^ + 2) = 0.
Whence, x^ = 6, or x* = - 2.
"From which, x = 216, or — 8. Result.
When tested both roots prove to be solutions.
4. Solve x* - 12 x~l = - 1.
Jk 12 _ i
With positive exponents, x 7 — *
x*
Clearing of fractions, x^ — 12 = — x*»
Transposing, x^ + x* — 12 = 0.
Factoring, (x^ + 4) (x* - 3) = 0.
Whence, x* = — 4, or x* = 3.
And, x s 256, or 81.
The equation is satisfied by the negative value of v / 256 and by the
positive value of VH.
329. For convenience of solution a single letter may be
substituted for a compound expression in equations in the
quadratic form. Care should be taken that no solution is lost
in the final substitutions.
288 THE QUADRATIC FORM. HIGHER EQUATIONS
Illustrations :
1. Solve (a 2 - 4) 2 - (a 2 - 4) = 20.
Transposing, (x 2 - 4) 2 - (x 2 -4) - 20 = 0.
Let (x 2 - 4) = y.
Then,
y 2 -y-
20 = 0.
Factoring,
(y - 5) (y +
4) =0-
Whence,
y = 5, or — 4.
If y = 5,
If*/ = -4,
x 2 - 4 = 5,
X 2 - 4 = - 4,
x 2 = 9,
x 2 = 0,
x = ± 3.
x = 0.
Therefore, x = ± 3 and
Result.
2. Solve
Let
X + IV , » + 1
X
x +
\)
+
= y-
x-l
Then, y 2 + y - 6 = 0.
Factoring, (j/ + 3) (y - 2) = 0-
And, y = — 3 and 2.
If 2/ = - 3,
x+1
x-l
Hy = 2,
x + l .
x-l"
= -3, x + l = -3(x-l), 4x = 2,
r Result.
2, x + 1 = 2(x - 1), - x = - 3, x = 3.
3. Solve 4a^ + 36 a;- 2 = 25.
Transposing, 4 x 2 + 36 x~ 2 - 25 = 0.
Then,
4z 2 +!^_25
x 2
Whence, 4 x 4 + 36 - 25 x 2 s= 0.
Or, 4x 4 -25x 2 + 36 =0.
Factoring, (x 2 - 4) (4 x 2 - 9) = 0.
Or, (x + 2) (x - 2) (2 x + 3) (2x - 3) = 0.
And, x = - 2, 2, - } , f. Result.
EQUATIONS IN THE QUADRATIC FORM
289
Exercise 111
Solve :
1. a; 4 -5ar> + 4 = 0.
2. z 4 -8ar 9 = 9.
3. a;- 2 - 3 a;- 1 = 10.
4. x 9 + 9a? + $ = 0.
5. a 6 -7ar* = 8.
6. ar* + 7a;- 2 = 144.
7. »- 1 + ^ = 20.
8. 2ar* + 5ar* + 2 = 0.
9. 3 a + \/3 = 2.
10. or 1 — 3ar* = 4.
11. 6aBr* + 13«r* + 6=aO.
12. 8ar°-3ar 2 = -10.
13. 2^ + ^ = 7.
14. </?+ 3^-10=0.
15. Va?-5^ = 24.
16. a* = l.
17. (3x + l)(x 2 + x-2) = 0.
18. 2(«» - 9) = 3(» - 3).
19. ^-8 + 3»(a;-2) = 0.
20. 3ar J + 15a; = a<3ar J + 17a;4-10).
21. 2(ar } -8)+7ar J -17a; + 6 = 0.
22. (a? 4- 3) 2 - 5(x + 3) = 14.
23. (ar 2 + 2) 2 - 6(a^ + 2) = 55.
24. (or* + a;) 2 - 8(a? + x) + 12 = 0.
25. (a + 5) -(aj + 5)^ =6.
26. (x 2 + 3a; + 6) - 2(ar> + 3 a; + 6)* - 8 = 0.
27. a^ + a;-f-2 = 7vV + a; + 2-10.
28. (x + — Y_3(W— ^ + 2 = 0.
ar> + 2
29.
a^-2
a^ + 2
4-8-6^
30.
SOM. EL. ALG
?-2
aT3T + °(a?T2j =6
19
290 THE QUADRATIC FORM. HIGHER EQUATIONS
EQUIVALENT EQUATIONS AND THE REJECTION OF ROOTS
330. If, in two equations involving the same unknown
number, the solutions of each include all the solutions of the
other, the equations are equivalent.
Thus, 3 x X 2 = x + 6. (1) Whence 2 x = 4. (2)
Equations (1) and (2) are equivalent, for x — 2 is a solution j
of each.
331. In the solution of fractional equations, and in the
solution of irrational equations also, it must be remembered
that the processes of simplification may introduce roots that |
will not satisfy the given equation. That is, equivalent
equations do not always follow in the successive steps of a process.
(a) Processes that do not change the Roots op an Equation I
6
Given the equation x + 3
x-2
Clearing, (x + 3) (x - 2) = 6.
From which, x 2 + x - 12 = 0.
And, x = 3, or — 4.
By trial both 3 and — 4 are found to be roots of the given equation.
Hence, multiplying by x — 2, the lowest common denominator, intro-
duced no new root.
The only root that could be introduced by the multiplier, x — 2, would
be the root of the equation, x - 2 = 0, or x = 2. And this root has not
been introduced.
In general :
332. The roots of a fractional equation are unchanged if:
(1) the given fractions having a common denominator are com-
bined, and
(2) the equation is multiplied by the loivest common multiple of
their denominators.
EQUIVALENT EQUATIONS 291
(b) Processes that change the Roots of an Equation
Given the equation x + 5 = 0.
Multiplying by as - 1, (se - 1) (x + 5) = 0.
From which, x = — 5, or 1.
The solution x = 1 fails on trial to satisfy the given equation.
In general :
333. A root is introduced if both members of an integral
equation are multiplied by an expression involving the unknown
number in the equation.
Given the equation x — 2 = 0.
Transposing, x = 2.
Squaring, x 2 = 4.
Whence, a 2 - 4 = 0.
Factoring, (x + 2) (x - 2) = 0.
Whence, x = - 2, or 2.
Of these two solutions only the solution, « = 2, satisfies the given
equation, the solution, x = — 2 being introduced by the process of squar-
ing.
In general :
334. A root is introduced if both members of an equation are
raised to the same power.
The Rejection of Roots.
Since it is clear that certain processes may affect the solu-
tions of involved types of equations, we make the following
conclusion :
335. Before accepting all the solutions of a given equation as
roots of that equation it is necessary that each solution be tested;
and all solutions not satisfying the given equation must be rejected.
292 THE QUADRATIC FORM. HIGHER EQUATIONS
IRRATIONAL EQUATIONS INVOLVING QUADRATIC FORMS
336. Illustration:
Solve V2 x + 7 — V#— 5 = VS.
Squaring, 2 x + 7 - 2 V(2x + 7)(x - 5) + x - 5 = x.
Transposing, — 2 V2 x 2 — 3 a; — 35 = — 2 a; — 2.
Dividing by - 2, V2 x 2 - 3 x - 35 = x + 1.
Squaring, 2 x 2 - 3 x - 35 = x 2 + 2 35 + 1.
From which, x 2 - 5 x - 36 = 0.
And, x = 9, or — 4. Result
The solution, x = 9, satisfies the given equation, but the solution,
x = — 4, does not satisfy, and is rejected.
Note. If the solution, x = — 4, is tested and, in extracting the square
root of the right member the negative value of the root is taken, we have : i
V2(-4)+ 7-V(-4)-6= V-4.
t/Zi - V^9 = */-!.
V^T-8 V^T = -2
But this process is at variance with the accepted condition that positive
square roots are to be consistently taken through the practice of ele-
mentary algebra.
Exercise 112
Solve and test the solutions of :
1. aj — l = V* + l. 4. x + V2 = Va + 2 x V2 + 4.
2. Vz 2 + « + l = V3a + 4. 5. V2(aj»-4) = (aj-2).
3. x + Va + l = 19. 6. ar 2 + a-3 = Va 4 -r-2ar J +l0.
7. Vx-f-2- Va-3 = V
.*•
8. (a: + 3)(a>-l) = Va>(a 8 -7) + 10 + 2a.
Vg + l--v^-l =!i 3 a . 4 . 2t
Va + 1 + Vx — 1
FORMULAS INVOLVING QUADRATIC EQUATIONS 293
PHYSICAL FORMULAS INVOLVING QUADRATIC EQUATIONS
Exercise 113
1. From the formula, E = -—, obtain an expression for v.
2. From the formula, H= .24<7V£, obtain an expression for C.
3. Find expressions for I and g when t = it -i/- .
4. Find an expression for t from the formula S = \gt 2 .
Wv 2
5. Given, J57 = -— — ; find an expression for v.
6. Given the formulas, V=gt, and S=^gt 2 . Find an expres-
sion for the value of t in terms of V and S.
7. Given the formula, F = — - — . Find a formula for t in
t •
terms of F, m, n, and r.
8. Given the formula, t = v\j- . Find the value of I when
Li. v »
9. From the formula, S = V t + igt 2 , obtain the value of t
in terms of V , g, and S.
10. Given the formulas, E=Fs, F=ma, and v= V2as;
obtain a formula for i? in terms of m and v.
v 2
11. If a = — , and t*=2 wr, find a formula for a in terms of
r
7r, r, and £.
12. E represents the energy of a moving body, m its mass,
7YIV
and v its velocity, in the formula E= -— • What is the rela-
z
tion of the energies of two bodies if one has twice the mass but
only two thirds the velocity of the other?
CHAPTER XXIII
SIMULTANEOUS QUADRATIC EQUATIONS. PROBLEMS
i
337. In the solution of simultaneous quadratic equations we
have particular methods for dealing with three common types ;
but no general method for all possible cases can be given.
SOLUTION BY SUBSTITUTION
1. Solve
338. When one equation is of the first degree and the other q
the second degree.
'2x*-3xy-y* = l, (1)
_5x+y=3. (2)
From (2), # y = S-6x. (3)
Substituting in (1), 2 x 2 - 3 x (3 - 5 x) - (3 - 5 a) 2 = 1.
2a; 2 - 9x + 15 x 2 - 9 + 30x - 25x 2 = 1.
8x 2 -21x + 10 = 0. (4)
Factoring, (8 x - 5) (x - 2) = 0.
From which, x = f , or 2.
Substituting in (3),
Ifx = f, y = 3-5(|) = -f
Ifx = 2, */ = 3 -5 (2) = -7.
Hence, the corresponding values of x and y are
,
**» • 1 Result.
Corresponding values must be clearly understood as to mean-
ing. From (4) in the above solution two values of x result.
Each value of x is substituted in (3), and two values for y result.
Therefore, we associate a value of x with that value of y result-
ing from its use in substitution.
294
SOLUTION BY COMPARISON AND FACTORING 295
Exercise 114
Solve :
1. x + 2y = 5, 6. « 2 + an/ + 2/ 2 = 19,
x* + 3xy = 7. x + y = 5.
2. x -3y=2, 7. x*-3xy -y 2 + 3 = 0,
xy + y 2 = ti. 3x + 2y — 5 = 0.
3. 2x + 3y = 12, 8. 3x + 2y = 5,
xy-6 = 0. 2-xy-y 2 = 0.
4. 3x + 2y = 9, 9. x? + y 2 + x +y = 6,
xy — x — 2. x — 4 y = 6.
5. a; + 22/ + l = 0, 10. 2 z + 3 # + 7 = 0,
3ay-2/ 2 = -4, 3aj*+2y*-a> + 4y=:12.
SOLUTION BY COMPARISON AND FACTORING
339. When both equations are of the second degree and both
are homogeneous.
Solve (**-*-* (*)
13^-^ = 11. (2)
A factorable expression in x and y results if the constant terms are
eliminated from (1) and (2) by comparison.
Multiplying (1) by 11,
22 x 2 - 11 xy = 110.
(3)
Multiplying (2) by 10,
30 x 2 - 10 y 2 = 110.
(4)
Equating left members,
30 x 2 -10y 2 = 22 x* -
-11
xy.
From which,
8x 2 + llxy - 10y 2 = 0.
(5)
Factoring (5),
(x + 2y)(8x-5y) =0.
Therefore,
,-f-
(6)
And,
Sx
V ~ 6 *
•
(7)
296
6 SIMULTANEOUS QUADRATIC EQUATIONS
By substitution in (1):
Again in (1) :
From (6) : If y = - 1,
From (7): tfys**,
5
»*-*(-§) = 10.
2z 2 -x(^Wo.
2x 2 + — = 10.
2i
2x2-^=10.
5
5a; 2 = 20.
2 x 2 = 50.
x 2 = 4.
x 2 = 25.
x =±2.
Hence, in (6),
x—± 5.
Hence, in (7),
"=-ih-2 1(±2)=T1 -
* = if=!<:k6)=±&
o
Therefore, x = ± 2,
y==Fl,
x
2/
= ±5;1
f Result.
= ±8.J
Note. The sign T is the result of a subtraction of a quantity having
the sign ±. Thus, — (± a) = — (+ a) or — (- a) = — a or + a, = T a.
Solve :
Exercise 115
1. x 2 -\-xy=6,
xy—y 2 = l.
6. tf-xy + y 2 = %,
x* + xy + y 2 = %.
2. x 2 — xy — 4,
7. x 2 — 3xy + y 2 = — 5,
2ic 2 +a*/ -2?/ 2 = -4.
3. x 2 -2xy = -8,
y 2 -3xy = -9.
8. 0^ + ^ — 4^ = 18,
a? 2 — a??/ — 3 y 2 = — 9.
4. x 2 + 2«2/ = 9,
3 xy — i/ 2 = — 4.
9. 10« 2 + 7«2/-2/ 2 = -ll,
12x 2 -^-9xy-y 2 = -15.
5. ar 2 + 5 xy — — 6,
ar> + 2/ 2 = 5.
10. 2x 2 + 4 : xy + 7y 2 -15 = 0,
5^ + 3x2/-32/ 2 -15 = 0.
SOLUTION OF SYMMETRICAL TYPES
340. When the given equations are symmetrical with respect to
x and y ; that is, when x and y may be interchanged without
changing the equations.
SOLUTION OF SYxMMETRICAL TYPES 297
1. Solve
(x + y = 7,
[xy = 10.
(1)
xy = 10. (2)
Squaring (1), x 2 + 2 xy + y 2 = 49.
Multiplying (2) by 4, 4 xy = 40.
By subtraction, x 2 — 2 xy + y 2 = 9.
Extracting square root, x — y — ± 3. (3)
Adding (1), x + y = 7.
2x=+3+7, or-3 + 7.
2 x = 10, or 4.
x = 5, or2;l Regult
Subtracting (1) from (3), y = 2, or 5. J
2. Solve (^= 12 > ft
Multiply (1) by 2, 2 set/ = 24. (3)
Add (2) and (3), x 2 + 2 xy 4 y 2 = 49. (4)
Subtract (3) from (2), x 2 - 2 xy + y 2 = 1 . (5)
Extract square root of (4), x 4 y = ± 7. (6)
Extract square root of (5), x — y = ± 1. (7)
Four pairs of equations result in (6) and (7), viz. :
x + y=7 x + y= 7 x + y = -l x + y = -7
x — y = 1 x - y =— 1 x- y = 1 x — y = — 1
x=4 x=3 x=-3 x = — 4
y =3 y= 4 y =-4 y =-3
Hence, the corresponding values for x and y are
x = ±4, y=±3,i
x = ±3; y=±4.j
Result.
Solve :
Exercise 116
1. a + y = 6, 3. ^ + 2/ 2 = 5,
ojy = 6. as + y = 3.
2. x + y = 5, 4. a#— 3=0,
zy-4 = 0. a^4-2/ 2 = 10.
298 SIMULTANEOUS QUADRATIC EQUATIONS
5. x 2 + xy + y 2 = 7, 7. x 2 + y 2 =5,
x + y = 3. i (x + y) 2 = 9.
6. x + y-5 = 0, 8. o 2 + 3a?/ + 2/ 2 = 59,
x 2 -xy + y 2 = 7. x 2 -f o?y + y 2 = 39.
SOLUTIONS OP MISCELLANEOUS TYPES
341. Systems of simultaneous quadratic equations not con-
forming to the three types already considered are readily
recognized, and the student will gradually gain the experience
necessary to properly solve such systems. No general method
for these types can be given.
It is frequently possible to obtain solutions of those sys-
tems in which a given equation is of a degree higher than
the second, derived equations of the second degree resulting
from divisions and multiplications.
. Illustrations :
l. Solve j^ + 3* + 3, = 28, (1)
lajy-6 = 0. (2)
Transposing and multiplying (2) by 2, 2 xy = 12. (3)
Adding (1) and (3), x 2 + 2 xy + y 2 + 3 x + 3 y = 40.
Whence, (x + y)* + 3 (x + y) - 40 = 0.
Factoring, (x + y + 8) (x + y — 5) = 0.
Hence, x + y =— 8, or x + y = 5.
Combining each of these results with (2), we have two systems :
(a) x + y = - 8, (6) x + y = 5,
xy - 6 = 0. xy - 6 = 0.
The two systems (a) and (&) may be readily solved by the principle
of Art. 340.
*. Solve mf**-* ■ I
[x — y = l. (*)
From (1), xiy* + 5 xy - 14 = 0,
(xy + 7 j (xy - 2) = 0.
SOLUTIONS OF MISCELLANEOUS TYPES
299
Whence, xy = 2, or xy = — 7.
Combining each of these derived equations with (2), we have,
(a) xy = 2, (6) xy=-l,
x — y = \. x — y = l.
Solving these two systems, we find
from (a), x = 2, or — 1,
y = 1, or - 2 ;
from (&), _ lj-3 V33 ^
*- 2
_l±3v3
y 2
3. Solve J
fl + l = ?
1 1 = 3
a y 2
Squaring (2),
Dividing (1) by (2),
Subtracting (4) from (3),
Whence,
x 2 xy y 2 4
xy y*
_3 = 6
xy 4'
J L = 1
xy 2
Subtracting (5) from (4) , — - — + - o = 1 .
x 2 scy y 2 4
x y 2
Extracting square root of (6) ,
x
Combining (2) and (7), we obtain, x = *' or J ; 1
y = 2, or 1. J
Result.
4. Solve
*« + y* = 82,
se+y=4
Let x = w + v,
and y = u — v.
Then in (1), (u + t>) 4 + (« - O 4 = 82.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(1)
(2)
800 SIMULTANEOUS QUADRATIC EQUATIONS
Or, m 4 + 4m 3 h 6 u 2 v 2 + 4 uv s + v 4 '
+ u l — 4 w 3 t? + 6 rt'V 2 — 4 2fy 3 + w 4
2 w 4 +12 ztV 2 + 2 1? 4 = 82
Whence, w 4 + 6 w 2 v 2 + v 4 = 41.
Substituting in (2), w + v + w ~ v = 4.
w = 2.
Substituting (5) in (4), (2) 4 + 6 (2)V + v i = 41 .
Or, tf + 24 1>2 _ 25 = 0.
Whence, v = ± 1, or ±5 V- 1.
Therefore, if u = 2 and t> ==± J, And, if u = 2 and v = ±5 y/~— 1,
x = w-fv = 2±l=3orl, x=u+v=2±5 V^T,
y — u — v = 2 + l=lor3. 2/ = w-<y = 2=F5 V— 1.
Hence, in brief form, x — 3, 1, or 2 ± 5 V— 1;1 R esu it
y = 1, 3, or 2 T5 V-1.1
r(a-j0 + (*-y)* = 12, (1)
5. Solve .
(z + 2/)-(z + 2,)* = 2. (2)
Considering (a; + yy and (x — y) * as unknown quantities and factoring :
From (1), (x - y) + (x - */)* - 12 = 0.
(x - yy = 3, and (x - */)* = - 4. (3)
From (2), (x + y) - (x - y)^ - 2 = 0.
(x + y)^ = 2, and (x + yy = -l. (4)
The negative roots, being extraneous, are rejected. Then,
From (3),
(x - y y = 3.
From (4),
(x + y)$ = 2.
Whence,
x-y = 9.
(5)
And,
x + y = 4.
(6)
From (5)
and
(6),
2 x - 13, and 2 y =
-5.
Therefore,
x = ^ and y =
5
1'
Result.
SOLUTIONS OF MISCELLANEOUS TYPES 301
Exercise 117
Solve :
1. a? + y 2 = 13, 16. ar 5 - ^ - 98 = 0,
2x = 3y. 2-x + y = 0.
2. x + y = 4, 17. x 2 -\-xy-i-y 2 = 91,
x 2 — 2xy + 3y — x = 3. x + Vxy + y = 13.
3. x + 2/ + l=%y, 18. x 2 + y 2 -\-x + y = l±,
5x-y-xy = l. x 2_ y 2 + x _ y=z xid.
4. ^ + 2/ 2 -f 0:2/ = 39, 19 X 2 + Xz + Z 2 = 30y
x-y = 3. x*-xz+* = l&.
5. x + y = 2, 2Q tT 5 + 2/5 = 2 44,
12-x*y* = ±xy. x + y = ±.
6. PV-» + ^ 21 . ,-,-12 = 0,
7. a? + y = 3a>y-l, ■
^ + / = 8^ 2 -3^-3. 22 « ^-^2/ + 2/ 2 = 3,
8. xy + x-y = b,
fcy(a - 2/) = 6 -
9. ^ 2 + 2/ 2 4-^ + 2/ = 2,
a*/-2 = 0.
10. IB 2 — 42/ 2 = a; + 2y,
x + 4 ?/ = 7.
11. 3*4-^=^+10,
x—y=xy+2.
12. x 4 +ary + 2/ 4 = 19,
ar 2 — a*/ 4- 2/ 2 = 7 -
13. x 4 + 2/ 4 -17=0,
a;4-y--3 = 0.
14. (z-?/) 2 -a; 2 ?/ 2 -5 = 0,
15 3a; 2 +2«?/-2?/ 2 =14(a;-2/),
2a? + xy-3y 2 = 7(x-y).
x — xy+y = l.
23.
it 12 6 y
.-i-l-o.
o,
24.
x 2 y 2 16
25.
1 1_7
x* f $'
1_1_1
x y 2'
1 1 1
+ ^ = o,
26.
x + 2/ » — y
2 3_12
x y xy'
+ 5 '
302
SIMULTANEOUS QUADRATIC EQUATIONS
GRAPHS OF QUADRATIC EQUATIONS IN TWO VARIABLES
TYPE FORMS OF EQUATIONS AND CORRESPONDING GRAPHS
(I) Type Form
7
3 J£Be
^•jT ^
-£ tn
2 -\
ZZZZt J
)r„± .x
-±- — 5
r 1
t V ^E
S^ -.^5
_5ta;~*2B
jegff
K
jr 2 +/ = c.
Illustration :
Plot the graph of
x 2 + y 2 = 25.
From the equation we have
If
y = ± V25 - P-
jc=0, y= ±' V25, or 5. P.
«s=X, y=±V24, or ±4.89+. Pi.
jc=2, y=± V2l", or ±4.06+. P 2 .
a; =3, y= ± Vie, or ±4. P 3 .
»=4, y=±V9, or ±3. etc.
Plotting these points, we ob*
tain a circle as the graph of the
equation, x 2 + y 2 = 25.
It will be seen that the
coordinates of any point (x, y)
are legs of a right triangle
whose hypotenuse is the dis-
tance from the origin to the
point (x, y). That is, for any
point on the curve we have
(see figure),
^ + 2/ 2 =5 2 ,
or, x 2 + y 2 = 25.
In general, therefore:
342. Tfie graph of any equation in two variables in the form
x 2 + y 2 = c is a circle.
GRAPHS OF QUADRATIC EQUATIONS
303
(II) Type Form ••• f = ax + c.
Illustration :
Plot the graph of
If
x=0, y = ± V8, or ±2.8+. P 2 .
x= l,y = ±V&, or ±3.4+. P 3 .
x=-l, y = ±V4, or ±2. Pl
35= -2, y=0; etc. P.
