tt .
NEW
ELEMENTARY ALGEBRA:
EMBRACING
THE FIRST PRINCIPLES OF THE SCIENCE.
PKOFE880K OF HIGHER MATH KM ATIO8, COLUMBIA COLLEOB.
NEW YORK :
BARNES & BURR, PUBLISHERS, 51 & 53 JOHN ST.
CHICAGO: GEORGE SHERWOOD. 118 LAKE ST.
CINCINNATI: RICKEY AND CARROLL.
ST. LOUIS : KEITH AND WOODS.
1804.
A. S. BARNES AND BUHIi'S PUBLICATIONS.
Da vies1 Course of Mathematics.
MATHEMATICAL WORKS,
IS A SERIES OF THREE PAKT8 :
ARITHMETICAL, ACADEMICAL, AND COLLEGIATE
DAVIES' LOGIC AND UTILITY OF MATHEMATICS.
Tins series, combining all that is most valuable in the various methods of Enropeas
Instruction, improved and matured by the suggestions of more than thirty years'
experience, now forms the only complete consecutive course of Mathematics. Ita
methods, harmonizing as the works of one mind, carry the student onward by the
same analogies, and the same laws of association, and are calculated to impart a com-
prehensive knowledge of the science, combining clearness in the several branches, and
unity and proportion in the whole ; being the system so long in use at West Point,
through which so many men, eminent for their scientific attainments, have passed,
and having been adopted, as Text Books, by most of the Colleges in the United States.
I.— THE ARITHMETICAL COURSE FOR SCHOOLS.
1. PRIMARY ARITHMETIC AND TABLE-BOOK.
2. INTELLECTUAL ARITHMETIC.
3. SCHOOL ARITHMETIC. (Key separate.)
4. GRAMMAR OF ARITHMETIC.
II.— THE ACADEMIC COURSE.
1. THE UNIVERSITY ARITHMETIC. (Key separate.)
2. PRACTICAL MATHEMATICS FOR PRACTICAL MEN.
3. ELEMENTARY ALGEBRA. (Key separate.)
4. ELEMENTARY GEOMETRY AND TRIGONOMETRY.
5. ELEMENTS OF SURVEYING.
III.— THE COLLEGIATE COURSE.
1. DAVIES' BOURDON'S ALGEBRA.
2. DAVIES' UNIVERSITY ALGEBRA.
3. DAVIES' LEGENDRE'S GEOMETRY AND TRIGONOMETRY.
4. DAVLES' ANALYTICAL GEOMETRY.
5. DAVIES' DESCRIPTIVE GEOMETRY.
6. DAVIES' SHADES, SHADOWS, AND PERSPECTIVE.
7. DAVIES' DIFFERENTIAL AND INTEGRAL CALCULUS.
8. MATHEMATICAL DICTIONARY, BY DAVIES & PECK.
ENTEBED, according to the Act of Congress, in the year one thousand eight hundred
and fifty-nine, by CHAELES DAVIES, in the .Clerk's Oflice of the District Court of the
United States, for the Southern District of New York.
1TILI.IAM DEXYSE, STEREOTYPES.
a
P K E F A C E.
ALGEBRA naturally follows Arithmetic in a course of scien-
tilic studies. The language of figures, and the elementary
combinations of numbers, are acquired at an early age.
When the pupil passes to a new system, conducted by
letters and signs, the change seems abrupt; and he often
experiences much difficulty before perceiving that Algebra
is but Arithmetic written in a different language.
It is the design of this work to supply a connecting link
between Arithmetic and Algebra ; to indicate the unity of
the methods, and to conduct the pupil from the arithmetical
processes to the more abstract methods of analysis, by easy
and simple gradations. The work is also introductory to
the University Algebra, and to the Algebra of M. Bourdon,
which is justly considered, both in this country and in
Europe, as the best text-book on the subject, which has yet
appeared.
In the Introduction, or Mental Exercises, the language
of figures and letters are both employed. Each Lesson is
so arranged as to introduce a single principle, not known
iii
IV PREFACE.
before, and the whole is so combined as to prepare the
pupil, by a thorough system of mental training, for those
processes of reasoning which are peculiar to the algebraic
analysis.
It is about twenty years since the first publication of the
ELEMKNTAEY ALGEBRA. Within that time, great changes
have taken place in the schools of the country. The sys-
tems of mathematical instruction have been improved, new
methods have been developed, and these require correspond-
ing modifications in the text-books. Those modifications
have now been made, and this work will be permanent in
its present form.
Many changes have been made in the present edition, at
the suggestion of teachers who have used the work, and
favored me with their opinions, both of its defects and
merits. I take this opportunity of thanking them for the
valuable aid they have rendered me. The criticisms of
those engaged in the daily business of teaching are invalu-
able to an author ; and I shall feel myself under special
obligation to all who will be at the trouble to communicate,
to me, at any time, such changes, either in methods or lan-
guage, as their experience may point out. It is only through
the cordial co-operation of teachers and authors — by joint
labors and mutual efforts — that the text-books of the country
can be brought to any reasonable degree of perfection.
COLUMBIA COLLEGE, NEW YORK, March, 1859.
CONTENTS.
CHAPTER I.
DEFINITIONS AND EXPLANATORY SIGHS.
PAQM.
Algebra — Definitions — Explanation of the Signs 83-41
Examples in writing Algebraic expressions 41
Interpretation of Algebraic language 42
CHAPTER II.
FUNDAMENTAL OPERATIONS.
Addition — Rule — Examples 48-50
Subtraction — Rule — Examples — Remarks 60-56
Multiplication — Monomials — Polynomials 56-63
Division — Monomials 63-68
Signification of the symbol «° 68-70
Division of Polynomials — Examples 71-76
CHAPTER III.
USEFUL FORMULAS. FACTORING, ETC.
Formulaa (1), (2), (3), (4), (5), and (6) 76-79
Factoring • 79-81
Greatest Common Divisor 81-84
Least Common Multiple 84-87
CHAPTER IV
FRACTIONS.
Transformation of Fractions .' 89
fo Reduce an Entire Quantity to a Fractional Form 90
V ^
VI CONTENTS.
To Reduce a Fraction to its Lowest Terms
To Reduce a Fraction to a Mixed Quantity
To Reduce a Mixed Quantity to a Fraction 93
To Reduce Fractions to a Common Denominator 94
Addition of Fractions 96-98
Subtraction of Fractions 98-99
Multiplication of Fractions 99-102
Division of Fractions 102-105
CHAPTER V.
EQUATIONS OF THE FIRST DEGRIB.
Definition of an Equation — Different Kinds 105-106
Transformation ot Equations — First and Second 106-110
Solution of Equations — Rule 110-1 14
Problems involving Equations of the First Degree 115-130
Equations involving Two Unknown Quantities 130-131
Elimination — By Addition — By Subtraction — By Comparison. . 131-143
Problems involving Two Unknown Quantities 143-148
Equations involving Three or more Unknown Quantities 148-159
CHAPTER VI.
FORMATION OF POWERS.
Definition of Powers 160-161
Powers of Monomials 161-163
Powers of Fractions 163-165
Powers of Binomials 1 65-1 67
Of the Terms— Exponents — Coefficients 167-170
Binomial Formula — Examples 170-172
CHAPTER VII.
SQUARE ROOT. RADICALS OF THE SECOND DEGREE.
Definition — Perfect Squares — Ride — Examples 173-179
Square Root of Fractions 179-181
Square Root of Monomials 181-183
Imperfect Squares, or Radicals 1 83-1 87
Addition of Radicals '. 187-189
Subtraction of Radicals .. 189-190
CONTENTS. Vll
Multiplication of Radicals ............................... 190-191
Division of Radicals .................................... 191-192
Square Root of Polynomials ............................. 193-197
CHAPTER VIII.
EQUATIONS OF THE SECOND DEGREE.
Equations of the Second Degree — Definition — Form ......... 198-200
Incomplete Equations ____ . .............................. 200-209
Complete Equations— Rule ............................... 209-211
Four Forms ........................................... 21 1-227
Four Properties ........................................ 227-229
formation of Equations of the Second Degree .............. 229-231
Numerical Values of the Roots ........................... 231-236
Problems .................. ............................ 236-240
Equations involving more than One Unknown Quantity . ..... 241—250
Problems ............................................. 250-254
CHAPTER IX.
ARITHMETICAL AND GEOMETRICAL PROPORTION.
Ways in which Two Quantities may be Compared ........... 255
Arithmetical Proportion and Progression .................. 256-257
Last Term ......................... . ................... 257-260
Sum of the Extremes— Sum of Series ..................... 260-262
The Five Numbers— To find any number of Means .......... 262-265
Geometrical Proportion ................................. 267
Various Kinds of Proportion ................... . ......... 268-278
Geometrical Progression ................................. 278-280
Last Term— Sum of Series ............................... 280-285
Progression having an Infinite Number of Terms ............ 285-288
The Five Numbers— To find One Mean .................... 288-289
CHAPTER X.
LOGARITHMS.
Theory of Logarithms ................................... 290-291
SUGGESTIONS TO TEACHERS.
1. THE Introduction is designed as a mental exercise. If
thoroughly taught, it "will train and prepare the mind of
the pupil for those higher processes of reasoning, which it
is the peculiar province of the algebraic analysis to develop.
2. The statement of each question should be made, and
every step in the solution gone through with, without the
aid of a slate or black-board ; though perhaps, in the begin-
ning, some aid may be necessary to those unaccustomed to
such exercises.
3. Great care must be taken to have every principle on
which the statement depends, carefully analyzed ; and equal
care is necessary to have every step in the solution distinctly
explained.
4. The reasoning process is the logical connection of dis-
tinct apprehensions, and the deduction of the consequences
which follow from such a connection. Hence, the basis of
all reasoning must lie in distinct elementary ideas.
5. Therefore, to teach one thing at a time — to teach that
thing well — to explain its connections with other things,
and the consequences which follow from such connexions,
would seem to embrace the whole art of instruction.
via
ELEMENTARY ALGEBRA.
INTRODUCTION.
MENTAL EXERCISES.
LESSON I.
1. JOHN and Charles have the same number of apples ;
both together have twelve : how many has each ?
ANALYSIS.— Let x denote the number which John has ;
then, since they have an equal number, x will also denote
the number which Charles has, and twice x, or 2aj, will
denote the number which both have, which is 1 2. If twice
x is equal to 12, x will be equal to 12 divided by 2, which
is 6 ; therefore, each has 6 apples.
WRITTEX. *
Let x denote the number of apples which John has;
then,
12
x + x = 2x = 12; hence, x = -- = 6.
2t
NOTE. — When x is written with the sign + before it,
it is read plu-s x : and the line above, is read, x phis ar
equals 12.
10 INTRODUCTION.
NOTE. — When x is written by itself, it is read one x»
and is the same as, la; ;
x or he, means once x, or one x,
2x, " twice x, or two x,
3x, " three times x, or three x,
4x, " four times x, or four «,
<fec., &c., &c.
2. What is x + x equal to ?
3. What is jc + 2x equal to ?
4. What is x + 2x + # equal to ?
5. What is a; + 5x + x equal to ?
G. What is x + 2x + 3x equal to ?
7. James and John together have twenty-four peaches,
and one has as many as the other : how many has each ?
ANALYSIS. — Let x denote the number which James has ;
then, since they have an equal number, x will also denote
the number which John has, and twice x will denote the
number which both have, which is 24. If twice x is equal
to 24, x will be equal to 24 divided by 2, which is 12 ;
therefore, each has 12 peaches.
•VVEITTKX.
Let x denote the number of peaches which James has ;
then,
24
x + x = 2x = £4 ; hence, x — — = 12.
VERIFICATION.
A Verification is the operation of proving that the num-
ber found will satisfy the conditions of the question. Thus,
James1 apples. John's apples.
12 + 12 = 24.
NOTE. — Let the following questions be analyzed, written,
and verified, in exactly the same manner as the above.
M E N T A I, K X K R 0 I S K S . 11
8. William and John together have 36 pears, and one has
as many as the other : how many has each ?
9. What number added to itself will make 20 ?
10. James and John are of the same age, and the sum of
their ages is 32 : what is the age of each?
11. Lucy and Ann are twins, and the sum of their ages
is 1 6 : what is the age of each ?
12. What number is that which added to itself will
make 30?
13. What number is that which added to itself will
make 50?
14. Each of two boys received an equal sum of money at
Christmas, and together they received 60 cents : how much
had each ?
15. What number added to itself will make 100?
16. John has as many pears as William; together they
have 72 : how many has each?
17. What number added to itself will give a sum equal
to 46?
1 8. Lucy and Ann have each a rose bush with the same
number of buds on each ; the buds on both number 46 :
how many on each?
LESSON H.
1. John and Charles together have 12 apples, and Charles
has twice as many as John : how many has each ?
ANALYSIS. — Let x denote the number of apples which
John has ; then, since Charles has twice as many, 2x will
denote his share, and x + 2aj, or 3oj, will denote the
number which they both have, which is 12. If 3x is equal
to 12, x will be equal to 12 divided by 3, which is 4;
therefore, John has 4 apples, and Charles, having twice as
many, has 8.
12 INTKOPUCTION.
W1UTTEN.
Let x denote the number of apples John has ; then,
2x will denote the number of apples Charles has ; and
x 4- 2a; = 3x = 12, the number both have; then,
12
x = — = 4, the number John has ; and,
3
2x = 2 x 4 = 8, the number Charles has.
VERIFICATION.
4 + 8 = 12, the number both have.
2. William and John together have 48 quills, and William
has twice as many as John : how many has each ?
3. What number is that which added to twice itself, will
give a number equal to 60 ?
4. Charles' marbles added to John's make 3 times as many
as John has; together they have 51 : how many has each?
ANALYSIS. — Since Charles' marbles added to John's make
three times as many as Charles has, Charles must have one
third, and John two thirds of the whole.
Let x denote the number which Charles has ; then 2x
will denote the number which John has, and x + 2#, or
So;, will denote what they both have, which is 51. Then, if
3x is equal to 51, x will be equal to 51 divided by 3,
which is 17. Therefore, Charles has 17 marbles, and John,
having twice as many, has 34.
WRITTEN.
Let x denote the number of Charles' marbles; then,
2x will denote the number of John's marbles ; and
3x = 51, the number of both ; then,
51
x — — = 17, Charles' marbles; and
17 X 2 = 34, John's marbles.
MENTAL EXERCISES. 13
5. What number added to twice itself will make 75 ?
6. "What number added to twice itself will make 57 ?
7. What number added to twice itself will make 39?
8. What number added to twice itself will give 90 ?
9. John walks a certain distance on Tuesday, twice aa
far on Wednesday, and in the two days he walks 27 miles •
how far did he walk each day ?
10. Jane's bush has twice as many roses as Nancy's: and
jn both bushes there are 36 : how many on each ?
11. Samuel and James bought a ball for 48 cents ; Samuel
paid twice as much as James : what did each pay ?
12. Divide 48 into two such parts that one shall be double
the other.
13. Divide 66 into two such parts that one shall be double
the other.
14. The sum of three equal numbers is 12 : what are the
numbers?
ANALYSIS. — Let x denote one of the numbers; then,
since the numbers are equal, x will also denote each of
the others, and x plus x plus #, or 3x will denote their
sum, which is 12. Then, if 3x is equal to 12, x will be
equal to 12 divided by 3, which is 4 : therefore, the numbers
are 4, 4, and 4.
WRITTEN.
Let x denote one of the equal numbers ; then,
x + x + x = 3x = 12; and
12
"
VERIFICATION.
4 + 4 + 4 = 12.
15. The sum of three equal numbers is 24 : what are the
numbers?
16. The sum of three equal numbers is 36 : what are the
numbers ?
I 2f T 14 O U U C T I O N .
17. The sum of three equal numbers is 54 : what .ire the
numbers ?
LESSON III.
1. What number is that which added to three times itself
will make 48 ?
ANALYSIS. — Let x denote the number; then, 3x will
denote three times the number, and x plus 3x, or 4»,
Avill denote the sum, which is 48. If 4x is equal to 48,
x will be equal to 48 divided by 4, which is 12; there-
fore, 12 is the required number.
WRITTEN.
Let x denote the number; then,
3x = three times the number ; and
x + 3x = 4x — 48, the sum : then,
x — -- = 12, the required number.
VERIFICATION.
12^ 34f-12 = 12 + 36 = 48.
NOTE. — All similar questions are solved by the same
form of analysis.
2. What number added to 4 times itself will give 40 ?
3. What number added to 5 times itself will give 42 ?
4. What number added to 6 times itself will give 63 ?
5. What number added to 7 times itself will give 84 ?
6. What number added to 8 times itself will give 81 ?
7. What number added to 9 times itself will give 100?
8. James and John together have 24 quills, and John has
three times as many as James : how many has each ?
9. William and Charles have 64 marbles, and Charles has
7 times as many as William : how many has each ?
M K N T A L K X K R C J S K S 15
10. James and John travel 96 miles, aud James travels
11 times as far as John : how far does each travel ?
11. The sum of the ages of a father and son is 84 years;
and the father is 3 times as old as the son : what is the age
of each ?
12. There are two numbers of which the greater is 7
times the less, and their sum is 72 : what are the numbers?
13. The sum of four equal numbers is 64: what are the
numbers ?
14. The sum of six equal numbers is 54 : what are the
numbers ?
15. James has 24 marbles ; he loses a certain number, and
then gives away 7 times as many as he loses which takes all
he has : how many did he give away ? Verify.
16. William has 36 cents, and divides them between his
two brothers, James and Charles, giving one, eight times as
many as the other : how many does he give to each ?
17. What is the sum of x and 3#? Of a; and 7a;?
Of x and 5x? Of x and I2x?
LESSOR IV.
1. If 1 apple costs 1 cent, what will a number of apples
denoted by x cost?
ANALYSIS. — Since one apple costs 1 cent, and since x
denotes any number of apples, the cost of x apples will be
as many cents as there are apples : that is, x cents.
2. If 1 apple costs 2 cents, what will x apples cost?
ANALYSIS. — Since one apple costs 2 cents, and since x
denotes the number of apples, the cost will be twice as many
cents as there are apples : that is 2x cents.
3. If 1 apple costs 3 cents, what will x apples cost ?
4. If 1 lemon costs 4 cent?, what will x lemons cost?
16 INTRODUCTION.
5. If 1 orange costs 6 cents, what will x oranges cost ?
6. Charles bought a certain number of lemons at 2 cents
apiece, and as many oranges at 3 cents apiece, and paid in all
20 cents : how many did he buy of each ?
ANALYSIS. — Let x denote the number of lemons ; then,
since he bought as many oranges as lemons, it will also
denote the number of oranges. Since the lemons were
2 cents apiece, 2x will denote the cost of the lemons ; and
since the oranges were 3 cents apiece, 3x will denote
the cost of the oranges ; and 2x + 3.r, or 5.r, will denote
the cost of both, which is 20 cents. Now, since 5x cents
are equal to 20 cents, x will be equal to 20 cents divided by
5 cents, which is 4 : hence, he bought 4 of each.
WRITTEN.
Let x denote the number of lemons, or oranges ; then,
2x = the cost of the lemons ; and
3x = the cost of the oranges ; hence,
2x + 3x — ox = 20 cents = the cost of lemons and
oranges ; hence,
20 cents f
x = - - — 4, the number of each.
5 cents
VERIFICATION.
4 lemons at 2 cents each, give, 4x2= 8 cents.
4 oranges at 3 cents each, " 4 x 3 =± 12 cents.
Hence, they both cost, 8 cents +12 cents = 20 cents.
7. A farmer bought a certain number of sheep at 4 dollars
apiece, and an equal number of lambs at 1 dollar apiece,
and the whole cost 60 dollars: how many did he buy of
each ?
8. Charles bought a certain number of apples at 1 cent
apiece, and an equal number of oranges at 4 cents apiece, and
paid 60 cents in all : how many did he buy of each ?
M K X T A I. E X E R C I S t S . 17
9. James bought an equal number of apples, pears, and
lemons ; he. paid 1 cent apiece fqr the apples, 2 cents apiece
for the pears, and 3 cents apiece for the lemons, and paid
72 cents in all : how many did he buy of each ? Verify.
10. A farmer bought an equal number of sheep, hogs,
and calves, for which he paid 108 dollars; he paid 3 dollars
apiece for the sheep, 5 dollars apiece for the hogs, and
4 dollars apiece for the calves : how many did he buy of
each?
11. A farmer sold an equal number of ducks, geese,
and turkeys, for which he received 90 shillings. The ducks
brought him 3 shillings apiece, the geese 5, and the turkeys
7 : how many did he sell of each sort ?
12. A tailor bought, for one hundred dollars, two pieces
of cloth, each of which contained an equal number of yards.
For one piece he paid 3 dollars a yard, and for the other
2 dollars a yard : how^many yards in each piece ?
13. The sum of three numbers is 28 ; the second is twice
the first, and the third twice the second: what are the
numbers ? Verify.
14. The sum of three numbers is 64 ; the second is 3 times
the first, and the third 4 times the second : what are the
numbers ?
LESSON V.
1. If 1 yard of cloth costs x dollars, what will 2 yards
cost ?
ANALYSIS. — Two yards of cloth will cost twice as much as
one yard. Therefore, if 1 yard of cloth costs x dollars,
2 yards will cost twice x dollars, or 2x dollars.
2. If 1 yard of cloth costs x dollars, what will 3 yards
cost ? Why ?
18 I N T K O I) U C T I O N .
3. If 1 orange costs x cents, what wil) i crai>g6b cost?
Why ? 8 oranges ?
4. Charles bought 3 lemons and 4 ora.ige?, for which he
paid 22 cents. He paid twice as much for an orange as for
a lemon : what was the price of each ?
ANALYSIS. — Let x denote the price of a lemon ; then, 2x
will denote the price of an orange ; 3x will denote the cost,
of 3 lemons, and 8x the cost of 4 oranges ; hence, 3x plus
8z, or lla;, will denote the cost of the lemons and oranges,
which is 22 cents. If lla: is equal to 22 cents, x is equal to
22 cents divided by 11, which is 2 cents: therefore, the
price of 1 lemon is 2 cents, and that of 1 orange 4 cents.
WRITTEN".
Let x denote the price of 1 lemon ; then,
2x = 1 orange ; and,
3o; -f 8x == llx = 22 cts., the cost of lemons and oranges;
22 cts
hence, x = — — — - = 2 cts., the price of 1 lemon ;
and, 2x2 = 4 cts., the price of 1 orange.
VERIFICATION.
3x2= 6 cents, cost of lemons,
4 x 4 = 16 cents, cost of oranges.
22 cents, total cost.
5. James bought 8 apples and 3 oranges, for which he
paid 20 cents. He paid as much for 1 orange as for 4 apples:
what did he pay for one of each ?
6. A farmer bought 3 calves and 7 pigs, for which he paid
1 9 dollars. He paid four times as much for a calf as for a
pig : what was the price of each ?
7. James bought an apple, a peach, and a pear, for which
he paid 6 cents. He paid twice as much for the peach as for
M ]•: N T A L EXERCISES. 19
the apple, and three times as much for the pear as for the
apple : what was the price of each ?
8. William bought an apple, a lemon, and an orange, for
which he paid 24 cents. He paid twice as much for the
lemon as for the apple, and 3 times as much for the orange
as for the apple : what was the price of each ?
9. A farmer sold 4 calves and 5 cows, for which he received
120 dollars. He received as much for 1 cow as for 4 calves :
what was the price of each ?
10. Lucy bought 3 pears and 5 oranges, for which she
paid 26 cents, giving twice as much for each orange as for
each pear: what was the price of each?
11. Ann bought 2 skeins of silk, 3 pieces of tape, and a
penknife, for which she paid 80 cents. She paid the same
for the silk as for the tape, and as much for the penknife as
for both : what was the cost of each ?
12. James, John, and Charles are to divide 56 cents
among them, so that John shall have twice as many as
James, and Charles twice as many as John: what is the
share of each ?
13. Put 54 apples into three baskets, so that the second
shall contain twice as many as the first, and the third as
many as the first and second : how many will there be in
each.
14. Divide 60 into four such parts that the second shall
be double the first, the third double the second, and the
fourth double the third : what are the numbers ?
LESSON VL
1. If 2x + x is equal to 3cc, what is Sx — x equal
to ? Written, 3x — x — 2x.
2. What is 4x — x equal to ? Written,
4x — x = 3x.
20 INTRODUCTION.
3. What is Sx minus Qx equal to ? Written,
8x — 6x = 2x.
4. What is 12a; — 9x equal to? Ans. 3%
5. What is 15x — Ix equal to ?
6. What is 11 x — ISx equal to? Ans. 4a\
7. Two men, who are 30 miles apart, travel towards each
other ; one at the rate of 2 miles an hour, and the other at
the rate of 3 miles an hour : how long before they will meet?
ANALYSIS. — Let x denote the number of hours. Then,
since the time, multiplied by the rate, will give the distance,
2x will denote the distance traveled by the first, and 3x
the distance traveled by the second. But the sum of the
distances is 30 miles ; hence,
2x + 3x = 5x = 30 miles ;
and if 5x is equal to 30, a; is equal to 30 divided by 6,
which is 6 : hence, they will meet in 6 hours.
WRITTEN".
Let x denote the time in hours ; then,
2* = the distance traveled by the 1st; and
3* = " " 2d.
By the conditions,
2a + 3x = 5x = 30 miles, the distance apart ;
30
hence, x = — = 6 hours.
5
VERIFICATION.
2x6 = 12 miles, distance traveled by the first.
3x6 — 18 miles, distance traveled by the second.
30 miles, whole distance.
8. Two persons are 10 miles apart, and are traveling ID
the same direction ; the first at the rate of 3 miles an hour,
and the second at the rate of 5 miles : how long, before the
second will overtake the first ?
M K N T A L ]•; X K U C I 8 E 8 . 21
ANALYSIS. — Let x denote the time, in hours. Then, Sx
will denote the distance traveled by the first in x hours;
and 5x the distance traveled by the second. But Avhen
the second overtakes the first, he will have traveled 10 miles
more than the first : hence,
5x — 3x — 2x = 10 ;
if 2x is equal to 10, x is equal to 5 : hence, the second will
overtake the first in 5 hours.
W1UTTEN.
Let x denote the time, in hours : then,
3x = the distance traveled by the 1st;
and, 5x = " " 2d;
and, 5x — 3x = 2x = 10 hours;
10
or, x = - - = 5 hours.
2i
VERIFICATION.
3x5 = 15 miles, distance traveled by 1st.
5 X 5 = 25 miles, "• " 2d.
25 — 15 = 10 miles, distance apart.
9. A cistern, holding 100 hogsheads, is filled by two
pipes ; one discharges 8 hogsheads a minute, and the other
12 : in what time will they fill the cistern ?
10. A cistern, holding 120 hogsheads, is filled by 3 pipes ;
the first discharges 4 hogsheads in a minute, the second 7,
and the third 1 : in what time will they fill the cistern ?
11. A cistern which holds 90 hogsheads, is filled by a pipe
which discharges 10 hogsheads a minute; but there is a
waste pipe which loses 4 hogsheads a minute : how long
will it take to fill the cistern ? m
12. Two pieces of cloth contain each an equal number of
yards ; the first cost 3 dollars a yard, and the second 5, and
both pieces cost 96 dollars : how many yards in each?
13. Two pieces of cloth contain each an equal number of
yards ; the first cost 7 dollars a yard, and the second 5 ; the first
22 IN T R O D C C T I O X .
cost CO dollars more than the second : how many yards in
each piece ?
14. John bought an equal number of oranges and lemons
the oranges cost him 5 cents apiece, and the lemons 3 ; and
he paid 56 cents for the whole: how many did he buy of
each kind ?
15. Charles bought an equal number of oranges and
lemons; the oranges cost him 5 cents apiece, and the
lemons 3 ; he paid 14 cents more for the oranges than for
the lemons : how many did he buy of each ?
16. Two men work the sa.me number of days, the one
receives 1 dollar a day, and the other two : at the end of
the time they receive 54 dollars : how long did they work ?
LESSON VIE.
1. John and Charles together have 25 cents, and Charles
has 5 more than John : how many has each ?
ANALYSIS. — Let x denote the number which John has ;
then, x + 5 will denote the number which Charles has, and
x -f- x + 5, or 2x + 5, will be equal to 25, the number
they both have. Since 2x -f 5 equals 25, 2x will be
equal to 25 minus 5, or 20, and x will be equal to 20
divided by 2, or 10: therefore, John has 10 cents, and
Charles 15.
WRITTEN-.
Let x denote the number of John's cents ; then,
x 4- 5 = " Charles' cents ; and,
* x + x + 5 = 25, the number they both have ; or,
2x + 5 = 25 ; and,
2x = 25 — 5 = 20 ; hence,
20
x = — =10, John's number; and,
I
10 + 5 z= 15, Charles' number.
MENTAL EXERCISES. 23
VERIFICATION.
John1!. Charles1.
10 +15 = 25, the sum.
Charles'. John's.
15 - 10 = '5, the difference.
2. James and John have 30 marbles, and John has 4 more
tnan James : how many has each ?
3. William bought 60 oranges and lemons ; there were
20 more lemons than oranges : how many were there of
each sort ?
4. A farmer has 20 more cows than calves; in all he has
36 : how many of each sort ?
5. Lucy has 28 pieces of money in her purse, composed
of cents and dimes ; the cents exceed the dimes in number
by 16 : how many are there of each sort ?
6. What number added to itself, and to 9, will make 29 ?
7. What number added to twice itself, and to 4, will
make 25 ?
8. What number added to three times itself, and to 12,
will make 60 ?
9. John has five times as many marbles as Charles, and
what they both have, added to 14, makes 44 : how many has
each ?
10. There are three numbers, of which the second is twice
the first, and the third twice the second, and when 9 is
added to the sum, the result is 30 : what are the numbers?
11. Divide 13 into two such parts that the second shall
be two more than double the first : what are the parts ?
12. Divide 50 into three such parts that the second shall
be tAvice the first, and the third exceed six times the first
by 4 : what are the parts?
13. Charles has twice as many cents as James, and John
24 I N T li O D U C T I O N .
has twice as many as Charles ; if 7 be added to what they
all have, the sum will be 28 : how many has each ?
14. Divide 15 into three such parts that the second shall
be 3 times the first, the third twice the second, and 5 over :
what are the numbers ?
15. An orchard contains three kinds of trees, apples, pears,
and cherries; there are 4 times as many pears as apples,
twice as many cherries as pears, and if 14 be added, the
number will be 40 ; how many are there of each ?
LESSON VIII.
1. John after giving away 5 marbles, had 12 left: how
many had he at first ?
ANALYSIS. — Let x denote the number ; then, x minus 5
will denote what he had left, which was equal to 12. Since
x diminished by 5 is equal to 12, x will be equal to 12,
increased by 5 ; that is, to 1 7 : therefore, he had 1 7 marbles.
WRITTEN.
Let x denote the number he had at first; then,
x — 5 = 12, what he had left; and
x = 12 + 5 = 17, what he first had.
VERIFICATION.
17 — 5 = 12, what were left.
2. Charles lost 6 marbles and has 9 left : how many had
he at first ?
3. William gave 15 cents to John, and had 9 left : how
many had he at first ?
4. Ann plucked 8 buds from her rose bush, and there
-?rere 1 9 left : how many were there at first ?
M E X T A L EXERCISES. 25
5. William took 27 cents from his purse, and there \verr
1 3 left : how many were there at first ?
6. The sum of two numbers is 14, and their difference is 2:
what are the numbers?
ANALYSIS. — The difference of two numbers, added to the
less, will give the greater. Let x denote the less number ;
then, x + 2, will denote the greater, and x + x + 2,
will denote their sum, Avhich is 14. Then, 2x + 2 equals
14; and 2x equals 14 minus 2, or 12: hence, x equals
12 divided by 2, or 6 : hence, the numbers are 6 and 8.
VERIFICATION.
6 + 8 = 14, their sum ; and
8 — 6 = 2, their difference.
7. The sum of two numbers is 18, and their difference 0 :,
what are the numbers ?
8. James and John have 26 marbles, and James has 4 more
than John : how many has each ?
9. Jane and Lucy have 16 books, and Lxicy has 8 more
than Jane : how many has each ?
10. William bought an equal number of oranges and
lemons ; Charles took 5 lemons, after which William had but
25 of both sorts : how many did he buy of each ?
11. Mary has an equal number of roses on each of two
bushes ; if she takes 4 from one bush, there will remain 24
on both : how many on each at first ?
12. The sum of two numbers is 20, and their difference
is 6 : what are the numbers ?
ANALYSIS. — If x denotes the greater number, x — 6 will
denote the less, and x + x — 6 will be equal to 20; hence,
2x equals 20 + 6, or 26, and x equals 26 divided by 2,
equals 13 ; hence the numbers are 13 and 7.
2
I N T It O U U O T I O N .
Let x denote the greater ; then,
x — 6 = the less ; and
x -j- x — 6 = 20, their sum ; hence,
2x = 20 + 6 = 26 ; or,
26
x = — — 13 ; and 13 — 6 = 7.
VERIFICATION.
13 + 7 = 20; and, 13 — 7 = 6.
13. The sum of the ages of a father and son is 60 yeais,
and their difference is just half that number : what are theij
ages?
14. The sum of two numbers is 23, and the larger lacks
1 of being 7 times the smaller : what are the numbers ?
15. The sum of two numbers is 50 ; the larger is equal to
10 times the less, minus 5 : what are the numbers ?
16. John has a certain number of oranges, and Charles
has four times as many, less seven ; together they have 53 :
how many has each ?
17. An orchard contains a certain number of apple trees,
and three times as many cherry trees, less 6 ; the whole num-
ber is 30 : how many of each sort?
LESSON IX.
1. If x denotes any number, and 1 be added to it, what
will denote the sum ? Ans. x -f 1 .
2. If 2 be added to x, what will denote the sum ? If 3
be added, what ? If 4 be added ? <fec.
If to John's marbles, one marble be added, twice his num.
ber will be equal to 10 : how many had he ?
ANALYSIS. — Let x denote the number ; then, x -\- 1 will
denote the number after 1 is added, and twice this number,
M K N T A L EXERCISKS. 27
or 2x + 2, will be equal to 10. If 2x + 2 is equal to 10,
2x will be equal to 10 minus 2, or 8 ; or x will be equal to 4.
WRITTEN;
Let x denote the number of John's marbles ; then,
x -f 1 = the number, after 1 is added ; and
2(oj + 1) = 2a? + 2 = 10; hence,
8
2x = 10 — 2 ; or x = - = 4.
2i
VERIFICATION.
2(4 + 1) = 2 X 5 = 10.
4. Write x -f 2 multiplied by 3. Ans. 3(x + 2).
What is the product ? Ans. 3a;+6.
5. Write x + 4 multiplied by 5. Ans. 5(x + 4).
What is the product ? Ans. 5x + 20.
6. Write x -f- 3 multiplied by 4. ,4ws. 4(ce -f 3).
What is the product? Ans. 4x -+- 12.
7. Lucy has a certain number of books ; her father gives
her two more, when twice her number is equal to 14 : how
many has she ?
8. Jane has a certain number of roses in blossom ; two
more bloom, and then 3 tunes the number is eqiial to 15 :
how many were in blossom at first ?
9. Jane has a certain number of handkerchiefs, and buys
4 more, when 5 times her number is equal to 45 : hoAV many
had she at first ?
10. John has 1 apple more than Charles, and 3 times
John's, added to what Charles has, make 15 : how many
has each ?
ANALYSIS. — Let x denote Charles' apples ; then x + 1 will
denote John's ; and x + 1 multiplied by 3, added to ce, or
!te + 3 + », will be equal to 15, what they both had ; hence,
4« + 3 equals 15; and 4a; equals 15 minus 3, or 12; and
gr. = 4. Write, and verify.
28 I N T K O 1) U C T ION.
] 1. James has two marbles more than William, and twice
his marbles plus twice William's are equal to 16 : how many
has each ?
12. Divide 20 into two such parts that one part shall ex-
ceed the other by 4.
13. A fruit-basket contains apples, pears, and peaches;
there are 2 more pears than apples, and twice as many
peaches as pears; there are 22 in all: how many of each
sort ?
14. What is the sum of x + 3x + 2(x + 1) ?
15. What is the snm of 2 (x + 1) + l(x + 1) + x?
16. What is the sum of x + 5(x + 8) ?
17. The sum of two numbers is 11, and the second is equal
to twice the first plus 4 : what are the numbers ?
18. John bought 3 apples, 3 lemons, and 3 oranges, for
which he paid 27 cents ; he paid 1 cent more for a lemon
than for an apple, and 1 cent more for an oi'ange than for a
lemon : what did he pay for each ?
19. Lucy, Mary, and Ann, have 15 cents; Mary has 1
more than Lucy, and Ann twice as many as Mary ?
LESSON X.
- 1. If x denote any number, and 1 be subtracted from it,
what will denote the difference? Ans. x — 1.
If 2 be subtracted, what Avill denote the difference ? If
3 be subtracted ? 4 ? &c.
2. John has a certain number of marbles ; if 1 be taken
away, twice the remainder will be equal to 12: how many
has he ?
ANALYSIS. — Let x denote the number ; then, x — 1 wilJ
denote the number after 1 is taken away ; and twice this
number, or 2(ar . — 1) — 2a- — 2, will be equal to 12. If 2x
M K N T A -L K X K H O I S 1C S . 29
diminished by 2 is equal to 12, 2x is equal to 12 plus 2, or
14 ; hence, x equals 14 divided by 2, or 7.
WRITTEN.
Let x denote the number ; then,
x — 1 = the number which remained, and
2(x — 1) = 2x — 2 = 12 ; hence,
14
2x = 12 4- 2, or 14 : and x = — = 7.
2
VERIFICATION.
2(7-1) = 14 - 2 = 12 ; also, 2(7 — 1) = 2 X 6 = 12
3. Write 3 times x — 1. Ans. 3(x — 1).
What is the product equal to ? Ans. 3x — 3.
4. Write 4 times x — 2. Ans. 4(x — 2).
What is the product equal to ? Ans. Ix — 8.
5. Write 5 times x — 5. Ans. 5(x — 5).
What is the product equal to ? Ans. 5x — 25.
6. If x denotes a certain number, will x — 1 denote a
greater or less number ? how much less ?
7. If x — 1 is equal to 4, Avhat will x be equal to ?
Ans. 4 + 1, or 5.
8. If x — 2 is equal to 6, what is x equal to ?
9. James and John together have 20 oranges ; John has
6 less than James: how many has each?
10. A grocer sold 12 pounds of tea and coffee ; if the tea
be diminished by 3 pounds, and the remainder multiplied by
2, the product is the number of pounds of coffee : how many
pounds of each ?
11. Ann has a certain number of oranges; Jane has 1 less,
and twice her number added to Ann's make 13 : how many
has each ?
AXALTSIS. — Let x denote the number of oranges which
Ana has; then, x — 1 will denote the number Jane has,
30 INTRODUCTION.
and x + 2x — 2, or 3x — 2, will denote the number both
have, which is 13. If 3x — 2 equals 13, 3x will be equal
to 13 + 2, or 15 ; and if 3x is equal to 15, x will be equal
to 15 divided by 3, which is 5 : hence, Ann has 5 oranges
and Jane 4.
WRITTEN.
Let x denote the number Ann has ; then,
x — I = the number Jane has; and
2(x — 1) = 2x — 2 = tAvice what Jane has; also,
* + 2x — 2 = 3x — 2 = 13; hence,
15
3x = 13 + 2 = 15; or x = — = 5.
3
VERIFICATION.
5—4 = 1 ; and 2x4 + 5 = 13.
12. Charles and John have 20 cents, and John has 6 less
thaii Charles : how many has each ?
13. James has twice as many oranges as lemons in his bas-
ket, and if 5 be taken from the whole number, 19 will re-
main : how many had he of each ?
14. A basket contains apples, peaches, and pears; 29 in
all. If 1 be taken from the number of apples, the remainder
will denote the number of peaches, and twice that remainder
will denote the number of pears : how many are there of
each sort ?
15. If 2x — 5 equals 15, what is the value of x?
16. If 4x — 5 is equal to 11, what is the value of a;?
17. If 5x — 12 is equal to 18, what is the value of x?
18. The sum of two numbers is 32, and the greater ex-
ceeds the less by 8 : what are the numbers ?
19. The sum of 2 numbers is 9 ; if the greater number
be diminished by 5, and the remainder multiplied by 3, the
product will be the less number : what are the numbers?
20. There are three numbers such that 1 taken from the
MENTAL EXERCISES. 31
first will give the second ; the second multiplied by 3 will
give the third ; and their sum is equal to 26 : what are the
numbers ?
21. John and Charles together have just 31 oranges; if
I be taken from John's, and the remainder be multiplied by
5, the product will be equal to Charles' number: how many
has each ?
22. A basket is filled with apples, lemons, and oranges, in
all 26 ; the number of lemons exceed the apples by 2, and
the number of oranges is double that of the lemons : how
many are there of each ?
LESSON XL
1. John has a certain number of apples, the half of which
is equal to 1 0 : how many has he ?
ANALYSIS. — Let x denote the number of apples ; then,
x divided by 2 is equal to 10 ; if one half of x is equal to
1 0, twice one-half of cc, or #, is equal to twice 10, which is
20 ; hence, x is equal to 20.
NOTE. — A similar analysis is applicable to any one of the
fractional units. Let each question be solved according to
the analysis.
2. John has a certain number of oranges, and one-third of
his number is 15 : how many has he?
3. If one-fifth of a number is C, what is the number?
4. If one-twelfth of a number is 9, what is the number?
5^ What number added to one-half of itself will give a
sum equal to 12?
ANALYSIS. — Denote the number by x ; then, x plus one-
half of a; equals 12. But x plus one-half of a; equals three
halves of x: hence, three halves of x equal 12. If three
halves of a; equal 12, one-half of x equals one-third of 12,
32 INTRODUCTION
or 4. If one-half of x equals 4, x equals twice 4, or 8 '
hence, x equals 8.
WRITTEN.
Let x denote the number; then,
1 3
x + -x = -x = 12; then,
-x — 4, or x = 8.
VERIFICATION.
8 4- I = 8 + 4 = 12.
6. What number added to one-third of itself will give a
sum equal to 12?
' 7. What number added to one-fourth of itself will give
a sum equal to 20 ?
8. What number added to a fifth of itself will make 24 ?
9. What number diminished by one-half of itself will
leave 4 ? Why ?
10. What number diminished by one-third of itself will
leave 6 ?
11. James gave one-seventh of his marbles to William,
and then has 24 left : how many had he at first ?
12. What number added to two-thirds of itself will give
a sum equal to 20 ?
13. What number diminished by three-fourths of itself
will leave 9 ?
14. What number added to five-sevenths of itself will
make 24 ?
15. What number diminished by seven-eighths of itself
will leave 4 ?
16. What number added to eight-ninths of itself wiJJ
make 34 ?
ELEMENTARY ALGEBRA.
CHAPTER I,
DEFINITIONS AND EXPLANATORY SIGNS.
1. QUANTITY is anything that can be measured, as num-
ber, distance, weight, time, &c.
To measure a thing, is to find how many times it contains
some other thing of the same kind, taken as a standard. The
assumed standard is called the unit of measure.
2. MATHEMATICS is the science which treats of the pro-
perties and relations of quantities.
In pure mathematics, there are but eight kinds of quantity,
and consequently but eight kinds of UNITS, viz. : Units of
Number ; Units of Currency ; Units of Length ; Units of
Surf act ; Units of Volume; Units of Weight; Units of
Time ; and Units of Angular Measure.
8. ALGEHRA is a branch of Mathematics in which the
quantities considered are represented by letters, and the
operations to be performed are indicated by signs.
1. What is quantity? What is the operation of measuring a thing ?
What is the assumed standard called ?
2. What is Mathematics ? How many kinds of quantity are there iu
the pure mathematics? Name the units of those quantities.
:5. What is Algebra?
1*
34: ELK M E X T A li Y A L G K B H A .
4. The quantities employed in Algebra are of two kinds,
Known and Unknown:
Known Quantities are those whose values are given ;
and
Unknown Quantities are those whose values are re-
quired.
Known Quantities are generally represented by the lead-
ing letters of the alphabet, as, a, #, c, &c.
Unknown Quantities are generally represented by the.
final letters of the alphabet ; as, x, y, z, &c.
When an unknown quantity becomes known, it is often
denoted by the same letter with one or more accents ; as,
ce', a;", x". These symbols are read: x prime ; x second;
x third, &c.
5. The SIGN OF ADDITION, +, is called plus. When
placed between two quantities, it indicates that the second
is to be added to the first. Thus, a + 5, is read, a plus #,
and indicates that b is to be added to a. If no sign is
written, the sign + is understood.
The sign +, is sometimes called \kepositive sign, and the
quantities before which it is written are called 2)ositiee quan-
tities, or additive quantities.
6. The SIGN OF SUBTRACTION, — , is called minus. When
placed between two quantities, it indicates that the second
is to be subtracted from the first. Thus, the expression,
4. How many kinds of quantities arc employed in Algebra ? How are
they distinguished ? What are known quantities ? What are unknown
quantities? By what are the known quantities represented? By what
are the unknown quantities represented? When an unknown quantity
becomes known, how Is it often denoted?
5. What is the sign of ailditiou called? When placed between two
quantities, what does it indicate ?
G. What is the sign of subtraction called ? When placed between two
quantities, what does it indicate ?
DEFINITION OK T K K M S . 36
c — d, rcxd c minus d, indicates that d is to be subtracted
from c. ff a stands for 6, and d for 4, then a — d is equal
to 6 — 4, which is equal to 2.
The v.gn — , is sometimes called the negative sign, and the
quantities before which it is written are called negative quan-
tities, or subtractive quantities.
7. The SIGN OF MULTIPLICATION, x , is read, multiplied
by, or into. When placed between two quantities, it indi-
cates that the first is to be multiplied by the second. Thus,
a X b indicates that a is to be multiplied by b. If a stands
for 7, and b for 5, then, a X b is equal to 7 X 5, which is
equal to 35.
The multiplication of quantities is also indicated by simply
writing the letters, one after the other ; and sometimes, by
placing a point between them ; thus,
a x b signifies the same thing as ab, or as a.b.
a X b x c signifies the same thing as abc, or as a.b.c.
§. A FACTOR is any one of the multipliers of a product.
Factors are of two kinds, numeral and literal. Thus, in the
expression, 5abc, there are four factors : the numeral factor,
5, and the three literal factors, «, b, and c.
9. The SIGN OF DIVISION, -f-, is read, divided by. When
written between two quantities, it indicates that the first is
to be divided by the second,
7. How is the sign of multiplication read ? When placed between two
quantities, what does it indicate ? In how many ways may multiplication
be indicated ?
8. What is a factor ? How many kinds of factors are there ? How
many factors are there in 3abc ?
9. How is the sign of division read ? When written between two quan-
tities, what does it indicate ? How many ways are there of indicating
division ?
36 E L E M E N T A K Y A L G K B U A .
There are three signs used to denote division. Thns,
a ~- b denotes that a is to be divided by b.
7 denotes that a is to be divided by b.
a | b denotes that a is to be divided by b.
10. The SIGN OF EQUALITY, =, is read, equal to. When
written between two quantities, it indicates that they are
equal to each other. Thus, the expression, a -f b = c, in-
dicates that the 'sum of a and b is equal to c. If a stands
for 3, and b for 5, c will be equal to 8.
11. The SIGN OF INEQUALITY, > < , is read, greater
than, or less than. When placed between two quantities,
it indicates that they are unequal, the greater one being
placed at the opening of the sign. Thus, the expression,
a > b, indicates that « is greater than b ; and the expres-
sion, c < d, indicates that c is less than d.
