GIFT OF
Dr. Horace Ivie
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AN
ELEMENTARY TREATISE
A L O E B E A
DESIGNED A3
FIRST LESSONS IN THAT SCIENCE.
H. N. ROBINSON, A. M.,
AUTHOR OF Ay UmrVTIRSITY EBITION OP ALaEBRA — AN ELEMENTART TREATISS
ON NATURAL PHILOSOPHY — A WORK ON GEOMETRY, CONTAINING PLANE
AND SPHERICAL TRIGONOMETRY; ALSO, AUTHOR OF A TEXT BOOK
ON ASTRONOMY, AJfD SEVERAL OTHER MATHEMATICAL WORKS.
NINTH ST AND AE-.D^ ]^J?PTliO;N'.__i
CINCINNATI:
PUBLISHED BY JACOB ERNST,
No. 112 MAIN STREET.
1856.
Entered according to Act of Congress in the year 1850, by
H. N. ROBINSON,
In the Clerk's Office of the District Conrt of tlie Uniteo States, for the
District of Ohio.
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';j>r» H i»r<ucc X vie
tOUCATlON DEPT
'., .. B£%»:BiiftTfeEf} BT JAMES * CO.,
CIMCmNATI.
PREFACE.
Every teacher is desirous of having as few textbooks in his school as
is consistent with efficient and sound instruction, and in accordance
with this object, great efforts have been made by several authors to pro-
duce a work on Algebra that would be a proper text book for all grades of
pupils. But in this all have failed, and given up the point in despair.
The student of adult age, and possessing a passably disciplined mind,
requires a different book from the mere lad, who is just commencing the
science. If we put a child's book into the hands of a young man, he
will, probably, become displeased with the book, and possibly imbibe
prejudice and distaste for the science itself ; and if we put a logical and
philosophical work into the hands of a child, he is sure not to comprehend
it, however well and fluently he may be made to repeat the contents of its
pages. But, nevertheless, as Algebra is the groundwork of all the
mathematical sciences, and is of itself a system of pure logic, it is
important that it should be commenced at an early age — eleven or
twelve, or if otherwise well employed, thirteen or fourteen is a more
suitable age.
It is a prevalent impression that Algebra should not be commenced
until the pupil has acquired a good knowledge of Arithmetic, but this is a
great error. The impression would be well founded, provided Arithme-
tic was the most elementary science, and Algebra was founded on Arith-
metic ; but the reverse is the fact — Algebra is elementary Arithmetic,
and no one can acquire a knowledge of Arithmetic in an enlarged and
scientific sense, without a previous knowledge of Algebra. Beyond nota-
tion, numeration, and the four simple rules. Arithmetic is not a science,
but a sequel to all sciences, it is numerical computation applied to any-
thing and to everything. Proportion, as a science^ is the comparison of
magnitudes, and belongs, properly speaking, to Algebra and Geometry;
and the rule of three, in Arithmetic, is but little more than some of its
forms of application. Problems in mensuration are very properly to be
found in books called Arithmetics, but mensuration is no part of the
9S4226
4 PREFACE.
■clenco of Arithmetic, it is a part of Geometry, and for a good under-
standing of it, geometrical science must be directly consulted.
So it is with many other parts of Arithmetic, the science is else-
where ; and to have a scientific comprehension of many parts of com-
mon Arithmetic, we must go to general Arithmetic, which is emphati-
cally Algebra ; and in preparing this work, we have given constant
attention to this branch of the subject, as may be seen in our treatment
of fractious, proportion, progression, the roots, fellowship, and interest.
All these subjects can be better illustrated by symbols than by num-
bers ; for numbers apply to everything, and, of course, can be made to
show no particular thing ; but not so with symbols, at every step the
particular elements are all visible, and the logic and the reason is as
distinct in every part of an operation as is the result. For these reasons.
Arithmetic should be studied by symbols, as it is in many parts of
Europe ; many of their books, entitled Arithmetics, are as full of signs
and symbols as any Algebra that ever appeared.
The prominent design of the author has been to adapt this treatise to
the wants of young beginners in Algebra, and at the same time not to
produce a mere childish book, but one more dignified and permanent,
and to secure this end, he has kept up the same tone and spirit as
though he were addressing mature and disciplined minds.
Great care has been taken in the selection of problems, and all very
severe ones have been excluded, and all such as might be difficult
when detached and alone, are rendered simple and easy by their con-
nection with other leading problems of kindred character.
To bring out the original thoughts of the pupil has been another
object which he designed to accomplish, and the illustrations are given
in such a way as to command the constant attention of the learner, and
if he learns at all, it will be naturally and easily, and what he learns
will become a part of himself.
In this work, great importance is attached to equations, not merely
in solving problems, but they are used as an instrument of illustrating
principles, and their application is carried further in this book than in
any other known to the author.
For instance, we have illustrated the nature of an equation by the^
aid of simple problems in subtraction and division ; and conversely, the
simple principle of equality is used to deduce rules for subtraction,
division, the reduction of fractions to a common denominator, the mul-
tiplication of quantities afTected by different fractional exponents, &-c.
Notwithstanding that this book is designed to be practical, it contains
more illustrations, and is more theoretical and scientific as far as it goes,
than any other book desiigned for the same class of pupils.
PREFACE. 5
We have not given demonstrations of the binomial theorem, nof
made any investigations of logarithms, or the higher equations, for these
subjects belong exclusively to the higher order of Algebra, and will be
found very clear and full in the University Edition of Algebra by the
same author.
In relation to great generalities, all books on the same science, are, in
substance, much alike, yet, in the clearness and distinctness with which
they present principles, they may be very different ; and to arrive at
perfection in this particular, is, and should be, the highest ambition of
an author.
For peculiarities in this work, the teacher is respectfully referred to
abbreviations generally in solving equations, to the philosophical uses
made of equations in demonstrating principles — the formation of prob-
lems, and the manner of arriving at arithmetical rules, which may be
found in various parts of this work.
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CONTENTS
Page
Introduction 9
Axioms 16
Simple problems for exercises 17
Definition of terms 21
SECTION I.
Addition 25
Subtraction 31
Subtraction illustrated by Equations 35
Multiplication 36
The product of minus by minus illustrated 37
Division 45
Negative exponents explained 49
Division in compound quantities 51
Factoring 55
Multiple and least common multiple 58
Algebraic Fractions 62
Complex fractions 68
Multiplication of fractions 70
Division of fractions 73
Division illustrated by Equations 76
Addition of fractions 78
Addition of fractions by Equations 79
Subtraction of fractions 85
Subtraction illustrated by Equations 86
SECTION II.
Equations 89
Transposition 93
General rule for reducing equations 93
Proportion, as applied to equations 98
8 CONTENTS.
Pagb.
Questions producing simple equations 101
How to propose convenient problems 107
How particular numerals are brought into problems Ill
Equations having compound fractions ' 116
Equations containing two unknown quantities 120
Three methods of elimination 122
Equations containing three or more unknown quantities — Rule
for elimination 130
Questions producing equations containing three or more un-
known quantities 134
Negative results, how understood 137
SECTION III.
Involution 140
Expansion of a binomial 144
Application of the binomial 148
Evolution 150
How to extract roots of polynomials 153
Approximate rule for cube root 1 63
Product of quantities affected by different fractional exponents —
Art 83 167
SECTION IV.
Equations of the second degree 171
Pure Equations 172
Problems producing pure equations 174
Rules for completing a square 180
Resolving a quadratic expression into two factors 186
Questions giving rise to quadratic equations 191
Homogeneous and symmetrical equations 193
SECTION V.
Arithmetical progression 204
Examples in arithmetical progression 211
Geometrical progression 213
Examples in geometrical progression 215
General problems that involve progression 220
Proportion, theoretically considered 224
Method of deriving certain arithmetical rules 236
Fellowship, theoretically considered 237
INTRODUCTION.
Algebra is the science of computation by mekiis of sym-
bols. Letters of the alphabet are generally used to represent
quantities or numbers, and conventional signs are employed
to represent operations, and to abridge and generalize the
reasoning in relation to propositions or problems.
We sometimes meet with persons who can readily solve
quite difficult problems, and yet are not able to explain
the steps in the process : they call their operations working in
the head — and, indeed, their reasoning, properly written out,
is Algebra; but not having a knowledge of the signs, and
possessing no skill in writing out the thoughts of the mind,
they do not know that it is Algebra.
This natural adaptation of the mind, to solve problems with-
out the aid of writing down the operation, is very essential to
success in this science. But the mind can only go a very
short distance, unaided by the pen ; nor is it important that
it should, for the aid given by that instrument is efficient and
complete, secures the ground gone over, and leaves the mind
free to advance indefinitely.
In a purely mental process, the mind must retain all the
results thus far attained, and continue the reasoning onward
at the same time. And this, carried to excess, breaks down
the mind rather than strengthens it ; and for this reason, a
mere mental Algebra must be regarded as one of the ephe-
meral efforts of the times. But let no reader construe these
sentiments into a disapproval of mental Algebra. Every Al-
gebra, properly understood, is mental Algebra ; for the mental
process — the reasoning power — must precede every operation.
10 ELEMENTARY ALGEBRA.
To compare the common operations of the mind with the
brief and refined language of science, we propose the follow-
ing problems. But before we use algebraical language, we
must explain some pi ?ts: §yiiibols, and here we insert only
those intended for immediate use.
J., .. I "' .',•■'-' ■„ _ » ,'., • (fc
, :;: . -THE SIGNS.
1. The perpendicular cross, thus +, caWed plus, denotes
addition.
2. The horizontal dash, thus — , called minusy denotes
subtraction.* These signs are written before the quantities
to which they are aflfixed.
3. The diamond cross, thus X, or a point between two
quantities, denotes multiplication. For example, 5X 4, or 5*4..
shows that 4 and 5 must be multiplied together.
4. A horizontal line with a point above and below, thus -^,
denotes division; also two quantities, one above another, as
numerator and denominator, as _ or ^, also indicates division,
7 h
and shows that 3 must be divided by 7, and a must be divided
hyb.
5. Double horizontal lines, thus =, represent equality, and
show that the quantities between which it is placed are equal.
6. A number or letter before any quantity shows how many
times the quantity is taken, and is called the coefficient of the
quantity, thus 3x, shows that the quantity x is taken 3 times,
and nx shows that the quantity represented by x is taken as
many times as there are units in n.
7. A vinculum or bar , or parenthesis ( ), is used
* The signs plus and minus, in general science, have a far more com-
prehensive meaning than is here expressed. Here they denote simply
what is to be done with the quantities to which they are attached ; but
in philosophical problems, they may denote the essential value of the
quantities, as credit and debt ; and in geometry tliey may represent jjosi-
tions, as north and sotrfA, or to the right or left of a zero line, &c.
INTRODUCTION.
11
to connect several quantities together. Thus, a4-6 or {a-\-b)
shows that a and b are there to be considered as connected, or
making but one quantity.
We now turn our attention to the problems — not for the
purpose of finding the answers to them, as the mere arith-
metical student might suppose — but for the purpose of teach-
ing the manner of solving them by the scieiice of Algebra.
1. A father divided 120 cts. among his three sons. Be gave
the youngest a certain number, the second 10 cerds more, and the
eldest 10 cents more than the second. What sum did, each
receive ?
By the use of cmnmon lan-
guage^ this question may be
solved thus :
The youngest son had a
share of the money ; the sec-
ond son had a like share and
10 cents more ; and the eldest
had also a like share as the
youngest, and 20 cents more.
Therefore the three boys had
3 shares and 30 cents ; but
the three boys had 120 cents,
hence, 3 shares and 30 cents
are equal to 120 cents, or the
three shares are worth 90
cents, and one share is worth
30 cents, which is the sum
given to the youngest.
By algebraical language,
thus:
Let X represent the share
of the youngest. Then by
the conditions
a;= 3d son's share,
a;-f 10= 2d "
ar-{-20= 1st " "
3a;-f30=120 by add.
This expression is called an
equation, and the quantities
on each side of the sign of
equahty are called members,
or sides of the equation.
It is an axiom, that equals
from equals the remainders
must be equal ; and in this
equation, if we *take 30 from
both members, we have
3a:= 120— 30=90
Dividing both members by
3 gives a;=30, tlie share of
tlie youngest.
1
12
ELEMENTARY ALGEBRA.
2. If 75 dollars he added to a share in a certain bridge coni'
pany, the sum will he the value of 4 shares. What is the
value of a share?
BY COMMON LANGUAGE.
One share and 75 dollars is
the same as 4 shares, there-
fore 75 dollars is 3 shares,
and one share is 25 dollars,
the third part of 3 shares.
ALGEBRAICALLY.
Let X represent the value
of a share.
Then x-^15=Ax.
Taking x an equal quantity
from both members and
75=3;r.
Dividing by 3 gives 25=ar.
3. A gentleman purchased a horse y a chaise and a harness, for
$230. The harness cost a certain sum, the chaise 3 times as
much as the harness, and the horse $20 more than the chaise.
Required the price of each.
BY COMMON LANGUAGE.
The harness cost a certain
share of the money, and the
chaise cost 3 such shares, and
the horse cost 3 such shares
and 20 dollars more ; there-
fore the whole cost 7 shares
and 20 dollars, which must
make 230 dollars. Take the
20 dollai's away, and the 7
shares is the same as 210 dol-
lars. Therefore 1 share is 30
dollars, the value of the har-
ness, and $90 is the value of
the chaise, and $1 10 the value
of the horse.
4. In a certain school ^ of the pupils are learning geometry ,
\ are learning Latin, and 10 more, which comprise all in the
school t are learning to read. What was the whole number.^
ALGEBRAICALLY.
Let x= the value
of the harness.
Then 3x- the
of the chaise.
And 3a;-[-20= the
of the horse.
value
value
Sum
Taking
both
7a;4-20=230.
equals from
members 7a:=210
Or a;=30 by division.
INTRODUCTION. 13
Br COMMON LANGUAGE.
One-third and one-fourth
added together make ■^-^.
ALGEBRAICALLY.
Let x=^ the number in
school.
Then by the conditions
X X
34-4+10=0;.
Therefore -^-^ is the number
learning to read, which, by
the problem, is 10; hence Jj
of the number in the school
is 2, and the whole number is
24, the number required.
This equation may be troublesome on account of the frac-
tions ; but in due time we shall give rules to clear equations
of fractions; however, fractions here, are just the same as
fractions elsewhere. One-third and one-fourth of anyth'mg is
j\ of that thing; therefore, ^|-f-10=Y¥. Now from the
two equals take ^f, and 10=ff ; dividing by 5, ^—j^\ mul-
tiplying by 12 gives 24=a?, the final result.
From these examples it will be perceived that Algebra is
but an artificial method of briefly writing out our mental
operations when we solve mathematical problems, and as such,
it may be extended and applied to almost every branch of the
mathematics ; and, therefore, the value of this science cannot
be over estimated.
The three following problems are extremely simple when
algebraic language is applied, but would be rather difficult by
common language.
5. On a certain day, a merchant jmid out ^2500 to three men,
A, B, and C; he paid to A a certain sum, to B $500 less than
the sum paid A, and to C he paid $900 moi'e than to A. He-
quired the sum paid to each.
Let .... x= the sura paid to A.
Then .... a:— 500= " *' '« B.
And .... a;-|-900= •* *' " C.
By addition, . . 32-+400=2500, the whole sum paid.
14 ELEMENTARY ALGEBRA.
Subtracting 400 from both members— that is, equals from
equals — and we have 3a;=2100.
Take J of each member, or divide both sides by 3 and
ar=700, the sum paid to A.
From this we determine that $200 was paid to B, and
$1600 to C.
In this example the sign before 500 shows what is to be
done with that quantity ; it shows that it is to be taken out,
and accordingly it was taken out, diminishing the 900 to 400.
This operation is called addition, but it is algebraic addition,
that is, writing the quantities according to their signs, and
finding the result.
6. It is required to divide the number 99 into Jive suck parts
thai the first may exceed the second by 3, be less than the third by
10, greater than the fourth by 9, and less than the fifth by 16.
Let ... .
x= the first part;
then . . . .
. X — 3= the second.
x-\-\0= the third.
X — 9= the fourth.
-
ir+16= the fifth.
Sum is . . .
5a;+14, but the sum of
99; therefore,
5ar+14=99.
By subtracting 14 from both members, and dividing the
remainders by 5, we have x=\l, the first part; and, there-
fore the parts are 17, 14, 27, 8, and 33.
7. Divide $1000 among four men, giving the second tidce
as much as the first, minus $200; the third double the sum of
the second, plus $400; the fourth three times as much as ike
first, plus $100. Required the share of each.
]jet .... rr= the sum paid to the 1st,
then .... 2ar— 200= ** " " 2d,
4.2r— 400+400= " " *' 3d,
3a;+100= " " ** 4th,
The sum is 10a:— 100= 1 000.
INTRODUCTION. 15
By adding 100 to both members we have
10a:=1100
By adding 100 to — 100 in the first member of the equa-
tion, makes 0, and then 10a; only is left in that member, which
must be equal to 1100, or a;=110, the sum paid to the first,
and the several sums are 8110, 820, 8440, and 8430.
The preceding remarks and problems serve to show, only
in same small degree^ the advantage of Algebra over common
language, and the learner should examine every problem, and
the reason of every step in the process of its solution, until
all is thoroughly understood ; then he will have no difficulty
in solving the examples that follow in this introduction. But
before we give additional problems, let us call the student's
mind to the precise idea of an
EQUATION.
An equation is simply what the word implies ; equality as
to value, weight, or measure ; and can be best understood by
comparing it to a pair of scales delicately balanced.
The halance can he preserved ly adding equal weights to
hoth sides; by taking equal weights from both sides; by multiply-
ing both sides by the same number, or by dividing both sides by
the same number, or by talcing like roots or like poivers of the
weights in both sides.
The object of working an equation is to bring the unloiown
quantity to stand alone as one member of the equation, equal
to known quantities in the other member. The unknown
quantity thus becomes known; and we may do anything to
accomplish this end, that the nature of the case may seem to
require, only taking scrupulous care to preserve equality
through every change.
It is usual to represent known quantities by their numerical
values, or by the first letters of the alphabet, as a, b, c, d, <fec.;
and unknown quantities by the last letters, as u, i, x, y, (fee.
16 ELEMENTARY ALGEBRA.
AXIOMS.
Axioms are self-evident truths, and of course are above
demonstration ; no explanation can render them more clear.
The following are those applicable to Algebra, and are the
principles on which the truth of all algebraical operations
finally rests.
Axiom 1. If the same quantity or equal quantities be
added to equal quantities, their sums will be equal.
2. If the same quantity or equal quantities be subtracted
from equal quantities, the remainders will be equal.
3. If equal quantities be multiplied into the same, or equal
quantities, the products will be equal.
4. If equal quantities be divided by the same, or by equal
quantities, the quotients will be equal.
5. If the same quantity be both added to and subtracted
from another, the value of the latter will not be altered.
6. If a quantity be both multiplied and divided by another,
the value of the former will not be altered.
7. Quantities which are respectively equal to any other
quantity are equal to each other.
8. Like roots of equal quantities arc equal.
9. Like powers of the same or equal quantities are equal.
Now suppose we have the following equation
xJ^a=^h
in which x is the unknown quantity, and a and b known quan-
tities. Before x can become known, a must be disensfacred
from it, that is, a must be subtracted from both members.
It must be subtracted from the first member, because it is our
object to have x stand alone, and we must subtract it from
the other member, to preserve equality. The equation then
stands x=b — a
Here we find the quantity a, whatever it may be, cm the other
side of the equation, with the contrary sign.
INTRODUCTION. 17
Now let us suppose we have an equation like
X — a-=b
In this equation we perceive that x is diminished by a;
therefore to have the single value of x we must add a to the
first member ; and, of course, to preserve equality, we must
add a to the second member, then we shall have
X — a+a=6-{-a
But — a and -]ra destroy each other, and the equation is in
brief x^=h-\-a
Here, also, we find a on the opposite side of the equation,
with its sign changed ; and from these investigations we draw
the following rule of operation.
Rule . — We may change any quantity from one member of
an equation to the other, if we change its sign.
The operation itself is called transposition.
For examples, transpose the terms so that the unknown
quantity x shall stand alone in the first member of the follow-
ing equations :
x-\-c — d=4g .... Ans. x==4g-\-d — c.
x-\-3 — a-\-m=30 .... Ans. ar=30 — m-\-a — 3.
SIMPLE PROBLEMS FOR EXERCISES.
1. A man bought a saddle and bridle for 45 dollars; the
saddle cost four times as much as the bridle. What was the
cost of each ? Ans. Bridle $9 ; saddle $36.
2. Three boys had 66 cents among them ; the second
had twice as many as the first, and the third three times as
many as the first. How many had each ?
Ans. 1st boy had 11 ; 2d, 22 ; 3d, 33 cents.
3. Two men had 1 00 dollars between them, and one had
3 times as many as the other. How many had each ?
Ans. One had ^25, th« other $75.
K. B. This last problem may be enunciated thus :
Two men had 100 dollars between them ; the first had
one -third as many as the other. How many had each ? .
2
18 ELEMENTARY ALGEBRA
4. Three men had 880 dollars among them ; the first had
~, the second had i as many as the third. Plow many had
each ? Let 6x= what the third had.
Ans. 1st had 160 ; 2d, 240, and the 3d, 480 dollars.
5. There are three numbers which together make 72, the
second is twice as much as the first, and the third is as much
as both the others. What are the numbers ?
A71S. 1st is 12; 2d, 24; 3d, 36.
6. Two men built 90 rods of fence in 3 days. The second
built twice as many rods in a day as the first. How many
rods did each build per day ? Ans. 1st built 10 rods, 2d, 20.
7. A man bought 3 oxen, 4 cows, and 6 calves, for 260 dol-
lars. He paid twice as much for an ox as he did for a cow,
and twice as much for a cow as for a calf. How much did
he give for each ?
Ans. For a calf, $10; cow, $20 ; and for an ox, 840.
8. A man bought a boat load of flour for 132 dollars, one-
half at 5 dollars per barrel, the other half at 6 dollars per
barrel. How many barrels did the boat contain ?
Let x= half the number of barrels. Ans. 24.
9. A boy bought an equal number of apples, oranges and
pears, for 96 cents : the apples at 3 cents apiece, the oranges
at 4, and the pears at 5. How many of each kind did he
buy? Ans. 8.
10. Two men bought a carriage for 86 dollars; one paid
five times as much as the other, and 26 dollars more. What
did each pay? Ans. One paid 10, the other 76 dollars.
11. If from 6 times a certain number we subtract 24, the
remainder will be 196. What is the number ? Ans. 44.
12. To the double of a certain number, if we add 18, the
sum will be 96. What is the number? Ans. 39.
1 3. What number is that whose double exceeds its
78 ? Let 2.v= the number.
1 4. A man had six sons, to whom ho gave T 20 dollars,^givin^
INTRODUCTION. 19
to eacli one 4 dollars more than to his next younger brother.
How many dollars did he give to the youngest ? Ans. f 10.
15. Three men received 65 dollars ; the second received 5
dollars more than the first, and the third 10 dollars more
than the second. What sum did the first receive ?
Ans. $15.
16. A man paid a debt of 29 dollars, in three different
payments ; the second time he paid 3 dollars more than at
first, and the third time he paid twice as much as at the sec-
ond time. What was the amount of his first payment ?
Ans. $5.
w. A man bought 6 pounds of coffee, and 10 pounds of
tea, for 360 cents, giving 20 cents a pound more for the tea
than for the coffee. What was the price of the coffee ?
Ans. 10 cents.
1 8. A man bought 6 barrels of flour, and 4 firkins of butter,
for 68 dollars. He gave 2 dollars more for a barrel of flour
than for a firkin of butter. What was the price of flour ?
Ans. $7.60.
19. A pound of coffee cost 5 cents more than a pound of
sugar, and for 3 pounds of sugar or for 2 pounds of coffee
you must pay the same sum. What is the price of sugar ?
Ans. 10 cents.
20. A person in market selling apples, peaches, and
oranges, asked 1 cent more for a peach than for an apple,
and 2 cents more for an orange than for a peach, and the
prices were such that 10 apples and 5 peaches cost as much
as 5 oranges. What was the cost of an apple ? Ans. 1 cent.
21. One-half of a post stands in the mud, one-third in the
water, and the remainder, which is 3 feet, is above the water.
What is the whole length of the post ? Ans. 18 feet.
2S.' One-third and one-half of a sum of money, and ten
dollars more, make the whole sum. What is the sum ?
Ans. 60 dollars.
20 ELEMENTARY ALGEBRA.
23. Divide 25 cents between two boys, and give one four
times as mucli as the other. Required the share of each.
Ans. 5 and 20 cents.
24. Divide 15 cents between two boys, and give one double
of the other. Required the share of each.
Ans. 5 and 10 cents.