(Note the enlarged scale.)
Plotting these points, we ob-
tain a parabola as the graph
of the equation, y 2 = 4 x + 8.
¥
^'
*'%
IS «3
X P
6^"
2L
y
£
x ir x
PI o
V
\
V
S^vp
1 ^s
s 5
^N
V
It will be seen that if x is less than — 2, y is imaginary ;
hence, no point in the curve lies to the left of P.
From this type form and graph we have, in general :
343. The graph of any equation in two variables in the form
y 2 = ax + c is a parabola.
(Ill) Type Form •••
ax 2 + by 2 = c.
Illustration :
Plot the graph of
±x 2 + 9y 2 = 36.
If y = 0, x 2 = 9, x = ±3. A.
x = 0, y 2 = 4, y = ±2. B.
Hence, x= + 3 and — 3, the
points where the graph cuts
XX ; and,
y = + 2 and — 2, the points
where the graph cuts YY'.
9 zn
g B P
^ j^
-7 V
+- t \
xi : o i *
A^ JA
\ Z
-S« ^
^._ -^
B B p
Y^
304
SIMULTANEOUS QUADRATIC EQUATIONS
For any other points let x = ± 1.
Then, 9 y 2 = 32, y = ± f i/% ttf f =s± 1 .9+.
Hence, for P, P if P 2 , and P 8 , we have (1, 1.9), (1, - 1.9), (- 1, 1.9),
(— 1, — 1.9) respectively.
Plotting these points, we obtain an ellipse as the graph of the equation
4x 3 + 9y 2 = 36.
By assuming values sufficiently greater or less than those by
which the points above are obtained, it can be shown that no
points in the graph can lie to the right or teft, above or below,
the intersections with the axes of reference.
From this type form and graph we have, in general :
344. The graph of any equation in two variables in the form
ax* -\-by 2 = c is an ellipse.
(IV) Type Forms • • • ax 2 — by 2 = c and xy = c.
Illustrations :
1. Plot the graph of
x 2 -4y 2 = l.
From the given equation
X
s. ^^**
^£ P + '
r -S ta -«£_
Hi 5I--A2 x
/- Q *s
>' s ^
+' £ _ U s s
*~
Y*~
Iix = ±l,y 2 = 0,y=0.
x = ±2,y*=%,y=±iy/3,
or ±.86+.
For the values of x we plot
A, (1, 0), and 5, (-1, 0).
AlsoP, (2, .86); P u (-2, .86);
P 2 ,(-2,-.86);P„(2,-.86).
With these points we obtain
an hyperbola as the graph of
the equation, x 2 — 4 y 2 = 1.
It will be found by trial that any value of x between -f- 1 and
— 1 gives an imaginary value for y, hence no part of the curve
can lie between the points A and B.
GRAPHS OF QUADRATIC EQUATIONS
305
2. Plot the graph of
If x = -4, ••• ••• + 4, etc.,
y = _l, ••• ±qo ». + l, etc.
Plotting the points, (—4, — 1),
etc., we obtain an hyperbola
whose branches lie in the angles
FOX and VOX*.
From the two cases :
345. TJie graph of any
equation in two variables in
the form ax* — by 2 — c, or in the form xy = c, is an hyperbola.
Exercise 118
= || | ||| *i
till
lllilll
HH-Iy- 1 1 1 1 1 1 -
Plot the graph of :
1. x 2 + y 2 = 16.
2. s 2 + y 2 = 49.
3. y 2 = 4a?4-4.
5. <c 2 -3 2 / 2 = 2.
6. xy — 10.
SOLUTION OF SIMULTANEOUS QUADRATIC EQUATIONS BY GRAPHS
m
/
, B
= 5 3E
\
Illustrations :
1. Given# 2 + 2/ 2 = 25,
Sx-2y = 6.
From the intersections of
the graphs of the given
equations obtain the roots,
and check the results by
solving the equations.
In the figure the graph of
x 2 -f- y 2 = 25 is a circle, and
the graph of 3x — 2y = 6
is a straight line.
SOM. EL. ALG. — 20
306
SIMULTANEOUS QUADRATIC EQUATIONS
Y ptm
S*
^ d>^
ii-zfiiiifc:::::::
i£^l&^
J!
-*>— ^PF-s-
E^S
o,
By measurement of the graphs we find the coordinates of
P % and P 2 to be (4, 3) and (— -J-|, — -f-f ). Solving the equations,
we obtain x — 4, y = 3 ; or x = — -}-|, y = — ^|. The accuracy
of this and of subsequent cases may be increased by plotting
on a larger scale.
2. Given
x 2 + y 2 + 9x + U
y 2 =4:X + 16.
From the intersections
of the graphs of the
given equations obtain
the roots, and check
the results by solving the
equations.
In the figure the graph
of x 2 + y 2 + 9 x + 14 =
is a circle and the graph of
2/ 2 = 4# + 16isa parabola.
By measurement of the
graphs we find the coordinates of P x and P 2 to be (— 3, 2) and
(—3,-2) respectively, and these coordinates correspond to
the real roots obtained from the solution. For the solution of
the system gives x = — 3, y = ±2, or x = — 10, y = ± V— 24.
We cannot find a point corresponding to the imaginary root,
and we make the important conclusion that :
346. Since there can be no point having one or both coordinates
imaginary, the graphs of two equations can have no intersections
corresponding to imaginary roots.
3. Given x 2 + 4 y 2 = 17,
xy = 2.
From the intersections of the graphs of the given equations
obtain the roots, and check the results by solving the equations.
GRAPHS OF QUADRATIC EQUATIONS
307
Iii the figure the graph of a? 2 + 4 ?/ 2 = 17 is an ellipse, and
the graph of xy = 2 is an
hyperbola.
By measurement of the
graphs we find the co-
ordinates of P lt P 2 , P 3 ,
and P 4 tobe'(4, £), (1,2),
(-4, -h), (-1, -2)
respectively. The solu-
tion of the equations gives
x= ±4, y= ±i, or x= ±1,
|= ±2.
It will be noted that this
last case is the first in
which both equations are
homogeneous and of the second degree, and that the graphs
serve to emphasize once more the importance that attaches to
the association of corresponding values of x and y in the
solution of a system of
simultaneous quadratic
equations.
4. Given
2a,- 2 + 5y 2 =125,
5 x + 2 y = 10.
From the intersections
of the graphs of the given
equations obtain the roots,
and check the results by
solving the equations.
In the figure the graph
of 2^+22/2 = 125 is
an ellipse, and the graph of 5 x + 2 y = 10 is a straight
line.
308
SIMULTANEOUS QUADRATIC EQUATIONS
Y
"^w
■\
^< "" m "™ ^ < P
z ~ qL
2 §\ T
^ J-^Sr
x: o s x
^
I 4 s
V -,£
^ ^
v -*». — tf^
SI
By measurement of the graphs we find the coordinates of
Pj and 1\ to be (0, 5) and (3.7+, 4.4-) respectively. The
solution of the equations
gives x = 0, y = 5, or
3 = 3.75+, ?/ =4.38+.
5. Given,
a; 2 + f = 100,
3a; + 4?/ = 50.
From the intersections
of the graphs of the given
equations obtain the roots,
and check the results by
solving the equations.
In the figure the graph
of x 2 + y 2 = 100 is a circle,
and the graph of 3 x + 4 y
= 50 is a straight line.
Solving the system, we find that the equation resulting from
substitution gives equal roots for y. Hence, for x we find but
one value, the solution of the system being x = 6, y = 8. It
will be seen from the ' graph that the circle and straight line
have one point only in common ; that is, the line is tangent to
the circle. The coordinates of the point of tangency are found
to be (6, 8) or P. In general :
347. If, in the solution of a system of equations, a derived
equation has equal roots, the graphs of the equations are tangent
to eadi other.
It will be found to assist the student in the exercise follow-
ing if the type forms of equations and the corresponding
graphs are remembered.
1 . ax + by — c, The Straight Line.
2. jr 2 +/ = c, The Circle.
3. / 2 = ax 4- c, The Parabola.
GRAPHS OF QUADRATIC EQUATIONS 309
4. ax 2 + by 2 = c, The Ellipse.
5. ax 2 -b/=c,ov\ Th e Hyperbola.
6. xy — c, J
Exercise 119
Plot the graphs of the following, and check the roots deter-
mined by a solution of each system :
1. x 2 +y 2 = 25, 5. x 2 + y 2 + x + y = 18,
x — y=l. xy + x + y = 11.
2. ^+2/2=74, 6. r = 4a + 8,
xy = 35.
« 4- y — 6 = 0.
3.
a; — y = 6.
7. 9 x 2 - 16y 2 = 144,
^4-^ = 36.
4.
a? + y = 12,
a*/ = 32.
8. 0^+4^ = 4,
x 2 + tf '■ = 17.
9.
a 2
a 2
4-2/ 2 =
-xy
= 32,
+ 2/ 2
= 28.
10. Solve the system, x + y = and a^/ = 4, and determine
if the solution, alone will show that the graphs intersect, or
do not intersect.
11. How do the graphs of xy = 4 and x l — y 2 — 16 differ in
their positions relative to the axes of reference ?
12. Can you describe the position of the graph of the equa-
tion, x 2 + y 2 — 2 x = 0, without plotting it ?
13. How do the graphs of the equations, x 2 -f 2y 2 = 32 and
4 # 2 + y 2 = 16, differ in their relative positions ?
14. Plot the graphs of x 2 — xy 4- y> = 28 and x — y = 0, and
show that their intersections check the roots found by solution.
15. In how many points may the graphs of x 2 + y 2 = 25 and
a,- 2 4- 2 y 2 = 64 intersect? Prove your answer by plotting.
810 SIMULTANEOUS QUADRATIC EQUATIONS
16. Show the positions of the graphs of x 2 + y 2 . = 100 and
x 2 + 4?/ 2 = 100, and determine the number of real roots.
17. Show by solution that y 2 — 4 x = 12 and x 2 + y 2 -f 3 x =
can have but one point in common, and prove your answer by
a graph of the system.
PROBLEMS PRODUCING QUADRATIC EQUATIONS
348. In the solution of a problem from which a quadratic
equation or equations result we retain only the solution that
satisfies the given conditions. As a rule negative results will
not ordinarily satisfy the conditions even if they satisfy the
equations.
Illustrations :
1. If 6 times the number of laborers in a field are increased
by the square of the number at work, there will be in all 55
men. How many laborers are there in the field ?
Let x t= the number of laborers in the field.
From the given conditions,
6cc + « 2 = 55.
From which, x 2 + 6 x — 55 = 0.
Solving, x as 5, or x = — 11.
Clearly the positive result only is retained. Hence, 5 laborers. Result.
2. A company of boys bought a boat, agreeing to pay for it
the sum of $ 60. Three of the boys failed to pay as agreed,
so each of the others was compelled to pay $1 more than he
had promised. How many boys actually paid for the boat?
Let x = the number of boys actually paying for the boat.
x + 3 = the number of boys first agreeing to share its cost.
— = the number of dollars paid by each boy.
= the number of dollars each had expected to pay.
x + S
PROBLEMS PRODUCING QUADRATIC EQUATIONS 311
Then, ■ S>— S»-«i
x x + 3
From which, x 2 + 3 x - 180 = 0,
and x = 12, or - 15.
That is, 12 boys actually shared in the cost of the boat.
Many of the problems in the following exercise must be
stated by the use of two unknown quantities, and simulta-
neous quadratic equations will result. But as far as pos-
sible the student should attempt to state most of the earlier
examples by means of one unknown number only.
Exercise 120
1. Find those two consecutive odd numbers the sum of
whose squares is 290.
2. Find a number that, added to 7 times its reciprocal,
equals 8.
3. Two factors of 48 are such that one exceeds the other
by 2. What are the factors ?
4. The sum of two numbers is 10, and the sum of their
squares is 52. Find the numbers.
5. Find two numbers such that their sum and the difference
of their squares are each 13.
6. Find the two factors of 600 whose sum is 49.
7. Find two numbers whose sum is 11, and whose product
is 17 less than 15 times their difference.
8. The sum of the cubes of two numbers is 126, and the
sum of the two equals 6. Find the numbers.
9. Find two numbers the sum of whose squares is 130 and
whose product is 63.
312 SIMULTANEOUS QUADRATIC EQUATIONS
10. How many yards of picture molding will be required
for a room whose ceiling area is 1200 square feet, the diagonal
of the ceiling being 50 feet ?
11. One of two numbers exceeds 30 by as much as the other
number is less than 30, and the product of the numbers is 875.
Ifind them.
12. The sum of the squares of two numbers equals 13 times
the smaller number, and the sum of the numbers is 10. What
are the numbers ?
13. If twice the product of the ages of two children is added
to the sum of their ages, the result is 13 years. One child is
3 years older than the other. Find the age of each.
14. The diagonal of a rectangle is 100 feet, and the longer
side is 80 feet. Find the area of the rectangle.
15. Find three consecutive numbers such that the sum of
their squares shall be 194.
16. The simple interest on $600 for a certain number of
years and at a certain rate is $ 120. If the time were two
years shorter and the rate 2 % more, the interest would be
$108. Find the time and the rate of interest.
17. The sum of the squares of two numbers is 2 a 2 + 2, and
the sum of the numbers is 2 a. What are the numbers ?
18. The combined capacity of two cubical tanks is 637 cubic
feet, and an edge of the one added to an edge of the other
equals 13 feet. Find the length of a diagonal on any one face
of each cube.
19. The product of two numbers is 15 greater than 5 times
the larger number, and is 6 less than 16 times the smaller
number. Find the numbers.
20. If a number of two digits is multiplied by the tens'
digit, the product is 96; and if the number is multiplied by
the units' digit, the product is 64. Find the number.
PROBLEMS PRODUCING QUADRATIC EQUATIONS 313
21. Divide 15 into two parts such that their product shall
equal 10 times their difference.
22. The difference of two numbers is 1, and the sum of the
numbers plus the product is 19. Find the numbers.
23. If the sum of two numbers is multiplied by the less,
the product is 5 ; and if the difference of the numbers is multi-
plied by the greater, the product is 12. Find the numbers.
24. A garden 40 feet long and 28 feet wide has around it a
path of uniform width. If the area of the path is 960 square
feet, what is its width ?
25. A dealer would have received $ 2 more for each sheep
in a drove if he had sold 6 less for $ 240. How many were
there in the drove, and at what price was each sold ?
26. In a number of two digits the units' digit is 3 times
the tens' digit, and if the number is multiplied by the sum of
the digits, the product is '208. Find the number.
27. A bicyclist starts on a 12-mile trip, intending to arrive at
a certain time. After going 3 miles he is delayed 15 minutes
and he finds he must travel the remainder of the journey at a
rate 3 miles an hour faster in order to arrive at his destination
on time. Find his original rate of speed.
28. The sum of the squares of the two digits of a number
is 13, and if the square of the units' digit is subtracted from
the square of the tens' digit and the remainder is divided
by the sum of the digits, the quotient is 1. Find the number.
29. The difference between the numerator and the denomina-
tor of a certain improper fraction is 2, and if both terms of
the fraction are increased by 3, the value of the fraction will
be decreased by -£$. Find the fraction.
30. From the formula, t = vxj- 9 find the length of a pen-
es'
dulum that vibrates once a second at a point where g = 32.16
feet.
314 SIMULTANEOUS QUADRATIC EQUATIONS
31. A body falls through a space of 3216 feet at a point where
g equals' 32.16 feet. From the formula, S = \ gt 2 , determine
the number of seconds required for the fall.
32. If an automobile traveled 3 miles an hour faster, it would
require 2 hours less time in which to cover a distance of 120
miles. What is the present rate of the automobile in miles per
hour ?
33. A certain floor having an area of 50 square feet can be
covered with 360 rectangular tiles of a certain size ; but if the
masons use a tile 1 inch longer and 1 inch wider, the floor can
be covered with 240 tiles. Find the sizes of the different tiles.
34. One leg of a right triangle exceeds the other leg by 2
feet, and the length of the hypotenuse is 10 feet. Find the
length of the legs of the triangle.
35. 168 feet of fence inclose a rectangular plot of land, and
the area inclosed is 1440 square feet. Find the dimensions of
the field.
36. The sum of the squares of two numbers is increased by
the sum of the numbers, and the result is 18. The difference of
the squares of the numbers is increased by the difference of the
numbers, and the result is 6. Find the numbers.
37. If the difference of the squares of two numbers is divided
by the smaller number, the remainder is 4 and the quotient 4. If
the difference of the squares of the numbers is divided by the
greater number, the remainder is 3 and the quotient 3. What
are the numbers ?
38. If the length of a certain rectangle is increased by 2
feet, and the width is decreased by 1 foot, the area of the rect-
angle will be unchanged. The area of the rectangle is the
same as the area of a square whose side is 3 feet greater than
one side of the rectangle. What are the dimensions of the
rectangle ?
CHAPTER XXIV
RATIO. PROPORTION. VARIATION.
RATIO
349. If a and b are the measures of two magnitudes of the
same kind, then the quotient of a divided by b is the ratio of
a to b.
Ratios are expressed in the fractional form, -, or with the
b
colon, a : b. Each form is read " a is to b."
350. In the ratio, a : b, the first term, a, is the antecedent,
and the second term, b, is the consequent.
THE PROPERTIES OF RATIOS
351. The properties of ratios are the properties of fractions;
for the ratios, -, m:n, (#,+ y):(x — y), etc., are fractions.
b
(a) The Multiplication and the Division of the
Terms of Ratios
352. The value of a ratio is unchanged if both its terms are
multiplied or divided by the same number.
353. A ratio is multiplied if its antecedent is multiplied, or if
Us consequent is divided, by a given number.
354. A ratio is divided if its antecedent is divided, or if its
consequent is multiplied, by a given number.
315
316 RATIO. PROPORTION. VARIATION
(6) Increasing or Decreasing the Terms of a Ratio
355. If a, b, and x are positive, and a is less than b, the ratio
a : b is increased when x is added to both a and b.
For a + x a = x Q*- a )
' b + x b b(b + x)
And, since a < b, the resulting fraction is positive and the
given ratio, - , is increased accordingly.
356. If a, b, and x are positive, and a is greater than b, the
ratio a:b is decreased when x is added to both a and b.
For the resulting fraction in Art. 355 is negative when a > b.
357. An inverse ratio is a ratio obtained by interchanging
the antecedent and the consequent.
Thus, the inverse ratio of m : n is the ratio n ; m.
358. A compound ratio is a ratio obtained by taking the
product of the corresponding terms of two or more ratios.
Thus, mx : ny is a ratio compounded from the ratios, m : n and x : y.
359. A duplicate ratio is a ratio formed by compounding a
given ratio with itself.
Thus : a 2 : 6 2 is the duplicate ratio of a i b.
In like manner, a z : 6 3 is the triplicate ratio of a : b.
Exercise 121
1. Write the inverse ratio of a : x ; of m : n ; of 7 : 12 ; of
3x:5x; of (2a + l) : (2a- 1); of (a? + xy + y 2 ) : (x 2 -xy + y 2 ),
2. Arrange in order of magnitude the ratios 2:5, 3:7,
4 : 9, 5 : 8, 10 : 17, 12 : 19, 21 : 27, 32 : 39, and 40 : 51.
3. Compound the ratios 3 : 7 and 10 : 17.
4. Find the ratio compounded of 3 : 8, 4 : 9, and 6 : 11.
PROPORTION 317
5. Compound the ratios (x 2 — 9) : (ic 3 + 8) and (x+2) : (x—3).
6. What is the ratio compounded from the duplicate of
2 : 3 and the triplicate of 3 : 2 ?
7. Find the value of the ratio (x + 6) : (x 2 + 7 x + 6).
8. Two numbers are in the ratio of 4:7, but if 3 is added
to each number, the sums will be in the ratio of 5 : 8. Find
the numbers.
9. The ratio of a father's age to his son's is 16 : 3, and the
father is 39 years older than the son. Find the age of each.
10. In a certain factory 5 men and 4 boys receive the same
amount for a day's work as would be paid if 3 men and 12
boys were engaged for the same time. What is the ratio of
the wages paid the men and the boys individually?
PROPORTION
360. A proportion is an equation whose members are equal
ratios.
Thus, the four numbers, a, 6, c, and d, are in proportion if - = -.
b d
361. A proportion may be written in three ways :
(1) 2-£. (2) a:b = c:d. (3) a:b::C:d.
b d
Each form is read " a is to b as c is to d." We understand the mean-
ing of a proportion to be that the quotient of a ■*■ b is the same in value
as the quotient of c -=- d.
362. The extremes of a proportion are the first and fourth
terms.
363. The means of a proportion are the second and third
terms.
318 RATIO. PROPORTION. VARIATION
364. The antecedents of a proportion are the first and third
terms, and the consequents the second and fourth terms.
In the proportion a:b = c:d
a and d are the extremes, a and c are the antecedents,
b and c are the means ; b and d are the consequents.
365. If the means of a proportion are equal, either mean is
a mean proportional between the first and fourth terms.
Thus : in a : b = b : c, b is a mean proportional between a and c.
366. The last term of a proportion whose second and third
terms are equal is a third proportional to the other two terms.
Thus, in a:b = b :c, cisa third proportional to a and b.
367. A fourth proportional to three numbers is the fourth
term of a proportion whose first three terms are the three given
numbers taken in order.
Thus, in a : b = c : d, diss, fourth proportional to a, 6, and c.
368. If, in a series of equal ratios, each consequent is the
same as the next antecedent, the ratios are said to be in con-
tinued proportion.
Thus : a :b = b:c = c:d=d:e = etc.
369. In the treatment of proportions certain relations are
conveniently discussed if we recall that - = -=?*. Whence
J b d
a = br, and c = dr. Substitutions of these values for a and c
will be of frequent service in practice.
370. In order that four quantities, a, b, c, and d, may be
in proportion, a and b must be of the same kind, and c and d
of the same kind. However, c and d need not be of the same
kind as a and b.
PROPERTIES OF PROPORTION 319
PROPERTIES OF PROPORTION
371 . Given a:b = c:d. Tlien ad = be.
Proof :
b~ d
Multiplying by bd, ad = be.
That is:
In any proportion, the product of the means equals the product
of the extremes.
372. Givena:b = c:d. Then a = — , b = — , etc.
d c
From the
equation ad =
be
we obtain by division
„ be
d
_bc
— »
a
o = — , c =
c
ad
b
That is :
Either extreme of a proportion equals the product of the means
divided by the other extreme; and either mean of a proportion
equals the product of the extremes divided by the other mean.
373. Given a:b = b:c. Then b = -Vac.
Proof :
a : b = b : c.
By Art. 371,
b 2 = ac.
Extracting square root,
b = y/ac.
That is:
The mean proportional between two numbers is equal to the
square root of their product.
374. Given ad = be. Then a : b = c : d.
Proof : ad = bc.
Dividing by bd, 9l = 9:.
b d
Or, a:b = c:d.
320 RATIO. PROPORTION. VARIATION
That is :
If the product of two numbers is equal to the product of two
other numbers, one pair may be made the extremes, and the other
pair the means, of a proportion.
In like manner we may obtain a proportion, a:c=b:d, etc.
375. Given
a
6 =
: c : d. Then b:a = d:c.
Proof :
a _c
b~ d
Then,
b d
Whence,
b_d
a c
Or,
b:a = d:c. .
That is :
If four numbers are in proportion, they are in proportion by
inversion.
376. Given a:b = c:d. Then a:c = b:d.
Proof: °l = c -.
b d
Multiplying by 5, «& = &£.
c be cd
Whence, ^ = -.
c d
Or, a : c = b : d.
That is :
If four 7iumbers are in proportion, they are in proportion by
alternation.
In applying alternation all four quantities considered must
be like in kind.
377. Given a:b = c:d. Then a+b:b = c +d:d.
Proof: 2«'5.
b d
PROPERTIES OF PROPORTION 321
Adding 1 to both members, - + 1 = - + 1.
b d
Whence,
a + V _c + d
b d
Or,
a + b :b = c + d:d.
That is :
If four numbers are in proportion; they are in proportion by
composition.