12. The sign . • . means, therefore, or consequently.
13. A COEFFICIENT is a number written before a quan-
tity, to show how many times it is taken. Thus,
a + a-\-a + a + a = 5a,
in which 5 is the coefficient of a.
A coefficient may be denoted either by a number, or a
letter. Thus, 5x indicates that x is taken 5 times, and ax
10. What is the sign of equality? When placed between two quanti-
ties, what does it indicate ?
11. How is the sign of inequality read ? Which quantity is placed on
rite side of the opening ?
1 2. What does . • . indicate ?
13. What is a coefficient? How many times is a taken in 5a. By
-.nnt may a coefficient be denoted? If no coefficient is written, what
coefficient is understood ? In 5ar, how many times is ax taken? How
many times is x taken ?
D K F I N 1 T I O X OF T K R M 8 . 37
indicates that x is taken a, times. If no coefficient is writ-
ten^ the coefficient 1 is understood. Thus, a is the same
as la.
14. AN EXPONENT is a number written at the right and
above a quantity, to indicate how many times it is taken as
a factor. Thus,
a x a is written a2,
a x a x a " a3,
a X a x a x a " a*,
&c., &c.,
in which 2, 3, and 4, are exponents. The expressions are
read, a square, a cube or a third, a fourth ; and if we have
am, in which a enters ra times as a factor, it is read, a to
the mth, or simply a, mth. The exponent 1 is generally
omitted. Thus, a1 is the same as a, each denoting that a
enters but once as a factor.
15. A POWER is a product which arises from the multi-
plication of equal factors. Thus,
a X « = a2 is the square, or second power of a.
a X a X a = a3 is the cube, or third power of a.
a X a x a x a = a4 is the fourth power of a.
a x a X . . . . = am is the mih power of a.
16. A ROOT of a quantity is one of the equal factors.
Flic radical sign, -y/ , when placed over a quantity, indi-
cates that a root of that quantity is to be extracted. The
root is indicated by a number written over the radical sign,
14. What is an exponent? In a3, how many times is a taken as a fac-
tor? When no exponent is written, what is understood?
1 5. What is a power of a quantity ? What is the third power af 2 ?
Of 4 » Of 6 ?
16. What is the root of a quantity? What indicates a root? What
indicates the kind of root? What is the index of the square root? Of
the cube root ? Of the mih root ?
38 ELEMENTARY ALGEBRA..
called an index. When the index is 2, it is generally omit-
ted. Thus,
^/a, or y/a, indicates the square root of a.
$/a indicates the cube root of a.
\/a indicates the fourth root of a.
'H/a indicates the with root of a.
17. An ALGEBRAIC EXPRESSION is a quantity written in
algebraic language. Thus,
j is the algebraic expression of three tunes
( the number denoted by a ;
2 1 is the algebraic expression of five times
( the square of a ;
is the algebraic expression of seven times
the the cube of a multiplied by the
( square of b ;
( is the algebraic expression of the differ-
Sa — 5b< ence between three times a and five
( times b ;
is the algebraic expression of twice the
. square of ff, diminished by three times
the product of a by £>, augmented by
four times the square of b.
18. A TERM is an algebraic expression of a single quan-
tity. Thus, 3a, 2a£, — 5a2£2, are terms.
19. The DEGREE of a term is the number of its literal
factors. Thus,
j is a term of the first degree, because it contains but
( one literal factor.
17. WThat is an algebraic expression
18. What is a term?
19. What is the degree of a term? What determines the degree of a term?
DEFINITION OF TEKMS. 39
_ 2 j is of the second degree, because it contains two lite-
( ral factors.
is of the fourth degree, because it contains four literal
factors. The degree of a term is deteraiined by
the sum of the exponents of all its letters.
20. A MONOMIAL is a single term, unconnected with any
other by the signs + or — ; thus, 3a2, 3 J3a, are monomials.
21. A POLYNOMIAL is a collection of terms connected
by the signs + or — ; as,
3a - 5, or, 2a3 — 3b + 4bz.
22. A BINOMIAL is a polynomial of two terms ; as,
a + b, 3a2 — c2, Gab — c2.
23. A TRINOMIAL is a polynomial of three terms ; as,
abc — a3 4- c3, ab — gh — f.
24. HOMOGENEOUS TERMS are those which contain the
same number of literal factors. Thus, the terms, abc, — a3,
4- c3, are homogeneous ; as are the terms, a£>, — gh.
25. A POLYNOMIAL is HOMOGENEOUS, when all its terms
aio homogeneous. Thus, the polynomial, abc — a3 + c3, is
homogeneous ; but the polynomial, ab — gh — f is not ho-
mogeneous.
i£6. SIMILAR TERMS are those which contain the same
literal factors affected with the same exponents. Thus,
lab -f- Bab — 2ab,
20. What is a monomial ?
21. What is a polynomial?
9.2. What is a binomial ?
23. What is a trincmial?
24. What are homogeneous terms ?
*>.5. When is a polynomial homogeneous?
26. What are similar terms?
40 ELEMENTARY ALGEBRA..
are similar terms ; and so also are,
but the terms of the first polynomial and of the last, are not
similar.
27. THE VrxcuLUM, - — , the Bar \ , the Paren-
thesis, ( ) , and the Brackets, [ ] , are each used to con-
nect several quantities, which are to be operated upon in the
same manner. Thus, each of the expressions,
6-f-cxcc, + b
(a + & -f c) x
and [a + # -f c] x x,
indicates, that the sum of a, £, and c, is to be multiplied
by x.
2§. THE RECIPROCAL of a quantity is 1, divided by that
quantity; thus,
1 1 c
a' o~+~ft' d'
are the reciprocals of
A d
a . a + o . -•
c
29. THE NUMERICAL VALUE of an algebraic expression,
is the result obtained by assigning a numerical value to each
letter, and then performing the operations indicated. Thus,
the numerical value of the expression,
ab + be + t7,
•when, a = 1, b = 2, c = 3, and d = 4, is
1X2 + 2X34-4 — 12;
by performing the indicated operations.
27. For what is the vincular used ? Point out the other ways ir. -whici
this may be done ?
28. "What is the reciprocal of a quantity?
29. What is the numerical value (fan algebraical expression?
ALGEBRAIC KXPKE8SION8. 41
EXAMPLES IX WHITING ALGEBRAIC EXPRESSIONS.
1. Write a added to b. Ans. a -f b.
2. Write b subtracted from a. Ans. a — b.
Write the following :
3. Six times the square of a, minus twice the square of b.
4. Six times a multiplied by £, diminished by 5 times c
cube multiplied by d.
5. Nine times a, multiplied by c plus d, diminished by
8 times b multiplied by d cube.
6. Five times a minus #, plus 6 times a cube into 3
cube.
7. Eight times a cube into d fourth, into c fourth, plus
9 times c cube into d fifth, minus 6 times a into £, into c
square. ,
8. Fourteen times a plus 5, multiplied by a minus b,
plus 5 times «, into c plus c?.
9. Six times a, into c plus </, minus 5 times 5, into a plus
<?, minus 4 times a cube b square.
10. Write #, multiplied by c plus <7, plus y minus </.
11. Write a divided by b + c. Three ways.
12. Write a — b divided by a + b.
13. Write a polynomial of three terms; of four terms; of
five, of six.
14. Write a homogeneous binomial of the first degree; of
the second ; of the third ; 4th ; 5th ; 6th.
15. Write a homogeneous trinomial of the first degree;
.with its second and third terms negative; of the second
degree ; of the 3rd ; of the 4th.
1G. Write in the same column, on the slate, or black-board,
a monomial, a binomial, a trinomial, a polynomial of four
terms, of five terms, of six terms and of seven terms, and all
of the same degree.
42 ELEMENT A 11 Y ALGEBRA
INTERPRETATION OF ALGEBRAIC LANGUAGE.
Find the nunierial values of the following expressions,
when,
a = 1, b = 2, c — 3, d — 4.
1. ab + be. Ans. 8.
2. a •+- be + d. Ans. 11.
3. «c? + b — c. Ans. 3.
4. aft + ftc — d. Ans. 4.
5. (a 4- ft) c2 — a*. .4ns. 23.
6. (a + ft) (a* — ft.) Ans. 6.
7. (aft + ad) c + d. Ans. 22.
8. (aft H c) (ad — a). Ans. 15.
9. 3a2ft2 - 2(a + d 4- 1). ^Ins. 0.
10. -- x (a + <?) ^?25. 10.
9
«2 + J2 + C2 a3 _J_ J3 + C3 _ J
11. ' — X • Ans. 32.
, ab* — c — a3 4az - b + d3
12.- -— -^ ^5. 4.
Find the numerical values of the following expressions,
when,
a, = 4, b = 3, c = 2, and <? = 1.
„ a ft
13. - - - + c -
- (7.
^ws. 2.
fab a —
aT\
14 *ii
^I?i5. 15.
I o
\ o o
/
15. [(a2ft + l)d]
! - (a2ft + d).
^71S. 1.
16. 4(aftc )
X (30c3 — ab3d3).
Ans. 11088.
. a + ft + c
abed 4a2 + ft2 - aT2
Ans. 14|.
' a - ft + d
aft ftc + ft
15(a+f?+ft)
T _l_ . v /Tr3/i3/>3/-73
Ana 14 CK
A DDITION.
CHAPTER II.
FUNDAMENTAL OPERATIONS.
ADDITION.
30. ADDITION is the operation of finding the simplest
equivalent expression for the aggregate of two or more
algebraic quantities. Such expression is called their SUM.
When the terms are similar and have like signs.
+
31. 1. What is the sum of a, 2a, 3a, and 4a?
Take the sum of the coefficients, and annex the
literal parts. The first term, or, has a coefficient, _|_
1, understood (Art. 13).
2. What is the sum of 2«#, 3a#, 6a£>, and ab.
When no sign is writtten, the sign -f is under-
stood (Art 5).
Add the following :
(3.) (4.) (5.)
a Sab lac -f-
a lab
- 10a
2ab
3ab
Gab
ab
I2ab
(6.)
I5ab
Sac
I2ac
3abc
+ 'Jabc
30. What is addition ?
SI. What is the rule for addition when the terms are similar and hare
like sicrns ?
44 ELEMENTARY ALGEBRA.
(7.) (8.) (9.) (10.)
— Sabc — Sad — 2adf — Qabd
— 2abc — 2ad — Gadf — I5abd
— 5abc — 5ad — 8adf — 24abd
Hence, when the terms are similar and have like signs :
RULE.
Add the coefficients, and to their sum prefix the common
sign ; to this, annex the common literal part.
EXAMPLES.
(11.) (12.) (13.)
Qab -{- ax 8«c2 — Sb* lo«J3c* — I2abc*
Sab + Sax lac2 — Sb2 12aft3c* — loabc*
I2ab + 4ax Sac2 — Qb2 ab3c* — abc2
When the terms are similar and have unlike signs.
32. The signs, + and — , stand in direct opposition to
each other.
If a merchant writes + before his gains and — before his
losses, at the end of the year the sum of the plus numbers
•wall denote the gains, and the sum of the minus numbers
the losses. If the gains exceed the losses, the difference,
which is called the algebraic sum, will be plus ; but if the
losses exceed the gains, the algebraic sum will be minus.
1. A merchant in trade gained $1500 in the first quarter
of the year, §4000 in the second quarter, but lost $3000 in
the third quarter, and $800 in the fourth : what was the re-
sult of the year's business ?
1st quarter, + 1500 3d quarter, — 3000
2cl " 3000 4th " — 800
+ 4500 — 3800
_|_ 4500 — 3800 = + 700, or $700 gain.
82. What is the rule when the terms are similar and hare unlike sitms ?
A I) I) ITI O N. 45
2. A merchant in trade gained $1000 in the first quarter,
and $2000 the second quarter ; in the third quarter he lost
$1500, and in the fourth quarter $1800 : what was the result
of the year's business ?
1st quarter, -f 1000 3d quarter — 1500
2d " + 2000 4th " - 1800
-f 3000 - 3300
+ 3000 — 3300 = — 300, or $300 loss.
3. A merchant in the first half-year gained a dollars and
lost b dollars ; in the second half-year he lost a dollars and
gained b dollars : what is the result of the year's business ?
1st half-year, + a — b
2d " — a -f b
Result, 0 0
Hence, the algebraic siim of a positive and negative quan-
tity is their arithmetical difference, icith the sign of the
greater prefixed. Add the following:
•Bab 4acbz
Sab — 8acbz
— (jab acb2 — 2azb2c2
5ab — 3acbz 0
Hence, when the terms are similar and have unlike signs :
I. Write the similar terms in the same column:
B.'. Add the coefficients of the additive terms, and also
the coefficients of the subtractive terms :
HI. Take the difference of these sums, prefix the sign
of the greater, and then annex the literal part.
EXAMPLES.
1. What is the sum of
46 ELEMENTARY ALGEBRA.
Having written the similar terms in the same
column, we find the sum of the positive coeffi-
cients to be 15, and the sum of the negative
coefficients to be — 16 : their difference is — 1 ;
hence, the sum is — azb3.
2. What is the sum of
5a25 — 3a2S + 4a25 — Qa*b — «2J ? Ans.
3. What is the sum of
2a3ic2— 4a3bc2+ 6a3£c2— 8a3bcz+ Ua3bc*? Ans. I7
4. What is the sum of
— 8a?b — Qazb + Uazb? Ans. —
5. What is the sum of
labc2 — abc2 — labc2 + Sabc2 + Gabcz? Ans. 13abcz.
6. What is the sum of
Qcb3- 5cb3— 8ac2+ 20c53+ 9ac2 — 24c53 ? Ans. + ac ,
To add any Algebraic Quantities.
33. 1. What is the sum of 3a, 55, and — 2c?
Write the quantities, thus,
3a + 5b — 2c;
which denotes their sum, as there are no similar terms.
2. Let it be required to find the sum of the quantities,
2a2 4ab
3a2 — Sab + b*
2ab — 5b*
5a2 — 5ab — 4b2
83. What is the rule for the addition of any algebraic quantities?
A 1) i) I T I O .V . 47
From the preceding examples, we have, for the addition
of algebraic quantities, the following
RULE.
I. Write the quantities to be added, placing similar terms
in the same column, and giving to each its proper sign :
IT. Add up each column separately and then annex the
dissimilar terms with their proper signs.
EXAMPLES.
1. Add together the polynomials,
3a2 — 2bz — 4ab, 5az — 52 -j- 2ab, and Sab — 3c2 — 2&3.
The term 3a2 being similar to C .
5a\ we write 8a2 for the result 3*
of the reduction of these two <|
terms, at the same time slightly ___ IT * ~_r".
crossing them, as in the first term. I &a~ + ab — 5bz — 3c2
Passing then to the term — 4ab, which is similar to
+ 2ab and + 3ab, the three reduce to + ab, which is
placed after So2, and the terms crossed like the first term.
Passing then to the terms involving b2, we find their sum
to be — 5b2, after which we write — 3c2.
The marks are drawn across the terms, that none of them
may be overlooked and omitted.
(2.) (3.) (4.)
*labc + 9ax Sax -} Sb I2a — 6c
— 3abc — 3ax Sax — 9b — 3a — Qc
4abc "f 6ax I3ax — Qb Qa — 15c
NOTE. — If a — 5, 5 = 4, c = 2, x = 1, what are the
numerical values of the several sums above found ?
48 ELEMENTARY ALGEBRA.
(5.) (6.) (7.)
9a + / Qax — Sao 3af + g + m
— 6a 4- <; — 7oKC — 9ac <z<7 — So/' — m
— 2o — / ace + I7ac ab — ag + 3g
(8.) (9.)
7ce + 3a5 + 3c 8a;2 + Oaca; -f 13a252c2
— 9a5 — 9c
(10.) (11.)
22/i — 3c — 7/ + 3/7 19a/*2 + 3a'
- 3 A + 8c — 2/ — 9.7 + 5a;
(12.) (13.)
7a? — 9y 4- 52 + 3 — g 8a + b
— x — 3y — 8 — g 2a - b + c
— x + y — 3z + l+7# -3a+& 4-2<?
— 2a 4- 6y + 82 — 1 — g - 6b — 3c + 3d
14. Add together — b + 3c — d — 115e 4- 6/ — 5<7, 35
»» 2c - 3c? - e + 27/, 5c — 8(7 + 3/ - 7^, - 75 - 6c
•f- J7<^+ 9e — 5/+ H<7, — 35 — 5<?— 2e4-6/— 9^ 4- h.
Ans. — 85 — 109e + 37/ - 10^ 4 A.
15. Add together the polynomials 7«25 — 3a5c — 852o
— 9c3 4- cd\ Babe — 5crb 4- 3c3 — 452c 4- cdz, and 4a25
— 8c3 4- 952c - 3d3.
Ans. Qa?b 4- 5abc — 352c — 14c3 4- 2cc72 — 3d3.
16. What is the sum of, 5a25c 4- Qbx — 4<7/, — 3a25c
— 6bx 4- 1 4a/, _ «/ + oftc + 2a25c, 4- 6 a/ — 85a: 4- 6a25e ?
ylws. 10a25c 4- bx + loaf.
17. What is the sum of a2^2 + 3«3m 4-5, - 6«2;z2
- Qa3m — 5, 4- 95 — Qa?m — 5a?n2 ?
Ans. - 10«2>?,2 - IZaPm 4 95.
ADDITION. 49
18. What is tlie sum of 4«3ft2c — IGa^x — 9ax3d,
f Ga3b2c - QaxPd + I7al«, + lGax3d — a*x — 9«3ft2c?
Ans. a?b~c + ax3d.
19. What is the sum of — Iff + 3ft + 4<? - 2ft 43^
- 3ft + 2ft? ^4ws. 0.
20. What is the sum of, aft 4 3xy — m — n, —
— 3m 4- lln 4- e(7, 4- 3.uy 4- 4m — 10;z- + fgt
Ans. aft + cd 4- /]?•
21. What is the sum of 4xy 4- n 4- 6aaj + 9am, — 6<ey
4- 6/1 — 6a# — 8am, 2xy — 7ft. 4- ax — am? Ans. 4- ax.
(22.) (23.) (24.)
2 (a + ft) 5 (a2 - c2) 9(c3 - a/3)
8 (a 4 ft) - 4(a2 - c2) 7(c3 — a/3)
2(a 4 ft) — l(a2 — c2) — 10(c3 — a/3)
7 (a 4- ft) G(c3 - a/3)
NOTE. The quantity within the parenthesis must be
regarded as a single quantity.
25. Add 3a(<72 — 7t2) - 2a(#2 - 7i2) 4- 4a(#2 — A2)
f- 8a(<72 — 7i2) — 2a(^/2 — 7i2). Ans. lla(^72 — 7i2).
26. Add 3c(a2c — ft2) - 9c(a2c - ft2) — 7c(a2c — ft2)
4- 15c(a2c - ft2) 4- c(a2c — ft2). Ans. 3c(a2c — ft2).
34. In algebra, the term add does not always, as in
arithmetic, convey the idea of augmentation ; nor the term
sum, the idea of a number numerically greater than any of
the numbers added. For, if to a we add — ft, we have,
a — ft, which is, arithmetically speaking, a difference be-
tween the number of units expressed by a, and the number
34. Do the words add and sum, in Algebra, convey the same ideas aa
in Arithmetic. What is the algebraic sum of 9 and — 4 ? Of 8 and
— 2 ? May an algebraic sum be negative ? What is the sum of 5 and
— 10? How are such sums distinguished from arithmetical sums?
50 ELEMENTARY ALGEBRA.
of units expressed by b. Consequently, this result is no*
merically less than a. To distinguish this sum from an
arithmetical sum, it is called the algebraic sum.
SUBTRACTION.
35. SUBTRACTION is the operation of finding the differ-
ence between two algebraic quantities.
36. The quantity to be subtracted is called the Subtra-
hend ; and the quantity from which it is taken, is called the
Minuend.
The difference of two quantities, is such a qxiantity as
added to the subtrahend will give a sum equal to the min-
uend.
EXAMPLES.
1. From I7a take 6a.
OPEEATION.
In this example, I7a is the minuend, and 6a
the subtrahend: the difference is lla; because,
lla, added to 6a, gives I7a.
6a
lla
The difference may be expressed by writing the quantities
thus:
17a — 6a = lla;
in which the sign of the subtrahend is changed from -f
to — .
2. From I5x take — 9x.
The difference, or remainder, is such a quantity,
as being added to the subtrahend, — Qx, will
givo the minuend, 15x, That quantity is 24x,
and may be found by simply changing the sign
SUBTRACTION. 51
of the subtrahend, and adding. Whence, we may write,
I5x — (— Qx) = 24aj.
3. From Wax take a — b.
The difference, or remainder, is such a quantity, as added
to a — J, will give the minuend, Wax: what is that quan
tity?
If you change the signs of both
terms of the subtrahend, and add,
OPKRAT10K.
lOax
+ a - b
Rem. lOax — a + b
add + a — b
IQax
you have, lOax — a + b. Is this
the true remainder ? Certainly.
For, if you add the remainder to
the subtrahend, a — b, you obtain
the minuend, lOax.
It is plain, that if you change the signs of all the terms
of the subtrahend, and then add them to the minuend, and
to this result add the given subtrahend, the last sum can be
no other than the given minuend ; hence, the first result is
the true difference, or remainder (Art. 36).
Hence, for the subtraction of algebraic quantities, we hava
the following
BT7LE.
I. Write the terms of the subtrahend under those of the
minuend, placing similar terms in the same column :
II. Conceive the signs of all the terms of the subtrahend
to be changed from + to —, or from — to +, and then
proceed as in Addition.
EXAMPLES OF MONOMIALS.
(1.) (2.) (3.)
From Sab 6ax Qabc
take 2ab Zax tabc
Rom. ab Sax Zabo
* 2LKMENTARY ALGEBRA.
(4.) (5.) (6.)
From IGaWc
taka 9a2&26'
Rem.
(7.) (8.) (9.)
From . Sax 4abx 2am
take 8c Qac ax
Rem. 3a« — 8c 4ate — 9ac 2am — ax
10. From 9«2i2 take 3azb2. Ans.
11. From !Ga2xy take — 15a2xy. Ans.
12. From 12a4?/3 take 8a4y3. Ans.
13. From 19c/5cc8y take — I8a5xsy. Ans.
14. From 3«2Z>3 take Sa3b2. -4«s. 3a2i3
15. From 7a254 take 6«4Z»2. Ans. 7a2^4
16. From Sab2 take «255. ^4ws. 3«5
17. From a:2?/ take y2a7. Ans. x^y — y~x.
18. From Sx^y3 take ccy. -4?w. 3x2y3 — xy.
19. From §cizy3x take a^7/2. -4?z5. 8a2y'Jx — xyz.
20. From 9a2#2 take — 3a2i2. Ans.
21. From 14a2y2 take — 20a2y2. J[n«.
22. From -- 24a455 take 16«4£5. Ans. —
23. From - ISx^y* take — 14cc2y4. Ans. xzy*.
24. From — 47a3#2y take — 5a3x2v Ans. — 42a3#2v.
* V *
25. From — 94«2x2 take 3a2ic2. Ans. — giazxz.
26. From a + cc2 take — y3. ^4^5. « -f x2 + y:i.
27. From a3 + i3 take — a3 — 53. ^tns 2a3 + 2b3.
28. From — 16a2a;3y take — 19a2x3y. Ans. + 3a2ic3y.
29. Froin a2 — x2 take a2 + r«2. Ans. — 2#3.
SUBTRACTION. 53
GENERAL EXAMPLES.
(1.) flj I1')
From 6«c — Sab + c2 6ac — 5ab + c2
take Sac + Sab + 7c *.f~ — Sac — Sab — 7c
rfl . '
Rem. 3ac — Sab + c2 — 7c.^,gj>g) 'Sac — 8ab + c2 — 7c.
(2.) (3.)
From Qax — a -f 3#2 6ya; — 3x2 + 5b
take 9a« — x + b2 yx — 3 -f a
Rem. — Sax — a + x + 2&2. 5ycc — 3<c2 -f 3 -t- 55 — a.
(4.) (5.)
From 5a3 — 4a26 + 352c 4ad — cc?+3a2
take — 2a3 + 3a"b — 8b*c oab - 4cd + 3a2 +
Rem. 7a3 — Ta26 4- 1.1 l~c. — ab
6. From a -f- 8 take c — 5. ^dns. a — c + 13.
7. From 6a2 — 15 take 9a2 + 30. Ana. — 3a2 — 45.
8. From Qxy — 8«2c3 take — *Ixy —
Ans.
9. From a + c take — a — c. -4?w. 2a + 2c.
10. From 4(a + b) take 2(o + b). Ans. 2(a + 5).
11. From 3 (a + x) take (a + a). -4ns. 2 (a + a).
12. From 9(a2 — a2) take — 2(a2 — cc2).
^na. 11 (a2 - x2)
13. From 6a2 — 1552 take — 3a2 + 9i2.
Ans. 9a2 — 2452.
14. -From 3am — 2bn take am — 2bn. Ans. 2am.
15. From 9c2m2 — 4 take 4 — 7c2?n2. -4n*. 16c2;n2 - 8.
16. From Gam -f y take 3a»i — x. Ans Sam + x + y.
17. From 3«jc take 3a« — y. Ans. + y.
54: ELEMENTARY ALGEBRA.
18. From — If + 3m — 8x take — 6/ — 5m — 2x +
3d + 8. Ans. ~ f + Sm — 6x — 3d — 8.
19. From — a — 5b + 7c -f- d take 4ft — c -f 2d + 2&.
Ans. — a — 9ft + 8c — d — 27c.
20. From — 3a + ft — 8c + 7e — 5/ + 3/i - 7a — 13</
take & + 2a — 9c + 8e — 'Jx + 7/ — y — 31 — A:
Ans. — 5a + ft + c — e — 12/ + 3/i - 12y + 3/.
21. From 2x — 4a — 25 -f- 5 take 8 — 55 + a + 6z.
Ans. — 4x — 5a + 3b — 3.
22. From 3a -{- b + c — d — 10 take c + 2a — c?.
uins. a + ft — 10.
23. From 3a + ft + c — d — 10 take ft — 19 + 3a.
Ans. c — d + 9.
24. From a3 + 3&2c + ab2 — abc take ft3 + ab* - ale.
Ans. a3 + 3b*c — b\
25 From 12as + 6a — 4J + 40 take 4ft — 3« -f 4z -f
6d — 10. Jbis. 8a; + 9a — 8ft — Qd + 50.
26. From 2x — 3a + 4ft -f 6c — 50 take 9a + « -f- 6ft
— 6c — 40. ^Ins. a; — 12a — 2ft + 12c — 10.
27. From 6a — 4ft — 12c + 12cc take 2a; — Sa + 4ft
— 6c. -4n*. 14a — 8ft — 6c + lOa.
38. In Algebra, the term difference does not always, as
in Arithmetic, denote a number less than the minuend. For,
if from a we subtract — ft, the remainder will be a -f- b ;
and this is numerically greater than a. "We distinguish
between the two cases by calling this result the algebraic
difference.
38. In Algebra, as in Arithmetic, does the term difference denote a
number less than the minuend ? How are the results in the two cases,
distinguished from each other?
SUBTRACTION. 55
39. When a polynomial is to be subtracted from an al-
gebraic quantity, we inclose it hi a parenthesis, place the
minus sign before it, and then write it after the minuend.
Thus, the expression,
6a2 — (Sab — 2&2 + 2fo),
indicates that the polynomial, Sab — 262 -f- 2ic, is to be
taken, from 6a2. Performing the operations indicated, by
the rule for subtraction, we have the equivalent expression :
6a2 — 3at> + 2b2 — 2bc.
The last expression may be changed to the former, by
changing the signs of the last three terms, inclosing them in
a parenthesis, and prefixing the sign — . Thus,
6«* — Sab + 252 — 2bc = Go? — (Sab - 2b2 + 25c).
In like manner any polynomial may be transformed, as in-
dicated below :
7a3 — 8a2& — 462c + Qb3 = 7a3 — (8a*b + We — 653)
= 7a3 — 9a?b - (4&2c - Qb3).
So,3 - W + c — d = 8a3 — (762 - c + d)
= So3 - 752 - (— c + d).
953 — a + 3a2 - d = Qb3 - (a — 3a2 -f d)
= 96s — a - (- 3a2 + d).
NOTE. — The sign of every quantity is changed when it is
placed within a parenthesis, and also when it is brought out.
40. From the preceding principles, we have,
a — (+ b) = a — b; and
a — (- b) = a + b.
39. How is the subtraction of a polynomial indicated ? How is thi*
indicated operation performed ? How may the result be again put under
the first form ? What is the general rule in regard to the parenthesis ?
40. What is the s'gn which immediately precedes a quantity called?
What is the sign which precedes the parenthesis called ? What is the
66 ELEMENTARY ALGEBRA.
The sign immediately preceding b is called the sign of the
quantity ; the sign preceding the parenthesis is called the
sign of operation ; and the sign resulting from the combin-
ation of the signs,is called the essential sign.
When the sign of operation is different from the sign of
the quantity, the essential sign will be — ; when the sign of
operation is the same as the sign of the quantity, the essen-
tial sign will be -f .
MULTIPLICATIOlSr.
41. 1. If a man earns a dollars in 1 day, how much will
he earn in 6 days?
ANALYSIS. — In 6 days he will earn six times as much as in
1 day. If he earns a dollars in 1 day, in 6 days he will earn
6 a dollars.
2. If one hat costs d dollars, what will 9 hats cost ?
A.ns. 9d dollars.
3. If 1 yard of cloth costs c dollars, what will 10 yards
cost? • Ans. lOc dollars.
4. If 1 cravat costs b cents, what will 40 cost ?
Ans. 40& cents.
5. If 1 pair of gloves coats b cents, what will a pairs
cost?
ANALYSIS. — If 1 pair of gloves cost b cents, « pairs will
cost as many tunes b cents as there are units in a : that is,
b taken a times, or ab ; which denotes the product of b
by a, or of a by b.
resulting sign called? "When the sign of opeiation is different from the
sign of the quantity, what is the essential sign ? When the sign of ope-
ration is the same as the sign of the quantity, what is thr. essential sign ?
41. What is Multiplication? What is the quantity to be multiplied
called? What is that called by which it is multiplied? What ia the
result called?
MULTIPLICATION. 57
MULTIPLICATION is the operation of finding the product
of two quantities.
The quantity to be multiplied is called the Midtiplicand ;
that by which it is multiplied is called the Multiplier ; and
the result is called the Product. The Multiplier and Multi-
plicand are called Factors of the Product.
6. If a man's income is 3a dollars a week, how much will
he receive in 45 weeks ?
3a X 46 = I2ab.
If we suppose a = 4 dollars, and b = 3 weeks, the pro-
duct will be 144 dollars.
NOTE. — It is proved in Arithmetic (Davies' School, Art. 48.
University, Art. 50), that the product is not altered by chang-
ing the arrangement of the factors ; that is,
I2ab = axbxl2 = bxaxl2 = axl2xb.
MULTIPLICATION OF POSITIVE MONOMIALS.
42. Multiply 3a2b2 by 2a2b. "We write,
3a?b2 X 2«2& = 3 X 2 X 0? X a? X b2 X b
— 3 x 2 a a a a b b b;
in which a is a factor 4 times, and b a factor 3 times ;
hence (Art. 14),
3«262 X 2a?b = 3 X 2a*b3 = 6a4i3,
in which ice multiply the coefficients together, and add the
exponents of the like letters.
The product of any two positive monomials may be found
in like manner; hence the
RULE.
I. Multiply the coefficients togetJier for a new coefficient:
II. Write after this coefficient all the letters in both mono-
id. V^hat is the rule for multiplying one monomial by another ?
3*
58 ELEMENTARY ALGEBRA.
mials, giving to each letter an exponent equal to the sum of
its exponents in the two factors.
EXAMPLES.
1. 8a?bcz X labd2 = 56a3b2c2d*.
2. 2la3b2cd X 8abc3 — 168«453c4<?.
3 4abc X Idf — 2Sabcdf.
(4.) (5.) (6.)
Multiply 3a2b I2a'2x Gxyz
by 2azb
(7.) (8.) (9.)
a*2-rp*i *^/77j2/i3
vJ\l OCtt/ t/
10 Multiply 5a3b2xz by 6c5ic6. ^4ns.
11. Multiply 10a455c8 by lacd.
12. Multiply 36a857c6c?5 by 20a£2c3d4.
13. Multiply 5am by 3a5".
14. Multiply 3am53 by 6azbn.
15. Multiply 6amin by 9a567. Ans. 54am+5Z»n+".
16. Multiply 5ambn by 2a^b^.
17. Multiply 5amb2c2 by 2a5nc.
18. Multiply Ga2bmcn by 3a3J2c2.
19. Multiply 20a5b5cd by I2a2x2y.
20. Multiply 14a456<?4y by 20a3c2«2y. ^4. 280a;b6c2d4x2y2.
21. Multiply 8a3i3y4 by 1a*bxy5. Ans. 56a~b*xy9.
22. Multiply I5axyz by
MULTIPLICAT1C N .
59
23. Multiply 64«3w5ar1?/2 by 8a5V. A. 512a4J2c3m5sc1ya.
24. Multiply 9«2£W3 by 12a3J*c6. Ans. W8a5beced3.
25. Multiply 216a£W8 by 3a3Z>V. ^4ws. 648a4#W8
26. Multiply 70a8£W2/« by 12
MULTIPLICATION OF POLYNOMIALS.
43. 1. Multiply a — b by c.
It is required to take the difference a — b
between a and 5, c times ; or, to c
take c, a — 6 times.
As we can not subtract b from c,
we begin by taking «, c times, which
is ac ; but this product is too large
by b taken c times, which is be ; 56 — 21 = 35
hence, the true product is ac — be.
If a, 5, and c, denote numbers, as a =. 8,
c = 7, the operation may be written in figures.
ac — be
8 -
7
= 5
7
= 3, and
Multiply a — b by c — d.
It is required to take a — b as
many times as there are units in
c - d.
If we take a — J, c times, we
have ac — bc\ but this product is
too large by a — b taken <? times.
But a — b taken c? times, is ad—db.
Subtracting this product from the
preceding, by changing the signs of
its terms (Art. 37), and we have,
a — b
c — d
ac— be
— ad + bd
ac— be — ad + bd
8-3
7-2
56 — 21
- 16 + 6
56 - 37 -f 6 = 2-3.
(a — b) X (a — c) = ab — be - ad + bd.
60 ELEMENTARY A L G E B K A .
Hence, we have the following
RULE FOR THE SIGXS.
I. "When the factors have like signs, the sign of their
product will be + •'
II. When the factors have unlike signs, the sign of their
product will be — :
Therefore, we say in Algebraic language, that + multi-
plied by + , or — multiplied by — , gives -f- ; — multi-
plied by -f-, or -f multiplied by — , gives — .
Hence, for the multiplication of polynomials, we have the
following
RULE.
Multiply every term of the multiplicand by each term of
the multiplier, observing that like signs give +, and unlike
signs — / then reduce the result to its simplest form.
EXAMPLES IN WHICH ALL THE TEEMS AEE PLUS.
1. Multiply .... 3a2 + 4a6 + b2
by ..... 2a + 5b
6a3 + 8a2J+ 2aS2
The product, after reducing, + I5a~b+ 20a52 + 5b3
becomes .... 6a3 -f 23o26+ 22a68 + 5b\
44. NOTE. — It Avill be found convenient to arrange the
terms of the polynomials with reference to some letter ; that
is, to write them down, so that the highest power of that
letter shall enter the first term ; the next highest, the
second term, and so on to the last term.
44. How are the terms of a polynomial arranged with reference to a
particular letter? "What is this letter called ? I 'the leading letter in the
Miultiplicand and multiplier is the same, which will be the leading letter
in the product ?
MULTIPLICATION. 61
The letter with reference to which the arrangemffit is
made, is called the leading letter. In the above example the
leading letter is a. The leading letter of the product will
always be the same as that of the factors.
2. Multiply x2 + 2ax + a2 by x 4- a.
Ans. x3 4- 3.ax2 4- 3a2a; 4- a3-
3. Multiply x3 4- y3 by x + y.
Ans. x* 4- xy* 4- o?y + y4-
4. Multiply 3a62 + 6«2c2 by 3ab2 + 3o2c2.
!c2 4- 18a4c*.
5. Multiply a2b2 + czd by a + b.
Ans. a3b2 + aczd + a2b3
6. Multiply 3aa;2 + 9a53 + cd5 by 6a2c2.
7. Multiply Qla3x3 + 27a2a; + 9ab by 8a3cc?."
x + >I2a4bcd.
8. Multiply a3 + 3a2« + Saw;2 + a^ by a + x.
Ans. a*
9. Multiply «2 + y2 by a; + y.
. x3 -f ay2 4- »2y 4- y3.
10. Multiply cc5 4- xy6 -\- lax by ax 4-
4-
11. Multiply a3 4- 3a2ft 4- Sab2 + b3 by a 4- b.
Ans. a4 4- 4a35 4- 6a252 '-f 4a53 + b4.
1 2. Multiply a3 4- x*y 4- »y2 + y3 by cc 4-. y.
4ns. cc4 4- 2z3y 4- 2x-2y2 4- 2i«y3 4- y4.
13. Multiply x3 + 2x2 + x + 3 by 3« 4- 1.
4ns. 3a* 4- Ta3 4- 5x2 + lOx + 3.
62 ELEMENTARY ALGEBRA.
«
GENERAL EXAMPLES.
1. Multiply ....... 2ax — Soft
by ........ 3x b.
The product ....... Gax2— Qabx
becomes after ...... — 2abx
reducing ........ 6axz— llabx + 3a52.
2. Multiply a* — 2b3 by a — b.
Ans. as — 2ab3 — a*b + 2b*.
3. Multiply xz — 3x — 7 by a; — 2.
Ans. x3 — 5a;2 — 85 + 14.
4. Multiply 3a2 — 5ab + 2b* by a2 — 7a&.
^1«5. 3a4 — 26a36 + 37a2*2 - Uab3.
5. Multiply b2 + b4 + b6 by 62 — 1. .4ns. *8 - J2.
6. Multiply x*— 2x*y + 4:X2yz—8xy3+ 16y* by x + 2y.
Ans. x5 + 32y5.
7. Multiply 4a;2 — 2y by 2y. -4ns. 8a2y — 4y2.
8. Multiply 2x -f 4y by 2« — 4y. ^4/15. 4a2 — 16y2.
9. Multiply x3 + azy + xyz + y3 by x — y.
Ans. x* — y*.
10. Multiply a2 -f %y + y2 by cc2 — ccy + y2.
-4ns. jc4 + #2y2 4- y4.
11. Multiply 2a2 — Sax + 4a^ by 5a2 — 6aa; — 2a2.
Ans. 10a4 — 27«3« -f 34a2it-2 — 18(7^ — 8x*.
12. Multiply 3x2 — 2xy + 5 by x"2 + 2xy — 3.
Ans. So4 + 4^y — 4x* - ±x~if + IGay — 15.
13. Multiply 3x3 + 2ar!y2 + 3y2 by 2X3 — 3z2y2 + oy3.
( Gx6 - Sa^y2 — 6^y4 +
HS' I5x3y3 - QaPy* + lOa^y5 + 15y5.
14. Multiply 8ax — Gab — c by 2ax + ab + c.
Ans. 16a2ce2 — 4a2to — 6a2J2 + Gacx — 7afto
DIVISION. (S3
1«. Mult.ply So2 — 5b2 + 3c2 by «2 —
^4n«. 3a* — 8aW + 3a2c2
16. 3a2 — bbd + r/
— 5a2 + 45d - 8c/.
Pro. red. —
17. Multiply amcc — a262 by a'cc".
18. Multiply «m + bn by am — 5". Ans. azm — b2*.
19. Multiply am + 5" by «m + bn.
Ans. a?m + 2ambn + 52lt.
DIVISION.
45. DIVISION is the operation of finding from two quan-
tities a third, which being multiplied by the second, will
produce the first.
The first is called the Dividend, the second the Divisor,
and the third, the Quotient.
Division is the converse of Multiplication. In it, we have
given the product and one factor, to find the other. The
rules for Division are just the converse of those for Multi-
plication.
To divide one monomial by another.
46. Divide 72a5 by 8a3. The division is indicated,
thus :
7205
~8a?'
The quotient must be such a monomial, as, being multiplied
by the divisor, will give the dividend. Hence, the coefficient
45. What is division ? What is the first quantity called? The second?
The third? What is given in division? What is required ?
46. What is the rule for the division of monomials?
64 ELEMENTART ALGEBRA.
of the quotient must be 9, and the literal part a2 ; fo;
quantities multiplied by 8a3 will give 72#s. Hence,
8«3
The coefficient 9 is obtained by dividing 72 by 3; and
the literal part is found by giving to <r, an exponent equal
to 5 minus 3.
Hence, for dividing one monomial by another, we have
the following
RULE.
I. Divide the coefficient of the dividend by the Coefficient
of the divisor, for a new coefficient :
H. After this coefficient write all the letters of the dividend^
giving to each an exponent equal to the excess of its <Fcpo~
ponent in the dividend over that in the divisor.
SIGNS IN DIVISION.
47. Since the Quotient multiplied by the Divisor must
produce the Dividend : and, since the product of two factors
having the same sign will be + ; and the product of two
factors having different signs will be — ; we conclude :
1. When the signs of the dividend and divisor are like,
the sign of the quotient will be 4- .
2. When the signs of the dividend and divisor are unlike,
the sign of the quotient will be — . Again, for brevity, we
say,
-f- divided by +, and — divided by — , give -f- ;
— divided by +, and + divided by — , give — .
+ ab — ab
+ a — b
— ab 4- ab
47. What is the rule fbr the signs, in division ?
DIVISION. 65
EXAMPLES.
(1.) (2.)
-f- \8cisbzc — 15<23cc2v
_ = + 2a2*. tr-^2 = + 3«^-
(3.) (4.)
— ~ = - 8a3.
+ 3abc
5. Divide 15ax*y3 by — Say. u4w«. —
6. Divide 84a£3a; by 1252. Ans. labx.
7. Divide — 36a465c2 by 9a352c. Ans. — 4aZ»3c.
8. Divide — 99a454.c5 by lla^x4. Ans. — Qab2x.
9. Divide lOSx6^3 by 54a^s. ^Iws. 2xy5z~.'
10. Divide 64ce7y526. by — le^y4^5. -4n*. — 4«yz.
11. Divide — Q6a~b6c5 by 12a2£c. Ans. — 8as5sc4.
12. Divide — 38a*ft6<?4 by 2a3S5a'. ^n«. — 19a5(?3.
13. Divide — 64a5£4c8 by 32a45c. ^?zs. — 2<zZ»3c7.
14. Divide 128a5a;6y7 by IQaxy4. Ans.
15. Dfvride — 256aiJ9c8^7 by lGa?bc6. Ans. —
16. Divide 200a8/n2;i2 by — 50a7mn. Ans. — 4amn.
17. Divide 300a;3y422 by 6Qxy*z. Ans.
18. Divide 27a552c2 by — 9abc. Ans. —
" 9, Divide 64a3y628 by 32ory537. Ans. 2a?yz.
20. Divide — 88a556c8 by Ila354c6. Ans. — 8a2£2c2.
21. Divide 77a4y324 by — Ila4y3z*. Ans. — 7.
22. Divide 84a4J2c2<7 by — 42a452c2c?. Ans. — 2.
23. Divide — 88a6J7c6 by 8a556c6. Ans. — llab.
24. Divide 16x2 by — 8x. Ans. — 2x.
25. Divide — 88anb2 by 1 la"*b. Ans. — 8a*-mb.
66 ELEMENTARY ALGEBRA.
26. Divide 77am&" by — 11«2&3. Ans. — *lam-ybn-*.
27. Divide 84a8£m by 42an&9. Ans. 2as-nbm-*.
28. Divide — 88a^? by Sanbm. Ans. — lla*- "&«-"•.
29. Divide QGabP by 48#"5?. Ans. 2al~nbP-*.
30. Divide IG8xayb by 12xnym. Ans. 14xa~nyb~m.
31. Divide 256a53c2 by 16
MONOMIAL FRACTIONS.
48. It follows from the preceding rales, that the exact
division of monomials will be impossible :
1st. When the coefficient of the dividend is not exactly
divisible by that of the divisor.
2d. When the exponent of the same letter is greater in
the divisor than in the dividend.
3d. When the divisor contains one or more letters not
found in the dividend.
In either case, the quotient will be expressed by a fraction.
A fraction is said to be in its simplest form, when the
numerator and denominator do not contain a common factor.
For example, 12a452cc7, divided by 8a25c2, gives
which may be reduced by dividing the numerator and de-
nominator by the common factors, 4, a2, #, and c, giving
Also,
48. Under what circumstances will the division of monomials be im-
possible ? How will the quantities then be expressed ? How is a mono-
mial fraction reduced to its simplest form ?
2c
5a
DIVISION. 67
Hence, for the reduction of a monomial fraction to its sim-
plest form, we have the following
KULE.
Suppress every factor, whether numerical or literal, that
is common to both terms of the fraction ; the result will be
the reduced fraction sought.
EXAMPLES.
(1.) (20
: and _ ... . m =
Sbce
(3.) (4.)
2«
also» 32 - : and 'T -
2a
2mn
5. Divide 49«2&2c6 by 14a3*c4. Ans.
, -r.. .,
6. Divide 6amn by 3aoc. ^3.^5.
be
7. Divide ISaWmn2 by
8. Divide 28a!Wd8 by
9. Divide 72aW by
10. Divide 100a8#5a»nn by 25a?b4d. Ans.
d
11. Divide 96a668cW by I5a?cxy. Ans.
25xy
12. Divide 85m2n3/a;2?/3 by ISam^/". ^Ins.
13. Divide 127d3o;2y2 by
127
68 ELEMENTARY ALGEBRA.
49. In dividing monomials, it often happens that the
exponents of the same letter, in the dividend and divisor,
are equal ; in which case that letter may not appear in the
quotient. It might, however, be retained by giving to it the
exponent 0.
If we have expressions of the form
a a2 a3 a* a5 .
a' a2'. a3' ^' ^' '
and apply the rule for the exponents, we shall have,
_ a _ a- _ a c>
a a a3
But since any quantity divided by itself is eqiial to 1, it fol-
lows that,
- = a° = 1, —, = a2-2 = a° = 1, &c. ;
a a2
or, finally, if we designate the exponent by m, we have,
/"flft
— = am~m =: a° = 1 ; that is,
am
The 0 power of any quantity is equal to 1 : therefore,
Any quantity may be retained in a term^ or introduced
into a term, by giving it the exponent 0.
EXAMPLES.
1. Divide 6«262c* by 2a2£2.
ftt&lfle*
= 3c*.
2. Divide 8a453c5 by — 4a453c. ^tna. — 2a°6°c4 = — 2c4.
3. Divide — 32m3>i2a2y2 by ±mzri*xy.
Ans. — 8m°n°xy = — Sxy.
49. When the exponents of the same letter in the dividend and divisor
are equal, what takes place ? May the letter still be retained ? Wi Ji
what exponent? What is the zero power of any quantity equal to?