Similar problems to the preceding might be framed indefi-
nitely, but it would be improper to propose any that involve
any difficulty until the pupil is better prepared to meet diffi-
culties. We only give the preceding to convince the learner
that he can find real utihty in the science ; but before he can
go into the subject to advantage, he must learn the nature of
algebraic expressions, and acquire the art of adding, subtract'
ing, multiplying, and dividing algebraic quantities, both whole
and fractional.
We now assure the young beginner that we will conduct
him through the elements of this science with as little delay
and trouble as possible ; and neither remarks nor examples
will be given which'are not, in the judgment of the author,
essential to the progress of the pupil.
With this assurance we close this introduction, and com-
mence Algebra, by giving more extended definitions of
terms.
%
ALGEBRA
DEFINITION OF TERMS.
The signs for addition, subtraction, multiplication, division,
and equality, liave already been explained. We have also
explained coefficient and vinculum. The word coefficient can
hardly be understood by a mere definition. It means any
factor connected with another, and may be simple or com-
pound, thus, ax; a is the coefficient of x, and in the term
3ax, 3a is the coefficient of x, and 3 is the coefficient of ax.
1. When a letter stands alone, as b, y, or any other letter,
one or unity may be considered its coefficient. In the expres-
sion (3a4-25 — c)x, (3a-|-25 — c) is a compound coefficient to
X. It is also a factor, and x is another factor. The word
factor has the same signification as in Arithmetic.
2. When we wish to note that two quantities are unequal y
we write this sign ^ between them. The opening of the
sign is always put toward the greater quantity, thus, a'^hy
signifies that a is greater than 6, and a<^h shows that h is
greater than a.
3. When we indicate the multiplication of numbers, with-
out actually performing the multiplication, we must write the
sign X or • between them, as 5X 5, or 3'4, because the mul-
tiplication could not be understood without the sign ; but
when we have letters in place of numbers, we may omit the
sign and write ah, in place of « X 6 ; ahx in place of a*h*x, <fec
22 ELEMENTARY ALGEBRA.
4. When factors are equal, and each equal to a, the pro-
duct of four such factors is aaaa; but in place of this, we
may write a^ which signifies that a is taken four times as a
factor ; a? indicates a product which is composed of a taken
seven times as a factor ; a;° indicates a product in which x is
taken as many times as a factor as there are units in n.
5. The smcdl number thus written to the right of a letter,
(or to any quantity) and a little above, is called its exponent.
Expmierds may be either whole numbers or fractions; but
this will be explained hereafter.
When an exponent is a whole number, it indicates the
power of the factors, or quantity to which it is attached.
When it is a fraction, it indicates a root of the quantity,
thus, c^ indicates the square or second root of a ; a^ indicates
the third root of a, and a* indicates the fourth root of a, &c.
Formerly, the sign of a root was indicated by the radical
sign J; and, for some purposes, this sign is still in use.
Whenever it is used, it is placed to the left of the quantity ;
thus, Ja. The nwmher of the root is denoted by a little
figure placed over the radical sign ; unless it is the second
root, when the figure 2 is omitted. Thus,
J a is the second or square root of a.
II a is the third or cube root of a.
lla\s the nt\\ root of a.
Ja-\-x is the second root of (a-j-a;).
It must be remembered, that if the radical sign is to affect
more than one factor, the vinculum must be used with it;
thus, Jba, Jab. ■^.
6. The number of literal factors which enter into any term,
is the degree of that term ; ah is of the second degree, cth of
the third, a6V of the fifth. In general, the degree of an
algebraic term is found by taking the sum of the exponents
of the letters which enter into that term. An algebraic quan-
tity which has all its terms of the same degree, is said to be
DEFINITION OF TERMS. 23
homogeneous: 4a'^+2a^5^ — '^ah(?-\-'b^c, is therefore homogene-
ous, and of the fifth degree. *
Similar terms, are those which contain tlie same letters in
the same powers, 2a^b and 5a^b are similar. But, 3ab^ and
Sab are not similar terms, for the letters, although the same,
are not in the same power.
7. Simple Quantities, are those which consist of one term
only. As 3a, or 5ab, or 6ab(^.
8. Compound Quantities, are those which consist of two or
more terms. As a-\-b, or 2a — 3c, or a-{-2b — 3c.
9. And when the compound quantity consists of two terms
it is called a Binomial, as a-^b ; when of three terms, it is a
Trinomial, as a-\-2b — 3c ; when of four terms, a Quadrino-
mial, as 2a — 3b-\-c — 4c? ; and so on. Also, a Multinomial or
Polynomial, consists of many terms.
10. A Residual Quantity, is a binomial having one of the
terms negative. As a — 2b.
11. Positive or Affirmative Quantities, are those which are
to be added, or have the sig-n +• As a or -\-a, or ab: for
when a quantity is found without a sign, it is understood to
be positive, or as having the sign 4" prefixed.
12. Negative Quantities, are those which are to be sub-
tracted. As — a, — 2ab, or — 8c^
13. Like Signs, are either all positive (+), or all negative
14. Unlike Signs, are when some are positive (+), and
others negative ( — ).
15. The reciprocal of any quantity is unity divided by that
quantity. Thus, i is the reciprocal of 4, - is the reciprocal
a
of a, and so on, of any other quantity.
16. The same letter, accented, is often used to denote
quantities which occupy similar positions in different equa-
tions or investigations. Thus, a, a', a", a'", represent four
24 ELEMENTARY ALGEBRA.
different quantities ; of wliicli a! is read a prime ; a" is read
a second ; a!" is read a third, and so on.
That the pupil may imbibe or catch the true spirit of an
algebraic expression, we give the following exercises in con-
verting common arithmetical operations into algebraic ex-
pressions, and finding the value of each under different
suppositions.
Express in algebraic langiuige the prodtict of three times a
into X, diminished by c, and the remainder divided by b.
. Zax — c
Ans. .
b
What is the value of this expression when a =2, x=S,
«=4, and 5=2 ? Ans. 7.
What is its value when a=3, x=5, c=9, and b=3 ?
Ans. 12.
Express in algebraic language 3 tim£S the square of a, dimin-
ished by 2b, and the difference divided by c. An^. .
c
What is the numerical value of this expression, when a=5,
5==10, and c=5? Ans. 11.
Wllatwhena=10, 5=7, and c=20? . . Ans. \^j\.
"What when a= 9, 5=0, and c= 1 ? . . Ans. 243.
What when a= 1, 5=1, and c= |? . . Ans. 2.
Write the followmg: Qa diminished by x, the diflPerence
increased by the square root of c, and the whole multiplied
by 5. Ans. (6a — x-]rc^)b.
What is the value of this expression, when a=3, a;=18,
c= 16, and 5=2? Ans. 8.
What when «=6, rr=9, c=9, and 5=7 ? Ans. 210.
What is the value of the expression a^-\-2>ah — c^, when
a=:6, 5=5, and c=4 ? Ans. 1 10.
With the same value to a, b, and c, what is the value of
the expression, 2a^ — 3a'^5-|-c^ ? Ans, — 44.
What of the expression a^(a-\-b) — 2a5c? Ans. 166.
ADDITION. ^
a?
What is the value of -r—--\-c^'l ^«s. 28.
What is the value of a^-^Jb^—<ic'> .... AJIk2,b.
w
SECTION I.
ADDITION.
(Art. 1.) Addition in Algebra is connecting quantities to-
gether by their proper signs.
Here the pupil should call to mind the fact that unlike
quantities cannot be added together. For instance, it would
be an absurdity to add dollars to yards of cloth, and so of
any other unlike quantities ; but dollars can be added to dol-
lars, yards to yards, &c.; so in Algebra, a may be added to
a, making 2a, or any number of a's may be added to any
other number of a's by uniting their coefficients ; but a can-
not be added to b or to any other dissimilar quantity : we
can write a-\-b, indicating the addition by the sign making a
compound quaiitity.
(Art. 2.) Addition in Algebra may be divided into three
cases : the first, when the quantities are alike and their signs
alike ; a second, when the quantities are alike and the signs
unlike ; and the third, when the quantities are unlike.
To discover a rule for^ase 1st, we propose the following
problem :
On Monday a merchant sent to a steamboat 17 barrels of
flour and 9 barrels of pork ; on Tuesday he sent 7 barrels
of flour and 10 of pork; on Wednesday, 20 barrels of flour
26 ELEMENTARY ALGEBRA.
and 6 of pork ; on Thursday, 10 barrels of flour and 10
of pork. How many barrels of each has he sent ?
Way write it thus :
17 barrels of flour + 9 barrels of pork
'J it <( it JL. 1Q ti tt it
20
tt
it
tt
+ 6
it
it
it
10
it
it
tt
+ 10
it
it
it
64 barrels of flour +35 barrels of pork
Now let b represent a barrel of flour, and p represent a
barrel of pork, then in place of writing out the words, we
write 175+ 9p
75+1 Op
205+ 6p
105+10;;
Sum is 545+35/;
From this example we perceive that to add together simi-
lar quantities, we have only to add their numeral coefiicients,
like simple numbers in Arithmetic.
Hence, the following rule will meet
Case 1 . When the quantities are similar and the siffns
alike, add the coefficients together, and set down the sum ; after
which set t/ie common letter or letters of the like quantities, and
prefix the common sign + or — .
EXAMPLES.
(0
(2)
(3)
(4)
, (5)
3a
— 3bx
bmj
3a+ 25— 5c
4a5— 2crf
9a
— bbx
2bxy
6a4- 65— c
7a5— cd
5a
— Abx
bbxy
7a-tll5— 8c
\5al>— 2cd
12a
— 25a;
— 75a;
bxy
Zbxy
a+ 5— 3c
ab—12cd
a
16a+205— 17c
27 ab— lied
2a
— bx
Qbxy
32a —225* 1850^
ADDITION. 27
(6) (7) (8)
3a-t- 3ax-\- c 4a6+3^ — 2b lOy — x-\- k
la-\- 5ax-\-5c lah-{- x — 36 ly — x^ 2>h
10a-|- lax-\-2c 121aZ>-f 2a;— h 2>y~ 2x-\- Ih
2a+10aa;+4c 99a64- a; — b y—\Ox-\-\Qh
Sum
(Art. 3.) Like quantities, of whatever kind, whether of
powers or roots, may be added together the same as more
simple quantities.
Thus 3a2 and Sa^ are 1 la^ and lW-^W=\Qb^. No matter
what the quantities may be, if they are only alike in kind. Let
the reader observe that 2(aH-5)+3(a-|-6) must be together
5(a4-6), that is, 2 times any quantity whatever added to 3
times the same quantity, must be five times that quantity.
Therefore, ^Jx-\-y-{-3jx-\-y=7jx-\-y, for ^x-{-y, which
represents the square root of ic+y, may be considered a single
quantity.
To illustrate these remarks we give the following
EXAMPLES.
4(a—x)
{^-\-y) Ja-\-^
6{a^-c)
7(a^x)
3(a;+2/) Qja^x
{a^-c)
I0(a—x)
20{x-\-y) \2,Ja+x
l{a^-c)
Sum 2\{a—x) 24(a;+y) ^O^a-^x
(Art. 4.) — Case 2. When the quantities are similar and
tJie siyns unlike, we have the following rule for addition.*
* In this rule, the word addition is not very properly used ; being
much too limited to express the operation here performed. The busi-
ness of this operation is to incorporate into one mass, or algebraic
expression, different algeltraio quantities, as far as an actual incorpora-
28 ELEMENTARY ALGEBRA.
Rule. — Add the affirmative coefficients into one sum and
the negative ones into another, and take their difference with the
sign of the greater, to which affix the common literal quantity.
Sum
EXAMPLES FOR
PRACTICE
(1)
(2)
(3)
^5a
+3tt;r«
+ 8af^-f 3y
+4a
+4aa:2
— '5^+4^
+6a ,
—8(^2
—16x'-\-5g
~3a
—Qax^
+ 3a:«— 7y
+ a
-\-Sa3^
+ 2:ir'— 22/
+3a
—^aj?
— 82^=^4- 3y
(4)
(5)
(6)
— So"
+ 3^y
+4ai+ 4
— bd'
4- 9iy
— 4a5-}-12
— lOa^
— lo^y
-|-7a5_14
4-10^2
—\Wf
+ a5 + 3
+ 14a2
— 2uy
—Bab —10
7. Add 2xy—2a\ JSa^-f-ary. a^-fay, 4a2_3;ry, 2xg—2a\
Ans. 4a^-{-3xy.
8. Add 8aV — 3xy, Box — 5xy, 9xg — box, 2a^a^-\-xj/,
Sax — 3xy. Ans. \0c^.t^-\-5ax — xy.
9. Add 3^2—1, 6am—2m:^-\-4, •7—Sam-{-2m\ and 6m^
+2am-\-l. Am. Om'+ll.
tion or union is possible ; and to retain the algebraic marks for doing
it, in cases where the former is not possible.
By using the word united in place of the word added, the reason of
the rule will become obvious.
Thus 3a united to — a makes 2a
Ix united to —2a; makes hx
ADDITION. 29
10. Add 12a—lSah-{-16ax, 8— 4m+2y, —Sa+TaS^^-
12y — 24, and lab — 16ax-i-4m.
Ans. Qa—Qah-\-\Ay-\-lalr—\Q.
1 1 . Uriite 4a^b — Qarb — 9a^b-\- 1 1 a/^b into one term if possible.
Ans. — 2a^b.
12. Ufiite 1ab(? — ab<? — lab^ — 8a5c^-|-9«5c^ into one term.
Ans. 0.
13. Add 3a(a4-6), la{a-\-b), —ba{a-\-b), and ^a{a-^b),
Ans. 8a(a+6).
14. Add 7(6.r4-y— 2)^ — 8(6a:+y— 2)2, (i^x-^y—zf, and
3(6ar+y— 2)'. ^tzs. 3(6x4- y— 2)2.
15. Add 3a5+4ii(62/+^), — 8a5— 9a(62/-}-Z>), 12a64-13a
(63/+5), «i+a(6?/+5), and 7a^4- 6a(6y4-5).
^ws. 15a64-15a(%+^)-
C A s E 3 . — When the quantities are unlike and the signs
ALIKE or UNLIKE, we havc the following rule to unite, or rather ' '
to reduce and condense the quantities. ,
R u L E. — Collect together all those terms that are similar, by
uniting their coefficients, as in the former cases : then lorite the dif-
ferent sums, one after another, udth their proper signs.
N. B. It is immaterial what quantity, in an aggregate sum,
stands first, for the whole of a thing is equal to the sum of
all its parts, whatever part may be first written. Thus,
ax-\-l}y-{-c is the same sum, whichever terai stands first.
EXAMPLES.
1. Add. 3a f, — 2xy^y — 3y^x, — Qct^y, and 2xy\
r3af — 2xy^
These terms may be arranged thus : < — 3xy^ — 8s^y
i +2a-f
Sum 3ay — 3xy — Sx'^y
2. Add together 15a^—Qbh-{-32ah'^—12bc, 19Pc—4a^-\-
lla^c^-\-2bc, a^— 29aV— 12^»''^c+56c, and 9a^c'^—14bc-\-b''c.
30 ELEMENTARY ALGEBRA.
a^— 12Zi2c— 29aV-i- bbc
+ bh-[- 9aV— 145c
12a2 * * 4-23aV— 19k, the ans.
3. What is the sum of 6a5-{-125c— 8cc?,t3cc?— 7a5— 95c,
and 12cd—2a7}—5bc^ Ans. 7cd—2al)—2bc.
4. What is the sum of 5j^—7jbc-\-M, ^Jd>^^jTc
-12c? and 1 Jab-\-2>Jbc-YM'> Ans. 15j'^-\-4jb^-\-5d.
5. Add 72ax^—8af, —SQaa;'^—Say*+7af, S-\-l2ay\
'-6af-\-12—34ax'^-i-5a7/—9a2/\ Ans. — 2a/+20.
Add a-\-b and 3a — 5b together.
Add 6x—5b-\-a-\-8 to -_5a— 4a;+45--3.
Add a4-25— 3c— 10 to 35— 4a+5c-|-10 and 65— c.
Add 3a-f5— 10 to c—d—a and — 4c+2a— 3&--7.
Add Sa^-\-2b^—c to 2a5— 3a2+Jc— 5. C
(Art. 5.) Let it be strictly observ^ed, that when we add sim-
ilar quantities together, as Sx, 4x, and 10a;, we perform it by-
writing the coefficients in one sum, 17, and writing the x, or
the quantity, whatever it may be, afterward, making in this
example 17ar. As principles never change, we must do the
same thing when the coefficients are literal; thus, the sum of
ax, bx, and ex must be [a-\-b-\-c)x, and ax — x may be written
{a—\)x.
EXAMPLES.
(1) (2)
Add ax-\-by^ ay-{'cx
2ca:+3ay* Say-\-2cx
Adx-\-li/ 4y -]rQx
Sum (a-i-2c+4c?)a.'-l-(5+3a+7)2/2 (4a-l-4)y-f (3c-f 6).r
ADDITION. ai
(3) (4)
AM . . . 3a;H-2d-2/ (tx-{-7y
hx -{-cxy lax — 3y
(a-\-h)x-\-9.cdxy — 9.x +4y
Sum {a-{-9b-\-2>)x-\-{'^cd-\-c-^9)ry (8a— 2)a;-|-8y
5. Add 8ai:+2(^+a)4-3i, Qax-^Q{x-\'a)—%, and 11a;
4-6^— 7aa;— 8(a;-fa). ^tw. lOax-^Ux.
6. Add {a-\-b)Jx and (c+2a — ^J^o: together.
^ns. (c+3a)^ar.
7. CTTii^e 3ax-{-7ax — 4ax — lx-\-3hx-{-4x, as far as possible,
and find the sum total of the coefficient of x.
Ans. 6ax-\'2bx-\-4x\
The sum total of the coefficient of x is (6a-|-26-f-4), and
the sum total of the whole expression may be written
(6a+254-4>.
SUBTRACTION.
(Art. 6.) Subtraction in Algebra is not, in all cases, taking
one quantity from another : ii is finding the difference between
two quantities.
What is the difference between 12 and 20 degrees of
north latitude ? This is subtraction. But when we demand
the difference of latitude between 6 degrees north and 3
degrees south, the result appears like addition, for the differ-
ence is really 9 degrees, the sum of 6 and 3. This example
serves to explain the true nature of the sign minus. It is
merely an opposition to the sign plus ; it is counting in another
direction^ and if we call the degrees north of the equator jy/w5.
3a ELEMENTARY ALGEBRA.
we must call those south of it minus, taking the equator as
the zero line.
So it is on the thermometer scale, the divisions above zero
are called pltcs, those below minus. Money due to us may be
called plus, money that we owe should then be called minus,
— the one circumstance is directly opposite in effect to the
other. Indeed, we can conceive of no quantity less than nothing^
as we sometimes express ourselves. It is quantity in oppo-
site circumstances or counted in an opposite direction ; hence
the difference or space between a positive and a negative quantity
is their apparent sum.
As a further illustration of finding diflferences, let us take
the following examples, which all can understand :
From . 16 16 16 16 16 16
Take .12 8 2 o — 2 -—4
Differ. . 4 8 14 16 Ts 20
Here the reader should strictly observe that the smaller
the number we take away, the greater the remainder, and
when the subtrahend becomes minus, its numeral value must
be added.
(Art. 7.) We cannot take a greater quantity from a less;
but we can, in all cases, Jind the difference between any two
quantities, and if we conceive a greater quantity taken from
a less, the diflference cannot be positive, but must be negative,
i. e. minus.
EXAMPLES.
From
12
12 12
12
12
12
Take .
30
20 16
12
10
6
Diflfer. .
—18
—8 —4
0
2
6
(Art. 8.) When we take any quantity from zero, the differ-
ence wUl he the same quantity with its sigri changed, as will be
obvious from the following examples :
SUBTRA.CTION.
{
From .
. 10a
3a 0
0
~5a
Take .
. 11a
6a 6a
—6a
—6a
Differ. .
. — a
— a — 6a
■^6a
a
(Art. 9.) Unlike quantities cannot be written in one sum,
(Art. 1,) but must be taken one after another with their
proper signs : therefore, the difference of unlike quantities
can only be expressed by signs. Thus, the difference be-
tween a and b is a — b, a positive quantity if a is greater than
b, otherwise it is negative. From a take b — c, (observe that
they are unlike quantities.)
OPERATION.
From a-f-O+O
Take 0-\-b—c
Remainder, or difference, a — b -\rc
This formal manner of operation may be dispensed with ;
the ciphers need not be written, and the signs of the subtra-
hend need only be changed.
From the preceding observation, we draw the following
GENERAL RULE FOR FINDING THE DIFFERENCE BETWEEN ALGE-
BRAIC QUANTITIES.
Rule . — Write the terms of the suUrahend, one after another,
with their signs changed ; and then unite terms, as far as j^ossl-
ble, by the rules of addition.
Or we may give the rule in the following words :
Conceive the signs in the subtrahend to be changed, and then
proceed as in addition.
EXAMPLES.
(1) (2) (3)
From . . . 4a-j-2.c — 3c 3aa;+2y a-\-h
Take . . . a-\-^x — 6c xy — 2y a — b
Remainder, . 3a,— 2a;4-3c 3aa: — xy-\-^y 26
34 ELEMENTARY ALGEBRA.
(4) (6) (6)
From . . ^^—^x-\-f 7a-}-2—5c ix-\-\y
Take . . — a? — Ax-\-a — a*\-'2.-\- c '^x — \y
Rem. . . 2>x'-\-x^f—a 8a * —^c 'y
(7) (8) ,
From . . . ^2^ — 2>xy-\-2'f-\- c ax-\-bx-\-cx
Take . . . x" — Qxy-\-'2>y" — 2c x-\-ax-\-hx
Difference, . lx^-^2,xy— y^-\-^c Jc—\)x
9. Find the difference between ^xy — 20 and — xy-\-^^'
Ans.^9xy — 32.
10. Find tlie difference between '7arx-\-a and 3a^x — 2a.
Ans. 4a^x-j-3a.
11. Find the difference between — Bx — 2y-\-3 and 10a;
— 3y+4. Ans. — 18^+y— 1.
12. Find the difference between 6y^ — 2y — 5 and — 83/^
— 5y4-12. Am. 14?/+3y— 17.
13. From 13aV/+lla— Sa^-f-G^*,
Take '7a—5a'^-\-6b—10a'b^ Remainder, 23a^b''-{-4a.
14. From 3a+b+c—d—10,
Take c-{-2a—d. Rem. a4-^— 10.
15. From 3a+^.-l-c— c?— 10,
Take 5— 19+ 3a. Rem. c— c/+9.
16. From 2ab-\-b^—4c-\-bc—b,
Take 3a:'—c-{-b\ Rem. 2ab—3c+bc— 30"— b.
17. From a!'+3b^c+ah^—abc,
Take P-\-ab^—abc. Rem. a"+3b''c—b\
(Art. 10.) From a take 5. The result is a — b. The
minus sign here shows that the operation has been performed :
b was positive before the subtraction ; changing the sign per-
formed the subtraction; so changing the sign of any other
quantity would subtract it.
SUBTRACTION S!^
18. From 3a take (ab-j-x — c — ?/), considering the terms in
the vinculum as one term, the difference must be 3a — (ab-]rx
— c — ^y), but if we subtract this quantity, not as a whole, but
term by term, the remainder must be 3a — ab — x-{-c-\-y.
That iSf when the vinculum is taken away, all the signs with-
in the vinculum must be changed.
EXAMPLES.
1. From 2>0xy, take (40a^— SS^+Sc— 4c?).
Kern. 9,b^—\0xy—3c-\-Ad.
From 3a2, take (3a — x-{-b). Rem. Sa^ — 3a-{-x—b.
From c? — a, take 4a — y — 3a^- — 1.
Rem. a^ — a — (4a — y — 3a^ — 1).
Or Aa^—5a-^y-\-\,
From a-\-b, take a — b.
From 4a4-4&, take b-\-a.
From 4a — 46, take 3a-l-5J.
From 8a — \9,x, take 4a — 3x.
(Art. 11.) It will be a useful exercise for the mind to look
at the principle of subtraction in Algebra, through the medi-
um of equations.
If we subtract 12 from 18, the remainder will be 6. Here
are three quantities.
1 . The minuend 1 8
2. The subtrahend 12
3. The remainder
In all cases, the remainder and the subtrahend, added to-
gether, must equal the minuend. N'ow let us suppose that
we do not know the value of the remainder, and, therefore,
represent it by the letter R. Then by the nature of the
case we have 72-|-12 = 18, an equation.
36 ELEMENTARY ALGEBRA.
Taking equals from equal quantities, that is, 12 from both
members of the equation, we have
Ii=6
Now let us take the third example under the last rule, and
call its remainder H.
Then we have Ji-\-a — b=a-\-b
Rejecting a from both members, and adding h, or (what is
the same thing), transposing — 5, (see page 16), and we find
R=^2b
Take example 9, and we have
R—xy-{- 1 2 = Sary— 20
By transposition, Ii=9xy — 32
In this manner we ma?/ perform all the examples in sub-
traction ; and in this manner perform the following examples :
From 2a-\-2b, take — a — b.