378. Given a:b = c:d. Then a — b:b = c — d:d.
Proof :
a _ c
b~ d
Subtracting 1,
|- 1 = ^-1.
b d
Whence,
a — b c — d
b d
Or,
a — b.b = c — d:d.
That is :
If four numbers are in proportion, they are in proportion by
division.
379. Given a:b = c:d. Then a-\-b:a—b = c + d:c — d.
Proof: a±b^c±d, (1)
b d
\„a a — b c — d /ON
And, — = — • (2)
Dividing (1) by (2), £±| = c - ± ^,-
a — b c — a
Or, a + b :a — b = c + d:c — d.
That is:
If four numbers are in proportion, they are in proportion by
composition and division.
80M. EL. ALG. 21
322 RATIO. PROPORTION. VARIATION
380. Given a:b = c:d. Then a n :b n = c n : d n .
Proof: Sfil
6 d
Raising both members to the nth power,
9- — SH.
b n ~ d n '
Or, a n :b n = c n : d n .
That is :
Like powers of the terms of a proportion are in proportion.
381. Given a:b = c:d = e:f= •••. T7iew (a + c -+- e + •••) :
(b + d + f+.>.)=a:b.
Proof: Since ® = -? = * = ...,
b d f '
then, - = r, - = r, - = r, etc.
6 d /
Whence, a = &r, c = dr, e =^fr, etc.
Adding, a -f c + e + ••• = br + dr +fr + •••.
Whence, a + c + e + ••• = (& + d + / + •••)»*.
And, q + c + e +- = r=g.
Or, ( + c+e + -) :(& + <* +/+•••) = a:&.
That is :
Jn a series of equal ratios, the sum of the antecedents is to the
sum of the consequents as any antecedent is to its consequent.
382. Given a
:& =
= b
c.
Then a :c—a 2 : b 2 .
Proof : Since
a b
6"c'
it follows that,
« v b _a a
b c~b b
Whence,
a_a 2
c & 2 '
Or,
a : c = a 2 : b 2 .
APPLICATIONS OF THE PROPERTIES OF PROPORTION 323
That is :
If three numbers are in continued proportion, the first is to the
third as the square of the first is to the square of the second.
383. Given a:b = b:c = c:d. Then a:d = a 8 :b s .
Proof : Since
it follows that, « x ^x- = ?x?x a
b~
_b_
c
c
= ~d'
c d
= <**
b
a
b
d~
~b*'
a
d =
= a*
b*
b c d b b b
Whence,
Or,
That is :
If four numbers are in continued proportion, the first is to the
fourth as the cube of the first is to the cube of the second.
384. Given a:b = c: d, e:f=g:h, k:l = m:n. Then aek :
bfl = cgm : dhn.
Proof: Since §«.£, 5 = 2, *«»«
b d f h I n
Multiplying, J7i=T-
bfl dhn
Or, aek : bfl = cgm : dhn.
That is :
The products of the corresponding terms of two or more pro-
portions are in proportion.
APPLICATIONS OF THE PROPERTIES OP PROPORTION
385. Of the important properties of proportion in common
use in the solution of problems involving certain relations, we
may note briefly the following :
(1) Since, in any proportion, the product of the means equals the
product of the extremes (Art. 371), we may readily find any one term
of a proportion when three terms are known.
324 RATIO. PROPORTION. VARIATION
(2) Since — = - = r, we have a — br, and c= dr ; and the substitution
b d
of these values for a and c will be of frequent service in reductions.
(3) Composition and division (Art. 379) are of frequent use in mini-
mizing the work necessary for the solution of certain types of equations.
(4) An assumed identity involving any four numbers, a, 6, c, and d,
is shown to be true if, by transformations, we obtain a proportion,
a:b = c:d.
Illustrations :
1. Find the ratio of x to y when 3a? + 2y = - .
In the form of a proportion we have
Zx + 2y.4:X-3y = 5:6.
By Art. 371, 6 (3* 4 2 y) = 5 (4 x - 3 y) .
Whence, 18x4l2*/ = 20x- 15 y.
2x = 27 y.
Therefore (Art. 374), x : y = 27 : 2. Result.
2. If a: 6 = c: d, show that a + 3c: 6 + 3<2 = 2a+c : 26+riL
Since a : & = c : d, we nave - = - — r. Whence, a = br, and c = dr.
6 d
Reducing each ratio separately,
a 4 3 c _ (6r) + 3 (dr) = r (b 4 3 <T) =
643d 6 + 3d ' 6 + 3d5
2g + c = 2(&r)+(dr) = r(264d) : , r
2&4d 2&4d ' 26+d
a 4 3c 2a+c
Therefore,
3. Solve the equation
b+Sd 2&4d
' 2iC — 1 X + 4:
x 2 + 2x — l x 2 4- a 4- 4'
By composition and division (Art. 379),
(2 x - 1) + (x 2 4 2 s - 1 ) ■ . . (x 4 4) + (x 2 4 x + 4)
(2x-l)-(x 2 4 2x-l) (x + 4)-(x' 2 + x+4)'
Simplifying, x 2 + 4x - 2 = x 2 4 2x48^
— x- — X 2
Whence, x 2 44x-2 = x 2 4 2x4 8.
2 a; = 10.
x = 5. Result.
APPLICATIONS OF THE PROPERTIES OF PROPORTION 325
4. Find two numbers whose sum is to their product as 3 is
to 20; and whose sum increased by 1 is to their difference
increased by 1 as 7 is to 1.
Let x and y = the two required numbers.
Then, x + y : xy = 3 : 20. (By the first condition.) (1)
And, x + y + l-.x — y + l=7:l. (By the second condition.) (2)
From (1), 20 x + 20 y = 3xy. (3)
From (2), 3z-4y = -3. (4)
x = *JL=l.
3
From (4),
Hence, in (3), 2o(±l^) + 20y = 3y (±£p^.
Simplifying, 12 y* - 149 y + 60 = 0. (6)
Solving (5), j y = 12, or T 5 2 .
x = 15, or f.
It will be found that the integral values only satisfy the given condi-
tions, and the fractional values are, consequently, rejected.
Therefore, 12 and 15 are the required numbers.
Exercise 122
Show that the following are true proportions :
1. 4:8 = 5:10. 4. 3:9.125 = 2:6^.
2. 2:7 = 2.5:8.75. 5 ' **: 3y = 8a>y : 6^.
3. 3:5* = 6:10x. 6 ' *>£-*^J?
Determine whether the following are true proportions :
7. 34:53 = 19:39. 9. 12.1 : 4.4 = 2.2 : .8.
8. 4£:4 = 8f:8J. 10. ar>-l : x + 1 = x-l : x.
Find the value of the unknown in each of the following :
11. 8:10 = 12:». 13. 32:12 = x:6.
12. 4: z = 16: 15. 14. y: 8 = 12: 4.
18. 4c:a = 3sc:6a.
19. 2 am : 3 = mx : 15.
1
20.
a-1
326 RATIO. PROPORTION. VARIATION
15. 5: z = 10.5:. 3.
13. 2a:5 = 6:3.
17. 3<c:16 = .9:.2.
Find the ratio of x to y in :
21. 5x = 7y.
22. 3# + 2?/ = 52/-3a;.
23. 2 ic + y : 3 x — y = f.
Find the value of x and ?/ in :
27. a- l:y + 1 = 2:3,
: a + l = x : a 2 — 1
24.
2x + ?/ m
3»— y ft
25. » 2 -9?/ 2 = 8iC2/.
26. x — y:3y = y — x:x +
30. a + 1 : y + 1 = 3 : 4,
a : 2 = y : 3.
w x + y+1 ^2 x+5 J
32.
a; 4- V = 5.
28. x + 2:y + 2 = 2:3,
3a-2/ + l = 0.
29. a; + 2/ : « — 2/ = 3 : 2,
a; + l:2/-l = 3:2.
Find a mean proportional between :
33. 20 and 5. 34. 3 and 27. 35. 4 a?x and 16 aV.
36. ^ + 3, + 2 ^ , 2 -,-2
a + 2/-l 3' 2/+2
a; + 2 y _ 5 a; + 1 _ 44
a;-22/~6' y + l~2l'
x 2 -3x + 2
37. ^ ~ 5 + * and
ar*+2a;-3
ar 2
a;-l
38.
2V2
and
a; 2 -l
5+4V2 4+3V2
Find a third proportional to :
39. 12 and 16. 42. (c + ar) 2 and c 2 - ar*.
43. a 8 - 1 and a? + a; + 1.
40. 3 a 2 and 2 a 3 .
44. ( m + ") 2 and( m2 - n2 ->.
41. 2:8 and 3:5. n 4 n 2
APPLICATIONS OF THE PROPERTIES OF PROPORTION 327
Find a fourth proportional to :
45. 4, 7, and 9. 48. x 2 — 1, x + 1, and x — 1.
46. 3a;, 2 y, and 62. 49. c 3 - d 3 , c 2 - a* 2 , and c 2 -f cd + cP.
47. |, |, and |. 50. 1 + V2, 2 + V2, and 2 - V2.
Change the form of each of the following proportions so that
the unknown quantity shall occur in but one term :
51. 2:5 = 3 — a?: a;. 56. m: 7i=p — z:z.
52. 5 : 3 = 4 — x : x. 57. 3 : 2 = x + 1 : x.
53. 6 : 7 = 12 — y : y. 58. 15 : 7 = x + 1 : a;.
54. 4a : 3a = 10 — a;: #. 59. 3a : 5 c = x 4- 1 : x — 1.
55. a 3 : a 2 c = c — a;: x. 60. a 2 +l : a 2 — l = #-}-l : &— 1.
61. c-\-d:c — d = c + y:c — y.
62. m 2 4-m + l:m 2 — m — l = a; — 2:a; — 8.
63. Find two numbers in the ratio of 2 : 3, such that if each
is decreased by 1 their ratio becomes as 3 : 5.
64. Find two numbers in the ratio of 4 : 7, such that if each
number is increased by 2 the ratio becomes as 5 : 8.
65. If ad = bc, write all the possible proportions whose
terms are a, b, c, and d.
66. Find two numbers whose sum is to their product as
8 : 15, and whose sum is to their difference as 4:1.
67. What number must be added to each of the numbers,
4, 14, 10, and 30, that the resulting sums may be in proportion ?
68. Separate 10 into two parts such that their product shall
be to the difference of their squares as 6 : 5.
69. Two rectangles have equal areas, and their bases are to
each other as 5 : 16. What is the ratio of their altitudes ?
70. The lengths of the sides of three squares are in the
ratio of 2, 3, and 4. Find a side of each if the sum of the
three areas is 725 square units of area.
328 RATIO. PROPORTION. VARIATION
71. If a : b = 2 : 3, and b : c = 3 : 5, find the ratio of a : c.
72. If x : y = 3 : 4, and y : z = 5 : 6, find the ratio of a? : z.
73. If m : n = 3 : k, and n : p = k : 4, what is the ratio of
m : p?
74. If c : d = 3.2 : 4.8, and d : fc = 3.2 : 8, find the ratio of
c : k.
75. If a : b = ft : c = 2 a; : 3 y, find the ratio of a to c in terms
of x and ?/.
76. If a : ft = 3 : 4, find (3a + 2&) : (2a + 36).
77. If m : w = 5 : 2, find (2 m + n) : (2 m — ?i).
78. If a : b = 2x : Sy, find (2 a + #) : (2 a — #) in terms of
b and y.
79. If a + b : b = x-\-y : y, find (# + 2/) : (®—y) in terms of
a; and y.
80. If a; -|- a : # — a = y + x : y — z, find (# -f ?/) : (x — y) in
terms of a and z.
If a : b — c : d prove that :
81. ab:cd = b 2 : d\
82. a + c : b + d = c : d.
83. a 2 d:b = be 2 : d.
84. 3a 2 -26 2 :26 2 = 3ac-26d: 2bd.
85. Va 2 -6 2 : Vc 2 -d 2 = 6 : d
86. a 2 + 2a& :a& = c 2 + 2cc?: cd
If a : b = b : c, prove that :
87. ab-b 2 : bc-c? = b 2 : c 2 .
88. a — b : 6 — c = 6 : c.
89. a\a + 6) : c(b + c) = a 2 (a - 6) : c(b- c) .
90. a + b : & + c=Va 2 -& 2 : V& -c 2 .
APPLICATIONS OF THE PROPERTIES OF PROPORTION 329
Solve the equations :
91. 2a + 3 : 2x + 5=3a+2 : 3a; + 4.
92. 2^ + 2^ + 1 : 2a 2 -2# + l:=a + l : x-1.
93. x + m : x — m = m-\-n : m — n.
94. x 2 -l:x + l=x 2 + l:x-l.
95. x 2 + 2x + 4 : : x*-x-l = x + 2 : x-2.
a? + x-l x?-x + 2
96.
97.
98.
99.
100.
a^-z + a ^ + x _2
a; + n + 1 _ 3#— n+1 '
x + n — 1 3 a; — n — 1
s 2 + 2a? + 3 == s 2 + 3a;-4
aj 2 + 3a; + 2 z 2 + 4a;-2'
3s s + o; + l = 3s 2 — a;— 1
4:X? + X+1 4ar*— a;— 1
5_Va7+2 6- Vaj + 1
4 + Va; + 2 3+Va + l
101. Separate 100 into three parts which shall be in the
Tatio of 2 : 3 : 5.
102. Three angles of a certain triangle are in the ratio of
1:2:3. If the sum of the angles of a triangle is 180°, find
the number of degrees in each of the angles.
103. The sum of three sides of a triangle is 240 feet, and
the ratio of the sides is as 3:4: 5. Find the length of each
side.
104. For what value of a will the quantity a — 1 be a mean
proportional between the quantities a + 3 and a + 2 ?
105. If 2 is subtracted from the smaller of two numbers
and 4 is added to the larger, the ratio is as 3 : 5, but if 1 is
subtracted from the greater while 1 is added to the smaller,
the ratio is as 10 : 9. Find the numbers.
330 RATIO. PROPORTION. VARIATION
106. Two trains approach each other between two points
100 miles apart, and their rates of traveling are as 4 : 5. How
many miles will each have traveled when they meet ?
107. Find two numbers whose sum is 10, such that the
ratio of the sum of their squares to the square of their sum is
13:25.
108. Find the length of the two parts into which a line I
inches long is divided if the parts are in the ratio of a : b.
109. What must be the value of x in order that 2x — 1,
2 x -f 1, 2 x + 5, and 2 (2 x + 1) may be in proportion.
110. Find each side of a triangle whose perimeter is n
inches, the ratio of the sides being as a : b : c. .
VARIATION
386. A variable is a quantity that, under the conditions of a
problem, may have many different values.
387. A constant is a quantity that, in the same problem, has
a single fixed value.
388. If the ratio of any two values of a given variable
equals the ratio of any two corresponding values of a second
variable, then the first quantity is said to vary as the second
quantity.
Illustration :
Suppose an automobile is running at a speed of 10 miles per hour.
The total distance traveled at the end of any hour depends upon the total
number of hours that have elapsed since the start was made.
If it runs 6 hours, the distance covered is 60 miles ;
If it runs 9 hours, the distance covered is 90 miles.
Clearly, therefore, the ratio of the two periods of time is the same as
the ratio of the two distances covered. That is :
6 : 9 = 60 : 90.
We may say, therefore, that the distance (d) varies as the time ($)» or
d cc t.
The symbol cc is read "varies as."
VARIATIONS UNDER DIFFERING CONDITIONS 331
389. If xccy, then x equals y multiplied by some constant
quantity.
Proof : Let a and b denote any one pair of corresponding values of
x and y.
Then by definition, - = -.
y b
From which, % = - y •
b
Denoting the constant ratio, -» by c, x = cy.
b
In general :
We may change any variation to an equation by the introduc-
tion of the constant factor or ratio.
390. If one pair of corresponding values for the variables in a
given variation is known, the constant ratio is readily obtained.
Illustration :
If x varies as y, and x = 12 when y = 3, find the value of x
when y = 10.
We have x — cy.
Substituting, 12 = c. x 8.
Whence, c = 4.
Hence, when y = 10, x = 4 y
= 4x10
= 40. Result.
VARIATIONS UNDER DIFFERING CONDITIONS
(a) Direct Variation
391. The simple form, xccy, is a direct variation.
Illustration :
The distance covered by a train moving at a uniform rate of speed
varies directly as the time elapsed. That is :
d x t or d = ct, where c is the constant ratio.
832 RATIO. PROPORTION. VARIATION
(6) Inverse Variation
392. A quantity is said to vary inversely as another quantity
when the first varies as the reciprocal of the second. ■
Illustration:
The time (£) required by an automobile going from New York to Phila-
delphia varies inversely as the speed (s) ; for if the speed is doubled, the
time required will be but one half the former time. That is :
1 c
tec- or t = -, where c is the constant ratio.
s s
I
(c) Joint Variation
393. If a quantity varies as the product of two quantities,
the first quantity is said to vary jointly as the other two
quantities.
Illustration:
The number of dollars (N) paid to a motorman for a certain number
of trips (£) varies jointly as the number of dollars paid to him for one
trip (w) and the number of trips made. That is :
Nccnt or ST as cnt, where c is the constant ratio.
(d) Direct and Inverse Variation
394. One quantity may vary directly as a second quantity,
and, also, inversely as a third quantity. In such a case the
quantities are said to be in direct and inverse variation.
Illustration :
In mowing a field, the time required for the work (t) varies directly as
the number of acres in the field (A), but inversely as the number of men
engaged at the task (n) . That is :
A cA
t oc — or t = — , where c is the constant ratio.
n n
VARIATIONS UNDER DIFFERING CONDITIONS 333
(e) Compound Variation
395. The result obtained by taking the sum or the differ-
ence of two or more variations is a compound variation.
Illustration :
If y equals the sum of two quantities, a and b, and if a varies directly
as x 8 while b varies inversely as x 2 , then
c'
a = ex 8 and b = —
x 2
c'
Adding (since y = a + 6), y = ex 8 + — .
x' 2 *
It is to be noted that in such cases tw% different factors are necessary.
(/) Important Principle
396. If x depends only upon y and z, and if x varies as y ivhen
z is constant, and x varies as z when y is constant, then x varies
as yz when both y and z vary.
Let x, y, z (1), Xi, yi, z (2), and x 2 , jfi, z\ (3) be three sets of corre-
sponding values. Then,
Since z has the same value in (1) and (2),
Since y x has the same value in (2) and (3),
Multiplying (I) by (II),
Or,
That is :
The ratio of any two values of x equals the ratio of the corre-
sponding values of yz, and, by definition, x varies as yz.
Illustration :
The area (A) of a rectangle varies as the base (B) when the height (H)
is constant, and the area varies as the height when the base is constant.
Therefore, when both the base and height vary, the area will vary as the
base and height jointly.
*i y\
(I)
X\ _ z
X 2 9%
(II)
x _ yz
x 2 y\z\
x-.x 2 = yz: y x z\.
334 KATIO. PROPORTION. VARIATION
I II III IV
8
10
8
. 12
5
10
5
IS
A = BH
= 105
= 50
A =[BH
= 12.5
= 60
A = BH
= 10-8
= 80
A = BH
= 12-8
= 96.
In II. B changes, H constant, A changes (AccB, H constant).
In III. H changes, B constant, A changes (AxH, B constant).
In IV. B changes, H constant, A changes (AccBH, Band H vary).
APPLICATIONS OP VARIATIONS
397. Illustrations:
1. If x varies as y 2 , and x = 2 when y — 4, find x when
y = 16.
Since
xocy 2 ,
we have
a; = my 2 .
And, if x = 2 and y =
= 4, 2 = m x 4 2 .
Whence,
m = £.
Then, when y = 16,
x = K16) 2 .
» = H*.
x = 32. Result.
2. If x varies inversely as y*, and x = 2 when y = 4, find
a; when y =2.
Since
xoci-,
y 8
we have
Whence,
2 =P'
and
m = 128.
Then, when y = 2,
23
x = 16. Result.
APPLICATIONS OF VARIATIONS 335
3. If s is the sum of two quantities, one of which varies
directly as x 2 and the other inversely as x, and if s = 6 when
x = 2, and s = 2 when x — — 2, find s when a; = — 1.
Let w and v represent the two quantities.
Then &
= w + t>;
uccx 2 ; floe-*
sacs 2 +--
X
From which,
s = mx 2 + -•
(1)
Substituting :
Ifs = 2,
2 zJ
(2)
If x = - 2,
2 = m(-2)2+i = 4m-
— 2
n
2
(8)
From (2) and
(3),
at =s 1, fi s 4.
If x = — 1, in
(1),
S = l(_l)2 + _£_
= 1-4
= - 3- Result.
4. The volume of a sphere varies as the cube of its diame-
ter. If three metal spheres whose diameters are 6, 8, and 10
inches, respectively, are melted and recast into a single sphere,
what is its diameter ?
Let V denote the volume of the required sphere, and D its diameter.
Then Foe 2)3.
Whence, V= mD 8 . (1)
Denote the volumes of the three given spheres by V u V 2 , and Vs.
We have, therefore, Vi = wi(6) 8 = 216 m,
F 2 = m(8) 8 = 512m,
F 3 = m(10) 8 = 1000 m,
Whence, by addition, V\ + V 2 + V s - 1728 m.
But, by the conditions, V x + V 2 + V* = V= m 7) 3 . (From 1.)
336 RATIO. PROPORTION. VARIATION
Hence, mi) 3 =1728 m,
D 3 = 1728,
D =12. Result.
That is, the diameter of the sphere obtained from the given spheres
is 12 inches.
Exercise 123
1. If a; varies as ?/, and x = 10 when y — 2, find, x when y = 5.
2. If a; varies as y, and # = 3.2 when y = 0.8, find a; when
y = 5.6.
3. If x + 1 varies as y — 1, and cc = 6 when y = 4, find 03
when y = 7.
4. If 2 a; — 3 varies as 3 2/ + 2, and ?/ = 2 when x — 0.2, find
y when a; = 1.5.
5. If x 2 varies as y 2 , and x = 3 when y = 2, find y when
a; = 4.
6. If x varies inversely as y, and x = 2 when 2/ = 4, find a;
when y = S.
7. If a; varies inversely as y 2 , and x = 2 when y = ^, find y
when a; = 1£.
8. If x varies jointly as y and z, and x = 3 when y = 4 and
2! = 2, find x when ?/ = 5 and 2 = 4.
9. If a; varies inversely as y 2 — 1, and a; = 4 when y =5,
find a; when ?/ = 15.
10. If a; varies as -, and y = — 2 when a? = 7, find the equa-
tion joining aj and y.
11. If the square of x varies as the cube of y, and if x— 6
when 2/ = 4, find the value of jf when x = 30.
12. If s is the sum of two quantities, one of which varies
as x while the other varies inversely as x ; and if s = 2 when
a? = £ and s = 2 when a; = — 1, find the equation between s and x.
APPLICATIONS OF VARIATIONS 337
13. If w varies as the sum of x, y, and z, and w = 4 when
x = 2, y = — 2, and 2 = 5, find a; if w = — 3, 2/ = 2, and
2= -6.
14. Given that s = the sum of three quantities that vary as
x, x 2 , and se 3 , respectively. If x = 1, s = 3 ; if a; = 2, s = 6 ; and
if x = 4, s = 16. Express the value of s in terms of #.
15. The area of a circle varies as the square of its diameter.
Find the diameter of a circle whose area shall be equivalent
to the sum of the areas of two circles whose diameters are 6
and 8 inches respectively.
16. The intensity of light varies inversely as the square of
the distance from the source. How far from a lamp is a certain
point that receives just half as much light as a point 25 feet
distant from the lamp ?
17. The volume of a sphere varies as the cube of its diameter.
If three spheres whose diameters are 3, 4, and 5 inches, respec-
tively, are melted and recast into a single sphere, what is
the diameter of the new sphere ?
18. The volume of a rectangular solid varies jointly as the
length, width, and height. If a cube of steel 8 inches on an
edge is rolled into a bar whose width is 6 inches and depth 2
inches, what will be the length of the bar in feet ?