DIVISION. 69
4. Divide — OGa'&V1 by — 24a4&5. Ans. 4a°J°cn = 4c*.
5. Introduce a, as a factor, into 6b5ci. Ans. 6a°b5c*.
6. Introduce ab, as factors, into Oc5^". Ans. 9a°b°c5dn.
7. Introduce ale, as factors, into Sd\fm. A. 8a°i°c°f?4/m.
5O. "When the exponent of any letter is greater in the
divisor than it is in the dividend, the exponent of that letter
in the quotient may be written with a negative sign. Thus,
— = —', also, — = a"-6 = a"3, by the rule;
a5 a3 a°
hence, a~3 = -5-
Since, a~3 = — =. we have, b x a~3 = — =;
a3 a3'
that is, « in the numerator, with a negative exponent, is
equal to a in the denominator, with an equal positive ex-
ponent; hence,
Any quantity having a negative exponent, is equal to the
reciprocal of the same quantity icith an equal positive ex-
ponent.
Hence, also,
Any factor may be transferred from the denominator to
the numerator of a fraction, or the reverse, by changing the
sign of its exponent.
EXAMPLES.
1. Divide 32a25c by 16«5i2.
16a5b2 a3b
60. When the exponent of any letter in the divisor is greater than in
the dividend, how may the exponent of that letter be written in the quo-
tient ? What is a quantity with a negative exponent equal to ? How
may a factor be transferred from the numerator to the denominator of a
frn.-tion ?
70 ELEilENTARY ALGEBRA.
2-
3. Reduce -rr-ns * Ans. -- , or
5 la;4?/3 3 ' 3
4. In 5ay~3x~z, get rid of the negative exponents.
5a
Ans. -—
K T ..
5. In _ 6 , get rid of the negative exponents.
4a?i8
Ans. - —
3x2
15a~yc~4d~s
6. In - - — - — i , get rid of the negative exponents.
-3-5~2'
— —
7. Reduce -^^r ^n*. - -^ -, or
8. Reduce 72a552 -f- 8a653. -4wa. Oa-1^-1. or —
9. In _2 l , get rid of the negative exponents.
_
10. Reduce -- --j — =- • ^Ins. 3aJ2c2.
— ~6~
To divide a polynomial by a monomial.
51. To divide a polynomial by a monomial :
Divide each term of the dividend, separately, by the
divisor ; the, algebraic sum of the quotients will be the quo-
tient sought.
EXAMPLES.
1. Divide 3a252 - a by a. Ans. 3ab2 — 1.
61. How do you divide a polynomial by a monomial ?
DIVISION. 71
2. Divide 5a3J2 — 25a"£2 by 5aW. Ans. 1 — 5a.
3. Divide 35a262 — 25ab by — Sab. Ans. — lab + 5.
4. Divide 10«5 — I5ac by 5o. -4ra. 25 — 3e.
5. Divide Gab — Sax + 4a2y by 2a.
6. Divide — 15aa2 + 6a3 by — 3x. Ans. 5ax — 2x2.
7. Divide — 2 lay2 + 35a262y - 7c2y by - 7y.
^4?i5. 3xy — 5a252 -f- c2.
8. Divide 40a85* + 8a457 — 32a4**c* by 8a*b4.
Ans. 5a* + 53 — 4c<.
DIVISION OF POLYNOMIALS.
52. 1. Divide — 2a + 6a2 — 8 by 2 + 2a.
Dividend. Divisor.
6a2 — 2a — 8 | 2a + 2
6a3 + 6a 3a — 4 Quotient.
— 8a — 8
— 8a — 8
0 Remainder.
We first arrange the dividend and divisor with reference
to a (Art. 44), placing the divisor on the left of the dividend.
Divide the first term of the dividend by the first term of
the divisor ; the result will be the first term of the quotient,
which, for convenience, we place under the divisor. The
product of the divisor by this term (6«2 -f- 6«), being sub-
tracted from the dividend, leaves a new dividend, which may
be treated hi the same way as the original one, and so on to
the end of the operation.
52. What is the rule for dividing one polynomial by another ? When
is the division exact ? When is it not exact ?
72 ELEMENTARY ALGEBRA.
Since all similar cases may be treated in the same way, we
Lave, for the division of polynomials, the following
K U L E .
I. Arrange the dividend and divisor with reference to the
same letter:
II. Divide the first term of the dividend by the first term
of the divisor, for the first term of the quotient. Multiply
the divisor by this term of the quotient, and subtract the
product from the dividend:
m. Divide the first term of the remainder by the first
term of the divisor, for the second term of the quotient.
Multiply the divisor by this term, and subtract the product
from the first remainder, and so on:
TV. Continue the operation, until a remainder is found
equal to 0, or one whose first term is not divisible by that
of the divisor.
NOTE. — 1. When a remainder is found equal to 0, the
division is exact.
2. "When a remainder is found whose first term is not
divisible by the first term of the divisor, the exact division
is impossible. In that case, write the last remainder after
the quotient found, placing the divisor under it, in the form
of a fraction.
SECOND EXAMPLE.
Let it be required to divide
by 4ab — 5a2 -f 3J3.
Wo first arrange the dividend and divisor with reference
to a.
DIVISION.
Dividend.
73
Divisor.
10a*-
t— Sa3b —
— 2a2+ 8a6 —
Quotient.
-f
25azbz—20ab3—l5b4
25a?b2—20ab3—15b*
+
(3.)
4- ccy2 — 2
+ y_
xy
4-
4-
4
4-
Here the division is not exact, and the quotient is frac-
tional.
(4.)
1 4- a
1 - a
1 - a
2a3 4 , &c.
+ 2a
4- 2a —
4-
4-
— 2a3
4- 2a3
In this example the operation does not terminate. It
be continued to any extent.
EXAMPLES.
1. Divide a2 4- 2ax + x2 by a + x.
Ans. a 4-
2. Divide a3 — 3a2y 4- 3ay2 — y3 by a — y,
Ans. a2 — 2ay 4- J/*.
74 ELEMENTARY ALGEBRA.
3. Divide 24a2£ — I2a*cb2 — Qab by — Gab.
Ans. *- 4 a -f 2a:cb -}-• 1.
4. Divide Gas4 — 96 by 3x — 6.
Ans. 2x? -f 4*2 + 8z + 16.
5. Divide a5 — 5a4a + 10a3cc2 — -lOa2*3 + Soa4 — ar"
by a2 —
6. Divide 48a^ — 76aa;- — 64a2a; + I05a3 by 2x — 3a.
Ans. 24a;2 — 2aa; — 35«2.
7. Divide y6 — 3y4a;2 + 3y2ic4 — a;6 by y3 — 3y2a +
— a?3. Ans. y3 + 3y*x + 3?/xz -f a^.
8. Divide 64«456 - 25a2i8 by 8a253 + 5ai4.
u4;?s. 8a2i3 — 5a54.
9. Divide 6a3+ 28a2J + 22a52+5J3 by
-4«*. 2a
10. Divide Qax* + Gaa2?/6 + 42a2a2 by aa; +
11. Divide — loa4 + S1azbd — 29aV/— 2052^ +
8c2/2 by 3a2 - 5bd + cf. Ans. - 5a2
12. Divide a^ + aj2^2 + y* by a;2 — ;ry + y2.
^L?25. a;2 + io// + y2.
13. Divide ar4 — y* by a; — y.
Ans. x3 4- a^y 4- %y~ + 2/3-
14. Divide 3a4— 8a2ft2+ 3a2c2+ 554— 352c2 by «2- i2.
^^5. 3a2 — 5bz + 3c2.
15. Divide 6a6 - Sa^y2— Ga^y4^- 6x3y24- 15a;3y3— 9.r2y4
-f 10a;2y5 + 15y5 by 3a^ + 2a;2y2 + 3y2.
Ans. 2x3 - 3a;V + 5y3.
16. Divide — c2+ 16a2a:2— tabc — Itfbx — 6a2J2+ Gaca:
by 8aa; — Qab — c. Ans. 2ax + ab -}- c.
•
17. Divide Bx* + 4a;3y — 4a;2 — 4cc2y2 + 16a;y — 15 by
2a;y + a;2 — 3. An-s. 3xz — 2a*y -f 5.
DIVISION. 75
18. Divide x5 + 3£y5 by x + 1y.
Ans. x* — 2x3t/ + 4ce2y2 — 8xy3 + 16y*.
19. Divide 3a4 — 26a3b — 14a53 + 37a252 by 2b2 — Sab
3a2. Ans. a2 — 7«£.
20. Divide a4 - 54 by a3 + «2# + «52 + b*.
Ans. a — b.
21. Divide «3 — 3a2y + y3 by x -\- y.
3y3
2
22. Divide 1 + 2a by 1 — a — a*.
Ans. 1 + 3a + 4a2 + fa3 + , &c.
76 ELEMENTARY ALGEBRA.
CHAPTER LIL
tJSEFtn FORMULAS. FACTORING. GREATEST COMMON DIVISOR.
LEAST COMMON MULTIPLE.
USEFUL FORMULAS.
53. A FORMULA is an algebraic expression of a general
rule, or principle.
Formulas serve to shorten algebraic operations, and are
also of muchjuse in the operation of factoring. When trans-
lated into common language, they give rise to practical rules.
The verification of the following formulas affords addi-
tional exercises in Multiplication and Division.
(1.)
54. To form the square of a + b, we have,
(a + by = (a + b) (a + b) = ct + 2ab + b\
That is,
The square of the sum of any two quantities is equal to
the square of the first, plus twice the product of the first by
the second, plus the square of the second.
1. Find the square of 2a -f 3b. "We have from the rule,
(2a + 35)2 = 4a2 + I2ab + 9bz.
53. What is a formula? What are the uses of formulas ?
64. What is the square of the sum of two quantities equal to ?
U 8 K F U L F t K M U L A 6 . 77
2. Find the square of 5ab + 3ac.
Ans. 25«2&2 -f ZQtfbc -f- 9cc2c2.
3. Find the square of 5a2 + 8a?b.
Ans. 25a* + 80a4& -f
4. Find the square of 6ax + 9a2ce2.
-f 108a3a;3 +
(2.)
55. To form the square of a difference, a — b, we have,
(a - J)2 = (a - b) (a - b) = a2 - 2ab + b\
That is,
The square of the difference of any two quantities is
equal to the square of the first, minus twice the product of
the first by the second, plus the square of the second.
1. Find the square of 2a — b. We have,
(2a - £)2 = 4«2 — 4«& + *2-
2. Find the square of 4ac — be.
Ans. 16a2c2 —
3. Find the square of 7a2Z>2 — 12aJ3.
Ans.
(3.)
56. Multiply a + b by a — b. We have,
(a + b) x (a - 5) = a2 — R Hence,
TAe s?m q/" ^wo quantities, multiplied by their difference,
is equal to the difference of their squares.
1. Multiply 2c + b by 2c — b. Ans. 4c2 — 62.
2. Multiply 9ac + 35c by 9ac — 35c.
. 81a2c2 -
55. What is the square of the difference of two quantities equal to ?
56. What is the sum of two quantities multiplied by the'.r difference
equal to ?
78 ELEMENTARY ALGEBRA.
3. Multiply 8a3 + 7aft2 by 8a3 — 7oft2.
Ans. 64a6 — 49a2ft
(4.)
57. Multiply a2 + ab + ft2 by a — b. We have,
(a2 + ab + ft2) (a - ft) = a3 - ft3.
(5.)
5§. Multiply a2 -- aft + ft2 by a -f ft. We have,
(a2 - aft + ft2) (a + ft) = a3 + ft3.
(6.)
59. Multiply together, a -f- ft, a — ft, and a2 -f ft2.
We have,
(a + ft) (a - ft) (a2 + ft2) = a4 - ft*.
60. Since every product is divisible by any of its factors,
each formula establishes the principle set opposite its number.
1. The sum of the squares of any two quantities, plus
twice their product, is divisible by their sum.
2. The sum of the squares of any two quantities, minus
twice their product, is divisible by the difference of the
quantities.
3. The difference of the squares of any two quantities
is divisible by the sum of the quantities, and also by their
difference.
4. The difference of the cubes of any two quantities is
divisible by the difference of the quantities ; also, by the
sum of their squares, plus their product.
5. The sum of the cubes of any two quantities is divisi-
60. By what is any product divisible ? By applying this principle, v,-hat
follows from Formula (1) ? What from (2)? What from (S) ? What from
(4) » What from (5) ? Wh at from (6) ?
FACTORING. 7»
ble by the sum of the quantities ; oho, by the sum of their
squares minus their product.
6. The difference between the fourth powers of any two
quantities is divisible by the sum of the quantities, by their
difference, by the sum of t/ieir squares, and by the dif-
ference of their squares.
FACTOEING.
61. Factoring is the operation of resolving a quantity
into factors. The principles employed are the converse of
those of Multiplication. The operations of factoring are
performed by inspection.
1. What are the factors of the polynomial
ac -f ab + ad.
We see, by inspection, that a is a common factor of all
the terms ; hence, it may be placed without a parenthesis,
and the other parts within ; thus :
ac + ab -f ad — «(c -f b + d).
2. Find the factors of the polynomial azbz + azd — azf.
Am. az(bz + <?—/).
3. Find the factors of the polynomial 3azb — 6azbz + bzd.
Ans. b(3az - Qazb + bd).
4. Find the factors of 3azb — 9azc — 18aVy.
Ans. 3a2(b — 3c — 6xy).
5. Find the factors of Sa2cx — ISacx* + 2acsy ~ 30a6c9.
Ans. 2ac(4ax — 9#2 4- c*y — 15a5c8).
6. Factor 3j&*52e - 6a3bzd* + 18aW.
Ans. 6«3J2(5«c — d3 + 3c2).
7. Factor 12cW3 — 15c3J* — 6c2f?3/.
Ans. 3czd3(4czb — 5cd - 2/).
61. What is factoring ?
80 ELEMENTARY ALGEBRA.
8. Factor 15a3bcf — lOabc4 — 25abcd.
Ans. 5abc(3a-f — 2c3 — 5d").
62. When two terms of a trinomial are squares, and
positive, and the third term is equal to twice the product of
their square roots, the trinomial may be resolved into factors
by Formula (l).
1. Factor a2 -f 2ab -f b*. Ans. (a + b) (a + b).
2. Factor 4a2 + I2ab -f 9&2.
Ans. (2a + 3b) (2a + 3b).
3. Factor 9a2 + I2ab -f 4b2.
Ans. (3a + 2b) (3a + 2b}.
4. Factor 4a;2 + 8x + 4. Ans. (2x + 2) (2x + 2).
5. Factor 9«2£2 + I2abc + 4c2.
Ans. (Sab + 2c) (Sab + 2c).
6. Factor lOa^y2 + 16a-y3 + 4^.
2y2) (4xy -f 2y2).
63. When two terms of a trinomial are squares, and
positive, and the third term is equal to minus twice their
square roots, the trinomial may be factored by Formula
(2).
1. Factor az — 2ab + b*. Ans. (a — b) (a — b).
2. Factor 4a2 — 4ab + b*. Ans. (2a — b) (2a — b).
3. Factor 9a2 — 6ac + c2. Ans. (3a — c) (3« — c).
4. Factor «2a;2 — 4ax + 4. Ans. (ax — 2) (ax — 2).
5. Factor 4&2 — 4.ry + y2. Ans. (2x — y] (2x — y).
62. When may a trinomial be factored ?
63. When mav a trinomial be factored bv this method*
FACTORING. 81
6. Factor 3Gx~ — 2-ixy -+- 4y2.
Ans. (6x — 2y) (Qx — 2y).
0
64. Wlien the two terms of a binomial are squares and
have contrary signs, the binomial may be factored by
Formula ( 3 ).
1. Factor 4c2 - &. Ans. (2c + b) (2c — b)
2. Factor 81«2c2 - 9&2c2.
Ans. (Qac + 3bc) (9«c — 3bc).
3. Factor 64a4J* — 25xzy2.
Ans. (8a?b2 + 5xy) (8a?bz — 5xy}.
4. Factor 25a2c2 — 9x*yz.
Ans. (5ac + 3x2y) (5ac — 3«2y).
5. Factor 36a4J4c2 - 9z6.
Ans. (GaWc + 3x3) (6a262c — Sx3).
6. Factor 49a^ - 36y4. Ans. (7z2 + 6y2) (7«2 — 6y2).
65. When the two terms of a binomial are cubes, and
have contrary signs, the binomial may be factored by
Formula (4 ).
<»
1. Factor 8a3 — c3. Ans. (2a — c) (4«2 + 2ac + c2).
2. Factor 27«3 — 64.
^4/zs. (3« — 4) (9a2 + 12a + 16).
3. Factor a3— 64R
^;i5. (a — 4i) (a2 + 4a& + 1652).
4. Factor a3 — .27i3. vl»s. (a — 3b) (a2 + Sab + 9^).
61. When may a binomial be factored ?
65. When mav a biuomial be factored by this method?
4*
82 E L E M E N T A K T A L G K B B A .
66. When the terms of a binomial are cubes and have
like signs, the binomial may be factored by Formula ( 5 ).
1. Factor 8a3 + c3. Ans. (2a + c) (4a2 — 2ac + c1).
2. Factor 27a3 + 64.
Ans. (3a + 4) (9a2 — 12a + 16).
3. Factor a? + 6463.
s. (a + 46) (a2- 4a6 + 1662).
4. Factor a3 + 2753. -4«*. (a + 3&) (a2 - 3aJ + 9J2).
67. When the terms of a binomial are 4th powers, and
have contrary signs, the binomial may be factored by
Formula (6).
1. What are the factors of a4 — 54?
Ans. (a + b) (a — b) (a2 + b2).
2. What are the factors of 81a* - 16&4 ?
Ans, (3a + 25) (3a — 26) (9a2 -f 4J2).
3. What are the factors of 16a*64 — 81c4<?4?
Ans. (2ab + Zed) (2ab — 3cJ) (4a262
GREATEST COMMON DIVISOK,
68. A COMMON DIVISOR of two quantities, is a quantity
that will divide them both without a remainder. Thus,
3«26, is a common divisor of 9«262c and 3a262 — 6a?b3.
66. When may a binomial be factored by this method?
67. When may a binomial be factored by this method?
68. What is the common divisor of two quantities ?
G K E A T K S T COMMON D I V I S O K . 83
69. A SIMPLE or PRIME FACTOR is one that cannot be
resolved into any other factors.
Every prime factor, common to two quantities, is a com-
mon divisor of those quantities. The continued product of
any number of prime factors, common to two quantities, is
also a common divisor of those quantities.
70. The GREATEST COMMON DIVISOR of two quantities,
is the continued product of all the prime factors which are
common to both.
71. "When both quantities can be resolved into prime
factors, by the method of factoring already given, the great-
est common divisor may be found by the following
RULE.
I. Resolve both quantities into their prime factors :
II. Find the continued product of all the factors which
are common to both ; it icitt be the greatest common divi-
sor required.
EXAMPLES.
1. Required the greatest common divisor of 75«262c and
25abcL Factoring, we have,
I5a-b'zc = 3 x 5 x 5aabbc
25abd = 5 x 5abd.
The factors, 5, 5, a and b, are common ; hence,
5x5xaxb = 25ab,
is the divisor sought.
69. What is a simple or prime factor ? Is a prime factor, common to
two quantities, a common divisor ?
70. What is the greatest common divisor ?
71. If both quantities can be resolved into prime factors, how do you
find the greatest common divisor ?
84 ELEMENTARY ALGEBRA.
VERIFICATION.
2c ^ 25ab = 3abc
25abd -=- 25ab = cZ;
and since the quotients have no common factor, they cannot
be further divided.
2. Required the greatest common divisor of «2 — 2ab +
b2 and a2 — b2. ./Ins. a - b.
3. Required the greatest common divisor of a2 + 2ab -f
62 and a + b. Ans. a + b.
4. Required the greatest common divisor of «2a;2 — 4ax
f- 4 and ax — 2. Ans. ax — 2.
5. Find the greatest common divisor of 3a?b — 9«2c
— 1 Sa?xy and &2c — 3bc2 — 65ca^. -4w«. b — 3c — Qxy.
6. Find the greatest common divisor of 4«2c — 4acx and
3«2<7 - - 3agx. Ans. a(a — x), or a2 — ax.
1. Find the greatest common divisor of 4c2 — 12cx -f- 9a;2
and 4c2 — 9x2. ^Iws. 2c — Sa1.
8. Find the greatest common divisor of x* — y* and
3.2 _ y-if Ans. x — y.
9. Find the greatest common divisor of 4c2 + 4bc -f b2
and 4c2 — bz. Ans. 2c + £>.
10. Find the greatest common divisor of 25a2c2 —
and 5acd2 -f 3dzx^yz. Ans. 5ac
NOTE. — To find the greatest common divisor of three
quantities. First find the greatest common divisor of two
of them, and then the greatest common divisor between this
result and the third.
1. What is the greatest common divisor of 4«ic2y, IQabx2,
and 24a«e2? Ans. 4ax°.
2. Of 3ic2— 6a, 2a;3— 4#2, and a;2//- 2xy ? Ans. x2— 2x.
72. When is one quantity a multiple of another ?
LEAST COMMON MULTIPLE. 85
LEAST COilMON MULTIPLE.
72. One quantity is a MULTIPLE of another, when it can
be divided by that other without a remainder. Thus, 8a25,
is a multiple of 8, also of a2, and of b. *
73. A quantity is a Common Multiple of two or more
quantities, when it can be divided by each, separately, with-
out a remainder. Thus, 24a3x3, is a common multiple of
6ax and
74. The LEAST COMMON MULTIPLE of two or more quan-
tities, is the simplest quantity that can be divided by each,
without a remainder. Thus, 12a2J2tc2, is the least common
multiple of 2a2cc, 4«52, and 6«2J2tc2.
75. Since the common multiple is a dividend of each of
the quantities, and since the division is exact, the common
multiple must contain every prime factor in all the quanti-
ties ; and if the same factor enters more than once, it must
enter an equal number of times into the common multiple.
When the given quantities can be factored, by any of the
methods already given, the least common multiple may bo
found by the following
EULE.
I. Resolve each of the quantities into its prime factors .
II. Take each factor as many times as it enters any ou6
of the quantities, and form the continued product of these
factors ; it will be the least common multiple.
73. When is a quantity a common multiple of several others?
74. What is the least common multiple of two or more quantities?
75 What does the common multiple of two or more quantities contain,
MS factors? How may the least common multiple be found?
* The }>i ultiph ol a quantity, is siraplj \ dividend which will give an exact quotient
86 ELEMENTARY ALGEBRA.
EXAMPLES.
1. Find the least common multiple pf 12a3&2c2 and 8a253.
I2a3b"*c2 = 2.2.3.aaabbcc.
8a?b3 =3 2.2.2.aabbb.
Now, since 2 enters 3 times as a factor, it must enter 3
times in the common multiple : 3 must enter once ; a, 3
times ; £, 3 times ; and c, twice ; hence,
2.2.2.3aaabbbcc = 24«353c2,
is the least common multiple.
Find the least common multiples of the following :
2. 6a, 5«25, and 25a£c2. Ans. loOa-bc2.
3. 3a2£, 9«5c, and 27a2»3. Ans. 2fla2bcxz.
4. 4a2ic2y2, 8a3tfy, 16a4y3, and 24a5y4a;.
5. ax — bx, ay — by, and cc2y2.
^Lws. (a — b}x.x.yy =
6. a + b, a2 — 52, and a2 + 2ab + J2.
u4?w. (a + 6)2 (a - 6).
7. 3a362, 9a2a;2, 18a*y3, 3a2y\ Ans. ISaWxty5.
8. 8a2(a — *), 15a5(a - £)2, and 12a3(a2 — bz).
Ans. 120a5(a — J)2 (a + J).
FRACTIONS. 37
CHAPTER IV.
FRACTIONS.
76. IF the unit 1 be divided into any number of equal
i>arts, each part is called a FRACTIONAL UNIT. Thus, - , -,
- , - , are fractional units.
77. A FRACTION is a fractional unit, or a collection of
fractional units. Thus, - , - , •- , -=• , are fractions.
2 4 7 o
78. Every fraction is composed of two parts, the De-
nominator and Numerator. The Denominator shows into
how many equal parts the unit 1 is divided ; and the Nu-
merator how many of these .parts are taken. Thus, in the
fraction - , the denominator J, shows that 1 is divided into
b equal parts, and the numerator <r, shows that a of these
parts are taken. The fractional unit, in all cases, is equal to
the reciprocal of the denominator.
76. If 1 be divided into any number of equal parts, what is each part
called ?
77. What is a fraction ?
78. Of how many parts is any fraction composed ? What are they
called? What does the denominator show? What the numerator?
What is the fractional unit equal to ?
88 ELEMENT A It T ALGEBRA.
79. An EXTIKE QUANTITY is one which contains no
fractional part. Thus, 7, 11, a*x, 4a;2 — 3y, are entire
quantities.
An entire quantity may be regarded as a fraction whose
denominator is 1. Thus, 7 = -, ab = — •
SO. A MIXED QUANTITY is a quantity containing both
bx
entire and fractional parts. Thus, 7^ , 8^ , a -\ , are
c
mixed quantities.
81. Let v denote any fraction, and q any quantity
ct
whateyer. From the preceding definitions, = denotes that
•=- is taken a times; also, ~ denotes that T is taken
0 o o
aq times ; that is,
aq a
-~ — - x q', hence,
Multiplying the numerator of a fraction by any quan-
tity, is equivalent to multiplying the fraction by that
quantity.
We see, also, that any quantity may be multiplied by a
fraction, by multiplying it by the numerator, and then
dividing the result by the denominator.
82. It is a principle of Division, that the same result will
be obtained if we divide the quantity a by the product
of two factors, p x q, as would be obtained by dividing it
79. What is an entire quantity ? When may it be regarded as a frao
tion ?
80. What is a mixed quantity ?
81. How may a fraction be multiplied by any quantity ?
82 How may a fraction be divided by any quantity ?
TRANSFORMATION OF FRACTIONS. 80
first by one of the factors, p, and then dividing that result
by the other factor, q. That is,
a ia\ a 1 a\
• — I ) -S- 2> or> - = I ) •*• PI hence,
pq \pl pq \q/
Multiplying the denominator of a fraction by any quan-
tity, is equivalent to dividing the fraction by that quantity.
83. Since the operations of Multiplication and Division
are the converse of each other, it follows, from the preced-
ing principles, that,
Dividing the numerator of a fraction by any quantity,
is equivalent to dividing the fraction by that quantity /
and,
Dividing the denominator of a fraction by any quantity,
is equivalent to multiplying the fraction by that quantity.
84. Since a quantity may be multiplied, and the result
divided by the same quantity, without altering the value,
it follows that,
Hoth terms of a fraction may be multiplied by any quan-
tity, or both divided by any quantity, without changing the
value of the fraction.
TRANSFORMATION OF FRACTIONS.
85. The transformation of a quantity, is the operation
of changing its form, without altering its value. The term
reduce has a technical signification, and means, to Trans-
form.
S3. What follows from the preceding principles ?
84. What operations may be performed without altering the value of
a fraction?
85. What is the transformation of a quantity ?
90 E I. £ M E N T A R \ ALGEBRA.
FIRST TRANSFORMATION.
To reduce an entire quantity to a fractional form having a
given denominator.
86. Let a be the quantity, and b the given denomi-
nator. We have, evidently, a = -j- ; hence, the
BULB.
Multiply the quantity by the given denominator, and
write the product over this given denominator.
SECOND TRANSFORMATION.
To reduce a fraction to its lowest terms.
87. A fraction is in its lowest terms, when the numerator
and denominator contain no common factors.
It has been shown, that both terms of a fraction may be
divided by the same quantity, without altering its value.
Hence, if they have any common factors, we may strike
them out.
EULE.
Resolve each term of the fraction into its prime fac-
tors / then strike out all that are common to both.
The same result is attained by dividing both terms of the
fraction by any quantity that will divide them, without a
remainder ; or, by dividing them by their greatest common
divisor.
86. How do you reduce an entire quantity to a fractional form having
a given denominator '?
87. How do you reduce a fraction to its lowest terms ?
TRANSFORMATION OF FRACTIONS. 91
EXAMPLES.
15<z2c2
1. Reduce — — -, to its lowest terms.
15a2c2 3.5aacc
Factoring, — = — - — -;
25acd 5.5acd
Canceling the common factors, 5, «, and c, we have,
15a2c* Sac
^ — 3 = TT* Ans.
25acd 5d
17
2. Reduce
60c<W5
3. Reduce ^ •
12c5ay9
- . -„ , a5 — ac . a
4. Reduce — : • Ans. - = a.
b — c 1
f T> 1 '^ *"i +1 . W 1
5. Reduce • Ans. -•
n2 — 1 7i + l
,,.,-. x3 — ax2 «?
6. Reduce — : — -• Ans.
x2 — 2ax + a2 x — a
9Qa3b2c 8
7. Reduce — ,^ .370 • Ans. —- = — 8.
8. Reduce ,„ ,-,, T---TTS- Ans.
a2 — b2 A a + b
9. Reduce — „ . . T0» jffu.
') r» 7 i IO — -*.*vv« _
a2 — 2ao + o2 a — o
5 a3 — 10a2& + 5ab2
10. Reduce ~..^.
8a3 — 8a26 8a
««a
11. Reduce
12a4 + 6a3c2 ' 4a2 +
a2 + 2aa + ce2
12. Reduce — ~-r—n ^ Ans.
3(a2 - x2) ' 3(a -
92 ELEMENTARY A L G E U R A. .
THIRD TRANSFORMATION.
To reduce a fraction to a mixed quantity.
88. When any term of the numerator is divisible by any
term of the denominator, the transformation can be effected
by Division.
RULE.
Perform the indicated division, continuing the opei'ation
as far as possible ; then write the remainder over the deno-
minator, and annex the result to the quotient found.
EXAMPLES.
ax — a2 a?
1. Reduce • Ans. a •
x x
ax — ar5
2. Keduce • Ans. a — x.
x
ab — 2a2 2a2
3. Reduce = • Ans. a =- •
b o
($, yZ
4. Reduce • An*, a + x.
a — x
5. Reduce ~ y • Ans. x2 + xy + y2.
x- y
10s2 - 5x + 3 3
6. Reduce • — • Ans. 2x — 1 + — •
5x 5x
7. Reduce - - '-T»^~ . . 4s2 - 8 +
9x y
T? ^ IBaef — Gbdcf — 2ad 60 lie 2
o» JL«/GCIU.CG r- """ _ ^ - - • • - r~ ' " ^— — •
3ac?^ d a 3/
3.2 _(_ g. ^ 2
9. Reduce -— — • Ans. x — 1 —
x + 2
88. How do you reduce a fractior to a mixed quantity "
TRANSFORMATION OF FRACTIONS. 93
10. Reduce =- • Ans. a — b -\ —.
a + b a +b
, _ , x2 + 3x - 25 3
11. Reduce • Ans. x -f- Y
- 4 'a;-4
FOURTH TRANSFORMATION.
To reduce a mixed quantity to a fractional form.
§9. This transformation is the converse of the preced-
ing, and may be effected by the following
RULE.
Multiply the entire part by the denominator of the frac-
tion^ and add to the product the numerator / write the result
over the denominator of the fraction.
EXAMPLES.
1. Reduce 6| to the form of a fraction.
6 X 7 = 42 ; 42 + 1 = 43 ; hence, 6} = ~ •
Reduce the following to fractional forms :
xz — (a? — a2) 2<e2 — az
v yf n a
,/JL/co.
xx x
ax + x2 ax — a2
3. x -- - -- Ans.
2a 2a
2s- 7 ifee - 7
4. 5 -i -- - -- Ans.
3£C 3x
x — a — 1 2a — x + 1
5. 1 --- Ans. -- '
a a
. , „ 05 — 3 lOcc2 + 4x + 3
6. 1 -f 2cc -- - -- Ans. - •— - ^—
5£C 5x
Sf*. How do yon reduce a mixed quantity to a fractional form?
94
7.
8.
o
ELEMENTARY
"alb 3C + 4
ALGEBRA.
16a + Sb — 3c?
4.
Qax + b
ft _i_ S/-/7>
8
Qa2x — ab
8
18a2x +
5«J
8 + t%'
4«
f W WUrw ~"|~* Ov/Cv V »<^
^1W5.
FIFTH TRANSFORMATION.
To reduce fractions having different denominators, to equi-
valent fractions having the least common denominator.
9O. This transformation is effected by finding the least
common multiple of the denominators.
13 5
1. Reduce -, -, and — , to their least common denomi-
94 J. +*
nators.
The least common multiple of the denominators is 12,
which is also the least common denominator of the required
fractions. If each fraction be multiplied by 12, and the result
divided by 12, the values of the fractions will not be changed.
- X 12 = 4, 1st new numerator ;
- X 12 = 9, 2d new numerator ;
4
— X 12 = 5, 3rd new numerator ; hence,
i *-
49 o
— , — , and — are the new equivalent fractions.
90. How do you reduce fractions having different denominators, to equl
valent fractions having the least common denominator ? When the nu-
merators have no common factor, bow do you reduce them ?
TRANSFORMATION OF FRACTIONS. 95
RULE.
I. Find the least common multiple of the denominators :
n. Multiply each fraction by it, and cancel the denom-
inator :
in. Write each product over the common multiple, and
the results will be the required fractions.
GENERAL RULE.
Multiply each numerator by all the denominators except
its own, for the new numerators, and all the denominators
together for a common denominator.
EXAMPLES.
ct c
1. Reduce — ^ and = to their least common
a2 — b2 a + b
denominator.
The least common multiple of the denominators is (a + b)
(a-b):
l-j- X (a + b) (a - b) = a
az-
c
'—=• X (a + b) (a — b) = c(a — b ; hence,
Ct ~f~ 0
c(a — b) ,, . ,
and /„ . v*w ' IM are the required
(a + b) (a - b) (a + b) (a - b)
fractions.
Reduce the following to their least common denominators :
Sx 4 12a2 45x 40 48^
2- 7' 6' and 15- An8' Weo'-eo-
3bz 5c3 12o 952 lOc3
3. a, -, and -. Ans. _,_,_.
, Sx 2b 9ccc 4ab 6acd
4. — - , — , and a. Ans. - — , - — , —
2a' 3c Qac Qac Gac
96 E L E M K X T ART A L G E B U A .
3 2« 2cc 9a Sax 12a2 + 24a?
a; a2 a;3
6. . - , TZ -- ^, and
1 -a;' (1 _a;)2» ^ (1 _a)3
as(l — a;)2 £C2(1 — as) x3
Ans. -± ~y -77 r^, and 7- TV
(1 — a;)3 ' (1 — a;)3 (1 — x)3
c c — b , c
> and
*• M ) ) €*.*-* V* —
5a c a + 0
ac2 + £c2 5«2c — 5«2o + 5abc —
5#2c -f* 5abc ' 5^2c 4~ 5abc ' 5o&2c + 5abc
ex dxz a;3
8. , — • — , and
a — x a + x a + x
: cx(a-{- x) dx'2(a — x) , x3(a — x)
i-iis. - — , — , and — — •
ADDITION OF FBACTIOKS.
91. Fractions can only be added when they have a com-
mon unit, that is, when they have a common denominator.
In that case, the sum of the numerators will indicate how
many times that unit is taken in the entire collection.
Hence, the
KTJLE.
L Jteduce the fractions to be added, to a common denom-
inator :
n. Add the numerators together for a new numerator^
and write the sum over the common denominator.
EXAMPLES.
64 2
1. Add -, -, and -, together.
23 5
91. What is the rule for adding fractions?
ADDITION OF FRACTIONS. 97
By reducing to a common denominator, we have,
6 x 3 x 5 i= 90, 1st numerator.
4 x 2 x 5 = 40, 2d numerator.
2x3x2 = 1 2, 3d numerator.
2 x 3 x 5 = 30, the denominator.
Hence, the expression for the sum of the fractions becomes
90 40 12 142 1
30 30 + 30 Z: ~30~;
which, being reduced to the simplest form, gives 4}J.
CL G &
2. Find the sum of -, - , and -•
b u j
Here, a X d x / = udf \
c X b x f = cbf > the new numerators.
e X b x d = eld )
and b x d x f = bdf the common denominator.
<*df , cbf , ebd adf -f- cbf + ebd Al
IIencc' wf + + = ^— ~ ' * sum'
Add the following :
3x2 2ax 2abx
3. a -- r- , and b -\ -- • Ans. a + b -\
, .
o c oc
. 85 X _ X . X
4. -, -, and -. Ans. * + -
, x — 2 . 4« 19a; — 14
5. ___ and y. Ans. — —
x—2 , , 2x — 3 , lOa; — 17
6. x H -- - — and 3x -\ -- - -- Ans: 4cc -\ -- — -
3 4 12
5x2 T x + a Sx3 + ax -+- a*
7. 4cc, -— , and — -!
. 2« 7a; _ 2a + 1 49o; + 12
8. T, T, and —g— • Ans. 2x + —^ --
YO* //• 44:35
9. 4«, --, and 2 H- - ylw*. 2 + 4* + —
5 45
98 ELEMENTARY ALGEBRA.
10. 3x -f ~ and x - ^. Ans. 3x + —
59 45
6b c
11. ac — -- and 1 ,•
8a d
wi/u. -f- Sac
Ans. 1 -f ac
Sad
±3? - 5x + 4
" 4(1 + a)' 4(1 -a)' "2(f~^) " l>
SUBTRACTION OF FEACTIOIfS.
92. Fractions can only be subtracted when they have
the same unit; that is, a common denominator. In that
case, the numerator of the minuend, minus that of the sub-
trahend, will indicate the number of times that the common
unit is to be taken in the difference. Hence, the
RULE.
I. Reduce the two fractions to a common denomi-
inator :
n. Then subtract the numerator of the subtrahend from
that of the minuend for a new numerator, and write the
remainder over the common denominator.
EXAMPLES.
3 2
1. What is the difference between - and - •
7 8
3 2 24 14 10 5
_ _ __ ,_ ._ ^__ — — ._ _ .
7 8 "" 56 56 " 56 ~" 28
92. What is the rule for subtracting fractions ?
MULTIPLICATION OF FRACTIONS. 9t»
3* — Cl 2l7 •— 4»J*
2. Find the difference of the fractions — r - and — — •
20 . 3c
j (cc — a) x 3c = Sex — Sac [ .
Here, i . v .j: rt. 0. > the numerators,
( (2a — 4aj) x 20 = 4ao — 8&c )
and, 2b X 3c = Qbc the common denominator.
3cx— Sac 4ab—8bx 3cx—3ac—4:ab+8bx
Hence, -- - --- -= - = -- — • Ans.
Qbc Qbo 6bc
, -D • j ^ *•# f * 3x A 39a;
3. Required the difference of -— and — • Ans. — — •
7 5 3o
4. Required the difference of 5y and — • Ans. — -
8 '^. 8
5. Required the difference of — and — • Ans. - — •
t \s Do
e. From Z±V subtract ^ - ^n*.
x — y x + y
__ _
7. From - subtract — - -• Ans. - — - -- ;- •
y — 2 y2 — s2 y2 — 22
Find the differences of the following :
3* + a , 2a; -f 7 24« + 8« — 105* — 35o
8. — -r — and — - — • Ans. -- —; -- •
ob 8 406
x , x — a , ex + bx — ab
9. 3x + - and x --- Ans. 2x -\ -- = -- •
be be
a — x , a + a 4a5
10. a -i — -. - r and —. — ! - :• Ans. a
(» ciu.v& f fc ^.m./tw# \Af ,. ,j
a + x) a(a — 05) a* — or
MULTIPLICATION OF FRACTIONS.
f* y»
93. Let v and -^, represent any two fractions. It has
O W
been shown (Art. 81), that any quantity may be multiplied
93. What 13 the rule for the multiplication of fractions ?
-*
lUU E L K M K X T A K Y A L G K B K A .
by a fraction, by first multiplying by the numerator, and
then dividing the result by the denominator.
CL G CIC
To multiply j- by -^, we first multiply by e, giving — ;
0 Ctr 0
then, we divide this result by d, which is done by multiply-
ac
ing the denominator by d; this gives for the product, •=— ;
OCl/
that is,
a c as ,
T X -, = r^; hence,
b d bd
RULE.
I. If there are mixed quantities, reduce them to a frac-
tional form ; then,
IL Multiply the numerators together for a new numera-
tor, and the denominators for a new denominator.
EXAMPLES.
bx , c _,. to a2 + to
1. Multiply a -\ by -• First, a H = ,
a J d a a
, «2 + to c a2c + bcx
hence, — x -, = -= — • Ans.
a a ad
Find the products of the following quantities :
, 2x Sab , Sac
2. — , — , and — j- Ans. 9ax.
a c 2b
to a ab -f to
3. b -\ and - • Ans. - — •
ax x
4. — and — Ans. ^ — -— r-=-
be b + c b2c + bo3
x + 1 , x — 1 ax* — ax + xz —1
5. x -\ , and 7 • Ant. •
a a + b a2 + ab
ax , a2 — xz a3 + a2x
6. a -i and • Ans. •
a — x x + ar x + x2
MULTIPLICATION OF FRACTIONS. 101
2d a? J2
7. Multiply — i by -^ — •
1 J a — b 3
la a* — b2 2a(ai — i2) 2a(a + b) (a - b~)
T X "
a — J 3 3 (a — 6) 3 (a —
After indicating the operation, we factored both numera-
tor and denominator, and then canceled the common factors,
before performing the multiplication. This should be done,
whenever there are common factors.
9. 0-2 — 4/2 9,(<r. 4- 1/\
o. uy — .
x - y a
a
x2 — 4 4cc
. 4a(a; — 2)
S ^ 3* 4- 2
O «C ~y^ *-
3
in Tw
-4ws. 2a;(a + 5).
2a; (a + 6)
(x - I)2 , (a; + l)y2
x nv /4/ .
3.2 i
y by « - 1
y
r, («2-*2) b i+«,
A a — x
1 _ & ^ a + x
1 — x
O'v»i/ Oo*)/
£ivul.i -. *-t< '/
Ans. a2.
x — y x + y
2a — b Ga — 2b
5 — 3(35
' "j 19 r» r
4a o2— 2a6
2a&
V2 x y
16. a; — — by - + -•
a4 — v4
Ans. — -Z-Z-
102 ELEMENTARY ALOEBKA.
DIVISION OF FRACTIONS.
P 1
94. Since •- = p x - , it follows that, dividing by a
quantity is equivalent to multiplying by its reciprocal. But
c d
the reciprocal of a fraction, -, is -- (Art. 28); conse-
Cv G
quently, to divide any quantity by a fraction, we invert the
terms of the divisor, and multiply by the resulting fraction.
Hence,
a c a d ad
_ ,_• T _ T ^ y ^L_ .
b d ~~ b c ~ ' be
Whence, the following rule for dividing one fraction by
another :
RULE.
I. Reduce mixed quantities to fractional forms :
II. Invert the terms of the divisor, and multiply the
dividend by the resulting fraction.
NOTE. — The same remarks as were made on factoring
and reducing, under the head of Multiplication, are appli
cable in Division.
EXAMPLES.
1. Divide a — — by £•
2c g
b 2ac — b
a =
2c 2c
„ b f 2ac — b fj 2acff — bg .
Hence, a j-^- = - - x 5 = - --?- -. Ans.
2c g 2c f 2cf
94. What is the rule for the division of fractions?
DIVISION OF FRACTIONS. 103
. ~. . , 2(» + y) . «- — y2
2. Divide v y' by ^--
a a
+ y) Y «_ _ 2(3 + y)
A o o — — s*
a (x + y) (x - y)
2
Ans,
/>• QI
x y
7«c 12 9 Is*
3. Let — be divided by :— • Ans. - — •
5 60
4. Let — — be divided by 5x. ,
7 3o
T a: + 1 ,-..,., 2a; a; + 1
5. Let be divided by — • Ans. — - —
6 3 4x
x x A 2
6. Let be divided by - • Ans. -
x — I 2 x — 1
c/j* 26R
7. Let — be divided by ~ •
3 30 2«
x — b . ., , , 3ca; x — I
8. Let =- be divided by — •= • Ans.
Divide the following fractions:
- **-** ' a2-4 Ans. 4X
3 3 * + 2
10 bv • Ans. x + —
iv/' «.9. nX^. I XS « » A /»•
_ _ _ X
4xz (a + o)2
11. 2a*(« + *) by -—=• Ans. ' >
a + o *x
'-- \- 1)7/2 (x — I)2
i2L . ^1«S. ..— —
- i y3
^2 fla. 3^ a;") 4a(a2 — cc2)
' 6c + bx bj 4(^+^) ' 4W*' 3i(c2 - a2)
104 ELEMENTARY A L G E B K A .
14.
15.
a
— x
by
1 +
X
X
t
y
Ans.
Ans. x -
a*
-as"
I — x
x2 by
a +
2xy
1
i_
— a;2
•
* +
1 x
-y
16.
b
— 3a
by
Qa -
-2b
Ans.
2a
- b
•
2ab
62 -
2ab
4a
T7
rt
-y4
l-nr
_L
y
Ana
X*
- y2
x'y y x
18. m2 + 1 -r -^ by m H hi.
m* J m
Ans. m -\ 1.
m
IP
(y — x \ I i
aj + f-r ) by (l - xf
I + xyJ J \ 1
(x + 2y x\ . /x + 2y a; \
20. I — 7—^ + - by I - — ^ -- • — )
V* + y y/ Vy * + y/
EQUATIONS OF THE FIRST DEGREE. 105
CHAPTER V.
EQUATIONS OP THE FIRST DEGREE.
95. AN EQUATION is the expression of equality between
two quantities. Thus,
x = b + c,
is an equation, expressing the fact that the quantity ic, is
equal to the sum of the quantities b and c.
9O. Every equation is composed of two parts, connected
by the sign of equality. These parts are called members :
the part on the left of the sign of equality, is called the first
member ; that on the right, the second member. Thus, in
the equation,
x -\- a = b — c,
x + a is the first member, and b — c, the second member.
97. An equation of the first degree is one which involves
only the first power of the unknown quantity ; thus,
Qx + 3x — 5 = 13; (1 )
and ax + bx + c =. d ; (2)
are equations of the first degree.
95. What is an equation ?
96. Of how many parts is every equation composed? How are the
parts connected ? What are the parts called ? What is the part on the
left called? The part on the right ?
97. What is an equation of the first degree ?
5*
103 ELEMENTARY ALGEBRA.
98. A NUMERICAL EQUATION is one in which the ^effi-
cients of the unknown quantity are denoted by numbers.
99. A LITERAL EQUATION is one in which the coefficients
of the unknown quantity are denoted by letters.
Equation ( 1 ) is a numerical equation ; Equation ( 2 ) is a
literal equation.
EQUATIONS OF THE FIRST DEGREE CONTAINING BUT ONE
UNKNTOWN QUANTITY.
100. The TRANSFORMATION of an equation, is the opera-
tion of changing its form without destroying the equality
of its members.
101. An AXIOM is a self-evident proposition.
102. The transformation of equations depends upon tin
following axioms:
1. If equal quantities be added to both members of an
equation, the equality icill not be destroyed.
2. If equal quantities be subtracted from both members
of an equation, the equality will not be destroyed.
3. If both members of an equation be multiplied by the
same quantity, the equality will not be destroyed.
4. If both members of an equation be divided by the same
quantity, the equality will not be destroyed.
5. Like powers of the two members of an equation are
equal.
6. Like roots of the two members of an equation are
equal.
98. What is a numerical equation ?
99. What is a literal equation ?
100. What is the transformation of an equation ?
101. What is an axiom ?
102. Name the axioms on which the transformation of an equation
depends.
CLEARING OF FRACTIONS. 107
103. T\vo principal transformations are employed in the
solution of equations of the first degree: Clearing of frac-
tions^ and Transposing,
CLEARING OF FRACTIONS.
1. Take the equation,
2# 3«5 £C
3~ ' ' 7 "f 6 =
The least common multiple of the denominators is 12. If
we multiply both members of the equation by 12, each term
will reduce to an entire form, giving,
Sx — Qx + 2» = 132.