From ax-\-bx, take ax — bx.
From a-]rc-[-b, take a+c — b.
From Sx-\-2y-\-2, take 5x-]-Sy-{-b.
From 6a-\-2x-{-c, take 5a-]rGx — 3c.
MULTIPLICATION.
(Art. 12.) The nature of multiplication is the same in
Arithmetic and Algebra. It is repeating one quantity as
many times as there are units in another ; the two quantities
may be called factors, and in abstract quantities either may
be called the multiplicand ; the other of course will be the
multiplier.
Thus, 4X5. It is indifferent whether we consider 4 re-
peated 5 times, or 6 repeated 4 times ; that is, it is indififerent
MULTIPLICATION. S7.
which we call the multiplier. Let a represent 4, and b repre-
sent 5, then the product is aXb; or with letters we may omit
the sign, and the product will be simply ab.
The product of any number of letters, as a, b, c, d, is abed.
The product of x, y, z, is xi/z.
In the product it is no matter in what order the letters are
placed, xy and yx is the same product.
The product oi axXby is axby or abxy. Now suppose a=6
and 5=8, then a5=48, and the product of axXby would be
the same as the product of 6a;X8y or 48a^. From this we
draw the following rule for multiplying simple quantities,
which may be called
Case 1 . Multiply the coefficients together y and annex the let-
ters, one after another, to tlie product.
EXAMPLES.
1. Multiply 3a; by 7a Product 21 oar.
2. Multiply Ay by ^ab Product l^aby.
3. Multiply 36 by 5c, and that product by 10a;.
Ans. \5Qbcx.
4. Multiply Qax by \9,by by lad. Ans. 504aaxydb.
5. Multiply Sax by 76 by 3y. Ans. 63abxy.
6. Multiply lOOaxy by lOa^cy by 2. Ans. 2000aabcxyy.
In the preceding examples no signs were expressed, and of
course plus was understood as belonging to every factor;
and a positive quantity, taken any number of times, must of
course be positive.
(Art. 13.) As algebraic quantities are liable to be affected
by negative signs, we must investigate the products arising
from them. Let it be required to multiply — 4 by 3, that is,
repeat the negative quantity 3 times, the whole must be neg-
ative, became a negative quantity taken any number of times
must be negative. Hence minus multiplied by plus gives
38 ELEMENTARY ALGEBRA.
minus, — aXh gives — ah; also a multiplied by — h must
give — ah, as we may conceive the minus h repeated a times.
!N"ow let us require the product of — 4 into — 3.
In all cases the multiplier shows how many times the mul-
tiplicand must be taken; — when the multiplier is plus, it
shows that the multiplicand must be added to zero as many
times as there are units in the multiplier; — when the mul-
tiplier is minus, it shows that the multiplicana muse ue suo-
traded from zero as many times as there are units in the
multiplier.
But to subtract — 4 from zero once, gives +4, (Art. 8,)
and to subtract it 3 times as the — 3 indicates, gives +12.
That is, minus multiplied into minus, gives plus.
This principle is so important that we give another mode
of illustrating it, making use of the following example.*
Required the product of a — h by a — c.
Here a — h must be repeated a — c times.
* There is also another method of showing that minus multiplied
into minus, must give plus ; and it rests on the principle that a times
0 gives 0 for a product, or 0 times any quantity must give 0. In short,
the product of two factors must be zero, if either one of them is zero.
Suppose we multiply . . a — a
By b
The product is .... ah— ah
Here a — a is in value 0. So in the product ah — ah is 0, as it should
be, and the whole subject is, thus far, very clear.
Now supppose we take
And multiply by ... — h
The product is . . . —HLh-\-ah
The first part of the product is clearly — ah, and the whole must be
zero ; therefore we must take the second part, -\-ah, to destroy the first,
that is, — h multiplied by ^—a, gives -\-ah.
The objection to this method is, that the reasoning at the last point is
rather m^clianical than intellectual ; we are forced to take ah as plus to
make a definite sum, giving no decided metaphysical reason that it
must be so.
MULTIPLICATION. 39
If we take a — 5, a times, we shall have too large a product,
as the multiplier a is to be diminished by c.
That is a—h
Multiplied by a
Gives . . . aa — ahy which is too great by a — 5 repeated
c times, or by ac — c5, which must be subtracted from the
former product ; but to subtract we change signs, (Art. 5,)
therefore the true product must be aa — ah — ac-{-cb.
That is, the product of minus h, by minus c, gives plus be,
and, in general, minus multiplied hy minus gives plus.
But plus quantities multiplied by plus give plus, and minus
by plus, or plus by minus, give minus; therefore we may
say, in short.
Thai quantities affected hy like signs, when multiplied together,
give plus, and when affected hy unlike signs, give minus.
(Art. 14.) The product of a into b can only be expressed
by ah or ha. The product of a, b, c, d, <fec., is ahcd; but if
b, c, and d are each equal to a, the product would be aaaa.
The product of aa into aaa is aaaaa; but for the sake of
brevity and convenience, in place of writing aaa, we write a^.
The figure on the right of the letter shows how many times
the letter is taken as a factor, and is called an exponent. The
product of a? into a* is a repeated 3 times as a factor, and 4
times as a factor, in all 7 times ; that is, write the letter and
add the exponents.
EXAMPLES.
What is the product of a^ by a^ ?
What is the product of rr"* by x^ ?
What is the product of 3/^ by y* by j^^
What is the product of a" by a^ ?
What is the product of 6V by bx ?
What is the product ©f <w by oc^ by aV ?
Ans.
a\
Ans.
x'\
9
Ans.
f\
Ans.
QU+m^
Ans.
b'x\
.¥?
Ans.
a'<?.
40 ELEMENTARY ALGEBRA.
What is the product of ar^ by a:^ by ar* ? . . Arts. sP.
What is the product of a^ by x^ by a;"* ? . Ans. a^+^ ^^.
What is the product of S:t^ hj2x^hj2'> . Ans. 12ar\
Find the product in each of the following examples :
4ac 9a^c — 3xy — 2xi/
— 3(ib — 4ay +9a:y — 6xi/
Product
-laij 210a;y 40rt —21p
3x7/ — Say 20pq — 3r
* (Art. 15.) The preceding examples are sufficient to illus-
trate the multiplication of simple factors — we now proceed to
Case 2 . When one of the factors is a compound quantity,
we have the following
Rule . — Multiply every term of the multijolicand, or com-
pound quardity, separately, hy the multiplier, a^ in the former
case; placing the products one after another, with the p)foper
signs; and the result will he the whole product required.
The reason of this rule is obvious from Case 1 .
EXAMPLES.
(1) (2) (3)
5a — 3c 3ac — 4^ 2a* — 3c+5
2a 3a he
10a*— 6ac 2ah—\^ah ^a^bc—3be-\-bbc
(4) (6) (6)
l,2ar — 2ac 26c — Ih Ax — h-^3ah
Aa — 2a 2aJ
MULTIPLICATION. 41
(7) (8)
3(?-^x 10^—3/
(9)
Axy — 4x^
^ax"
10. Multiply 36— 2c by 56. . .
. . Ans. 156^— 106c.
11. Multiply Axy—d by Qx. .
. . Ans. 24ar''y — 3'ix.
12. Multiply a^— 2a;-M by A:t^.
Ans. 4a2a;2_8^3_|_4^2^
13. Multiply llaW— 13^y by 2,ax.
Ans. SSa'bc'x—S9ax^y.
14. Multiply 42c2— 1 by —4.
^725. — 168c2+4.
15. Multiply —2>Q)a^ha^y-\-\2, by
16. Multiply 26— 7a— 3 by 4a6
—5a\ ...
Ans. -{-150a^bx^y~65aK
Ans. Qab^—2Qa^b—'i2ab.
17. Multiply a+36— 2c by — 3a6.
Ans. — 3a26— 9a6'+6a6c.
18. Multiply 13a'— 62c by —4c. Ans. —52ah+4b'c\
19. Multiply ISary- 36 by —25^:2^
Ans. —325x^y—75bx\
Case 3 . When both the factors are comjjound quantiiies,
we have the following
Rule . — Multiply" every term of the multiplicand by evei-y
term of the multiplier, separately ; setting down the products one
after or under another, with their proper signs ; and add the
several lines of products all together for the whole product
required.
EXAMPLES.
(1) (2)
Multiply . . 2a-}-36 Qa^y —2z
By . . . . a-f ^ Sax —bd
Product by a 2a^-\-Sab
Product by 6 2a6-|-362
Entire product 2a'-|-5a6-f-36'
4
1 ^aa^y—Qaxz—SQ^dxy^ 1 Qdz
42
ELEMENTARY ALGEBRA.
3. Multiply a-\-h-{-c by x-{-y-{-x, that is, repeat a-\-h-\-c, x
times, then y times, then z times, and the operation stands
thus:
a-\-h-]rc
x-\-y-\-z
Product by x
Product by y
Product by s
Entire product
4. Multiply
By
Partial product
2d partial product
Whole product
5. Multiply 3a2
ax-\-hx-\-cx
ay^hyArcy
az-\-bz-\-cz
ax-\-bx-\-cx-\-ay-\-by-{-cy-\raz-{-bz-{-cz.
2a^'-{-xy — 2y^
3x — 3y
ex^-j-Sx'y—Gxf
—Sx'y—SxfJf-ef
ex'—^x^y—dxy'^+Gf
■2ab—b'^ by 2a— 45.
Frod. 6a^—16a^b-{-6ab'^+4b\
6. Multiply x^ — xy-\-'f by a;-hy. . . . Prod. aP-\-y^.
7. Multiply 3a-l-4c by 2a — 5c. Arts. Ga^ — 7ac — 20c^.
8. Multiply a^-j-ay — y^ by a — y. Ans. a? — 2ay'^-\-y^.
9. Multiply a^-\-ay-\-'f by a — y.
10. Multiply a^ — ay+y^ by a-\-y.
11. Multiply a'^+aV+ay^+y^ by a — y.
12. Multiply f — y-[-\ by y+1.
13. Multiply a^-Vf by ar^— /. .
14. Multiply a2_3a_|_8 by a+3. ,
15. Multiply 5^4-5V-fa;^ by i^— a:^^
16. Multiply rt'^+i"* by a+Z*.
17. Multiply a;^+a;'*+^ by or — 1. . Ans. a? — x^.
18. Multiply 771+71 by ^m — 9w. . . Ans. 9m^ — 9n^.
Ans. a^ — y^.
Ans. a'+y*.
Ans. a^ — y*.
Ans. y^-j- 1 .
An^. x^ — y'^.
Ans.
5-1-24.
A71S. b^ — x^..
MULTIPLICATION. 43
19. Multiply 2/2_20 by t/'-^+SO.
20. Multiply a+5 by a-\-b. .
21. Multiply rr+y by x-\-y. .
22. Multiply a — h by a — h. .
23. Multiply x — y by x — y. .
-4n5. y^ — 400
Ans. a2+2a5+62
Arts. ^Ar^xy-T'if-
Ans. c? — ^ah-\-W-
Ans. a? — 9,xy-\-'t^
(Art. 16.) When a number is multiplied by itself, the
product is called its square, the square of one of the fac-
tors, and by inspecting the last four examples, we perceive
that the square of any binomial quantity, (that is, the square
of any two terms connected together by the sign plus or
minus), the result must be the squares of the two parts,
AND twice the PRODUCT OF THE TWO PARTS.
N. B. The product of the two parts will be plus or minus,
according to the sign between the terms of the binomial.
By this summary process perform the following examples :
1. Square (3a+5) or multiply this quantity, by itself con-
sidered as a numeral quantity.* Ans. ^d?-\-Qah-\-l?.
2. Square 2a; — y. Ans. 4x'^ — 4xy+y^.
We write the product, in the second place, in the answer,
because it naturally falls there when the multiplication is
formally made ; but this is not essential.
Write out the following squares as indicated by ike exponent.
(a--3cy=a'—6ac-\-9c^ .
(3a__c)2=9a2_l-6ac+c2
(2a;4-3y)==4a;2-{- 1 2:ry+92/2
(20x-]-yy= 400a;2-f 40xy-]-7/
* We make this last remark because things, arithmetically, cannot ])e
multiplied by things. For instance, dollars cannot be multiplied by
dollars, &c. In fact, every multiplier is always a number ; and when
we demand the square or any other power of a quantity, it always
means the power of its numeral value considered abstractly.
44 ELEMENTARY ALGEBRA.
(Art. 17.) The product of the sum and difference of
two quantities is equal to the difference of their squares,
as will be seen by inspecting the followiny products :
The first example should be multiplied in full to establish
the principle.
What is the product of {a-\-b) by (a — h) ?
Ans. a? — i^.
What is the product of 2m4-2w by 2m — 2« ?
Ans. 4m? — 4n^.
What is the product of x-\-i/ by x — ^y? Ans, x^ — y^.
What is the product of %x-\-2>y by 3a; — 3y ?
Ans. 92;2— 92/2.
What is the product of la-\-b by la — b ?
Ans. 49a2— 52.
What is the product of 1 + lOa by 1— 10a?
Ans. 1—1 OOa^.
Observation. — By attention to principles much labor may be
saved in the common operations of Algebra. For instance,
if the product of three equal binomial factors were required,
as (x'\-^)[x-\-2>){x-\-^), we may first write out the product
of two of those factors. by (Art. 16); then multiply that
product by the other factor.
Thus, .... x''-\-e>x-\-9
a: +3
?+^6?+9a:
3a:2_j-i8a:-f-27
Product . . . "^9^+27^+27
If the product of the four factors, {x — A){x — 5)(a:-}-4)
(^x-\-b), were required, we would take the product of ihe first
and third factors, then of the second and fourth, by (Art.
17), then the product of those two products would be the
final product required.
DIVISION.
45
Thus, the required product is the product of (x^ — 16) by
(a;2_25) =2:^—4 la;2-l-400.
What is the product of (a-\-c)(a-{-d)(a — c)(a — d) ?
Ans. a^ — a^c^ — a^d^-{-cW.
DIVISION.
(Art. 18.) Division is the converse of multiplication, the
product being called a dividend, and one of the factors a
divisor. If a multiplied by b give the product ab, then ab
divided by a must give b for a quotient, and if divided by 5,
give a. In short, if one simple quantity is to be divided by
another simple quantity, the quotient must be found, by in-
sjpection, as in division of numbers.
EXAMPLES.
1. Divide 1 6a5 by 4a *Ans.
4b.
2. Divide 2\acd by 7c. .
, . Ans.
Sad.
3. Divide ab^c by ac.
. Ans.
b\
4. Divide 6abc by, 2c.
. Ans.
3ab.
6. Divide cui^ by ax^.
, Ans.
X.
6. Divide Smx^ by mx. .
. Ans.
Z'.'^.
7. Divide 2\0c^b by 7cb. .
, Ans.
SOc\
8. Divide 42a^ hy xy. , ,
. Ans.
42.
9. Divide 3xy by ax.
, Ans.
^1.
a
* The term quotient would be more exact and technical here ; but,
U.'. results hereafter, we shall invariably use the term Ans., as more brief
and elegant, and it is equally well understood.
46
ELEMENTARY ALGEBRA.
Remark. — In this last example we cast out the equal
factor X from both the dividend and divisor, and set the other
factor a of the divisor under the dividend as a denominator.
(Art. 19.) When the dividend and divisor have no factors
in common, we can only indicate the division by setting the
divisor under the dividend for a denominator, as in the follow-
ing example :
Divide Zahc by 2xy,
. . . Ans,
Sabc
2xy
Divide Aaxy by Zay,
. , . Ans.
4x
Divide 2)Qaby by 4a5y. .
. . . Ans.
9.
Divide 9.1 aby by Wabx.
. . . Ans.
27y
llx
Divide 19J}''x by 2,abx.
/ * f\r\ \ T_ J.!. - J
. . . Ans.
9b
a
(Art. 20.) In the preceding examples no signs were ex-
pressed, and, of course, every term and every factor is under-
stood to be positive ; but as algebraic quantities may have
negative signs, and unlike signs, we must investigate and
decide upon the sign to prefix to the quotient. This can be
done by merely observing what sign must be put to the quo-
tient so that the product of the divisor and quotient will give
the same sign as in the dividend, according to the principles
laid down in multiplication, (Art. 13).
For example, divide — 9y by 3y, the quotient must be — 3 ;
so that 3y multiplied by — 3 will give — 9y, the dividend.
Divide — Py by — 3y, the quotient must be ~\-3.
Divide 4-9y by — 3y, the quotient must be — 3.
From these examples we draw the following rule for the
signs :
Rule. — When the dividend and divisor have like signs,
hoik -j- or both — , then the quotient must be plus.
DIVISION. 47
When the dividend and divisor have unlike signs, the quotient
must he minus.*
EXAMPLES,
Divide — 2 lac by — la Ans. +3c.
Divide — \9,xy by 4-3y Ans. — 4a:.
Divide I'^abc by — 8c Ans. — 9aZ>.
Divide \Umn by +8ac Ans. —^.
• * N o T E . — We address this note to those only who are fond
of the Tnetaphysique of science. Division, considered in its most
elementary sense, is not merely the converse of multiplication ;
it is a short process of finding how many times one quantity
can be subtracted from another of the same kind. When the
suhraction is possible, and diminishes the humeral value of the
minuend, and brings it nearer to zero, the operation is real and
must be marked plixs. When the subtraction is not possible
without going farther from zero, we must take the converse oper-
ation, and the converse operation we must mark minus.
Thus, divide 18a by 6a. Here, it is proposed to find how many
times 6a can be subtracted from 18a ; and as we can actually sub-
tract it 3 times, the quotient must be -|-3.
Divide — 18a by — 6a. Here, again, the subtraction can actu-
ally be performed, and the number of times is 3, and, of course,
the quotient is -|-3.
Divide — 18a by 6a. Here, subtraction will not reduce the
dividend to zero ; but addition will, and must be performed 3
times ; but the operation is the converse of the one proposed,
and therefore must be marked by the converse sign to plus, that
is —3.
Again, divide 18a by — 6a. Here, if we sub, — Qa it will
not reduce 18a ; but the converse operation will, and therefore
the quotient must be minus, that is, — 3.
Now let us inspect the common operation of division, by the
help of the following example : Divide 24 by 8. Let the oper-
ation stand thus :
48 ELEMENTARY ALGEBRA.
(Art. 21.) The product of a^ into a^ is a^ (Art. 14), that
is, in multiplication we add the exponents ; and as division is
the converse of multiplication, to divide powers of the same
letter, we must subtract the expone7it of the divisor from that of
the dividend.
1. Divide 2a^ by a'' Ans. 2a
2. Divide — a' by a^ Ans. — c
.3. Divide \6x^ hj 4x Ans. 4a^.
4. Divide IBaxy^ by — Say Ans. — 5xi/^
5. Divide 63a"* by 7a" Ans. 9a'"-''
6. Divide I2ax^ by — 3aa; Ans. — 4x'^~^
7. Divide 28ay by 4ac2^ Ans. 1^
c
Divisor. Divi. Quotient.
8)24(3
The product of the divisor and quotient, in all cases, equals
the dividend. Let d represent any divisor, D any dividend, and q
the corresponding quotient, then
dq=D
Or d^jE^
9
In the above numeral example, let us suppose the divisor 8 to
be — 8, and the quotient — 3. Then the dividend must be the
product of ( — 8)X( — 3); but suppose that we do not know
whether this is plus or minus, we will therefore represent it by D.
Then (— 8)(— 3)=Z>
By dividing both members by either factor, as ( — 3), we have
-8=A or J?_= -8
—3 —3
Here D cannot be minus, for minus divided by minus must give
plus in the quotient ; (as we have just determined in this note),
but the quotient is actually ( — 8), therefore D must be plus. That
is, the product of minus into minus gives plus ; corresponding to
(Art. 13).
DIVISION. 40
8. Divide — ISa^x by — 6ax. . . . Ans. 3a*,
9. Divide ^acdxif- by SLadxif-. . . . Atis, 3c.
10. Divide ^b{a—xf by \b{a—xf. . Ans. 3(a— a;).
In this last example consider {a — x) as one quantity.
11. Divide 45y^ by \tAf- Ans. 3y.
Examples 10 and 11 are exactly alike, if we conceive
(a — x) equal to y.
12. Divide 12aV by — ^o^x Ans. — 4a?.
13. Divide \bmf- by Say Ans. — by.
14. Divide — -18aa:^y by — 8ac2. . . . Ans. ^,
15. Divide 7a^b by 2l;^b\ . .
16. Divide —5a^a^ by — 7aV.
17. Divide 117a'bY by TSa^^c^
18. Divide ^ hj ^ Ans.
19. Divide (x—^f by (a;— y)'. . . . Ans. (x-^yf.
Observe, that example 18 and 19 are essentially alike.
20. Divide {a-\-by by (a+b). . . . Ans. (a+b)*.
21. Divide (a—cy'* by (a — c)». . . Ans. (a — c)'»-«.
To perform example 21 we adhere to the principle of per-
forming 18, 19, and 20.
(Art. 22.) In the process of division, exponents may be-
come negative, and it is the object of this article to explain
their import.
To explain this, take a* and divide successively by a, form-
ing the following series of quotients :
Ans
7a^b
Slla'b''
1
^?iab
•
Ans.
Ans.
5
70,^'
36'
2c'
,
Ans.
'^.
5
cf, a, 1, _, _^, , dfc.
1 -2,
1 -'
_=a '
a"-
a«
50 ELEMENTARY ALGEBRA.
Divide a'* successively by a again, rigidly adhering to the
principle that to divide any power of a by a, the exponent
becomes one less, and we have
a', a', aS a", a~S a~^ a~^, &c.
Now these quotients must be equal, that is, a? in one series
equals a' in another, and
1 ~*
a^=c^y a=a}, l=a°, -=a ,
a
Another illustration. We divide exponential quantities by
subtracting the exponent of the divisor from the exponent of
the dividend. Thus, d^ divided by a^ gives a quotient of
c^^=a^. aP divided by a''= a^~'=a~^. We can also divide
by taking the dividend for a numerator and the divisor for a
c^ 1 1
denominator, thus, _=_, therefore, _=a~^ (Axiom 7).
From this we learn, that exponential terms may be changed
frum a numerator to a denominator, and the reverse ^by changing
the signs of the exponents.
Thus, %==a^^ -^::=-l3 ^=:.-
Divide cc'hc by c^h^c'^ Ans. a'^lr^c.
Observe, that to divide is to subtract the exponents.
Divide aV by aVy^. . . . Ans. or a~^or^y~^.
a^xy"^
3 3
Divide 2)ay'^ by bc^x^y"^. . . . Ans. — ^— or -a~^x~^,
(Art. 23.) When the dividend is a compound quantity,
and the divisor a simple (or single) quantity, we have the fol-
lowing rule, the reason of which will be obvious if the pre-
ceding part of division has been comprehended.
Rule. — Divide each lei'm of the dividend by the divisor,
and the several results connected together by their proper signs
wUl be the quotient sought.
DIVISION 51
EXAMPLES.
1. Divide \5ab — l^ax hj 3a Ans. 5b — 4x,
2. Divide — 25a^x-\-15aa^ by — Box. . . Ans. 5a — 3x.
3. Divide 10ab-\'15ac hj 5a Ans. 2^-1- 3c.
4. Divide SOax — 54x by 6a: Ans. 5a — 9.
5. Divide 8a;3-j-12a:2 |jy 4^ ^^^_ 2^:4-3,
6. Divide 3bcd-{-\2bcx — %h by 35c. . Ans. d-\-Ax — 3h,
7. Divide lax-\-3ay — 75c? by — lad,
S Am. — ^_?2^-{-i
d Id a
8. Divide 3ao^-\-Qx^-\-3ax — 15a; by 3a;.
Ans. a.7^-{'2x-\-a — 5.
9. Divide 3a5c+12a5a; — 3a^5 by 3a5. Ans. c-\-4x — a.
10. Divide 25a^bx — 15a^ca;^-t-5a5c by — 5ax.
Ans. — 5a5-|-3aca; — bcx~^.
11. Divide 20a5^+15a62+10a5+5a by 5a.
Ans. 453+352+25-1-1.
(Art. 24.) We now come to the last and most important
operation in division, the division of one compound quantity
by another compound quantity.
The dividend may be considered a product of the divisor
into the yet unknown factor, the quotient ; and the highest
power of any letter in the product, or the now called divi-
dend, must be conceived to have been formed by the highest
power of the same letter in the divisor into the highest power
of that letter in the quotient. Therefore, both the divisor and
the dividend must be arranged according to the regular powers
of some letter.
After this, the truth of the following rule will become
obvious by its ^eat similarity to division in numbers.