19. If the amount earned while erecting a certain wall
varies jointly as the number of men engaged and the number
of days they work, how many days will it take 4 men to earn
$ 100 when 6 men working 9 days earn $ 135 ?
20. The pressure of the wind on a plane surface varies
jointly as the area of the surface and the square of the wind's
velocity. The pressure on a square foot is one pound when
the wind is blowing at a rate of 15 miles an hour. What
will be the velocity of a wind whose pressure on a square yard
is 81 pounds ?
SOM. EL. ALG. — 22
CHAPTER XXV
• PROGRESSION
ARITHMETICAL PROGRESSION
398. A series is a succession of terms formed in accordance
with a fixed law.
399. An arithmetical progression is a series in which each
term, after the first, is greater or less than the preceding term
by a constant quantity. This constant quantity is the common
difference.
400. We may regard each term of an arithmetical progres-
sion as being obtained by the addition of the common difference
to the preceding term; hence,
An increasing arithmetical progression results from a positive
common difference, and a decreasing arithmetical progression re-
sults from a negative common difference. Thus :
1, 5, 9, 13, . . . etc., is an increasing series in which the common differ-
ence is 4.
7, 4, 1, —2, . . . etc., is a decreasing series in which the common differ-
ence is —3.
401. In general, if a is the first term, and d the common
difference,
a, a + d, a + 2 d, a 4- 3 d, a + 4 d, . . . etc.,
is the general form of an arithmetical progression.
338
ARITHMETICAL PROGRESSION 339
402. The nth Term of an Arithmetical Progression.
In the general form,
a,a + cf.fl + 2d,a + 3d,a + 4d ) . . . etc.,
it will be seen that the coefficient of d in any term is less by 1 than the
number of the term. Hence, the coefficient of d in the nth term is n — 1.
Therefore, if a = the first term of an arithmetical progression,
d = the common difference,
n = the number of terms in the series,
I = the last term (that is, any required term),
then, 1 = a + (n - l)d. (I)
403. The Sum of the Terms of an Arithmetical Progression.
If s denotes the sum of the terms of an arithmetical progression, we
may write the following identities, the second being the first written in
reverse order :
s = a + (a + d) + (a + 2 d) + + (I - 2d) + (* - &) + I
8 = 1 + (l-d) + (l-2d) + • + (a + 2d) + (a + d) + a.
2 s = (a + + (a + I) + (a + I) + .-• + (a + 1) + (a + I) + (a + I).
Whence, 2s = n(a + l)>
Or, s = 5(a + l). (II)
404. Combining the two formulas, (I) and (II), we obtain
the following convenient formula for finding the sum of an
arithmetical progression when the first term, the common
difference, and the number of terms are known. That is :
s = 5[2a + (n-l)d]. (Ill)
z
405. The first term (a), the common difference (d), the
number of terms (n), the last term (I), and the sum of the
terms (s), are the elements of an arithmetical progression.
APPLICATION OF THE FORMULAS FOR ARITHMETICAL PROGRESSION
406. Illustrations :
1. Find the 10th term and the sum of 10 terms of the
arithmetical progression, 1, 4, 7, 10, . . .
340 PROGRESSIONS
We have given, a — 1, d = 3, n = 10.
In the formula, I = a + (n — l)d.
Substituting, I = 1 + (10 - 1) (3)
= 28, the required 10th term.
In the formula, s = - (a + Z)-
2
Substituting, s = — (1 + 28)
z
= 5 x29
= 145, the required sum of 10 terms.
2. The first term of an arithmetical progression is 3, the last
term, 38, and the sum of the terms, 164. Find the series.
We have given, a = 3, I = 38, s = 164.
Then, s = - (a + Z) or 164 = | (3 + 38), whence n = 8.
Also, Z = a + (w - 1) d, or 38 = 3 + (8 - 1) d, whence d = 5.
Therefore, 3, 8, 13, 18, 23, 28, 33, 38, is the required series.
3. How many terms of the series, 2, 5, 8, •••, will be re-
quired in order to give a sum of 126 ?
We have a - 2, d = 3, s = 126.
Then Z = a+(n-l)d = 2 + (»-l)3 = 2 + 3n-3 = 3n-l.
This value for Z in terms of n is substituted in the formula
S =|(a + Z).
Whence, 126 = 2 (2 + 3 n - 1), or 3 n 2 + w = 252.
2
Solving this quadratic equation in n, we have (using the quadratic
formula),
n = -l ± VrT3024 - n = -L±JL5, M = 9)0 r-9|.
6 6
Therefore, the required number of terms for the given sum is 9.
Formula III (§ 404) may be used here without finding Z.
ARITHMETICAL PROGRESSION 341
Exercise 124
1. Find the 16th term of 7, 10, 13, ....
2. Find the 10th term of 2, - 1, - 4, ....
3. Find the 12th term of 4, 3.2, 2.4, •••.
4. Find the 10th term of J, f, J, ....
5. Find the -20th term of f, f, |j .....
6. Find the 10th term of - 12, -9.5, - 7, -4.5, ....
7. Find the 14th term of x + 1, x + 3, x + 5, •••.
8. Find the 11th term of x — 5 a, x — 4 a, #— 3 a, •••.
9. Find the 10th term of 4 + V2, 3 + 2^2, 2 + 3V2, ....
10. Find the 358th term of .0075, .01, .0125, .015, ....
Find the sum of :
11. 3, 8, 13, 18, ... to 24 terms.
12. -2, -9, -16, ... to 12 terms.
13. .25, .3, .35, ... to 40 terms.
14 - i>lbih -to 16 terms.
15. 20V2-10V3, 18V2-9V3, 16V2-8V3, *.- to 21 terms.
Find the first term and the common difference when :
16. s = 297, w = 9, 1=16. 19. Z = 0, s = 100, n = 25.
17. s = 294, 7i = 12, 1 = 41. 20. n = 6, s = 20 A, I = 4.9.
18. 7i = 13, s = 260, Z = -f. 21. s = 0, Z = 34.5, w=24.
Find the common difference when:
22. a = 4, ? = 40, n = 13. 24. I = 2, ti = 7, s = 19.25.
23. s = -27.5, = 4.5,71=11. 25. Z = .97, a = .8, s = 35.4.
26. a = .08, s = — 25, 7* = 25.
27. 7i = 30, a=-13V2, Z = 16V2.
342 PROGRESSIONS
Find the first term when :
28. n == 12, I = 10, s = 60.
29. d = - 1, n = 10, s = - 100.
30. d = - .6, n = 14, Z = - .83.
Find the number of terms when:
31. a = 7, d = -3, Z = -23.
32. Z = f, a = }, s = 17.
33. a = -|,Z = -A,d = -^.
How many consecutive terms of
34. 2, - 3, - 8, •-. will give a sum of - 205 ?
35. 1, 1J, If, ••• will give a sum of 831 ?
36. — 3, - 3 }, — 4, • • • will give a sum of — 97£ ?
37. 0.36, 0.32, 0.28, ••• will give a sum of -.4 ?
38. 12V3, 10V3, 8V3, will give a sum of 0?
THE DERIVATION OF GENERAL FORMULAS FROM THE FUNDAMENTAL
FORMULAS
407. From the fundamental formulas (I) and (II) in Arts
402 and 403 we may derive a formula for any desired element
in terms of any other three elements.
Illustrations :
1. Given a, w, and s; derive a formula for d.
From the fundamental formulas,
l = a+(n — l)d and s=-(a + J),
Ss
we must eliminate the element I ; and by substituting the value of I from
the first formula for I in the second formula, we have
s =-[a + a + (n- l)d].
ARITHMETICAL PROGRESSION
343
From which, 2 s = ?i[2 a + (n — l)d],
2 s = 2 a?i + w(w — l)a*,
2 s — 2 an = n(» — l)d",
d _ 2(g-an)^ the required formula for d.
n(n — 1)
2. Given d, I, and s, find a formula for n.
From (I), Art. 402, I = a + (n - l)rZ.
Hence, a = Z — (ft — l)a\
Substituting this value for I in (II), Art 403, we have,
« = |[2 a+ (n- l)d]
* =^[2Z-2(n-l)d+(n-l)d].
2
2s = 2 »Z-n(n-l)d.
on 2 - (2Z + d>-f 2s = 0.
Simplifying,
Whence,
Solving for n,
2 Z + d ± V(2 Z + d) 2 - 8 ds.
_
1. Given
2. Given
3. Given
4. Given
5. Given
6. Given
7. Given
8. Given
9. Given
10. Given
11. Given
12. Given
Exercise 125
dj ft, and I ; derive a formula for a.
n, Z, and s
a, ft, and s
a, I, and n
a, ft, and s
a, I, and s
dj I, and n
n, ?, and s
a, Z, and s
dj ft, and s
a, c?, and s
a, d, and s
derive a formula for a.
derive a formula for I.
derive a formula for d.
derive a formula for d.
derive a formula for d.
derive a formula for s.
derive a formula for d.
derive a formula for n.
derive a formula for a.
derive a formula for I.
derive a formula for ft.
344 PROGRESSIONS
ARITHMETICAL MEANS
408. If we know a and b, the first and last terms respec-
tively in an arithmetical progression, we may form an arith-
metical progression of m + 2 terms by inserting m arithmetical
means between a and b.
Illustration :
Insert 7 arithmetic means between 3 and 43.
We seek an arithmetical progression of (m + 2) = (7 + 2) = 9 terms.
(For two terms, the first and the last, are given.)
Therefore, a = 3, I = 43, n = 9. It remains to find d.
In the formula, 1= a+ (n — l)d.
Substituting, 43 = 3 + (9 — 1)<?.
Whence, d = 5.
Therefore, 3, 8, 13, 18, 23, 28, 33, 38, 43 is the required series.
409. To insert a single arithmetic mean, m, between two
numbers a and b, we have as a result the series a, m, I.
Therefore, m — a = l — m.
And, 2m = a+ I.
a + I
Or, the arithmetic mean between two numbers equals one half
the sum of the numbers.
Exercise 126
1. Insert 5 arithmetical means between 13 and 37.
2. Insert 9 arithmetical means between 6 and 11.
3. Insert 7 arithmetical means between f and f .
4. Insert 15 arithmetical means between — f and f.
5. Insert the arithmetical mean between 12.4 and 13.2.
6. Insert the arithmetical mean between a + 2 and a — 2.
7. Insert the arithmetical mean between and •
a+ c a — c
ARITHMETICAL PROGRESSION 345
PROBLEMS INVOLVING ARITHMETICAL PROGRESSIONS
410. Illustrations:
1. The 4th term of an arithmetical progression is 11, and
the 9th term 26. Find the first 3 terms of the progression.
The fourth term is a + 3 d, and the ninth term a + 8 d.
Therefore, a + 8 d = 26.
And, a + 3 d = 11.
Whence, 5 d = 15.
d = 'S.
a = 2.
Hence, the required terms of the progression are 2, 5, 8. Result.
2. Find 5 numbers in arithmetical progression, such that
the product of the 1st and 5th shall be 28, and the sum of
the 2d, 3d, and 4th shall be 24.
Let x — 2 y, x — y, x, x + y, and x + 2 y represent the numbers.
Then, (x - 2 y) (x + 2 y) = 28. (1)
And, x-y + x + x + y = 24. (2)
(3)
(4)
Hence, the required numbers are 2, 5, 8, 11, 14 ; or 14, 11, 8, 5, 2.
The symmetrical forms, x — 2 y, x — y, x, x + y, x + 2y, are chosen
merely for convenience.
Exercise 127
1. How many numbers between 50 and 500 are exactly-
divisible by 6 ?
2. Find the sum of all the numbers of two figures that are
exactly divisible by 7.
From (2),
3z = 24.
x = S.
From (1),
X 2_±y2 = 28.
Substituting from (3),
82 _ 4 y2 = 28.
4 y2 = 36.
y=±3
346 PROGRESSIONS
3. Find the sum of the first 20 odd numbers.
4, Find x so that 2 x — 1, 2 x + 2, 3 x — 2, and 3 a + 1 shall
be in arithmetical progression.
5 Are the 3 numbers, 5x — 3y, x + 2y, and 7y — 3x, in
arithmetical progression ?
6. What will be the value of x if the numbers 2 x — 3, x y
and 10 — x are in arithmetical progression ?
7. Find the sum of 1 + 2 + 3 + 4 + 5+ ..-ton terms.
8. Find the sum of 1+3 + 5 + 7+ • • • to n terms.
9. Find the sum of 21 terms of an arithmetical progression
whose middle term is 23 and whose common difference is 2.
10. Find the sum of 2 + 4 + 6 + 8+ •••to n terms, and
compare the result with that of example 8.
11. Find the sum of the first 25 numbers that are divisible
by 7.
12. The sum of 3 numbers in arithmetical progression is
24, and the product of the 2d and 3d is 88. What are the
numbers ?
13. The 5th term of an arithmetical progression of 49
terms is 3, and the 15th term, 63. Find the 34th term.
14. The 6th term of an arithmetical progression is —19,
and the sum of the first 18 terms, 36. Find the common dif-
ference and write the first 5 terms of the series.
15. Prove that the sum of n consecutive odd numbers begin-
ning with 1 is n 2 .
16. A clerk's salary is increased $ 50 every 6 months for a
period of 8 years. At the end of the 3d year he was receiving
$1000. At what salary did he begin and what will he receive
during the last half of his 8th year ?
17. Prove that the sum of the terms of an arithmetical pro-
gression in which a, n, and d are all equal, is equal to — ~T -
ARITHMETICAL PROGRESSION 347
18. Find 4 numbers in arithmetical progression such that
the sum of the 1st and 3d shall be 44, and the product of
the 2d and 4th, 572.
19. How many strokes are sounded in 24 hours by a clock
striking hours only ? ' #
20. A boy saves 25 cents the first week of a new year, and
increases his savings 5 cents each week through the entire
year. How much will he have saved by December 31st ?
21. The sum 'of three numbers in arithmetical progression is
27, and their product, 693. Find the three numbers.
22. Show that if every alternate term of an arithmetical
progression is removed, the remaining terms will be in arith-
metical progression.
23. A laborer agreed to fill 40 tanks with water, the tanks
being placed in a straight line and at a uniform distance of
10 feet from each other. Each tank holds 18 gallons and he
carries at each trip 2 pails holding 3 gallons each. If the
source of supply is 10 feet from the first tank in the row, how
far does he travel before the 40 are filled ?
24. A man travels 210 miles. The first day he goes 12
miles, and he increases each succeeding day's distance by 2
miles. How many days will he require to complete the
journey, and how far will he be obliged to go on the last day ?
25. There are 2 arithmetical progressions, 13, 15, 17, •••
and 37, 35, 33, •••, in one of which d = 2, and in the other
d = — 2. What must be the number of terms for both series
in order that the sums for both series shall be equal ?
26. A contractor failing to complete a bridge in a certain
specified time is compelled to forfeit $ 100 a day for the first
10 days of extra time required, and for each additional day
beginning with the 11th the forfeit is daily increased by $ 10.
He loses in all $2550. By how many days did he overrun
the stipulated time ?
348 PROGRESSIONS
GEOMETRICAL PROGRESSION
411. A geometrical progression is a series in which each
term, after the first, is obtained by multiplying the preceding
term by a constant quantity. , This, constant number is the
common ratio.
Thus, 3, 6, 12, 24, 48,-
is a geometrical progression in which the common ratio is 2.
2, |, b i'-
is a geometrical progression in which the common ratio is \.
412. In general, if a is the first term of a geometrical pro-
gression, and r the common ratio,
a, ar, ar 2 , ar*, ar*, ar 5 , ••• etc.,
is the general form of a geometrical progression.
413. The /7th Term of a Geometrical Progression.
In the general form,
a, ar, ar 2 , ar z , ar*, ar 5 , ••• etc.,
it will be seen that the exponent of r in any term is less by 1 than the
number of the term. Hence, the exponent of r in the nth term is n — 1.
Therefore, if a = the first term of a geometrical progression,
r = the common ratio,
n = the number of terms in the series,
I = the last term (that is, any required term).
Then, l = ar»- 1 . (I)
414. The Sum of the Terms of a Geometrical Progression.
If s denotes the sum of the terms of a geometrical progression, we
may write
s = a + ar-\- ar 2 + ••• ar n ~ s + ar n ~ 2 + ar 11 ' 1 . (1)
Multiplying (1) by r,
rs = ar + ar 2 + ar*-\ — ar n ~ 2 -f ar"- 1 + ar". (2)
Subtracting (1) from (2),
rs — s = ar* 1 — a.
From which, s = arW ~ a . (II)
GEOMETRICAL PROGRESSION 349
415. Combining the two formulas (I) and (II), we obtain
the following convenient formula for finding the sum of a
geometrical progression when the first term, the last term, and
the common ratio are known. That is :
s=7fr cm)
416. The first term (a), the common ratio (r), the number
of terms (n), the last term (I), and the sum of the terms (s),
are the elements of a geometrical progression.
APPLICATIONS OF THE FORMULAS FOR GEOMETRICAL PROGRESSION
Illustrations :
1. Find the 8th term and the sum of 8 terms of the pro-
gression, 2, 6, 18, 54, ... .
We have given, a = 2, r = 3, n = 8.
In the formula, I = ar n ~\
we substitute I = (2)(3) 7 .
Whence, = 2 • 2187,
Or, = 4374, the required 8th term.
rl — a
In the formula,
r-l
we substitute g = (3) (4374) - 2 #
3-1
Whence, s = 6560, the required sum of 8 terms.
2. Find the 12th term and the sum of 12 terms of 6, — 3,
We have given, a = 6, r = — \, n = 12.
In the formula, I = at— \ \ = 6(- |)H = 6(-^) = - T & ? .
In the formula, s = ^-=1^.
r-l
-i-l - | 1024 " t7f¥ "
Hence, the required 12th term is - y^j, and the sum of 12 terms,
3mt-
350 PROGRESSIONS
Exercise 128
1. Find the 7th term of 2, 6, 18, ... .
2. Find the 8th term of 3, - 6, 12, ... .
3. Find the 6th term of - 3, - 6, - 12, ... .
4. Find the 9th term of J, £j 1, ... .
5. Find the 9th term of — f, f , — £, ... .
6. Find the 8th term of 3, 3^2, 6, ... .
7. Find the 6th term of 2.4, 0.24, 0.024, ... .
8. Find the 10th term of 0.0001, 0.001, 0.01,
9. Find the 15th term of a 4 x, aV, a 2 x\ ... .
Find the sum of :
10. 2, 4, 8, ... to 8 terms.
11. 40, 20, 10, ... to 8 terms.
12. 120,12,1.2, ...to 6 terms.
13. -§, I, -§, ... to 5 terms.
14. - 27, 9, - 3, ... to 6 terms.
15. 2J, 1£, fy ... to 10 terms.
16. 1 + m + m 2 + m 3 + ... to n terms.
17. 64-32 + 16-8 + 4+ ... ton terms..
Find the common ratio when :
18. a = 4, Z = 1024, n = 9.
19. I = - 729, a = 3, n = 6.
20. a = - 4, 2 = - 512, s = - 1020.
21. a = 4, Z = -Ti ¥ , * = ffi.
22. a = 4, l=s\, n = 9.
23. = 3,* = ^,* = ^.
24. n = 10,l = ^,a = 2.
25. ,9 = -V^a = 3,*=-- T k-
GEOMETRICAL PROGRESSION 351
Find the number of terms when:
26. a=-S, r=-2, $ = 1023.
27. r = — 2, s = — 22, a = — 2.
28. a=-2, Z=-128, r = 2.
29. Z=-160, a = 5, r=-2.
30. s=,m^,r = i,l = ^.
31. a = .2, s = 204.6, Z = 102.4.
How many terms of the series :
32. f, \ y y 1 ^, ••• will make a sum of ffj-?
33. 18, — 6, 2, ••• will make a sum of 4^ 4 - ?
34. V2, 2, 2 V2, .- will make a sum of 62 + 31 V2 ?
GENERAL FORMULAS DERIVED FROM THE FUNDAMENTAL FORMULAS
417. From the fundamental formulas (I) and (II) in Arts.
413 and 414 we may derive a formula for any desired element
in terms of any other three elements.
Illustrations :
1. Given a, I,
and s;
derive
a formula for
r.
From (III) (Art. 415),
r — 1
we obtain
rs -
-8 = rl — a.
Whence,
rs -
r(s -
rl = s — a,
Z) = s — a,
s -I
2. Given a, n,
and Z;
derive
a formula for
s
From Ex. 1,
s-l
From (I) (Art. 4
US), l =
I n-1 /
ar nl , r" -1 =-, r — a/
I
(1)
(2)
352
From (1) and (2),
PROGRESSIONS
a_ n ~i/l
■\Ta
C~t/a
s- I
*V+ = s n ~i r l - n -Vl»,
n -y/i) = n -l/a»- n ~\/T»,
n ~l/a
-in
Note. The general formulas for n involve logarithms, and are ordi-
narily reserved for advanced students.
Exercise 129
1. Given r,
2. Given r,
3. Given a,
4. Given a,
5. Given r,
6. Given r,
7. Given r,
8. Given a,
9. Given n,
10. Given n,
n, and I
n, and s
n, and I
r, and s
n, and s
I, and s
n, and I
n, and s
I, and s
I, and s
derive a
derive a
derive a
derive a
derive a
derive a
derive a
derive a
derive a
derive a
formula for a.
formula for I.
formula for r.
formula for. I.
formula for a.
formula for a.
formula for s.
formula for r.
formula for a.
formula for r.
THE INFINITE DECREASING GEOMETRIC SERIES
418. If the absolute value of r in a geometrical progression,
a, ar, ar 2 , •••, ar 71 " 1 , is less than 1, the successive terms become
numerically less and less; hence, by taking a sufficiently great
number of terms, that is, by making n sufficiently great, the nth
term becomes as small as we may choose, although never equal
»to zero.
GEOMETRICAL PROGRESSION 353
Thus, if a = 1 and r = £, the series, £, J, |, 1 " 5 , •••, may be continued
until the nth term is so small as to have no assignable value. Hence, we
may say that is the limiting value of the nth term.
It follows, therefore, that if I approaches 0, rl approaches 0. Hence,
$ = rl ~ a , or a ~ approaches -^— as a limiting value.
r — 1 1 — r 1 — r
That is, s = — ^— is the formula for the sum of the terms of an infinite
1-r
decreasing geometric series. v
Illustration :
Find the sum of the infinite series, l+i + -g- + -^H •
We have a s= I, r = |.
Hence, s = -i— = - = - • Result.
THE RECURRING DECIMAL
419. A recurring decimal is the sum of an infinite decreasing
geometrical progression in which the ratio is 0.1 or a power of
0.1.
Thus, .272727 ... = .27 + .0027 + .000027 + ••• is a geometrical pro-
gression in which a = .27 and r = .01.
Illustration :
Find the value of .292929 — .
We have .292929 ••• = .29 + .0029 + .000029 + ....
From which, a = .29 and r = .01.
Hence, s = _«_ = -J*?— = i 2 ^ - 2?. Result-
l_ r 1-.01 .99 99
In case the recurring portion of the decimal does not include
the first decimal figures, the sum of the recurring portion is
found as above and then added to the other. Thus :
2.23189189 ••• = 2.23 + .00189189 ... .
For the sum of .00189 + .00000189 + ■ •-,
™w* « . 00189 .00189 189 7
we nave s = = = = = .
l_ r 1-.001 .999 99900 3700
Hence, 2.23189189-.. =2 + ^ + 77 fo = 2#&. Result.
SOM. EL. ALG. 23
354 PROGRESSIONS
GEOMETRICAL MEANS
420. If we know a and I, the first and last terms respectively
in a geometrical progression, we may form a geometrical pro-
gression of m-f-2 terms by inserting m geometrical means
between a and I.
Illustration :
1. Insert 4 geometrical means between 2 and -g-J^.
We seek a geometrical progression of (m + 2) = (4 + 2) = 6 terms.
Hence, a = 2, Z = ^f 3, n = 6.
Then, I = Of*-*, ^ = 2 . r6-i, - 2 l ¥ = r6, * = r.