Any equation may be reduced to entire terms in the same
manner.
104. Hence for clearing of fractions, we have the fol-
lowing
RULE.
I. Find the least common multiple of the denominators:
II. Multiply both members of the equation by it, reduc-
ing the fractional to entire terms.
NOTE. — 1. The reduction will be effected, if we divide the
least common multiple by each of the denominators, and
then multiply the corresponding numerator, dropping the
denominator.
2. The transformation may be effected by multiplying
each numerator into the product of all the denominators
except its own, omitting denominators.
103. How many transformations are employed in the solution of equa-
tions of the first degree ? What are they ?
104. Give the rule for clearing an equation of fractions? In what throe
«vays may the reduction be effected?
108 ELEMENTAKY ALGEBRA.
3. The transformation may also be effected, by multiplying
both members of the equation by any multiple of the de-
nominators.
EXAMPLES.
Clear the following equations of fractions :
1. f + f — 4 = 3. Ans. ?x + 5x — 140 = 105.
5 7
2. '; + - — — — 8. Ans. 925 + Qx — 2x = 432.
O J — /
x x x x
3. - H ----- 1 -- = 20.
•2^3 9 ^ 12
Ans. I8x + 12a; — 4x + 3* = 720.
•C £C SB
4. - + - — - = 4. .4ws. 14a; + lOa; — 3ox = 280.
o i Z
5. — + = 15. ,4w*. 15aj — 12a; + lOaj = 900.
x — 4 x — 2 5
6 "
. — 2a; + 8 — cc -f- 2 = 10.
x 3
7. - -- h 4 = - • -4ns. 5z + 60 — 20a = 9 — 3x.
x «,a5,«_19
8' 4 ~ 6 + 8 + 9 - 12'
-4ns. 18« — 12a; + 9x + 8« = 864.
ft ft
9. - — -j -\- f = g. Ans. ad — be -f bdf = bdg
o u
axe, 2c2ai , 4bo-x 5a3 <*<?
10. -r -- =r- + 4a = — - --- 75- H --- 30.
b ab a3 b2 a
The least common multiplt of the denominators is cr-3^2 \
TRANSPOSING. 109
TRANSPOSING.
105. TRANSPOSITION is the operation of changing a term
from one member to the other, without destroying the
equality of the members.
1. Take, for example, the equation,
5x — 6 ; = 8 + 2x.
If, in the first place, we subtract 2x from both members
the equality will not be destroyed, and we have,
5x — 6 — 2x = 8.
Whence we see, that the term 2ce, which was additive in
the second member, becomes subtractive by passing into
the first.
In the second place, if we add 6 to both members of
the last equation, the equality will still exist, and we have,
5x — 6 — 2x + 6 = 8 + 6,
or, since — 6 and + 6 cancel each other, we have,
5x — 2x = 8 + 6.
Hence, the term which was subtractive in the first member,
passes into the second member with the sign of addition.
106. Therefore, for the transposition of the terms, we
have the following »
BULK.
Any term may be transposed from one member of an
equation to the other, if the sign be changed.
105. What is transposition?
106. What is the rule for the transposition of the terms of an equation ?
110 ELEMENTARY ALGEBRA.
EXAMPLES.
Transpose the unknown terms to the first member, and
thp known terms to the second, in the following :
1. 3x -f 6 — 5 = 2x — 7. Ans. 3x — 2x = — 7 — 6 -f 5.
2. ax + b = d — ex. Ans. ax + ex = d — b.
3. 4x — 3 = 2x + 5. Ans. 4x — 2x — 5 + 3.
4. Qx + c — ex — d. Ans. dx — ex = — d — c.
5. ax + / = dx + b. Ans. ax — dx — b — f.
6. 6# — c = — 02 + 5. ^l?zs. 62; -f ax =. b + c.
SOLUTION OF EQUATIONS.
1O7. The SOLUTION of an equation is the operation of
finding such a value for the unknown quantity, as will
satisfy the equation ; that is, such a value as, being sub-
stituted for the unknown quantity, will render the two mem-
bers equal. This is called a ROOT of the equation.
A Hoot of an equation is said to be verified, when being
substituted for the unknown quantity in the given equation,
the two members are found equal to each other.
1. Take the equation,
Clearing of fractions (Art. 104), and performing the opera-
tions indicated, we have,
12* — 32 = 4x — 8 + 24.
107. What is the solution of an equation? What is the found value
of the unknown quantity called ? When is a root. of an equation said to
he verified.
SOLUTION OF EQUATIONS. Ill
Transposing all the unknown terms to the first member,
and the known terms to the second (Art. 106), we hare,
12x — 4x — — 8 + 24 + 32.
Reducing the terms in the two members,
8z = 48.
Dividing both members by the coefficient of *,
48
: T :
VERIFICATION.
3X6 4(6 — 2) ,
— 4 = - — i + 3 ; or,
L o
+ 9 — 4 = 2 + 3 = 5.
Hence, 6 satisfies the equation, and therefore, is a root.
1O8. By processes similar to the above, all equations of
the first degree, containing but one unknown quantity, may
be solved.
KULE.
I. Clear the equation of fractions, and perform all the
indicated operations :
II. Transpose all the unknown terms to the first member,
and all the known terms to the second member :
III. ^Reduce all the terms in the first member to a single
term, one factor of which will be the unknown quantity,
and the other factor will be the algebraic sum of its coeffi-
cients :
IV. Divide both members by the coefficient of the unknown
quantity : the second member will then be the value of the
unknown quantity.
108. Give the rule for solving equations of the first degree with one
unknown quantity.
112 ELEMENTARY ALGEBRA.
EXAMPLES.
1. Solve the equation,
5x 4x 7 13a;
13 = •
12 3 86
Clearing of fractions,
lOx — 32a; — 312 = 21 — 52x.
By transposing,
lOa — 32x + 52x - 21 + 312.
By reducing, 30a = 333 ;
333 111
hence, x : — = — = 11.1;
a result which may be verified by substituting it for x in
the given equation.
2. Solve the equation,
(3a — x) (a — b) + 2ax = 4J(a -f a).
Performing the indicated operations, we have,
3a2 — ax — Sab + bx + 2ax = 4&c + 4«5.
By transposing,
— ax + bx + 2ax — 4&B = 4a5 + Sab — 3a*.
By reducing, ax — 3bx = *ldb — 3a2 ;
Factoring, (a — 3b)x = lab — 3a2.
Dividing both members by the coefficient of x,
lab - 3a2
a — 3b
3. Given 3x — 2 + 24 = .31 to find x. Ans. x = 3.
4. Given x + 18 = 3x — 5 to find x. Ans. x= 11 J-
SOLUTION OF EQUATIONS. 113
6. Given 6 — 2x + 10 ;= 20 — 3x — 2, to find a.
Ans. x = 2.
6. Given x + IJB -f \x = 11, to find x. Ans. x = 6.
7. Given 2x — %x + 1 = 5x — 2, to find x.
Ans. x = %.
Solve tlie following equations:
a 6 — 8a
8. Sax -\ --- 3 = Ix — a. Ans. x = -- =••
2 Qa — 25
x — 3 x x — 19
9. — — + - = 20 -- - • Ans. x = 231 .
• 0 .6
x + 3 , a; x — 5
10. -- + = 4 -- - -- ^W5. x = 3-
x 3a; . 4»
11. T -- — + x = — — 3. Ans. x = 4.
V 2 O
12. --- ^ -- 4 = f. Ans,. x = —^ - =-
c d Sad — 2bc
x — a 2x — 35 a — x
.
245.
13.
a 8— a; 5 + ajll
14. - - _ --- — + - = 0. ^. « = 12.
a + c , a — c 252 a2 —
I >% ^_^^^___^ I _______ — _
iw. — „ _
a + a a — cc a2 — x2
Sax — b 35 — c
16. -^ ---- 2— = 4-5.
56 + 95 —
. a; = - --—
16a
x x — 2 x 13
17 5 -~3- + 2 = T- ^' »=
114 ELEMENTARY ALGEBRA.
?: X — f
c~ d~f'
Ans. x =
,
oca — aca + aoa — abc
NOTE. — What is the numerical value of «, when a = 1,
= 2, c = 3, <? — 4, and / = 6 ?
19. - -
3x — 5 , 4x — 2
20. a; ---- (-• -- = a; + 1. J.ns. x = 6, .
13 11
21. a; H --- 1 ---- = 2« — 43. ^4^5. a; = 60.
456
4x — 2 3a; — 1 .
A. ^ • ^»C ~~~ ' — — ^^— — ^-^ • jfjLftS* *C — O»
5 2
, bx — d 3a + d
23. 3x -J --- . = x 4- «. -4^5. « = - -- 7- •
3 6 -t- b
ax — b a bx bx — a
24. — = •
4^32 3
A 35
Ans. x = - —
3a — 26"
4a5 20 - 4x _ 15 2
JbO« - ' ~~~ ' - - * -ii/(O« »C — _ O •
5 — JB JB X 11
2a; + 1 402 — 3x 471 — 6z
20. ---- — 9
29 12 2
Ans. x = 72.
„* fa + b}(x-b) 4ab-b* a?-bx
—^^r~ ~^+b~ ~T~
a4 + 3a35 + 4a~bz — Gab3 + 25*
~ aft - 52 "
PROBLEMS. 115
PROBLEMS.
1O9. A PKOBLEM is a question proposed, requiring a
solution.
The SOLUTION of a problem is the operation of finding a
quantity, or quantities, that will satisfy the given conditions.
The solution of a problem consists of two parts :
I. The STATEMENT, which consists in expressing, algebra-
ically, the relation between the known and the required
quantities.
IE. The SOLUTION, which consists in finding the values
of the unknown quantites, in terms of those which are
Jcnoicn.
The statement is made by representing the unknown
quantities of the problem by some of the final letters of the
alphabet, and then operating upon these so as to comply
with the conditions of the problem. The method of stating
problems is best learned by practical examples.
1. What number is that to which if 5 be added, the sum
will be equal to 9 ?
Denote the number by x. Then, by the conditions,
x + 5 = 9.
This is the statement of the problem.
To find the value of x, transpose 5 to the second member ;
then,
x =. 9 — 5 = 4.
This is the solution of the equation.
VERIFICATION.
x + 5 = 9.
I'i9. What is a problem? Wait is the solution of a problem? Of
how many parts does it consist* What are they? What is the state-
niwit ? What is the solution ?
116 E L E M E N T A K V A L G E B U A .
2. Find a number such that the sum of one-half, one-third,
and one-fourth of it, augmented by 45, shall be equal to 448
Let the required number be denoted by x.
M
Then, one-half of it trill be denoted by — ,
2
o*
one-third " by -,
O
B
one-fourth " by -;
and, by the conditions,
This is the statement of the problem.
Clearing of fractions,
6x + 4x + Sx + 540 = 5376 ,
Transposing and collecting the unknown terms,
13x = 4836;
4836
hence, x = — — = 372.
Id
VERIFICATION.
^*72 ^5*72 ^72
- + — - + - - + 45 = 186 + 124 + 93 + 45 = 448.
±t O 4
3. What number is that whose third part exceeds its
fourth by 16 ?
Let the required number be denoted by x. Then,
-x — the third part,
3
-x = the fourth part ;
P It () Ii L K M S . 117
and, by tlie conditions of the problem,
\x - \x = 16.
Tbis is the statement. Clearing of fractions,
4a; — 3x = 192,
and hence, x = 192.
VERIFICATION.
4. Divide $1000 between A, .Z?, and (7, so that A shall
have $72 more than JB, and C $100 more than A.
Let x denote the number of dollars which S received.
Then, x = B's number,
x + 72 = A's number,
and, a;+l72 = C's number;
and their sum, 3x + 244 — 1000, the number of dollars.
This is the statement. By transposing,
Sx = 1000 — 244 = 756 ;
and, x = • - = 252 = IPs share.
Hence, x + 72 = 252 -f 72 = 324 = A's share,
and, x + 172 = 252 + 172 = 424 = <7's share.
VERIFICATION.
252 + 324 -f 424 = 1000.
5. Out of a cask of wine which had leaked away a third
part, 21 gallons were afterwards drawn, and the cask being
then guaged, appea7%ed to be half full : how much did it
hold ?
118 ELEMENTARY ALGEBRA.
Let x denote the number of gallons.
fly
Then, - = the number that had leaked away.
3
and, - + 21 = what had leaked and been drawn.
3
x x
Hence, by the conditions, - + 21 = - •
3 2
This is the statement. Clearing of fractions,
2x + 126 = 3x,
and, - x = — 126 ;
and by changing the signs of both members, which does not
destroy their equality (since it is equivalent to multiplying
both members Jby — 1), we have,
x = 126.
VERIFICATION.
- + 21 = 42 + 21 = 63 i= -
6. A fish was caught whose tail weighed 9 Ibs., his head
weighed as much as his tail and half his body, and his body
weighed as much as his head and tail together : what was
the weight of the fish ?
Let 2x = the weight of the body, in pounds.
Then, 9 + x = weight of the head ;
and since the body weighed as much as both head and tail,
2x = 9 + 9 -f x,
which is the statement. Then,
2x — x = 18, and x = 18.
PROBLEMS. 119
Hence, we have,
2x = 36$. = weight of the body,
9 + x = 27$. — weight of the head,
9$. = weight of the tail ;
hence, 72$. = weight of the fish.
7. The sum of two numbers is 67, and their differer ce 19
what are the two numbers ?
Let x denote the less number.
Then, x -f- 1 9 = the greater ; and, by the conditions,
2x + 19 = 67.
This is the statement. Transposing,
2x = 67 — 19 = 48;
hence, x = — = 24, and x + 19 = 43.
VERIFICATION.
43 + 24 = 67, and 43 — 24 = 19.
ANOTHER SOLUTION.
Let x denote the greater number.
Then, x — 19 will represent the less,
and, 2x — 19 = 67 ; whence 2ce = 67 -f 19.
o/>
Therefore, x = - - :_ 43 ;
2i
and, consequently, x — 19 = 43 - 19 = 24.
GENERAL SOLUTION OF THIS PROBLEM.
The sum of two numbers is s, their difference ia d: what
are the tAVO numbers ?
120 ELEMENTARY ALGEBRA. ~
Let x denote the less number.
Then, a; + d will denote the greater,
and 2x + d = s, their sum. Whence,
s — d s d
~2~ = 2~2*'
and, consequently,
s d , s d
«+*=j-5+*=5+V
As these two results are not dependent on particular
values attributed to s or d, it follows that :
1. The greater of two numbers is equal to half their sum,
plus half their difference :
2. The less is equal to half their sum, minus half their
difference.
Thus, if the sum of two numbers is 32, and their differ-
ence 16,
32 16
the greater is, — + -- ==16 + 8 = 24 ; and
2t 2i
32 16
the less, — — =16 — 8= 8.
2i 2i
VERIFICATION.
24 + 8 = 32; and 24 — 8 = 16.
8. A jv*rson engaged a workman for 48 days. For each
day that he labored he received 24 cents, and for each day
that he vras idle, he paid 12 cents for his board. At the
end of the 48 days, the account was settled, when the laborer
received 504 cents. Required, the number of working days,
and the number of days he was idle.
If the number of working days, and the number of idle
days, were known, and the first multiplied by 24, and the
r u o B L E M s . 121
second by 12, the difference of these products would»be.
504. Let us indicate these operations by means of algebraic
signs.
Let x denote the number of working days.
Then, 48 — x = the number of idle days,
24 x x = the amount earned,
and, 12(48 — x} = the amount paid for board.
Then, 24a - 12(48 — a;) = 504,
what was received, which is the statement.
Then, performing the operations indicated,
24* — 576 + 12as = 504,
or, 3Qx — 504 + 570 = 1080,
and, x = - - = 30, the number of working days ;
3G
whence, 48 — 30 =18, the number of idle days.
VERIFICATION.
Thirty days' labor, at 24 cents ) n
J ' 1 30 X 24 = 720 cents,
a day, amounts to )
And 18 days' board, at 12 cents )
J }• 18 X 12 = 216 cents,
a day, amounts to ) •
The difference is the amount received . 504 cents.
GENERAL SOLUTION.
* This problem may be made general, by denoting the whole
number of working and idle days, by n ;
The amount received for each day's work, by a ;
The amount paid for board, for each idle day, by b ;
And what was due the laborer, or the balance of the
account, by c.
6
122 K L E M K N T A R Y ALGEBRA.
^ft before, let the number of working days be denoted
by x.
The number of idle days will then be denoted by n — x.
Hence, what is earned will be expressed by ax, and- the
sura to be deducted, on account of board, by b(n — x).
The statement of the problem, therefore, is,
ax — b(n — x) = c.
Performing indicated operations,
ax — bn -f bx — c, or, (a 4- f>}x = c -f- bn ;
whence, x = - -— — number of working days ;
(Ju ~*|~ (s
c + bn an-\-bn—c—bn
and, n — x = n —j- = - —— v ,
a+ b .« + o
or, n — x = — = number of idle davs.
a + b
Let us suppose n = 48, a — 24, b = 12, and c — 504 ;
these numbers will give for x the same value as before
found.
9. A person dying leaves half of his property to his wife,
one-sixth to each of two daughters, one-twelfth to a servant,
and the remaining $600 to the poor ; what was the amount
of the property ?
Let x denote the amount, in dollars,
M
Then, ^ — what he left to his wife,
2t
JM
- = what he left to one daughter, *
O/k* /£
and, — = - what he left to both daughters,
u 3
X
also, — = what he left to his servant,
and, $600 = what he left to the poor.
PROBLEMS. 123
Then, by the conditions,
/j« /•« />•
- + - 4- — + 600 — x, the amount of the property,
— . t > 1 w
which gives, x = $7200.
10. A and J3 play together at cards. A sits down with
$84, and _Z> with $48. Each loses and wins in turn, wrhen
it appears that A has five times as much as J5. How much
did A win ?
Let x denote the number of dollars A won.
Then, A rose with 84 + x dollars,
and J? rose with 48 — x dollars.
But, by the conditions, we have,
84 + x = 5(48 — aj),
hence, 84 + x — 240 — 5x;
and, Qx = 156,
consequently, x = 26 ; or A won $26.
VERIFICATION.
84 -f 26 = 110 ; 48 — 26 = 22;
110 = 5(22) = 110.
11. A can do a piece of work alone in 10 days, JB in 15
days ; in what time can they do it if they work together ?
Denote the time by &, and the work to be done, l»y 1.
Then, in
1 day, A can do — of the work, and
JB can do — of the work ; and in
13
•C
x days, A can do — of the work, and
•C
B can do ~ of the work.
13
124 K L E M E N T A 14 Y ALGEBRA.
Hence, by the conditions,
•C *C
•- (- — = 1, which gives, 13x + Wx — 130;
10 13
130
hence, 23x =130, x = — = 5£f days.
23
12. A fox, pursued by a hound, has a start of 60 of his
own leaps. Three leaps of the hound are equivalent to 7 of
the fox ; but while the hound makes 6 leaps, the fox makes
9 : how many leaps must the hound make to overtake the
fox?
There is some difficulty in this problem, arising from the
different units which enter into it.
Since 3 leaps of the hound are equal to 7 leaps of the fox,
7
1 leap of the hound is equal to - fox leaps.
3
Since, while the hound makes 6 leaps, the fox makes 9,
9 3
while the hound makes 1 leap, the fox will make - , or -
0 —
leaps.
Let x denote the number of leaps which the hound makes
before he overtakes the fox ; and let 1 fox leap denote the
unit of distance.
7
Since 1 leap of the hound is equal to - of a fox leap, x
7
leaps will be equal to -x fox leaps ; and this will denote the
0
distance passed over by the hound, in fox leaps.
' 3
Since, while the hound makes 1 leap, the fox makes ^
3
leaps, while the hound makes x leaps, the fox makes -x leaps ;
m
and this added to 60, his distance ahead, Avill give
g
-x + CO, for the whole distance passed over by the fox.
I
PROBLEMS. 125
Hence, from the conditions,
7 3
-a; = -x + 60 ; whence,
Ux = Qx + 360;
x = 72.
The hound, therefore, makes 72 leaps before overtaking
le fo
leaps.
Q
the fox; in the same time, the fox makes 72 x - = 108
m
VERIFICATION.
108 + 60 =: 168, whole number of fox leaps,
72 X I - 168.
o
13. A father leaves his property, amounting to $2520, to
four sons, A, _Z?, (7, and D. C is to have $360, J3 as much
as C and D together, and A twice as much as .#, less $1000 :
how much do A, B-, and D receive ?
Am. A, $760; .#, $880; Z>, $520.
14. An estate of $7500 is to be divided among a widow,
two sons, and three daughters, so that each son shall receive
twice as much as each daughter, and the widow herself $500
more than all the children : what was her share, and what
the share of each child ?
{Widow's share, $4000.
Each son's, 1000.
Each daughter's, 500.
15. A company of 180 persons consists of men, women,
and children. The men are 8 more in number than the
women, and the children 20 more than the men and women
together : how many of each sort in the company ?
Ans. 44 men, 36 women, 100 children.
126 E L E M E N T A II Y A L „ E BRA.
16. A father divides $2000 among five sons, so that each
elder should receive $40 more than his next younger bro-
ther : what is the share of the youngest? Ans. $320.
17. A purse of $2850 is to be divided among three per-
sons, A, J3, and C. A's share is to be to jB's as 6 to 1 1 ,
and C is to have $300 more than A and B together : what
is each one's share? A>s, $450 ; B's, 8825 ; (7's, $1575.
18. T\vo pedestrians start from the same point and travel
in the same direction ; the first steps twice as far as the
second, but the second makes 5 steps while the first makes
but one. At the end of a certain time they are 300 feet
apart. Now, allowing each of the longer paces to be 3 feet,
how far will each have traveled ?
Ans. 1st, 200 feet ; 2d, 500.
19. Two carpenters, 24 journeymen, and 8 apprentices
received at the end of a certain time $144. The carpenters
received $1 per day, each journeyman, half a dollar, and
each apprentice, 25 cents : how many days were they em-
ployed? Ans. 9 days.
20. A capitalist receives a yearly income of $2940 ; four-
fifths of his money bears an interest of 4 per cent., and the
remainder of 5 per cent. : how much has he at interest ?
Ans. $70000.
21. A cistern containing 60 gallons of water has three
unequal cocks for discharging it ; the largest will empty it
in one>hour, the second in two hours, and the third, in three:
in what time will the cistern be ehipticd if they all run to-
gether ? Ans. 32T8T min.
22. In a certain orchard, one-half are apple trees, one-
fourth peach trees, one-sixth plum trees; there are also, 120
cherry trees, and 80 pear trees : how many trees in the
orchard? Ans. 2400.
23. A farmer being asked ho\\ many sheep ho had,
P ii ) B L K M 8 . 127
answered, that he had them in five fields ; in the 1st he had
|, in the 2d, £, in the 3d, |, and in the 4th, TV, and in the
5th, 450 : how many had he ? Ans. 1200.
24. My horse and saddle together are worth $132, and
the horse is worth ten times as much as the saddle : what
is the value of the horse ? Ans. $120.
25. The rent of an estate is this year 8 per cent, greater
than it was last. This year it is $1890: what was it last
year? Ans. $1750.
26. What number is that, from which if 5 be subtracted,
| of the remainder will be 40 ? Ans. 65.
27. A post is | in the mud, £ in the water, and 10 feet
above the water : Avhat is the whole length of the post ?
Ans. 24 feet.
28. After paying | and i of my money, I had 66 guineas
left in my purse : how many guineas were in it at first ?
Ans. 120.
29. A person was desirous of giving 3 pence apiece to
some beggars, but found he had not money enough in his
pocket by 8 pence; he therefore gave them each 2 pence
and had 3 pence remaining : required the number of beg-
gars. Ans. 11.
30. A person, in play, lost % of his money, and then won
3 shillings ; after which he lost £ of what he then had ; and
this done, found that he had but 12 shillings remaining:
what had he at first ? Ans. 20s.
31. Two persons, A and J5, lay out equal sums of money
in trade; A gains $126, and B loses $87, and A's money is
then double of B's : what did each lay out? Ans. $300.
32. A person goes to a tavern with a certain sum of
money in his pocket, where he spends 2 shillings : he then
borrows as much money as he had left, and going to another
tavern, he there spends 2 shillings also; then borrowing
128 ELK MEN TAUT ALGEBRA.
again as much money as was left, he went to a third tavern,
where likewise he spent 2 shillings, and borrowed as much
as he had left : and again spending 2 shillings at a fourth
tavern, he then had nothing remaining. What had he at
first ? Ans. 3s. 9 d.
33. A tailor cut 19 yards from each of three equal pieces
of cloth, and 17 yards from another of the same length,
and found that the four remnants were together equal to
142 yards. How many yards in each piece ? Ans 54.
34. A fortress is garrisoned by 2600 men, consisting of
infantry, artillery, and cavalry. Now, there are nine times
as many infantry, and three times as many artillery soldiers
as there are cavalry. How many are there of each corps ?
Ans. 200 cavalry; COO artillery ; 1800 infantry.
35. All the journeyings of an individual amounted to 2970
miles. Of these he traveled 3| times as many by water as
on horseback, and 2| times as many on foot as by water.
How many miles did he travel in each way ?
Ans. 240 miles; 840 m. ; 1890 m.
36. A sum of money was divided between two persons,
A and J5. A's share was to J?'s in the proportion of 5 to 3,
and exceeded five-ninths of the entire sum by 50. "What
was the share of each? Ans. A's share, 450; JFs, 270.
37. Divide a number « into three such parts that the
second shall be n times the first, and the third m times as
great as the first.
a na ma
2d» T-T-; rrrr, '•> 3d>
± OC« « ^*U»* .5 wi«
1 + m + n 1 + m + n 1 + m + n
38. A father directs that $1170 shall be divided among
his three sons, in proportion to their ages. The oldest ia
t\vice as old as the youngest, and the second is one-third
older than the youngest. How much was each to receive?
A/tv. $'270, youngest ; *:!UO, second ; $540, < Ulest.
PROBLEMS. 129
39. Three regiments are to furnish 594 men, and each to
furnish in proportion to its strength. Now, the strength of
the first is to the second as 3 to 5 ; and that of the second
to the third as 8 to 7. How many must each furnish ?
Ans. 1st, 144 men ; 2d, 240 ; 3d, 210.
40. Five heirs, A, J?, (7, D, and E, are to divide an inher-
itance of $5600. -B is to receive twice as much as A, and
8200 more ; C three times as much as A, less $400 ; D the
half of what -Z> and C receive together, and 150 more ; and
E the fourth part of what the four others get, plus $475.
How much did each receive ?
A>s, $500; .#'s, 1200; C"a, 1100; D's, 1300; E's, 1500.
41. A person has four casks, the second of which being
filled from the first, leaves the first four-sevenths full. The
third being filled from the second, leaves it one-fourth full,
and when the third is emptied into the fourth, it is found to
fill only nine-sixteenths of it. But the first Avill fill the third
and fourth, and leave 15 quarts remaining. How many
gallons does each hold ?
Ans. 1st, 35 gal. ; 2d, 15 gal. ; 3d, 11} gal. ; 4th, 20 gal.
42. A courier having started from a place, is pursued by
a second after the lapse of 10 days. The first travels 4
miles a day, the other 9. How many days before the
second will overtake the first ? Ans. 8.
43. A courier goes 3H miles every five hours, and is fol-
lowed by another after he had been gone eight hours. The
second travels 221 miles every three hoxirs. How many
hours before he will overtake the first ? Ans. 42.
44. Two places are eighty miles apart, and a person leaves
one of them and travels towards the other at the rate of Si-
miles per hour. Eight hours after, a person departs from
6*
130 ELEMENTARY ALGEBRA.
the second place, and travels at the rate of 5| miles per hour.
How long before they will be together?
A.ns. 6 hours.
EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES.
11O. If we have a single equation, as,
2x + 3y = 21,
containing two unknown quantities, % and y, we may find
the value of one of them in terms of the other, as,
21 - 3
x =
(1.)
Now, if the value of y is unknown, that of x will also be
unknown. Hence, from a single equation, containing two
unknown quantities, the value of x cannot be determined.
If we have a second equation, as,
5x + 4y = 35,
we may, as before, find the value of x in terms of y, giving,
35 — 4y
"--— (2°
Now, if the values of x and y are the same .in Equations
( 1 ) and ( 2 ), the second members may be placed equal to
each other, giving,
21 — 3y 35 — 4y
t = ^-2 , or 105 - 15y = 70 - Sy ;
£. O
from which we find, y = 5.
110. In one equation containing two unknown quantities, can you find
the value of either ? I£ you hare a second equation involving the same
two unknown quantities, can YOU find their values? What are such equa-
tions called ?
ELIMINATION. 131
Subtituting this value for y in Equations (1) or (2), we
6iid x = 3. Such equations are called Simultaneous
equations. Hence,
111. SIMULTANEOUS EQUATIONS are those in which the
Talues of the unknown quantity are the same in both.
ELIMINATION.
112. ELIMINATION is the operation of combining two
equations, containing tAvo unknown quantities, and deducing
therefrom a single equation, containing but one.
There are three principal methods of elimination :
1st. By addition or subtraction.
2d. By substitution.
3d. By comparison.
We shall consider these methods separately.
Elimination by Addition or Subtraction.
1. Take the two equations,
3x — 2y = 7,
8x -f- 2y = 48.
If we add these two equations, member to member, we
obtain,
llaj = 55;
which gives, by dividing by 11,
x = 5;
and substituting this value in either of the given equations,
we find,
y = 4.
111. What are simultaneous equations?
112. What is elimination? How many methods of elimination are
there ? What are they ?
132 ELEMENTARY ALGEBRA.
2. Again, take the equations,
Sx + 2y = 48,
3z + 2y = 23.
If we subtract the 2d equation from the 1st, we obtain,
5x =. 25;
which gives, by dividing by 5,
x — 5;
and by substituting this value, we find,
V = 4.
3. Given the sum ot two numbers equal to s, and theif
difference equal to <7, to find the numbers.
Let x = the greater, and y the less number.
Then, by the conditions, x + y = g.
and, x — y = d.
By adding (Ait. 102, Ax. 1), 2x = s + d.
By subtracting (Art, 102, Ax. 2), . . . 2y = * — d.
Each of these equations contains but one unknown quantity.
From the first, we obtain, x = S-
o '
and from the second, y —
These are the same values as were found in Prob. 7, page
120.
4. A person engaged a workman for 48 days. For each
day that he labored he was to receive 24 cents, and for each
day that he was idle he was to pay 1 2 cents for his board.
At the end of the 48 days the account was settled, when the
laborer received 504 cents. Required the number of work-
ing days, and the number of days he was idle.
ELIMINATION. 133
Let a; = the number of working days,
y — the number of idle days.
Then. 24a = what he earned,
and, 12y = what he paid for his board.
Then, by the conditions of the question, we have,
x + y = 48,
and, 24« — 12y = 504.
This is the statement of the problem.
It has already been sho~wn (Art. 102, Ax. 3), that the two
members of an equation may be multiplied by the same num-
ber, without destroying the equality. Let, then, the first
equation be multiplied by 24, the coefficient of x in the
second ; we shall then have,
24<e + 24y = 1152
24jc — 12y = 504
and by subtracting, 36y = 648
•1 '"''' ' •'• " = i? = 18- '
Substituting this value of y in the equation,
24x — 12y = 504, we have, 24* — 216 = 504;
which gives,
720
24a? — 504 + 216 = 720, and x = — - = 30.
VERIFICATION'.
x + y = 48 gives 30 + 18 = 48,
24« — 12y = 504 gives 24 X 80 — 12 X 18 = 504.
134 ELEMENTARY ALGEBRA.
113. In a similar manner, either unknown quantity may
be eliminated from either equation ; hence, the folio Aving
RULE.
I. Prepare the equations so that the coefficients of the
quantity to be eliminated shall be numerically equal:
II. If the signs are unlike, add the equations, member
to member / if alike, subtract them, member from member.
EXAMPLES.
Find the values of x and y, by addition or subtraction,
in the following simultaneous equations :
., j 4x - >ly = — 22 )
6. •{ Ans. x = 5, y = 6.
( 5x + 2y = 37 )
„ j 2x + Gy = 42 )
7. -j Ans. x = 44, y = 54.
( 8x — Qy = 3 )
j 8x — 9y = 1 )
8. i Ans. x — 4, y — i.
( 6cc — 3 = 4z )
. J a; = 6, y = 9.
11. X 7 V 4ns. < x = 14, y = 16.
113. What is the rule fa? elimination by addition or subtraction?
ELIMINATION. 135
12. Says A to .Z?, you give me $40 of jour money, and
I shall then have five times as much as you will have left.
Xow they both had $120 : how much had each?
Ans. Each had $60.
13. A father says to his son, " twenty years ago, my age
was four times yours ; now it is just double : " what were
their ages ? / A i Father's, 60 years.
Ans. S ,.,
( Son's, 30 years.
•
14. A father divided his property between his two sons.
At the end of the first year the elder had spent one-quarter
of his, and the younger had made $1000, and their property
was then equal. After this the elder spent $500, and the
younger made $2000, when it appeared that the younger had
just double the elder: what had each from the father?
. j Elder, $4000.
' ( Younger, $2000.
15. If John give Charles 15 apples, they will have the
same number; but if Charles give 15 to John, John will
have 15 times as many, wanting 10, as Charles will have left.
How many has each ? j j J°^n> 50.
S- 1 Charles, 20.
16. Two clerks, A and _Z?, have salaries which are together
equal to $900. A spends T^ per year of what he receives,
and _B adds as much to his as A spends. At the end of the
year they have equal sums: what was the salary of each?
A>s = $500.
Elimination by Substitution.
114. Let us again take the equations,
Sa; + ty = 43, ( 1.)
Ux + 9y = 69. (2.)
1 1 4. Give the rule for eliminaticm by substitution. When is this method
t>sed to the greatest advantage?
136 ELEMENTAKY ALGEBRA.
Find the value of x in the first equation, which gives,
Substitute this value of x in the second equation, and w«
have,
11 x -L_J£+ gy = 69;
or,, 473 — 77y -f 45y = 345 ;
or, — 32y = — 128.
Here, ar has been eliminated by substitution.
In a similar manner, we can eliminate any unknown quan-
tity ; hence, the
RULE.
I. Find from either equation the value of the unknown
quantity to be eliminated:
II. Substitute this value for that quantity in the other
equation.
NOTE. —This method of elimination is used to great advan-
tage when the coefficient of either of the unknown quantities
is 1.
EXAMPLES.
Find, by the last method, the values of x and y in the
following equations :
1. 3cc — y = 1, and 3y — 2x — 4.
Ans. x = 1, y = 2.
2. 5y — 4x = — 22, and 3y + 4x = 38.
Ans. x = 8, y •=. 2.
3. x + 8y = 18, and y — Sx = — 29.
Ans. x = 10, y = 1.
ELIMINATION. 137
2
4. 5x -• y = 13, and Sx -f -y = 29.
y
Ans. x = 3£, y =
5. 103 — = 69, and lOy — " '= 49.
5 7
6. x + * - = 10, and + = 2.
x = 8, y = 10.
7. | - | + 5 = 2, a; + | = I7f
Ans. x = 15, y = 14.
8- 1 + 5 + 3 = 8*. and 7~* = I'
•. « = 3|, y = 4.
y
n _ i « . . K
8 4 H 5) 12 16
Ans. x = 12, y = 16.
10. | - y - 1 = - 9, and 5a: - ^ = 29.
^Iws. a; = 6, y = 7.
11. Two misers, -4 and jB, sit down to count over their
money. They both have $20000, and B has three times as
much as A : how much has each ?
, A
A \ ^3.,
' (^,
$15000.
12. A person has two purses. If he puts 87 into the first,
the whole is worth three times as much as the second purse :
but if he puts $7 into the second, the whole is worth five
times as much as the first: what is the value of each purse?
Ans. 1st, $2 ; 2d, $3.
138 K L E M K N T A Ii Y A L G K B K A .
13. Two numbers have the following properties : if the
first be multiplied by 6, the product will be equal to the
second multiplied by 5 ; and 1 subtracted from the first
leaves the same remainder as 2 subtracted from the second :
what are the numbers ? Ans. 5 and 6.
14. Find two numbers with the following properties : the
first increased by 2 is 3£ times as great as the second ;
and the second increased by 4 gives a number equal to half
the first : what are the numbers ? Ans. 24 and 8.
15. A father says to his son, "twelve years ago, I was
twice as old as you are now: four times your age at that
time, plus twelve years, will express my age twelve years
hence : " what were their ages ?
( Father, 72 years.
AnS' \ Son, 30 «
Elimination by Comparison.
115. Take the same equations,
5x + 1y = 43
llx + Qy = 69.
Finding the value of x from the first equation, we have,
43 — iy
iyt _ __ 5_ •
~T~
and finding the value of x from the second, we obtain,
69 — 9y
115. Give the rule for climinatnn by comparison.
ELIMINATION. 139
Let these two values of x be placed equal to each other,
and we have,
43 — 7y __ 69 — 9y
5 11
Or, 473 — 77y = 345 - 45y;
or, — 32y = — 128.
Hence, y — 4.
69 — 36
And, x = — — — = 3.
This method of elimination is called the method by com-
parison, for which we have the following
BULE.
I. Find, from each equation, the value of the same
unknown quantity to be eliminated:
II. Place these values equal to each other.
EXAMPLES.
Find, by the last rule, the values of x and y, from the
following equations,
1. 3a5 4- ^ -f 6 = 42, and y — ^- = 14£-
Ans. 35 = 11, y = 15.
2. | - | + 5 = 6, and | -f 4 = 2- + 6.
Ans. x — 28, y = 20.
•> «c 22
3' 15 - 4 + T = '' and 3y ~ * = 6'
,4ns. 35 = 9, y = 5.
4. y — 3 = -a + 5, and — ~= = y — 3£.
A ^
^lw*. a; = 2, y = 9,
14:0 ELEMENTARY ALGEBRA.
y-x x y_
Ans. x = 16, y = 7.
2y
= as - -|, and s + y = H.
. a; = 10, y = 6.
— —
7. - --S = a> - 2f, BB - L— = 0.
. a; = 1, y — *•
8. 2y + 3« = y + 43, y - ^=— = y - ^-
o o
Ans. x — 10, y = 13.
9. 4y — x ~ y = x + 18, and 27 — y — X + y + t.
2
. a; = 9, y =. 7.
f-2 =|.
. a; = 10, y = 20.
116. Having explained the principal methods of elimina-
tion, we shall add a few examples which may be solved by
any one of them ; and often indeed, it may be advantageous
to employ them all, even in the same example.
GENERAL EXAMPLES.
Find the values of x and y in the following simultaneous
equations :
1. 2a; 4- 3y = 16, and 3x — 2y = 11.
Ans. x = 5, y = 2.
ELIMINATION.
HI
to ty _ j
2 5 4 - 20'
3x 2y 61
T 5" = T2~6"
Ans. x = -, y = -
3. + 7y = 99, and + 7s = 51.
7 7
4 *_i2-y
* 2 "4
.4ns. a; = 7, y = 14.
?L+y + 1 _ 8 = 2_y^ + 27.
5.
4x
5
x —
— V x 4- V
6., -^rV^-8*
j«-
7.
8.
9.
3y - as 2a; - y _ g
e — y +
• o*K ~~~ o
4
Sx — 3 —
o
6 - y
4a; — 4 y — 5
!-i*-iy +
= 79
6 =
f_
Ans. x = CO, y •=. 40.
a; = 6.
y-5.
E L E M E N T A li Y A L G K B K A .
c + ab — bd
I x =
Ans.
IT.
P li O B L K M 8 . 143
PROBLEMS.
1. What fraction is that, to the numerator of which if 1
be added, the value will be - , but if 1 be added to ita
1
denominator, the value will be - ?
m
Let the fraction be denoted by - •
y
Then, by the conditions,
x -f 1 1 x I
— ;»> :"d' y + i--r
whence, Sx + 3 = y, and 4x = y +1.
Therefore, by subtracting,
x — 3 = 1, and x = 4.
Hence, 12 + 3 = y;
.-. y = 15.
2. A market-woman bought a certain number of eggs at
2 for a penny, and as many others at 3 for a penny ; and
having sold them all together, at the rate of 5 for 2c?, found
that she had lost 4d: how many of both kinds did she buy ?
Let 2x denote the whole number of eggs.
Then, x — the number of eggs of each sort.
Then will, -x = the cost of the first sort,
•" • «
and, -x = the cost of the second sort.
3
But, by the conditions of the question,
hence, — will denote the amounts for which the eggs
O
were sold.
144 ELEMENTARY ALGEBBA.
But, by the conditions,
1 1 • 4«
S"+s?:-T--41
therefore, 15« -f 10x — 24ce = 120;
.*. x = 120 ; the number of eggs of each sort.
3. A person possessed a capital of 30,000 dollars, for
which he received a certain interest ; but he owed the sum
of 20,000 dollars, for which he paid a certain annual interest.
The interest that he received exceeded that which he paid
by 800 dollars. Another person possessed 35,000 dollars, for
which he received interest at the second of the above rates ;
but he owed 24,000 dollars, for which he paid interest at the
first of the above rates. The interest that he received, an-
nually, exceeded that which he paid, by 310 dollars. Re-
quired the two rates of interest.
Let x denote the number of units in the first rate of
interest, and y the unit in the second rate. Then each may
be regarded as denoting the interest on $100 for 1 year.
To obtain the interest of $30,000 at the first rate, denoted
by «, we form the proportion,
100 : 30,000 : : x : — '—- — , or 300a;.
And for the interest of $20,000, the rate being y,
100 : 20,000 : : y : - Or 200y.
But, by the conditions, the difference between these two
amounts is equal to 800 dollars.
We have, then, for the first equation of the problem,
SOOaj — 200y = 800.
PROBLEMS. 145
By expressing, algebraically, the second condition of the
problem, we obtain a second equation,
350y — 240o; = 310.
Both members of the first equation being divisible by 100,
and those of the second by 10, we have,
$x — 2y = 8, 35y — 24« = 31.
To eliminate jc, multiply the first equation by 8, and then
add the result to the second ; there results,
19y = 95, whence, y = 5.
Substituting for y, in the first equation, this value, and
that equation becomes,
3x — 10 = 8, whence, x = 6.
Therefore, the first rate is 6 per cent, and the second 5.
VERIFICATION.
$30,000, at 6 per cent, gives 30,000 X .06 = $1800.
$20,000, 5 " " 20,000 X .05 — $1000.
And we have, 1800 — 1000 = 800.
The second condition can be verified in the same manner.
4. "What two numbers are those, whose difference is 7,
and sum 33 ? Ans. 13 and 20.
5. Divide the number 75 into two such parts, that three
times the greater may exceed seven times the less by 15.
Ans. 54 and 21.
6. In a mixture of wine and cider, | of the whole plus 25
gallons was wine, and £ part minus 5 gallons was cider : how
many gallons were there of each ?
Ans. 85 of wine, and 35 of cider.
7
146 ELEMENTARY ALGEBRA.
7. A bill of £120 was paid in guineas and moidores, and
the number of pieces used, of both sorts, was just 100. If
the guinea be estimated at 21s, and the moidore at 2Vs, how
many pieces were there of each sort ? Ans. 50.
8. Two travelers set out at the same time from London
and York, whose distance apart is 150 miles. One of them
travels 8 miles a day, and the other 7 : in what time will
they meet? Ans. In 10 days.
9. At a certain election, 375 persons voted for two candi-
dates, and the candidate chosen had a majority of 91 : how
many voted for oaeli ?
Ans. 233 for one, and 142 for the other.
10. A person has two horses, and a saddle worth £50.
Now, if the saddle be put on the back of the first horse, it
makes their joint value double that of the second horse ;
but if it be put on the back of the second, it makes their
joint value triple that of the first : what is the value of each
horse ? Ans. One £30, and the other £40.
11. The hour and minute hands of a clock are exactly to-
gether at 12 o'clock : when will they be again together?
Ans. Ih. 5/Tni'
12. A man and his wife usually drank out a cask of beer
in 12 days ; but when the man was from home, it lasted the
woman 30 days : how many days would the man alone be
in drinking it ? Ans. 20 days.
13. If 32 pounds of sea-water contain 1 pound of salt, how
much fresh water must be added to these 32 pounds, in order
that the quantity of salt contained in 32 pounds of the new
mixture shall be reduced to 2 ounces, or | of a pound ?
Ans. 224 Ibs.
14. A person who possessed 100,000 dollars, placed the
greater part of it out at 5 per cent interest, and the other
PUOBLKMS. 14:7
at 4 per cent. The interest which he received for the whole,
amounted to 4640 dollars. Required the two parts.
Ans. $64,000 and $36,000.
15. At the close of an election, the successful candidate
had a majority of 1500 votes. Had a fourth of the votes of
the unsuccessful candidate been also given to him, he would
have received three times as many as his competitor, want-
ing three thousand five hundred : how many votes did each
receive? • i 1st, 6500.
2d, 5000.
•
- \
16. A gentleman bought a gold and a silver watch, and a
chain worth $25. When he put the chain on the gold watch»
it and the chain became worth, three and a half times more
than the silver watch ; but when he put the chain on the
silver watch, they became worth one-half the gold watch
and 15 dollars over : what was the value of each watch ?
j Gold watch, $80.
(Silver « $30.
1 7. There is a certain number expressed by two figures,
which figures are called digits. The sum of the digits is 11,
and if 13 be added to the first digit the sum will be three
times the second: what is the number? Ans. 56.
18. From a company of ladies and gentlemen 15 ladies
retire; there are then left two gentlemen to each lady.
After which 45 gentlemen depart, when there are left 5
ladies to each gentleman : how many were there of each at
first ? ( 50 gentlemen.
Ans. 1 . f ,.
} 40 ladies.
19. A person wishes to dispose of his horse by lottery.
If he sells the tickets at $2 each, he will lose $30 on his
horse ; but if he sells them at $3 each, he will receive $30
14:8 KLKMKNTAIIY ALGEBRA.
more than his horse cost him. "What is the value of the
horse, and number of tickets? . j Horse, $150.
' (No. of tickets, CO.
20. A person purchases a lot of wheat at $1, and a lot of
rye at 75 cents per bushel ; the whole costing him §1 17.50.
He then sells £ of his wheat and 1 of his rye at the same rate,
and realizes $27.50. How much did he buy of each ?
80 bush, of wheat.
Ans. -, , ,
50 bush, of rye.
21. There are 52 pieces of money in each of two bags. A.
takes from one, and IB from the other. A takes twice as
much as J3 left, and J? takes 7 times as much as A left.
How much did each take? ( A. 48 pieces.
Ans. < '
( B, 28 pieces.
22. Two persons, A and B, purchase a house together,
Worth $1200. Says A to _Z?, give me two-thirds of your
money and I can purchase it alone ; but, says J5 to A, if
you will give me three-fourths of your money I shall be able
to purchase it alone. How much had each ?
Ans. A, $800 ; J5, $600.
23. A grocer finds that if he mixes sherry and brandy in
the proportion of 2 to 1, the mixture will be worth 78s. per
dozen ; but if he mixes them in the proportion of 7 to 2, he
can get 79s. a dozen. "What is the price of each liquor per
dozen? Ans. Sherry, 81s. ; brandy, 72s.
Equations containing three or more unknown quantities.
Let us now consider equations involving three or
more unknown quantities.
Take the group of simultaneous equations,
117. Give the rule for solving any group of simultaneous equations?
EXAMPLES. 149
5x - Qy + 42 = 15, . . (1.)