Rule . — Divide the first term of the dividend by the first
term of the divisor, and set the result in the quotient.*
* Divide the first term of the dividend and of the remainders by the
first term of the divi8<>r ; b© not troubled about other term*.
52 ELEMENTARY ALGEBRA.
Multiply the whole divisor hy the quotient thus found, and
subtract the product from the dividetid.
The remainder will form a new dividend, with which proceed
as before, till the first term of the divisor is no longer contained
in the first term of the remainder.
The divisor and remainder, if there he a remainder, are then
to he written in the form of a fraction, as in division of numbers.
EXAMPLES.
Divide a^-{-2ah+h^ by a-\-b.
Here, a is the leading letter, standing first in both dividend
and divisor : hence no change of place is necessary.
OPERATION.
a-\-h)a''-\-2ab+h\a-\-h
a^-jr ah
ah-\-l>'
ab-\-h^
That the pupil may perceive the close connection between
multiplication and division, we
Multiply a^-\-2ab-\-4b^
By 2a^—2ah-\-b'^
2a'-\-4a:'h-{-Qa^h^ (1)
+d'b^-h2ab^-\-4b^ (3)
Prod, is 2a'-{-2a''h-\-5d'h^'-6ab'-{-4h*
Now take this product for a dividend, and one of the fac-
tors, (c^-{-2ah'{-4h^), for a divisor, and of course the other
factor, (2a^ — 2ah-\-h^), will be the quotient, and the operation
will stand thus r
DIVISION. 53
(1)
(2)
(3)
2a'-{-4a'b+8a'b^
—2a'b—3a^b^-
—2a^b—4a'b'-
-6ab^
-Qab^
a'b'+2ab'+4b'
a'b^-\-2ab'-{-4b*
The several partial products which make up the dividend,
'^ and marked (1), (2), (3), are again found in the operation of
division, and there marked (1), (2), (3), the same as in
Arithmetic.
Some operators put the divisor on the right* of the divi-
dend, as in the following example :
Divide a^ — b^ by a — b.
a — h
-b' ) a—b
a'b—ab^
ab'^—b^
ab^—b'
b-\^b\ Quo.
GENERAL EXAMPLES.
1. Divide a^+2aa;+ar* by a-^-x. Ans. a-\-x.
2. Divide a? — 3a?y-\-3ay'^ — y^ by a — y.
m Ans. a? — 2ay-l-3/'«
3. Divide 9,Aa''b—nahb'^—Qab by — 6a5.
Ans. — 4a-\-2a?cb-\-'[.
4. Divide a?-[-ba?b^-5ab'^-\-W by a+5.
Ans. a?-{-4ab-\-b'^.
* N o T E . — This is in imitation of the French, and being a mere matter
of taste, involving no principle, we have no right to find fault with those
who adopt it ; and others must not complain of us because we prefer
tlie English custom.
54 ELEMENTARY ALGEBRA.
5. Divide a^+^a'^b-\-2aL^+P by a^+ai+i^. Ans. a+L.
0. Divide ic^— 9a;2+27a;— 27 by ar— 3. Ans. x^—6x-{-d.
7. Divide 6a;*— 96 by 6x— 12. Ans. a^+2a;2+4a;+8.
(Art. 25.) When a factor appears in every term of both
dividend and divisor, it may be cast out of every term with-
out affecting the quotient ; thus, in the last example, the fac-
tor 6 may be cast out by division ; and x* — 16 divided by
X — 2 will give the same quotient as before.
8. Divide ea-'+Qa^— 15a by 3a^—3a.
Ans. 2a2+2a4-5.
(Observe Art. 25).
9. Divide 252r^— ar^— 2a;2— 8.^ by 5x^—4x.
Ans. 5a^+4a:2+3a:+2.
10. Divide ISa^—Sb^ by 6a-f 45. Ans. 3a— 2b.
11. Divide 22;^— 19a;2+26a;— 16 by x—S.
Ans. 2x^ — 3a;4-2.
12. Divide 2/'+l by y-\-l. Ans. y"^ — y^-\-y^ — y+1.
13. Divide /—I by y—\. Ans. f-\-y'^^-f-\-y^-Vy-\-\.
14. Divide ar^ — a^ by x — a. Ans. x-\-a.
15. Divide Qa?—3a~b—2a-^b by 3a2— 1. Ans. 2a— h.
16. Divide /— 3yV+3yV— a;*' by f—3y'^x-\-3ijx'^—x\
Ans. 2/'*+3y2a;4-3ya;2+ar3.
17. Divide 64a*6«— 25a=Z»« by ^a%^-\-bah\
Ans. Sa-b^-Sab^
18. Divide 2a'* — 2a;* by a—x.
Ans. 2a^-l-2a=^a;4-2aar'4-2ar».
19. Divide (a — xy by (a — xy. Ans. (a — xf.
20. Divide a^— 3a2a:4-3aa;2— a;^ by a—x.
Ans. a^ — 2aa;+a-^.
21. Divide a'^+1 by a+l. Ans. a*— a^'+a^— a-f-l.
22. Divide b^—\ by b—\. Ans. lf'^b''^b^-\-b--\-b-\-\.
DIVISION. 55
23. Divide 4Sa^—92a^x—40ax^-\-100x^ by 3a— 5x.
Ans. lCia^—4ax—20x^.
24. Divide 4d'—9d^-{-6d—l by ^d'^+Zd—l.
Ans. 2^2— 3c^4-l.
25. Divide 10ab+15ac by 2b-\-3c.
26. Divide SOao;— 54a; by 5a — 9.
27. Divide Sx^-^nx" by 2a;-f3.
28. Divide — 25a^x-{-15ax^ by 5a — 3x.
Observe that these last four examples are the same as some
m (Art. 23.)
If more examples are desired for practice, the examples in
multiplication may be taken. The product or answer may be
taken for a dividend, and either one of the factors for a divi-
sor ; the other will be a quotient.
Also, the examples in division may be changed to examples
in multiplication ; and these changes will serve to impress on
the mind of the pupil the close connection between these two
operations.
(Art. 26.) The operation of division is the art of finding
one of two factors of a product, when the product itself
and one factor is given. When the product only is pre-
sented, and its factors required, the operation is properly
called
FACTORING.
Factors of a number are such numbers as may be multi-
plied together to produce the number ; and factors of an al-
gebraic expression are such quantities as being multiplied
together will produce the expression. Thus, 2 and 3 are
the factors of 6, because 2X3=6, and 3, a, and c, are fac-
tors of 3ac, because by their multiplication they form that
product.
But some numbers /taw wo/adors, (except! and the number
56 ELEMENTARY ALGEBRA.
these should not be considered factors), and such numbers
are called prime numbers.
Also, some algebraic expressions have no factors, and such
expressions are called prime quantities. Thus, 5a-^c is a
prime quantity.
The following is a list of the prime numbers up to 100 :
1, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
All the other intermediate numbers are composite numbers.
Any number whatever is either a prime number, or composed
of the product of 2^rime factors.
Knowing this fact enables us to decompose any number into
its prime factors by the following rule :
Rule . — Divide the given number by any prime number that
will divide it without a remainder, and divide that quotient again
by any prime number that will exactly divide it, and so continue
until the last quotient is a prime number. The divisors and
last quotient are the factors required.
N. B. Use the smallest prime divisors first.
EXAMPLES.
1. Required the factors composing 102. Ans. 2, 3, 17.
OPERATION.
2)102
"~17
2. Find the prime factors in the number 112.
Ans. 2, 2, 2, 2, 7.
3. Find the prime factors in the number 126.
Ans. 2, 3, 3, 7.
4. Find the prime factors in the number 12769.
Ans. 113, 113.
FACTORING. 57
5. Find the prime factors in the number 1156.
Ans. 2, 2, 17, 17.
6. Find the prime factors in the number 1014.
Ans. 2, 3, 13, 13.
(Art. 27.) As the combination of numbers is endless, it is
impossible to give any definite rule that will decide in each
and every case -whether a number is a prime or a composite
number ; and as the practical utility of factoring is limited, it
is proper to confine our investigations to small numbers, and
from observations on numbers, we deduce the following prin-
ciples for finding these factors :
1st. That any number ending with an even number, or a
cipher, can be divided by 2.
2d. Any number ending with 5 or 0, is divisible by 5.
3d. If the right hand place of any number be 0, the
Avhole is divisible by 10 ; if there be two ciphers, it is divisi-
ble by 100; if three ciphers, by 1000, and so on, which is
only cutting off those ciphers.
4th. If the two right hand figures of any number be divis-
ible by 4, the whole is divisible by 4 ; and if the three right
hand figures be divisible by 8, the whole is divisible by 8,
and so on.
5th. If the sum of the digits in any number be divisible by
3 or by 9, the whole is divisible by 3 or by 9.
6th. If the right hand digit be even, and the sum of all
the digits be divisible by 6, then the whole is divisible by 6.
7th. A number is divisible by 11, when the sum of the 1st,
3d, 5th, &c., or all the odd places, is equal to the sum of the
2d, 4th, 6th, &c., or of all the even places of digits.
8th. If a number cannot be divided by some quantity less
than the square root of the same, that number is a prime,
or cannot be divided by any number whatever.
9th. All prime numbers, except 2 and 5, have either 1, 3,
58 ELEMENTARY ALGEBRA.
7, or 9, in the place of units; and all other numbers are
composite numbers, and can be divided.
(Art. 28.) The multiple of a number is some exact number
of times that number. Thus, 6 is a multiple of 2, 3 ; 12 is
also a multiple of 2, 3 ; but not so small a multiple as 6 is,
therefore 6 is the least common multiple of 2 and 3.
The least common multiple of several numbers is the least
number that is divisible by these numbers without a remainder,
A COMMON MULTIPLE IS FOUND BY MEANS OF PllIME FACTORS.
For example, find the least common multiple of the num-
bers 24, 20, and 15.
That is, find the least number which is divisible by 24, 20,
and 15. First find the prime factors to these numbers, (Art.
27), (2, 2, 2, 3,) (2, 2, 5,) (3, 5).
That the number required may be divisible by the first
number 24, it must have all the factors in that number; that
is, 2, 2, 2, 3 ; and to be divisible by the second number, 20,
it must contain the factor 5 ; putting in this factor we have
2, 2, 2, 3, 5. This number is divisible also by 1 5, because it
contains the factors 3, 5. The least common multiple re-
quired is, therefore, 120.
The least common multiple of the numbers 3, 7, 19, is
their product, because the numbers are prime, and there is
no common factor that can be cast out.
On these principles the following rule for finding the com-
mon multiple will be easily comprehended :
Rule. — Write the numbers one after the other, and draw a
line beneath them ; then, take any prime number which will di-
vide two or more of them vnthout remainder, and divide all the
numbers that will so divide — writing the quotients beneath, and cdl
the numbers that are not divisible by it. Find a prime numher
that will divide two or more numbers in this second line, and
proceed as before. Continue the operation until there are no two
FACTORING.
59
numbers left having a common divisor : then, multiply all the
divisors and remaining numbers together, and their product will
he the least common multiple sought.
EXAMPLES.
1 . Let it be required to find tlie least common multiple of
12, 15, 7, 18, 3, 5, and 35.
7
12,
15, '
r, 18,
3,
5,
35,
6
12,
15, ]
i, 18,
3,
5,
5,
3
12,
3, 1
I, 18,
3,
1,
1,
2
4,
1,
I, 6,
1,
1,
1,
2,
1,
I, 3,
1,
1,
1,
7X5X3X3X2X2=1260.
2. Find the least number that can be divided by 9, 1 2, 1 6,
24, 36, without remainders. Ans. 144.
3. Find the least number that is divisible by each of the
nine digits. Ans. 2520.
4. Find the least number divisible by 75, 50, 15, 20, 30,
and 45. Ans. 900.
(Art. 29.) A. prime quantitij in Algebra, like z, prime num-
ber, is divisible only by itself and unity. Thus, a, h, a-\-b,
are prime quantities ; and ah, and ab-\-ac, are composite quan-
tities, the first is composed of the factors a, h, and the other of
the factors a, and {b-\-c).
The prime factors of a purely algebraic quantity consisting
of a single term, are visible to the eye, and this is one of the
principal advantages of an algebraic expression.
Thus, in the expression obex, we perceive at once the prime
factors, a b, c, and x; the expression o?b'^x has three prime
factors, each equal to a, two prime factors equal to b, and one
equal to x.
60 ELEMENTARY ALGEBRa.
(Art. 30.) When the algebraic expression is a. polynomial y
and has prime factors that are monomials, such monomial
factors are visible, as in the following expressions :
Factors.
1. x-]rax (^\-\-a)x
2. am-\-an-]r(ix {m-\-n-\'x)a
3. bc^-^-hcx-^rhcy (^c-\-x-\-y)hc
4. 4a;2+6a:y (2a;+3y)2a:
Thus in the first expression, x is visible in every term, it
is, therefore, a common factor to every term, and ( 1 -\-a) is
the other factor, and the product of these two factors makes
the expression ; and so for the other expressions. •
The examples in division (Art. 23), are analagous to these,
except that in that article the divisor is given, and may not
be contained in every term, as in example 7, (Art. 23).
(Art. 31.) When all the prime factors composing any
algebraic expression consist of binomials or polynomials, they
are not visible in the expression like a monomial, and we can
find them only from our general knowledge of algebraic
expressions.
For instance, the prime factors in the expression (a^-|-2ad
'\-l?) we know to be (a+5) and {a-\-h) by (Art. 16), and
all other expressions that correspond to a binomial squared,
is immediately recognized after a little experience in algebraic
operations.
Also, any expression which is the difference of two squares,
as (a^ — 52^ jg instantly recognized as the product of the two
prime factors, («-!-&), and (a — h), (Art. 17).
The expression ax-^ay-\-hx-\-hy can be resolved into two
prime factors, by inspection, thus, a{x-\-y)-\-b{x-{-y) is
merely a change in the form of the expression. Now put
{x-\-y) — S. Then the next change is aS-]rbS; the next is
{a-\-b)S, Restoring the value of S, we have (a-\-b){x'{-y)
for the prime factors i:^ the original expression.
FACTORING. 61
(Art. 32.) Any trinomial expression in the form of ax^-{'
hx-\-c, can be resolved into two binomial factors ; but the art
of finding tlie factors is neither more nor less than resolving
an equation of the second degree, a subject of great impor-
tance and some difficulty, which will be examined very closely
in a subsequent part of this work ; therefore it is improper to
treat upon this subject at present. (See Art. 95).
(Art. 33.) Common multiple, and least common multiple,
have the same signification in Algebra as in Arithmetic, and
are found by the same rule, except changing the words num-
ber and numbers in the rule for quantity and quantities.
Or, we may take the following rule to find the least common
multiple in algebraic quantities.
Rule . — 1 . Resolve the numbers into their prime factors.
2. Select all the different factors which occur, observing, when
the same factor has different pmvers, to take the highest power.
3. Multiply together the factors thus selected, and their pro-
duct will be the least common multiple.
EXAMPLES.
1 . Find the least common multiple of Wx^y, and 1 9.a^I^x.
Resolving them into their prime factors,
8aVy=2'^Xa2Xa;2Xy
nan'x=^^''Xa^XxXb^X^
The different factors are ^^, d^, x^, y, b^, 3, and their product
is ^\c?ly^x'^y, which is the least common multiple required.
2. Required the least common multiple of 27a, 155, 9a5,
and 3a2. Ans. I'iba'^b.
3. Find the least common multiple of (a^ — x^), 4(a — x)^
(a+ar). Ans. M^a^ — x^),
4. Required the least common multiple of a?{a—^x), and
ax\Gi? — 7?), Ans. a^x^a? — x*).
62 ELEMENTARY ALGEBRA.
5. Required the least common multiple of x^(x — y), aV,
and 12axf. " Am. 12aV/(a;_y).
6. Required the least common multiple of 10aV(a — 5),
15a^(a-\-b), and nioJ^—l^). Ans. 60a^a^(a^—P),
ALGEBRAIC FEACTIONS.
The nature of fractions is the same, whether in Arithmetic
or Algebra, and of course those who understand fractions in
Arithmetic, can have no difficulty with the same subject in
Algebra.
(Art. 33.) A fraction is one quantity divided by another
when the division is indicated and not actually performed.
Hence every fraction consists of two parts, the dividend and
divisor, which take the name of numerator and denominator.
The numerator is written above a Une, and the denominator
below it, thus, _, and is read, a divided by h.
h
For illustration, we may consider any simple fraction as | ;
here we consider one or unity divided into 5 parts, and 3 of
these parts are taken. The 5 denotes the parts that the unit
is divided into, hence it is properly named denominator, and
the 3, numbers the parts taken, and is, therefore, properly
called the numerator. So in the fi-action -, b denotes the
b
parts into which unity is divided, and a shows the number of
parts taken.
In a numeral fraction, as |, it is evident that if we double
both numerator and denominator, we do not chanofe the value
of the fraetien ; thus> f is the same part of the whole unit as
ALGEBRAIC FRACTIONS. 63
|, and thus it would be if we multiplied by any other num-
ber ; and conversely, we may divide both numerator and de-
nominator by the same number, without changing tha value
of the fraction. Hence, if any fraction contains any factors
common to both numerator and denominator, we may sup-
press them by division, and thus reduce the terms of the frac-
tions to smaller quantities.
Hence, to reduce fractions to lower terms when possible,
we have the following rule :
Rule . — Divide both terms by their greatest common divisor.
Or, resolve the numerator and denominator into their prime fac-
tors, and then cancel those factors common to both terms.
EXAMPLES.
1. Reduce to its lowest terms.
21a6c-2
Here labc is the common divisor, and dividing according
2a5^
to the rule, gives , the fraction reduced.
3c
2. Reduce — ^_ Ans. ~.
55a^x^ 5x
3. Reduce to its lowest terms Ans.
18a5 33
4. Reduce H^^^ to its lowest terms. . . . Ans. ^l.
2\ax^ 3
6. Reduce il^^ to its lowest terms. . Ans. ?^.
^ rf A Bla^b—Q^a^b"^ , ., ,
6. Reduce — --— to its lowest terms.
2>Qa'b^—2ab
Ans.
12a'b-~3
7. Reduce i^!z:i^' to its lowest terms. Ans. K^ZI^.
3(a4-a:) 3
8. Reduce ^ ^ ■ to its lowest terms. . Ans. _— .
64 ELEMENTARY ALGEBRA.
^ \
9. Eeduce to its lowest terms. . . Ans.
X-
10. Reduce \ to its lowest terms. Ans. — ~ .
acx-f-abx ac-rob
11. Divide icV'+ar'y' by or^y-fory^. . . . Ans, ^.
a
12. Divide4a+45by2a*— 2S* Ans.—.
a — b
«
13. Divide %^ — 2n^ hj n^ — 4w-{"4. . . . Ans.
(Art. 34.) Fractions in Algebra, as in Arithmetic, may be
simple or complex, proper or improper, and the same defini-
tions to these terms should be given, as well as the same rules
of operation ; for in fact this part of Algebra is but a gener-
alization of Arithmetic, and in some cases we give arithmeti-
cal and algebraical examples side by side.
A mixed quantity in Algebra is an integer quantity and a
fraction ; and to reduce these to improper fractions, we have
the following rule :
Rule . — Multiply the integer hy the denominator of ihefrac-
tioUj and to the 'product add the numerator, or connect it with its
proper sign -\- or — ; then the denominator being set under
this suniy will give tJie improper fraction required*
EXAMPLES.
1. Reduce 2f and a-\-- to improper fractions.
Ans. V and ^±1.
These two operations, and the principle that governs them,
are exactly ahke.
2. Reduce 6} and a-f— to improper fractions.
A7i>. Y and 5i+:f!.
ALGEBRAIC FRACTIONS. 65
3. Reduce 7| and ax-\-- to improper fractions.
c
4. Reduce 3 — | and x^ — - to improper fractions.
Ans. 4 and ^tZ^.
y
6. Reduce y — \-\-ZIIK to a fractional form. Ans. ^ ^.
1+y y+1
6. Reduce x-\-y-\- to tlie form of a fraction.
x-{-y
Ans. ^+^^y+y^+g
h ' ^+^
7. Reduce 4+2a;-j-- to an improper fraction.
c
8. Reduce 5x — — Z_ to an improper fraction.
3^2 2Q
9. Reduce 3a — 9 — to a simple fraction.
a-1-3 ^
-4n5.
(a-l-37
The converse of this operation must be true, and, there-
fore, to reduce an improper fraction to a mixed quantity, we
have the following
Rule . — Divide the numerator by the denominator, as far as
possible, and set the remainder, (if any), over the denominator
for the fractional part ; the two joined together with their proper
sign, will be the mixed quantity sought.
EXAMPLES.
1 . Reduce y and to mixed quantities.
X
Ans. 5 1 and a-\-i*
2. Reduce \^ and —L — to mixed quantities.
bx
An^. 2| and a-\ — •
66 ELEMENTARY ALGEBRA.
3. Reduce — ^ "* to mixed quantities.
y
Ans. 5a-\-^ «
2^2 QJ2
4. Reduce 1- to a whole or mixed quantity.
a — b
Ans. 2a-{-2b.
5. Reduce lil — to a mixed quantity. Ans. 3a-\-—.
bdf- bet
6. Reduce ^ ~^^ ' to a mixed quantity.
a
Ans. a-\-h-\--~.
a
7. Reduce X to a mixed quantity,
4a
Ans. Sa-\-l——.
4a
(Art. 35.) A fraction is an expression for unperformed
division. Thus, 2 divided" by 5, is written |. The double of
this is f , 3 times f is |, &c. That is, to multiply a fraction
by any number, we multiply tlie nwnerator of the fraction hy
the nwmhery vnthovt changing the denominator.
The nature of division is the same, whatever numbers rep-
resent the dividend and divisor. Hence, for the sake of
simplicity, let us consider the result of dividing 24 by 6.
Here 24 is the dividend and 6 the divisor, and the division
expressed and unperformed, must be written \* , and the value
of this expression, or quotient, is 4. Now observe, that we
can double the quotient by doubling 24, or by taking the half
of 6. We can find 3 times the value of this quotient, by
multiplying the numerator 24 by 3, or hy dividing the denom-
inator 6 by 3.
Hence, to multiply a fraction by a whole number, we have
the following rule :
ALGEBRAIC FRACTIONS.
at
Rule . — Multiply the nvmeraior hy the whole number ; or,
when you can, divide the denominator by the whole nianber.
EXA
1. Multiply f by 5.
2. Multiply I by 3., .
3. Multiply If by 4. .
4. Multiply j\ by 100.
5. Multiply ij by 18.
6. Multiply i-f by 19.
7. Multiply 1 by 24. .
8. Multiply ii by 105.
9. Multiply f by 63. .
10. Multiply -^ by c.
b
(Art. 36.) When we multiply a fraction by its denomina-
tor, we merely suppress the denominator. Thus, multiply ^
by 3, the result is 1, the numerator of the fraction ; multiply
I by 5, and we have 2, the numerator for the product.
MPLES.
. Ans. y=2|
. Ans. -1=1^
Ans. |f=3ii.
. Ans. V4°=64f
, . . Ans. 7-J
. . . Ans. 5if
. \ Ans. 56
. . Ans. 85
. . Ans. 27
. . Ans. — .
EXAMPLES.
1. Multiply -f by 7 Ans. 3.
2. Multiply -hyb Ans. a.
b
3. Multiply yV by 1 1 ■^^' 4.
4. Multiply -^ by 11 Ans. x.
5. Multiply Jr by 55 Ans. Sax.
5b
St-
6. Multiply _ by 7 Ans. 3x.
7. Multiply ?^^ by 20.
Ans. 6a — 2x.
68 ELEMENTARY ALGEBRA.
8. Multiply ^l^^H^by 6bx, . . . Ans. 42ax—^b.
Sbx
9. Multiply t-\-- by 6 Ans. 3x+2x.
A) O
10. Multiply ?f +?f by 3 Am, 2ar+^.
o ^ 2,
U. Multiply 3| by 3 Ans. 10.
(Art. 37.) As a fraction is an expression for unperformed
division, we may express the division of %\ by 6|-, in tlie
following form :
But this is certainly a complex frcLction ; so are -1 and
complex fractions ; hence complex fractions may be defined
thus :
A complex fraction is (yne in which the numerator or denom.'
inator, or loth, are fractions or mixed quantities.
To simplify a ccmplex fraction, we multiply loth numerator
and denominator hy the denominators of the fractional parts :
or by their product, or by their least common multiple.
For example, let us simplify the fraction -?. If we mul-
tiply both numerator and denominator by 2, the numerator
will contain no fraction, and the result will be Multiply
3
numerator and denominator of this fraction by 3, and the
denominator will contain no fraction ; and the final result will
be _» a simple fraction, equal in value to the complex fraction.
But we could have arrived at this result at once, by multi-
plying both terms by 6, the product of 2»3. Hence, the rule
just given.
ALGEBRAIC FRACTIONS. 69
EXAMPLES.