Hence, the required progression is 2, f, f, A, ^ 2 r , Y | f .
421. To insert a single geometrical mean, m, between the two
numbers, a and I, we require the value of m in
a :m = m :l.
From which, m = Val.
The geometrical mean between two numbers equals the square
root of their product.
PROBLEMS INVOLVING GEOMETRICAL PROGRESSION
422. Illustrations:
1. The 3d term of a geometrical progression is 2, and the
7th term 162. Find the 5th term.
We have ar 2 - 2, (1)
ar 6 = 162. (2)
Dividing (2) by (1), r 4 = 81, r = 3.
Therefore, ar 2 = 2, a(3) 2 = 2, a = f .
Hence, the 5th term, ar 4 , = (f ) (3)* = (2) (&) = 18. Result.
2. Find 3 numbers in geometrical progression, such that the
sum of the 1st and 3d decreased by the 2d shall be 7, and the
sum of the squares of the 3 shall be 91.
GEOMETRICAL PROGRESSION 355
Let a, ar, and ar 2 represent the 3 numbers.
Then, a — ar + ar 2 = 7, (1)
a 2 + a*r* + a¥ = 91. (2)
Dividing (2) by (1), a + ar + ar 2 = 13. (3)
Subtracting (1) from (3), 2 ar = 6.
Whence, ar = 3,
3
r = -.
a
Substituting in (1), a - alA + «f- J =7,
a_3 + ?_ = 7.
a
a = 1 or 9.
Hence, the required numbers are 1, 3, and 9. Result.
i
Exercise 130
1. Find the first 3 terms of a geometrical progression
whose 3d term is 9 and whose 6th term is 243.
2. Find the first 2 terms of a geometrical progression
whose 5th term is £ and whose 10th term is 16.
3. Insert 4 geometrical means between 1 and 243.
4. Determine the nature, whether arithmetical or geometri-
cal, of the series, \, i, -J, ... .
5. Find the first 2 terms of a geometrical progression in
which the 5th term is £ and the 12th term 16.
6. Which term of the geometrical progression, 3, 6, 12,
..., is 3072 ?
7. Find to n terms the sum of the series, 1, 3, 9, 27, ... .
8. Insert 4 geometrical means between -^ and 32.
9. Find, to infinity, the sum of 2, 1|, |, ... .
10. Find the value of the recurring decimal, 0.1515 ... .
11. Find the value of x such that x — 1, x + 3, x -f- 11 may-
be in geometrical progression.
356 PROGRESSIONS
12. Insert a single geometrical mean between 6f and 5\.
13. Find the value of the recurring decimal, 2.214214 ... .
14. Find 4 numbers in geometrical progression such that
the sum of the 1st and 3d shall be 15, and the sum of the 2d
and 4th, 30.
15. Find to infinity the sum of 4, — f, £, ... .
16. Find to infinity the sum of 1.2161616 ... .
17. Insert a single geometrical mean between 3 V2 — 2 and
3V2 + 2.
18. The 1st term of a geometrical progression is 10, and
the sum of the terms to infinity is 20. Find the common
ratio.
2 11
19. Find to infinity the sum of — — > — => — —•> ••• .
J V2 V2 2V2
20. Insert 3 geometrical means between a~* and a 4 , and
find the sum of the resulting series.
o x
21. Insert 5 geometrical means between — 3 and — and find
the sum of the series. *
22. What must be added to each of the numbers, 5, 11, 23,
that the resulting numbers may be in geometrical progression ?
23. The sum of the first 3 terms of a decreasing geomet-
rical progression is to the sum to infinity as 7 : 8. Find the
common ratio.
24. If the numbers, x — 2, 2x — l, and 5 x + 2, are in geo-
metrical progression, what is the common ratio of the series ?
25. The sum of the first 8 terms of a geometrical pro-
gression is equal to 17 times the sum of the first 4 terms.
Find the common ratio.
26. The population of a certain city is 312,500, and it has
increased uniformly by 25 % every 3 years for a period of
12 years. What was the population 12 years ago ?
GEOMETRICAL PROGRESSION 357
27. A ball on falling to the pavement rebounds ^ of the
height from which it was dropped, and it continues to succes-
sively rebound ^ of each preceding distance until it is at rest.
If the height from which it originally fell was 60 feet, through
how great a distance does it pass in falling and rebounding ?
28. What is the condition necessary that a + 1, a-j-S, a+ 7,
and a + 15 shall be in geometrical progression? For what
value of a is this condition true ?
29. Show that if 4 numbers, m, n, x, and y, are in geo-
metrical progression, then m + n, n + a?, and x + y are also in
geometrical progression.
30. A sum of money invested at 6 % compound interest will
double itself in 12 years. What will be the amount of $10
invested at compound interest at the end of 60 years ?
31. If 4 numbers are in geometrical progression, the sum
of the 2d and 4th divided by the sum of the 1st and 3d is
equal to the common ratio.
32. Find 3 numbers in geometrical progression whose sum is
21, and the sum of whose squares is 189.
33. Find an arithmetical progression whose first term is 2,
and whose 1st, 3d, and 7th terms are in geometrical pro-
gression.
34. If the alternate terms of a geometrical progression are
removed, the remaining terms are in geometrical progression.
35. Find to infinity the sum of the series
X + 1 \X + 1J \x + 1
36. Find the 4th term of an infinite decreasing geometric
series the sum of whose terms is ff> and whose first term
is .25.
CHAPTER XXVI
THE BINOMIAL THEOREM. POSITIVE INTEGRAL
EXPONENT
423. A finite series is a series having a limited number of
terms.
424. The binomial theorem is a formula by means of which
any power of a binomial may be expanded into a series.
425. By actual multiplication we may obtain :
(a + by = a 2 + 2 ab + 6 2 .
(a + 6)3 = a 3 + 3 a 2 6 + 3 a& 2 + b s .
(a + by = a* + 4 a 3 & + 6 a 2 6 2 + 4 a& 3 + 6* ; etc.
In the products we observe :
1. The number of terms exceeds by 1 the exponent of the
binomial.
2. The exponent of a in the first term is the same as the
exponent of the binomial, and decreases by 1 in each succeed-
ing term.
3. The exponent of b in the second term is 1, and increases
by 1 in each succeeding term until it is the same as the expo-
nent of the binomial.
4. The coefficient of the first term is 1, the coefficient of the
second term is the same as the exponent of the binomial.
5. If the coefficient of any term is multiplied by the exponent
of a in that term, and the product is divided by the exponent
of b in that term increased by 1, the result is the coefficient of
the next following term.
358
THE BINOMIAL THEOREM 359
426. By observing these laws we may write the expansion
of (a + 6) 4 thus:
/ , ni i ( .-ijn'*'*Ji'j 4.3-2^,3 . 4.3.2.1 M
(a + 6)4 = a 4 + 4 «3 6 + __ a 2 6 2 + ^^a&B + j^-M
In like manner, we may write the expansion of (a + b) n in
the form :
(a + &)• = a» + na-ift + n (" - 1) a n-2 6 2 + n(n-l)(n- 2) g ^ ft , + ...
1 «2 1 • 2« o
This expression is the binomial formula, and we will now
prove that it is a general expression for any power of (a -+- 6),
for positive integral values of n.
PROOF OP THE BINOMIAL THEOREM FOR POSITIVE INTEGRAL EXPONENTS
427. We have shown by actual multiplication that the laws
governing the successive expansions of (a-f b) are true up to
and including the fourth power.
If, now, we assume that the laws of Art. 425 are true for
any power, as the nth power, and if, furthermore, we show the
laws to hold for the (n -f- l)th power, then the truth of Art. 425
for all positive integral values of n is established. This method
of proof is known as mathematical induction.
Both members of the formula (1) below are multiplied by (a + 5).
(a + by = a» + nfl»-*6+ n(n " ^ «-%* 4- n < n ~ &g ~ 2) a»-'6»+ - (1)
1 • 2 1 * 2 • o
(a + 6) (a + 6)
(a + &)•+! = *h-i + na»6 + "("-^ o-iy + ^- 1 )(^- 2 ) a a-2 & 3 + ...
1 • 2 1 « 2 • o
+ a*6 + wa"- 1 ^ 2 + w(n-l) a«- 2 6 3 + -
1*2
(a + 6)«+i = tfH-i + (n + l)a»6 + f" 71 ^- 1 ) + »1 a »-i&»
r n (n-l)(n-2) n(n -1)"1 w _ 2&8 , ...
^L 1-2-3 1-2 J
360 THE BINOMIAL THEOREM
m tfH-1 + (n + l )a - 6 + QL±^a»-W + (» + 1 M;- 1) ^268. + .... (2 )
1 • 2 1 • 2 * O
It will be observed that (2) is identical with (1) excepting
that every n of (1) is replaced by n -j- 1 in (2). That is, we
assumed the laws of 425 to be true for n, and have shown them
to be true for n + 1. Similarly, we might show the laws to be
true for n -f- 2, and so on, indefinitely. Hence, the laws of Art.
425 being true for the 4th power may be shown to hold true for
the 5th power ; and holding for the 5th power, may be shown
to hold for the 6th power.
Therefore, for any positive integral values of n :
(a + b)» = a« + na*-*b + HCEzJQ a «-2 b 2 + n(n-l)(n-2) ari _ 31)8
1-2 1-2-3
428. The Factorial Denominator.
In practice it is convenient to write (_3_ f or 1 • 2 • 3, [6_ for
1 . 2 • 3 • 4 • 5 • 6, etc.
We read [3 as "factorial 3," [6_ as "factorial 6," etc. In
general, \n_ means the product of the natural numbers
1 . 2 • 3-4-5 • • • n inclusive.
429. An expansion of a binomial is a finite series when n is
a positive integer. For, in the coefficients, we finally reach a
factor, n — n, or 0. And the term containing this factor
disappears. Moreover, every following term contains this
factor, hence each term following disappears.
430. The Signs of the Terms in an Expansion.
If both a and b are positive in (a + 6) n , the signs of all the
terms in the expansion are positive.
If b is negative, that is, given (a— b), all terms involving
even powers of — b are positive, while all terms involving
odd powers of — b are negative. Therefore, the signs of the
terms of the expansion of (a— b) n are alternately + and — .
APPLICATIONS OF THE BINOMIAL FORMULA 361
APPLICATIONS OF THE BINOMIAL FORMULA
431. Illustrations :
1. Expand (2a + 3 6) 5 .
(2 a + 3 by = (2 ay +5(2 a)* (8 &) + ^J (2 a )«(8 6) 2 + f^| (2 a)«(S 6)3
1 • J 1 • 1 • o
1.2.8*4 N yv 1.2.3.4.5^ '
= 32 a 5 + 240 a 4 6 + 720 a 8 6 2 + 1080 a*& 4- 810 a6* + 243 b 5 . Result.
2. Expand /^-^Y.
Changing to a form best suited to the binomial formula,
(2 er*-af)«=(2 a-!) 6 -6(2 a-*)*(rft) + 15(2 a- 1 )*(a*) 2 -20(2 a- 1 ) 8 ^)*
+ 15 (2 a-i)2( a t)4 - 6 (2 a- 1 ) (a*) 5 + (a§)«
= 64 a" 6 - 192 a- 5 at + 240 a-*a% - 160 a~ 3 a 2 + 60 a~H% - 12 a-^r+a*
= 64_192 + 240_160 + 60a |_ 12a | + a4 ^
a<J aV a§ a
432. To find Any Required Term of (a + b) n .
In finding the rth or general term in an expansion, we ob-
serve the general formation of any term of (a -4- b) n .
If r be the number of the term required :
1. The exponent of b is less by 1 than the number of the
term.
2. The exponent of a is n minus the exponent of b.
3. The last factor in the numerator of the coefficient is
greater by 1 than the exponent of a.
4. The last factor in the denominator of the coefficient is
the same as the exponent of b.
362 THE BINOMIAL THEOREM
Therefore, In the rth term :
The exponent of b = r — 1.
The exponent of a = n — (r — 1) = n — r + 1.
The last factor of the numerator in the coefficient = n — r + 2.
The last factor of the denominator in. the coefficient = r — 1.
Or, the rth term = n(n - l)(n-2)... (n -r + S) ^-^^
IL=1
Illustration :
1. Find the 7th term of (2 a 2 - a;" 1 ) 10 .
We have r = 7, n = 10.
Then, (2 x 2 - ar 1 ) 10 = [ (2 x 2 ) + ( - ar*) ] 10 ,
whence, for the formula of Art. 360,
a = (2 ce 2 ) and 6 = (- ar 1 ).
The exponent of6 = r-l = 7-l=6.
The exponent ofa = w — r+l = 10 — 7 + 1=4.
The last factor of the numerator of the coefficient
= w~r-f2 = 10-7+2 = 5.
The last factor of the denominator of the coefficient
= r _l = 7_l = 6.
Then the 7th term = 10 • 9 ■ 8 . 7 ■ 6 • 5 (2 x 2)4(_ x -i)6
1.2.3.4.5.6 v : •■ ' v }
= 210 • 16 x 8 • x-*
= 3810 x 2 . Result.
The same principle enables us to find the number of a term containing
a required power of a letter when that letter occurs in both terms of the
given binomial.
2. Find the term containing a 11 in [2 sc 8 ] .
Let r = the number of the required term. Disregarding all but the
literal factors of the term, we may write :
X 11 = (z 3 ) 12 -'-+ 1 (ar- 2 ) r - 1 = (a;86-8r+8) (a;-2r+2) _ ^l-Sr..
Therefore, r s= 6, the number of the required term.
That is, on writing the 6th term, we shall find the exponent of x to be 11.
APPLICATIONS OF THE BINOMIAL FORMULA S6S
Exercise 131
Expand :
1. (a + x)\ 16. (ax- 1 — VoaF 1 ) 5 .
2 - 0-5) 4 . 1 7 . (2af*-3aTt)<\
3. (1-2 a) 5 - g y _
4. (3V-2y)» 18 ' (2^-W-l) 5 .
5. (4a-<. ^ 4
6. (2m 2 -3^. 19 - (^— ~5j'
7. (5**+3*)» /3Va;__2a^
8. (2a^-32/) 7 . ^ ^ a v|,
9. (a+Va) 6 . /aV=l , 3
22.
10. («Va5 4-^) 6 .
11. (3V^l-2) 5 .
12. (4V2-3V3)«.
13. (2aVx-Vx 2 )*. 23. U-*yJ?-x-W-a)
/ oV-1 3 \§
\2Vic y
- Nl- 2 ^) 4
14. (3 a -2-V- a 2 ) 4 .
15. (2mV2n-3) 5 .
Find the
25. 5th term of (a + a) 7 . 26. 8th term of (aV- ax) 1 *.
27. 7th term of ( V3a + » V2a) 10 .
28. 6th term of (f V* - x^V^l) 9 .
Find the number of the term containing
29. a 8 in (x> + x- 1 ) 10 . 32. X s in (^ + 3 aj"Y-
30. a 12 in (2 *" -jy ' 33. ar 16 in A* or 2 - |Y° •
31. xmfex- ?Y 3 • 34. x~» in f 3 a;" 1 - ~r J •
CHAPTER XXVIT
LOGARITHMS
433. By means of the exponents, 2, 3, 4, etc., we may express
certain numbers as exact powers of 10.
Thus, 100 = 10 2 ,
1000 = 103,
10000 = 10*, etc.
434. Suppose, now, that 10 is given any real exponent, as x.
Then some positive number, N } results as the #th power of 10.
That is, N=ltf.
435. If, therefore, we knew the necessary approximate values
for x, we might express any number as an approximate power
of 10.
436. By a method of advanced algebra, these approximate
values for x have been obtained. For example, it has been
found that
180 = io 2 - 2663 ,
4500 = 103-6532,
19600 = lO*^ 23 , etc.
437. These exponents are called the logarithms of the num-
bers they produce.
438. The exponent that must be given 10 in order to produce
a required number, JV", is called the logarithm of N to the base 10.
The expression 10* = N
is usually written x = logic -ZV»
and is read, M x is the logarithm of N to the base 10."
364
THE PARTS OF A LOGARITHM 365
The object and use of logarithms is to simplify numerical
work in the processes of multiplication, division, involution,
and evolution.
439. Logarithms to the base 10 are known as common log-
arithms, and are in universal use for numerical operations.
Unless otherwise stated, the discussions of this and subsequent chap-
ters refer to common logarithms.
Any positive number, except unity, may be taken as the base of a
system of logarithms.
A negative number is not considered as having a logarithm.
THE PARTS OF A LOGARITHM
440. Consider the results in the following :
IO 3 = 1000, log 1000 = 3,
IO 2 = 100, log 100 = 2,
101 = io, log 10 =1,
10° =1, logl =0,
10-i = .1, log.i = _l,
IO" 2 = .01, log .01 = - 2,
IO" 8 = .001, log .001 = - 3 ; etc.
From these results it is evident that
(1) The common logarithm of a number greater than 1 is
positive.
(2) The common logarithm of a number between and 1 is
negative.
» (3) The common logarithm of an integral or a mixed number
between 1 and 10 is + a decimal,
between 10 and 100 is 1 + a decimal,
between 100 and 1000 is 2 + a decimal, etc.
(4) The common logarithm of a decimal number
between 1 and 0.1 is — 1 -f- a decimal,
between 0.1 and 0.01 is — 2 + a decimal,
between 0.01 and 0.001 is — 3 4- a decimal, etc.
366 LOGARITHMS
441. The integral part of a logarithm is the characteristic.
The decimal part is the mantissa.
In log 352 = 2.5465, 2 is the characteristic and .5465 is the mantissa.
I. To obtain the characteristic of a logarithm.
(a) When the given number is integral or mixed.
442. By Art. 440 (3), the characteristic of the logarithm of a
number having one digit to the left of the decimal point is 0,
of a number having two digits to the left of the decimal point
is 1, and of a number having three digits to the left of the
decimal point is 2. In general,
The characteristic of the logarithm of a number greater than
unity is 1 less than the number of digits to the left of the decimal
point.
(b) When the given number is a decimal.
443. By Art. 440 (4), the characteristic of the logarithm of a
decimal having no cipher between its decimal point and its
first significant figure is — 1, of a decimal having one cipher
between its decimal point and its first significant figure is — 2,
and of a decimal having two ciphers between its decimal point
and its first significant figure is — 3.
In order to avoid writing these negative characteristics — 1,
— 2, — 3, etc., it is customary to consider that
- 1 = 9 - 10,
- 2 = 8 - 10,
- 3 = 7 - 10, etc.
With the negative results written in this form, we have, in
general,
The characteristic of the logarithm of a number less than unity
is obtained by subtracting from 9 the number of ciphers between
its decimal point and the first significant figure, annexing — 10
after the mantissa.
USE OF FOUR-PLACE TABLES 367
II. To obtain the mantissa of a logarithm.
(a) Important principle governing the finding of all mantissas.
444. It has been computed that the logarithm of 35,200 is
4.5465, and from our discussion we have seen that the logarithm
of 100 is 2. We may write, therefore,
104.5465 = 35200 (1)
and 10 2 = 100. (2)
These logarithms being, by definition, exponents, we may
treat them as such in the following operation.
Dividing (1) by (2), we have :
104.5466 35200
10 2 100
From which,
102.5465 = 352.
That is,
log 352 = 2.5465.
Clearly, therefore, the mantissas of the logarithms of 352
and 35,200 are equal.
In like manner, we may show that
In general :
log 35.2 = 1.5465,
log 3.52 = 0.5465,
log .352 = 9.5465 - 10, etc.
If two numbers differ only in the position of their decimal points,
their logarithms have the same mantissas.
THE USE OF THE FOUR-PLACE TABLE
445. The table of logarithms is used for two distinct and
opposite operations.
(1) Given a number, to find the corresponding logarithm.
(2) Given a logarithm, to find the corresponding number.
368 LOGARITHMS
I. To find the logarithm of a given number.
446. (a) Numbers having three figures.
Illustrations :
1. What is the logarithm of 247 ?
On page 370, in the column headed "N," we find "24," the first two
figures of the given number.
In the same horizontal line with 24, under the heading corresponding
to the last figure of the given number, "7," we find the mantissa, 3927.
Since the given number has three figures to the left of the decimal
point, the required characteristic is 2 (Art. 442).
Therefore, log 247 = 2.3927. Result.
2. What is the logarithm of .0562 ?
Opposite "56" of the "N" column (p. 371), and under "2," we
find the mantissa 7497.
Since the given number is a decimal and has one cipher between its
decimal point and its first significant figure, we subtract 1 from 9, and
annex — 10 to the mantissa (Art. 443).
Therefore, log .0562 = 8.7497 - 10. Result.
(b) Numbers having two figures.
Illustrations :
1. What is the logarithm of 76 ?
Opposite "76" of the "N" column, and under "0," we find the
mantissa 8808. The characteristic is 1 (Art. 442).
Therefore, log 76 = 1.8808. Result.
2. What is the logarithm of .0027 ?
Opposite " 27 " of the " N " column, and under "0," we find the man-
tissa 4314. The characteristic is (9—2), or 7, with —10 annexed
(Art. 443).
Therefore, log .0027 = 7.4314 - 10. Result.
INTERPOLATION 369
(c) Numbers having onejlgure.
Illustrations :
1. What is the logarithm of 7 ?
Opposite "70" of the"N" column, and under "0," we find the
mantissa 8451. The characteristic is (Art. 442).
Therefore, log 7 = .8451. Result.
2. What is the logarithm of .00008 ?
Opposite "80" of the "N" column, and under "0," we find the
mantissa 9031. The characteristic is (9 — 4), or 5, with — 10 annexed
(Art. 443).
Therefore, log .00008 = 5.9031 - 10. Result.
447. It is evident that the logarithms of numbers having
one or two figures have mantissas from the " " column of the
table.
Exercise 132
K
nd the
logarithms of the following numbers :
l.
124.
7. 84.2.
13. 6.73.
19.
.0642.
2.
283.
8. 39.6.
14. .829.
20.
.0006.
3.
589.
9. 2.85.
15. .342.
21.
.00016.
4.
676.
10. 6.76.
16. .676.
22.
.0676.
5.
643.
11. 3.70.
17. .037.
23.
.00809.
6.
540.
12. 5.89.
18. .0681.
24.
.00000734.
INTERPOLATION
448. Interpolation is based upon the assumption that the dif-
ferences of logarithms are proportional to the differences of
their corresponding numbers. While the assumption is not
absolutely correct, the results obtained are exceedingly close
approximations. The process is necessary in obtaining the
logarithms of numbers having four places by means of the
four-place table.
SOM. EL. ALG. — 21
370
LOGARITHMS
N
1
2
3
4
5
6
7
8
9
10
0000
0043
0086
0128
0170
0212
0253
0294
0334
0374
11
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1106
13
1139
1173
1206
1239
1271
1303
1335
1367
1399
1430
14
15
1461
1492
1523
1553
1584
1614
1644
1673
1703
1732
1761
1790
1818
1847
1875
1903
1931
1959
1987
2014
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
20
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
24
25
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
.4200
4216
4232
4249
4265
4281
4298
27
4314
4330
4346
4362
4378
4393
4409
4425
4440
4456
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
29
30
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4942
4955
4969
4983
4997
5011
5024
5038
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
34
35
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
36
5563
5575
5587
5599
5611
5623
5635
5647
5658
5670
37
5682
5694
5705
5717
5729
5740
5752
5763
5775
5786
38
5798
5809
5821
5832
5843
5855
5866
5877
5888
5899
39
40
5911
5922
5933
5944
5955
5966
5977
5988
5999
6010
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
44
45
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
49
50
6902
6911
6920
6928
6937
6946
6955
6964
6972
6981
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093
7101
7110
7118
7126
7135
7143
7152
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7235
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
54
7324
7332
7340
7348
7356
7364
7372
7380
7388
7396
1
2
3
4
5
6
7
8
9
FOUR-PLACE TABLE
371
N
1
2
3
4
5
6
7
8
9
55
7404
7412
7419
7427
7435
7443
7451
7459
7466
7474
50
7482
7490
7497
7505
7513
7520
7528
7536
7543
7551
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
60
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
7853
7860
7868
7875
7882
7889
7896
7903
7910
7917
62
7924
7931
7938
7945
7952
7959
7966
7973
7980
7987
63
7993
8000
8007
8014
8021
8028
8035
8041
8048
8055
64
65
8062
8069
8075
8082
8089
8096
8102
8109
8116
8122
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
8215
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8287
8293
8299
8306
8312
8319
68
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
69
70
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
72
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
75
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
79
80
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
9085
9090
9096
9101
9106
9112
9117
9122
9128
9133
82
9138
9143
9149
9154
9159
91(55
9170
9175
9180
9186
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
85
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
89
90
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
92
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
93
9685
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
95
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
97
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
N
1 o
1
2
3
4
5
6
7
8
9
372 LOGARITHMS
Illustrations :
1. What is the logarithm of 6874 ?
Since the table gives mantissas for numbers having but three places, we
will consider that we are seeking the logarithm of 687.4. (See Art. 444.)