7a + 4y — 3z = 19, . . (2.)
2x + y -f- 62 = 46. ... (3.)
To eliminate z by means of the first two equations, multi-
ply the first by 3, and the second by 4 ; then, since the
coefficients of z have contrary signs, add the two results
together. This gives a new equation :
43z — 2y = 121 . . . . . (4.)
Multiplying the second equation by 2 (a factor of the
coefficient of z in the third equation), and adding the result
to the third equation, we have,
16a + 9y = 84 ' (5.)
The question is then reduced to finding the values of x
and y, which will satisfy the new Equations (4) and (5).
Now, if the first be multiplied by 9, the second by 2, and
the results added together, we find,
419a; = 1257; whence, x = 3.
"We might, by means of Equations ( 4 ) and ( 5 ) deter-
mine y in the same way that we have determined x ; but
the value of y may be determined more simply, by substi-
tuting the value of x in Equation ( 5 ) ; thus,
04 40
48 + Qy = 84. .-. y = - — = 4.
In the same manner, the first of the three given equations
becomes, by substituting the values of x and y,
15 — 24 -f 4s = 15. .'. z = — = 6.
4
In the same way, any group of simultaneous equations
may be solved. Hence, the
150 ELEMENTARY ALGEBRA.
EULE.
1. Combine one equation of the group icith each of the
others, by eliminating one unknown quantity; there will
result a neio group containing one equation less than the
original group :
II. Combine one equation of this new group with each
of the others, by eliminating a second unknown quantity ;
there will result a new group containing two equations less
than the original group :
HI. Continue the operation until a single equation is
found, containing but one unknotcn quantity :
IV. find the value of this unknown, quantity by the
preceding rules ; substitute this in one of the group of
two equations, and find tlie value of a second unknown
quantity ; substitute these in either of the group of three,
finding a third unknown quantity ; and so on, till the
values of all are found.
NOTES. — 1. lu order that the value of the unknown quan-
tities may be determined, there must be just as many inde-
pendent equations of condition as there are unknown quan-
tities. If there are fewer equations than unknown quantities,
the resulting equation will contain at least two unknown
quantities, and hence, their values cannot be found (Art. 110).
If there are more equations than unknown quantities, the
conditions maybe contradictory, and the equations impossible.
2. It often happens that each of the proposed equations
does not contain all the unknown quantities. In this case,
with a little address, the elimination is very quickly per-
formed.
Take the four equations involving four unknown quanti-
ties :
2x — 3y + 2z = 13. (1.) 4y + 2z = 14. (3.)
4u — 2x = 30. (2.) 5y + 3u — 32. (4.)
EXAMPLES. 151
By inspecting these equations, we see that the elimination
of s in the two Equations, ( 1 ) and ( 3 ), will give an equa-
tion involving x and y\ and if we eliminate u in Equa-
tions ( 2 ) and ( 4 ), we shall obtain a second equation, in-
volving x and y. These last two unknown quantities may
therefore be easily determined. In the first place, the
elimination of z from ( 1 ) and ( 3 ) gives,
ty - 2x = 1 ;
That of u from ( 2 ) and ( 4 ) gives,
20y + Gx = 38.
Multiplying the first of these equations by 3, and adding,
41y = 41;
Whence, y = 1.
Substituting this value in 1y — 2x = 1, we find,
x = 3.
Substituting for x its value in Equation ( 2 ), it becomes
4u — 6 = 30.
Whence, u = 9.
And substituting for y its value in Equation (8), there
results,
z — 5.
EXAMPLES.
X + y + 2 = 29
x + 2y + 3z = 62
1. Given < > to find «, y, and z.
+ ^ + f = :o
Ans. x = 8, y = 9, z — 12.
152
ELEMENT AKY ALGEBRA.
f 2x + 4y — 33 = 22 ~\
2. Given < 4x — 2y + 5z — 18 I to find cr, y, and a.
[ 6z + 7y - z = 63 J
Ans. x = 3, y = 7, z = 4.
3. Given •
^B + -V 4- -8 =
L f
to find a?, y, and z.
2 = 12
. a; = 12, y = 20, z = 30.
4. Given < x + y — z = 18$ > to find a, y, and z.
x - y + z = 13|J
^ln«. x = 16, y — V£, s = 5^
{3a; + 5y = 161 ^
7* + 23 = 209 V to find a, y, and 2.
2y + z - 89 J
. x = 17, y = 22, z = 45.
6. Given •«
x =
1 1
— I — = a
x y
-1 + 1 = 6
X Z
to find a;, y, and z.
a + b — c'
a + c — £'
z =
b+c —
NOTE. — In this example we should not proceed to clear
the equation of fractions; but subtract immediately the
second equation from the first, and then add the third : we
thus find the value of y.
PROBLEMS. 153
PKOBLEMS.
1. Divide the number 90 into four such parts, that the
first increased by 2, the second diminished by 2, the third
multiplied by 2, and the fourth divided by 2, shall be equal
each to each.
This problem may be easily solved by introducing a new
unknown quantity.
Let a;, y, z, and u, denote the required parts, and desig-
nate by m the several equal quantities which arise from the
conditions. We shall then have,
u
x + 2 = m, y — 2 = m, 2« = m, - = m.
2
From which we find,
x = m — 2, y = m + 2, z = — , u = 2m.
And, by adding the equations,
x -{- y -\- z + u = m -f m H -f 2m = 4£m.
And since, by the conditions of the problem, the first
member is equal to 90, we have,
4|m = 90, or fw — 90;
hence, m = 20.
Having the value of m, we easily find the other values;
viz.:
a; — 18, y = 22, z = 10, u = 40.
2. There are three ingots, composed of different metals
mixed together. A pound of the first contains 7 ounces of
silver, 3 ounces of copper, and 6 of pewter. A pound of
the second contains 12 ounces of silver, 3 ounces of copper,
and 1 of pewter. A pound of the third contains 4 ounces
of silver, 7 ounces of copper, and 5 of pewter. It is required
7*
EL K M K x T A 11 y A L G K B K A .
to find how much it will take of each of the three ingots to
form a fourth, which shall contain in a pound, 8 ounces of
silver, 3J of copper, and 4£ of pewter.
Let a, y, and 2, denote the number of ounces which it
is necessary to take from the three ingots respectively, in
order to form a pound of the required ingot. Since there
are 7 ounces of silver in a pound, or 16 ounces, of the first
ingot, it follows that one ounce of it contains T^- of an ounce
of silver, and, consequently, in a number of ounces denoted
1x
by a*, there is -- ounces of silver. In the same manner,
16
12v 4 z
we find that, - — , and — , denote the number of ounces
16 16
of silver taken from the second and third ; but, from the
enunciation, one pound of the fourth ingot contains 8 ounces
of silver. We have, then, for the first equation,
"tx 12 y 42
16 4' T6~ " 16 =
or, clearing fractions,
Va; + 12y + 42 = 128.
As respects the copper, we should find,
3aj -f 3y + 72 = 60 ;
and with reference to the pewter,
6« + y + 02 = 68.
As the coefficients of y in these three equations are the
most simple, it is convenient to eliminate this unknown
quantity first.
Multiplying the second equation by 4, and subtracting the
first from it, member from member, we have,
5x + 242 r= 112.
PROBLEMS. 15ft
Multiplying the third equation by 3, and subtracting the
second from the resulting equation, we have,
I5x + Sz = 144.
Multiplying this last equation by 3, and subtracting the
preceding one, we obtain,
40a; = 320;
whence, x = 8.
Substitute this value for x in the equation,
I5x + Sz = 144;
it becomes, 120 -f 8z = 144,
whence, 2 = 3.
Lastly, the two values, x = 8, z = 3, being substituted
in the equation,
6aj + y + 5z = 68,
give, 48 + y + 15 = 68,
whence, y = 5.
Therefore, in order to form a pound of the fourth ingot,
we must take 8 ounces of the first, 5 ounces of the second,
and 3 of the third.
VERIFICATION.
If there be 1 ounces of silver in 16 ounces of the first
ingot, in eight ounces of it there should be a number of
ounces of silver expressed by
7x8
16
In like manner,
12 x 5 ,4x3
onri _
~T6 ' ' 16 '
will express the quantity of silver contained in 5 ounces of
the second ingot, and 3 ounces of the third.
156 E L K M E N T A K Y ALGEBRA.
Now, we have,
7X8 12 X 5 4x3 128
16 16 16 16
therefore, a pound of the fourth ingot contains 8 oiui ;es of
silver, as required by the enunciation. The same conditions
may be verified with respect to the copper and pewter.
3. A>s age is double .Z?'s, and IPs is triple of <7's, and the
sum of all their ages is 140 : what is the age of each?
Ans. A's - 84; B's = 42; and C's = 14.
4. A person bought a chaise, horse, and harness, for £60 ;
the horse came to twice the price of the harness, and the
chaise to twice the cost of the horse and harness : what did
he give for each? ( £13 6s. 8d. for the horse.
Ans. i £6 135. 4d. for the harness.
( £40 for the chaise.
5. Divide the number 36 into three such parts that 1 of
the first, i of the second, and 1 of the third, may be all
equal to each other. Ans. 8, 12, and 16.
6. If A and B together can do a piece of work in 8 days,
A and C together in 9 days, and B and C in ten days, how
many days would it take each to perform the same work
alone? Ans. A, 14|f ; B, I7ff; C, 23/T.
7. Three persons. A, B, and (7, begin to play together,
having among them all $600. At the end of the first game
A has won one-half of _B's money, which, added to his own,
makes double the amount B had at first. In the second
game, A loses and B wins just as much as C had at the be-
ginning, when A leaves off with exactly what he had at first :
how much had each at the beginning ?
Ans. A, $300 ; B, §200 ; C $100.
8. Three persons, A, B, and C, together possess $3640.
PROBLEMS. 157
If B gives A $400 of his money, then A will have $320
more than B\ but if B takes §140 of C 's money, then B
and C will have equal sums : how much has each ?
Ans. A, $800 ; B, $1280; <7, $1560.
9. Three persons have a bill to pay, which neither alone
is able to discharge. A says to B, " Give me the, 4th of
your money, and then I can pay the bill." B says to (7,
" Give me the 8th of yours, and I can pay it." But C says
to A) " You must give me the half of yours before I can
pay it, as I have but $8 " : what was the amount of their
bill, and how much money had A and B ?
. ( Amount of the bill, $13.
S' (A had $10, and .B $12.
10. A person possessed a certain capital, which he placed
out at a certain interest. Another person, who possessed
10000 dollars more than the first, and who put out his capital
1 per cent, more advantageously, had an annual income
greater by 800 dollars. A third person, who possessed
15000 dollars more than the first, putting out his capital 2
per cent, more advantageously, had an annual income greater
by 1500 dollars. Required, the capitals of the three per-
sons, and the rates of interest.
. j Sums at interest, $30000, $40000, $45000.
' ( Rates of interest, 4 5 6 pr. ct.
11. A widow receives an estate of $15000 from her de-
ceased husband, with directions to divide it among two sons
and three daughters, so that each son may receive twice as
much as each daughter, and she herself to receive $1000
more than all the children together : what was her share,
and what the share of each child ?
( The widow's share, $8000
Ans. < Each son's, $2000
' Each daughter's, $1000
158 ELEMENTARY ALGKBRA.
12. A certain sum of money is to be divided between
three persons, A, B, and C. A is to receive $3000 les?
than half of it, B $1000 less than one-third part, and C to
receive $800 more than the fourth part of the whole : what
is the sum to be divided, and what does each receive ?
{Sum, $38400.
A receives $16200.
B " $11800.
G " $10400.
13. A person has three horses, and a saddle which is worth
$220. If the saddle be put on the back of the first horse, it
will make his value equal to that of the second and third ;
if it be put on the back of the second, it will make his value
double that of the first and third ; if it be put on the back
of the third, it will make his value triple that of the first
and second : what is the value of each horse ?
Ans. 1st, $20 ; 2d, $100 ; 3d, $140.
14. The crew of a ship consisted of her complement of
sailors, and a number of soldiers. There wrere 22 sailors to
every three guns, and 10 over ; also, the whole number of
hands was five times the number of soldiers and guns to-
gether. But after an engagement, in which the slain were
one-fourth of the survivors, there wanted 5 men to make
13 men to every two guns: required, the number of guns,
soldiers and sailors.
Ans. 90 guns, 55 soldiers, and 670 sailors.
15. Three persons have $96, which they wish to divide
equally between them. In order to do this, A, who has the
most, gives to B and C as much as they have already ; then
B divides with A and C in the same manner, that is, by
giving to each as much as he had after A had divided with
them • C then makes a division with A and J9, when it is
PROBLK MS. 159
found that they all have equal sums: how ui.ich had each
at first? Ans. 1st, $52; 2d, $28; 3d, $16.
16. Divide the number a into three such parts, that the
first shall be to the second as m to n, and the second to the
third as p to q.
amp anp anq
mp+np-rnq mp+np+nq mp+np+nq
17. Three masons, A, J3, and C", are to build a wall. A
and B together can do it in 12 days ; B and C in 20 days ;
and A and C in 15 days : in what time can each do it alone,
and in what time can they all do it if they work together ?
Ans. A, in 20 days ; J5, in 30 ; and C, in 60 ; all, in 10.
160 ELEMENTARY ALGEBRA.
CHAPTER VI.
FORMATION OF POWERS.
118. A POWER of a quantity is the product obtained by
taking that quantity any number of times as a factor.
If the quantity be taken once as a factor, we have the first
power ; if taken twice, we have the second power ; if three
tunes, the third power; if n tunes, the nth power, n being
any whole number whatever.
A power is indicated by means of the exponential sign
thus,
a = a1 denotes first power of a.*
axa = a? " square, or 2d power of a.
a X a x a = a3 " cube, or third power of a.
axaxaxa = a* " fourth power of a.
axaxaxaxa = a5 " fifth power of a.
axaxaxa — = am " m'* power of a.
In every power there are three things to be considered :
1st. The quantity which enters as a factor, and which is
called the first power.
2d. The small figure which is placed at the right, and
a little above the letter, is called the exponent of the
» Since a° = 1 (Art. 49), a9 X a = 1 X a = a1 ; so that the two
factors of a1, are 1 and a.
118. What is a power of a quantity? What is the power when the
quantity is taken once as a factor ? When taken twice ? Three times ?
n times ? How is a power indicated ? In every power, how many things
are considered ? Name them.
POWERS OF MONOMIALS. 161
power, and shows how many times the letter enters as a
factor.
3d. The power itself, which is the final product, or result
of the multiplications.
POWERS OF MONOMIALS.
1 19. Let it be required to raise the monomial 2a3£>2 to
the fourth power. We have,
(2a3S2)4 = 2a352 x 2a3bz X 2a?I>* X 2a3&2,
which merely expresses that the fourth power is equal to
the product which arises from taking the quantity four
times as a factor. By the rules for multiplication, this pro-
duct is
from which we see,
1st. That the coefficient 2 must be raised to the 4th
power ; and,
2d. That the exponent of each letter must be multiplied
by 4, the exponent of the power.
As the same reasoning applies to every example, we have,
for the raising of monomials to any power, the following
RULE.
I. JRaise the coefficient to the required power :
IT. Multiply the exponent of each letter by the exponent
of the power.
EXAMPLES.
1. What is the square of 3o2y3? Ans. Oa4^6.
119. What is the rule for raising a monomial to any power? When
the monomial is positive, what will be the sign of its powers ? When
negative, what powers will be plus? what minus?
162 ELEMENTARY ALGEBRA.
2. What is the cube of 6a5y-x ? Ans. 216aiyx3.
3. What is the fourth power of 2aV** ? 16a12y12£20.
4. What is the square of a?bsy3 ? ^4.ns. a4510y6.
5. What is the seventh power of a?bcd3?
Ans.
6. What is the sixth power of a2£3c2<??
Ans.
7. What is the square and cube of — 2a2J2 ?
Square. Cube.
— 2azb2
By observing the way in which the powers are formed,
we may conclude,
1st. When the monomial is positive^ all the, powers icill
be positive.
2d. When the monomial is negative, all even powers icill
be positive, and all odd will be negative.
8. What is the square of — 2a*b5 ? Ans. 4a8£10.
9. What is the cube of — 5anb2 ? Ans. — 125a3nb*.
TO. What is the eighth power of — «3a;y2 ?
Ans. +'a2'ix8y16.
11. What is the seventh power of — ambnc ?
Ans. — a!mb~*cl.
12. Whot is the sixth power of 2ab6y5 ?
Ans.
P O W E B 8 OF F li A C T I O N S . 163
13. What is the ninth power of — anbci ?
Ans. — a9n69c18.
14. What is the sixth power of — Sabzd?
Ans. 729a6J12d8.
15. What is the square of — 10amZ»nc3 ?
Ans. 100a2m62nc6.
16. What is the crie of — 9ambnd3f* ?
Ans. —
17. What is the fourth power of — 4aflft
Ans. 256a20£12c16c?20.
18. What is the cube of — 4«2m52nc3dr?
Ans. —
19. What is the fifth power oflaWxy ?
Ans.
20. What is the square of 20xni/mc5? Ans. 400a52ny2mc10.
21. What is the fourth power of 3an£2"c3?
Ans.
22. What is the fifth power of — cndZmy?\f ?
Ans. — c5n
23. What is the sixth power of — c^b2"?* ?
Ans.
24. What is the fourth power of — 2azc2d3.
Ans.
POWERS OF FRACTIONS.
1S4O. From the definition of a power, and the rule for
the multiplication of fractions, the cube of the fraction ^, is
written,
(a\3 _ a a a _ a?
b) ~-~~ b x b x b == ^ ;
120. What is the rule for raising a fraction to any power*
16:1 ELEMENTARY ALGEBRA.
and since any fraction raised to any power, may be written
under the same form, we find any power of a fraction by
the following
BT7LE.
Raise the numerator to the required power for a new
numerator, and the denominator to the required power for
a new denominator,
The rule for signs is the same as in the last article.
EXAMPLES
Find the powers of the following fractions :
a — c \z a2 — 2ac + c2
/a-_cV
\b+c)
1. ^— - • AM.
b2 + 2bc + c2
2 lXy\.
2' I
,3Jc/
3. 1-gM- ^«.
4. (-fp-j- -4w- ^
M
cfa\3
5. ( — r-^1- ^s. —
6- (1
/ 3aw4 .
1 — ^vf- I • -4ns.
8. Fourth power of — — - • Ans.
9. Cube of X- y--
x H- y ce3 + oxly -{-
POWERS OF BINOMIALS. 165
2 (Zmx^ ct^ mx^ n
10. Fourth power of -• Ans. - -— -
**
„. f ,
11. Fifth power of — - — • Ans. —
POWERS OF BINOMIALS.
121. A Binomial, like a monomial, may be raised to any
power by the process of continued multiplication.
1. Find the fifth power of the binomial a 4- ft.
a 4- ft ...» ....... 1st power.
a 4- b
a2 4- ab
4- ab + ft2
a 4- ft
a3 4- 2a2ft 4-
aft2
2«ft2 4- ft3
a3 4- 3a2ft 4-
a 4- ft
3aft2 + ft3 . .
. . 3d power.
a* 4~ 3a3ft 4~
3a2ft2 4- aft3
3a2ft2 4- 3aft3 4-
» " '
a* 4- 4a3ft 4-
a 4- ft
6a2ft2 4~ 4aft3 4"
ft4 4th power.
a5 + 4a4ft + 6a3ft2 + 4a2ft3 + aft*
+ a4ft + 4a3ft2 + 6a2ft3 + 4aft* + ft8
a8 + 5a4ft 4- 10a3ft2 4- 10a2ft3 4- 5aft* 4- ft5
121. How may a binomial be raised to any power?
122. How does the number of multiplications compare with the ex-
ponent of the power? If the exponent is 4, what is the number of
multiplications? How many when it is ?n? How many things are con«
•idorod in the raising of powers ? Xame them.
166 ELEMENTARY ALOE BE A.
NOTE. — 122. It Avill be observed that tbe number of
multiplications is always 1 less than the units in the expo-
nent of the power. Thus, if the exponent is 1, no multipli-
cation is necessary. If it is 2, we multiply once ; if it is 3,
twice ; if 4, three times, &c. The powers of polynomials
may be expressed by means of an exponent. Thus, to
express that a + b is to be raised to the 5th power, we
write
(<• + *)5;
if to the mih power, we write
(a + b}m.
2. Find the 5th power of the binomial a — b.
a — b 1st power.
a - b
a? — ab
— ab + b2
a? — lab + b2 2d power.
a — b
1^ - 2 ($"J) \r Cllfi
— a*b + 2ab* — b3
_ 53 .... 3d power.
a - b __
a* — 3a3b + 3azb2 — ab3
4a53 + b* . 4th power.
a - b
-f 4a5* - b*
10a362 — 10a2i3 + Sab* — b5 Ans.
POWEH8 OF BINOMIALS. 167
In the same way the higher powers may be obtained. By
examining the powers of these binomials, it is plain that four
things must be considered :
1st. The number of terms of the power.
2d. The signs of the terms.
3d. The exponents of the letters.
4th. The coefficients of the terms.
Let us see according to what laws these are formed.
Of the Terms.
123. By examining the several multiplications, we shall
observe that the first power of a binomial contains two terms ;
the second power, three terms ; the third power, four terms ;
the fourth power, five ; the fifth power, six, &c. ; and hence
we may conclude :
That the number of terms in any power of a binomial,
is greater by one than the exponent of the power.
Of the Signs of tJie Terms.
124. It is evident that when both terms of the given
binomial are plus, all the terms of the power icill be plus.
If the second term of the binomial is negative, then all
the odd terms, counted from the left, will be positive, and
all the even terms negative.
123. How many terms does the first power of a binomial contain? The
second? The third? The nth power?
• 124. If both terms of a binomial are positive, what will be the signs
of the terms of the power? If the second term is negative, how are the
signs of the terms ?
168 ELEMENTARY ALGEBRA.
Of the Exponents.
125. The letter which occupies the first place in a bino-
mial, is called the leading letter. Thus, a is the leading
letter in the binomials a + ft, and a — b.
1st. It is evident that the exponent of the leading letter
in the first term, will be the same as the exponent of the
power; and that this exponent will diminish by one in each
term to the right, until we reach the last term, when it will
be 0 (Art. 49).
2d. The exponent of the second letter is 0 in the first
terra, and increases by one in each term to the right, to the
last term, when the exponent is the same as that of the given
power.
3d. The sum of the exponents of the two letters, in any
term, is equal to the exponent of the given power. This
last remark will enable us to verify any result obtained by
means of the binomial formula.
Let us now apply these principles in the two following
examples, in which the coefficients are omitted :
(a + b)6 . . . a6 + «5ft + «4ft2 + «3ft3 + «2ft4 -f aft5 + b6,
(a — by . . .a6 - a5b + a*ft2 — a3ft3 + a?b* — aft5 + ft6.
As the pupil should be practised in writing the terms with
their proper signs, without the coefficients, we will add a
few more examples.
125. Which is the leading letter of a binomial? What is the exponent
of this letter in the first term ? How does it change in the terms towards
the right ? What is the exponent of the second letter in the second term ?
How does it change iu the terms towards the right ? What is it in th«
last term ? What is the sum of the exoonents in any term equal to ?
POWERS OF BINOMIALS. 169
1. (a-f b)3 . . «3-fa26+a&2.f b\
2. (a — by . . a*-a3b+a?bz— ab3 + b*.
3. (a + b}5 • • as+a*&+a3&2+a2&3+aft* + b5.
4. (a - b)1 . . a?-aebi-a5b'i—atb3+a:ibl—azb5+ab*—b'1.
Of the Coefficients.
126. The coefficient of the first terra is 1. The coeffi
cient of the second term is the same as the exponent of the
given power. The coefficient of the third term is found by
multiplying the coefficient of the second term by the expo-
nent of the leading letter in that term, and dividing the
product by 2. And finally :
If the coefficient of any term be multiplied by the expo-
nent of the leading letter in that term, and the product
divided by the number which marks the place of the term
from the left, the quotient will be the coefficient of t/te
next term.
Thus, to find the coefficients in the example,
(a - 5)7 . . . a1- aeb + a5b*- a*b3+ a3b*- azb5 + ab6— b\
we first place the exponent 7 as a coefficient of the second
term. Then, to find the coefficient of the third term, we
multiply 7 by 6, the exponent of a, and divide by 2. The
quotient, 21, is the coefficient of the third terra. To find the
coefficient of the fourth, we multiply 21 by 5, and divide
the product by 3 ; this gives 35. To find the coefficient of
the fifth term, we multiply 35 by 4, and divide the product
by 4 ; this gives 35. The coefficient of the sixth term, found
126. What is the coefficient of the first terra ? What is the coefficient
of the second term ? How do you find the coefficient of the third term ?
How do you find the coefficient of any term ? What are the coefficients
of the first and last terms ? How are the coefficients of the exponents
of any two terms equally distant from the two extremes?
8
170 ELEMENTARY ALGEBKA.
in the same way, is 21 ; that of the seventh, 7 ; and that of
the eighth, 1. Collecting these coefficients,
(a - by =
-f 21a562— 35a453 + 35a354 — 21a?b5
NOTE. — We see, in examining this last result, that th*
coefficients of the extreme terms are each 1, and that the
coefficients of terms equally distant from the extreme terms
are equal. It will, therefore, be sufficient to find the coeffi-
cients of the first half of the terms, and from these the
others may be immediately written.
EXAMPLES
1. Find the fourth power of a + b.
Ans. a4 + 4a3b + 6a252 + 4a&3 + b4.
2. Find the fourth power of a — b.
Ans. a* — 4a3b + 6a252 — 4a£3 + b*.
3. Find the fifth power of a + b.
Ans. a5 + 5a45 -f 10a352 + 10a2&3 + 5a5* + #>.
4. Find the fifth power of a — b.
Ans. a5 — 5a4b + Wa3b2 — 10a2ft3 + 5ab* - b5.
5. Find the sixth power of a + b.
Gab5 + bs.
«8. Find the sixth power of a — b.
a6 — 6a55 + 15a452 — 20a3J3 + 15aW - 6ab5
127. When the terms of the binomial have coefficients,
we may still write out any power of it by means of the
Binomial Formula.
X Let it be required to find the cube of 2c + 3d.
(a + b)3 — a3 + Sazb + 3abz + b3.
POWERS OF BINOMIALS. 171
Here, 2c takes the place of a in the formula, and 3d the
place of b. Hence, we have,
. (I.)
and by performing the indicated operations, we have,
(2c + 3d)3 = 8c3 + 36c9-J
If we examine the second member of Equation ( 1 ), we
see that each term is made up of three factors: 1st, the
numerical factor ; 2d, some power of 2c ; and 3d, some
power of 3d. The powers of 2c are arranged in descend-
ing order towards the right, the last term involving the 0
power of 2c or 1 ; the powers of 3d are arranged in ascend-
ing order from the first term, where the 0 power enters, to
the last term.
The operation of raising a binomial involving coefficients,
is most readily effected by writing the three factors of each
term in a vertical column, and then performing the multipli-
cations as indicated below.
Find, by this method, the cube of 2c + 3d.
OPERATION.
1+3 +3 +1 Coefficients.
8c3 -f 4c2 + 2c +1 Powers of 2c
1 + 3d + Qd2 + 2ld3 Powers of 3d
(2c + d)3 = 8c3 + 3Qczd + 54cc?2 + 27d3
The preceding operation hardly requires explanation. In
the first line, write the numerical coefficients corresponding
to the particular power ; in the second line, write the de-
scending powers of the leading term to the 0 power ; in the
third line, write the ascending powers of the following term
from the 0 power upwards. It will be easiest to commence
172 K I. K M E N T A U Y ALGEBRA.
the second line on the right hand. The multiplication should
be performed from above, downwards.
8. Find the 4th power of 3a2< — 2bd.
(a + b)< = a4 + 4«35 + 6«2&2 + 4a53 + b*.
1+4 +0 +4 '+ 1
81a8c4 + 27a6c3 + 9a*c2 + 3a2c + 1
1 2bd + 4b*d2 - 8b3d3 + 16b*d .
BlaV —
9. What is the cube of 3x - Gy ?
Ans. 27cc3 — 162cc2y
10. What is the fourth power of a — 35?
Ans. a* — I2a3b + 54a2i2 — 108«i3 + Sib4.
11. What is the fifth power of c —
Ans. c5 — lOcV? + 40c3t?2 — 8Qc"d3 + BOcd* — 32f7».
12. What is the cube of 5a — 3d?
Ans. I25a3 — 225a?d + I35adz - 2
* This Ingenious method of writing the development of a binomial Is due to
fr-.ofewor WILLIAM G. PECK, of Columbia College.
EXTRACTION OF ROOTS. 173
CHAPTER VH.
SQUARE ROOT. RADICALS OF THE SECOND
DEGREE.
12 §. THE SQUARE ROOT of a number is one of its two
equal factors. Thus, 6 X 6 = 36; therefore, 6 is the square
root of 36.
The symbol for the square root, is y' , or the fractional
exponent \ ; thus,
r *
ya, or a ,
indicates the square root of a, or that one of the two equal
factors of a is to be found. The operation of finding sucli
factor is called, Extracting the Square Hoot.
129. Any number which can be resolved into two equal
integral factors, is called a perfect square.
The following Table, verified by actual multiplication, in-
dicates all the perfect squares between 1 and 100.
TABLE.
1, 4, 9, 16, 25, 36, 49, 64, 81, 100, squares.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, roots.
128. What is the square root of a number? Wha is the operation of
finding the equal factor called ?
129. What is a perfect square ? How many perfect squares are there
between 1 and 100, including both numbers ? What arc they?
174 ELEMENTARY ALGEBRA.
"We may employ this table for finding the square root of
any perfect square between 1 and 100.
Look for the number in the first line / if it is found
there, its square root will be found immediately under it.
If the given number is less than 100, and not a perfect
square, it will fall between two numbers of the ujyper line, and
its square root will be found between the ttco numbers directly
below ; the lesser of the two will be the entire part of the
root, and will be the true root to within less than 1.
Thus, if the given number is 55, it is found between the
perfect squares 49 and 64, and its root is 7 and a decimal
fraction.
NOTE.— There are ten perfect squares between 1 and 100,
if -we include both numbers ; and eight, if we exclude both.
If a number is greater than 100, its square root will be
greater than 10, that is, it will contain tens and units. Let
JV denote such a number, x the tens of its square root, and
y the units ; then will,
XT = (x + yY = a2 + 2a:y + t/2 = «2 + (2x + y)y.
That is, the number is equal to the square of the tens in its
roots, plus twice the product of the tens by the units, plus
the square of the units.
EXAMPLE.
1. Extract the square root of 6084.
Since this number is composed of more than
two places of figures, its root will contain more GO 84
than one. But since it is less than 10000, which
is the square of 100, the root will contain but two figures j
that is, units and tens.
SQUARE ROOT OF NUMBERS. 175
Now, the square of the tens must be found in the two
left-hand figures, which we will separate from the other two
by putting a point over the place of units, and a second over
the place of hundreds. These parts, of two figures each,
are called periods. The part 60 is comprised between the
two squares 49 and 64, of which the roots are 7 and 8 ; hence,
7 expresses the number of tens sought / and the required
root is composed of 7 tens and a certain number of units.
The figure 7 being found, we
write it on the right of the given 60 84 78
number, from which we separate 49
it by a vertical line : then we 7 X 2 = 14 8|118~4~
subtract its square, 49, from 60, 118 4
which leaves a remainder of 11, 0
to which we biing down the two
next figures, 84. The result of this operation, 1184, con-
tains twice the product of the tens by the units, plus the
square of the units.
But since tens multiplied by units cannot give a product
of a less unit than tens, it follows that the last figure, 4, can
form no part of the double product of the tens by the units ;
this double product is therefore found in the part 118, which
we separate from the units' place, 4.
Now if we double the tens, which gives 14, and then
divide 118 by 14, the quotient 8 will express the units, or a
number greater than the units. This quotient can never be
too small, since the part 118 will be at least equal to twice
the product of the tens by the units ; but it may be too
large, for the 118, besides the double product of the tens by
the units, may likewise contain tens arising from the square
of the units. To ascertain if the quotient 8 expresses the
right number of units, we write the 8 on the right of the 14,
which gives 148, and then we multiply 148 by 8. This
multiplication being effected, gives for a product, 1184, a
176 ELEMENTARY ALGEBRA.
number equal to the result of the first operation. Hav-
ing subtracted the product, we find the remainder equal
to 0; hence, 78 is the root required. In this operation,
we form, 1st, the square of the tens ; 2nd, the double
product ot the tens by the units ; and 3d, the square of
the units.
Indeed, in the operations, we have merely subtracted from
the given number GO 84 : 1st, the square of 7 tens, or of 70 ;
2d, t\vice the product of 70 by 8 ; and, 3d, the square of 8 ;
that is, the three parts which enter into the composition of
the square, 70 -f 8, or 78 ; and since the result of the sub-
traction is 0, it follows that 78 is the square root of 6084.
ISO. The operations in the last example have been per-
formed on but two periods, but it is plain that the same
methods of reasoning are equally applicable to larger num-
bers, for by changing the order of. the units, AVC do not
change the relation in which they stand to each other.
Thus, in the number 60 84 95, the two periods 60 84,
havo the same relation to each other as in the number
60 84 ; and hence the methods used in the last example are
equally applicable to larger numbers.
131. Hence, for the ex-traction of the square root of
numbers, we have the following
E u L E .
I. Point off the given number into periods of two figures
each, beginning at tlie right hand: .
H. Note the greatest perfect square in the first period on
tJie left, andpkice its root on the right, after the manner of
131. Give the rule for the extraction of the square root of numbers?
What is the first step ? What the second ? What the third ? What the
fourth? What the filth?
SQUARE KOOT OF NUMBERS. 177
a quotient in division y then subtract the square of this
root from the first period, and bring down the second period
for a remainder:
III. Double the root already found, and place the result
on the left for a divisor. Seek hoic many times the divisor
is contained in the remainder, exclusive of the right-hand
figure, and place the figure in the root and also at the right
of the divisor :
IV. Multiply the divisor, thus augmented, by the last
figure of the root, and subtract the product from the re-
mainder, and bring down the next period for a new remain-
der. J3ut if any of the products should be greater than
the remainder, diminish the last figure of the root by one :
V. Double the ichole root already found, for a new di-
visor, and continue the operation as before, until all the
periods are brought down.
132. NOTE. — 1. If, after all the periods are brought
down, there is no remainder, the given number is a perfect
square.
2. The number of places of figures iu the root will always
be equal to the number of periods into which the given
number is divided.
3. If the given number has not an exact root, there will
be a remainder after all the periods are brought down, in
which case ciphers may be annexed, forming new periods,
for each of which there will be one decimal place in the root.
132. What takes place v.-hen the given number id a perfect square !
How many places of figures will there be in the root? If the given num-
ber is not a perfect square, what may be done after all the periods are
brought dowii ?
178
ELEMENTARY ALGEBKA
EXAMPLES.
1. What is the square root of 36729 ?
In this example there are
two periods of decimals,
and, hence, two places of
decimals in the root.
3 67 29 191.64 +
1
29
261
381
3826
629
381
24800
22956
38324
31104 Rem.
2. To find the square root of 7225. Ans. 85.
3. To find the square root of 17689. Ans. 133.
4. To find the square root of 994009. Ans. 997.
5. To find the square root of 85673536. Ans. 9256.
6. To find the square root of 67798756. Ans. 8234.
7. To find the square root of 978121. Ans. 989.
8. To find the square root of 956484. Ans. 978.
9. What is the square root of 36372961 ? - Ans. 6031.
10. What is the square root of 22071204 ? Ans. 4698.
11. What is the square root of 106929? Ans. 327.
12. What of 12088868379025? Ans. 3476905.
13. What of 2268741 ? Ans. 1506.23 +.
14. What of 7596796? Ans. 2756.22 +,
15. What is the square root of 96 ? Ans. 9.79795 +.
16. What is the square root of 153? Ans. 12.36931 -f .
17. What is the square root of 101 ? Ans. 10.04987 -K
SQUARE ROOT OF FRACTIONS. 179
18. What of 285970396644 ? Ans. 534762.
19. What of 41605800625 ? Ans. 203975.
20. What of 48303584206084? Ans. 6950078.
EKTRACTION OF THE SQUARE ROOT OF FRACTIONS.
133. Since the square or second power of a fraction is
obtained by squaring the numerator and denominator sepa-
rately, it follows that „
The square root of a fraction will be equal to the square
root of the numerator divided by the square root of tJie
denominator.
For example, the square root of 7-5- is equal to r: for,
a a __ a^
V ^ T TO *
b b bz
1. What is the square root of -? Ans. -•
2.
What is the square root
of
16'
Ans. - -
4
8.
What is the
square root
of
81
8
Ans. -•
4.
What is the
square root
of
256
361
16
Ans. -.
5.
6.
What is the
What is the
square root
square root
of
of
64*
4096
Ans. I.
64
A<w o
61009
• 247
7.
What is the
square root
of
582169 ?
763
A -JT «
956484 '
* 978
183. To what is the square root of a fraction equal ?
ELEMENTAKY ALGEBKA.
134. If the numerator and denominator are not perfect
squares, the root of the fraction cannot be exactly found.
We can, however, easily find the approximate root.
RULE.
Multiply both terms of tlie fraction by the denominator :
Then extract the square root of the numerator, and divide
this root by the root of the denominator ; the quotient will
be the approximate root.
g
1. Find the square root of - •
5
Multiplying the numerator and denominator by 5
\/f - v/i =
hence, (3.8729 -f ) -f- 5 = .7745 -f = Ana.
7
2. What is the square root of - ? Ans. 1.32287 f.
14
3. What is the square root of — ? Ans. 1.24721 +.
9
4. What is the square root of 11—? Ans. 3.41869 +.
16
13
5. What is the square root of 7—? Ans. 2.71313 +.
36
6. What is the square root of 8—? Ans. 2.88203 -f.
g
7. What is the square root of — ? Ans. 0.64549 -f .
Q
8. What is the square root of 10 — ? Ans. 3.20936 +.
134. What is the rule when the numerator and denominator are not
perfect squares ?
SQUARE ROOT OF MONOMIALS. 181
135. Finally, instead of the last method, we may, if wo
please,
Change the common fraction into a decimal, and continue
tho> division until the number of decimal places is double
the number of places required in the root. TJien extract
the root of the decimal by the last rule.
EXAMPLES.
1. Extract the square of — to within .001. This num-
ber, reduced to decimals, is 0.785714 to within 0.000001 ; but
the root of 0.785714 to the nearest unit, is .886; hence,
0.886 is the root of — to within .001.
14
/ 1 3
2. Find the \/2~ to within 0.0001. Ans. 1.6931 +.
3. What is the square root of — ? Ans. 0.24253 +.
7
4. What is the square root of -? Ans. 0.93541 +.
8
g
5. What is the square root of - ? Ans. 1.29099 +.
3
EXTRACTION OF THE SQUARE ROOT OF MONOMIALS.
136. In order to discover the process for extracting the
square root of a monomial, we must see how its square ia
formed.
By the rule for the multiplication of monomials (Art. 42),
we have,
x 5a2£3c = 25a4i6c2 ;
135. What is a second method of finding the approximate root?
136. Give the rule for extracting the square root of monomials?
182 ELEMENTARY ALGEBRA.
that is, in order to square a monomial, it is necessary to-
square its coefficient and double the exponent of each oftht
letters. Hence, to find the square root of a monomial, we
have the following
RULE.
I. Extract the square root of the coefficient for a new
coefficient :
H. Divide tJie exponent of each letter by 2, and then
annex all the letters with their new exponents.
Since like signs in two factors give a plus sign in the pro-
duct, the square of — a, as well as that of -f a, will be
+ a2; hence, the square root of a2 is either + «» 01
— a. Also, the square root of 25«254, is either + 5aJ2,
or — 5abz. Whence we conclude, that if a monomial is
positive, its square root may be affected either with the sign
+ or — ; thus, 1/9O4 = ± 3a2 ; for, + 3a2 or — 3a2,
squared, gives -'- 9a4. The double sign ± , with which the
root is affected, is read plus and minus.
EXAMPLES.
1. What is the square root of 64a6&4?
and, -y/64a664 = — 8a352; for — 8a352x — 8a352=
Hence, -y/64a664 = ± Sa3b*.
2. Find the square root of 625a258c6. ± 25a£4c3.
3. Find the square root of 576a456c8. ± 24a2£3c4.
4. Find the square root of 196a;6y2s4. ± 14x3yz2.
5. Find the square root of 441a8£6c10J16. ± 21a453c5J8.
6. Find the square root of 784a12&14c16d'2. ± 2Saeb'ced.
7. Find the square root of 8la*b*c*. ± 9o4£2c3.
IMPERFECT 8 Q \I A K E S . 1 83
NOTES. — 137. 1. From the preceding rtuo it follows,
that when a monomial is a perfect square, its numerical
coefficient is a perfect square, arid all its exponents even
numbers. Tims, 25«4&2 is a perfect square.
2. If the proposed monomial were negative, it would be
impossible to extract its square root, since it has just been
shown (Art. 136) that the square of every quantity, whether
positive or negative, is essentially positive. Therefore,
are algebraic symbols which indicate operations that cannot
be performed. They are called imaginary quantities, or
rather, imaginary expressions, and are frequently met with
in the resolution of equations of the second degree.
IMPERFECT SQUARES.
13§. When the coefficient is not a perfect square, or
when the exponent of any letter is uneven, the monomial is
an imperfect square : thus, QSab* is an imperfect square.
Its root is then indifcated by means of the 'adical sign ; thus.
Such quantities are called, radical quantities, or radicals of
the second degree : hence,
A RADICAL QUANTITY, is the indicated root of an imperfect
power.
137. When is a monomial a perfect square ? What monomials are
these whose square roots cannot be extracted ? What are such expres-
sions called?
138. When is a monomial an imperfect square ? What are such
'Jtios called ? What is A radical quantity ?
184 ELEMENTARY ALGEBRA.
TRANSFORMATION OF RADICALS.
139. Let a and b denote any two numbers, and J5
the product of their square roots: then,
V^ X V* = P CD
Squaring both members, we have,
a X b = p* .... (2.)
Then, extracting the square root of both members of (2),
i/ao = p (3.)
And since the second members are the same in Equations
( 1 ) and ( 3 ), the first members are equal : that is,
The square root of the product oftico quantities is equal
to the product of their square roots.
140. Let a and b denote any two numbers, and q
the quotient of their square roots ; then,
^ = q (1.)
Squaring both members, we have,
a . .
•=• = q2 (2.)
o
then extracting the square root of both members of (2),
.....
and since the second members are the same in Equations ( 1 )
and (3 ), the first members are equal ; that is,
139. To what is the square root of the product of two quantities equal?
140. To what is the square root of the quotient of two quau tiling
ual?
equal?
TRANSFORMATION CF RADICALS. 185
The, square root of the quotient of two quantities is equal
to the quotient of their square roots.
These principles enable us to transform radical expres-
sions, or to reduce them to simpler forms ; thus, the expres-
sion,
98«&* = 495* X 2a ;
hence, yOSab* = -v/49i4 x 2a;
and by the principle of (Art. 139),
x 2a =
In like manner,
X 5bd =
•V/864a2£5cn = -v/144a264c10 x Qbc — 12ab*c5
The COEFFICIENT of a radical is the quantity without the
sign ; thus, in the expressions,
the quantities 752, 3abc, 12a52c5, are coefficients of the
radicals.
141. Hence, to simplify a radical of the second degree,
we have the following
RULE.
I. Divide the expression under the radical sign into two
factors, one of which shall be a perfect square :
II. ^Extract the square root of the perfect square, and
then multiply this root by the indicated square root of the,
remaining factor.
141. Give the rule for simplifying radicals of the second degree. How
do you determine whether a given number has a factor which is a perfect
square ?
186 ELEMENTAKY A L G K B E A .
-To determine if a given number has any factor
which is a perfect square, we examine and see if it is divi-
sible by either of the perfect squares,
4, 9, 1C, 25, 36, 49, 64, 81, &c. ;
if it is not, we conclude that it does not contain a factor
•vhich is a perfect square.
EXAMPLES.
Reduce the following radicals to their simplest form :
Ans. 5a-\/3abc.
2. Vl2Sb5a6dz. Ans.
3* / QO/797j8/» A ,vj o
• \t O — C(, U C« -/A/co.
4. i/256a2&4c8. Ans.
5. -v/1024a967c*. Ans. 32alb3c2yabc.
6.
7. \/675a755c2<?. Ans.
Ans.
9. V1008a9tf7ra8. Ans.
10. V215Ga10Z»8c6.
11.
142. NOTES. — 1. A coefficient, or a factor of a coeffi-
cient, may be carried under the radical sign, by squaring it.
Thus,
2. 2abid = 2
142. How may a coefBcient or factor be carried under the radical sign
To what is the square root of a negative quantity equal ?
ADDITION OF RADICAL
187
3.
4.
c2).
2. The square root of a negative quantity may also
simplified; thus,
x - 1 = -y/9 X -y/^1 — 3-/^~l,
^ 4a2 — -\/4a? x •/— 1 — Say'— 1 ; also,
X
and,
that is, the square root of a negative quantity is equal to
the square root of the same quantity with a positive sign,
multiplied into the square root of — 1.
Reduce the following :
2. V— 128«465.
3.
Ans. 6a2 J3c3
4. '- 48a3£c5.
ADDITION OF EADICALS.
143. SIMILAR RADICALS, of the second degree, are those
in which the quantities under the sign are the same. Thus,
the radicals 3-y/7>, and 5c^/b are similar, and so also are
2, and
144. Radicals are added like other algebraic quantities ;
hence, the following
143. \V h;;t are similar radicals of the second degree ?
144. Give the rule for the addition of radicals of the second degree ?
188 ELEMENTARY ALGEBKA
BULE.
L If the radicals are similar, add their coefficients, and
to the sum annex the common radical :
II. If the radicals are not similar, connect them together
icith their proper signs.
Thus, 3ay^ + Sc-y/J = (3a 4-
In like manner,
li/2a + 3-/2a = (7 + 3)-v/2
NOTES. — 1. Two radicals, which do not appear to be sim-
ilar at first sight, may become so by transformation (Art.
141.)
For example,
2-V/45 + 3-v/5 = 6-v/S + 3-/5 = 9^5-
2. When the radicals are not similar, the addition or sub-
traction can only be indicated. Thus, La order to add 3 -y/J
to 5-v/a, we write,
5-v/a + 3-/J.
Add together the following :
2. -/SOo^2 and yY2a4Z>2.
3. - - and
5
7^5
\/ — •
V 15
4. -y/125 and v^OOo2. ^/w. (5 + 10a)-v/5
/
V
oo 10
294
11.
SUBTRACTION OF RADICALS. 189
'9Sazx and
Ans. 1a^/2x + 6 i/x2 — a2.
12. -v243 and 10-/363. Ans.
13. -v/320o2i2 and -\/245a8b6. Ans. (Sab + 7a*53) i/5.
14. y^oa6*7 and -v/300a665. Ans. (5a353
SUBTRACTION OF RADICALS.
145. Radicals are subtracted like other algebraic quan-
tities ; hence, the following
'RULE.
I. If the radicals are similar, subtract the coefficient of
the subtrahend from that of the minuend, and to tJie differ-
ence annex the common radical :
II. If the radicals are not similar, indicate the operation
by the minus sign.
EXAMPLES.
1. What is the difference between 3a-\/b and a
Here, 3a\/b — a^fb = 1a-\/b. Ans.