1. Reduce -J to a simple fraction Ans» |f .
5
2. Reduce — to a simple fraction. . . . Ans. |=1|.
m
3. Reduce to a simple fraction. . Ans, . III — .
, c nbd — en
4. Reduce to a simple fraction. . . . Ans. -xt*
h 36
5. Reduce ifJif to a simple fraction. . . Ans. J^_f_.
y 4a;-f2y
a
6. Divide - by -, that is, simplify the complex fraction —'
b d _
d
Here the division is expressed, but unperformed, and by
the rule to simplify the fraction, we find its value to be — .
be
From this result -we can draw a rule for dividing one frac-
tion by another ; and the rule here indicated, when expressed
m words, is the rule commonly found in Arithmetic.
7. Simplify the fraction Ans.
, 1 ac-fl
8. Simplify the fraction Ans.
m n-{-m
rm ELEMENTARY ALGEBRA.
MULTIPLICATION OP FRACTIONS.
(Art. 38.) We have already given a rule to multiply a
fraction by a whole number; (Art. 35) ; but when two frac-
tions are multiplied together, the result is the same, whichever
we consider the multiplier. That is, | multiplied by 4, and
4 multiplied by |, is the same product. Also, - multiplied
b
by X is _— , therefore, x multiplied by - is also —. Hence to
b b b
multiply a quantity by a fraction, observe the following rule :
B u L E . — Multiply the quantity by the numerator of the frac-
iian, and set the denominator under the result.
EXAMPLES.
1. Multiply 7 by f.
. . . Ans. V-
2. Multiply a by -.
. . . Ans.^J..
y
3. Multiply 5 by 4.
. . . Ans. Y.
Now, in this last example write (me under the 5, which will
give it a fractional form without changing its value. Then it
will be f- X 4 ; and if we multiply the numerators together, and
the denominators together, we have Y» as before. Again,
we may take \ and multiply both numerator and denom-
inator by any number, say 3, and we have y, which is
really 5 as at first. We have now to multiply y by 4, and
if we multiply numerators and denominators as before, we
shall have |^ for the product, which is in value Y» as it
ought to be.
4. Multiply ahyi. . Prod. ^.
d d
MULTIPLICATION OF FRACTIONS. TL
5. Multiply ^ by 1 Frod. 1'.
la d
As a=^ we can thus change the form of the first factor
n
without changing its value, then the example will be to
Multiply!?? by £. .% Prod.'"^^^!,
n d nd d
From these examples we have the following rule for multi-
plying fractions together :
E, u L E. — Multijply the numerators together for a new nume-
rator ^ and the denominators, for a new denominator.
N. B. Equal factors in numerators and denominators may
be canceled out, which will save the reduction of the product
to lower terms.
To find such equal factors, separate the quantities into their
prime factors (Art. 27), before multipHcation.
EXAMPLES.
1. Multiply ?? by ^ Ans. ii^.
2. Multiply 2? by ?-^ Ans. -?.
^ ^ by ^ dx 5x
3. Multiply ?^ by ^ Ans. 1
^ "^ lOy "^ 9a; 6
A -KIT n- 1 « — ^ T, 25a; — 25 ,1 .5
4. Multiply by — — _ by . . Ans.
5 •' a'—b^ " x—\ a-\-h
In this example we separate the second fraction into its
prime factors, (Art. 27), and the operation stands thus :
a—h^ 25(a;— 1) ^ 1
5 {a-\-b){a—b) x—\
Suppressing all the factors which are found common in the
5
numerator and denominator, and the result is , ans.
a+b
T9
5. Multiply
ELEMENTARY ALGEBRA.
X
a-^x
, and — ^ together. Ans. -
ar a — x
6. Multiply ?? by 2^. . .
7. Multiply 1^^ by ?^. . .
8. Multiply ?f by ^X^'.
Ans, ^,
xz
Ans.
Atis.
4a
z
a
9. Multiply — - into — ^. . .
X z
10. Multiply -, —, ^ together.
11. Multiply (^±^ by ;^^ .
^•^30 ^ 3(a+x)
12. Multiply ?^±i^ by ??. .
13. What is the product of -—, l^ and ?? Ans. 12:j
y 2a X
14. What is the product of — _ into ^f^lt
^ 2>b-\-c 5ab
Ans.
X
9ax
2b
Ans. a^x
Ans. 9ax.
18
Ans. ^±5^
Ans.
4ac—2bc
I5b'-i-5bc
15. Multiply 5-I-?5 by 1 ..... Ans. ^^^fl.
^ 7,2
16. Multiply fZZ- by
be
X
Ans.
bh-^b^
17. Multiply ^!i:^ by ^ Am. i^Z^,
2y a-\-x ■ y
18. Multiply —t, JL. and -^ Ans. a.
X x-\-y X — y
MULTIPLICATION OF FRACTIONS.
19. Multiply 3a, fill, and ^ together.
2a a-i-b
20. Multiply ^"^""^^ by —2^.
^"^ 14 '' 2a^—3x
21. Multiply -Jf- by 1^^=??.
^•^ 6;^;— 10 •" 2a;
Ans.
Am.
73
3(a:^~-l)
2(a+6) •
3ax — 5a
4a;2~6"
Ans. £f.
22. Multiply ^- by i- Am,!,
DIVISION IN FRACTIONS.
(Art. 39.) When we multiply a fraction by a whole num-
ber, we multiply the numerator by that number, or divide
the denominator, (Art. 35) ; and as division is the converse
of multiplication, therefore, conversely, when we divide a
fraction by a whole number, we divide the numerator (when
possible), or multiply the denominator by that number.
Thus, -f divided by 3, would be |, and divided by 4, would
be -y^ ; jn the first case the division is actually performed, in
the second it is only expressed.
EXAMPLEJ
3.
1. Divide | by 3
Ans. J.
2. Divide f by 9
^^. ^.
3. Divide f by 6
..... Ans, \.
4. Divide j-f by 13
^^'^'
a. Divide tV by 8
^«*. iV
74 ELEMENTARY ALGEBRA.
6. Divide — by 3c Ans. . —
b ^ ch
In this example we divide first by 3, and that quotient by c.
7. Divide ?^by^ Ans. ^^.
(Art. 40.) Let us now consider division when the divisor
is a fraction. We must now go back to the elementary prin-
ciple of division. It is the art of discovering how many times
a number or qtuintity, called the divisor, can be subtracted from
another number or quantity of the same kind, called the dividend.
For example, we require the division of 6 by -J. The un-
disciplined and inconsiderate often understand this as demand-
ing the third of 6 ; but it is not so, it is demanding how many
times -J is contained in 6, or how many times -J can be svh-
traded from 6.
To. arrive at the true result, we consider that -J is contained
in 1 three times ; therefore, it must be contained in 6, 18 times.
Now, |- must be contained in 6, 9 times ; and we may arrive
6*3
at this result, thus,
2
Again, suppose we divide the number a by |.
The divisor | is contained in one unit 7 times, therefore, it
is contamed in a units, la times.
To make this more general, we will suppose the denomina-
tor of the divisor to be any other number as well as 7 ; there-
fore, suppose it n, the quotient will then be na. To make
the example still more general^ let us suppose a to be divided
by - , m being a wliol^ njimber.
n
The divisor can be resolved into two factors, - and m. Di-
n
viding a by the factor -, we have already shown the quotient
DIVISION IN FRACTIONS. 75
to be na ; dividing this by tbe whole number m, (Art. 39),
the residt must be — -.
m
This shows that when the divisor is a fraction, the quotient
is found hj the following rule :
Rule . — Multiply the dividend (whatever it may he) hy the
denominator of the divisor^ and divide thai product hy the
numerator.
In the result last given, let the dividend a be a fraction
_, aud in the place of a write _.
d\ ^ d
Then the problem will be to divide £. by —, that is, one frac-
d n
tion by another, and the result must be
c
n-
d
This is a complex fraction, and simplifying it (by Art. 37),
1 nc
we have —
md
From this result we draw the following rule for dividing
one fraction by another :
Rule . — Invert the terms of the divisor, and proceed as in
(Art. 41.) For the purpose of illustrating the nature of
an equation, and showing the power and simplicity of alge-
braic operations, we will arrive at this rule by another course
of reasoning.
Let us again consider the nature of division, and for this
purpose, divide 32 by 8.
Divisor. Dividend. Quotienl.
8 ) 32 ( 4
76 ELEMENTARY ALGEBRA.
Here it is visible that the product of the divisor and quo-
tient is equal to the dividend ; and this is a general principle,
true in every possible case.
Now let us divide - by -. There will be a certain quotient
b d
which we will represent by Q. Then the product of the
divisor and quotient will be equal to the dividend ; that is, we
shall have the following equation :
cQ a
Coth members of this equation are fractional, and if we
multiply the first member by d, the denominator, the product
will be the numerator, (Art. 35) ; but if we take d times one
member, we must take d times the other member, to preserve
equality. (Ax. 3).
Therefore, multiplying by d, we have
e* ad
Dividing both members by c, then Q will stand alone.
And «=^
oc
This equation shows that when we divide one fraction by
another, the value of the quotient is found by inverting the
terms of the divisor, and then multiplying the numerators
together for a new numerator, and the denominators together
for a new denominator ; or more briefly, we say
Invert the terms of the divisor, and proceed as in multiplication.
EXAMPLES.
1. Divide f by f Ans. \l=^\^.
DIVISION IN FRACTIONS.
77
2. Divide by _.
1— a "^ 5
3. Divide ?f by ^.
a5 a6
4. Divide . by —
ah ah
5. Divide 'l^^-t byf±f!f.
5a5/ 5as^
6. Divide ^ by _£_. .
c a-\-h
7. Divide ~ by -.
a c
8. Divide by ..
a — X (t — ^
Ans.
. Ans. —
3y
3a— 5
. Ans.
Ans.
Ans.
2d'
Trxj-a"
4b-\-ax
(a+hf
Operation,
15ah^(a-\.x)(a~^x)
a — X lOac
>... 3J(«+.)
2c
9. Divide
a^ — x^ a — X
. Ans. 2«+- .
10. Divide
14^-3 . lO.r-4
6 -^ 25
102'— 4
11. Divide
^^-^^by^. . .
. 9^—3
• 1 t Ans.
5 ^5 • '
x
12. Divide
60.-7, o:-!
A.. 1S.;-21
x+l ^3
x^—\
13. Divide
16a^ by ^"^
5 -^ 15 ' * *
. . . . Ans. 12a.
14. Divide
6,+4 3.-I-2 ^
, , . , Ans ^y.
5 -^ 4y
5
15. Divide
7-^ by ^^
.. 21
3 ^ 6
6a;
* Separate into factors wherever separation
is obvious.
78
ELEMENTARY ALGEBRA.
. 16. Divide «±1 by 2^. . .
6 -^ 3
. 1 7. Divide by - . . .
x—l ^ 2
18. Divide ^ti:^y±l by ^"Z^
ab he
19. Divide
im;
by
in-\-n
5x
20. Divide _ by — . .
21. Divide ^ by ^i?.
Set? -^ 4d
22. Divide
a:-*— 5^
a:"-^— 26a?-l-&2
by
x^-\-bx
X — b
Ans.
Ans.
4a
2
x—\
Ans. f^=fZ
a
^?^5. 2m — 27i
2a
a; — 5»
Atis. x-\-
Qcrx
a:
Operation, (^!±?Mz?) >< --^- -
oNrf
X
ADDITION OF FRACTIONS.
(Art. 42.) When fractions have a common denominator,
they can be readily added together by adding their numera-
4 7
tore, because f and ^ is obviously ^, and _ and - is obviously
n n
11 a J 6 . a-\-h -
. — , or - and - is — \ — , &c.
n n n n
But when the denominators are unlike, we cannot directly
add the fractions together, because we cannot add unlike
things, as dollars and cents, or units and tens, &c.
ADDITION OF FRACTIONS. 79
In all such cases we can only indicate the addition by signs,
unless we first reduce the quantities to like denominations, or
(as applied to fractions), to common denominators.
We shall investigate a rule for the addition of fractions
through the medium of equations.
For example, we require the s^ of ^, |, and |. By the
summary process of Algebra, we pronounce the sum to be S.
Then we have the following equation :
^=i+t+-f (1)
The first member of this equation is a symbol merely ;
and in the second member the addition is only indicated^ not
performed ; and to perform it, the fractional form of the equa-
tion must be changed to whole numbers, or the denoihiriators
made common.
If we multiply every term of both members hy 2, the first
fraction will he removed , (Art. 35), and the equation will stand
thus :
25=1+1+1 (2)
If we multiply every term hy 3, the second fraction vrill he
removed, and the equation will stand thus :
65=3+4+ V (3)
In the same manner we can remove the third fraction by
multiplying by 5 ; then we have
305=15+20+18 (4)
Now, if we divide every term of equation (4) by 30, we
shall have
Here we have the sum 5 equal to fractions having a com-
mon denominator, and that common denominator is the pro-
duct of the denominators of the given fractions 2, 3, and 5.
In equation (4), we may add the numbers 15, 20. and 18
directly, making 53, and the equation will be
305=53
^ ELEMENTARY ALGEBRA.
Dividing by 30, and S=^ or Iff.
Also, the sum of the fractions in equation (5) is /S^=f f .
That the operation may be more distinct, we will require
the sum of the literal fractions -, t, and ?..
b d h
Assume >S^ to be their sSn as before.
a c
9 (1)
Then /S'=_j___l-
Remove the fractions first, by multiplying by 6, then by c?,
then by A y or by mvltiplying the whole at once by the product
Idh.
Multiplying by b, gives bS=a^--\'it (2)
d h
Again by d, gives dbS==adJrcb-\-2^ (3)
Again by A, gives hdbS=adh-\-cbh-\-gbd (4)
Dividing both members of equation (4) by hdb, and we
f^_^adh cbh gbd r_x
Mb'^hdb'^hdb ^ ^
But these fractions in the second member of equation (5),
have a common denominator, and, therefore, it need not be writ-
ten under every numerator, if it be written under their sum.
Ti„3 . . . ^^M+,^t£^ (6)
Here, then, we have the sum of the fractions in one quantity.
By inspecting the second member of equation (5), and
comparing it with the original fractions to be added, we per-
ceive that the numerator of the first fraction, a, is multiplied
by the denominators of the other fractions ; and the numera-
tor of the second fraction, c, is also multiplied by the denom-
inators of the other fractions : and the same is true of tlio
third fraction, and so on.
ADDITION OF FRACTIONS. 81
The common denominator is made up — or is tlie product —
of all the denominators.
Hence, we derive the following rule for reducing fractions to a
common denominator :
Rule . — Multiply each numerator into all the denominators
except its mun, for the new numerators; and all the denominators
together, for a common denominator.
And to add fractions, we have the following rule :
Rule . — Reduce the fractions to a common denominator ;
and the sum of the numerators, written over the common denom-
inator, will he the sum of the fractions.
EXAMPLES.
1. Add —, — and - together.
^ '^ ^ ^^^ 63a:4-30ar-f35a; 128a?
105 105
2. Add « and "-±-^ Ans. ''J±^±l
be
X
3. Add *, - and t together Ans. .^•4-T?.
2 3 4 ^2
4. Add ^— and — together Ans, l?^tlli
3 7 * 21
5. Add and together. . . . Ans. ~-
a-j-6 a — b a~ — b"^
6. Add and -J—, together. . . . Ans. — ^-^~
x-\-y x—y x'^—y'-
3x ^b
7. Reduce — , — , and d, to fractions having a common
2a 3c
. 9 ex 4ab •, 6acd
denommator. ^^^*- -r—' 3— > ^^^^ -5 —
oac 6ac oac
3 2a; 2x
8. Reduce _, — , and «+ — , to fractions having a com-
4 3 a
1 . , . 9a Qax . \2a^-\-24x
mon denommator. Ans. , — , and ^ .
12a 12a 12a
B2 ELEMENTARY ALGEBRA.
(Art. 43.) The preceding rules are general, and corres-
pond to quantities that are prime to each other ; but in cases
of multiple denominators, the general rule would carry the
operator through a much longer process than necessary.
We will, therefore, investigate a more convenient practical
rule, which will apply to fractions having multiple denomina-
tors. For example, we require the sum of the fractions
a c d
b nh nib
As before, we designate the sum by S, which gives the
equation ,
Multiplying every term by h, then we have
hS=a-\--~{-^ (2)
n m
Multiplying by n, and then by m, or multiply at once by
WW, then we have
nmhS=anm-]rcm-{-dn (3)
Dividing equation (3) by nmh, and we have
anm-\-c'm-\-dn
nmb
The product nmb is composed of all the different factors in
the denominators, and no more ; it is, therefore, the least com-
mon multiple of the denominators, (Art. 32).
To find the numerators, we divide this product by the de-
nominator of any one of the fractions, and multiply the quotient
by the numerator. For instance, take the first fraction, _.
^ b
Divide nmb by b, and we have nm ; multiply this by a, and
we have anm, the new numerator for the first fraction ; and
by the same operation we find the numerators for the other
fractions.
ADDITION OF FRACTIONS. 83
Hence, we have the following rule for reducing fractions to
equivalent fractions having a least common denominator, and
thence finding their sum.
R u L E 1 . — Find the least cmnmon multiple of all th.e denom-
inators, which will he the least common denominator.
2. Divide the common denominator hy the denominator of the
first given fraction, and multiply the quotient hy the numerator,
the product will he the first of the required numerators.
3. Proceed in like manner to find each of the required
numerators.
4. The sum of the fractions will he the algehraic sum of
these numerators, with the common denominator under them.
Note . — The fractions should be reduced to their lowest
terms before this or the preceding rules are applied.
OTHER EXAMPLES.
1. Add i, -|, |, and y^ together. . . . Ans. ^=-1^.
The least common multiple of these denominators is obvi-
ously 12; therefore, multiply both members of the equation
by 12, and we have
125=6-{-8+10-{-7 (2)
Dividing again by 12, and we have
'S'=/^+A+1|+t\ (3)
The second member of equation (3) is composed of equiv-
alent fractions to those in equation (1), as may be seen by
reducing these fractions to their lowest terms.- In equation
(3), the fractions have a common denominator, composed of
the least common multiple of the original denominators.
The sum of these fractions is, of course, the sum of the
84 .. ELEMENTARY ALGEBRA.
numerators with tlie common denominator under it ; tliua, f |,
and might have been taken for equation (2), thus :
12>S'=31
Or S=^
2. A.dd , and together. . . Ans. — ! —
35c 7c ^ 5c
3. Add , — ^, and ^ together. . Ans.
1-fa i_-a i+a ^ l_a
4. Add -, — , and — together. . . Ans.
b 3b 4a ^ I2ab
5. Add ! , and tosfether.
125c 35 °
Ans.
4c
Note. — Examples 2, 3, and 5, and all others like them,
had better be performed by solving an equation. If not so
performed, multiply the numerator and denominator of the
second fraction in example 2, by 5, and in example 5, by 4c,
and thus make the denominators common by inspection. Then
unite the numerators, and reduce to lowest terms.
In examples like the following, consisting of entire quanti-
ties and fractions, make two examples of the operation, by
first uniting the entire quantities, and then the fractions, and
lastly uniting the two sums together by their proper signs.
6. Add 2x, Sxj^ — and a;_j together. Ans. 6a:_j
7. Add 5xj^ ZL and 4ar__ — ~~ together.
• 15a;
8. Add and together. Ans. .
(^On-bXa+b) a+5 ^ a— 5
ADDITION OF FRACTIONS. 85
9. Add , -HI and JH^ together. , . Ans. 0
ab be ac
10. Add "-— ^ and ?II^ together. . . . Ans. ?IJ?
ax X a
11. Add 5±5, 5=^ and 1 together.
y ay 3a 15a4-5y+9
12.
Add ^^ and ^ together. . , Am. ^1^11
a~-6 a+6 ° a'—b^
13. Add -, and together.
b cd bed J Aii \ 9
Am. '!±:^±^.
bed
14. Add _fL and -1- together. . . . Ans. ^H'
a-{.h a-^ ° a2_^»
16. Add and — ^—„ together. . . . Ans. — ^-,
16. Add J^ and ln?- together. . . . Ans. ^"*"'^*
l__a4 l+a2
1+a' ° 1— a'
17. Add 1+1 and 1— (^) together. . . An^, 1
SUBTRACTION OF FRACTIONS.
(Art. 44.) We would remind the pupil that, in addition,
we took the sum of the numerators, after the fractions were
reduced to a common denominator. Hence, the difference of
the two fractions must be found by taking the difference of
their numerators, when the denominators are alike. For ex-
ample, the difference between /^ and /j, must be fV=i> and
86 ELEMENTARY ALGEBRA.
the difference between | and f , must be f , &c. These obser-
vations must give us the following
Rule 1 . — Reduce the fractions to a common denominator.
2. Subtract the numerator of the subtrahend frcm the numer-
ator of the minuend, and place the difference over the commrni
denominaior.
(Art. 45.) We may also find the result of any proposed
example by means of equations, as in addition.
For example, from | take -f. The remainder is some num-
ber, which we may represent by R.
Then -^=f--f (1)
Or, we may consider that in every possible example, the
remainder and subtrahend added together, must equal the
minuend ; that is*
Ii+^=i (2)
Equation (2) is the same as equation (1), except the frac-
tion ^ is transposed, according to the rule of transposition
on page 1 7.
Multiply equation ( 1 ) by 7, and we have
7i2=V— 5
Multiplying by 4, and 28i2=2 1—20=1
By division . , , . B=-^^
EXAMPLES,
1. From !^ take ^-^l An., 21^Z^f±?_l!^:
2 3 6 ~" 6
2. From _L take — . Eq. fractions ^"^^ , ^t.
x—y ■ x-\-y ^—y^ ^—^^
Difference or Ans. "-^ .
* We take this view of the subject to show the pupil the nature of
equations ; not that it is, or is not, a better method of solving the
problems.
SUBTRACTION OF FRACTIONS. 87-
3. From - take —., Difif. —
4. From ?e^ take ^. . ..'... ^Ans. ^H^
3 2, 6
5. From take — IZ — Ans.
a-l-1 a2— a+1 l+a^
6. From — take — Ans.
4a 2x 4ax
7. From — take ^-
4x. 3a
12ax
8. From ^'''^^ take «'*^ ....
4a2— 5^ 2a+b^
A 0,'b'
. Ans.
4a^—b'
OPERATION.
39 J .
2a-\-b'^~(2a+b^)(2a—b')
{4a^—b')R-^2a^b'' -<i%^=2a%^
Dropping from both members (2a^5), and transposing a^5*,
and we have
{4a^—b')R=a^b''
d'b'^
By division, .... jR=
4d'—b'
9. From — take -1_ ^n*. tz^.
x—\ x-\-\ s^—1
10. From 2a—2x-{-^.II^ take 2a— 4a:+?II^.
Ans. 2ar-f
n. From t±^ fake ^fzf. .... Ans. 1?^.
5c 7c 35ff
12. From ^±1 take !if±?. . Am. ( -lUfZll^
ELEMENTARY ALGEBRA.
13. From ^i:^ take ^. . . . Ans. ( ^^11^)
14. From 1±5_' take izi^' Ans, ^^'
1—d' 1+a* l—a*
15. From x-^^Zl- take -f^. . . Arts. x-J^.
:^-\-xy ar — xy ar — if
16. From ?=? take 5^r±«. ^m. Sf*z:Si*-4Jc+8ac
2(; 5rf lOcrf
17. From !(?!±?1) take ?=? ^n.. ?+*.
a^ — 6^ a-t-6 a — 1>
18. From JL. take ^ ^W5. ^
a; — 3 a; ar — 3a;
19. From 6a+ll^^=H take Aa^^^±.
^w*. 2a+^±r.
6
SECTION II
EQUATIONS.
The most interesting and the most essential part of Alge-
bra is comprised in equations ; and nearly all of our previous
preparations have been with a view to a more ready under-
standing of equations.
The use of equations is the solving of problems in almost
every branch of mathematical science, and also in the inves-
tigation of scientific truths.
For instance, we have already investigated rules for the
addition, subtraction, and division of fractional quantities,
by means of algebraical equations, and we have thus, inci-
dentally, given some explanations concerning the nature of
equations ; but now coming to the subject in order, we shall
disregard all this, and commence on the supposition that the
pupil must yet learn every particular.
(Art. 46.) An equation is an algebraical expression, mean-
ing that certain quantities are equal to certain other quantities.
Thus, 34-4=7; a-\-b=G; x-\-4 = \0, are equations, and
express that 3 added to 4 is equal to 7, and in the second
equation, that a added to b is equal to c, &c. The signs are
only abbreviations for words.
The quantities on each side of the sign of equality are
called members. Those on the left of the sign form the Jirst
member, those on the right, the second member.
90 ELEMENTARY ALGEBRA.
(Art. 47.) As unlike things can neither be added to, nor
subtracted from each other, it follows that a member of an
equation must consist of the same kind of quantities ; and as
it is absurd to suppose one kind of quantity equal to another
in any other sense than a numerical one, it also follows that the
members of an equation must be equal in kind as well as in
number. That is. Dollars = Dollars,
Or .... Pounds = Pounds, &c., &c.