From the table we find
the mantissa of 687 = .8370
the mantissa of 688 = .8376
Difference = .0006
That is, an increase of 1 in the number causes an increase of .0006 in
the mantissa. Therefore, an increase of .4 in the number (for 687.4 is
.4 greater than 687) will cause an approximate increase in the mantissa
of .0006 x .4 = .0002.
(The fifth decimal place being disregarded if less than .5 of the fourth.)
Hence, for the mantissa of the logarithm of 687.4, we have
.8370 + .0002 = .8372.
Therefore, with the proper characteristic,
log 6874 = 3.8372. Result.
2. What is the logarithm of 23.436?
Consider the number to be 234.36.
From the table we find
the mantissa of 234 = .3692
the mantissa of 235 = .3711
Difference = .0019
Then, as above, .36 x .0019 = .000684,
or, approximately, = .0007.
Therefore, the mantissa of the logarithm of
234.36 = .3692 + .0007 = .3699.
Whence, log 23.436 = 1.3699. Result.
Exercise 133
Eind the logarithms of the following numbers :
1. 2762. 5. 8625. 9. 9824. 13. 68.741.
2. 3894. 6. 8.642. 10. .06421. 14. 430.05.
3. 2007. 7. 726.4. 11. .005432. 15. .0006941.
4. 6492. 8. 54.29. 12. 44.212. 16. .0004682.
THE ANTILOGARITHM 373
II. To find the number corresponding to a given logarithm.
449. The number to which a given logarithm corresponds
is called its antilogarithm.
450. When the mantissa of a given logarithm is found in
the four-place table, the antilogarithm is readily found, and if
the mantissa given is not found in the table, a close approxi-
mation for the required antilogarithm is obtained by a process
of interpolation.
Illustrations :
1. What is the number whose logarithm is 1.4669 ?
Look in the table for the mantissa .4669.
In the same horizontal line with this mantissa and in the " N " column,
we find "29," the first two figures of the required number.
The mantissa is found under the "3."
Therefore, the required sequence of figures is 293.
Now the given logarithm has a characteristic, 1. Therefore (Art. 442),
there must be two figures to the left of the decimal point in the required
number. Hence, the antilogarithm of 1.4669 = 29.3. Result.
2. What is the antilogarithm of 7.9112 -10 ?
The mantissa, .9112, lies in the horizontal line with " 81 " and under " 5."
The sequence of figures is, therefore, 815.
Since the characteristic is 7 — 10, the number has two ciphers between
the decimal point and the first significant figure (Art. 443).
Therefore, the antilog of 7.9112—10 = .00815. Result.
3. What is the antilogarithm of 2.8828 ?
The mantissa, .8828, is not found in the table, but lies between
the mantissa of log 763 = .8825
and the mantissa of log 764 = .8831
The difference of these mantissas = .0006
That is, an increase of .0006 in the mantissas causes an increase of 1 in
their antilogarithms.
Now between the given mantissa and the next lowest mantissa there
exists a difference of
.8828 - .8825 = .0003-
374 LOGARITHMS
Therefore, an increase of .0003 in the mantissas will cause an approxi-
mate increase in the antilogarithms of
.0003 _ 5
.0006""' '
Hence, .8828 is the mantissa of the log 763.5.
Or, antilog 2.8828 = 763.5. Result.
Exercise 134
Find the antilogarithms of the following logarithms :
1. 1.6085. 8. 0.1088. 15. 8.7219-10.
2. 1.7284. 9. 3.9799. 16. 7.9007-10.
3. 2.9345. 10. 2.6897. 17. 6.7370-10.
4. 3.8317. 11. 4.9733. 18. 9.6477-10.
5. 0.5509. 12. 9.8987-10. 19. 5.8566-10.
6. 3.5974. 13. 9.7306-10. 20. 9.9995-10.
7. 2.0095. 14. 8.9099-10. 21. 4.9543-10.
THE PROPERTIES OF LOGARITHMS
451. In any system, the logarithm of 1 is 0.
For, a = 1. (Art. 241)
Whence, log 1=0. (Art. 438)
452. In any system, the logarithm of the base is 1.
For, a 1 = a.
Whence, log a = 1. (Art. 438)
While, for convenience, the base of the common system is
used in the following discussions, each theorem is a general
one for any base, a, hence, for any system.
453. The logarithm of a product is equal to the sum of the
logarithms of its factors.
THE PROPERTIES OF LOGARITHMS 375
Let 10* = to (1) and 10^ = n. (2)
Multiplying (1) by (2), \v x +v = mn.
Whence, log mn = x + y. (3) (Art. 438)
From (1) and (2), x = log to and y = log n. (Art. 438)
Substituting in (3), log mn = log to + log n.
454. The logarithm of a quotient is equal to the logarithm of
tlie dividend minus the logarithm of the divisor.
Let 10* = m (1) and 10*/ = n. (2)
Dividing (1) by (2),
iv _ m
10» ~~ n
That is,
10*-y = Tl.
n
Whence,
log^ = x-y. (3)
n
(Art. 438)
But, from (1) and (2),
x = log m and y = log n.
(Art. 438)
Hence, in (3),
log — = log m — log n.
ft
455. The logarithm of a power of a number is equal to the
logarithm of the number multiplied by the exponent of the power.
Let 10* = m. (1)
Raising both members of the equation to the pth. power,
IOp* = mP.
Whence, log mP = px.
But, from (1), x = log m, (Art. 438)
Hence, . log m" =p(log m).
456. TJie logarithm of a root of a number is equal to the
logarithm of the number divided by the index of the root.
Let 10* == w. (1)
Raising both members to the power indicated by-,
x 1
10* = TO*.
1
Whence, log m q = -. (Art. 438 )
But, from (1), x = log to, (Art. 438)
- _ log to
Therefore, log m q — — - —
376 LOGARITHMS
457. These demonstrations may be briefly summarized as
follows :
(1) To multiply numbers, add their logarithms.
(2) To divide numbers, subtract the logarithm of the divisor
from the logarithm of the dividend.
(3) To raise a number to a power, multiply the logarithm of
the number by the exponent of the required power.
(4) To extract a root of a number, divide the logarithm of the
number by the index of the required root.
The antilogarithm of a result obtained by one of these processes
is the required number.
THE COLOGARITHM
458. The cologarithm of a number is the logarithm of the
reciprocal of the number. By its use an appreciable saving
of labor is made in computations with logarithms.
By definition, I
cologiV=log—
2v
= log 1 — log N
(Art. 454)
= 0- log .2V
(Art. 451)
= -log2V. .
To avoid this negative form we may write,
colog JV= 10 - log 2V— 10.
459. In general, to obtain the cologarithm of a number :
Subtract the logarithm from 10 — 10.
In practice the subtraction is usually accomplished by be-
ginning at the characteristic and subtracting each figure from
9, excepting the last significant figure, which is subtracted from
10.
LOGARITHMS IN COMPUTATIONS 377
Illustration :
Find the cologarithm of 16.
The log of 1 is 0, which we may write as 10 — 10.
Then, colog 16 = log 1 — log 16.
log 1 = 10 - 10
log 16 = 1.2041
Subtracting, colog 16 = 8.7959 — 10. Result.
460. If the characteristic of a logarithm is greater than 10
but less than 20, we use in like manner,
colog 2V = 20 - log N— 20.
Similarly, 30 — 30, 40 — 40, etc. , may be used if necessary.
461. Negative factors in computations with logarithms. The
logarithms of negative factors in any group are found without
regard to the negative signs, and the result is positive or nega-
tive according as the number of negative factors is even or
odd.
USE OP LOGARITHMS IN COMPUTATIONS
462. Illustrations :
1. Multiply 6.85 by 37.8.
log 6.85 = 0.8357
log 37.8 = 1.5775
(Art. 453), 2.4132 = log of product,
antilog 2.4132 = 258.9. Result.
2. Divide 30,400 by 1280.
log 30400 = 4.4829
log 1280 = 3.1072
(Art. 454) , 1 . 3757 = log of quotient,
antilog 1.3767 = 23.75. Result.
3. Divide 2640 by 36,900.
log 2640 = 3.4216
(Art. 458), colog 36900 = 5.4330 - 10
8.8546 - 10 = log of quotient,
antilog 8.8546 - 10 = 0.07155. Result.
378 LOGARITHMS
4. What is the value of ^846"?
log 846 = 2.9274.
(Art. 456) 3)2.9274
.9758 = log of cube root,
antilog .9758 = 9.458. Result.
5. Find the value of ^.00276.
log .00276 = 7.4409 - 10.
When a negative logarithm occurs in obtaining a root, the characteris-
tic must be written in such a form that the number subtracted from the
logarithm shall be 10 times the index of the root. The divisor in this
case being 6, we change the form of the logarithm by adding 50 — 50,
and the subsequent division gives the required negative characteristic,
log .00276 = 7.4409 - 10.
Adding, 50 - 50
57.4409-60
(Art. 456) 6 )57.4409-60
9.5735 - 10 = log of sixth root,
antilog 9.5735 - 10 = .3745. Result.
432 2 x27.1 3
6. Find the value of
35.1 4
(Art. 455) 2 log 432 = 2 (2.6355) = 5.2710
(Art. 455) 3 log 27 . 1 = 3 (1 .4330) =4 .2990
(Arts. 455, 458) 4 colog 35.1 = 4 (8.4547 -10) = 3.8188 - 10
3.3888 = log of result,
antilog 3.3888 = 2448. Result.
7. Simplify V/ 2720* X V288 X 432* .
.068 7 X V27.8 3 x 624 2
\ log 2720 a | (3.4346) = 1.7173
| log 288 z= i(2.4594) =1.2297
-flog 432 =|(2.6355) =3.5140
7 colog .068 = 7 (1.1675) =8.1725
f colog 27.8 = | (8.5560 - 10) = 7.8340 - 10
2 colog 624 = 2 (7.2048 -10 ) = 4.4096 - 10
6.8771
(Art. 456) 4)6.8771
1.7193, log of result,
antilog 1.7193 = 52.40. Result.
MISCELLANEOUS APPLICATIONS OF LOGARITHMS 379
1. 4600 x. 85.
2. 72x380.
3. .28 x. 00012.
4. .017 X .0062.
5. 4.96 x 58.4.
6. .00621 x .000621
7. 73400 x .00811.
8. .0293 x .000602.
9. 691 x .0000131.
624 x 372 x 891
19
20.
21.
Exercise 135
e value of :
10.
1280 -r
-.0064.
11.
68.5 +
6.12.
12.
2.741 -
- .00822.
13.
.00461
- .0931.
14.
.07241
-r- .3623.
15.
(2741
K 3.623) -
-242.
16.
(4.625
X .5821) - 2.067.
17.
34.74 -
- (2.851 x
4.309).
18.
6.904-
- (3.676 x
.00275^
.0372
X .584 x .
30027
457 x 196 x 583
43.2 x 3.28 x .246
.537x3.41x56.8'
630 x 2100 x .007
23.
24.
3.25 X 472 x 6500
25. 4.52 2 . 28. ^377.
26. 2.74 5 . 29. VWM.
27. .0276 4 . 30. a/,00724.
.273 x .00042 x .0121
.007 x .07 x .7 X 7
35.7 x 7.14 x .1428*
.643 X. 0468x2760
346 x .0072 x .01 '
31. 3.207f.
32. 5.602*.
33. .0007544.
34.
4
,0027 3 x5.82 4 x762 3
3.207 4 x6.42 3 x26.7 2
35.
4
V.2809 x ^.003241
V^962X ^.000011
MISCELLANEOUS APPLICATIONS OF LOGARITHMS
463. I. Changing the base of a system of logarithms.
Let a and b represent the bases of two systems of logarithms, and m
the number under consideration.
We are to show that
log b m
l0ga&
380 LOGAKITHMS
Let a* = m (1)
and bv = m. (2)
Then, ' x = \og a m and y = log b m. (Art. 438)
From (1) and. (2), & = W .
X
Extracting the yth root, a* = b.
Therefore, log«& = - or y = — —
V log a b
That is, lo g6 m=^^.
log a 6
Illustration:
1. Find the logarithm of 12 to the base 5.
By Art. 463, log 6 12=^ "
logi 5
= 1.0792
.6990
= 1.5439. Result.
II. Equations involving logarithms.
464. An equation in which an unknown number appears as
an exponent is called an exponential equation. Thus : a x = b
is a general form for such equations.
Illustrations :
1. Solve the equation, 4 X = 64.
If, in an equation in the form of a x — b, b is an exact power of a, the
solution is readily obtained by inspection.
From 4* = 64,
we have 4* = 4 3 .
Therefore, x = 3. Result.
2. Find the value of x in 5* = 13.
Using logarithms, * log 5 = log 13.
x
log 5
1.1139
.6990
1.5935. Result.
MISCELLANEOUS APPLICATIONS OF LOGARITHMS 381
3. Solve the. equation 2 s/x = VS.
8 /<r 3$
2 Vce = y/3, \Jx = , x = — (changing form and squaring) .
A 4
Then,
log x = | log 3 + colog 4
f log3 = f (.4771) = .3181
colog 4= 9.3979-10
. Therefore, 9.7160 - 10 =
: log Of X
x = .52. Result.
465. Certain forms of equations involving logarithms may-
be so transformed as to give results without the aid of loga-
rithm tables.
1. Find a if log x 33 = 5. 2. Find x if log 3 x = 4.
By logarithms, 32 = x 5 . By logarithms, x = 3 4 .
Whence, x = 2. Result. Whence, x = 81. Result.
(That is, the base of that system (That is, the number, whose log
in which the log of 32 is 5, is 2.) to the base 3 is 4, is 81.)
III. Use of formulas for compound interest and annuities.
466. If P is a given principal, n the number of years during
which the interest is compounded annually, and r the given
rate per cent, the amount, A, can be obtained from the formula
A = P(l + r)\
Illustration :
1. Find the amount of $ 1500 for 8 years at 5 %, compounded
annually.
In the formula
A = P(l + r)».
By substitution,
A = 1500 (1.05)8.
By logarithms,
log A = log 1500 + 8 (log 1.05)
log 1500 = 3.1761
8 (log 1.05) = 8 (.0212) = .1696
log .4= 3.3457
A = $2216.50. Result.
382
LOGARITHMS
467. An annuity is a fixed sum of money payable yearly,
or at other fixed intervals.
If the amount of an annuity is represented by A, the number
of yearly payments by % and the rate of money at the present
time by r, the present value of the annuity, P, can be obtained
from the formula
p=A
1
Illustration :
2. Find the present value of annuity of $ 900 for 20 years
at 4%.
In the formula,
By logarithms,
Hence,
Then,
By logarithms,
r L (1 + r-)2°J .04 L
(1.04)'
20 (log 1.04) = 20 (.0170)
(1.04)20 - 2.1879.
P = 9oor 1 1
.04 L 2.1879
-.3400.
3=
900 x 1.1879
.04 x 2.1879 '
log 900 =2.9542
log 1.1879= .0747
colog.04 =1.3979
colog 2.1879 = 9.6600 -10
4.0868 = log P
P= $12,211, approximately. Result.
Calculate :
1. log, 12.
2. log 6 28.
3. log 7 42.
4. log 12 54.
Solve :
13. a; = 12 VR
14. xV5 = 70.
Exercise 136
5. log 15 60.
6. log 3 256.
7. log 8 9.6.
8. log 16 .7.
15. x-V3=</7.
16. 12vz = 19.
9. log L6 .48.
10. lo glM .416.
11. log. 5 .875.
12. log. 9 .007.
17. 2V3^=5V^
18. 10 a;"* = 64.
MISCELLANEOUS APPLICATIONS OF LOGARITHMS 383
19. -v^TaJ 2 =VI(J. 23. 5 Z = 20. 28. 5 X " 1 = 35.
24. 24* = 100. 29. 16* = 64- 1 .
25. 3.5* = 170. 30. (-J )- = 16.
21. 5* = 125. 26 12.4* = .124. 31. 25* +2 = 30*" 1 .
22. 3- = 729. 27. 3* +1 = 12. 32. 1.62 x " 1 =3.24.
20. (3a>)*=</20.
Given: log 2 = .3010, log 3 = .4771; find logarithms of the
following :
33. log 6, log 18, log 72, log 2.88.
34. log 3 V2, log 4 V3, log V60, log ^045.
35. log 270, logl6f, log2i logl2 2 .
36. log^ -, log^a, log (2* -*- 3* x 6*).
.Find x if
37. log x 27 = 3. 40. log x 125 = -3.
38. log z 16 = 4. 41. log, 64 = If
39. log x 216=3. 42. log.JV = -f
Find x if
43. logs & = 5. 46. log 49 a? = — J.
44. log 4 a; = 6. 47. log 8 » = l|.
45. log25a; = i 48. log 27 a; = --|.
What is the expression for x in :
49. a x = c 2 . 52. a x ~ 1 = c\
50. 2a x = mw a! . 53. ac x = mn x+2 .
51. 4c 8 = oflf. _ 54. c*m = n i y +1 .
55. Find the amount of $ 600 for 10 years at 4 % compound
interest.
56. Find the amount of $ 7000 for 15 years at 4 % compound
interest.
57. In what time will $ 5000 amount to $ 7500 at 5 % com-
pound interest?
384 LOGARITHMS
58. A man buys an annuity of $ 600 for 20 years. If money
is worth 4 %, what amount should he pay for it ?
59. What is the cost of an annuity of $ 1000 per year for 10
years, money being worth 5 % ?
60. Find the present value of an annuity of $1200 for 15
years at 5%.
61. Write as a polynomial, log x (x 2 — l) 2 (x + l) -2 .
62. Change 2 log a -f log b — log 3 — \ log c to an expression
of a single term.
63. Find an expression for x in m** 1 =p 2 .
64. If log 429 = 2.6325, and log 430 =2.6335, find log 429.3.
What is the log .004293?
65. If log 864 = 2.9365, and log 8.65 = 0.9370, find log
.08642. What is the log 864.2 ?
66. How may you obtain log 3, log 9, log 243, and log .9 if
you know log 81 ?
67. How many digits in 25 25 ?
68. Find the values of x and y in the equations 3 X+V = 27
and S 2x ~» = 512.
69. If a, b, and c are the sides of any triangle, and s is one
half their sum, the area of the triangle may be obtained from
the formula, A = Vs (s — a)(s — b)(s — c). Write this expres-
sion in logarithmic form.
70. With the formula of example 69 find the area of a
triangle whose sides are 65 feet, 70 feet, and 75 feet respect-
ively.
71. The diameter of a circle circumscribing an equilateral
triangle equals f the altitude of the triangle. If h is the alti-
tude of an equilateral triangle, the area may be obtained from
the formula, A = |7i 2 V3. Using logarithms, find the area of
the three segments cut from a circle by an equilateral triangle
whose altitude is 9 feet.
CHAPTER XXVIII
SUPPLEMENTARY TOPICS
THE REMAINDER THEOREM
468. If any polynomial of the form, C^ + C^c n ~ l + C<p?~ 2 -\
C n , be divided byx — a, the remainder will be Qa" + C 2 a n ~ l + C s a n ~ 2
+ • • • C n ; which expression differs from the given expression in
that a takes the place of x.
Proof: Let Q denote the quotient and B the remainder when
Cix n + C2X 11 - 1 + Czx n - 2 H C n is divided by x - a.
Continue the division until the remainder does not contain x.
Then, Q(x - a)+ B = dx* + C 2 x n ~ l + C 3 x n - 8 + ••• C„.
Since this identity is true for all values of x, let x = a.
Then, Q(a - a)+ B = da n + C^a"- 1 + C 3 a n - 2 + ••• C„.
And, B = Cia n + C 2 a n_1 + C s a n -* + ••• C„.
Application :
1. Without division obtain the remainder when 7^ + 3^
— 13 x -|- 8 is divided by x — 2.
We have given, $e — a = x — 2, hence a = 2.
Hence, 22 = 7 • 2^ + 3 • 2 2 - 13 -2 + 8 = 60. Result.
THE FACTOR THEOREM
469. If any rational and integral expression containing x be-
comes . equal to when a is substituted for x, the expression is
exactly divisible by x — a.
Proof : Let E be the given expression, and let E be divided by x — a
until the remainder no longer contains x. Let Q denote the quotient ob-
tained and B the remainder.
som. el. alg. — 25 386
386 SUPPLEMENTARY TOPICS
Then, E = Q(x -a) + B.
This identity being true for all values of x, we may assume that x = a.
By the hypothesis the substitution of a for x makes E equal to 0.
Therefore, = E(a - a) + B, = + B, B = 0.
Therefore, the remainder being 0, the given expression is exactly
divisible by x — a.
Or, x — a is a factor of the given expression, E.
Illustrations :
1. Factor X s - 19 x + 16.
By trial we find that the expression, x 8 — 12 a; +16, equals when x = 2.
Therefore, if x = 2, x — a = x — 2, and x — 2 is a factor.
Then (z 3 - 12 x + 16) -h (x - 2) = x 2 + 2 g - 8.
The factors of x 2 + 2 x — 8 are found to be (x + 4) and (x — 2).
Therefore, x 3 - 12 x + 1 6 = (x - 2)(x + 4)(x - 2). Result.
2. Factor x*+9x* + 23x -f- 15.
(In an expression whose signs are all plas, it is evident that no positive
number can be found the substitution of which will make the expression
equal to 0.)
By trial we find that if x = — 1, the expression becomes 0.
Hence, if x = — 1, x — a=[x — ( — 1)] = x + 1, a factor required.
Then, (x 8 + 9x 2 + 23 x + 15) + (z + 1) = x 2 + 8x + 16.
Furthermore, x 2 + 8 x + 15 = (x + 3) (x + 5).
Therefore, x 8 + 9 x 2 + 23 x + 15 = (x + 1) (x + 3) (x + 5) . Result.
Exercise 137
Without division show that
1. a 3 + 3 a 2 4- 3 a + 2 is divisible by a + 2.
2. ^-8^ + 24^-3205 + 16 is divisible by x — 2.
3. c 4 -c 8 -8c 2 + 9c-9is divisible by c + 3.
4. 27a 3 + 9a 2 -3a-10 is divisible by 3a-2.
5. m 5 — 5 m 4 + 9 m 3 — 6 m 2 — m + 2 is divisible by m — 2.
DIVISORS OF BINOMIALS 387
Factor :
6. c 8 + c-2. 12. c 8 + c 2 -5c + 3.
7. a?-7x + 6. 13. rf + Zrf-bx-Z.
8. m 3 -8m + 3. 14. m 3 + 5m 2 -13m + 7.
9. a 8 + 3a 2 + 7a + 5. 15. 2 2/ 3 + 3 1/ 2 — 3 y — 2.
10. ^ + 4^ + 5^ + 2. 16. 3a 3 -10a 2 -f 4a + 8.
11. m 3 -2m 2 -7m-4. 17. x^-x^-Sx 2 -\-5x-~2.
THE THEORY OF DIVISORS OF BINOMIALS
470. The following proofs depend directly upon the prin-
ciples established in Arts. 468 and 469.
If n is a positive integer, we may establish as follows the
divisibility of the binomials, x n — y n and af -j- y n , by the bino-
mials, x — y and x-^-y.