14 K. Give the rule for the subtraction of radicals.
190 ELEMENTARY ALGEBRA.
2. From Oav/2762 subtract 6ay2752.
First, Qa-^27b2 — 27a£-y/3, and Ga-v/2762 = 1
and, 27aJy/3 — l8ab\/3 — 9a6-v/3. ^4.ns.
Find the differences between the following :
3. -1/75 and
4.
and
. (2ab —
27
and
j_
45
363 and
and
6.
7.
8.
**\ and
10. -/32002 and -SOa2.
Ans.
. 4a-/5".
12.
13. -v/ll2a866 and -v/28a866.
MULTIPLICATION OF RADICALS.
146. Radicals are multiplied like other algebraic quan-
tities ; hence, we have the following
EULE.
L Multiply the coefficients together for a new coefficient:
146. Give the rule for the multiplication of radicals.
DIVISION OF KADICAL8-. 391
II. Multiply together the quantities under the radical
signs:
HI. Then reduce the result to its simplest form.
1. Multiply 3a\/bc by 2\/ab.
3a^/bc X 2<\/ab = 3a X 2 x -\/bc x ^/ab.
which, by Art. 139, = 6a\/b2ac = §dbi/ac.
Multiply the following :
2. 3\/5ab and 4y^0a. Ans. 120«v^-
3. 2a-[/bc aad Sa^bc. Ans. 6a25c.
4. Za^/aT+W and — 3a<^a?~+&. A. -
5. Zabi/a + 6 and
6. 3-V/2 and 2-v/8. ^ws. 24.
7. fy^«^ and TSO
8. 2x + -\/b and 2« — ^/b. Ans. 4cc2 — b.
9.
10. 3«-/27a3 by
DIVISION OF EADICALS.
147. Radical quantities are divided like other algebraic
quantities ; hence, we have the following
RULE.
I. Divide the coefficient of the dividend by the coefficient
of the divisor, for a new coefficient :
147. Give the rule for the division of radicals.
192
ELEMENTARY ALGEBRA.
II. Divide the quantities under the radicals, in the same
manner :
HI. Then reduce the result to its simplest form.
1. Divide
EXAMPLES.
by <
'
— ' = 2, new coefficient.
Art. 140,
hence, the quotient is 2 X - = -
2. Divide
by 25-y/c.
Ans. V/--
3. Divide 12ac-y/6Jc by
4. Divide 6a\X%6* by
5. Divide 4a2-/5065 by
6. Divide 26a35-/81a2&2 by 13a\/Qab. A.
1. Divide 84a354v^7ac by 42a5-v/3ff- ^L
8. Divide -v/i^ ^7 V^-
9. Divide 6a2J2-/20a3 by 12-v/5«.
10. Divide 6a-v/10i2 by 3-v/sl
11. Divide 485* -/15 by 2J2y^
12. Divide Sa2^3-/^3 by 2a-v/28d
13. Divide 96a4c3/98i5 by
4a5v/3".
a3i2.
360R
. 2ai*c3J.
. 14a3ic2.
SQUARE BOOT OF POLYNOMIALS. 193
14. Divide 27a566-/2«"3 by ^Ja. Ans.
15. Divide 18aBb6^/8a* by 6ab</a?. Ans.
SQUARE ROOT OF POLYNOMIALS.
148. Before explaining the rule for the extraction of the
square root of a polynomial, let us first examine the squares
of several polynomials : we have,
(a + b)2 = a? + 2ab + b\
(a + b + c)2 = a2 + lab + b2 + 2(a + fyc + c2,
(a -f- b + c + (I)* = a2 + 2ai + b2 + 2(a + % + c3
+ 2(a + J + c)c? -f- f?2.
The /ate by which these squares are formed can be enun
ciated thus :
The square of any polynomial is equal to the square of
the first term, plus twice the product of the first term by the
second, plus the square of the. second; plus twice the first
two terms multiplied by the third, plus the square of the
third ; plus twice the first three terms multiplied by the
fourth, plus the square of the fourth ; and so on.
149. Hence, to extract the square root of a polynomial,
we have the following
RULE.
I. Arrange the polynomial \oith reference to one of its
letters, and extract the square root of tJie first term : thi*
witt give the first term of the root :
148. What is the square of a binomial equal to? What, is the square
of a trinomial equal to ? To what is the square of any polynomial equal ?
149. Give the jule for extracting the square root of a polynomial ?
What is the first step? Wl at the second ? What the third ? What the
fourth ?
9
194; KLKMKNTAKY ALGEBEA.
II. Divide the second term of the polynomial by double
the first term of the root, and the quotient will be the second
term of the root :
HI. Then form the square of the algebraic sum of the
two terms of the root found, and subtract it from the first
polynomial, and then divide the first term of the remainder
by double the first term of the root, and the quotient will be
the third term :
IV. Form the double product of the sum of the first and
second terms by the third, and add the square of the third /
then subtract this result from the last remainder, and divide
the first term vf the result so obtained, by double the -first
term of the root, and the quotient will be the fourth term.
Then proceed in a similar manner to find the other terms.
EXAMPLES.
1. Extract the square root of the polynomial,
49«252 — 24a£3 + 25a* — 30a3b
First arrane it with reference to the letter a.
165'
25a4 — 3Qa3b
5a2 — Sab
10a2
IQb* . . 1st Rem.
0 . 2c? Hem.
After having arranged the polynomial with reference to
a, extract the square root of 25«4 ; this gives 5a2, which
is placed at the right of the polynomial : then divide the
second term, - SQa3b, by the double of 5 a2, or lOa2;
the quotient is — Sab, which is placed at the right of 5a2.
Hence, the first two terms of the root are 5 a2 — Sab.
Squaring this binomial, it becomes 25a4 — 30«3J + Qa2b2,
which, subtracted from the proposed polynomial, gives a
remainder, of which the first term is 40a2ia. Dividing thia
'SQUARE ROOT OF POLYNOMIALS. 195
first term by 10a2, (the double of 5«2), the quotient is
-f- 4b2 ; this is the third term of the root, and is written on
the right of the first two terms. By forming the double
product of 5a2 — Sab by 4£>2, squaring 4£»2, and taking
the sum, we find the polynomial 40a2£2 — 24ab3 + 16&4,
which, subtracted from the first remainder, gives 0. There-
fore, 5a2 — Sab + 4£2 is the required root.
2. Find the square root of a*+ 4a3«+6a2a;2+4cfa^+ cc4*
-4ns. «2+ 2a« + a2.
3. Find the square root of a4— 4a3a+6a2cc2— 4a#3+ JB*.
Ans. a2 — 2ax + «2.
4. Find the square root of
4cc6 + 12x5 + So4 - 2x3 + Va:2 — 2aj + 1.
Ans. 2x? + 3a;2 — x + 1.
5. Find the square root of
3a2 — 2ab + 4i2.
6. What is the square root of
a4 — 4ax3 + 4a2a2 — 4a2 + 8aa; + 4 ?
-4;i5. a2 — 2aa — 2.
7. What is the square root of
9«2 — 12aj + 6xy + yz — 4y + 4 ?
-4ws. 3a; + y — 2.
8. What is the square root of y4 — 2y2a;2 + 2x2 — 2y2
1 + a* ? ^tn5. y2 — & — 1.
9. What is the square root of 9a4£4 — 30as£3+ 25a252?
Ans. 3a2b2 — Sab.
10. Find the square root of
48a52c3 f 3652c*
- 36a25c3 + 9a*c2.
Ans. 5a?b - 3«2c — 4abc -f
196 ELEMENTARY ALGEBKA.
I5O. Wo will conclude this subject with the following
remarks :
1st. A binomial can never be a perfect square, since we
know that the square of the most simple polynomial, viz.,
a binomial, contains three distinct parts, which cannot ex-
perience any reduction amongst themselves. Thus, the
expression a2 -f bz, is not a perfect square ; it wants the
term ± 2a£, in order that it should be the square of a ± b.
2d. In order that a trinomial, when arranged, may be a
perfect square, its two extreme -terms must be squares, and
the middle term must be the double product of the square
roots of the two others. Therefore, to obtain the square
root of a trinomial when it is a perfect square : Extract the
roots of the two extreme terms, and give these roots the same
or contrary signs, according as the middle term is positive
or negative. To verify it, see if the double product of the
two roots is the same as the middle term of the trinomial.
Thus,
9«6 — 48a4£2 + 64a2J4, is a perfect square,
since, y^6 = 3c/3, and -/64«2i4 — — Sab2 ;
and also,
2 x 3a3 X — Sab2 = — 4Sa*b2 = the middle term.
But, 4a2 4- 14a# + 952 is not a perfect square : for,
although 4a2 and + 9bz are the squares of 2a and 3#,
yet 2 x 2a X 35 is not equal to I4ab.
3d. In the series of operations required by the general
rule, when the first term of one of the remainders is not
exactly divisible by twice the first term of the root, we may
150. Can a binomial ever be a perfect power? Why not? When is
a trinomial a perfect square ? When, in extracting the square root, we
find that the first term of the remainder is not divisible by twice the root,
is the polynomial a perfect power or not?
SQUARE ROOT OK PCLYNOMIAI, S. 197
conclude that the proposed polynomial is not a perfect
square. This is an evident consequence of the course of
reasoning by which we have arrived at the general rule for
extracting the square root.
4th. When the polynomial is not a perfect square, it may
sometimes be simplified (See Art. 139).
Take, for example, the expression, <^d*b + 4a252 -f- 4ab3.
The quantity under the radical is not a perfect square ;
but it can be put under the form ab(a? + 4a5 + 4&2.)
Now, the factor within the parenthesis is evidently the
square of a -f 25, whence, we may conclude that,
+ 4a252 + 4a63 = (a + 25)
2. Reduce -y/2a26 — 4a52 + 253 to its simplest form.
Ans. (a — b)
1#S ELEMENTARY ALGEBRA
CHAPTER VILL
EQUATIONS OF THE SECOND DEGREE.
EQUATIONS CONTAINING ONE UNKNOWN QUANTITY.
151. AN EQUATION of the second degree containing bu^
one unknown quantity, is one in which the greatest exponent:
is equal to 2. Thus,
a2 = a, ax2 + bx = c,
are equations of the second degree.
152. Let us see to what form every equation of the
second degree may be reduced.
Take any equation of the second degree, as,
(1 +«)>-?*- 10 = 5 -? + £- .
Clearing of fractions, and performing indicated operations,
we have,
4 + 8x + 4«2 — 3x — 40 = 20 — « + 2x\
Transposing the unknown terms to the first member, the
known terms to the second, and arranging with reference to
the powers of #, we have,
4a2 — 2a;2 + 8x — 3x + x = 20 + 40 — 4 ;
151. What is an equation of the second degree ? Give an example.
152. To what form may every equation of the second degree be reduced?
EQUATIONS OF THE SECOND DEGREE.
and, by reducing,
2x2 + 6x = 56 ;
dividing by the coefficient of #2, we have,
x2 + 3x = 28.
If we denote the coefficient of x by 2p, and the second
member by q, we have,
xz + 2px = q.
This is called the reduced equation.
153. When the reduced equation is of this form, it con-
tains three terms, and is called a complete equation. The
terms are,
FIRST TERM. — The second power of the unknown quan-
tity, with a plus sign.
SECOND TERM. — The first power of the unknown quantity,
Math a coefficient.
THIRD TERM. — A known term, in the second member.
Every equation of the second degree may be reduced to
this form, by the following
BULK.
I. Clear the equation of fractions, and perform all the
indicated operations :
n. Transpose all the unknown terms to the first member,
and all the known terms to the second member :
153. How many terms are there in a complete equation ? What is the
first term ? "What is the second term ? What is the third term ? How
many operations are there in reducing an equation of the second degree
to the required form ? What is the first ? What the second ? What the
third ? What the fourth ?
200 ELEMENTAKY ALGEBKA.
TTT. Reduce all the terms containing the square of the
unknown quantity to a single term, one factor of which is
the square of the unknown quantity ; reduce, also, all thd
terms containing the first power of the unknown quantity,
to a single term :
IV. Divide both members of the resulting equation by
the coefficient of the square of tJie unknown quantity.
154. A ROOT of an equation is such a value of the un-
known quantity as, being substituted for it, will satisfy the
equation ; that is, make the two members equal.
The SOLUTION of an equation is the operation of finding
its roots.
INCOMPLETE EQUATIONS.
155. It may happen, that 2/>, the coefficient of the first
power of x, in the equation x + 2px = q, is equal to 0.
In this case, the first power of x will disappear, and the
equation will take the form,
*=! (1.)
This is called an incomplete equation ; hence,
AN INCOMPLETE EQUATION, when reduced, contains but
two terms; the square of the unknown quantity, and a
known term.
156. Extracting the square root of both members of
Equation ( 1 ), we have,
x = ±i/q.
154. What is the root of an equation ? What is the solution of an
equation ?
155. What form will the reduced equation take when the coefficient ol
a: is 0 ? What is the equation then called ? How many terms are there
iu an incomplete equation ? What are they ?
156. What is the rule for the solution of an incomplete equation T
How many roots are there in every incomplete equation ? How do tb«
roots compare with each other ?
EQUATIONS OF THE SECOND DEGREE. 201
Hence, for the solution of incomplete equations :
RULE.
I. Reduce the equation to the form x2 = q:
II. Then extract the square root of both members.
NOTE. — There will be two roots, numerically equal, but
having contrary signs. Denoting the first by a;', and the
second by «", we have,
x' — +i/q, and x" = — *<fq.
VERIFICATION.
Substituting + -y/^, or — i/q, for JB, in Equation ( 1 ),
we have,
(+V5)2 = q; and, (-V?)2 = <?;
hence, both satisfy the equation ; they are, therefore, roots.
(Art. 154.)
EXAMPLES.
1. What are the values of x in the equation,
Sx2 -f 8 = 5x2 — 10 ?
By transposing, 3x2 — 5x2 = — 10 — 8.
Reducing, — 2x2 = — 18.
Dividing by — 2, »2 = 9.
Extracting square root, x = ± -y/9 = + 3 and — 8.
Hence, x' = +3, and x" = — 3.
2. What are the roots of the equation,
3«2+ 6 = 4a2- 10?
Ans. x' = +4, x" = — 4,
202 ELEMENTARF ALGEBKA.
8. What are the roots of the equation,
Ans. x' — + 9, x" = - 9.
I. What are the roots of the equation,
4<e2 + 13 — 2x2 = 45 ?
Ans. x' = +4, x" = — 4.
5. What are the roots of the equation,
Gx" — 7 = 3x2 + 5 ?
Ans. . x' = + 2, x" = - 2,
6. What are the roots of the equation,
x2
8 4- 5x2 = - + 4x2 -|- 28 ?
5
Ans. x' — +5, *" = — 5.
7. What are the roots of the equation,
^+5 _ *2+29 = m_5a.2?
Ans. .x'= +5, x" = — 6.
8. What are the roots of the equation,
jB2 + ab = So2 ?
. a =
9. What are the roots of the equation,
a;2 = 5 -f a^ ?
'« - 25 Va - 2*
PROBLEMS.
PROBLEMS.
1. What number is that which being multiplied by itself
the product will be 144 ?
Let x = the number : then,
x X x = xz = 144.
It is plain that the value of x will be found by extracting
the square root of both members of the equation : that is,
T/x* = yT*4: that is, x = 12.
2. A person being asked how much money he had, said,
if the number of dollars be squared and 6 be added, the sum
will be 42 : how much had he ?
Let x = the number of dollars.
Then, by the conditions,
xz + 6 = 42 ;
hence, x2 = 42 — 6 = 36,
and, x = 6. Ans. $6.
3. A grocer being asked how much sugar he had sold to
a person, answered, if the square of the number of pounds
be multiplied by 7, the product will be 1575. How many
pounds had he sold ?
Denote the number of pounds by x. Then, by the con-
ditions of the question,
7«3 =' 1575 ;
hence, x2 = 225,
and, x = 15. Ans. 15.
4. A person being asked his age, said, if from the square
204 ELEMENTARY ALGEBKA.
of my age in years, you take 192 years, the remainder
be the square of half my age : what was his age ?
Denote the number of years in his age by x.
Then, by the conditions of the question,
/I \ x2
y? — 192 = (-a;]2 = — ,
and by clearing the fractions,
402 — 768 — x2 ;
hence, 4ce2 — cc2 = 768,
and, 3JC2 = 768,
a? = 256,
x — 16. Ans. 16 years.
5. What number is that whose eighth part multiplied by
its fifth part and the product divided by 4, will give a quo-
tient equal to 40 ?
Let x = the number.
By the conditions of the question,
hence, = 40 ;
by clearing of fractions,
x2 = 6400,
x = 80. Ans. 80.
6. Find a number such that one-third of it multiplied by
one fourth shall be equal to 108. Ans. 36.
7. What number is that whose sixth part multiplied by
its fifth part and the product divided by ten, will give a
quotient equal to 3 ? Ans. 30.
PROBLEMS. 205
8. "What number is that whose square, plus 18, -will be
equal to half the square, plus 30£ ? Ans. 5.
9. What numbers are those which are to each other as
1 to 2, and the difference of whose squares is equal to 75 ?
Let x = the less number.
Then, 2x = the greater.
Then, by the conditions of the question,
4a52 - y? = 75 ;
hence, 3«2 = 75,
and by dividing by 3, y? = 25, and x = 5,
and, 2« = 10.
Ans. 5 and 10.
10. What two numbers are those which are to each other
as 5 to 6, and the difference of whose squares is 44 ?
Let x = the greater number.
Then, -x = the less.
6
By the conditions of the problem,
3.2 3.2 A A .
36
by clearing of fractions,
36a;2 — 25ar» = 1584 ;
hence, lla;2 = 1584,
and, aJ2^ 144 ;
hence, x = 12,
and, -x = 10.
6
Ans. 10 and 12.
206 ELEMENTARY ALGEBRA.
11. What two numbers are those which are to each other
as 3 to 4, and the difference of whose squares is 28 ?
Ans. 6 and 8.
12. What two numbers are those which are to each other
as 5 to 11, and the sum of whose squares is 584 ?
Ans. 10 and 22-.
13. A says to .Z?, my son's age is one quarter of yours,
and the difference between the squares of the numbers
representing their ages is 240 : what were their ages ?
I Eldest, 16,
Ans. \ „
( lounger, 4.
Tico unJcnoicn quantities.
15 7. When there are two or more unknown quantities:
I. Eliminate one of the unknown quantities by Art.
113:
IL Then extract the square root of both members of the
equation.
PEOBLEMS.
1. There is a room of such dimensions, that the difference
of the sides multiplied by the less, is equal to 36, and the
product of the sides is equal to 360 : what are the sides?
»
Let x — the length of the less side ;
y = the length of the greater.
Then, by the first condition,
(y — x)x = 36 ;
and by the 2d, xy = 360.
157. How do you proceed when there are two or more unknown quan-
tities ?
PROBLEMS. 207
From the first equation, we have,
xy — xz = 36 ;
and by subtraction, x2 = 324.
Hence, x = -y/324 = 18;
360
y = - - = 20.
18
Ans. a; = 18, y — 20.
2. A merchant sells two pieces of muslin, which together
measure 12 yards. He received for each piece just so many
dollars per yard as the piece contained yards. Now, he gets
four times as much for one piece as for the other : how many
yards in each piece ?
Let x = the number of yards in the larger piece ;
y = the number of yards in the shorter piece.
Then, by the conditions of the question,
x + y = 12.
x X x = x2 = what he got for the larger piece ;
y x y = y2- = what he got for the shorter ;
and, as2 = 4y2, by the 2d condition,
x = 2y, by extracting the square root.
Substituting this value of x in the first equation, we have,
y + 2y = 12 ;
and, consequently, y = 4,
and, x = 8.
Ans. 8 and 4.
3. What two numbers are those whose product is 30, and
the quotient of the greater by the less, 3i ? Ans. 10 and 3.
4. The product of two numbers is a, and their quotient
b : what are the numbers ?
Ans. yob, and
208 ELEMENTAET ALGEBKA.
5. The sum of the squares of two numbers is 117, and the
difference of their squares 45 : what are the numbers?
Ans. 9 and 6.
6. The sum of the squares of two numbers is a, and the
difference of their squares is b : what are the numbers ?
A fa + b fa — b
Ans. x = v / , y = \ / .
V 2 V 2
7. What two numbers are those which are to each othei
as 3 to 4, and the sum of whose squares is 225 ?
Ans. 9 and ^2.
8. What two numbers are those which are to each other
as m to n, and the sum of whose squares is equal to a2 ?
ma na
Ans.
9. What two numbers are those which are to each other
as 1 to 2, and the difference of whose squares is 75 ?
Ans. 5 and 10.
10. What two numbers are those which a*re to each other
as m to w, and the difference of whose squares is equal to b2 ?
mb nb
Ans. —
11. A certain sum of money is placed at interest for six
months, at 8 per cent, per annum. Now, if the sum put at
interest be multiplied by the number expressing the interest,
the product will be $562500 : what is the principal at in-
terest? Ans. $3750.
12. A person distributes a sum of money between a num-
ber of women and boys. The number of women is to the
number of boys as 3 to 4. Now, the boys receive one-half
as many dollars as there are persons, and the women, twice
as many dollars as there are boys, and together they receive
COMPLETE EQUATIONS. 209
J38 dollars: how many women were there, and how many
boys?
36 women.
48 boys.
COMPLETE EQUATIONS.
Ans. j
158. The reduced form of the complete equation (Art.
153) is,
a2 + 2px = q.
Comparing the first member of this equation with the
square of a binomial (Art. 54), we see that it needs but the
square of half the coefficient of #, to make it a perfect square.
Adding p2 to both members (Ax. 1, Art. 102), we have,
y? + 2px -f- P" = q + pz.
Then, extracting the square root of both members (Ax. 5),
we have,
x + p = ± iq + p2.
Transposing p to the second member, we have,
x = — p ± -\/q + pz.
Hence, there are two roots, one corresponding to the plus
sign of the radical, and the other to the minus sign. De-
noting these roots by x' and x", we have,
x' — — p + ^q + p2, and x" = — p — -\/q -f p2.
The root denoted by x' is called the first root ; that de-
noted by x" is called the second root.
1 58. What is the form of the reduced equation of the second degree ?
What is the square of the binomial x + p ? How many of those terms
are found in the first term of the reduced equation? What must be
added to make the first member a perfect square ? How many roots are
there in every equation of the first degree? What is the first root equal
to ? What is the second equal to ?
210 ELEMENTARY ALGEBUA
159. The operation of squaring half the coefficient of
x and adding the result to both members of the equation, is
called Completing the Square. For the solution of every
complete equation of the second degree, we have the fol-
lowing
KULE.
I. Reduce the equation to the form, xz + 2px = q:
IE. Take half the coefficient of the second term, square
it, and add the result to both members of the equation :
HI. Then extract the square root of both members ; after
ivhich, transpose the known term to the second member.
NOTE. — Although, in the beginning, the student should
complete the square and then extract the square root, yet
he should be able, in all cases, to write the roots immediately,
by the following (See Art. 158)
ETTLE.
I. The first root is equal to half the coefficient of the
second term of the reduced equation, taken with a contrary
sign, plus the square root of the second member increased
by the square of half the coefficient of the second term :
II. The second root is equal to half the coefficient of the
second term of the reduced equation, taken with a contrary
sign, minus the square root of the second member increased
by the square of half the coefficient of the second term.
160. We will now show that the complete equation of
159. What is the operation of completing the square? How many
operations are there in the solution of every equation of the second de-
gree ? What is the first ? What the second? What the third ? Give
the rule for writing the roots without completing the square?
160. How many forms will the complete equation of the second degree
assume? On what will these forms depend? What are the signs of 2p
COMPLE1E EQUATIONS. 211
the second degree will take four forms, dependent on the
signs of 2p and q.
1st. Let us suppose 2p to be positive, and q positive; we
shall then have,
a2 + 2px = q. . . . . (l.)
2d. Let us suppose 2p to be negative, and q positive ;
we shall then have,
x7 — 2px = q (2.)
3d. Let us suppose 2p to be positive, and q negative ;
\ve shall then have,
«2 + 2px = — q. . . . ( 3.)
4th. Let us suppose 2p to be negative, and q negative ;
we shall then have,
x2 — 2px = — q. ... (4.)
As these are ah1 the combinations of signs that can take
plane between 2p and q, we conclude that every complete
equation of the second degree will be reduced to one or the
other of these four forms :
jc2 + 2px = + <7, . . 1st form.
x* — 2px = + §-, . . 2d form.
xz -f 2pa; — — y, ' . . 3d form.
a2 — 2px = — q, , . 4th form.
EXAMPLES OF THE FIRST FOEM.
1. What are the values of x in the equation,
2x> + 8x = 64 ?
If we first divide by the coefficient 2, we obtain
«2 -f- 4x = 32.
and q in the first form? What in the second? What in the third?
WViat in the fourth ?
212 ELEMENTARY ALGEBKA.
Then, completing the square,
x2 + 4x + 4 = 32 + 4 = 36.
Extracting the root,
x + 2 = ± -v/36 = + 6, and — 6.
Hence, x' = — 2 + 6 = +4;
and, x" — — 2 — 6 = — 8.
Hence, in this form, the smaller root, numerically, is positive,
and the larger negative.
VERIFICATION.
If we take the positive value, viz. : xf = + 4,
the equation, a;2 + 4& = 32,
gives 42 + 4 x 4 = 32 ;
and if we take the negative value of », viz. :«,"=.— 8,
the equation, a2 + 4x = 32,
gives (— 8)2 + 4(— 8) = 64 - 32 = 32 ;
from which we see that either of the values of x, viz.:
x' = + 4, or x" — — 8, will satisfy the equation.
2. What are the values of x in the equation,
3«2 + 12aj — 19 = — a2 — 12» + 89?
By transposing the terms, we have,
3a2 + y? + 12cc + I2x = 89 + 19 ;
and by reducing,
4z2 + 24o; = 108;
and dividing by the coefficient of jc2,
a;2 + Qx = 27.
C O M P L K T K EQUATIONS. 213
Now, by completing the square,
x2 + 6x + 9 = 36 ;
extracting the square root,
x + 3 = ±-/36 = + 6, and -• 6;
hence, xr =+6 — 3 = -f 3;
and, x" — — 6 — 3 = — 9.
VERIFICATION.
If we take the plus root, the equation,
cc2 -f 6z = 27,
gives (3)2+ 6(3) = 27;
and for the negative root,
xz + Qx = 27,
gives (- 9)2 + 6(— 9) = 81 — 54 = 27.
3. What are the values of a; in the equation,
xz - lOaj + 15 = ^ - 34a; + 155 ?
O
By cleaiing of fractions, we have,
5xz — 50x +75 = xz — I70a + 775;
by transposing and reducing, we obtain,
4xz + 12Cte = 700 ;
then, dividing by the coefficient of «2, we have,
xz+ 30x = 175;
and bj completuig the square,
y? + 30a -f- 225 = 400 ;
214 ELEMENTARY ALOEBKA.
and by extracting the square root,
x + 15 = ±-y/40() = + 20, and — 20.
Hence, xf = + 5, and x" = — 35.
VERIFICATION.
For the plus value of #, the equation,
x2 + 30x = 175,
gives, (5)2 -f 30 X 5 = 25 + 150 = 175.
And for the negative value of £C, we have,
(_ 35)2 _j_ 30(_ 35) _ 1225 — 1050 — 175.
4. What are the values of x in the equation,
s*-3»+! = '-!— *+T?'
Clearing of fractions, we have,
10a2 - 6x + 9 = 96 — Sx — I2xz + 273;
transposing and reducing,
22a2 + 2x = 360 ;
dividing both members by 22,
2 360
£C2 H X = •
r 22 22
(1 \2
— I to both members, and the equation becomes,
2 / 1 \2 360 / 1 \2
/>•* _l _,__/>• J, | __ _ I •
^22* h \22/ ' 22 hl22/'
whence, by extracting the square root,
- J_ /360 / 1 y
h 22 Z : V 22 "h \22/'
COMPLETE EQUATIONS. 215
therefore,
/ 1 \2
\22/'
and, «" = __
It remains to perform the numerical operations. In the
first place,
860/1 \2
22 "h \22/'
must be reduced to a single number, having (22)2 for its
denominator. Now,
360 /J_\2 _
"22" \22/ :
360 X 22 + 1 7921
22 \22/ (22)2 " (22)2'
extracting the square root of 7921, we find it to be 89;
therefore,
89
22 \22
Consequently, the plus value of x is,
-1 L — —
o2 + 22 ~ 22 ~~ '
and the negative value is,
1 _ 89 45
22 ~ 22 ~ ~ TI
that is, one of the two values of x which will satisfy the
proposed equation is a positive whole number, and the other
a negative fraction.
NOTE. — Let the pupil be exercised in writing the roots, in
Jie last five, and in the following examples, without com-
pleting the square.
216 ELEMENTARY ALQEBKA.
6. What are the values of x in the equation,
SaJ2 + 2x — 9 = 76 ?
x' = 5.
6. What are the values of a; in the equation,
K/v. /vt2
2x2 + 8a; + 7 = — - ^ + 197 ?
4 8
'
= 8.
7. What are the values of x in the equation,
=: 9.
' ( x"— — 64}.
8. What are the values of a; in the equation,
x2 5x x
T 4~ " 2 ~ ' *'
' = 2.
9. What are the values of a; in the equation,
9? . x xz x 13
EXAMPLES OF THE SECOND FORM.
1. What are the values of x in the equation,
a-2 — Sx -f- 10 = 19?
COMPLETE EQUATIONS. 217
By transposing,
«2 - Sx = 19 — 10 = 9;
then, by completing the square,
a;2 - 8x + 16 = 9 + 16 = 25 ;
and by extracting the root,
x — 4 = ± i/25 = + 5, or — 5.
Hence,
x' = 4 + 5 — 9, and x" = 4 — 5 =. • — 1.
That is, in this form, the larger root, numerically, u
positive, and the lesser negative.
VERIFICATION.
If we take the positive value of x, the equation,
a2 — 8x = 9, gives (9)2 — 8x9 = 81 — 72 = 9;
and if we take the negative value, the equation,
jc2 - 8a; = 9, gives (- I)2 — 8 (— 1) = 1 + 8 = 9;
from which we see that both roots alike satisfy the equa-
tion.
2. What are the values of x in the equation,
By cleaiing of fractions, we have,
6*2 + 4x — 180 = 3xz + 12* — 177 ,
and by transposing and reducing,
3x2 — 8x = 3 ;
and dividing by the coefficient of a;2, we obtain,
*•-!* = '•
218 ELEMENTARY ALGEBRA.
Then, by completing the square, we have,
^LV+21 h^---
Vit **J ^ — .1 1 — •
39 99
and by extracting the square root,
4 /25 5
Hence,
= 3 + 3 = 3) "3 3 ~ S
VEKIFICATION.
For the positive root of a;, the equation,
Q
gives 32 -- x3 = 9 — 8 = 1;
3
and for the negative root, the equation,
xz- ~x = 1,
1\2 8 1 1,8
3J - 3 X ~ 3 = 9 + 9 = l
3. What are the values of x in the equation,
Clearing of fractions, and dividing by the coefficient of
x2, we have,
*•-• = n-
COMPLETE EQUATIONS. 219
Completing the square, we have,
2 - - + - I " 1?.
3 9 ~ 9 ~~ 36 '
then, by extracting the square root, we have,
1 /49 7 7
3~~ V 36 ~ 6' 6 J
hence,
,_17 9 „ _ \ 7 5
O O O O O O
VERIFICATION.
If we take the positive root of JB, the equation,
x ^jc =. 14,
2
gives (H)2 X 1£ = 2J — 1 = Ij.;
O
and for the negative root, the equation,
52 2 5 25 10 45
4. What are the values of a; in the equation,
4a2 — 2a2 + 2craj = I8at> —
By transposing, changing the signs, and dividing tfj- 2,
the equation becomes,
a2 — ax = 2a2 — 9a& + 952 ;
220 ELEMENTARY ALGEBRA.
whence, completing the square,
- ** + 7 = -T ~
extracting the square root,
/y» — .
VJ
Now, the square root of — — Qab + 95'^, is evidently
i
35. Therefore,
s' = 2a — 3b.
fj'-- — a + 3b.
What will be the numerical values of cc, if we suppose
a = 6, and 5 = 1?
5. What are the values of a: in the equation,
4
xz — 45 —
5
within
(x' = 7.12 Ho withi
AnS'\x»= -5.73f 0.01.
6. What are the values of x in the equation,
8xz — Ux +10 = 2<e + 34?
7. What are the values of x in the equation,
c' = 8.
- 30 + x = 2x - 22 ?
4
A
Ans.
x" = — 4.
8. What are the values of x in the equation,
C O M 1' L E T E EQUATIONS.
9. What are the values of x in the equation,
2ax — x2 = — 2ab — bz?
j x' = '2a + ft.
10. "What are the values of x in the equation,
EXAMPLES OF THE THIKD FOKM.
1. What are the values of a; in the equation,
x2 + 4x — — 3 ?
First, by completing the square, we have,
£C2 + 4x + 4 = — 3 + 4 = 1;
and by extracting the square root,
as+2=±-v/I = +l> and — 1 ;
hence, x' = — 2 + 1 = — 1; and x" = — 2 — 1 = — 8.
That is, in this form both the roots are negative.
VERIFICATION.
If we take the first negative value, the equation,
x2 + 4x = — 3,
gives (— I)2 + 4(— 1) = 1— 4=— 3;
and by taking the second value, the equation,
222 E L E M K N T A R Y ALGEBRA.
gives (— 3)2 +4( — 3) -9 — 12 = — 3;
hence, both values of x satisfy the given equation.
2. What are the values of a; in the equation,
— Y — 5x — 16 = 12 + -z2 + 6z?
By transposing and reducing, we have,
— x2 — lice i= 28;
then, dividing by — 1, the coefficient of £2, we have,
xz + Hxi= — 28;
then, by completing the square,
x* + llx + 30.25 = 2.25;
hence, x + 5.5 = ±-y/2.25 = + 1.5, and — 1.5;
consequently, x' = — 4, and x" = — 7.
3. What are the values of a; in the equation,
- |2 - 2x - 5 = Ix* + 5a; + 5 ?
x' = — 2.
4. What are the values of a; in the equation,
O/J.2 I Oy, 02. _ . -Cf 9
ZK -f- SX — L% A I
..
as" = — 4.
5. What are the values of x hi the equation,
4a;2 + \x + 3a; = - 14» - 3} -
5
COMPLETE EQUATIONS. 223
6. What are the values of a: in the equation,
— y? — 4 - -x = ^ + 24a; + 2 ?
( x' = — J-
Ans. \
(x" = — 8.
7. What are the values of a; in the equation,
ic2 + to + 20 = - |e2 — lla; - 60 ?
' ( x" = - 10.
8. What are the values of x in the equation,
9. What are the values of a; in the equation,
4 T 1 3
,^/y»2 I K rt/t I ^— ^_/>»2 CI JL'T* -
»O [^ t7*O ^^ ^^ f*^ """"" 1 "0"
5 45 4
Ans.
10. What are the values of x in the equation,
x - y? — 3 — Qx + 1 ?
j x' = — 1.
AnS' \ x" = - 4!
11. What are the values of * in the equation,
x* + 4a; - 90 = — 93 ?
EXAMPLES OF THE FOURTH FOKM.
1. What are the values of x in the equation,
x2 — 8a; = — 7 ?
224: ELEMENTARY ALGEBRA.
By completing the square,' we have,
x2 - Sx + 16 = — 7 + 16 = 0;
then, by extracting the square root,
x — 4 = ±i/9 = + 3, and — 3 ;
hence, x' = + 7, and x" = + !•
That is, in this form, both the roots are positive.
VERIFICATION.
If we take the greater root, the equation,
a;2 — 8a; = — 7, gives, 72 — 8x7 = 49 — 56 : z — 7 ;
and for the lesser, the equation,
x* - Sx - - 7, gives, I2 -8x1 = 1 — 8= -7;
hence, both of the roots wih1 satisfy the equation.
2. What are the values of x in the equation,
40
— lia;2 + 3x - 10 = Ha2 - IBx + — ?
2
By clearing of fractions, we have,
— So;2 + Qx — 20 — Sx2 - 36z + 40 ;
then, by collecting the similar terms,
_ 6cc2 + 42a; = 60 ;
then, by dividing by the coefficient of cc2, which is — 0,
we have,
a:2 - Vaj = - 10.
By completing the square, we have,
x7' — 7z + 12.25 =- 2.25,
COMPLETE EQUATIONS. 225
and by extracting the square root of both members,
x — 3.5 = ±y^25 = + 1.5, and — 1.5;
hence,
x' = 3.5 + 1.5 = 5, and x" = 3.5 - 1.5 = 2.
VERIFICATION.
If we take the greater root, the equation.
xz — fce = - 10, gives, 52 - 7 X 5 = 25 — 35 •— - 10;
and if we take the lesser root, the equation,
x2 — 7x = — 10, gives, 22 — 7 X 2 = 4 —'14 =• - 10.
3. What are the values of x in the equation,
- Bx + 2a2 + 1 = 17fe - 2z2 - 3 ?
By transposing and collecting the terms, we have,
4«2 — 20|a; = — 4 ;
then dividing by the coefficient of *2, we have,
£C2 - 5}X = - I.
By completing the square, we obtain,
, 169 169 144
»-i4«,+ _== -i + -5- = — ;
and by extracting the root,
hence,
12 12 1
* = 2f -f — = 5, and, x" = 2J - -- = -.
VERIFICATION.
If we take the greater root, the equation,
a;2 — 5$x = — 1, gives, 52 — 5£ x 5 = 25 — 26 = — •
226 ELEMENTARY ALGEBRA.
and if we take the lesser root, the equation,
* -5lx=- 1, gives, Q)2- 6jxJ=^-! = -
4. What are the values of x in the equation,
fa' = 3.
5. What are the values of a in the equation,
— 4a2 — x + 1 = — 5a2 + 8a?
6. What are the values of a in the equation,
7. What are the values of a in the equation,
a2 — 10Tyc = — 1 ?
(a' =
AllS- \ vn _
I a =
8. What are ;he values of a in the equation,
= 10.
2a2
— 27a ^ — — + 100 = -— + 12a — 26 ?
5 5
f = '
Ans. i „
( x" = 6.
9. What are the values of a in the equation,
— — 22a -f 15 = — — - -f 28a — 30 ?
8 3
x' = 9.
PROPERTIES OF EQUATIONS. 227
10. What are the values of x in the equation,
2»2 -30a;-|-3 = - »2 -f- 3-frx — ^- ?
f — 11
fc — J. A»
PROPERTIES OF EQUATIONS OF THE SECOlfD DEGREE.
FIRST PROPERTY.
161. We have seen (Art. 153), that every complete
equation of the second degree may be reduced to the form,
«2 + 2px = q ..... ( 1.)
Completing the square, we have,
x2 + 2px + p2 — q -f pz ;
transposing q -f p2 to the first member,
tc2 + 2px + p* — (q + j(?2) = 0. . ( 2.)
Now, since as2 + 2px + ^>2 is the square of x + ^?, and
q + j92 the square of ^/g~+ pz, we may regard the first
member as the difference between two squares. Factoring,
(Art. 56), we have,
(x+p + V 3 + Pz) (x+p— V<1 + P*) = 0. . ( 3.)
This equation can be satisfied only in two ways :
1st. By attributing such a value to x as shall render the
first factor equal to 0 ; or,
161. To what form may every equation of the second degree be re-
duced? What form will this equation take after completing the square
and transposing to the first member ? After factoring ? In how many
wajc may Equation ( 3 ) be satisfied ? What are they ? How many roots
has every equation of the s^ond degree?
228 ELEMENTARY ALGEBRA.
2d. By attributing such a value to x as shall raider ih«
second factor equal to 0.
Placing the second factor equal to 0, we have,
x+p — -\/q+pz = 0; and x' = — p+ \/Q +P2- (4-)
Placing the first factor equal to 0, we have,
" = —p —
Since every supposition that will satisfy Equation ( 3 ),
also satisfy Equation ( 1 ), from which it was derived, it fol-
lows, that x' and x" are roots of Equation ( 1 ) ; also, that
Every equation of the second degree has two roots, and
only two.
NOTE. — The two roots denoted by x' and »", are the
same as found in Art. 158.
SECOND PROPERTY.
•
162. We have seen (Art. 161), that every equation ol
the second degree may be placed under the form,
(x + p + Vg + PZ) (x + P — V<i +P2) = °-
By examining this equation, we see that the first factor
may be obtained by subtracting the second root from the
unknown quantity x ; and the second factor by subtracting
the first root from the unknown quantity x; hence,
Every equation of the second degree may be resolved into
two binomial factors of the first degree, the first terms, in
both factors, being the unknown quantity, and the second
terms, the roots of the equation, taken with contrary signs.
162. Into how many binomial factors of the firs', degree may every
equation of the second degree be resolved ? What a: e the first terms cf
these factors ? What the second ?
FORMATION OF EQUATIONS. 229
THIRD PROPERTY.
163. It* we add Equations (4) and (5), Art. 161, we
have,
xf — — p + -\/q + p2
x" = — p — -\fq + p2
x' + x" = — 2p ; tliat is,
In every reduced equation of the second degree, the sum
of the two roots is equal to the coefficient of the second term,
taken with a contrary sign,
FOURTH PROPERTY.
164. If we multiply Equations (4) and (5), Art. 161,
member by member, we have,
x' x x = - p +
= p2- (q +p2) = - q; that is,
In every equation of the second degree, the product of
the two roots is equal to the known term in the second mem-
ber, taken with a contrary sign.
FORMATION OF EQUATIONS OF THE SECOND DEGREE.
165. By taking the converse of the second property,
(Art. 162), we can form equations which shall have given
roots ; that is, if they are known, we can find the corre-
sponding equations by the following
RULE.
I. Subtract each root from the unknown quantity :
163. Whai is the algebraic sum of the roots equal to in every equation
of the second degree ?
164. What is the product of the roots equal to?
165. How will you find the equation when the roots are known ?
230 ELEMENTARY ALGEBRA.
II. Multiply the results together, and place their product
equal to 0.
EXAMPLES.
NOTE. — Let the pupil prove, in every case, that the roots
will satisfy the third and fourth properties.
1. If the roots of an equation are 4 and — 5, what is the
equation ? Ans. x2 + x = 20.
2. What is the equation when the roots are 1 and — 3 ?
Ans. x2 + 2x = 3.
3. What is the equation when the roots are 9 and — 10 ?
. Ans. x2 -{- x = 90.
4. What is the equation whose roots are 6 and — 10?
Ans. x2 + 4x = 60.
5. What is the equation whose roots are 4 and — 3 ?
Ans. x2 — x = 12.
6. What is the equation whose roots are 10 and — TV ?
Ans. x2 — 9TVe = 1.
7. What is the equation whose roots are 8 and — 2 ?
Ans. x2 — Qx = 16. .
8. What is the equation whose roots are 16 and — 5 ?
Ans. x2 — lice = 80.
9. What is the equation whose roots are — 4 and — 5 ?
Ans. x2 -f- Qx = — 20.
10. What is the equation whose roots are — 6 and — 7 ?
Ans. x2 + 13x = — 42.
1 1 . What is the equation whose roots are — - and — 2 ?
g
Ans. x2 -f 2$x = •
12. What is the equation whose roots are — 2 and — 3 ?
Ans. x2 -{- 5x = — 6.
NUMERICAL VALUKS OF TIIK BOOTS. 231
13. What is the equation whose roots are 4 and 3 ?
Ans. xz — 7x = — 12.
14. What is tho equation whose roots are 12 and 2?
Aiis. x~ — 14a; = — 24,
15. What is the equation whose roots are 18 and 2?
Ans. x2 — 20x = — 36.
16. What is the equation whose roots are 14 and 3 ?
Ans. x2 — l*ix = — 42.
4 9
17. What is the equation whose roots are - and — -?
Ans. xz -\ x = 1,
36
2
18. What is the equation whose roots are 5 and ?
3
13 10
Ans. x- x = — •
3 3
19. What is the equation whose roots are a and 5?
Ans. y? — (a + b)x = — ab.
20. What is the equation whose roots are c and — d?
Ans. x2 — (c — d) x = cd.
TRINOMIAL EQUATIONS OF THE SECOND DEGEEE.
165.* A trinomial equation of the second degree con-
tains three kinds of terms :
1st. A term involving the unknown quantity to the second
degree.
2ct. A term involving the unknown quantity to the first
degree ; and
3d. A known term. Thus,
x2 — 4x — 12 = 0,
is a trinomial equation of the second degree.
232 ELEMENTARY ALGEBRA.
FACrORIKG.
165.** What are the factors of the trinomial equation,
y? — 4x — 12 = 0?
A trinomial equation of the second degree may always be
reduced to one of the four forms (Art. 160), by simply trans-
posing the known term to the second member, and then
solving the equation. Thus, from the above equation, we
have,
a3 — 4x = 12.
Resolving the equation, we find the two roots to be +6
and — 2 ; therefore, the factors are, x — 6, and x 4- 2
(Art. 162).
Since the sum of the two roots is equal to the coefficient
of the second term, taken with a contrary sign (Art. 163) ;
and the product of the two roots is equal to the known
term in the second member, taken with a contrary sign, or
to the third term of the trinomial, taken with the same
sign : hence it follows, that any trinomial may be factored
by inspection, when two numbers can be discovered whose
algebraic sum is equal to the coefficient of the second term,
and whose product is equal to the third term.
EXAMPLES
1. What are the factors of the trinomial, xz — 9x — 36 ?
It is seen, by inspection, that — 12 and + 3 will fulfil the
conditions of roots. For, 12 — 3 = 9 ; that is, the co-
efficient of the second term with a contrary sign ; and
12 X — 3 = — 36, the third term of the trinomial ; hence,
the factors are, x — 12, and x 4- 3.
2. What are the factors of & — tx - 30 = 0 ?
Ans. x — 10, and x + 3
TRINOMIAL KQDATIONS. 233
8. What are the factors of x~ + I5x + 36 =: 0?
Ans. x + 12, and x + 3.
4. What are the factors of x2 — I2x — 28 = 0 ?
Ans. x — 14, and x + 2.
5. What are the factors of x2 — Ix — 8 = 0 ?
Ans. x — 8, and x + 1.
TRINOMIAL EQUATIONS OF THE FORM
x~n -f 2pxn — q.
In the above equation, the exponent of &, in the first term,
is double the exponent of x in the second term.
x6 — 4x3 = 32, and x* + 4xz = 117,
are both equations of this form, and may be solved by the
rules already given for the solution of equations of the
second degree.
In the equation,
we see that the first member will become a perfect square,
by adding to it the square of half the coefficient of xn ; thus,
in which the first member is a perfect square. Then, ex-
tracting the square root of both members, we ha\ e,
xn + p = ± -\/q + p2 ;
hence, xn — — p ± -\fq + p2 ;
then, by taking the nth root of both members,
and x" = v — p — -\/—~p + pz>
234 ELEMENTARY ALGEBRA.
EXAMPLES.
1. What are the values of jc in the equation,
x6 + Qx3 = 112?
Completing the square,
x6 + Qx3 + 9 = 112 + 9 = 121 ;
then, extracting the square root of both members,
x3 + 3 = ± -/121 = ± 11 J hence,
x' = *f— 3 + 11, and x" = y— 3 — 11 ; hence,
x' = ^/8 = 2, and x" = */— 14 = --
2. What are the values of a; in the equation,
x* — 8xz = 9 ?
Completing the square, we have,
x* — 8x2 + 16 = 9 + 16 = 25.