It is true we may say that a farmer has as many dollars
in his purse as he has sheep and cows on his farm.
Here we cannot say that his sheep and cows are eqval to
his dollars ; but the number of his sheep added to the number
of his cows, are equal to the number of his dollars.
That is, . . Kumber = Number.
Indeed, when dollars equal dollars, or yards equal yards,
it is but really a number of dollars equal to a number of
dollars, cfec; that is, universally, number equal to number.
(Art. 48.) In the solution of problems, every equation is
supposed to contain at least one unknown quantity ; and the
solution of an equation is the art of changing and operating
on the terms by means of addition, subtraction, multiplica-
tion, or division, or by all these combined, so that the un-
known quantity may stand alone as one member of the equa-
tion, equal to known quantities in the other member, by which
it then becomes known.
Every equation is to be regarded as the statement, in alge-
braic language, of a particular question.
Thus, X — 3=4, may be regarded as the statement of the
following question: To find a number from which, if 3 be
subtracted, the remainder wuU be equal to 4.
An equation is said to be verified, when the value of the
unknown quantity being substituted for it, the two members
are rendered equal to each other.
EQUATIONS. 91
Thus, in the equation x — 3=4, if 7, the value of a?, be
substituted instead of it, we h^ve 7 — 3=4,
Or 4=4.
(Art. 49.) Equations are of the fir sty second, third and
higher degrees, according to the highest power of the unknown
quantity involved.
' _i_z ~ !• are equations of the first degree.
,T , _T >■ are equations of the second degree.
^ 9 , , __ [- are equations of the third degree, <fec.
Equations of the first degree are also called simple equa-
tions, and equations of the second degree are called quadratic
equations ; but quadratic equations may include many other
equations of any even degree, according to certain relations
that may exist between the several parts of the equation,
which will be explained hereafter. At present we shall con-
fine our investigations to simple equations.
(Art. 60.) Equations are either numeral or literal. Nu-
meral equations contain numbers only, excepting the un-
known quantity. In literal equations, the given quantities
are represented by letters, in whole or in part.
An identical equation, is one in which the two members are
identical ; or, one in which one of the members is the result
of the operations indicated in the other.
Thus, 2r— l=2a;— 1 ) -a ,• ^ +•
6^+3.:= 8a: [ are identical equations.
(Art. 51.) The unknown quantity of an equation may be
united to known quantities, in four different ways ; by addi-
tion, by subtraction, by multiplication, and by division, and
further by various combinations of these four ways, as shown
by the foUowiog equations, both numeral and literal :
ELEMENTARY ALGEBRA.
NUMERAL.
LITERAL.
1st.
By addition, . .
. ^a;+6=10
x+a=b
2d.
By subtraction,
. a:— 8=12
X—C=d
3d.
By multiplication, .
. 20a:=80
ax=e
A 4.1.
■D-, J :,„•„:
X
X
4tli. By division _16 5=^+«
5th. a;+6— 8+4=10-1-2— 3, x-\-a—h-hc=d-\-c, &c., are
equations in which the unknown is connected with known
quantities, both by addition and subtraction.
2a;+-=21, ax-\--=^c, are equations in which the unknown
3 0
is connected with known quantities, by both multiplication and
division.
Equations often occur, in solving problems, in which all of
these operations are combined.
(Art. 52.) Let us now examine and discover, if possible,
how the unknown quantity can be separated from known
quantities, and be made to stand alone as one member of the
equation. For this purpose, let us take the equation
x-\-a=^h
Take equals from equals, a=a
Remainders are equal, x=^h — a (AlX. 2).
Here the quantity a, connected to x, appears on the other
side of the equation, with its oj^posite sign.
Again, suppose we have the equation
a;— 8=10
Add equals to equals, 8=8
Sums will be equal, a:=104-8 (Ax. 1).
Here, again, the quantity connected with x appears on the
opposite side of the equation, with its opposite sign. ,
EQUATIONS. $f
From this we derive tlie following operation, which operation
is called
TRANSPOSITION.
Rule. — Any quantity may he changed from one memler of
an equation to the other, if, in so doing, we change its sign.
Now, suppose we have an equation in the form of
ax=c
Here, x is united to a by multiplication ; it can be disunited
by division. Dividing by a, gives
c
x=-
a
Again, suppose an equation appears in the form of
X
Here, x is united to a known quantity by division, and it
can be disunited by multiplication ; that is, multiply by a, and
we have x=^ag
From these observations, we deduce this general principle :
That to separate the unknown quantity from additional tei^mSy
we must use sid^traction ; from subtracted terms, we must use
addition; from multiplied terms, we must use division; from
division, we must use multiplication.
In all cases take the opposite operation.
(Art. 53.) In many practical problems, the unknown
quantity is often combined with the known quantities, not
merely in a simple manner, but under various fractional and
compound forms. Hence, rules can only embody general
principles, and skill and tact must be acquired by close atten-
tion and practical application ; but from the foregoing princi-
ples, we derive the following
General Rule . — Connect and unite, as much as possi-
ble, all the terms of a similar kind on both sides of the equation.
li ELEMENTARY ALGEBRA.
Then, Id clear of fractions, multiply both sides by the denomina-
lors, erne after another, in succession. Or, multiply by their
continued product, or by their least common multiple, (when
such a number is obvious), and the equation will be free of
fractions.
Then transpose the unknown terms to the first memher of the
equation, and the known terms to the other. Then unite the sim-
ilar term^, and divide by the coefficient of the unknoum term, and
ike equation is solved.
EXAMPLES.
1. Given 3x — 2-\-5=2x-\-l2, to find x. . Ans. xz=9.
By transposition, Sx — 2ar=12+2 — 5
Unitino: terms, x=9
In place of transposing, we may drop equals from both
Bides, or add equals to both sides, as the circumstances may
require.
In the present example, we d7vp 2x from both sides, and
conceive — S-f-S united, then we have
x-\-3=n
Drop 3 from both sides, and we have
x—9, as before.
Dropping and transposing is one and the same operation,
differing only in form.
2. Given 6— 2a;+10=:20— 3a;— 2, to find a:. Ans. a.-=2.
Uniting similar terms in both members, we have
.—2a;-}- 16 = 18— 3a:
Adding 3a; to both sides, and dropping 16 from both, we
have a;=2
3. Given |__^-|-a;=9a;— 3, to find x. . . Ans. x^4.
Drop X from both members, then we liave
X X
Arts.
x= 3.
Am.
x== 5.
Ans.
x= 9.
Arts.
a;=12.
Ans.
x=^ m.
EQUATIONS. 95
Multiply every term by 4, and we have
2;r— a:~4a; — 12
Transpose Ax^ and unite.
Then .... — 3a;= —12
Divide both members by — 3, and x=A.
4. Given 5a;+22 — 2a;==31, to find x. .
5. Given 4a;4-20— 6=34, to find x. .
6. Given 3a;-l-12-|-7a;=102, to find ar.
7. Given 10a;— 6a;-}- 14= 62, to find ar.
8. Given ax-\rbx='ma-]r'^i, to find x. .
Separate both members into their prime factors.
Thus, .... {a-{-h)x={a-{-h)m.
Dividing both members by (a+J), gives x=m.
a — c
9. Given ax-\-dx=a — c, to find x. . . Ans. x==--j^,'
10. Given 3(a;-{-l)-i-4(a;-f2)=6(a;+3), to find x.
Ans. a;=7.
Perform the multiplication indicated, then reduce.
Sa* X X
11. Given — 4- 1 6=-+ -+17, to find ar. . Ans. a;=8.
In the first place, drop 16 from both members, according
to the general rule. Then
3a; 2; a;
T"^2"^8"^^
Multiply both members by 8, the least common multiple
of the denominators, and we have
, 6a;=4a;+a;-|-8, or, a;=8
12. Given 5--3-{--=5— 3, to find x. . . Ans. a;= 6.
2 3
13. Given ?— ?+2=3, to find a-. . . . Ans. a;=12.
3 4
96 ELEMENTARY ALGEBRA.
14. Given ? 4- -_-_-, to find a;. . . . Ans, x=2.
4^8 6~12
5x 1 11 7x
15. Given "8'~^4~"6"~^T2* *^ ^^^ ^' * ' ^^' ^~^^*
a; X — 5 2ab-\-5a
16. Given --1-— ^+25=3^, to find iP. ^W5. a;=— ^x
3x X
17. Given y-l-2j4-n=^+17, to find x. Ans. a;=10.
18. Given ^x~{-ix-\-^x=S9, to find the value of x.
Here are no scattering terms to collect, and clearing of
fractions is the first operation.
By examination of the denominators, 12 is obviously their
least common multiple, therefore, multiply by 12.
Hence, . . . Gx-{-4x+3x=39X12
Collect the terms, 13a;=39 X 12
Divide by 13, and x= 3X12=36, Ans.
19. Given |a;+-Ja;+ia;=a, to find x.
This example is essentially the same as the last. It is
identical if we suppose a=39.
Solution, . . . 6x-]-4x+3x=l2a
Or, 13a;=12a
12a
Divide and ^"13"
Kow if a be any multiple of 13, the problem is easy and
brief in numerals.
20. Given irr— 5-f ia;+8+ia;— 10=100— 6— 7 to find
the value of x.
Collecting and uniting the numeral quantities, we have
ia:+{«+i(r=94
f
EQUATIONS. 97
Multiply every term by 60, and we have
20x-{-\5x-]-nx==94'60
Collecting terms, 47a:=94*60
Divide both sides by 47, and a?= 2-60=120, Ans.
21. Given ia;-|-|a;-{-ia;-l-ia;= 77, to find ar. Ans, x=: 60.
22. Given ia:-l-ia;-|-ia:=130, to finda;. . Ans. x=120,
23. Given ^x-\-}z-{-j'jX=90, to find x. . Ans. a:=120.
24. Given iy+|y+4y=82, to find y. . Ans. y= 84.
25. Given 5x~\-^x-r-^x=34, to find x. . . Ans. x= 6.
N. B. In solving 21, 22, 23, 24 and 25, take 19 for a
model, and write a to represent the second members of the
equations, to save numeral multiplications.
3a? a:— 1 20a:-fl3
26. Given -j — —^ =6a; — r to find x.
Multiply by 4, to clear of fractions, and
3a: — 2x-\-2=24x — 20a; — 13. Reduced, x=5.
(Art. 54.) When a minus sign stands before a compound
quantity, it indicates that the whole is to be subtracted ; but
we subtract by changing signs, (Art. 5). The minus sign
before in the last example, does not indicate that the x
2
is minus, but that this term must be subtracted. When the
term is multiplied by 4, the numerator becomes 2a; — 2, and
subtracting it, we have — 2x-\-2.
Q. 3 9 a;-l-4
27. Given x — ~o~=q — ~3~* *° ^^^ ^* ' ^^* ^~ ^'
J. I O />« g X— 1
28. Given — g— — ^-+2=3;— — g-, to find x.
Ans. x= 7.
29. Given ^•^"~^_,^^~^=1, to find a:. . Ans. ar= 6.
11 13
30. Given ^ — —^=—54-0-, to find a;. . Ans. x=]0.
98 ELEMENTARY ALGEBRA.
PROPORTION.
(Art. 65.) Sometimes an equation may arise, or a prob-
lem must be solved through the aid of proportion.
Proportion is nothing more than an assumption that the
same relation^ or the same ratio exists between two quantities
as exists between two other quantities.
Quantities can only be compared when they are alike in
kind, and one of them must be the unit of measure for the
other.
Thus, if we compare A and ^, we find how many times y
or part of a time, A is contained in By by dividing B by Ay
thus, B
-j=r, or l>=rA
That is, a certain number of times A is equal to B.
Now if we have two other quantities, C and J), having the
same relation or ratio as A to B, that is, if JD=rO,
Then ^ is to -B as 0 is to J).
But in place of writing the words between the letters, we
write the signs that indicate them.
Thus, . . . . A:B::C:J)
But in place of B and J), write their values rA, and r(7.
Then, . . . . A:rA::C:rO
Multiply the extreme terms, and we have rCA.
Multiply the mean terms, and we have rA 0.
Obviously the same product, whatever quantities may be
represented by either A, or r, or C.
Hence, to convert a proportion into an equation, we have
the followinor
o
Rule . — Place the produd of the extremes equal to the pro^
duct of i/ie means.
EQUATIONS. 99
(Art. 56.) The relation between two quantities is not
changed by multiplying or dividing both of them by the
same quantity. Thus, a:b: :2a:2b, or more generally,
a:b: :na:nh, for the product of the extremes is obviously
equal to the product of the means.
That is, a is to 5 as any number of times a is to the same
number of times b.
We shall take up proportion again, but Articles 55 and 56
are sufficient for our present purpose.
EXAMPLES.
\. If 2> 'pounds of coffee cost 25 ce7iis, what will a bag of 60
pounds cost ? Ans. 500 cents.
Ans. It will cost a certain number of cents, which I desig-
nate by X, and the numerical value of x can be deduced from
the following proportion : Pounds compare with pounds, as
cents compare with cents. That is, these different kinds of
quantities must have the same numerical ratio.
Thus, .... 3:G0::25:x
Without the x, this is the rule of three in Arithmetic, be-
cause there are three terms given to jSnd the fourth ; and in
Algebra we designate the fourth term by a symbol before
we know its numerical value, which makes the proportion
complete.
By the rule (Art. 55), 3a;=60*25
60*25
Or,
3
Hence, when the first three terms of a proportion are given
to find the fourth, multijdy the second and third together, and
divide by the first.
In Arithmetic it requires more care to state a question than
it does in Algebra, because in the former science we have not
so much capital at command as in the latter.
In Algebra it is immaterial what position the nidinown term
100 ELEMENTARY ALGEBRA.
has in the proportion, if the comparison is properly made.
Thus, in the foregoing question the demand is money, and
money must be compared with money ; and the statement may
be made thus, . . 25 : a; : : 3 : 60
Or thus, . . . ic : 25 : : 60 : 3
From either one of these proportions the value of x is found
by multiplying and dividing by the same numbers.
2. If 2 cords of wood cost 5 dollars, what will 48 cords
cost? Ans. ^120.
Given 5 : a; : : 2 : 48, to find x.
3. Given 2:x: : 6 : 5x — 4, to find x. . . Ans. x=2.
The equation, -. . 10a: — 8=6a;
4. Given (li::}}(^l} : ^^ : 2a: : 1, to find x.
ScT^ 3a
Divide the first two terms by (x-{-l), (Art. 56). Also
multiply by 3a.
6. Given x-\-2 :a: :b:c, to find the value of x.
Ans. x=— — 2.
c
6. Given 2a; — 3 : x — 1 : : 2a; : a:-f 1, to find the value of x.
Ans. x=3.
7. Given x-\-6 : 38 — x : : 9 : 2, to find x. Ans. a;=30.
8. Given x-^4 : x—1 1 : : 100 : 40, to find x. Ans. x=2\.
9. Given x-\-a : x — a : :c:d, to find x. Ans. x= ^ 1.
c — d
10. Given a; :2a; — a: :a :5, to find a;. . Ans. x= —
^a—h
1\. Given a :b: :2y:d, to ^ndy. . . . Ans. y—- '
26.
12. Given a^ — ac '.ax\:\: (d—b), to find x.
Ans. x=(d-'b){a—<).
18. Given x : 76 — x : : 3 : 2, to find x. . . Ans. x=45.
EQUATIONS. 101
/.
QUESTIONS PRODUCING SIMPLE EQUATIONS.
(Art. 57.) We now suppose flie p,i?p|l '^can readily' t^duce
a simple equation containing Ji^tone iinknbwn qua'ntfty, and
he is, therefore, prepared to ^§ol\t .the; j©libWii^_;qiiefeti^ns.
The only difficulty he can experience, is the want' of tact to
reason briefly and poAverfully with algebraic symbols ; but
this tact can only be acquired by practice and strict attention
to the solution of questions. We can only give the following
general direction :
Represent the unknovm quantUy hy some symhU or letter, and
really consider it as definite and known, and go over the same
operations as to verify tlie answer when known.
EXAMPLES.
1. A merchant paid $480 to two men, A and B, and he
paid three times as much to B as to A. How many dollars
did he pay to each ? Ans. To A, $120, to B, $360.
Let ar= the sum to A,
Then ...... 3a:= the sum to B,
Sum Ax= the sum paid to hoth.
But 4Q0= the sicm paid to hoth.
Thus, when any question has been clearly and fully stated,
it will be found that some oondition has been represented in
two ways ; one having the unknovm quantity in it, and the
other having a known quantity. These two expressions must
be put in the same line, with the sign = between them, so as
to form an equation. And then, by reducing the equation,
the required result will be found.
Thus, . . 4a;= 480, therefore, a;= 120.
Observe that the problem would be essentially the same,
whatever number of dollars were paid out. It is not n^es-
sary that the number should have been 48C, any more than
102 ELEMENTARY ALGEBRA.
48, or any otlier number. Therefore, to make the problem
more general, we may represent the number of dollars paid
out ti^av^d the 6Cftia<|o^ yiiiW then be 4a;=a. And a;=-.
Again, the problem w-9u>d h^-ve been the same in charac-
ter,.'an J "eqlfaBy -as "stmf^leVhjJd- the merchant paid 4 times,
or 5 times, or n times as much to B as to A.
We may therefore make it general by stating it in the
following words :
A merchant paid a dollars to two men, A and B, and he paid
n times as many dollars to B as to A. What did he pay to
each?
a;= the sum to A,
nx^= the sum to B,
Let .
Then .
By add.
Or .
Also, .
Therefore
. x-\-nx
(]+w)rr= the sum to both.
a= the sum to both.
Cl"i~wV=«> or a:=
l4-7^•
This shows that the sum paid to A was dollars, and
^ \-\-n
as B had n times as many, the sum to B was .
^ \^n
For proof, 4~ must equal a. As the denomina-
, ,1 X? ii X • a-\-7ia (\4-n)a
tors are common, the simi of the two is — i — or i — ! — I
or a, by suppressing the common factors in numerator and
denominator
2. My horse and saddle are worth $100, and my horse is
worth 7 times my saddle. What is the value of each ?
Ans. Saddle, ^12^; horse, 887-^^.
EQUATIONS. 108
3. My horse and saddle are worth a dollars, and my horse
IS worth n times my saddle. What is the value of each ?
Ans. Saddle, _^ ; horse, ^^.
4. A farmer said he had 4 times as many cows as horses,
and 5 times as many sheep as cows ; and the nimiber of
them all was 100. How many horses had he ? Ans. 4.
5. A farmer said he had n times as many cows as horses,
and m times as many sheep as cows ; and the number of
them all was a. How many horses had he ?
Ans. horses.
\-\-n-\-mn
6. A school-girl said that she had 120 pins and needles;
and that she had seven times as many pins as needles. How
many had she of each sort ? Ans. 15 needles, and 105 pins.
7. A teacher said that her school consisted of 64 scholars ;
and that there were three times as many in Arithmetic as in
Algebra, and four times as many in Grammar as in Arithme-
tic. How many were there in each study ?
Ans. 4 in Algebra ; 12 in Arithmetic ; and 48 in Grammar.
8. A certain school consisted of a number of scholars ; a
certain portion of them studied Algebra ; n times as many
studied Arithmetic, and there were m times as many in Gram-
mar as in Arithmetic. How many were in Algebra ?
Ans. ^L_
\-\-n-\-mn
9. A person said that he was $450 in debt. That he
owed A a certain sum, B twice as much, and C twice as much
as to A and B. How much did he owe each ?
Ans. To A, $50 ; to B, $100 ; to C, $300.
10. A person said that he was owing to A a certain sum ;
to B four times as much ; and to C eight times as much ; and
to D six times as much ; so that $570 would make him even
with the world. What was his debt to A ? Ans. $30.
104 ELEMENTARY ALGEBRA,
11. A person said that he was in debt to four individuals,
A, B, 0, and D, to the amount of a dollars ; and that he was
indebted to B, « times as many dollars as to A ; to C, m times
as many dollars as to A ; and to D, j9 times as many dollars
as to A. What was his debt to A ?
Ans. — dollars.
l+w-j-m+jo
12. If $75 be divided between two men in the proportion
of 3 to 2, what will be the respective shares ?
Ans. $45 and $30.
Let x-=- the greater share.
Then . . . 75 — x=^ the other.
To answer the demands of the problem, we must have
X : 75— ar : : 3 ; 2
see example 13, (Art. 56). Observe the following method
of solution : ■
Let 3a;= the greater share, and 2a? the smaller share,
Then bx the two shares, must equal the whole sum.
That is, 5ar=75 or a:=15. Therefore, 3.r=45, the greater
share.
13. Divide $150 into two parts, so that the smaller may
be to the greater as 7 to 8. Ans. $70; and $80.
14. Divide $1235 between A and B, so that A's share
may be to B's as 3 to 2. Ans, A's share, $741 ; B's, $494.
N. B. When proportional numbers are required, it is gen-
erally most convenient to represent them by one unknown
term, with coefficients of the given relation. Thus, numbers
in proportion of 3 to 4, may be expressed by 2>x and Ax, and
the proportion of a to h may be expressed by ax and hx.
15. Divide d dollars between A and B so that A's share
may be to B's as m is to » ^ „^
Ans. As share, ; B s,
m-\-n m-\-n
16. A gentleman is now 25 years old, and his youngest
EQUATIONS. 105
brother is 15. How many years must elapse before tlieir
ages will be in the proportion of 5 to 4 ? Ans. 25 years.
25-l-a;:15+a;::5:4
1 7. Two men commenced trade together ; the first put in
S40 more than the second ; and the stock of the first was to
that of the second as 5 to 4. What was the stock of each ?
Ans. $200; and 8160.
18. A man was hired for a year for $100, and a suit of
clothes ; but at the end of 8 months he left, and received his
clothes and $60 in money, as full compensation for the time
expired. What was the value of the suit of clothes?
Ans, $20.
19. Three men trading in company gained $780, which
must be divided in proportion to their stock. A's stock was
to B*s as 2 to 3, and A's to C's was in the proportion of 2 to
5. What part of the gain must each receive ?
Ans. A, $156 ; B, $234; C, $390.
Let 2x= A's share of the gain,
Then 3x= B's "
And 5x=: C's << » «
Therefore, . . . 10a:=780, or a:=78.
20. A field of 864 acres is to be divided among three farm-
ers, A, B, and C ; so that A's part shall be to B's as 5 to 11,
and C may receive as much as A and B together. How
much must each receive ?
Ans. A, 136; B, 297; C, 432 acres.
21. Three men trading in company, put in money in the
following proportion ; the first 3 dollars as often as the second
7, and the third 5. They gain $960. What is each man's
share of the gain ? Ajis. $192 ; $448; $320.
22. A man has two flocks of sheep, each containing the
same number ; from one he sells 80, from the other 20 ;
then the number remaining in the former is to that in the hit-
106 ELEMENTARY ALGEBRA.
ter as 2 to 3. How many sheep did each flock originally
contain ? Ans. 200.
23. There are two numbers in proportion of 3 to 4 ; but
if 24 be added to each of them, the two sums will be in the
proportion of 4 to 5. What are the numbers ?
Ans. 72 and 96.
24. A man's age when he was married was to that of his
wife as 3 to 2 ; and when they had lived together 4 years, his
age was to hers as 7 to 5. What were their ages when they
were married? Ans. His age, 24; hers, 16 years.
25. A certain sum of money was put at simple interest,
and in 8 months it amounted to $1488, and in 15 months it
amounted to ^1630. What was the sum ? Ans. $1440.
Let x= the sum. The sum or principle subtracted from
the amount will give the interest: therefore 1488 — x repre-
sents the interest for 8 months, and 1530 — x is the interest
for 15 months.
Now, whatever be the rate per cent, double time will give
double interest, &c. Hence, 8:15:: 1488 — x: 1630 — x,
N. B. To acquire true delicacy in algebraical operations, it
is often expedient not to use large numerals, but let them be
represented by letters. In the present example, let a=1488.
Then a-l-42=1530, and the proportion becomes 8: 15:: a
— X : a4-42 — x.
Multiply extremes and means, then
8a+8'42— 8afel5a~15a7
HDrop 8a and — 8x from both members, and we have
8«42=7a— 7a;
Dividing by 7, and transposing, we find
a;=a— 48=1440, Ans.
26. A certain sum of money was put at simple interest for
2|- years, and in that time it amounted to $3526, and in 38
months it amounted to $3606. What was the sum put at
interest? Ans. $3226.
EQUATIONS. , 107
(Art. 58.) The object of solving problems should be to
acquire a knowledge of the utility and the power of the
science, and this knowledge cannot be attained to the fullest
extent by merely solving problems ; we must also learn how-
to propose them, and to propose such as are convenient and
proper for instruction.