I. x n — y n is always divisible by x — y.
For, if y is substituted for x in x n — y n , we have
3jn _ y» = yn __ yn — Q^
Therefore, x — y is always a divisor of cc n — y n .
II. £c n — ?/ n is divisible byx + yifn is even.
For, if — y is substituted for xm x n — y n , we have
x n — y n =(—y) n — y n = y n — y n = 0.
Therefore, x + y is a divisor of x n — y n when n is even.
III. a? + y n is never divisible by x — y.
For, if y is substituted for x in cc n -f y», we have
x n _j_ yn — yn .J. yn — 2 y«.
which result does not reduce to by the substitution.
Therefore, x — y is never a divisor of x n + y n .
IV. x n -f- ?/ w ?'s divisible byx + yifnis odd.
For, if — y is substituted for x in se n 4- 2/*, we have
£« _j_ ^n _ (_ y)n _|_ yn _ 0.
Therefore, x + 2/ is a divisor of # + ?/ when n is odd.
388 SUPPLEMENTARY TOPICS
These results may be summarized with the corresponding
quotients as follows :
— — IC ss x"- 1 + x n ~ 2 y + x ro -% 2 -|- . . • + w"- 1
x-y
x n — y n _ ^ n _ 1 _ x n-2y _j_ x n ~ 8 y 2 — • • . — y*- 1
x + y
x — y
x + y
when n is even.
>, when n is odd.
Illustrations :
1. Divide a?° + y w by tf + tf.
We may write x 10 + y™ = (x 2 ) 6 + (y 2 ) 5 .
Then, (x 2 ) 6 + (y 2 ) 6 is divisible by a 2 + y 2 . (Art. 470, IV.)
Whence,
(S 2 )6+y) 6 =(a . 2)4 _ (a . 2) 3 (y2) + (^ )2(2/2) 2 _ (^)(y«)8 + (y2)4
a;-* + y-*
= x 8 — xV -I- x 4 */ 4 - xV + f. Result.
2. Divide 64 a 12 - tf by 2 a? - */.
We may write 64 x 12 — y* = (2 x 2 ) 6 — y*.
Then,
(2 x 2 )6 - y6 _ ^ + ^ + ^ ^^ + ^ ^ g + ^ x2)y4 + y6
2^ — y
= 32 x 10 + 16 x 8 y + 8 xfy 2 + 4 x 4 y 3 + 2 xV + y 5 . Result.
Exercise 138
Divide :
1. a? + 32y 5 byx + 2y.
2. a? 7 -128bya;-2.
3. 32^ + 243 by 2o; + 3.
4. ^-729 by --S.
x 6 x
5. What are the exact binomial divisors of 81 — a* ?
HIGHEST COMMON FACTOR 389
6. What are the exact binomial divisors of x 10 — y 10 ?
7. What are the exact binomial divisors of x 2 — 64 c 6 ?
8. Obtain the three factors of x 9 — y 9 .
9. Obtain the four factors of X s — y 8 .
10. . Obtain the six factors of x u — 4096.
11. Obtain the ten factors of a 12 — 64 a 6 .
THE HIGHEST COMMON FACTOR OF EXPRESSIONS NOT READILY
FACTORED
471. The method of obtaining the highest common factor of
two expressions that cannot be factored by inspection is analo-
gous to the process of division used in arithmetic for obtaining
the greatest common divisor of two numbers.
The principles involved in the algebraic process are estab-
lished as follows :
Let A and B represent two expressions, both arranged in the descend-
ing powers of some common letter ; and let the degree of A be not higher
than the degree of B. Let B be divided by A, the quotient being Q and
the remainder B.
Thus, A)B (Q
A$
B
It follows, therefore, that B = AQ + B, (1)
and B = B-AQ. (2)
Now a factor of each of the terms of an expression is a factor
of the expression.
We make, therefore, three important statements :
(1) Any common factor of A and R in (1) is a factor of
AQ + E, hence a factor of B. Or,
A common factor of A and R is a common factor of B and A.
(2) Any common factor of B and A in (2) is a factor of
B— AQ, hence a factor of R. Or,
A common factor of B and A is a common factor of A and R.
390
SUPPLEMENTARY TOPICS
(3) Hence, the common factors of B and A are the same as
the common factors of A and R. Or,
The H. G. F. of A and B is the H. G. F. of A and R.
Each succeeding step of the process may be proved in a
similar manner.
If, at any step, the remainder becomes 0, the divisor is a
factor of the corresponding dividend and is, therefore, the
H. G. F. of itself and the corresponding dividend. Or,
The last exact divisor is the required highest common factor.
472. We may multiply or divide either given expression, or may
multiply or divide any resulting expression, by a monomial that
is not a common factor of both expressions.
For the process refers to polynomial expressions only, and
the introduction or the rejection of monomial factors not com-
mon to both expressions cannot affect the common polynomial
expression sought.
Such factors are introduced or rejected merely for con-
venience, as will be shown in the following illustrations.
In finding the highest common factor of three or more expres-
sions, as A, B, and G, we first find F 1} the highest common
factor of A and B. It remains to find the highest common
factor of F 1 and G. Let this result be F 2 . Then F 2 is the
required highest common factor of A, B, and G.
Illustrations :
Find the H. C.F. of 2 a*-a*+ x -6 and 4 a*+2 a^-lO x-3.
The following arrangement is universally accepted as most satisfactory :
2 as 8 — x 2 + x- 6
2
4x 3 + 2 a; 2 -10 x - 3
4 a; 8 - 2 x' 1 + 2 x - 12
X
4 a; 8 - 2x 2 + 2 a; -12
4 a 8 - 12 a; 2 + 9 a;
+ 4 x 2 - 12 x + 9
5
5x
+ 10 x 2 - 7 a; -12
+ 10 a: 2 - 15 x
20 x a - 60 x + 45
20 a; 2 -14 a; -24
-23)- 46 a; + 69
H.C. F. 2x- 3
4
+ 8 a; -12
+ 8 a; -12
HIGHEST COMMON FACTOR 391
Explanation :
1. Dividing 4 a; 3 + 2 x 2 — 10 x - 3 by 2x* — x 2 + x — 6, we obtain a
remainder, 4 x 2 — 12 x + 9, an expression lower in degree than the divisor
just read.
2. Dividing 2x 3 - x 2 + x — 6 by 4 oj 2 — 12 x + 9, we obtain a remainder
having such coefficients that the next succeeding division does not reduce
the degree of the remainder. To obtain a division that will reduce the
degree of the remainder we multiply 4 x 2 — 12 * + 9 by 5, and the product,
20 x 2 — 60 x + 45, contains 10 x 2 — 7 x — 12, with a remainder as desired,
— 46 se + 69. From this remainder we may discard the factor, —23,
since this factor is not common to both expressions at this point. •
3. The remainder, 2 x — 3, divides 10 x 2 — 7 x — 12 exactly.
Therefore, 2 x — 3 is the required H. C. F.
It may be noted that the process might have been completed by multi-
plying 10 x 2 — 7 x — 12 by 2, the resulting product being used as a divi-
dend with 4 x 2 — 12 x + 9 as a divisor. Division either way is possible
when both expressions are like in degree.
Exercise 139
Find the H.C.F. of:
1. ^-3^ + 4a-2and^- r -o 2 -4a; + 2.
2. a 3 -2a 2 -a + 2anda 3 -2a 2 + 3a-2.
3. c 3 + 5c 2 + 7c-f-3andc 3 + c 2 -5c + 3.
4. ^_^_5a;_|_2 and a^ + 4 » 2 -f 3 a; - 2.
5. m 3 + 3m 2 -f-5ra + 3 and m 3 + 2m 2 -2m-3.
6. a 3 -5a 2 + 7a-3anda 3 -a 2 -7a + 3.
7 . c 4 + c 2 + 2candc 4 -c 3 -3c 2 -c.
8. 2a 6 + 4ar i + 6a 4 -f-4ar J + 2a^ and 4a 5 - 4 a 3 -8^-40;.
9. m 3 + ra 2 — m + 15 and m 3 + 6 m 2 + 5 m — 12.
10. 2c 3 + 5c 2 + 5c + 6and3c 3 + 5c 2 -c + 2.
11. 6a; 4 + 24(» 3 + 30a! 2 -f-36a;and3a; 4 -f-6a5 3 -12ic 2 -9a;.
12. 2 a 4 + a 3 + 3 a 2 + a + 2 and a 4 + 2 a 3 +4 a 2 + 3 a + 2.
13. c 3 -3c 2 + 5c-3and2c 4 -3c 3 -3c 2 + 10c-6.
14. 2» 4 +a^-4^ + 3a;-2and3a; 4 + 4^-3ar 2 -4.
392 SUPPLEMENTARY TOPICS
15. 4a 4 -4 a 3 - 5 a 2 + 6a- land 6a 4 -5a 3 -6a 2 + 3a + 2.
16. I m 5 -3m 4 -2m 3 +5m 2 -2m and 3m 6 -7m 5 + 6m 4 - 3m 3
+m 2 .
Reduce the following to lowest terms :
1? m 3 -4m 2 + 2m + l , 20 c 3 -c 2 -5c-3
m 3 -2m 2 + 3ra-2 c 3_4 c 2_n c _ 6
3-5a + 7a 2 + 3a 3 , gl 3 a 3 + 14 or 2 -5 a -56
6-7a + 3a 2 + 2a 3 6a 3 + 10 a 2 + 17 a + 88
10 4a 2 + 9a-9 00 2m 4 -2m 2 + m-l
4a 4 -f-10a 3 -7a 2 + 9 m *_ m 3 + 2 m 2 -m-l
Find the H. C. F. of:
23. a 3 -2a + l, a 3 -2a 2 + l, and a 3 -2 a 2 + 2a-l.
24. a 3 + 4 a 2 +5 x+2, ^+3^+4^+4, and ^+3^+3 a +2.
25. 2a? 4 -2af'4-4a; 2 -4ic, 3 a; 4 -f- 9x 2 -12a;, and 4a 4 -8a; 3
-20« 2 + 24a-.
26. a 4 -4a 2 -a + 2, a 5 - 3a 3 + 3a 2 -l, and a 4 -f-3a-2.
THE LOWEST COMMON MULTIPLE OP EXPRESSIONS NOT READILY
FACTORED
473. In finding the lowest common multiple of two ex-
pressions not readily factored we first find the highest common
factor by Art. 471, after which the process is established by the
following.
Let A and B represent any two expressions whose highest common
factor we find to be H. Dividing both A and B by H, we obtain
^ = a and ^ = 6.
II H
Then, A-Hxa, (1)
B = Hxb. (2)
Now, since H is the highest common factor of A and 2?, the two quo-
tients, a and 6, can have no common factor. Hence, for the lowest com-
mon multiple of A and B we write
L.C.M.=Rxaxb. (3)
Writing (3) thus, L. C. M.=5ax 6.
LOWEST COMMON MULTIPLE 393
Multiplying and dividing by J9",
L.C.M. -Hax^-
H
Substituting Ha = A from (1), and Hb — B from (2),
L.C. M. = ^x— .
H
In general, therefore:
To find the lowest common multiple of two expressions, divide
one of the expressions by their highest common factor , and multi-
ply the other expression by the quotient.
The product of two expressions is equal to the product of their
highest common factor by their lowest common multiple. For,
Multiplying (1) by (2), we obtain AB = Hx ax Hxb
= H(Hab).
From (3) L. C. M. = Hab. Hence, AB = H(L. C. M.).
In finding the lowest common multiple of three or more expressions as
A, B, and C, we first find L\, the lowest common multiple of A and B.
It remains to find the lowest common multiple of L\ and C. Let this
result be'-L 2 . Then L 2 is the required lowest common multiple of A, B,
and G.
Illustrations :
1. Find the L. C. M. of :
2 ra 3 — ra 2 + m — 6 and 4 m 3 + 2 ra 2 — 10 m — 3.
By the method of division we find the H.C.F.=2m-3.
Then, L. C. M. = ( *™*-m* + m-Q \ (4 m s + 2 w2 _ 10 m _ 3)
V 2i»-3 ) K J
= (m? + m + 2)(4m 8 + 2m 2 - 10m - 3). Result.
Exercise 140
Find the L. C. M. of:
1. ^ + 5^ + 705 + 3 and<B 8 + <B 2 -3a> + 9.
2. a 3 + a 2 -10a + 8 and a 3 -4a 2 + 9a- 10.
3. 2m 3 + 5m 2 -f-2m — 1 and 3 m 3 + ra 2 — ra + 1.
4. 4c 3 -8c 2 -3c + 9and 6c 3 + c 2 -19c + 6.
5. 4ar 5 -7a + 3and2;c 3 + ar 2 -3a; + l.
394 SUPPLEMENTARY TOPICS
6. 2a 3 -7a 2 -16a + 5anda 3 -9a 2 + 18a + 10.
7. c 3 + 5c 2 -c-5 and c 4 -c 3 -f-c 2 + c-2.
8. a^ + 2^-a; + 6and2ar J -f-llar J + 16a; + 3.
9. m 4 -3m 2 + l andm 3 + 2m 2 -4m-3.
10. 4^-2a; 2 + 6a + 4and6a; 4 + 9x 3 -l5ar 2 -9a.
11. 2a 4 + 3a 3 + 4a 2 + 2a + l and 3a 4 -f-2a 3 + a 2 -2a-l.
12. m 5 - 3 m 4 + 3 m 3 -ll ra 2 +6 m and 2 m 4 -18 ra 2 +8 m-24.
13. 4a 4 -a 3 6 + 2a 2 6 2 + 2a6 3 + & 4 and3a 4 -5a 3 &+9a 2 6 2
-6a6 3 + 46 4 .
14. a^+2a; 2 4-2a;4-l, ^+2^+305+2, and a 3 -^ ^+4^+3.
15. 4 m 3 — 4 m 2 4- 3 m — 1, 6 m 3 — m 2 -f- m — 1, and 8 m 3 — 2 m 2
+ m-l.
THE CUBE ROOT OF POLYNOMIALS
474. If a binomial, (a+6), is cubed, we obtain (a 3 +3a 2 6
-f3a& 2 -f-& 3 ). We have in the following process a method for
extracting the cube root, (a + b), of the given cube, (a 3 + 3 o?b
+ 3a& 2 + 6 3 ).
a 3 + 3 a 2 b + 3 ab 2 + 6 8 la + 6 Cube Root.
Trial Divisor Complete Divisor a?
3a 2 (3a 2 + 3ab + b 2 )
b
+3a 2 6+3a& 2 +& 8
+ 3a 2 &+3a6 2 +6 8
The first term of the root, a, is the cube root of the first term of the
given expression, a 3 . Subtracting a 3 from the given expression, the re-
mainder, 3 a 2 b + 3 ab 2 + ft 3 , results.
The second term of the root, b, is obtained by dividing the first term of
the remainder, 3 a 2 b, by three times the square of a. This divisor, 3 a 2 ,
is the Trial Divisor.
The complete divisor, (3 a 2 + 3 ab + b 2 ) , consists of (the trial divisor)
+ 3 (the trial divisor) (the last quotient obtained) + (the square of the last
quotient).
The product of the complete divisor by the last quotient obtained,
(3 a 2 + 3 ab + b 2 )b, completes the cube.
The process may be repeated in the same order with any polynomial
perfect cube whose root has more than two terms.
CUBE ROOT OF POLYNOMIALS
395
Illustration :
1. Find the cube root of
8 a 6 -36 a 5 + 102 a 4 -171 a 3 + 204 a 2 -144a + 64.
The polynomial is first arranged in descending order.
|2q 2 -3q + 4
8a 6 -36a5+102a*-171a 3 +204«2_i44a+64
8«s
3(2a 2 ) 2 =12a 4
3(2 a 2 ) (-3 a) = -18 a 9
(-3q) 2 = +9a'
(12a*-18a 3 +9« 2 )(-3q):
-36 a* + 102 a*- 171 a 3
-36 a 5 + 54 a*- 27 a 3
3(2a 2 -3a) 2 = 12a 4 -36a 3 +27a 2
3(2a 2 -3a)(4) = 24a 2 -36a
4 2 = +16
(12 a*- 36 a 3 +51 a 2 - 36 a +16) (4)
48 a 4 - 144 a 3 + 204 a 2 - 144 a +64
48 a 4 -144 a 3 +204 a 2 -144 a + 64
475. The distribution of the volume of a cubic solid whose
edge is a + 6, and the relation of the separate portions to the
separate terms of the polynomial representing its cube, can
be readily seen from the following illustrations. The planes
cutting the cube pass at right angles to each other and parallel
to the faces of the cube ; each at a distance of b units from the
faces of the cube.
I n m is?
In the figure let the edge of the cube in (I) be a and the
edge of the complete cube in (IV) be a + 6.
I. a 3 .
II. a 3 + 3a 2 6.
III. a 3 + 3a 2 6 + 3a6 2 .
IV. a 3 + 3a 2 6 + 3a6 2 +6 3 .
396 SUPPLEMENTARY TOPICS
„ M
THE CUBE ROOT OF ARITHMETICAL NUMBERS
476. It will be observed that
l 8 = 1 A number of one figure has not more than three
9 s as 729 figures in its cube.
10 8 = 1000 A number of two figures has not more than six
99 8 = 970299, etc. figures in its cube.
Conversely, therefore:
If an integral cube has three figures, its cube root has one figure.
If an integral cube has six figures, its cube root has two figures, etc.
Hence :
477. Separate any integral number into groups of three figures
each, beginning at the right, and the number of groups obtained
is the same as the number of integral figures in its cube root.
It is to be noted that the process of cube root of numbers as
given in. arithmetical practice is based directly upon the al-
gebraic principles learned in Art. 474.
Illustration :
1. Find the cube root of 421875.
Separating the number into groups of three figures each, we have 42i875.
42J875 |70 + 5 = 75. Result.
a 8 = 70 8 = 343000 a + b
3(a) 2 = 3(70) 2 = 14700
3(a)(6) =3(70) (5)= 1050
(6)2 - (5)2 = 25
(15775) (5) =
'8875
78875
The analogy between the formation of the cube of (a + 6) and the cube
of 75, or (70 + 5), may be seen in the following :
(a + b) s = a 3 + 3a 2 6 .+ 3«6 2 + 6 8
(70 + 5) 8 = 70 8 + 3(70)2(5) + 3(70) (5) 2 + 6 8
= 343000 + 73500 + 5250 + 125
= 421876.
CUBE ROOT OF ARITHMETICAL NUMBERS
397
2^Find the cube root of 12812904.
In practice the process is usually abbreviated as follows:
. 12812904 |234. Result.
2» =
8
3(20)2 _
1200
4812
3(20) (3) =
180
32 =
9
3(1389) =
4167
3(230) 2 =
3(230) (4) =
42 =
158700
2760
16
4(161476)
645904
645904.
Exercise 141
Find the cube root of :
1. a 3 + 6a 2 + 12a+ 8.
2. 27 c 3 - 54c 2 + 36 c -8.
3. 8a 3 -36a 2 + 54a-27.
4. 27 a 6 - 135 a 4 + 225 a 2 -125.
5. a 6 + 6a 5 + 15a 4 + 20a 3 +15a 2 + 6a + l.
6 . ^-6^ + 21a 4 -44a 3 + 63a 2 -54a-f27.
7. a 6 + 9a 5 + 21a 4 -9a 3 -42a 2 + 36a-8.
8. 102a 4 + 204 x 2 - 171a 3 -144a + 64 -36a 5 + $x«.
9. 27m 10 - 35 m 6 + 12m 2 - 8 + 30m 4 - 45m 8 + 27m 12 .
10. c 9 -3c 8 + 6c 7 -4c 6 + 6c 4 -2c 3 + 3c + l.
11. 74088. 16. 3112.136.
12. 389017. 17. 1470612$.
13. 658503. 18. 48.228544.
14. 912673. 19. .559476224.
15. 1953125. 20. .000138991832.
398 GENERAL REVIEW
GENERAL REVIEW - |
Exercise 142
1. Find the numerical value of 81"*, 16^", (i)~ 4 , 5°, 25~t
Define briefly the law that governs each reduction.
2. Solve x + V.25 jc + 0.06 = 0.
3. Form the equation whose roots shall be a — V — 1
and a + V—1.
4. Solve for x : 2(4 - 3a)^ - 12 = s&
5. Factor 125 ar 6 -Va7.
6. Solve a? -f 10 aT +7 = 0, and test the roots obtained.
7. By inspection determine the nature of the roots of
Sx 2 -7x= -5.
8. Solve 4 x 2 — 4 ax + a 2 — c 2 = 0.
9. Solve and verify both solutions of (x + T)\x — 2) =
a,(a>-3)(&-l)-4.
10. Solve ?— - = a?(aj + 2) + 1.
0* - !)
11. Find the square root of
g*-s» + l 6(0* + 1) 15a? 20 .
(a^ + 1)- 1 a;- 1 (a^ + l)" 1
12. Find the value of m for which the equation, 3x 2 — 6x +
m = 0, has equal roots.
13. Factor 2[2 a 2 - (x - a) - 2 a(2 x - a)].
14. Write the 8th term of the quotient of x l5 —y 15 divided
by x-y.
15. Solve x* + 2xi- 3 = 0.
16. Solve *- 2m + 3m = * + « ■
c + d a? -j- 2m a? + 2m
17. Find the five roots ofar 5 -4ar J -a? 2 + 4 = 0.
GENERAL REVIEW 399
18. Factor 15 x 15 — 15.
19. Solve x(x ~y) — 2, (x + y) 2 = 9.
20. Form that quadratic equation whose roots shall be *
- i and - f
21. Obtain the factors of x 2 — 5# + 2 by solving the equa-
tion, x 2 — 5 x + 2 = 0.
22. Divide 16 into 3 parts in geometrical progression, so
that the sum of the 1st and 2d shall be to the difference of the
2d and 3d parts as 3 : 2.
23. Find log of 7, given log 5 = .6990 and log 14 = 1.1461.
24. Solve and test the solutions of
V2 m + x — V2 m — x — V2# = 0.
,.-2
25. Factor 4 + 4i? + m +
mx~
x p~ x
26. Form the quadratic equation whose roots shall be
l+V^andl-V^T.
27. Solve x 2 + xy + y 2 = 7 x,
x 2 — xy-{-y 2 — Sx.
28. What is the sum of the first n numbers divisible by 5 ?
„ ^.(H+fXf-M)-*.
30. Solve 2 s " 1 x 2* = 40.
31. Givenl+V«:l + 3V^ = 2:5; find a.
32. Factor ar 4 - 2 -f a 4 .
33. If 7i is an odd integer, which of the following indicated
divisions are possible ?
x n + y n xF+y" x n — y n x n — y n
x + y ' x — y ' x + i/ ' # — ?/
34. Form the quadratic equation whose roots are 1 -+- a yi
and 1 — a V#.
400 GENERAL REVIEW
35. Under what condition will the roots of ax 2 -+- bx -f- c =
be imaginary ? Prove your answer.
36. If the length and breadth of a certain rectangle are
each increased by 2 rods, the area will become 48 sq. rd. ; but
if each dimension were decreased by 2 rods, the area would
become 8 sq. rd. Find the dimensions of the rectangle.
37. Factor x* + 2x 2 -5x-6.
38. Write in logarithmic form 27 x = 81, and find x.
39. Solve for n : -^ 2 c =
n n — 1
40. What is the price of candles per dozen when 3 less for
36 cents raises the price 12 cents per dozen ?
41. Show that the product of the roots of x 2 — 5x — 2 =
is —4.
42. Prove that — -, 3, — - — , are in geometrical progression,
and find the sum of 10 terms.
43. If m : x = n : y, show that
mx — ny : mx -\-ny = m 2 —n 2 :m 2 + n 2 .
44. Find an expression for the nth odd number, and illus-
trate your answer by a numerical substitution.
45. Find n in the formula I = ar n ~ l .
46. Solve for a and m : 2 a 2 — am — 12,
2 am — m 2 = 8.