Extracting the square root of both members,
x*— 4 = ± -y/25 = ± 5 ; hence,
x' = ± -y/4 + 5, and cc" = ± -y/4 — 5 ; hence,
= + 3 and — 3 ; and a" = + -/— 1 and — -/—
3. What afe the values of x in the equation,
SB6 + 20Z3 = 69 ?
Completing the square,
a;6 + 20a:3 + 100 = 69 + 100 = 169.
Extracting the square root of both members,
a3 + 10 = ± yl69 = ± 13 ; hence,
x' = \/- 10 + 13, and x" = 3^/— 10 - 13.
x' = 3/3, and a" = ^/- 2o.
TRINOMIAL, EQUATIONS. 235
4. Wha: are the values of x in the equation,
«* — 2xz = 3 ?
Ans. x' = ± y's, and a" = ± y'— i.
5. What are the values of x in the equation,
x* + 8a;3 — 9 ?
. x' = 1, and SB" = \~ $•
6. Given x + -/Gee + 4 = 12, to find cc.
Transposing cc to the second member, and then squaring,
Qx + 4 = cc2 — 24a; + 144 ;
.-. a;2 — 33a; = —140;
and, x' = 28, and x" = 5.
7. 4a; -f 4 + 2 = 7. ^ln«. SB' = 4J-, x" —
8. a; + y5x +10 — 8. Ans. x' = 18, ce" = 3.
NUMERICAL VALUES OF THE ROOTS.
166. We have seen (Art. 160), that by attributing all
possible signs to 2p and <?, we have the four following
forms :
a3+ Zpx = q (1.)
xz — 2px — q (2.)
xz + 2px = — q (3.)
a2 — 2px =• — q (4.)
1C6. To how many forms may every equation cf the second degree be
reduced ? What are they ?
236 ELEMENT A U^ ALGEBRA.
First Form.
167. Since q is positive, we know, from Property
Fourth, that the product of the roots must be negative ;
hence, the roots have contrary signs. Since the coefficient
22) is positive, we know, from Property Third, that the alge-
braic sum of the roots is negative ; hence, the negative root
is numerically the greater.
Second Form.
168. Since q is positive, the product of the roots must
be negative; hence, the roots have contrary signs. Since
2p is negative, the algebraic sum of the roots must be posi-
tive ; hence, the positive root is numerically the greater.
Third Form.
169. Since q is negative, the product of the roots is
positive (Property Fourth) ; hence, the roots have the same
sign. Since 2p is positive, the sum of the roots must be
negative ; hence, both are negative.
Fourth Form.
170. Since q is negative, the product of the roots is
positive ; hence, the roots have the same sign. Since 2p is
negative, the sum of the roots is positive ; hence, the roots
are both positive.
167. What sign has the product of the roots in the first form? How
are their signs? Which root is numerically the greater ? Why ?
168. What sign has the product of the roots in the second form ? How
are the signs of the roots ? Which root is numerically the greater?
169. What sign has the product of the roots in the third form ? How
are their signs ?
170. What sign has the product of the roots in the fourth form ? How
are the signs of the roots ?
NUMERICAL VALUE OF THE ROOTS. 237
First and Second Forms.
171. If we make q = 0, the first form becomes,
«2 + 2px = 0, or x(x + 2p) = 0 ;
which shows that one root is equal to 0, and the other to — 2jo.
Under the same supposition, the second form becomes,
xz — 2px = 0, or x(x — 2p) — 0 ;
which shoAvs that one root is equal to 0, and the other to
2p. Both of these results are as they should be ; since, when
q, the product of the roots, becomes 0, one of the factors
must be 0 ; and "hence, one root must be 0.
Third and Fourth Forms.
172. If, in the Third and Fourth Forms, q>pz, the
quantity under the radical sign will become negative ; hence,
its square root cannot be extracted (Art. 137). Under this
supposition, the values of x are imaginary. How are these
results to be interpreted ?
If a given number be divided into two parts, their pro-
duct will be the greatest possible, when the parts are equal.
Denote the number by 2p, and the difference of the parts
by d', then,
p + - == the greater part, (Page 120.)
2
and,
and,
p-
P2-
2
dz
4
the less part,
P, their product.
171. If we make q = 0, to what does the first form reduce? What,
then, are its roots ? Under the same supposition, to what does the second
form reduce ? What are, then, its roots?
1*72. If q > pa, in the third and fourth forms, what takes place?
If a number be divided into two parts, when will the product be the
greatest possible ?
238 ELEMENTARY ALGEBRA
It is plain, that the product P will increase, as d dimin-
ishes, and that it Avill be the greatest possible when d •= 0 ;
for then there will be no negative quantity to be subtracted
from p2, in the first member of the equation. But when
d = 0, the parts are equal ; hence, the product of the two
parts is the greatest when they are equal.
In the equations,
x2 + 2px = — <7, x2 — 2px = — q,
2p is the sum of the roots, and — q their product; and
hence, by the principle just established, the product q,
can never be greater than p7. This condition fixes a limit
to the value of q. If, then, we make q > ^>2, we pass thie
limit, and express, by the equation, a condition which cannot
be fulfilled ; and this incompatibility of the conditions is
made apparent by the values of x becoming imaginary.
Hence, we conclude that,
When the values of the unknown quantity are imaginary,
the conditions of the proposition are incompatible with
each other.
EXAMPLES.
1. Find two numbers, whose sum shall be 12 and pro
duct 46.
Let x and y be the numbers. .
By the 1st condition, x + y = 12 ;
and by the 2d, xy = 46.
The first eqiiation gives,
x = 12 — y.
Substituting this value for x in the second, we have,
12y - y2 = 46 ;
and changing the signs of the terms, we have,
» _ 12 = - 46
NUMERICAL VALUE OF THE KOOTS. 239
Then, by completing the square,
2/3 — 12y + 36 = —46 + 36 = — 10;
which gives, y' = 6 + ^/— 10,
and, y" = 6 — •/— 10 ;
both of which values are imaginary, as indeed they should
be, since the conditions are incompatible.
2. The sum of two numbers is 8, and their product 20:
what are the numbers ?
Denote the numbers by x and y.
By the first condition,
x + y = 8;
and by the second, xy = 20.
The first equation gives,
x = 8 — y.
Substituting this value of x in the second, we have,
8y - y2 = 20 ;
changing the signs, and completing the square, we have,
2/2 - 8y + 16 = - 4 ;
and by extracting the root,
y' = 4 + •>/— 4, and y" = 4 — y' — 4.
These values of y may be put under the forms (Art. 142),
y = 4 -f- 2-v/— ^j and y = 4 — 2^— 1.
3 What are the values of x in the equation,
x* 2x = — 10?
( 85' rr —1+3 <J — 1.
yl».<?. -j '.
( ic" =r: — 1 - 3V— 1.
240 ELEMENTARY ALGEBKA.
PEOBLEMS.
1. Find a number such, that twice its square, added to
three times the number, shall give 65.
Let x denote the unknown number. Then, the equation
of the problem is,
2«2 + 3x = 65 ;
whence,
3
/65 9
3 23
X
= ~4±\
'T + fi =
-[-
4 3- 4
Therefore,
, 3
X — — -
, 23 _
and «" =
3 23
13
4
~ 4 ~ T :
" ~2~
Both these values satisfy the equation of the problem.
For,
2 X (5)2 +3x5 = 2x25 + 15 = 65;
13\2 , 13 169 39 130
- ^r) + 3 X - — =— — = — - =65.
NOTES. — 1. If we restrict the enunciation of the problem
to -its arithmetical sense, in which "added" means numer-
ical increase, the first value of x only will satisfy the con-
ditions of the problem.
2. If we give to " added," its algebraical signification
(when it may mean subtraction as well as addition), the
problem may be thus stated :
To find a number such, that twice its square diminished
by three times the number, shall give 65.
The second value of x will satisfy this enunciation ; for,
/13V 13
HT) - 3 x T =
169 39
-r ~ T = 66-
P K O B L E M 8 .
241
3. The root which results from giving the plus sign to the
radical, is, generally, an answer to the question in its arith-
metical sense. The second root generally satisfies the pro-
blem under a modified statement.
Thus, in the example, it was required to find a number,
of which twice the square, added to three times the num-
ber, shall give 65. Now, in the arithmetical sense, added
means increased ; but in the algebraic sense, it implies dimi-
nution when the quantity added is negative. In this sense,
the second root satisfies the enunciation.
2. A certain person purchased a number of yards of cloth
for 240 cents. If he had purchased 3 yards less of the same
cloth for the same sum, it would have cost him 4 cents more
per yard : how many yards did he buy ?
Let x denote the number of yards purchased.
240
Then, - - will denote the price per yard.
•B
If, for 240 cents, he had purchased three yards less, that
•s, x — 3 yards, the price per yard, under this hypothesis,
240
would have been denoted by - — - • But, by the condi-
tions, this last cost must exceed the first by 4 cents. There-
fore, we have the equation,
240 240
x — 3 x '
whence, b
and,
therefore,
y reducing
r.i —
y~
itij
3x =
180,
3 -t 21
N/! + >
15, and
RCl
~ 2 ±
x' =
oU —
x" =
2
- 12.
NOTES. — 1. The value, x' = 15, satisfies the c nunciation
in its arithmetical sense. For, if 15 yards cost 240 cents,
11
242 ELEMENTARY ALGEBKA.
240 -4-15 = 16 cents, the price of 1 yard ; and 240 -f- 12 = 20
cents, the price of 1 yard under the second supposition.
2. The second value of x is an answer to the following
Problem :
A certain person purchased a number of yards of cloth
for 240 cents. If he had paid the same for three yards more,
it would have cost him 4 cents less per yard : how many
yards did he buy ?
This would give the equation of condition,
240 240
-i^'iTi = 4; or'
xz — 3x = 180;
the same equation as found before ; hence,
A single equation will often state two or more arith-
metical problems.
This arises from the fact that the language of Algebra is
more comprehensive than that of Arithmetic.
3. A man having bought a horse, sold it for $24. At the
sale he lost as much per cent, on the price of the horse, as
the horse cost him dollars : what did he pay for the horse ?
Let x denote the number of dollars that he paid for the
horse. Then, x — 24 will denote the loss he sustained. But
£C
as he lost x per cent, by the sale, he must have lost — —
upon each dollar, and upon x dollars he lost a sum denoted
xz
b\ : we have, then, the equation,
* 100
— - = x — 24; whence, a3 — lOOa z= — 2400,
100
PROBLEMS. il4:i)
and, x = 50 ± -v/2500 — 2400 = 50 ± 10.
Therefore, x' = 60, and «" = 40.
Both of these roots will satisfy the problem.
For, if the man gave $60 for the horse, and sold him for
$24, he lost $36. From the enunciation, he should have lost,
60 per cent, of $60 ; that is,
60 60 X 60
100 °f6° -loo" >6'
therefore, $60 satisfies the enunciation.
Had he paid $40 for the horse, he would have lost by the
sale, $16. From the enunciation, he should have lost 40 per
cent, of $40 ; that is,
40 40 X 40
of 40 = = 16 :
100 100
therefore, $40 satisfies the enunciation.
4. The sum of two numbers is 11, and the sum of their
squares is 61 : what are the numbers? Ans, 5 and 6.
5. The difference of two numbers is 3, and the sum of their
squares is 89 : what are the numbers ? Ans. 5 and 8.
6. A grazier bought as many sheep as cost him £60, and
after reserving fifteen out of the number, he sold the re-
mainder for £54, and gained 25. a head on those he sold :
how many did he buy ? Ans. 75.
7. A merchant bought cloth, for which he paid £33 15s.,
which he sold again at £2 8s. per piece, and gamed by the
bargain as much as one piece cost him : how many pieces
did he buy? Ans. 15.
8. The difference of two numbers is 9, and their sum,
multiplied by the greater, is equal to 266 : what are the
numbers? Ans. 14 and 5.
24-i ELEMENTARY A L G E B K A .
9. To find a number, such that if you subtract it from 10,
and multiply the remainder by the number itself, the pro-
duct will be 21. Ans. 7 or 3.
10. A person traveled 105 miles. If he had traveled 2
miles an hour slower, he would have been 6 hours longer in
completing the same distance : how many miles did he travel
per hour ? Ans. 7 miles.
11. A person purchased a number of sheep, for which he
paid $224. Had he paid for each twice as much, plus 2 dol-
lars, the number bought would have been denoted by twice
what was paid for each : how many sheep were purchased ?
Ans. 32.
12. The difference of two numbers is 7, and their sum
multiplied by the greater, is equal to 130: what are the
numbers? Ans. 10 and 3.
13. Divide 100 into two such parts, that the sum of their
squares shall be 5392. Ans. 64 and 38.
14. Two square courts are paved with stones a foot square ;
the larger court is 12 feet larger than the smaller one, and
the number of stones in both pavements is 2120 : how long
is the smaller pavement? Ans. 26 feet.
15. Two hundred and forty dollars are equally distributed
among a certain number of persons. The same sum is again
distributed amongst a number greater by 4. In the latter
case each receives 10 dollars less than in the former: how
many persons were there in each case. Ans. 8 and 12.
16. Two partners, A and _B, gained 360 dollars. A1 a
money was in trade 12 months, and he received, for prin-
cipal and profit, 520 dollars. JPs money was 600 dollars,
nnd was in trade 16 months: how much capital had A ?
Ans. 400 dollars.
MOKE THAN UNK UNKNOWN QUANTITY. 2-i;r>
EQUATIONS INVOLVING ilOKE THAN ONE UNKNOWN QUANTITY.
173. Two simultaneous equations, each of the second
degree, and containing two unknown quantities, will, when
combined, generally give rise to an equation of the fourth
degree. Hence, only particular cases of such equations can
be solved by the methods already given.
FIRST.
Two simultaneous equations, involving two unknown
quantities, can always be solved when one is of the first
and the other of the second degree.
EXAMPLES.
1. Given , , , ^ r to find x and y.
( xz + y = 100 )
By transposing y in the first equation, we have,
x = 14 - y;
and by squaring both members,
x2 = 190 — 28y + y2.
Substituting this value for x2 in the second equation, we
have,
196 — 28y + y2 + y2 = 100 ;
from Avhich we have,
By completing the square,
y2 - 14y + 49 = 1 ;
173. When may two simultaneous equations of the second degree be
solved ?
246 ELEMENTARY ALGEBRA.
and by extracting the square root,
y — 7 — ± -v/1 = + 1> and — 1 ;
hence, y' — 7 + 1 = 8, and y" — 1 — . 1 = 6.
If we -take the greater value, we find x = 6 ; and if we
fake the lesser, we find x = 8.
( v' — s »•" — fi
j 1 »'. — U» *o — • U*
AnS'\y>= 6, y"^ 8
VERIFICATION.
For the greater value, y = 8, the equation,
a; + y = 14, gives 6 + 8 — 14 ;
and, a2 + y2 = 100, gives 36 + 64 = 100.
For the value y — 6, the equation,
x + y = 14, gives 8 + 6 = 14 ;
and, a8 + y2 = 100, gives 64 + 36 = 100.
Hence, both sets of values satisfy the given equation.
2. Given j [ to find x and y.
( xz — y2 = 45 )
Transposing y in the first equation, we have,
x = 3 + y;
then, squaring both members,
a;2 = 9 + 6y + y2.
Substituting this value for a;2, in the second equation, we
have,
9 + 6y + y2 — y2 = 45 ;
whence, we have,
6y = 36, and y =• 6.
SIMULTANEOUS EQUATIONS. 247
Substituting this value of y, in the first equation, we have,
x — 6 = 3,
and, consequently, x' = 3 + 6 = 9.
VERIFICATION.
x — y = 3, gives 9 — 6 = 8 ;
and, x2— y2 = 45, gives 81 — 36 = 45.
Solve the following simultaneous equations :
+ y = 12 r «''= 7.
g
' = 5.
ix — y = 3) . ( a;' = 9, x"= — 6.
4. J 2,0 , , H J- -A7W. -I , „
( a;2+ y2 =z 117 f ( y' = 6, y"= — 9.
r />• 4. «/ — Q \ /•/>•' *j " «;
_4^-ry — y • A i x — o, — o.
5. 4 , , , L -4^5. 4 . ..
( a;- — 2xy -\- y2 = 1 $ ) y = 4, y == 4.
raj-y = 5 i
( a$2+ 2a?y + y2 = 225 [
yf = 10, x"= - 5.
y' = 5, y"= - 10.
SECOND.
174. Two simultaneous equations of the second degree,
which are homogeneous with respect to the unknown quan-
tityr, can always be solved.
EXAMPLES.
r. j a? + Sxy = 22 (1.)
' (a?+ 3a:y + 2y2 = 40 (2.)
to find x and y.
174. When may two simultaneous equations of the second degree be
solved ?
2i8 ELEMENTARY ALGEBRA.
Assume x = ty, t being any auxiliary unknown quantity.
Substituting this value of x in Equations ( 1 ) and ( 2 ),
we have,
<V + atf = 22, .-. 3,2 = fTqrisJ (3->
40
«y+ 3ty*+w = 40, .-. 7/2 = 2 ; (*0
22 40
hence,
t2 + 3t t* + 3t + 2 '
hence, 22£2 + 66J + 44 = 40f2 + 120«;
22
reducing, <2 _j_ 3^ _ — .
y
2 11
whence, «' = -, and t"=
Substituting either of these values in Equations ( 3 ) or
( 4 ), we find,
y' — +3, and y" — — 3
Substituting the plus value of y, in Equation ( 1 ), we
have,
y? + 9x = 22 ;
from which we find,
x' = + 2, and x" = — 11.
If we take the negative value, y" = — 3, we have,
from Equation ( 1 ),
xz — 9« = 22 ;
from which we find,
x' = + 11, and x" = — 2.
VERIFICATION.
For the values y' = +3, and x' = +2, the giveii
equation,
*2 4- 3ay = 22,
SIMULTANEOUS EQUATIvNS. 249
gives, 22 +3X2X3 = 4+18 = 22;
and for the second value, x" •— — 11, the same equation,
x2 + Zxy = 22,
»
gives, (— 11)2+ 3 X - 11 X 3 = 121 - 99 = 22.
If, now, we take the second value of y, that is, y" = — 3,
and the corresponding values of ic, viz., x' = +11, and
x" = — 2; for xf = +11, the given equation,
x2 + Sxy = 22,
gives, II2 + 3 x 11 x — 3 = 121 — 99 = 22;
and for x" = — 2, the same equation,
xz + 3xy = 22,
gives, (— 2)2 + 3 X - 2 X — 3 ^4 + 18 = 22.
The verifications could be made in the same way by em-
ploying Equation ( 2 ).
NOTE. — In equations similar to the above, we generally
find-but a single pair of values, corresponding to the values
hi this equation, of y' = +3, and x' — +2.
The complete solution would give four pairs of values. '
X2 - y2 = _ 9 j j_. -|» » *•
y = 5.
a; = 6.
-<9L7^. -{
( y = 7.
= 470
I .o — « — — a I (23 — 4»
2. .J „ _ t 4n». J
/* - xy = 5 f ( y = 5.
3. V? 7 ^o . i -4»w.
rz: HO
4. ^ T "•7 '
5 — xy
( 5xy — 3y2 — 32 .
1 i -L. 2 o , 5- -4*1'*'
11*
250 ELEMENTARY ALGEBRA.
THIRD. PARTICULAR CASES.
175. Many other equations of the second degree may be
so transformed, as to be brought under the rules of solution
already given. The seven following formulas will aid in
such transformation.
(1.)
When the sum and difference are known:
x + y = s
x — y — d.
Then, page 132, Example 3,
s + d 1,1, s — d 1 1 ,
x = _- = -s + -d, and y = -^- = -s - -d.
(2.)
When the sum and product are known:
x -{- y = s (l.)
xy = p (2.)
»2 + 2#y + y2 = s2, by squaring ( 1 ) ;
4xy = 4p, by mult. ( 2 ) by 4.
K2 — 2xy + yz = s2 — 4/>, by subtraction.
x — y = ± yV — 4p, by ext. root.
But, x + y — s\
hence
and,
I7i5. What is the first formula of this article? What the second?
Third? Fourth? Fifth? Sixth? Seventh?
SIMULTANEOUS EQUATIONS.
251
(3.)
When the difference and product are known:
« - y = d ....... . ( 1.)
xy = p . . % f . . . (2.)
x2 — 2xy + y2 = of2, by squaring ( 1 ) •
4xy = 4p, mult. ( 2 ) by 4.
«2 -f 2;cy + y7 = <?2 + 4p, by adding.
a; + = ±
4.)
When the sum of the squares and product are known .
a2 + 2,2= «..(!.) ay=i>..(2.) .-. 2ay = 2^> . . (3.)
Adding ( 1 ) and ( 3 ), x2 + 2«y + y2 = s + 2p ;
hence, SB + y= ± yT+~2^ (4.)
Subtracthig (3) from ( 1 ), xz — 2xy + y2 = 5 — 2p ;
hence, x — y = ± •/« — 2p (5.)
Combining (4) and (5), x = i-y/s + 2p + %-Js — 2p,
(5.)
When the sum and sum of the squares are known :
x + y = s ..... ( 1.)
cc2 + y2 = s' ..... (2.)
x2 + 2xy + y2 = s2 by squaring ( I )
2cey = sz — s'
s2- s'
*y = —7T-
= p.
(3-)
ELEMENTABY ALGEBKA.
By putting xy = p, and combining Equations ( 1 ) and
(3 ), by Formula (2), we find the values of x and y.
(6.)
"When the sum and sum of the cubes are known :
x + y = 8 .... (1.)
3.3 _|_ y3 — i52 . . . . (2.)
JB3 + 3x2y + 3xyz + y3 = 512 by cubing ( 1 ).
3a:2y + 3xy2 =360 by subtraction.
3xy(x + y) = 360 by factoring.
3ay(8) = 360 from Equa. ( 1 ).
24xy = 360
hence, xy — 15 .... (8.)
Combining ( 1 ) and ( 3 ), we find a; = 5 and y = 3
|»0
When we have an equation of the form,
(» + yY + (x + y) = q.
Let us assume x 4- y = 2.
Then the given equation becomes,
+ 2 = ?; and s = - - ± \ q + -
EXAMPLES.
2 t ~i \ \
ir I
1. Given •{ cc + y + z = 7 ( 2 )> to find a, y, and as,
' + 2,2+22 = 21 (3))
SIMULTANEOUS EQUATIONS. 253
Transposing y in Equation ( 2 ), we have,
x + z = 7-y; ... (4.)
then, squaring the members, we have,
K2 -f 2cez + z2 = 49 — 14y + y2.
If now we substitute for 2a:z, its value taken from Equa-
tion ( 1 ), we have,
a;2 + 2y2 + z2 = 49 — 14y -f y2 ;
and cancelling y2, in each member, there results,
«2 -f y2 + z2 = 49 — 14 y.
But, from Equation ( 3 ), we see that each member of the
last equation is equal to 21 ; hence,
49 — 14y = 21,
and, 14y = 49 — 21 — 28 ,
28
hence, y = — = 2.
Substituting this value of y in Equation (1 ), gives,
xz = 4 ;
and substituting it in Equation (4 ),• gives,
x + z = 5, or a; = 5 — z.
Substituting this value of «, in the previous equation, we
obtain
5z - zz =4, or zz — fiz = — 4 ;
and by completing the square, we have,
22 - 5z + 6.25 = 2.5.
and, 2 — 2.5 = ± \/2^ = + 1.5, or — 1.5;
hence, z = 2.5 -f 1.5 = 4, and 2 = -f 2.5 — 1.6 = 1
254 ELEMENTARi' ALGEBRA.
_ \
2. Given x + yxy + y = 19 )• ~ , ,
, , • , „ f to find x and y.
and x2 + ay + y2 = 133 )
Dividing the second equation by the first, AVC have,
x — i/xy + y = 7
but, a; + Va-y + y = 19
hence, by addition, 2« + 2y = 26
or, a; + y = 13
and substituting in 1st Equa., -v/ay + 13 = 19
or, by transposing, V^y — ^
and by squaring, ay = 36.
Equation 2d, is a2 + xy + y2 = 133
and from the last, we have, 3a;y =108
Subtracting, x2 — 2xy + y2 = 25
hence, x — y = db 5
but, x + y = 13
hence, x = 9 and 4 ; and y = 4 and 9.
PROBLEMS.
1. Find two numbers, such that their sum shall be 15 and
the sum of their squares 113.
Let x and y denote the numbers ; then,
x 4- y = 15, (1.) and a2 + y2 = 113. (2.)
From Equation ( 1 ), \ve have,
a;2 = 225 — 30y -f y2
Substituting this value in Equation ( 2 ),
225 -80y + y2 + y2 = 113;
PROBLEMS. 256
hence, 2y2 — 30y = — 112;
2/2 - I5y = - 56, .
hence, ' y' =. 8, and y" — 7.
The first value of y being substituted in Equation ( 1 ),
gives x' = 7 ; and the second, x" = 8. Hence, the num-
bers are 7 and 8.
2. To find two numbers, such that their product added to
their sum shall be 1 7, and their sum taken from the sum of
their squares shall leave 22.
Let x and y denote the numbers; then, from the con-
ditions,
(x + y) + xy = 17. ... (1.)
x* + y> - (x + y) = 22. . . . (2.)
Multiplying Equation ( 1 ) by 2, we have,
<2xy + 2(x + y) = 34. ... (3.)
Adding ( 2 ) and ( 3 ), we have,
a2 + Ixy + y2 + (a? + y) = 56 ;
hence, (x + y}z + (x + y) - 66. . . (4.)
Regarding (x -\- y) as a single unknown quantity (page
248),
Substituting this value in Equation ( 1 ), we have,
7 + xy = 17, and y = 5.
Hence, the numbers are 2 and 5.
3. What two numbers are those whose sum is 8, and sum
of their squares 34 ? Ans. 5 and 3.
256 ELEMENTARY ALGEBIIA.
4. It is required to find two such numbers, that the first
shall be to the second as the second is to 16, and the sum oi
whose squares shall be 225 ? Ans. 9 and 12.
5. "What two numbers are those which are to each othei
as 3 to 5, and whose squares added together make 1666 ?
Ans. 21 and 35.
6. There are two numbers whose difference is 7, and half
their product plus 30 is equal to the square of the less
number: what are the numbers? Ans. 12 and 19.
7. What two numbers are those whose sum is 5, and the
sum of their cubes 35 ? Ans. 2 and 3.
*
8. What two numbers are those whose sum is to the
greater as 11 to 7, and the difference of whose squares is
132? Ans. 14 and 8.
9. Divide the number 100 into two such parts, that the
product may be to the sum of their squares as 6 to 13.
Ans. 40 and 60.
10. Two persons, A and _B, departed from different places
at the same time, and traveled towards each other. On
meeting, it appeared that A had traveled 18 miles more
than £ ; and that A could have gone I$'s journey in 15£
days, but JB would have been 28 days in performing A's
journey : how far did each travel ? $ A, 72 miles.
Ans. j ' .
( J3, 54 miles.
11. There are two numbers whose difference is 15, and
half their product is equal to the cube of the lesser number :
what are those numbers ? Ans. 3 and 18.
12. What two numbers are those whose sum, multiplied
by the greater, is equal to 77 ; and whose difference, multi-
plied by the less, is equal to 12 ?
Am. 4 and 7, or f \/2 and y V^-
PEOBLKMS. 257
13. Divide 100 into two such parts, that the sum of their
square roots may be 14. Ans. 64 and 36.
14. It is required to divide the number 24 into two such
parts, that their product may be equal to 35 times their dif-
ference. Ans. 10 and 14.
15. The sum of two numbers is 8, and the sum of their
cubes is 152 : what are the numbers ? Ans. 3 and 5.
16. Two merchants each sold the same kind of stuff; the
second sold 3 yards more of it than the first, and together
they receive 35 dollars. The first said to the second, " I
would have received 24 dollars for your stuff;" the other
replied, "And I should have received 12^ dollars for yours :"
how many yards did each of them sell ?
1st merchant x' = 15, x" = 5.
17. A widow possessed 13,000 dollars, which she divided
into two parts, and placed them at interest in such a manner
that the incomes from them were equal. If she had put out
the first portion at the same rate as the second, she would
have drawn for this part 360 dollars interest ; and if she
had placed the second outsat the same rate as the first, she
would have drawn for it 490 dollars interest : what Avere
the two rates of interest ? Ans. 7 and 6 per cent.
18. Find three numbers, such that the difference between
the third and second shall exceed the difference between the
• second and first by 6 ; that the sum of the numbers shall be
83, and the sum of their squares 467.
Ans. 5, 9, and 19.
19. "What number is that which, being divided by the
product of its two digits, the quotient will be 3 ; and if 1 8
be added to i$ the resulting number will be expressed by
the digits inverted? Ans. 24.
258
ELEMENTARY A L G E Ji K A .
20. What two numbers are those which are to each other
as m to n, and the sum of whose squares is b ?
myb n-\fb
Ans.
21. What t\\ro numbers are those which are to each other
as m to ft, and the difference of whose squares is b ?
Ans. — — — . - ,
22. Required to find three numbers, such that the product
of the first and second shall be equal to 2 ; the product of
the first and third equal to 4, and the sum of the squares
of the second and third equal to 20. Ans. 1, 2, and 4.
23. It is required to find three numbers, whose sum shall
be 38, the sum of their squares 634, and the difference
between the sQcond and first greater by 7 than the difference
between the third and second. Ans. 3, 15, and 20.
24. Required to find three numbers, such that the product
of the first and second shall be equal to a ; the product of
the first and third equal to b ; and the sum of the squares
of the second and third equal to c.
Ans.
x =.
y =
25. What two numbers are those, whose sum, multiplied
by the greater, gives 144; and whose difference, multiplied
by the less, gives 14 ?
Ans. 9 and 7.
PROPORTIONS AND PROGRESSIONS. 259
CHAPTER IX.
OF PROPORTIONS AND PROGRESSIONS.
176. Two quantities of the same kind may be compared,
the one with the other, in two ways :
1st. By considering how much one is greater or less than
the other, which is shown by their difference ; and,
2d. By considering hoic many times one is greater or less
than the other, which is shown by their quotient.
Thus, in comparing the numbers 3 and 12 together, with
respect to their difference, we find that 12 exceeds 3, by 9;
and in comparing them together with respect to their quo-
tient, we find that 12 contains 3, four times, or that 12 is 4
tunes as great as 3.
The first of these methods of comparison is called Arith-
metical Proportion, and the second, Geometrical Propor-
tion.
Hence, Arithmetical Proportion considers the relation of
quantities with respect to their difference, and Geometrical
Proportion the relation of quantities with respect to their
quotient.
176. In how many ways may two quantities be compared the one with
the other? What does the first method consider? What the second?
What is the first of these methods called? What is the second called?
How then do you define the two proportions?
260 ELEMENTARY ALGEBRA.
OP AlimniETICAL I'KOrOKTION AXD PIIOGKESSION.
If we have four numbers, 2, 4, 8, and 10, of which
the difference betM'een the first and second is equal to
the difference between the third and fourth, these numbers
are said to be in arithmetical proportion. The first term 2
is called an antecedent, and the second term 4, with which
it is compared, a consequent. The number 8 is also called
an antecedent, and the number 10, with which it is com-
pared, a consequent.
When the difference between the first and second is equal
to the difference between the third and fourth, the four
mimbers are said to be in proportion. Thus, the numbers,
2, 4, 8, 10,
are in arithmetical proportion.
178. When the difference between the first antecedent
and consequent is the same as between any two consecutive
terms of the proportion, the proportion is called an arith-
metical progression. Hence, a progression by differences,
or an arithmetical %>rogression, is a series in which the suc-
cessive terms are continually increased or decreased by a
constant number, which is called the common difference of
the progression.
Thus, in the two series,
1, 4, 7, 10, 13, 16, 19, 22, 25, ...
60, 56, 52, 48, 44, 40, 36, 32, 28, ...
177. When are four numbers in arithmetical proportion ? What is the
first called? What is the second called? What is the third called?
What is the fourth called ?
178. What is an arithmetical progression ? What is the number called
by which the terms are increased or diminished ? What is an increasing
progression? What is a decreasing progression? Which term is only
an antecedent ? Which only a consequent ?
A K I T II M K T I C A L P It O G K K S S I O N. 26 1
the first is called an increasing progression, of which the
common difference is 3, and the second, a decreasing pro-
gression, of which the common difference is 4.
In general, let a, #, c, d, e, f, ... denote the terms of
a progression by differences ; it has been agreed to write
them thus :
a . b . c . d . e . f . g . h . i . k . . .
This series is read, a is to b, as b is to c, as c is to d, as d
is to e, &c. This is a series of continued equi-differences, in
which each term is at the same time an antecedent and a
consequent, with the exception of the first term, which is
only an antecedent, and the last, which is only a consequent.
179. Let d denote the common difference of the . pro-
gresion,
a.b.c.e.f.g.h. &c.,
which we will consider increasing.
From the definition of the progression, it evidently fol
lows that,
b = a + d, c = b + d — a -f 2d, e = c + d = a + 3d;
and, in general, any term of the series is equal to the first
term, plus as many times the common difference as there are
preceding terms.
Thus, let I be any term, and n the number which marks
the place of it ; the expression for this general term is,
I = a + (n — l)d.
Hence, for finding the last term, we have the following
179. Give the rule for finding the last term of a series when the pro-
gression is increasing.
262 ELEMENTARY ALGEBRA.
KULE.
I. Multiply the common difference by the number of
terms less one:
IT. To the product acid the first term y the sum will be
the last term.
EXAMPLES.
The formula,
I = a + (n — !)<£,
serves to find any term whatever, without determining all
those which precede it.
1. If we make n = 1, we have, I = a ; that is, the
series will have but one term.
2. If we make n = 2, we have, I = a + d ; that is,
the series will have two terms, and the second term is equal
to the first, plus the common difference.
3. If a = 3, and d — 2, what is the 3d term?
Ans. 7,
4. If a = 5, and d = 4, what is the 6th term?
Ans. 25.
5. If a — 7, and d = 5, what is the 9th term ?
Ans. 47.
6. If a = 8, and d = 5, what is the 10th term ?
Ans. 53.
7. If a = 20, and d — 4, what is the 12th term?
Ans. 64.
8. If a = 40, and d = 20, what is the 50th term ?
Ans. 1020.
0 If a = 45, and d = 30, what is the 40th term?
Ans. 1215.
ARITHMETICAL PROGRESSION. 263
10. If a = 30, and d = 20, what is the 60th term?
Ans. 1210.
11. If a — 50, and d = 10, what is the 100th term?
Ans. 1040.
12. To find the 50th term of the progression,
1 . 4 . 7 . 10 . 13 . 16 . 19 . . .
<ve have, I = 1+49x3 = 148.
13. To find the 60th term of the progression,
1 . 5 . 9 . 13 . 17 . 21 . 25 . . .
we have, ^=1 + 59x4 = 237.
18O. If the progression were a decreasing one, we
should have,
I = a — (n — l)d.
Hence, to find the last term of a decreasing progression, we
have the following
BULE.
I. Multiply the common difference by the number of terms
less one :
IT. Subtract the product from tJie first term ; ike re-
mainder will be the last term.
EXAMPLES.
1. The first term of a decreasing progression is 60, the
number of terms 20, and the common difference 3 : what is
the last term ?
l = a—(n — i)di gives 1= 60 — (20 — 1)3 = 60 -57 = 3.
180. Give the rule for finding the last term of a series, when the pro-
gression is decreasing.
264 ELEMENTARY ALGEBRA.
2. The first term is 90, the common difference 4, and the
number of terms 15 : what is the last term ? Ans. 34.
3. The first term is 100, the number of terms 40, and the
common difference 2 : what is the last term ? Ans. 22.
4. The first term is 80, the number of terms 10, and the
common difference 4 : what is the last term ? Ans. 44.
5. The first term is 600, the number of terms 100, and
the common difference 5 : what is the last term ?
Ans. 105.
6. The first term is 800, the number of terms 200, and
the common difference 2 : what is the last term ?
Ans. 402.
181. A progression by differences being given, it is pro-
posed to prove that, the sum of any two terms, taken at
equal distances from the tico extremes, is equal to the sum
of the two extremes.
That is, if we have the progression,
2 . 4 . 6 . 8 . 10 . 12,
we wish to prove generally, that,
4+10, or 6 + 8,
is equal to the sum of the two extremes, 2 and 12.
Let a.b.c.e.f... i . k . /, be the proposed
progression, and n the number of terms.
"We will first observe that, if x denotes a term which has
p terms before it, and y a term which has p terms after it,
we have, from what has been said,
181. In every progression by differences, what is the sum of the two
extremes equal to ? What is the rule for finding the sum of an arith
metical series?
ARITHMETICAL PRtGRESSION. 265
X = a + p X <?,
and, y = I — p X c?;
whence, by addition, a; -f y = a -f J,
which proves the proposition.
Referring to the previous example, if we suppose, in the
first place, x to denote the second term 4, then y will de-
note the term 10, next to the last. If x denotes the third
term 6, then y will denote 8, the third term from the last.
To apply this principle in finding the sum of the terms
of a progression, write the terms, as below, and then
again, in an inverse order, viz. :
a . b . c . d . e ./...*. k . /.
/ . Jc . i ......... c . b . a.
Calling S the sum of the terms of the first progression,
2S will be the sum of the terms of both progressions, and
we shall have,
Now, since all the parts, a + ?, b + k, c + i . . . are
equal to each other, and their number equal to ?z,
2S = (a + 1) X n, or S = (— j-J x n-
Hence, for finding the sum of an arithmetical series, we
have the following
KULE.
I. Add the two extremes together, and take half their sum :
II. Multiply this half-sum by the number of terms ; tJie
product imtt be the sum of the strip*.
12
ELEMENTARY ALGEBRA.
i: X A M P L E S .
1. The extremes are 2 and 16, and the number of terms
8 : what is the sum of the series ?
I Ct l~ ' \ • ** I J- v
;ives S — — — — X 8 = 72.
2i
2. The extremes are, 3 and 27, and the number of terms
12 : what is the sum of the series ? Ans. 180.
3. The extremes are 4 and 20, and the number of terms
10: what is the sum of the series? Ans. 120.
4. The extremes arc 100 and 200, and the number of
terms 80: what is the sum of the series? Ans. 12000.
5. The extremes are 500 and 60, and the number of terras
20 : what is the sum of the series ? Ans. 5600
6. The extremes are 800 and 1200, and the number of
terms 50 : what is the sum of the series? Ans. 50000.
1§2. In arithmetical proportion there are five members
to be considered :
1st. The first term, «.
2d. The common difference, d.
3d. The number of terms, n.
4th. The last term, /.
5th. The sum, S.
The formulas,
I = a + (n — l)d, and S = {- — J x n,
contain five quantities, o, d, ??, /, and /S>, and consequently
give rise to the following general problem, viz. : Any three
182. How many numbers are considered in arithmetical proportion?
What are they ? In every arithmetical progression, \vhat is the common
difference equal to ?
ARITHMETICAL PROGRESSION. 2(17
of these Jive quantities briny given, to determine the other
tico.
We already know the value of S in terms of a, n, and I.
From the formula,
I = a + (n — I)c7,
we find, a = I — (n — \}d.
That is : The first term of an increasing arithmetical pro-
gression is equal to the last term, minus the product of the
common difference by the number of terms less one.
From the same formula, we also find,
I — a
d =
n — 1
That is : In any arithmetical progression, the common dif-
ference is equal to the last term, minus the first term, divided
by the number of terms less one.
The last term is 10, the first term 4, and the number of
terms 5 : what is the common difference ?
The formula,
n — 1
16-4
gives, d = - - — 3.
4
2. The last term is 22, the first term 4, and the number
of terms 10 : what is the common difference? Ans. 2.
183. The last principle affords a solution to the follow-
ing question :
To find a number m of arithmetical means between two
given numbers a and b.
183. How do you find any number of arithmetical means between two
given numbers ?
268 E 1 K M K X T A K Y A L G K B R A .
To resolve this question, it is first necessary to find the
common difference. Now, we may regard -a as the first
term of an arithmetical progression, b as the last term, and
the required means as intermediate terms. The number of
terms of this progression will be expressed by m + 2.
Now, by substituting in the above formula, b for ?, and
m + 2 for n, it becomes,
, b — a b — «
m + 2 — 1 ' ~ m + 1 '
that is : The common difference of the required progression
is obtained by dividing the difference betioeen the given
numbers, a and b, by the required number of means 2>hts one.
Having obtained the common difference, d, form the second
term of the progression, or the first arithmetical mean, by
adding d to the first term a. The second mean is obtained
by augmenting the first mean by d, &c.
1. Find three arithmetical means between the extremes
2 and 18.
The formula, d = - .
m + 1
18-2
gives, £ = — — = 4 ;
hence, the progression is,
2 . 6 . 10 . 14 . 18.
2. Find twelve arithmetical means between 12 and 77.
The formula, d =
m
77 - 12
gives, d = -- - — = 5 ;
hence, the progression is,
12 . 17 . 22 . 27 . . 77.
A R I T H M K T I C A L 1' K O G R K S S I O N . 209
184. REMARK. — If the same number of arithmetical
means are inserted between all the terms, taken two and
two, these terms, and the arithmetical means united, will
form one and the same progression.
For, let a . b . c . e . f . . . be the proposed progression,
and m the number of means to be inserted between a and
£>, b and c, c and e . . . . &c.
From what has just been said, the common difference of
each partial progression Avill be expressed by
b — a c — b e — c
m + 1 ' m + 1 ' m + 1 '
expressions which are equal to each other, since «, 5, c . . .
are in progression ; therefore, the common difference is the
same in each of the partial progressions ; and, since the last
term of the first forms the first term of the second, &c., we
may conclude, that all of these partial progressions form a
single progression.
EXAMPLES.
1. Find the sum of the first fifty terms of the progression
2 . 9 . 16 . 23 ...
For the 50th term, we have,
I = 2 + 49 X 7 = 345.
50
Hence, 8 = (2 + 345) x -- = 347 X 25 = 8675.
£
2. Find the 100th term of the series 2 . 9 . 16 . 23 ....
Ans. 695.
3. Find the sum of 100 terms of the series 1.3.5.7.
9 . Ans. 10000.
270 K L ]-: M K X T A K Y ALGEBRA.
4. The greatest term is 70, the common difference 3, nnd
the number of terms 21 : what is the least term and the
sum of the series ?
Ans. Least term, 10 ; sum of series, 840.
5. The first term is 4, the common difference 8, and the
number of terms 8 : what is the last term, and the sum of
the series? Ans. j Last term, GO.
( Sum = 25G.
6. The first term is 2, the last term 20, and the number
of terms 10 : what is the common difference ? Ans. 2.
7. Insert four means between the two numbers 4 and 1 9 :
what is the series? Ans. 4 . 7 . 10 . 13 . 16 . 19.
8. The first term of a decreasing arithmetical progression
is 10, .the common difference one-third, and the number of
terms 21 : required the sum of the series. Ans. 140.
9. In a progression by differences, having given the com-
mon difference 6, the last term 185, and the sum of the
terms 2945 : find the first term, and the number of terms.
Ans. First term = 5 ; number of terms, 31.
10. Find nine arithmetical means between each antecedent
and consequent of the. progression 2. 5. 8. 11. 14...
Ans. Common diff., or d = 0.3.
11. Find the number of men contained in a triangular
battalion, the first rank containing one man, the second 2,
the third 3, and so on to the nth, which contains n. In other
words, find the expression for the sum of the natural num-
bers 1, 2, 3 . . ., from 1 to n inclusively.
Ans. S = *±-L>.
2
12. Find the sum of the n first terms of the progression
of uneven numbers, 1.3.5.7.9,... Ans. 8 = ;i2.
GEOMETRICAL PROFOKTION. 271
13. One hundred stones being placed on the ground in a
straight line, at the distance of 2 yards apart, how far will
a person travel who shall bring' them one by one to a basket,
placed at a distance of 2 yards from the first stone ?
Ans. 11 miles, 840 yards.
GEOMETRICAL PROPORTION AND PROGRESSION.
185. jRatio is the quotient arising from dividing one
quantity by another quantity of the same kind, regarded
as a standard. Thus, if the numbers 3 and 6 have the same
unit, the ratio of 3 to 6 will be expressed by
And in general, if A and It represent quantities of the same
kind, the ratio of A to J5 will be expressed by
B
A'
1§6. The character cc indicates that one quantity is
proportional to another. Thus,
A cc .B,
is read, A proportional to JB.
If there be four numbers,
2, 4, 8, 16,
having such values that the second divided by the first is
equal to the fourth divided by the third, the numbers are
185. What is ratio ? What is the ratio of 3 to G ? Of 4 to 12 ?
186. What is proportion? How do you express that four numbers
are in proportion ? What are the numbers callnd? What are the first
•ud fourth terms cilled ? What the second and third ?
272 ELEMENTAKY ALGEBRA
said to form a proportion. And in general, if there be four
quantities, A, J?, <7, and .Z), having such values that,
B D
A '"'' <?'
then, A is said to have the same ratio to B that C has to D\
or, the ratio of A to B is equal to the ratio of C to D.
When four quantities have this relation to each other, com-
pared together two and two, they are said to form a geo-
metrical proportion.
To express that the ratio of A to B is equal to the ratio
of C to _D, we write the quantities thus,
A : B :: C : D;
and read, A is to B as C to D.
The quantities which are compared, the one with the
other, are called terms of the proportion. The first and last
terms are called the two extremes, and the second and third
terms, the two means. Thus, A and D are the extremes,
and B and C the means.
187. Of four terms of a proportion, the first and third
are called the antecedents, and the second and fourth the
consequents ; and the last is said to be a fourth proportional
to the other three, taken in order. Thus, in the last pro-
portion A and C are the antecedents, and B and D the con-
sequents.
188. Three quantities are in proportion, when the first
has the same ratio to the second that the second has to the
"187. In four proportional quantities, what are the first and third called?
What the second and fourth ?
188. When are three quantities proportional? What is the middle one
called ?
GEOMETRICAL PROPORTION. 273
third ; and then the middle term is said1 to be a mean pro-
portional between the other two. For example,
3 : 6 : : 6 : 12;
and 6 is a mean proportional between 3 and 12. -
189. Four quantities are said to be in proportion by in-
version, or inversely, when the consequents are made the
antecedents, and the antecedents the consequents.
Thus, if we have the proportion,
3 : 6 : : 8 : 16,
the inverse proportion would be,
6 : 3 : : 1C : 8.
190. Quantities are said to be in proportion by alterna-
tion, or alternately, when antecedent is compared with ante-
cedent, and consequent with consequent.
Thus, if we have the proportion,
3 : 6 : : 8 : 16,
the alternate proportion would be,
3 : 8 : : 6 : 16.
191. Quantities are said to be in proportion by compo-
sition, when the sum of the antecedent and consequent is
compared either with antecedent or consequent
Thus, if we have the proportion,
2 : 4 : : 8 : 16,
189. When are quantities said to be in proportion by inversion, or in
versely ?
190. When are quantities in proportion by alternation?
191. When are quantities in proportion by composition?
12*
ELEMENTAKY ALGEBRA.
the proportion by composition would be,
2 + 4 : 2 : : 8 + 16 : 8;
and, 2 + 4 : 4 : : 8 + 16 : 16.
192. Quantities are said to be in proportion by division,
when the difference of the antecedent and consequent is
compared either with antecedent or consequent.
Thus, if we have the proportion,
3 : 9 : : 12 : 36,
the proportion by division will be,
9 — 3 : 3 : : 36 — 12 : 12;
and, 9 — 3 : 9 : : 36 — 12 : 36;
193. Equi-multiples of two or more quantities are the
products which arise from multiplying the qiiantities by the
same number.