Problem 25 is extracted from an English work ; and let the
reader observe that the two amounts, $1488, 8 1530, and
$1440, the sum put at interest, are all whole numbers, no
fraction of a dollar in any of them, which makes the problem
a neat and convenient one.
The question now is, how the proposer discovered these
numbers ? Did he happen upon them ? Did he find them
by repeated trials ? or did he deduce them naturally and easily
from a scientific process ?
We can best answer these questions by showing how we
found the numbers to form problem 26.
Wanting another example of the same kind as 25, but of
different data, I wrote on a slip of paper thus :
A sum of money was put at interest for 2^ years, and the
amount for that time was a dollars ; and for 38 months the
amount was a-\-d dollars. What was the sum ?
The amount for 38 months must be greater than the
amount for 30 months, therefore c? is a positive number.
Let X represent the sum lent. Then a — x= the interest
for 30 months, and a-{-d — x= the interest for 38 months.
Hence, . . 30 : 38 : : a — x : a-^d — x. ,^
Product of extremes and means gives
30a -f- 30«?— 30a;= 3 8a— 3 8a;
Dropping 30a and — 30a; from both members, we have
30d=Sa — Sx
Dividing by 8, and transposing, gives
a- a _-
108 , ELEMENTARY ALGEBRA.
Here a stands alone, and any whole number greater than
— can be written in its place ; and if we take d of such a
8 ^
value as to render it divisible by 8, the fraction — will be a
^ 8
whole number, and cause a? to be a whole number also.
In preparing the example, I took d equal 80, then
8
is in value 300 ; and I took «, hap-hazard at $3526 ; there-
fore, a-l-^=3606, and a;=3226, the numbers given in the
problem. By taking different values to a and d, we may
form as many numeral problems as we please like problem
25 or 26 ; and if, in every instance, we take care to take d
of such a value as to render it divisible by 8, no fractions
will appear in the problems.
Again, observe the expression x=za — The numera-
tor of the fraction has 30 for a coefficient, and that is the
number of months that the sum of money was out at interest
before the first amount was rendered ; and 8, the denomina-
tor, is the number of months between the times of rendering
the two amounts.
Observing these facts, we may solve another problem of
the like kind without going through the steps of the process.
For example.
27. A certain sum of money was put at simple interest, and
in 13 months it amounted to a dollars, and in 20 months it
amounted to a-j-d dollars. What was the sum? Ans. x.
And ... . . x=^a — — (1)
7 ^ ^
To form a numerical problem from equation (1), such as
shall contain only whole numbers, and correspond to the
times here mentioned, we must take c?=7, or some midiiple oj 7.
Suppose we take c?= 1 4 ; then
a-=a— 26 (2)
EQUATIONS. 109
Now it is ray object to form another numerical problem of
this kind, corresponding to the times mentioned in 27, having
such numbers that the answer — the sum put at interest, shall
be just 100 dollars.
Take equation (2), and in place of x write 100, transpose
— 26, and we have a=126, the first amount; and as c?=14,
a-\-d, the second amount, must be 140.
Hence, we may write the problem thus :
28. A certain sum of money was put at interest, and in 13
months the amount due was ^126, and if continued at inter-
est for 20 months, the amount due would have been $140.
What was the sum put at interest ? Ans. $100.
In equation (1), the fraction — is the interest on the sum
for 13 months, because it is the sum, which, if added to the
principal, will give the amount.
Here in these problems, d and a are perfectly arbitrary ; we
pay no attention to the rate of interest ; and if we take d of
any great value, there will be an unreasonable quantity of
interest ; and if d is taken very small in relation to a, the rate
of interest will be small ; but the algebraist can adjust the
13d
rate by putting , equal to any given rate of interest ; but
in a work like this, it is not proper to carry these investiga-
tions any further.
•
EQUATIONS CONTINUED.
(Art. 59.) Problems in which fractions mostly occur.
1. The number 12 is | of what number ?
Ans. It is I of the number x»
To determine the numerical value of Xy we solve the follow-
3x
ing equation, —=12. Hence, 16 is the number.
no ELEMENTARY ALGEBRA.
2. The number a is | of what number ? Am. ~.
3
3. The number 21 is -^ of what number ? Ans. 49.
4. The number 21 is the ^th part of what number ?
Ans. ^.
5. The number a is the _ th part of what number ?
n
an
m
6. If you add together } and | of a certain number, the
sum will be 130. What is the number? Ans. 420.
The following solution is taken from another book, and it
IS a fair specimen of the manner of teaching Algebra, both
in this country and in England ; but in this particular we
insist on improvement.
♦ Let x= the number,
Then ^-1-~130
Multiplying both members by 7 and 6, or by 42, we have
6x'\''7x=5460
lSx=5460
a:=420
This is but half Algebra. An algebraist never multiplies num-
bers together, except in final results, or in some rare cases where it
is impossible to do otherwise.
To avoid this, let numbers be represented by letters ; and in
place of 130 in the equation, write a to represent it, as taught in
(Art. 63.)
Then -+--a
Clearing of fractions, 6a;+7a;=42a
Or 18a;=i42a
EQUATIONS. HI
Now, as a is divisible by 13, and the quotient 10, dividing
both members by 13, gives a:=420, without the least effort
at numerical computation.
It is not, in fact, necessary to write a ; we may retain the
number as a factor, or what is better, take its obvious factors.
Thus, .... 6a;4-7^=42-13«10
Uniting and suppressing the factors common to both mem-
bers, and a;=420
We extract a solution to the following problem :
7. A farmer wishes to mix 116 bushels of provender, con-
sisting of rye, barley, and oats, so that it may contain f as
much barley as oats, and \ as much rye as barley. How
much of each must there be in the mixture ?
Stating the question, Xz=z oats ; and — = barley.
Then, -i- of _ is — = rye.
Forming the equation, a;_j -j- — =116
Multiplying by 1 4, . .1 4a;-}- 1 Qx-\- 5a;= 1 624
Uniting terms, 29a;=1624
Dividing by 29, ..... x=^6Q the Arts.
If we keep the factors separate, we have
29a:=n6'14
Dividing by 29, gives a;=4* 14=56.
Here we find a reason why the farmer wished to mix 116 bush-
els— not 100, or 115, or 117 — it must be some multiple of 29 to
have the different kinds of grain come out in whole numbers.
Indeed, the numbers in all numeral problems are so chosen that
the final coefficient of the unknown quantity shall he some factor
in the other member ; therefore it is worse than useless to hide the
factors (as is often done), by laborious multiplication.
112 ELEMENTARY ALGEBRA.
8. Divide 48 into two such parts, that if the less be divided
by 4, and the greater by 6, the sum of the Quotients will be 9.
Am. 12 and 36.
9. A clerk spends §■ of his salary for his board, and | of
the remainder in clothes, and yet saves $150 a year. What
is his yearly salary ? Ans. $1350.
10. An estate is to be divided among 4 children, in the
following manner :
The first is to have $200 more than i of the whole.
The second is to have $340 more than ^ of the whole.
The third is to have $300 more than } of the whole.
And the fourth is to have $400 more than i of the whole.
"What is the value of the estate ? Am. $4800.
•11. Of a detachment of soldiers, | are on actual duty, |
of them sick, i of the remainder absent on leave, and the
rest, which is 380, have deserted. What was the number of
men in the detachment ? Am. 2280 men.
12. A man has a lease for 99 years, and being asked how
much of it was already expired, answered that | of the time
past was equal to | of the time to come. Required the tiipe
past and the time to come.
Assume a=99. Am. Time past, 54 years.
13. It is required to divide the number 204 into two such
parts, that 2. of the less being taken from the greater, the
remainder will be equal to ^ of the greater subtracted from 4
times the less. Am. The numbers are 154 and 50.
Put a=204, and resubstitute in the result.
14. In the composition of a quantity of gunpowder
The nitre was 10 pounds more than | of the whole.
The sulphur 4^ pounds less than ^ of the whole.
The charcoal 2 pounds less than ■} of the nitre.
What was the amount of gunpowder ? Am. 69 pounds.
EQUATIONS. lis
Let ir= the -whole.
Then .... ?^-f 10= the nitre,
3
- — 4i= the sulphur,
22; 10
2j+-^ — 2= the charcoal.
By addition, — _l-4_ — \ .l^^i—^
Multiply both members by 6, and
4ar4-a;+y+^+21=6a;
Drop bx from both members, then
4^ . 60 , „^
-;^ + y4-21=:r
Multiply by 7, and drop 4a; from both members.
And .... 60+21 •7=3a;
Dividing by 3, 204-7«7=a;
Or, 69=a;
15. Divide $44 between three men. A, B, and C, so that
the share of A may be | that of B, and the share of B, |
that of C. Ans. A, $9 ; B, $15, C, $20.
Will the student find the reason why the problem requires
the division of the number 44 ; why not 45, 47, or any other
number, as well as 44 ?
Let 3a;=A's share, 5a;=B's, and y=C's share,
3y
Then -t=^^
4
20a;
Or y=—~-=z C's share.
o
Hence, . . 3ar-l-5a;+— -=44
o
10
114 ELEMENTARY ALGEBRA.
Clearing of fractions, and uniting terms, we have
44a:=44*3, or x==3
If the problenx had required the division of any other
number of dollars, for instance, a dollars, the value of z
would have been — dollars. Taking a equal 44, or any num-
ber of times 44, gives whole numhers for the respective shares.
16. What number is that, to which, if we add its ^, i, and
1, the sum will be 50 ? Ans. 24.
17. What number is that, to which, if we add its i, ^, and
A, the sum will be a .^ . 24a
ns. -_-.
18. If A can build a certain wall in 10 days, and B can
do the same in 14 days, what number of days will be re-
quired to build the wall, if they both work together ?
Ans. 5f days.
Let X represent the days required. If A can do the work
in 10 days, in one day he will do -^^ of it, and in x days he
X • X
will do — of the whole work. By the same reasoning, '—
is the part of the work done by B.
Therefore, —-+—- = 1. (1 is the whole work).
10 14 ^ '
19. If A can do a piece of work in a days, and B can do
the^same in b days, how long will it take them, if they both
work together? ^^^ _^ days.
I now wish to propose a numerical problem in all respects
like problem 18, except that the number of days shall be a
whole number, and the answer shall be 8.
The answer to 1 9 is a r/eneral answer ; and now if we
EQUATIONS. 115
require a particular answer, 8, we simply require the verifi-
cation of the following equation.
Or ah=Sa-]-Sh
In this equation, if we assume a, the equation will give b,
or if we assume b, the equation will give a corresponding
value to a. But whichever letter we assume, it must be
assumed greater than 8 ; because it requires either man more
than 8 days to do the work, for they together do it in 8 days.
Now, assume a=12, then the equation becomes
125=8'12+85
46=8'12
5=24
We can now write out our numerical problem thus :
20. A can do a piece of work in 1 2 days ; B can do the
same in 24 days. How many days will be required, if they
both work together ? Ans. 8.
2t. A young man, who had just received a fortune, spent
f of it the first year, and ^ of the remainder the next year ;
when he had $1420 left. What was his fortune ?
Ans. 811360.
22. If from ^ of my hight in inches, 12 be subtracted, i
of the remainder will be 2. What is my hight ?
Ans. 5 feet 6 inches.
23. A laborer. A, can perform a piece of work in 5 days,
B can do the same in 6 days, and C in 8 days ; in what time
can the three together perform the same work ?
Ans. 2/^ days.
Let a;= the number of days in which all three can do it.
24. After paying out i and } of my money, I had remain-
ing 66 guineas. How many guineas had I at first ?
Ans. 120.
116 ELEMENTARY ALGEBRA.
25. In a certain orchard, i are apple trees, | peach trees,
J- plum trees, 100 cherry trees, 100 pear trees. How many
trees in the orchard ? Ans. 2400.
26. A farmer has his sheep in jfive different fields, viz : ^
in the first field, ^ in the second, } in the third, ^j in the
fourth, and 45 in the fifth field. How many sheep in the
Jock? Ans. 120.
27. A person at play, lost ^ of his money, and then won 3
shilhngs ; after which he lost i of what he then had ; and, on
counting, found that he had 12 shillings remaining. What
had he at first? Ans. 20 shilhngs.
(Art. 60). When equations contain compound fractions, and
simple ones, clear them of the simple fractions first, and unite,
as far as possible, all the simple terms.
We give a few examples to show the advantage of observ-
ing this expedient.
1. Given ; — = to find the value of a;.
9 6x-\-3 3
Multiply all the terms by the smallest denominator, 3.
That is, divide all the denominators by 3, and
3 ^2a:-{-l ^
Multiplying again by 3, and dropping 6a;-{-7 from both
members, we have
21a;— 39_
~2x-{-l"~
Clearing of fractions, transposing, (fee, we find a;=4.
^ ^. 7a:+16 ar-f 8 ,a:
2. Given-^-=^^-^j+-tofind..
Multiply by 21, and from both members drop 7x, then
^ 21a:-|-21'8
16= !
4a^— 11
Clearing of fractions
And . . . 64a:— llM6=21a;+21'8
EQUATIONS. II?
For the purpose of showing something of the spirit of
Algebra, we will put a=8: after dropping 21a; from both
members,
Then . . 43a;— 11 -2^=2 la
. Or . . . 43a;— 22a=21a
Or . . . . 43a;=43a or ar=a=8.
„ ^. 9a;4-20 4a; — 12 , x
3. Given — ^-~ =— +-, to find x,
36 5x — 4 4
Multiplying by 36, and dropping 9a; from both members.
Then 20=??(i^l-l)
5x — 4
4. Given -\ p^ =^ -^^ — ■ ^ to find x.
X ax-f-ox a-\-b
Multiply by a;.
Then . . a-6+?^=?=(5^?^(.5!f±!^
a-\-b a-J-5
Multiply by (a-\-b), and unite known quantities,
Then ... 3a^—2b^=(3a^—2b^)(3a'+2b^)x
1
^~3a^-f2b^'
5x-\-5 9 6x — 12
5. Given ■ — r-^-r 1 = — to find a;. . Ans. a?=2.
x-\-2 4 X — 2
^ -,. 2a;+l 402— 3a; 471— 6a; ^ . ,
6. C^iven —^- ig""^"^ 2 ^^ ^^^ ^•
-4ws. a;=72.
^ _. 18a;— 19 , lla;-|-21 9a;+15 ^ . ,
Ans. a;=7.
^ ^. 20a;+36 , 5a;+20 4a; , „ ■ ,
8. Given _^_+^_=_+3xx tofind^..
u4w5. x=^4.
« ^. 6 3a; , 2 2a;-}-5 3 , ,
9. Given -^---+-=~3 to find a;.
Ans. x=^.
118 ELEMENTARY ALGEBRA.
I Sar-f- 1 8
10. Given \-5-{-x=z — - — to find a:. . Ans. x=6.
X — 5 3
11. Divide the number 48 into two such parts, that 7
divided by one part shall be equal to 5 divided by the other
part. Required the parts. Ans. 28 and 20.
1 2. Divide the number 48 into two such parts, that one may
be to the other as 7 to 5. Required the parts. Ans. 28 and 20.
13. A person in play, lost a fourth of his money, and then
won back 3 shillings ; after which he lost a third of what he
now had, and then won back 2 shillings ; lastly, he lost a
seventh of what he then had, and then found he had but 1 2
shillings remaining. What had he at first ?
Ans. 20 shillings.
14. A shepherd was met by a band of robbers, who plun-
dered him of half of his flock and half a sheep over. After-
ward a second party met him, and took half of what he had
left, and half a sheep over ; and soon after this, a third party
met him and treated him in like manner ; and then he had 5
sheep left ? How many sheep had he at first ?
Ans. 47 sheep.
15. A man bought a horse and chaise for 341 (a) dollars.
Now, if f of the price of the horse be subtracted from twice
the price of the chaise, the remainder will be the same as if
4 of the price of the chaise be subtracted from 3 times the
price of the horse. Required the price of each.
Ans. Horse, $152; chaise, $189.
N. B. Let 8a;= the price of the horse.
Or let . 7a:= the price of the chaise. ^
Solve this question by both of these notations.
16. A laborer engaged to serve for 60 days^ on these con-
ditions : That for every day he worked he should have 75
cents and his board, and for every day he was idle he should
forfeit 26 cents for damage and board. At the end of the
time a settlement was made, and he received $25. How
many days did he work, and how many days was he idle ?
EQUATIONS. 119
The common way of solving sucli questions is to let a'=
the days he worked ; then 60 — x represents the days he was
idle. Then sum up the account and put it equal to $25.
Another method is, to consider that if he worked the whole
60 days, at 75 cents per day, he must receive ^45. But for
every day he was idle, he not only lost his wages, 75 cents,
but 25 cents in addition. That is, he lost %\ every day he
was idle.
Now, let rr= the days he was idle. Then, x=- the dollars
he lost. And 45 — a:=25 or a;=20, the days he was idle.
17. A person engaged to work a days on these conditions:
For each day he worked he was to receive h cents ; for each
day he was idle he was to forfeit c cents. At the end of
a days he received d cents. How many days was he idle ?
. ah — d .
Ans. -rr-i — days.
Let x= the number of days he was idle.
Had he worked every day he must have received ah cents.
But for every idle day we must diminish this sum by (6+c)
centsr; and for x days, the diminution must be (b-\-c)x cents.
That is, al) — (b-\-c)x=d by the question.
,T cib — d
Hence, x=:—-, —
18. A boy engaged to convey 30 glass vessels to a certain
place, on condition of receiving 5 cents for every one he de-
livered safe, and forfeiting 12 cents for every one he broke.
On settlement, he received 99 cents. How many did he
break? Ans. 3.
19. A boy engaged to carry n glass vessels to a certain
place, on condition of receiving a cents for every one he deliv-
ered, and to forfeit b cents for every one he broke. On set-
tlement he received d cents. How many did he break ?
Ans. The number represented by —ri"
120 ELEMENTARY ALGEBRA.
SIMPLE EQUATIONS
CONTAINING TWO UNKNOWN QUANTITIES.
(Art. 61.) We have thus far considered such equations
only as contained but one unknown quantity ; but we now
suppose the pupil sufficiently advanced to comprehend equa-
tions containing two or more unknown quantities.
There are many simple problems which one may meet with
in Algebra, which cannot be solved by the use of a single
unTcnown quantity, and there are also some which may he
solved by a single letter, that may become much more simple
by using two or more unknown quantities.
When two unknown quantities are used, two independent
equations must exist, in which the value of the unknown let-
ters must be the same in each. When three unknown quan-
tities are used, there must exist three independent equations,
in which the value of any one of the unknown letters is the
same in each.
In short, there must he as mxiny independent equations as
unknown quantities used in the question.
An independent equation may be called a primitive or prime
equation — one that is not derived from any other equation.
Thus, x-\-2>y=a, and 2a;+6y=2a, are not independent equa-
tions, because one can be derived from the other ; but x-\-2)y
=a, and Ax-\-by=h, are independent equations, because
neither one can be reduced to the other by any arithmetical
operation.
The reason that two equations are required to determine
two unknown quantities, will be made clear by considering
the following equation :
a;+y=20
This equation will be verified if we make x—\, and y=19,
ora:=2, and 2/= 18, or *=^, and y=19i, &c., &c., without
limit. But if we combine another equation with this, as
EQUATIONS. 121
gc — ^y=4, then we have to verify two equations with the same
values to x and y, and only one value for x and one value for
y will answer both conditions.
Thus, ..... a;4-y=20
And X — y— 4
By addition . . . 2a:=24
x=\2y and y=8.
That is, we have found a value for x and another to y (12
and 8), so that their sum shall be 20, and their difference 4 ;
and no other possible numbers will answer.
A merchant sends me a bill of 16 dollars for 3 pairs of shoes
and 2 pairs of hoots ; afterward he sends another bill of 23
dollars for 4 pairs of shoes and 3 pairs of boots, charging at the
same rate. What was his price for a pair of shoes, and what
for a pair of boots ?
This can be resolved by one unknown quantity, but it is far
more simple to use two.
Let x= the price of a pair of shoes.
And y= the price of a pair of boots.
Then by the question 3a;+2z/=16
And ..... 4a:+3y=23
These two equations are independent ; that is, one cannot
be converted into the other by multiphcation or division, not-
withstanding the value of x and of y are the same in both
equations.
Equations are independent when they express dififerent con-
ditions, and dependent when they express the same conditions
under different forms.
To reduce equations involving two unknown quantities, it
is necessary to perform some arithmetical operation upon
them, which will cause one of the unknown quantities to dis-
appear. These operations are called ehmination.
There are three principal methods of eliminaHcn.
11
122 ELEMENTARY ALGEBRA.
1. By comparison. 2. By substitution. 3. By addition or
^subtraction.
All the operations rest on the axioms.
FIRST METHOD.
(Art. 62.) Transpose the terms containing y to the right
hand sides of the equations, and divide by the coefficients of
X, and
From equation {A) we have x= ( C)
o
And from {B) we have . a:= (B)
Put the two expressions for x equal to each other (Ax 7),
, , 16— 2y 23— 3y
And -=
3 4
An equation which readily gives ^=-5, which, taken as the
value of y in either equation ( 0) or (B), will give x=2.
SECOND METHOD.
(Art. 63.) To explain the second method of elimination,
resume the equations
3x+2y=16^ (A)
4a;-l-3y=23 (B)
The value of x from equation (A) is ic=i(16 — 2y).
Substitute this value for x in equation (B), and we have
4X1(16 — 2y)-\-3y=23, an equation containing only y.
Reducmg it, we find y=5, the same as before.
Observe, that this method consists in finding the value of
one of the unknown quantities from one equation, and substi-
tuting that value in the other. Hence, it is properly called
the method by substitution.
^ EQUATIONS. 1'2S
THIRD METHOD OF ELIMxNATION.
(Art. 64.) Resume again 3x-\-2y = 16 (A)
4a;+3y=23 (JB)
When the coefficients of either a; or y are the same in both
equations, and the signs alike, that term will disappear by
subtraction.
When the signs are unlike, and the coefficients equal, the
term will disappear by addition.
To make the coefficients of x equal, multiply each equation hy
the coefficient of x in tJte other.
To make the coefficients of y equal, multiply each equation by
i/ie coefficient of y in the other.
Multiply equation (A) by 4 and 12ii;+8y=64
Multiply equation (B) by 3 and 12a?+9y=69
Difference y=5, as before.
To continue this investigation, let us take the equations
2ar-f3y=23 (A)
5x—2y=10 (B)
Multiply equation (A) by 2, and equation (B) by 3, and
we have 4x-\-6y=46
15x—6y—30
Equations in which the coefficients of y are equal, and the
signs unlike. In this case add, and the y's will destroy each
other, giving 19a;=76
Or x=4
Of these three methods of ehmination, sometimes one is
preferable and sometimes another, according to the relation
of the coefficients and the positions in which they stand.
No one should be prejudiced against either method ; and in
practice we use either one, or modifications of them, as the
124 ELEMENTARY ALGEBRA.
case may require. The forms may be disregarded when the
principles are kept in view.
EXAMPLES.
1. Given ] *"i/=^^^ (^U to find, and y.
From (A) .... x=-^^
From (JB) .... a;=16— 4y
Therefore . . — ^^=16— 4y (Ax. 7).
Ans. y=2, x=S.
( 7x-{-4y=5S (A) )
2- Given -j 9^_4^_33 ^^^ [ to find a: and y.
Ans. x=6, y=4.
Here, it would be very inexpedient to take the first method
of elimination.
Observe that the coefficients of y are alike in number, but
opposite in signs.
A skillful operator takes great advantage of circumstances,
and very rarely goes through all the operations of set rules ;
but this skill can only be acquired by observation and
practice.
Add the two equations. Why ?
( 5x+6y=5S (A) ) , ,
3. Given j ^^_^q^^^^ ^^'^ \ to find x and y.
Ans. ir=8 ; y=3.
Subtract ( J5) from (A). Why ?
4. Given j Tll'^^\o iiU^^-^---^y-
Ans. x=4; y=2.
Add (^^and (i?). Why?
5. Given j f i^^=^f 1 p| [ to find . and y.
( 3a:+4y=88 (B) f
Ans. x=S ; y=16.
Multiply (B) by 2. Why ?
EQUATIONS. 125
C. Given i „ ; ..^ r to find x and z,
Ans. x=7 ; z=8.
^ _. C 4x-\- 6y= 46 )
7. Given S ^ ^ . ^ r to nnd a; and y.
( 5a:— 2?/= 10 j ^
„ ^. ( 2x+ 3y= 31 ) ^ ^
8. Given 1 ^ „ . „ f to find a? and y.
( 4a; — 3y= 17 )
Ans. x=S ; y=5.
( 4y-|- 2?=102 ) , ,
9. Given i , , ,^ h to find y and z.
( 3/4- 42= 48 ) ^
Ans. ?/=24 ; 2=6
« I .^ ^^ r to find a; and y.
8a;+10y=26 ) ^
^715. a:=2 : y=l.