47. Factor , A •*" ■ - - — ±— — •
(V + 3)- 1 (1+3 a)" 1
48. Rationalize the denominator of — i L^l — .
2-V6 + 2V2
49. Insert 6 geometrical means between T \ and 12|.
C a -cv a • xy/S -v/18
50. Find x in — ^— = ~
12 y/i
GENERAL REVIEW 401
51. Without solving, prove that the roots of 6 x 2 + 5 x — 21
are real and rational.
52. Solve m-f \Z3x-\-x 2 : m + l=m- V3 #-}-#* : m — 1.
53. The difference between the 5th and the 7th terms of an
arithmetical progression is 6, and the sum of the first 14 terms
is — 105. Find the first term and the common difference.
54. Without solving, determine the nature of the roots of
16ar> + l=8a;.
55. Find the 5th term of (x 2 - ar 2 ) 15 .
56. Show that the sum of the squares of the roots of
»2-3iC + l = is 7.
57. If m 2 — n 2 varies as x 2 , and if x = 2 when ra = 5 and
ra = 3, find the equation between m, n, and x.
58. Find x and y if 2 x+ » = 16 and S x ~ v = 9.
59. Find a 4th proportional to x* — 1, x 2 -f 1, and cc 2 — 1.
60. Construct the quadratic equation, the product of whose
roots shall be twice the sum of the roots of x 2 — 7 sc -f- 12 = ;
and the sum of whose roots shall be 3 times the product of
the roots of x 2 -f 2 x = 3.
61. Factor 3 x*-2 + x 2 + x 4 -3 x.
62. Find by logarithms the value of -y/2 x (£)* X .01 x 3*.
63. The sum of two numbers is 20, and their geometrical
mean increased by 2 equals their arithmetical mean. What
are the numbers ?
i _
64. What is the interpretation of x 1 = Vsc?
65. Write the 5th term of (a -f b) m .
66. Solve c 2 O 2 + 1) = m 2 + 2 c 2 x.
67. The sum of the last 3 terms of an arithmetical progres-
sion of 7 terms equals 3 times the sum of the first 3 terms.
The sum of the 3d and 5th terms is 32. Find the 1st term
and the common ratio.
SOM. EL. ALG. — 26
402 GENERAL REVIEW
68. How many digits in 35 s5 ?
69. Expand and simplify (a _1 Va — aVa -1 ) 4 .
70. What must be the equation between m and n if the roots
of mxP+nx+p are real ? if equal ?
71. Find the (r + l)th term of (1 -x) 20 .
72. Find two numbers in the ratio of 3:2, such that their
sum has to the difference of their squares the ratio of 1 : 5.
73. What is the sum and product of the roots of ,>
5 x~ 2 x~ l
74. Solve x 2 + y 2 = 26,
5 x + y = 24.
Plot the graphs of the system and verify the solutions.
75. Show that either root of x 2 — c = is a mean propor-
tional between the roots of x 2 + bx + c = 0.
76. Solve x 2 — 2 x + 3 == V&- 2 — 2 a; + 5. Are all the solu-
tions roots of the given equation ?
77. What is the value of the 6th term of f x ] when
x = 2?
78. The intensity of light varies inversely as the square of
the distance from its source. How far must an object that is
8 feet from a lamp be moved so that it may receive but \ as
much light?
79. Find an expression for x in a 2x = 3 c.
80. Solve x — 1.3 = .3 x~\
81. If a : b = c > : d, show that
(ma — nb)(ma +nb)~ l _ (mc — nd)(mc + wcT)" 1
mn win
10
82. Solve V7 x - V3 a + 4 =
V3a+4
GENERAL REVIEW 403
83. If r x and r 2 represent the roots of x 2 -f bx -f c = 0, find
r* -f- r 2 2 an( i ^i 2 **2 2 »
84. Solve x~% + af* + 1 = 0.
85. Find the geometrical progression whose sum to infinity
is 4^, and whose 2d term is .002.
86. Expand ( V2 - V^) 6 .
87. Solvefors: -2 - 2 s 2 + V7 +2 s + 4 s 2 - 2s = 2s 2 + 5.
88. Find the value of k in order that the equation
(k + 6)x 2 -2k(x 2 -l)-2kx-3 =
may have equal roots.
89. The floor area of a certain room is 320 sq. ft., each
end wall 128 sq. ft., and each side wall 160 sq. ft. What
are the dimensions of the room ?
90. For what values of m are the roots of the equation
(m + 2)x 2 + 2ma;+l = equal?
91. Given a:b = c: dj prove that 3a + 2c:3a — 2c
= 12 b + 8 d : 12 b - 8 d.
92. Plot the graph of 3 x 2 -f- 10 x = 12, and check the result
by solving.
93. Find the ratio between the 5th term of f 1 + - ) and
the 4th term of (l + |Y 2 -
94. Calculate by logarithms the fourth proportional to 3.84,
2.76 and 4.62, and, also, the mean proportional between -\/T2
and V12.
95. Solve for sand t: sP + P = 91, s = 7 — t.
96. Findnins = arW ~ a »
404 GENERAL REVIEW
97. The velocity of a body falling from rest varies directly
as the time of falling. If the velocity of a ball is 160 feet
after 5 seconds of fall, what will it be at the end of the 10th
second ?
98. In an arithmetic progression, a— — V— 1, cZ=l+ V^T,
n a* 20. Find I and s.
99. Write the (r + l)th term of (a + b) n .
100. Find the middle term of ( — X~3Y°.
W-l ,* J
101. Plot the graphs of 4 x 2 + 9 f = 36, x + 2 y = 3. Check
by solution.
102. Insert 4 geometrical means between V— 1 and — 32.
103. Prove the formula for I in each of the progressions.
104. Form the quadratic equation which has for one root
the positive value of V J 4- 4V3, and for the other root the
arithmetic mean between 4 — 2 V3 and zero.
105. If m : n — n : s = s : t, show that n + s is a mean pro-
portional between m + n and t -\- s.
106. Solve xy = 2 m 2 + 5 m + 2, ar> + ?/ 2 = 5 ra 2 + 8 m + 5.
107. Solve and test the solution : V# 2 — mx -f» *- x*s m,
f 2\ 12
108. Find the term of (x 4 J that does not contain x.
109. Solve for sand t: s 2 + st + 2t 2 = 46,
2s 2 -st + t 2 =29.
110. Plot the graphs of a^ + y 2 + x + y =34, x + y-7 = 0.
111. What is the value of 1.027027 •;• ?
112. Find x if 3 2 *~ 2 j== (9~ l y-\
113. If a -f b — 61, and a^ — 6^ = 1, find the values of a
and b.
INDEX
(Numbers refer to pages.)
Abscissa, 197.
Addends, 22.
Addition, 22.
Affected quadratic equation, 2
Aggregation, signs of, 29.
Algebraic expression, 20.
Algebraic fraction, 122.
Algebraic number, 16.
Alternation, 320.
Annuity, 382.
Antecedent, 315, 318.
Antilogarithm, 373.
Arithmetical means, 344.
Arithmetical progression, 338.
Ascending order, 44.
Associative law, 22, 39.
Axes of reference, 197.
Axiom, 13, 65.
Base of logarithm, 364. *
Binomial, 21.
Binomial formula, 359.
Binomial theorem, 358.
Brace, 29.
Bracket, 29.
Characteristic of logarithm, 3
Checking results, 25, 46, 59.
Clearing of fractions, 147.
Coefficient, 12.-
compound, 35.
detached, 51, 61.
Cologarithm, 376.
Collecting terms, 25.
Common difference, 338.
Common factor, 119.
Common multiple, 127.
Common ratio, 348.
Commutative law, 22, 39.
Complex fraction, 144.
Complex number, 255.
Composition, 321.
Compound ratio, 348.
Comoound variation, 333.
Conditional equation, 64.
Conjugate imaginary, 260.
Consequent, 315, 318.
Constant, 200, 300.
Continued proportion, 318.
Coordinates, rectilinear, 197.
Cube root, 99, 394, 396.
Definite number symbols, 9.
Degree, of an expression, 44.
of a term, 44.
Denominator, 126.
factorial, 360.
lowest common, 129.
Density, 164.
Descending order, 44.
Difference, 32.
common, 338.
Direct and inverse variation, 332.
Direct variation, 331.
Discriminant, 278.
Discussion of a problem, 193.
Distributive law, 39, 52.
Dividend, 52.
Division, 52, 321.
Divisor, 52.
Divisors, theory of, 387.
Duplicate ratio, 316.
Elements, of an arithmetical 'progres-
sion, 339.
of a geometrical progression, 349.
Elimination, 165.
Equality, 12.
Equation, 64.
affected quadratic, 268.
complete quadratic, 268.
conditional, 64.
equivalent, 174, 290.
exponential, 380.
identical, 64.
in the quadratic form , 286.
incomplete quadratic, 266.
inconsistent, 174.
405
406
INDEX
Equation, independent, 173.
irrational, 252.
linear, 65.
pure quadratic, 266.
quadratic, 266.
simple, 65.
simultaneous, 174.
solution of an, 64.
systems of, 174.
Equilibrium, 165.
Equivalent equation, 174.
Exponent, 41.
in the fractional form, 223.
negative, 222.
zero, 221.
Exponential equation, 380.
Expression, algebraic, 20.
homogeneous, 45.
integral, 99.
mixed, 125.
prime, 100.
rational, 99.
Extraneous roots, 181.
Extremes, of a proportion, 317.
Factor, 12, 99.
common, 119.
highest common, 119.
theorem, 385.
Formula, 87.
binomial, 359.
physical, 164, 293.
quadratic, 273.
Fourth proportional, 318.
Fourth root, 99.
Fraction, algebraic, 122.
complex, 144.
terms of a, 122.
Fractions, clearing of, 147.
General number symbols, 9.
Geometric means, 354.
Geometric progression, 348.
Graph, of a linear equation, 200.
of a point, 197.
of a quadratic equation in one vari-
able, 282.
of a quadratic equation in two vari
ables, 302.
Grouping, law of, 22, 39.
Identical equation, 64.
Identity, 64.
Imaginaries, conjugate, 260.
unit of, 254.
Imaginary number, 254.
pure, 254.
Inconsistent equation, 174.
Independent equation, 174.
Indeterminate equation, 173.
Index law, 42, 53, 205, 210.
Indicated operations, 12.
Integral expression, 99.
Interpolation, 369, 373.
Inverse ratio, 316.
Inverse variation, 332.
Inversion, 320.
Involution, 205.
Irrational equation, 252.
Joint variation, 332.
Law, associative, 22, 39.
commutative, 22, 39.
distributive, 39, 52.
Linear equation, 65.
Literal expression, 20.
Literal number symbols, 9.
Logarithm, 364.
base of a, 364.
characteristic of a, 366.
common, 365.
mantissa of a, 366.
Lowest common denominator, 129.
Lowest common multiple, 127.
Mantissa of logarithm, 366.
Mean proportional, 318.
Means, arithmetical, 344.
geometrical, 354.
of a proportion, 317.
Mixed expression, 125.
Moment, 165.
Monomial, 21.
Multiple, common, 127.
lowest common, 127.
Multiplicand, 38.
Multiplication, 38.
Multiplier, 38.
Negative exponents, 222.
INDEX
407
Negative numbers, 14.
Number, algebraic, 16.
negative, 14.
Numerator, 126.
Numerical expression, 20.
Order, ascending, 44.
descending, 44.
law of, 22, 38.
Ordinate, 197.
Origin, 197.
Parenthesis, 13, 26.
Physical formulas, 164, 293.
Polynomial, 21.
Power, 41.
Prime expression, 100.
Principal root, 237.
Problem, 70.
discussion of a, 193.
solution of a, 71.
Progression, arithmetical, 338.
geometrical, 348.
Proportion, 317.
continued, 318.
extremes of a, 317.
means of a, 317.
Pure imaginary, 254.
Pure quadratic equation , 266.
Quadrants, 198.
Quadratic equation, 266.
Quadratic form, 286.
Quadratic formula, 273.
Quality, signs of, 14.
Quotient, 52.
Radical, 236.
index of a, 209, 236.
sign of, 209.
similar, 242.
Radicand, 236.
Ratio, common, 348.
compound, 316.
duplicate, 316.
inverse, 316.
Rational expression, 99.
Rationalization, 246.
Real number, 254.
Reciprocal, 140.
Rectilinear coordinates, 197.
Remainder, 32.
Root, 41, 65.
cube, 99, 394, 396.
square, 99.
Roots, rejection of, 291.
Series, finite, 258.
infinite decreasing, 352.
Signs, of aggregation, 29.
of quality, 14.
Similar terms, 20.
Simple equation, 65.
Simultaneous equations, 174.
Solution of equations, 64.
Square root, 99.
Substitution, 84.
Subtraction, 32.
Subtrahend, 32.
Sum, 22.
Surd, 236.
coefficient of a, 236.
entire, 236.
mixed, 236.
Surds, conjugate, 247.
Symbols of operation, 11.
Systems of equations, 174.
Term, 20.
degree of a, 44.
Terms, of a fraction, 122.
similar, 20.
Third proportional, 318.
Transposition, 66.
Trinomial, 21.
Unit of imaginaries, 254.
Variable, 200, 330.
Variation, compound, 333.
direct, 331.
direct and inverse, 332.
inverse, 332.
joint, 332.
Verification of a root, 67.
Vinculum, 17.
| Zero exponent, 221.
ROBBINS'S GEOMETRY
By EDWARD RUTLEDGE ROBBINS, A.B., Senior
Mathematical Master, The William Benn Charter School,
Philadelphia.
Plane and Solid Geometry, $1.25 Plane Geometry . . . $0.75
Solid Geometry . . . #0.75
THIS text-book is intended to meet the needs of all
secondary- schools and the requirements in geometry
for entrance to all colleges and universities. It is clear,
consistent, teachable, and sound.
^j So suggestively and comprehensively is the work outlined
that the teacher is saved many explanations, while the pupil
receives the help he desires wherever it is required. The
preliminary matter is intentionally brief and simple, so that
the theorems and their demonstrations may be reached as early
as possible. Each theorem is employed in the demonstration
of other theorems as promptly as is practicable and desirable,
^j The book has been so constructed that it meets the various
degrees of intellectual capacity and maturity in all ordinary
classes. The successive truths in a demonstration are stated,
and the pupil is asked the reasons. But the latter is not left
in ignorance if he is unable to perceive the correct reason,
because the numbers of the paragraphs containing these desired
truths are cited wherever experience has shown that the
pupil is likely to require this assistance.
^j The conciseness and rigor of demonstration, the great wealth
of original exercises, both classified and graded, the economy
of arrangement, the full treatment of measurement, and the
superior character of the diagrams distinguish the book. Pre-
ceding the earlier collections of original exercises are summaries,
while useful formulas are grouped both in the Plane and in
the Solid. All the theorems in the Plane Geometry are
stated at the beginning of the Solid.
AMERICAN BOOK COMPANY
C65)
ELEMENTS OF
TRIGONOMETRY
By ANDREW W. PHILLIPS, Ph.D., and WENDELL
M. STRONG, Ph.D., Professors in Yale University
Plane and Spherical Trigonometry. With Tables $i-4Q
The same. Without Tables 90
Logarithmic and Trigonometric Tables 1.00
IN this text-book full recognition is given to the rigorous
ideas of modern mathematics in dealing with the funda-
mental series of trigonometry. Both plane and spherical
trigonometry are treated in a simple, direct manner, free from
all needless details. The trigonometric functions are defined
as ratios, but their representation by lines is also introduced
at the beginning, because certain parts of the subject can be
treated more simply by the line method, or by a combination
of the two methods, than by the ratio method alone.
^J Many valuable features distinguish the work, but attention
is called particularly to the graphical solution of spherical
triangles, the natural treatment of the complex number, and
the hyperbolic functions, the graphical representation of the
trigonometric, inverse trigonometric, and hyperbolic functions,
the emphasis given to the formulas essential to the solution of
triangles, the close and rigorous treatment of imaginary quan-
tities, the numerous cuts which simplify the subject, and the
rigorous chapter on computation of tables,
•fl" Carefully selected exercises are given at frequent intervals,
affording adequate drill just where it is most needed. An
exceptionally large number of miscellaneous exercises are
included in a separate chapter.
^[ The tables include, besides the ordinary five-place tables,
a complete set of four-place tables, a table of Naperian
logarithms, tables of the exponential and hyperbolic functions,
a table of constants, etc.
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(70)
ELEMENTS OF DESCRIPTIVE
GEOMETRY
By CHARLES E. FERRIS, Professor of Mechanical
Engineering, University of Tennessee
#1.25
THE leading engineers and draughtsmen, as investigation
shows, do nearly all their work in the third quadrant or
angle. It seems reasonable, therefore, that the subject
of descriptive geometry should be taught in technical and
scientific schools as it will be used by their graduates,
•[f Many years of experience in teaching descriptive geometry
have proved to the author that the student can learn to think
with his problem below the horizontal, and behind the vertical
and perpendicular planes, as well as above and in front of
those planes.
^[ This volume forms an admirable presentation of the subject,
treating of definitions and first principles ; problems on the
point, line, and plane ; single curved surfaces ; double curved
surfaces ; intersection of single and double curved surfaces by
planes, and the development of surfaces ; intersection of solids ;
warped surfaces ; shades and shadows ; and perspective.
^| Besides dealing with all its problems in the third angle
instead of in the first, the book presents for each problem a
typical problem with its typical solution, and then gives
numerous examples, both to show variations in the data, and to
secure adaptability in the student. In consequence, no sup-
plementary book is necessary.
*H To show the projections on the horizontal, and on the
vertical planes, it uses v and h as exponents or subscripts
instead of the usual method of prime, etc.
^[ In scope the treatment is sufficiently broad, and yet it is
not so abstruse as to make the book difficult for the average
college course. Both text and plates are bound together, thus
being very convenient for use. There are 1 1 3 figures.
AMERICAN BOOK COMPANY
(63)
PLANE SURVEYING
#3.00
By WILLIAM G. RAYMOND, C. E., member Ameri-
can Society of Civil Engineers, Professor of Geodesy,
Road Engineering, and Topographical Drawing in
Rensselaer Polytechnic Institute.
IN this manual for the study and practice of surveying the
subject is presented in a clear and thorough manner; the
general method is given first and afterward the details.
Special points of difficulty have been dwelt on wherever
necessary. The book can be mastered by any student who
has completed trigonometry, two formulas only being given,
the derivation of which requires a further knowledge. The
use of these is, however, explained with sufficient fullness.
^| In addition to the matter usual to a full treatment of land,
topographical, hydrographical, and mine surveying, par-
ticular attention is given to system in office-work, labor-saving
devices, the planimeter, slide-rule, diagrams, etc., coordinate
methods, and the practical difficulties encountered by the
young surveyor. An appendix gives a large number of
original problems and illustrative examples.
^J The first part describes the principal instruments and deals
with the elementary operations of surveying, such as measure-
ment of lines, leveling, determination of direction and measure-
ment of angles, stadia measurements, methods of computing
land surveys, etc.
•[f In the second part are treated general surveying methods,
including land surveys, methods adapted to 'farm surveys,
United States public land surveys, and city surveys, curves,
topographical surveying, ordinary earthwork computations,
hydrographic and mine surveying, etc.
^[ Both four-place and five-place tables are provided. They
are unusually numerous and practical, and are set in large,
clear type. The illustrations are particularly helpful.
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(76)
ESSENTIALS IN HISTORY
ESSENTIALS IN ANCIENT HISTORY . . $i.;o
From the earliest records to Charlemagne. By ARTHUR
MAYER WOLFSON, Ph.D., First Assistant in History,
DeWitt Clinton High School, New York.
ESSENTIALS IN MEDIEVAL AND MODERN
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From Charlemagne to the present day. By SAMUEL
BANNISTER HARDING, Ph.D., Professor of Euro-
pean History, Indiana University.
ESSENTIALS IN ENGLISH HISTORY . . fi.50
From the earliest records to the present day. By
ALBERT PERRY WALKER, A.M., Master in His-
tory, English High School, Boston.
ESSENTIALS IN AMERICAN HISTORY . $1.50
From the discovery to the present day. By ALBERT
BUSHNELL HART, LL.D., Professor of History,
Harvard University.
THESE volumes correspond to the four subdivisions re-
quired by the College Entrance Examination Board,
and by the New York State Education Department.
Each volume is designed for one year's work. Each of the
writers is a trained historical scholar, familiar with the con-
ditions and needs of secondary schools.
^[ The effort has been to deal only with the things which
are typical and characteristic; to avoid names and details
which have small significance, in order to deal more justly
with the forces which have really directed and governed man-
kind. Especial attention is paid to social history, as well as
to the movements of sovereigns and political leaders.
^| The books are readable and teachable, and furnish brief
but useful sets of bibliographies and suggestive questions.
No pains have been spared by maps and pictures, to furnish
a significant and thorough body of illustration, which shall
make the narrative distinct, memorable, and clear.
AMERICAN BOOK COMPANY
[130)
A BRIEF COURSE IN
GENERAL PHYSICS
$1.20
By GEORGE A. HOADLEY, A.M., C.E.,
Professor of Physics, Swarthmore College
A COURSE, containing a reasonable amount of work for
an academic year, and covering the entrance require-
ments of all of the colleges. It is made up of a reliable
text, class demonstrations of stated laws, practical questions
and problems on the application of these laws, and laboratory
experiments to be performed by the students.
^J The text, which is accurate and systematically arranged,
presents the essential facts and phenomena of physics clearly
and concisely. While no division receives undue prominence,
stress is laid on the mechanical principles which underlie the
whole, the curve, electrical measurements, induced currents,
the dynamo, and commercial applications of electricity.
•J]" The illustrative experiments and laboratory work, intro-
duced at intervals throughout the text, are unusually numerous,
and can be performed with comparatively simple apparatus.
Additional laboratory work is included in the appendix, to-
gether with formulas and tables.
HOADLEY'S PRACTICAL MEASUREMENTS IN
MAGNETISM AND ELECTRICITY. #0.75
THIS book, which treats of the fundamental measurements in elec-
tricity as applied to the requirements of modern life, furnishes a satis-
factory introduction to a course in electrical engineering for secondary
and manual training schools, as well as for colleges. Nearly ioo experiments
are provided, accompanied by suggestive directions. Each experiment is
followed by a simple discussion of the principles involved, and, in some
cases, by a statement of well-known results.
AMERICAN BOOK CO.MPANY
C*59)
CHEMISTRIES
By F. W. CLARKE, Chief Chemist of the United States
Geological Survey, and L. M. DENNIS, Professor of
Inorganic and Analytical Chemistry, Cornell University
Elementary Chemistry . $1.10
Laboratory Manual . . $0.50
THESE two books are designed to form a course in
chemistry which is sufficient for the needs of secondary
schools. The TEXT- BO OK is divided into two parts,
devoted respectively to inorganic and organic chemistry.
Diagrams and figures are scattered at intervals throughout the
text in illustration and explanation of some particular experi-
ment or principle. The appendix contains tables of metric
measures with English equivalents.
^| Theory and practice, thought and application, are logically
kept together, and each generalization is made to follow the
evidence upon which it rests. The application of the science
to human affairs, its utility in modern life, is also given its
proper place. A reasonable number of experiments are in-
cluded for the use of teachers by whom an organized laboratory
is unobtainable. Nearly all of these experiments are of the
simplest character, and can be performed with home-made
apparatus.
^f The LABORATORY MANUAL contains 127 experi-
ments, among which are a few of a quantitative character. Full
consideration has been given to the entrance requirements of
the various colleges. The left hand pages contain the experi-
ments, while the right hand pages are left blank, to include
the notes taken by the student in his work. In order to aid
and stimulate the development of the pupil's powers of observa-
tion, questions have been introduced under each experiment.
The directions for making and handling the apparatus, and
for performing the experiments, are simple and clear, and are
illustrated by diagrams accurately drawn to scale.
AMERICAN BOOK COMPANY
(162)
UNIVERSITY OF CALIFORNIA LIBRARY
BERKELEY
Return to desk from which borrowed.
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OCT 2 91953
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^66
THE UNIVERSITY OF CALIFORNIA LIBRARY
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