Thus, if we have any two numbers, as 6 and 5, and mul-
tiply them both by any number, as 9, the equi-multiples will
be 54 and 45 ; for,
6 X 9 = 54, and 5 X 9 = 45.
Also, m x A, and m x JB, are equi-multiples of A and
1?, the common multiplier being m.
•
194. Two quantities A and J?, which may change their
values, are reciprocally or inversely proportional, when one
is proportionaljo unity divided by the other, and then their
product remains constant.
192. When are quantities in proportion by division ?
193. What arc equi-multiples of two or more quantities?
194. When are two quantities said to be reciprocally proportional?
GEOMETRICAL PROPORTION. 275
"We express this reciprocal or inverse relation thus,
A ocl.
in which A is said to be inversely proportional to 1$.
195. If we have the proportion,
A : B : : C : D,
T) J\
we have, f = -^, (Art. 186);
•A. C
and by clearing the equation of fractions, we have,
BG = AD.
That is : Of four proportional quantities, the product of
the two extremes is equal to the product of the two means.
Tliis general principle is apparent in the proportion be-
tween the numbers,
2 : 10 : : 12 : 60,
which gives, 2 x 60 — 10 X 12 = 120.
196. If four quantities, A, B, (7, D, are so related to
each other, that
A x D = B x C,
we shall also have,
A C
and hence, A : B : : C : D.
That is : If the product of two quantities is equal to the
product of two other quantities, two of them may be made
the extremes, and the other tico the means of a proportion.
195. If four quantities are proportional, what is the product of the two
means equal to ?
196. If the product of two quantities is equal to the product of two
^ther quantities, may the four be placed in a proportion ? How ?
276 ELEMENTARY ALGEBRA.
Thus, if we have,
2x8 = 4x4,
we also have,
2 : 4 : : 4 : 8.
197. If we have three proportional quantities,
A : B : : B : C,
B C
we have, -j = -^ ;
A J$
hence, Bz = AC.
That is: If three quantities are proportional, the square of
the middle term is equal to the product of tJie two extremes.
Thus, if we have the proportion,
3 : 6 : : 6 : 12,
we shall also have,
6 X 6 = 62 = 3 X 12 = 36.
198. If we Lave,
7? Tl
A : B : : C : D, and consequently, : T = -^ ,
.4 C
multiply both members of the last equation by -= , and
we then obtain,
CD
A " B'
and, hence, A : C : : B : D.
That is : If four quantities are proportional, they icitt be
in proportion by alternation.
197. If three quantities are proportional, what is the product of the
extremes equal to ?
198. If four quantities are proportional, will they oe in proportion by
alternation ?
GEOMETRICAL PROPORTION. 277
Let us take, as an example,
10 : 15 : : 20 : 30.
We shall have, by alternating the terms,
10 : 20 : : 15 : 30.
199. If we have,
A : B : : C : D, and A : B : : E : F,
we shall also have,
B D B F
— = -^ , and - T = -= ;
A C A E
~T\ 77T
hence, -^ = — , and C : D : : E : F.
L> xL
That is : If there are two sets of proportions having an an-
tecedent and consequent in the one, equal to an antecedent
and consequent of the other, the remaining terms will be
proportional.
If we have the two proportions,
2 : 6 : : 8 : 24, and 2 : 6 : : 10 : 30,
we shall also have,
8 : 24 : : 10 : 30.
200. If we have,
7? J\
A '. B : : C : D, and consequently, -j = -^,
we have, by dividing 1 by each member of the equation,
A C
-= = _ , and consequently, B : A : : D : C.
199. If you have two sets of proportions having an antecedent and con-
sequent in each, equal ; what will follow ?
200. If four quantities are in proportion, will they be in proportion
when taken inversely ?
278 ELEMENTARY ALGEBRA.
That is : Four proportional quantities will b$ in proportion^
when taken inversely.
To give an example iu numbers, take the proportion,
7 : 14 : : 8 : 16;
then, the inverse proportion will be,
14 . 7 : : 16 : 8,
hi which the ratio is one-half.
2O1. The proportion,
A : B : : C : D, gives, A x D = B x C.
To each member of the last equation add H x D. "We
shall then have,
(A + J?) x D = (C + D) x JB;
and by separating the factors, we obtain,
A + B : B : : C + D : D.
If, instead of adding, we subtract JB x D from both
members, we have,
(A - B) x D = (C- D) x J?;
which gives,
A - B : JB : : C - D : Z>.
That is: If four quantities are proportional, they will be
in proportion by composition or division.
Thus, if we have the proportion,
9 : 27 : : 16 : 48,
21'!. If four quantities are in proportion, will they be in proportion by
composition ? Will they be in proportion by division ? What is th«
difl'ereiice between composition and division ?
GEOMETRICAL PROPORTION. 279
tre shall have, by composition,
9 + 27 : 27 : : 16 + 48 : 48;
that .is, 36 : 27 : : 64 : 48,
in which the ratio is three-fourths.
The same proportion gives us, by division,
27 — 9 : 27 :: 48 — 16 : 48;
that is, 18 : 27 : : 32 : 48,
in which the ratio is one and one-half.
2O2. If we have,
and multiply the numerator and denominator of the first
member by any number m, we obtain,
j = -= , and mA : mJ3 : : C : D.
mA C
That is : Equal multiples of two quantities have the same
ratio as the quantities themselves.
For example, if we have the proportion,
5 : 10 : : 12 : 24,
and multiply the first antecedent and consequent by 6, we
have,
30 : 60 : : 12 : 24,
in which the ratio is still 2.
2O3. The proportions,
A : B : : C : D, and A : B : : E : F,
202. Have equal multiples of two quantities the same ratio as the
quantities ?
203 Suppose the antecedent and consequent be augmented or dimin-
ished by quantities having the same ratio ?
280 ELEMENTARY ALGEBRA.
give, A x D = JB x <7, and AxF=JSxE\
adding and subtracting these equations, we obtain,
A(D+F) = E(C±E], or A : B : : G± E : D± F.
That is : If C and D, the antecedent and consequent, be
augmented or diminished by quantities E and F, which
have the same ratio as C to D, the resulting quantities will
also have the same ratio.
Let us take, as an example, the proportion,
9 : 18 : : 20 : 40,
in which the ratio is 2.
If we augment the antecedent and consequent by the
numbers 15 and 30, which have the same ratio, we shall
have,
9 -f 15 : 18 + 30 : : 20 : 40;
that is, 24 : 48 : : 20 : 40,
in which the ratio is still 2.
If we diminish the second antecedent and consequent by
these numbers respectively, we have,
9 : 18 :: 20 — 15 : 40 — 30;
that is, 9 : 18 : : 5 : 10,
in which the ratio is till 2.
2O4. If we have several proportions,
A : B : : C : D, which gives A x D — B x (7,
A : B : : E : F, which gives A x F = B x E,
A : B : : G : H, which gives A x H = B x G,
&c., &c.,
204. In any number of proportions having the same ratio, how viD
any one antecedent be to its consequent?
GEOMETRICAL PROPORTION. 281
we shall have, by addition,
A(D + F+ II) = B(C + E + G);
and by separating the factors,
A : B : : C + E + G : D + F + H.
That is: In any number of proportions having the same
ratio, any antecedent will be to its consequent as the sum
of the antecedents to the sum of the consequents.
Let us take, for example,
2 : 4 : : 6 : 12, and 1 : 2 : : 3 : 6, &c.
Then 2:4::6 + 3:12 + 6;
that is, 2 : 4 : : 9 : 18,
in which the ratio is still 2.
2O5. If we have four proportional quantities,
A : B : : C : D, we have, -. — -^ ;
^3L 0
and raising both members to any power whose exponent is
n, or extracting any root whose index is n, we have,
j?» D»
-^ =-. -£-, and consequently,
An : Bn : : Cn : Dn.
That is : If four quantities are proportional, their like
powers or roots will be proportional.
If we have, for example,
2:4 : : 3 : 6,
we shall have, 22 : 42 : : 32 : 62 ;
205. In four proportional quantities, how arc like powers or roots?
ELEMENTARY A L G .E B H A .
that is, 4 : 16 : : 9 : 36,
in which the terms are proportional, the ratio being 4.
2O6. Let there be two sets of proportions,
7? T)
A : J5 : : C : D, which gives r = -~ ;
jfl 0
F II
E : F : : G : JET, which gives -=, = -^ •
.& Or
Multiply them together, member by member, we have,
E x F _ D x H
A x E z: G x~£'
A x E : B x F:: C x G : D x H.
Tliat is : In two sets of proportional quantities, the product*
of the corresponding terms are proportional.
Thus, if we hav^ the two proportions,
8 : 16 : : 10 : 20,
and, 3 : 4 : : 6 : 8,
we shall have, 24 : 64 : : 60 : 160.
GEOMETRICAL PROGRESSION.
2O71. We have thus far only considered the case in which
the ratio of the first term to the second is the same as that
of the third to the fourth.
20C. In two sets of proportions, how arc the products of the correspond-
ing terms ?
207. What is a geometrical progression? What is the ratio of the
progression ? If any term of a progression be multiplied by the ratio,
what will the product be ? If any term bo divided by the ratio, w hat
G E U M K T K I C A L I' K O G B K 6 S I O N. 283
If we 1m e the farther condition, that the ratio of the
second term to the third shall also be the same as that of
the first to the second, or of the third to the fourth, we shall
have a series of numbers, each one of which, divided by
the preceding one, will give the same ratio. Hence, if any
term be multiplied by this quotient, the product will be the
succeeding term. A series of numbers so formed, is called
a geometrical progression. Hence,
A Geometrical Progression, or progression, by quotients,
is a series of terms, each of which is equal to the preceding
term multiplied by a constant number, which number is
called the ratio of the progression. Thus,
1 : 3 : 9 : 27 : 81 : 243, &c.,
is a geometrical progression, in which the ratio is 3. It is
written by merely placing two dots between the terms.
Also, 64 : 32 : 16 : 8 : 4 : 2 : 1,
is u geometrical progression in which the ratio is one-half.
In the first progression each term is contained three times
in the one that follows, and hence the ratio is 3. In the
second, each term is contained one-half times in the one
which follows, and hence the ratio is one-half.
The first is called an increasing progression, and the
second a decreasing progression.
Let a, b, c, d, e, f, ... be numbers, in a progression by
quotients ; they are written thus :
a : b : c : d : e : f : g . . .
and it is enunciated in the same manner as a progression by
differences. It is necessary, however, to make the distinc-
will the quotient be? How is a p-ogression by quotients written? Which
of the terms is only nu antecedent? Which only a consequent? Hovt
laav each of the others be considered?
284 ELEMENTARY ALGEBRA..
tion, that one is a series formed by equal differences, and
the other a series formed by equal quotients or ratios. It
should be remarked that each term is at the same time an
antecedent and a consequent, except the first, which is only
an antecedent, and the last, which is only a consequent.
2O8. Let r denote the ratio of the progression,
a : b : c : d . . .
r being > 1 when the progression is increasing, and r< 1
when it is decreasing. Then, since,
b _ c d e .
a b c d ~
we have,
b = ar, c = br — ar2, d = cr = ar3, e = dr = ar*,
f = er = ar5 . . .
that is, the second term is equal to ar, the third to ar2, the
fourth to ar3, the fifth to ar4, &c. ; and in general, the nth
term, that is, one which has n — I terms before it, is ex-
pressed by arn ~ 1.
Let I be this term • we then have the formula,
I = arn~\
by means of which we can obtain any term without being
obliged to find all the terms which precede it. Hence, to
find the last term of a progression, we have the following
E u L E .
I. Raise the ratio to a power whose exponent is one less
than the number of terms.
H. Multiply the power thus found by the first term : the
product will be the required term.
208. By what letter do we denote the ratio of a progression? In an
increasing progression is r greater or less than 1 ? In a decreasing pro-
GEOMETRICAL PROGRESSION. 285
EXAMPLES.
1. Find the 5th term of the progression,
2 : 4 : 8 : 16 . . .
in which the first term is 2, and the common ratio 2.
5th term = 2 X 2* = 2 X 16 = 32. Ans.
2. Find the 8th tetm of the progression,
2 : 6 : 18 : 54 ...
8th term = 2 X 37 = 2 X 2187 = 4374. Ans.
3. Find the 6th term of the progression,
2 : 8 : 32 : 128 ...
6th term = 2 X 45 = 2 X 1024 = 2048. Ant
4. Find the 7th term of the progression,
3 : 9 : 27 : 81 . . .
7th term = 3 x 36 = 3 x 729 = 2187. Ans.
5. Find the 6th term of the progression,
4 : 12 : 36 : 108 ...
6th term = 4 X 3s — 4 x 243 = 972. Ans.
6. A person agreed to pay his servant 1 cent for the first
day, two for the second, and four for the third, doubling
every day for ten days : how much did he receive on the
tenth day? Ans. $5.12.
grcssiou is r greater or less than 1 ? If a is the first term and r the
ratio, what is the second term equal to ? What the third ? What the
fourth ? What is the la?t term equal to ? Give the rule for finding the
last terra.
286 ELK MJ:> ;TA i: Y ALGEBRA.
7. What is the 8th term of the progression,
9 : 36 : 144 : 576 . . .
8th term = 9 X 47 = 9 x 16384 - 147456. Ans.
8. Find the 12th term of the progression,
64 : 16 : 4 : 1 : 7 .
4
/l\u 43 1 1
12th term — 64-1 = — = - = — — - • Ans.
\4/ 411 48 60086
2O9. We will now proceed to determine the sum of n
terms of a progression,
a : b : c : d : e : f : . . . : i : Jc : /;
I denoting the nih term.
We have the equations (Art. 208),
b — ar, c = br, d = cr, e — dr, . . . k = ir, I = AT,
and by adding them all together, member to member, we
deduce,
Sum of 1st members. Sum of M members.
b+c+d+e+ . . . +&+l=(a + b + c
in which we see that the first member contains all the terms
but «, and the polynomial, within the parenthesis in the
second member, contains all the terms but I. Hence, if we
call the sum of the terms S, we have,
S - a = (S - l)r = Sr - Ir, .-. Sr - S = Ir - a;
„ Ir — a
whence, o = -- •
r — 1
209. Give the rule for finding the sum of the series. What is the first
sttp? What the second? What the third ?
G E O M K T li I C A L PROGRESSION. 287
Therefore, to obtain the sum of all the terms, or sum of the
series of a geometrical progression, we have the
KULE.
I. Multiply the last term by the ratio :
II. Subtract the first term from the product :
III. Divide the remainder by the ratio diminished by 1
and the quotient will be the sum of the series.
1. Find the sum of eight terms of the progression,
2 : 6 : 18 : 54 : 1C2 . . . 2 x 37 = 4374.
= lr_-a = 13IM-. _ 656fl_
' r — 1 2
2. Find the sum of the progression,
2 : 4 : 8 : 16 : 32.
8= lr~Cl - 64~2 -- 62
' r - 1 1
3. Find the sum of ten terms of the progression,
2 : 6 : 18 : 54 : 16^ . . . 2 X 39 = 39366.
Ans. 59048.
4. What debt may be discharged in a year, or twelve
months, by paying $1 the first month, $2 the second month,
$4 the third month, and so on, 'each succeeding payment
being double the last ; and what will be the last payment ?
i Debt, . $4095.
Ans. *\ -.
( Last payment, $2048.
5. A daughter was married on New-Year's day. Her
father gave her Is., writh an agreement to double it on the
first of the next month, and at the beginning of each succeed-
ing month to double what she had previously received. How
much did she receive ? Ans. £204 15*.
288 ELEMENTARY ALGEBRA.
6. A man bought ten bushels of wheat, on the condition
that he should pay 1 cent for the first bushel, 3 for the second,
9 for the third, and so on to the last : what did he pay for
the last bushel, and for the ten bushels ?
j Last bushel, $196 83.
HS' ( Total cost, $295 24.
7. A man plants 4 bushels of barley, which, at the first
harvest, produced 32 bushels ; these he also plants, which,
in like manner, produce 8 fold ; he again plants all his crop,
and again gets 8 fold, and so on for 16 years : what is his
last crop, and what the sum of the series ?
. s j Last, 140737488355328 bush.
m' ( Sum, 160842843834660.
21O. When the progression is decreasing, we have,
r < 1, and ?< a ; the above formula,
_ lr-g
-
for the sum, is then written under the form,
a — Ir
o — ^ - 1
1 — r
in order that the two terms of the fraction may be positive.
1. Find the sum of the terms of the progression,
32 : 16 : 8 : 4 : 2
32 — 2 X
0 a — Ir 2 31
S = - - = -- - -- zr — — 62.
1 — r 1 1
i 210. What is the formula for the sum of the series of a decreasing
' progression ?
GEOMETRICAL PROGRESSION. 289
2. Find the sum of the first twelve terms of the pro-
gression,
64 : 16 : 4 : 1 :
4
84/7 ), or — l—
\4/ 65536
X- 256--
S — a"~lr _ 65536 4 _ 65536 _ 65535
= 1 - r ~ '~3~~ ~~ *" 196008
4
211. REMARK. — We perceive that the principal difficulty
consists in obtaining the numerical value of the last term, a
tedious operation, even when the number of terms is not
very great.
3. Find the sum of six terms of the progression,
512 : 128 : 32 ...
AM. 682f
4. Find the sum of seven terms of the progression,
2187 : 729 : 243 ...
Ans. 3279.
5. Find the sum of six tenns of the progression,
972 : 324 : 108 ...
Ans. 1456.
6. Find the sum of eight terms of the progression,
147456 : 36864 : 9216 . . .
Ans. 196605.
OF PROGRESSIONS HAVING AN INFINITE NUMBER OF TERMS.
212. Let there be the decreasing progression,
a : b : o : d : e : f : . . .
212. When the progression is decreasing, and the number of termg In-
finite, what is the expression for the value of the sum of the series ?
18
290 ELEMENTARY A L O E B K A .
containing an indefinite number of terms. In the formula,
„ a - Ir
S = T^T'
substitute for I its value, arn~l, (Art. 208), and we have,
a — arn
S =
1 — r
which expresses the sum of n terms of the progression.
This may be put under the form,
„ a ar*
& == i~
1 — r I — r
Now, since the progression is decreasing, r is a proper
fraction ; and r* is also a fraction, which diminishes as n
increases. Therefore, the greater the number of terms we
take, the more will - - X rn diminish, and consequently,
-L ^™* /
the more will the entire sum of all the terms approximate
to an equality with the first part of S, that is, to - —
Finally, when n is taken greater than any given number,
or n — infinity, then - - x r* will be less than any
L — ~ T
given number, or will become equal to 0 ; and the expres-
t will then represent the true value of the sum
1 ™"~ 7*
of all the terms of the series. Whence we may conclude,
that the expression for the sum of the terms of a decreasing
progression, in which the number of terms is infinite, is,
that is, equal to the first term, divided by 1 minus the ratio.
GEOMETRICAL P KOU KE8SI ON . L".»l
This is, properly speaking, the limit to which the partial
Bums approach, as we take a greater number of terms in the
progression. The difference between these sums and - — - ,
may be made as small as we please, but will only become
nothing when the number of terms is infinite.
EXAMPLES.
1. Find the sum of
"We have, for the expression of the sum of the terms,
* = r^ = — i=l = * *&
~3
The error committed by taking this expression for the
value of the sum of the n first terms, is expressed by
X r*
= -(-V-
2\3/
1 — r
First take n = 5 ; it becomes,
3/l\5_ 1 JL_
2W := 2 . 3* " 162*
When n = 6, we find,
3/l\6_ _1_ 1 1
2\3/ : = 162 X 3 == 486'
q
Hence, we see, that the error committed by taking - for
m
the sum of a certain number of terms, is less in proportion
as this number is greater.
29*3 K L K M K X 'C A K Y A I. Ci E FJ K A .
2. Again, take the progression,
• - - l- •-• — •—'&
: 2 : 4 : 8 : 16 : 32 :
We have, S = --^— = -— : = 2.
I — r \
Ans.
_
~~ 2
3. What is the sum of the progression,
111 1
1, — , - , - , - , &c., to infinity.
10' 100' 1000' 10000'
K - a l - i1 Av<
8 - r=~r - ~r V Ans-
10
a 13. In the several questions of geometrical progres-
sion, there are five numbers to be considered :
1st. The first term, . . a.
2d. The ratio, . . . . r.
3d. The number of terms, n.
4th. The last term, . . /.
5th. The sum of the terms, S.
214. We shall terminate this subject by solving this
problem :
To find a mean proportional between any two numbers,
as m and n.
Denote the required mean by x. We shall then have
(Art. 197),
a;2 = m x n ;
and hence, x = \/m x n.
213. How many numbers are considered in a gcometrlcnJ progression?
What are they?
214. How do you find a mean proportional between two numbers ?
GEOMETRICAL PROGRESSION. 293
That is : Multiply the two numbers together, and extract the
square root of the product.
1. What is the geometrical mean between the numbers
2 and 8 ?
Mean = 18 x 2 = /lQ — 4. Ans.
2. What is the mean between 4 and 16 ? Ans. 8.
3. What is the mean between 3 and 27 ? Ans. 9.
4. What is the mean between 2 and 72 ? -4ws. 12.
5. What is the mean between 4 and 64 ? ^1«*. 16.
therefore, $40 satisfies the enunciation.
294: ELEMENTARY ALGEBRA.
CHAPTER X.
OF LOGARITHMS.
215. THE nature and properties of the logarithms in
common use, will be readily understood by considering
attentively the different powers of the number 10. They
are,
10° = 1
101 1= 10
102 = 100
103 = 1000
104 = 10000
10s = 10000*
&c., &c.
It is plain that the exponents 0, 1, 2, 3, 4, 5, <fcc., form an
arithmetical series of which the common difference is 1 ; and
that the numbers 1, 10, 100, 1000, 10000, 100000, <fcc., form
a geometrical progression of which the common ratio is 1 0.
The number 10 is called the base of the system of logarithms ;
and the exponents 0, 1, 2, 3, 4, 5, &c., are the logarithms of
215. What relation exists between the exponents 1, 2, 3, &c. ? How
are the corresponding numbers 10, 100, 1000? What is the common
difference of the exponents ? What is the common ratio of the corre-
sponding numbers ? What is the base of the common system of loga-
rithms ? What are the exponents ? Of what number is the exponent J
the logarithm ? The exponent 2 ? The exponent 3 ?
OF LOGARITHMS.
the numbers which are produced by raisiug 10 to the powers
denoted by those exponents.
216. If we denote the logarithm of any number by ?n,
then the number itself will be the mth power of 10 ; that is,
if we represent the corresponding number by M,
10m = M.
Thus, if we make m = 0, 3/will be equal to 1 ; if m = 1,
M will be equal to 10, <fcc. Hence,
The logarithm of a number is the exponent of tJie power
to which it is necessary to raise the base of the system in
order to produce the number.
217. If, as before, 10 denotes the base of the system
of logarithms, m any exponent, and M the corresponding
number, we shall then have,
10"1 = Jf, (1.)
in which m is the logarithm of M.
If we take a second exponent n, and let N denote the
corresponding number, AVC shall have,
10" = Ar, (2.)
in which n is the logarithm of N~.
If, now, we multiply the first of these equations by the
second, member by member, we have,
10"1 x 10" = 10Bl+" = M X N;
but since 10 is the base of the system, m + n is the loga-
rithm M x .2V; hence,
216. If we denote the base of a system by 10, and the exponent by
m, what will represent the corresponding number? What is the logarithm
of a number ?
217. To what is the sum of the logarithms of any two numbers equal ?
To what, then, will the addition of logarithms correspond ?
296 ELEMENT AKT ALGEBRA.
The sum of the logarithms of any tico numbers is equal
to the logarithm of their product.
Therefore, the addition of logarithms correspond* to the
multiplication of their numbers.
218. If we divide Equation ( 1 ) by Equation ( 2 ), mem-
ber by member, we have,
JO1
10
but since 10 is the base of the system, m — n is the loga-
rithm of -=^.; hence,
If one number be divided by another, the logarithm of
the quotient will be equal to the logarithm of the dividend,
diminished by that of the divisor.
Therefore, the subtraction of logarithms oorretponA to
the division of their numbers.
219. Let us examine further the equations,
10° = 1
101 = 10
102 = 100
10s = 1000
&c., &c.
It is plain that the logarithm of 1 is 0, and that the loga-
rithm of any number between 1 and 10, is greater than
218. If one number be divided by another, what will the logarithm
of the quotient be equal to ? To what, then, will the subtraction of loga-
rithms correspond?
•J19. What is the logarithm of 1 ? Between what limits are the loga-
rithms of all numbers between 1 and 10? Row are they generally ex-
pressed ?
OF LOGARITHMS. 297
0 and less than 1. The logarithm is generally expressed by
decimal fractions ; thus,
log 2 = 0.301030.
The logarithm of any number greater than 10 and less
than 100, is greater than 1 and less than 2, and is expressed
by 1 and a decimal fraction ; thus,
log 50 = 1.698970.
The part of the logarithm which stands at the left of the
decimal point, is called the characteristic of the logarithm.
The characteristic is always one less than the number of
places of figures in the number whose logarithm is taken.
Thus, in the first case, for numbers between 1 and 10,
there is but one place of figures, and the characteristic is 0.
For numbers between 10 and 100, there are two places of
figures, and the characteristic is 1 ; and similarly for other
numbers.
TABLE OF LOGARITHMS.
22O. A table of logarithms is a table in which are writ-
ten the logarithms of all numbers between 1 and some other
given number. A table showing the logarithms of the
numbers between 1 and 100 is annexed. The numbers are
written in the column designated by the letter N, and the
logarithms in the column designated by Log.
How is it with the logarithms of numbers between 10 and 100? What
is that part of the logarithm called which stands at the left of the char-
acteristic? What is the value of the characteristic?
22<>. What is a table of logarithms? Explain the manner of finding
the logarithms of numbers between 1 and 100?
13*
298
ELEMENTARY ALGEBRA.
TABLE.
N.
Log.
N.
Log.
N.
Log.
N.
Log.
,1
2
3
4
5
0.000000
0.301030
0.477121
0.602060
0.698970
26
27
28
29
30
1.414973
1.431364
1.447158
1.462398
1.477121
51
52
53
54
55
1.707570
1.716003
1.724276
1.732394
1.740363
7(5
77
78
79
80
1.880814
1.886491
1.892095
1.897627
1.903090
6
7
8
1
10
0.778151
0.845098
0.903090
0.954243
1.000000
31
32
33
34
35
1.491362
1.505150
1.518514
1.531479
1.544068
56
57
58
59
60
1.748188
1.755875
1.763428
1.770852
1.778151
81
82
83
84
85
1.908485
1.913814
1.919078
1.924279
1.929419
11
12
13
14
15
1.041393
1.079181
1.113943
1.146128
1.176091
36
37
38
39
40
1.556303
1.568202
1.579784
1.591065
1.602060
61
62
63
64
65
1.785330
1.792392
1.799341
1.806180
1.812913
86
87
88
89
90
1.934498
1.939519
1.944483
1.949390
1 . 954243
16
17
18
19
20
1.204120
1.230449
1.255273
1.278754
1.301030
41
42
43
44
45
1.621784
1.623249
1.633468
1.643453
1.653213
66
67
68
C9
70
1.819544
1.826075
1 . 832509
1.838849
1 . 845098
91
92
93
94
95
1.959041
1.963788
1.968483
1.973128
1.977724
21
22
23
24
25
1.322219
1 . 342423
1.361728
1.380211
1.3979-10
46
47
48
49
' 50
1.662758
1.672098
1.681241
1.690196
1.698970
71
72
73
74
75
1.851258
1 . 857333
1.863323
1.869232
1.875061
96
97
98
99
100
1.982271
1.986772
1.991226
1.995635
2.000000
EXAMPLES.
1 . Let it be required to multiply 8 by 9, by means of
logarithms. We have seen, Art. 216, that the sum of the
logarithms is equal to the logarithm of the product. There-
fore, find the logarithm of 8 from the table, which is 0.903090,
ami then the logarithm of 9, which is 0.954243 ; and their
sum, which is 1.857333, will be the logarithm of the product.
In searching along in the table, we find that 72 stands oppo-
site this logarithm ; hence, 72 is the product of 8 by 9.
OF LOGABITI1MS.
299
2. What is the product of V by 12?
Logarithm of 7 is,
Logarithm of 12 is, . . .
Logarithm of their product,
and the corresponding number is 84.
3. What is the product of 9 by 11 ?
Logarithm of 9 is,
Logarithm of 11 is,
Logarithm of their product,
and the corresponding number is 99.
0.845098
1.079181
1.924279
0.954243
1.041393
1.995636
4. Let it be required to divide 84 by 3. We have seen
in Art. 218, that the subtraction of Logarithms corresponds
to the division of their numbers. Hence, if we find the
logarithm of 84, and then subtract from it the logarithm of
3, the remainder will be the logarithm of the quotient.
The logarithm of 84 is,
The logarithm of 3 is,
Their difference is, ...
and the corresponding number is 28.
5. What is the product of 6 by 7?
Logarithm of 6 is,
Logarithm of 7 is,
Their sum is, 1.623249
and the corresponding number of the table, 42.
1.924279
0.477121
1.447158
0.778151
0.845098
RECOMMENDATIONS OF DAVIES' MATHEMATICS.
DAVIKS' COUBKE OF MATHEMATICS are the prominent Texl-Jiooks in ntott
of the Golltgts of the United Sltite.i, and also^in the various Scho< is in. I
Acitdttinics throughout the Union.
YORK, PA., Auy.'2\ !«•>.
Davits' Seriet of Mathematict 1 deem the very best I ever saw. From a nunihei
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Hie Teachers attending the sessions of the Turk Co. Normal School— believing It also
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Mlv believe that in » very short time the Teachers of our country en ma»*e will i*
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1 .-itijtlysls of numbers. A. E. BLAIR,
Principal of York Co. Normal Scb»m
JACKSON UNION SCHOOL. MICIIIGAX. Xei>t. 2.\ K,-.
Mrc*i:8 A. S. BARSRS & Co. :— I take pleasure in adding my testimony in favor ol
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over any other Series, that we 'neither contemplate making, iior de-ire to make, HHJ
change 'in that direction. Yours truly, E. L. IJIl'LKY.
NKW UKITAIK. June I'M. 1V.S.
MI-.SHKS. A. S. P,AI:XI:S & Co . :— I h:ive eNamincd I><irl- x' Sf, -<V« <if .t>-if!i.- , '"
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DAVID N. C.\Ml\f'riii(.-ij><il<>fConn. State Normal Scho»'.
I have long regarded Davits' Series of Mathematical Tert-Books as far superioi
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Union Schools: and his success is complete, and undeniable. 1 know of no Arith-
metics which exhibit so clearly the philosophy of numbers, and at the same time lead
the pupil surely on to readiness and practice. A. 8. WELCH.
From PBOF. G. W. PLVMPTON, late of the Sbite Normal School, X. Y.
Out of a great number of Arithmetics that I have examined during the past year, 1
find none that will compare with Datie/f Intellectual and Dar.iej Analytical and
Practical Arithmetics, in clearness of demonstration or philosophical arrangement.
I shall with pleasure recommend the use of these two excellent works to those who
go from our institution to teach.
From C. MAY, JK., School Commissioner, Keene, 2V. H.
I bare carefully examined Da-tie*' Seriei of Arithmetics, and Higher Mat'if-
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I am acquainted.
ffiim Joim L CAMPBBLL, Professor uf Jfitt?ifmatic*, Natural PhUovophy, an/I
Astronomy, in Wabash College, Iwlidna.
W ABASH COLI.KOK. Jntif 22. !-.>
M«SRS. A. S. BAKHES & Co. : — GF.NTLKMKN : Every text-book on Science pvo|" r'y
•< IINJSIS of two jiarts — the philosophical and the illuxtratitt. A proper cointiitiiiior.
.• -i pstract reasoning and practical illustration is the chief excellence in Prof. l)»vi»-
Malhcmatical Works. I prefer his Arithmetics, Algebras. Geometry, and Trisronoii:
•try. to all others now in use. and cordially recommend them to all who d<-Miv i> ».
vhancemcnt of sound learning, Yours, very truly, JOHN L. CAMPBELL
PK.IKKSSORS MAH.SN, BAHTLETT, and CtirROii. of the United States Military Aca.l<-n.»
West 1'oint. HfvtDnvfa? L'niei-ryiti/ Arithmetic: —
" In the distinctness wi^h whieh the various definitions are given, the clear «nd
strictly mathematical demonstration of the rules, the convenient form and well-chosen
matter of the tables, as well as in the complete and much-desired application of all to
the business of the country, th*> Uv>rf»ity Arithmetic of Prof. Davits is st perior to
*!!}• other work of the kind wi;h •« h'.-h ' c »rp sc'jnainti-.i "
RECOMMENDATIONS
OF
PARKER & WATSON'S READERS
from PROF. FREDERICK S. JKWKLI., of the _\><c York. State Jiormal School
It gives me pleasure to find in the National Series at School Readers ample i. )i>
&>r commendation. From n brief examination of them. I urn led to believe tin.' <ri
have none equal to them. I hope they will prove as popular as they are excellent
f>-nm IIox. TiiKonoRK FRKUXGIIUYSKN, President of Uutgeri? College, 2f J.
.\ cursory examination leads me to the conclusion that the system contained ID
>.li«>.»e volumes deserves the patronage of our schools, and I have no doubt that il w!t
W-come extensively used in the education of children and youth.
fnan N. A. HAMILTON, President of Teacher*' Union, Whitewater, n'i*.
Tin- National Readers and Speller I have examined, and carefully compared witb
others, and must pronounce them decidedly superior, in respect to literary merit,
style. :in<l pri. e. The gradation is more complete, and the series much more Owtr»bl«
for use in our schools than Sanders' or McGuffey's.
f'nnn PROF. T. F. THICKS-IT s, Principal of Academy and Normal Scliool,
Meadrille, Pa.
1 am much p'eased with the National Series of Readers after having canvassed
tlieir merits pretty thoroughly. The first of the series especially pleases me, because
it affords the means of teaching the •• word-method" in an appropriate and natural
manner. They a'l arc progressive, tbe rules of elocution are stated with clearness,
and the selection of pieces is such as to pk-itJc at the s:iine time that they in*triirt
Fiom J. \V. SciiF.iiMERiioKN, A. 15., Principal Coll. Institute, Stiddletown, S. J
I consider tliem emphatically the Readers of the present day, and I believe thu
,heir ii.rrinsic merits will insure for them a full measure of popularity.
From PETER Rouoicr, Principal Public School Xo. 10, Brooklyn.
It gives me great pleasure to be able to bear my unqualified testimony to the >-xcel
lence of the National Series of Readers, by PAKKF.K and WATSON. The gradation of
the books of the series is very fine : we have reading in its elements and in its highest
style. The fine taste displayed in the selections ami in the collocation of the piece?
it'serves much praise. A distinguishing feature of the series is the variety of thf
subject-matter and of the style. The practical teacher knows the value of this charm:
terisik for the development of the voice. The authors seein to have kept constantly
in view tlie fact that a reading-book is designed for children, and therefore they have
surceeded in forming a very interesting and improving collection of reading-matter,
highly adapted to the wants and purposes of the school-room. In short, I look upon
the National Scries of ReaUers as a great success.
from \. P. HARRINGTON, Principal of Union School, MaraUion, N. }'.
These Renders, in my opinion, aie the best I have ever examined. The rhetorical
exercises, in particular, are superior Ui any thing of the kind I have ever seen. I have
iiad tn-tter success with my reading classes since I commenced training them i.n those
ilnii 1 ever met with before. The marked vowels in the reading exori-ises coiury 10
'.lie reaiicr's iniiid ut once the astonishing f«ct that he has been accustomed to im-pro-
:.»nnve more than one-third of the words of the English language.
f"rrm OriAr.i.KS S. HALSF.V. Prinrijml C<>Ufyi<itt Inntitute, Nnrtun. A". J.
In ih< simplicity and clearness with which the principles are stated, in the appro
•r;ati-n<*s of the selections for reading, and i:i the happy adaptation of the dilf-ient
-.•rt.« V, ihe series to oach .-ther. these works are s-nperior to any other text books on
i , si tiji-ct which I have examined.
t'ntin WILLIAM TRAVIS, Principal of Union School, flint, J/icA.
i b .v exan'ined the Nc.tiona! Series of Readers, and am delighted to find 'l so fat
. •, ani'H of most other serii-s now in use, and so wtll adapted to the wains ..f ik»
i ) .: Srhoo;». It is in, equaled in the skillful arrangement of the m.-iti-rihl r.>< it
V-ij ii'u! typiwriipby. and the general neat and inviting appearance of it* several
!H>. I.* I predict tor it a cordial Ve'.coine and a general introduction by many of 0111
nv>st *t,tei(.] isinj ton hers.
RECOMMENDATIONS
CLARK'S ENGLISH GRAMMAR.
We cannot better set forth the merits of this work than by quoting a part of a com-
munication from Prot F. 8. JEWKLL, of tlie New York State Normal School, in wblct
school this Grammar is now used as the text book on this subject : —
•; CT.ARK'S SYSTKM or GRAMMAR is worthy of the marked attention of the friends 01
jilucntion. Its points of excellence are of the most decided character, and will nc*.
*>'>n be surpassed. Among them are —
1st. '-The justness of its around principle of classification. There is no simple, phi!-
isophieal. and practical classification of the elements of language, other than thnt bui't
;, iheir use or i;fltce. Our tendencias hitherto to follow tho analogies of the classical
.«!!gu.-is:«-s. and classify extensively according to forms, have been mischievous and ab-
minl. It is time we corrected them.
2d. '• Its thorough and yt simple and transparent analysis of the elements of the
language according to its ground principle. Without such an analysis, no broad and
comprehensive view of the structure and power of the language can be attained. The
al.setx-e of this analysis has hitherto precipitated the study of Grammar upon a surface
of dry details and bare authorities, and useless technicalities.
3d." " Its happy method of illustrating the relations of elements by diagrams. These,
however uncouth they may appear to the novice, sre really simple and philosophical.
Of their utility there can be no question. It is supported by the usage of other sci-
ences, and has bc«-n demonstrated by experience in this.
4th "The tenden'-y of the system, when rightly taught and faithfully carried out,
to cultivate habits of nice discrimination and close reasoning, together with skill in
illustrating truth. In this it is not excelled by any, unless it be the mathematical sci-
ences, and even there it hn> this advantage. th:it it deals with elements more within
tire present gra>p of the intellect On this point I speak advisedly.
5th. -'The system is thoroughly progressive and practical, and as such. American in
[•f character. "It does not adhere to old usages, merely because thev are veneraliy
musty; anil yet it does ma discard things merely because they are old. or are in un-
Importxni minutie not prudishly perfect. It does nut overlook details and technicali-
ties, nor does it allow them to interfere with plain philosophy or practical utility.
'• Let any clear-headed, independent-minded teacher master the system, and then
give it a fa'ir trial, and there.wiil be no doubt as to his testimony."
A Testimonial from tlie Pr in ci juris of the Public School* of Rochester, N. Y.
We reg.-ird CLARK'S GRAMMAR as the clearest in its analysis, the most natural and
logical itt its arrangement, the most concise and accurate in its definitions, the mos-.
M .-lemittic in design, and the best adapted to the use of schools of any Grammar »tth
which we are acquainted.
C C. MKSKRVE, WM. C. FEGLES.
M D. ROWLEY, OHN ATWATKK,
C. R. BUBRICK, EDWARD WEBSTER,
J. R. VOSBURG, 8. W. STARKWEATHER,
E. K. ARMSTRONG PHILIP CURTISS.
LAWEF.XCR INSTITUTE, Brooklyn, Jan 15, 1859.
MESSRS. A. S. BARXFS & Co: — Having used Clark's New Grammar since Its publica-
tion, i do most unhesitatingly recommend it as a work of cupvrior merit By the ut*
of ii-i other work, ami I have used several, have I been enabled to advance uiy pupil*
v. njiidly and thoroughly.
The author has, by an "Etymological Chart and a system of Diagrams, made Gram
:iar the study that it ought to be, interesting as well as nseful.
MARGARET S. LAWRENCE, Principal.
WELCH'S ENGLISH SENTENCE,
frtttK P*or. J. R BOISE, A. M., Profetxor of the Latin and Greek Langvaget etna
Literature in the University of Michigan.
Tliis work belongs to a new era in the grammatical study of our own language. \V r
n»/.Mr<l nothing. In expressing the opinion, that for severe, searching, and ~ezlianstiv«
m»!ysis. the work of 1'rofessor Welch is second to none. His book Is not Intended fo«
beginners, but only f--r advanced students, snd by inch only It will be understood tnl
appreciated.
MONTEITH AND PIcNALLY'S GEOGRAPHIES
THE MOST SUCCESSFUL SERIES EVER ISSUED.
RECOMMENDATIONS.
A. B. CLARK, Principal of one of the largest Public Schools In Brooklyn (*yt:—
M have used over a thousand copies of Monteith's Manual of Geography sine* Iti
adoption by :he Board of Education, and am prepared to say it is the beat wuik fo»
Junior and intermediate classes in our schools 1 have ever seen."
The Series, in whole or in part. hrt« I/fen adopted in th«
New York State Normal School.
New York City Normal School.
New .Jersey Stale Normal Si-hool.
Kentucky State Normal School.
Imlimia State Normal School.
Ohio ^rate Normal School.
Michigan Stale Normal School.
York County (Pa.) Normal School.
Brooklyn 1'olytechnio, Institute.
Cleveland Female Seminary.
Public Schools of MilwMiikie.
Public Schools of Pittsburgh
Public Schools of Lancaster, Pa.
Public Schools of New Orleans.
Public Schools of New York.
Public Schools of Brooklyn, L. I.
Public Schools of New liaven.
Public Schools* of Toledo, Ohio.
Public Schools of Norwalk, Conn.
Public Schools of Richmond. Va.
Public Schools of Madison, Wig.
Public Schools of Indiananc'is.
Public Schools of Springfield, Maw.
Public Schools of Columbus. Ohio.
Public Schools of Hartford. Conn.
Public Schools of Cleveland, Ohio.
And other places too numerous to
mention.
They have also been recommended by the State Superintendents of ILLINOIS,
INDIANA, WISCONSIN. Missouiti, Nor.ru CAROLINA, ALABAMA, and by numerous
Teachers' Associations and Institutes throughout the country, ami are in successful
use in a multitude of Public and Private Schools throughout the United States.
From PROF. WM. F. PIIELPS, A. M., Principal ofth« New Jersey Staff
jformal School.
TRKNTOK. Junf 17. iSfH.
MF.SRXS. A. S. BARNKS «fc Co.: — GBNTLF.MKN: It gives me much pleasure to state
th.tt MrNally's Geography has been used in this Institution from its organization in
1SS5. with great acceptance. The author of this work has avoided on one hand the
extreme of being too meager, and on the other of going too much into detail, while
he has presented, in a clear and concise manner, all those leading facts of Descriptive
Geography which it is important for the young to know. The mtps are accurate and
well executed, the type clear, and indeed the entire work Is a decided success. I most
cheerfully commend it to the profession throughout the country.
Very *ruly yours, WM. F. PHELP8.
From W. V. DAVIS, Piindpnl of High School, Lancaster, Pa.
LANCASTER, PA., Junf 26. 1S58.
DKAR SIRS:— I have examined yonr National Geographical S<*rien with much
care, and find them most excellent works of their kind. They have been used in the
various Public Schools of this city, ever since their publication, with treat succee* and
satisfaction to both pupil and teacher. All the Geographies embraced in your series
are weil adapted to school purposes, and admirably calculated to impart to the pupil,
in a very attractive manner, a complete knowledge of a science, annually becoming
more useful and important Their maps, illustrations, mid typography, are unsur-
passed. One peculiar feature of McNally's Geography — and which will recommeud
it. at once to every practical teacher — is the arrangement of i»s maps and lessons;
each map fronts the particular lesxm which it is designed to illustrate — thus enabling
the scholar to prepare his ta>k without that constant turning over of leave*, or refrr-
•nce to a separate book, as Is necessary with most othor Geographies. Yours. &c.
Messrs. A. S. BAKNKS A Co., New York. V. W. DAVIS.
From CHARLES BARNRS, late Preaiilfnt State Ttavktr*' Amouiutiim, and Stijifhi-
tendent qfth,« Public Softools at Neic Albany, Indiana.
MKSSHS. A. S. BAKNES <fe Co.: — DK.AK SIRS: I have examined with eonrider:ib'«
care the Series of Geographies published by you, and have no hesitktion in saying
that it Is altogether the best with wllch I am acquainted. A trial of more than •
in ih« Public School
ools .if this city has demonstrated that CornfU is utterlv unflt
Tours'. Ac. C. BARNES.
RECOMMENDATIONS
or
MONTEITH'S HISTORY OF THE UNITED STATES.
Fhis volume is designed for youth, and we think the author has been unusual /
»'iccessfiil in its arrangement and entire preparation. Books of the same design a i
too often beyond the full understanding of the scholar. As history is so much no-,,-
lected in all our schools, the publication of such a work as this should be hailed wiih
pleasure; for if scholars find their first studies of history pleasant, it will become •
pleasure rather than a task. This is a book of SS pages, and finely illustrated. It i> in
every way worthy of a place in every Public School in the State. — Maine Tettcker.
This is a most capital work : just the thing for children. Our boy commenced ,he
"tudy of it the day it came to hand. It is arranged in the catechetical form, and is
fiiK-ly illustrated with maps, with special reference to the matter discussed in the text,
It begins with the first discoveries of America. »nd comes down to the laying of the
Atlantic Telegrnph Cable. Many spirited engravings are given to illustrate the work.
It also contains brief Biographies of all prominent men who have identified them-
selves with the history of this country. It is the best work of the kind we have
Been.— Cluster County Times.
WILLARD'S HISTORIES.
from RET. HOWARD MALCOLM, D. D., President <tf th« University of Leirisburg.
I have examined, during the thirteen years that I have had charge of a College,
many School Histories of the United States, and have found none, on the whole, so
proper for n text-book as that of Mrs. Willard. It is neither too short nor too long,
«11 the space given to periods, events, and persons, is happily proportioned to their
importance. The style is attractive and lucid, and the narrative so woven, as both
to sustain the interest and aid the memory of the student. Candor, impartiality, and
accuracy, are conspicuous throughout. I think no teacher intending to commence a
history class will be disappointed in adopting this book.
MBS. L. H. SIGOUBNKT, Vt4 diilinguithtd Authoress, writes:
Mrs. Willard should be considered as a benefactress not only by her own sex, of
whom she became in early years a prominent and permanent educator, but by the
country at large, to whose good she has dedicated the gathered learning and faithful
labor of life's later periods. The truths that she has recorded, and the principles that
she has impressed, will win, from a future race, gratitude that cannot grow old, aud a
garland that will never fade.
DANIEL WEBSTBR wrote, in a letter to the Author:
I cannot better express my sense of the value of your History of the United Statf-t,
than by saying I ieep it near me as a book of reference, accurate in fact* and dates.
DWIGHT'S MYTHOLOGY.
The mythology of the Grecians and Romans Is so closely interlinked with the his-
• TV »nd literature of the world, thtt some knowledge ot It is indispensable to any
,-nn!arly familiarity with either that history or literature. We have seon no book so
!••>.. vrnu-nt in (size that contains so full and elegant an exposition of mythology ax thi
one before us. It will be found at once a most interesting and a most useful book to
any one who wiskos an acquaintance with the splendid myths and fables with whiob
the great masters of ancient learning amused their leisure and cheated tbtir faith -
tficMgan Journal offduca'ien. ,
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