( 6y+ 3a?=93 ) , ^ ^
11- Given] ^^_^^^._g^^tofindyanda..
Ans. y=\2; a?=ll.
12. Given Aar-|-^y==14, and Ja:-1-Jy=ll, to find x and y,
^725. a;=24; 3^=6.
13. Given x-\-y=S, and lx-\-y=7, to find a; and y.
Ans. x=s^6; y=4.
14. Given ja;+7y=99, and ^y-f 7a;=51, to find x and y.
Ans. x=7 ; y — 14.
PEOBLEMS PRODUCING EQUATIONS OF TWO UNKNOWN QUANXITIES.
1. A man bought 3 bushels of wheat and 5 bushels of rye
for 38 shillings ; and at another time, 6 bushels of wheat and
3 bushels of rye for 48 shillings. What was the price for a
bushel of each ?
Let x= price of wheat, and y= price of rye.
By the first condition, 3a;-l-5y=38 (A)
By the second, . . 6a:+3y=48 {J3)
Ans. x=6 ; y='U
126 ELEMENTARY ALGEBRA.
2. A gentleman paid for 6 pairs of hoots, and 4 pairs of
shoes, $44 ; and afterward, for 3 pairs of boots, and 7 pairs
of shoes, $32. What was the price of each per pair ?
Ans. Boots, $6 ; shoes, $2.
3. A man spends 30 cents for apples and pears, buying his
apples at the rate of 4 for a cent, and his pears at the rate of
5 for a cent. He afterward let his friend have half of his
apples and one-third of his pears, for 13 cents, at the same
rate. How many did he buy of each sort ?
Let x= the number of apples,
y= « « of pears,
A cent = the price of 1 apple ; hence, x apples are worth
- cents. Therefore,
4
"By the first condition, . .
^ l=-
(^)
By the second, . . .
i+f5=-
(-8)
Multiplying (B) by 2,
I+S=-
(1)
Subtracting (1) from (A),
5 15
(2)
Multiplying (2) by 15, .
Sy— 22/=60,
or y=60
This value of y put in (^), gives --f 12=30, or ar=72.
4. What fraction is that, to the numerator of which, if 1 be
added, its value will ^, but if 1 be added to the denominator,
its value will be ^ ?
Let cc= the numerator.
And y== the denominator,
X
Then — ==• the fraction,
y
EQUATIONS, 127
And we shall have the two equations.
Clearing of fractions, 3a;+3=y (1)
And 4a;— l=y (2)
Taking (1) from (2), ic— 4=0, or x=4.
Hence, from 1 we find y=15, and the fraction is j-\.
5. What fraction is that, to the numerator of which, if 4 he
added, the value is i, but if 7 be added to its denominator,
its value will be | ? Ans. j\.
G. A and B have certain sums of money : says A to B,
** Give me ^15 of your money, and I shall have five times as
much as you have left." Says B to A, ** Give me ^5 of
your money, and I shall have exactly as much as you have
left." How many dollars had each ?
Ans. A had ^35 ; and B, ^25.
7. What fraction is that whose numerator being doubled,
and its denominator increased by 7, the value becomes |;
but the denominator being doubled, and the numerator
increased by 2, the value becomes -j ? Ans. a.
8. If A give B ^5 of his money, B will have twice as much
money as A has left ; and if B give A ^5, A will have thrice
as much as B has left. How much had each ?
Ans. A had $13; B, $11.
9. A merchant has sugar at 9 cents and at 13 cents a
pound, and he wishes to make a mixture of 100 pounds tliat
shall be worth 12 cents a poimd. How many pounds of each
quality must he take ? '
Ans. 25 pounds at 9 cents, and 75 pounds at 13 cents.
10. A person has a saddle worth ;i650, and two horses.
When he saddles the poorest horse, the horse and saddle are
worth twice as much as the best horse ; but when he saddles
128 ELEMENTARY ALGEBRA.
the best, the horse and saddle are together worth three times
the other. What is the value of each horse ?
Ans. Best, MO ; poorest, ^30.
1 1 . One day a gentleman employs 4 men and 8 boys to
labor for him, and pays them 40 shillings ; the next day he
hires at the same rate, 7 men and 6 boys, for 50 shillings.
What are the daily wages of each ?
Ans. Man's, 5 shillings ; boy's, 2 shillings 6 pence.
1 2. A merchant sold a yard of broadcloth and 3 yards of
velvet, for $25; and, at another time, 4 yards of broadcloth
and 5 yards of velvet, for ^65. What was the price of each
per yard? Ans. Broadcloth, $10; velvet, $5.
13. Find two numbers, such that half the first, with a third
part of the second, make 9, and a fourth part of the first, with
a fifth part of the second, make 5. Ans. 8 and 15.
14. A gentleman being asked the age of his two sons, an-
swered, that if to the sum of their ages 18 be added, the
result will be double the age of the elder ; but if 6 be taken
from the difference of their ages, the remainder will be equal
to the age of the younger. What were their ages ?
Ans. 30 and 12.
15. A says to B, ** Give me 100 of your dollars, and I
shall have as much as you." B replies, **Give me 100 of
your dollars and I shall have twice as much as you. How
many dollars has each ? ^725. A, $500; B, $700.
16. Find two numbers, such that f of the first and f
of the second added together, will make 1 2 ; and if the first
be divided by 2, and the second multiplied by 3, | of the sum
of these results will be 26. Ans. 15 and 10^.
17. Says A to B, "^ of the difference of our money is
equal to yours ; and if you give me $2, I shall have five
times as much as you." How much has each ?
Ans. A, $48; B, $12.
18. A market-woman bought eggs, some at the rate of 2
for a cent, and some at the rate of 3 for 2 cents, to the amount
EQUATIONS. 129
of 65 cents. She afterward sold them all for 100 cents,
thereby gaining half a cent on each egg. How many of each
kind did she buy ? Ans. 50 of one ; 60 of the other.
19. What two numbers are those, whose sum is a and dif-
ference b ?
Let x== the greater.
Ans. The greater is
a b
2+2
y= the less.
The less is
a b
£-2
Sum is . .
a
Difference is
h.
(Art. 64.) From the result of this problem, we learn one
important fact, which will be of use to us in solving other
problems.
The fact is this : That the half sum of any two nmnbers,
added to the half difference, is the greater of the two numbers ;
and the half sum, diminished by the half difference ^ gives the
less.
20. There are two numbers whose sum is 100, and three
times the less taken from twice the greater, gives 150 for
remainder. What are the numbers ? Ans. 90 and 10.
The half sum of the two numbers is 50. Now let xz;= the
half difference. Then 50-|- ar= the greater number.
And 50 — a;= the less number.
Twice the greater is \00-\-2x
Three times less is 1 50 — 3.r
Difference is . — 50-}-52;=150.
21. What two numbers are those whose sum is 12, and
whose product is 35 ? A71S. 7 and 5.
Let Q-\-x = the greater,
Then 6 — x = the less.
Product, .... 36— a''=35. Hence, x=^\.
130 ELEMENTARY ALGEBRA.
22. What two numbers are those whose difference is 4, and
product 96 ? Ans. 12 and 8.
Let x= the half sum ; 2= the half difference.
23. The difference of two numbers is 6, and the sum of
their squares is 50. What are the numbers ? Ans. 7 and 1.
24. The difference of two numbers is 8, and their product
is 240. What are the numbers ? A7is. 12 and 20.
(Art. 65.) To reduce equations containing three or more
unknoAvn quantities, we employ the same principles as for two
unknown quantities, and no more ; and from these principles
we draw the following
Rule . — Combine any one of the equations vnih each of the
others, so as to eliminate the same unknown quantity ; there will
thus arise a new set of equations containing one less unknown
quantity.
In the same manner combine one of these new equations with
each of the others, and thus obtain another set of equations con-
taining one less unknown quantity than the last set ; and so
continue, until an equation is found containing one unknown
quantity ; solving this equation, and substituting the value of iis
unknovm quantity in the other equations, the other unknown
quantities are easily found.
EXAMPLES.
( x-\- y-f- ^= 9 "^
1. Given < a;+2y4- 3^=16 > to find x, y, and z.
( x-\-3y+4z=21 )
By the first method, transpose the terms containing y and
z in each equation, and
x= 9 — y — z
x=zl6—2y—3z
x=2\—Sy—4z
EQUATIONS. iM
Then putting the 1st and 2d values equal, and the 2d and
3d values equal, gives
9 — y— ^=16 — 2y— 32
1 6—2?/ — 3z = 2 1 — 3y — 4z
Transposing and condensing terms,
And y=7 — 2z
A-lso, y=5 — z
Hence, .... 5 — z=7 — 2z, or z=2.
Having z=2, we have y=5 — 2=3, and having the values
of both z and y, by the first equation we find x=4. -
r 2a?+4y— 32=22 ^
2. Given I 4x — 2?/4-52=18 > to find x, y and z.
( 6ir-f7y— 2=63 )
Multiplying the first equation by 2, . 4a;+8y — 62=44
And subtracting the second, . . . 4x — 2y-f- 52=18
The result is, (.4) lOy— -112=26
Then multiply the first equation by 3, 6x-\-12y — 92=66
And subtract the third, 6x-\- 'ly — 2=63
The resxilt is, (^) 5y— 82= 3
Multiply the new equation, {B), by 2, lOy — 162= 6
And subtract this from equation (A)y \0y — 1 12=26
The result is, 52=20
Therefore, 2=4.
Substituting the value of z in equation {B), and we find
y=7.
( 3:r+9y+82=41 \
3. Given \ 5x-\-Ay — 22=20 \ to find x, y, and 2.
( ILT+Ty— 62=37 )
Ans. x=2; y=S ; 2=1.
(Art. 66.) When three, four, or more unknown quantities,
w^ith as many equations, are given, and their coefficients are
132
ELEMENTARY ALGEBRA.
all prime to each other, the operation is necessarily long.
But when several of the coefficients are multiples, or mea-
sures of each other, or unity, several expedients may be re-
sorted to for the purpose of facilitating calculation.
No specific rules can be given for mere expedients. Exam-
ples alone can illustrate. Some few expedients will be illus-
trated by the following
EXAMPLES.
■x-\-y-{-z=zm-\
1. Given •{ x-\-y — 2=25 )■ to find x, y, and z.
X — y — z= 9 J
Subtract the 2d from the 1st, and 22=6.
Subtract the 3d from the 2d, and 2?/= 16.
Add the 1st and 3d, and . . 2;r=40.
2. Given \ x — y = 4 V to find ar, y, and z.
yx — z = 6J
Add all three, and 3:r=36, or a:=12.
r x—y—z= 6-^
3. Given \ 3y — x — z— 1 2 V to find x, y, and z. . .
I 'lz—y—x=^^]
4. Given a;4-iy=100, y-\-\z^\m, 2+^0;= 100, to find
X, y, and z.
Puta=100. . Ans. a:=64; y=72 ; and 2=84.
' u-\-v-\-x-\-y=\0
u-\-v-\-z-\-x=:^ 1 1
6. Given ■{ u-\-v-\-z-\-x=12 S- to find the value of each.
u-\-x-{-y-\-z—13
v-]-x-\-y-\-z='i4^
Here are Jive letters and five equations. Each letter has
EQUATIONS. 133
the same coefficient, one understood. Each equation has 4
letters, z is wanting in the 1st equation, y in the 2d, &c.
Now assume u-]rv-\-x-]ry-]rz=^s.
Then 5—2=10 {A)
s—y=\\
s—x—l^
5— v=13
s — w=14
Add, and .... 5s—s=Q0 Or «= 15.
Put this value of « in equation {A), and 2=5, &c.
6. Given ir+y=a, a;-l-2=6, 2^4-0=c.
Add the 1st and 2d, and from the sum subtract the 3d.
Am. a;=i(a+5 — c), y=l{a-\-c — 5), z—\(J)-{-c — a).
5. Reduce the
equations
6. Reduce the
equations
' x-\-y=5^
y+ 2=82
2-1- 2^=68
u-\-x=32
ix-^3y=23
.+ ^=8
J/+ 32=31
x-\-y +2-l-2it'=39
Ans.
7. Reduce the
equations
8, Reduce the
equations
r4ar-|-2y— 32= 4
} 3a;— 5y+22=22
I x-{- y-{- 2=12
•i.r+iy-f 2=46 >|
\x— y-hzi= 9 I
a^+|y—i2=19J
-4w«.
Ans.
Ans,
a:=20,
y=32,
2=50,
m;=18,
w=12.
a;=6,
I 2=8,
lw=9.
ar=5,
y-1,
2=6.
a:=20,
y=12,
2=32.
134 ELEMENTARY ALGEBRA
QUESTIONS PRODUCING EQUATIONS
CONTAINING THREE OR MORE UNKNOWN QUANTITIES
1. There are three persons, A, B, and C, whose ages are as
follows : If B's age be subtracted from A's, the difference
will be C's age ; if five times B's age and twice C's, be added
together, and from their sum A's age be subtracted, the
remainder will be 147. The sum of all their ages is 96.
What are their ages? Ans. A's, 48; B's, 33; C's, 15.
2. Find what each of three persons. A, B, and C, is worth,
from knowing, 1st, that what A is worth added to 3 times
what B and C are worth make 4700 dollars; 2d, that what
B is worth added to 4 times what A and C are worth make
5800 dollars ; 3d, that what C is worth added to 5 times
what A and B aie worth make 6300 dollars.
Ans. A is worth 8500; B, 8600; C, 8800.
3. A gentleman left a sum of money to be divided among
four servants, so that the share of the first was ^ the sura
of the shares of the other three ; the share of the second, J-
of the sum of the other three ; and the share of the third, J-
the sum of the other three ; and it was found that the share
of the last was 14 dollars less than that of the first. What
was the amount of money divided, and the shares of each
respectively ?
Ans. The sum was 8120 ; the shares, 40, 30, 24 and 26.
4. The sum of three numbers is 59; | the difference of
the first and second is 5, and ^ the difference of the first and
third is 9; required the numbers. Ans. 29, 19, and 11.
5. There is a certain number consisting of two places, a
unit and a ten, which is four times the sum of its digits, and
if 27 be added to it, the digits will be inverted. What is the
number? Ans. 36.
Note. — Undoubtedly the reader has learned in Arithme-
tic that numerals have a specific and a local value, and every
EQUATIONS. 135
remove from the unit multiplies by 10. Hence, if x repre-
sents a digit in the place of tens, and y in the place of units,
the number must be expressed by 10x-\-y. A number con-
sisting of three places, with x, y, and z to represent the digits,
must be expressed by 100a;-l-]0?/+s.
6. A number is expressed by three figures ; the sum of
these figures is 1 1 ; the figure in the place of units is double
tliat in the place of hundreds, and when 297 is added to this
number, the sum obtained is expressed by the figures of this
number reversed. What is the number ? Ans. 326.
7. Divide the number 90 into three parts, so that twice
the first part increased by 40, three times the second part
increased by 20, and four times the third part increased by
10, may be all equal to one another.
A71S. First part, 35; second, 30; and third, 25.
If the object is merely to solve the seventh example, it
would not be expedient to use three unknown symbols.
Let ^(x — 40)= the first part, &c.
8. Find three numbers, such that the first with J- of the
other two, the second with 1 of the other two, and the third
with I of the other two, shall be equal to 25.
Ans. 13, 17, and 19.
9. A man with his wife and son, talking of their ages, said
that his age, added to that of his son, was 16 years more
than that of his wife ; the wife said that her age, added to
that of her son, made 8 years more than that of her husband ;
and that all their ages added together amounted to 88 years.
What was the age of each ?
Ans. Husband, 40, wife, 36, and son 12 years.
10. There are three numbers, such that the first, with i
the second, is equal to 14; the second, with i part of the
third, is equal to 18; and the third, with i part of the first, is
equal to 20; required the numbers. Ans. 8, 12, and 18.
136 ELEMENTARY ALGEBRA.
11. Find three members, such that i of the first, i of the
second, and } of the third shall be equal to 62; i of the first,
I of the second, and i of the third equal to 47; and i of the
first, i of the second, and ^ of the third equal to 38.
Ans. 24, 60, and 120.
12. There are two numbers, such that i the greater added
to 1 the lesser, is 13; and if i the lesser is taken from ^ the
greater, the remainder is nothing. Required the numbers.
Ans. 18 and 12.
12. Find three numbers of such magnitude, that the first
with the ^ sum of the other two, the second with ^ of the
other two, and the third with i of the other two, may be the
same, and amount to 51 in each case. Ans. 16, 33, and 39.
14. A said to B and C, ** Give me, each of you, 4 of your
sheep, and I shall have 4 more than you will have left." B
said to A and C, **If each of you will give me 4 of your
sheep, I shall have twice as many as you will have left." C
then said to A and B, "Each of you give me 4 of your
sheep, and I shall have three times as many as you will have
left." How many had each ? Ans. A, 6; B, 8; C, 10.
16. A person bought three silver cups ; the price of the first,
with ^ the price of the other two, was 26 dollars ; the price
of the second, with i of the price of the other two, was 26
dollars ; and the price of the third, with i the price of th*»
other two, was 29 dollars ; required the price of each.
Ajis. $8, $18, and $16
16. A's age is double that of B's, and B's is triple that of
C's, and the sum of all their ages is 140; what is the age of
each? Ans. A's =84; B's =42; C's =14.
17. A man wrought 10 days for his neighbor, his wife 4
days, and son 3 days, and received 1 1 dollars and 60 cents ;
at another time he served 9 days, his wife 8 days, and his son
6 days, at the same rates as before, and received 1 2 dollars ;
a third time he served 7 days, his wife 6 days, his son 4 days,
EQUATIONS. 137
at the same rates as before, and received 9 dollars. What
were the daily wages of each ?
A71S. Husband's wages, 81.00; wife, 0; son, 50 cts.
(Art. 67.) In this last example we put x to represent the
daily wages of the husband, y the wages of the wife, and z
the wages of the son, and in conclusion, y was found equal to
0 ; but it might have came out a minus quantity, and if it had,
it would have shown that the presence of the wife was not a
source of income, but expense ; and if correct results are
given at the settlements, the si^ns to the different quantities
will show whether any particular individual received wages
or was on expense for board, as in the following problems :
18. A man worked for a person ten days, having his wife
with him 8 days, and his son 6 days, and he received 10 dol-
lars and 30 cents as compensation for all three ; at another
time he wrought 12 days, his wife 10 days, and son 4 days,
and he received 13 dollars and 20 cents ; at another time he
wrought 15 days, his wife 10 days, and his son 12 days, at
the same rates as before, and he received 4 3 dollars 85 cents.
What were the daily wages of each ?
Ans. The husband 75 cts.; wife, 50 cts. The son 20 cts.
expense per day.
Here the language of the problem is improper, as It im-
plies that all received wages ; but the solution shows that this
could not be the case ; for the value of the son's wages comes
out minus, which is opposition to plus or to positive wages, that
is, expense.
A stronger illustration of this principle will be shown by
the following problem :
19. Two men, A and B, commenced trade at the same
time ; A had 3 times as much money as B, and continuing in
trade, A gains 400 dollars, and B 150 dollars; now A has
twice as much money as B. How much did each have at
first?
12
138 ELEMENTARY ALGEBRA.
Without any special consideration of the question, it imphes
that both had money, and asks how much. But on resolving
the question with x to represent A's money, and y B's, we
find x= — 300
And 2/=— 100 dollars.
That is, they had no money, and the minus sign in this
case indicates debt; and the solution not only reveals the nu-
merical values, but the true conditions of the problem, and
points out the necessary corrections of language to correspond
to an arithmetical sense.
That is, the problem should have been written thus :
A is three times as much in debt as B ; hut A. gains 400 dol-
lars, and B 150 ; now A has twice as much money as B. How
much were each in debt ?
As this enunciation corresponds with the real circumstances
of the case, we can resolve the problem without a minus sign
in the result. Thus :
Let x= B's debt, then 3ar= A's debt.
150 — x= B's money, 400 — 2>xz= A's money.
Per question, 400— 3a; =300— 2a:. Or a;=]00.
20. What number is that whose fourth part exceeds its
third part by 12 ? Ans. —144.
But there is no such abstract number as — 144, and we
cannot interpret this as debt. It points out error or impos-
sibUity, and by returning to the question we perceive that a
fourth part of any number whatever cannot exceed its third
part ; it must be, its third part exceeds its fourth part by 1 2,
and this enunciation gives the positive number, 144. Thus
do equations rectify subordinate errors, and point out special
conditions.
21. A man when he was married was 30 years old, and his
EQUATIONS. 139
wife 15. How many years nmst elapse before his age will be
three times the age of his wife ?
Ans. The question is incorrectly enmiciated ; 7|- years
before the marriage, not after, their ages bore the specified
relation.
22. What fraction is that which becomes | when 1 is
added to its numerator, and becomes f when 1 is added to
its denominator ?
A71S. In an arithmetical sense, there is no such fraction.
The algebraic expression, Zxj> "^iH give the required results.
23. Divide the number 10 into two such parts that their
product shall be 50.
Let ^4-y= the greater number.
And x — y= the less.
Then 22^=10, or x=5.
The product of the two numbers is x^ — y^, and by the
question must be equal to 50.
That is, .... x^—y^—50.
But, x=5; hence, a^=25, which, drop from both members.
And —f=25
Or y*=— 25
That is, the question calls for two equal factors whose pro-
duct is minus 25; but equal factors will never give a minus
product; there is no 2d root of — 25, and the value of the
unknown quantity in such cases is said to be imaginary,
which shows that the problem is impossible.
Here, y=±5^ — 1, a value that has no existence in
numbers.
SECTION III.
INTOLUTION.
(Art. 68.) Equations, and the resolution of problems pro-
ducing equations, do not always result in the first powers of
the unknown (Quantities, but different powers are frequently in-
volved, and therefore it is necessary to investigate methods of
resolving equations containing higher powers than the first ;
and preparatory to this we must learn involution and evolution
of algebraic quantities.
(Art. 69.) Involution is the method of raising any quan-
tity to a given power. Evolution is the reverse of involution,
and is the method of determining what quantity raised to a
proposed power will produce a given quantity.
As in Arithmetic, involution is performed by multiplication,
and evolution by the extraction of roots.
The first power is the root or quantity itself.
The second power, commonly called the square, is the
quantity multiplied by itself.
The third power is the product of the second power by the
quantity.
The fourth power is the third power multiphed into the
quantity, &c.
Or we may consider the second power of a quantity to be
the quantity taken twice as a factor.
The third power is the quantity taken three tirties as a factor.
INVOLUTION. 141
The fourth power is tlie quantity taken four times as a
factor.
The tenth power is the quantity taken ten times as a factor ;
and so on for any other power.
The nth. would be the quantity taken n times as a factor.
Thus, let a represent any quantity.
Its frst power is . . . a=a
Its second power is .
a*a=c^
The third power is .
. a'a'a=a^
The fourth power is .
a*a*a*a=^a^
The fifth power is a*a*a'a*a=^c^
In general terms a to the n\h power is a'a, &c., =a", and
n may be any number whatever.
(Art. 70.) When the quantity is negative, all the odd
powers will be minus, and all the even powers will be plus.
For, by the rules of multiplication,
— aX— a=+a*
And . . . — aX — aX — a= — a^
&c.=<&c.
(Art. 71.) When we require the 5th power of a, we
simply write a^; when the 8th power we write a^, &c., for
any other power.
That is, a is the same as a\ the exponent 1 is. understood ;
and when we require the nth. power of a, we conceive its
exponent 1 understood multiplied by n, which makes a".
The second power of a? is a^Xc^=a^.
The third power of a? is a^'a^*a^=a^.
The tenth power of a^ is a^^.
The nth power of a? is a^".
That is, it is the exponent of the quantity repeated as many
times as there are units in the index of the power.
Thus, the 7th power of d^ has the exponent of a (4) re*
peated 7 times, and the result is a*.
142
ELEMENTARY ALGEBRA,
From thig we derive the following rule to raise a single
quantity to any power.
Rule . — Midtiply the exponent of iJie qxtaniity hy the index
qf the required power.
EXAMPLES
Raise a^ to the 3(1 power. .
Raise i/* to the 4th power. .
Raise i" to the 5th power. .
Raise a? to the 4th power. .
Raise y' to the 3d power. .
Raise x^ to the 6th power. .
Raise x^ to the mth power. .
Raise aa? to the 3d power. .
Raise ah^x'^ to the 2d power. .
Raise cV to the 5th power. .
. Ans. 2^
. Am. y^.
. Ans, I^\
. Ans. x^^
. Ans. y^V
. Ans. a^^
, Ans. a;""^
. Ans. aV
Ans. d^b'^3?.
Ans. c^y
(Art. 72.) By the definition of powers, the second power
is any quantity multiplied by itself; hence the second power
of ax is a^x^, the second power of the coefficient a, as well as
the other quantity x; but a may be a numeral, as 6x, and its
second pcywer is 36x^. Hence, to raise any simj^le quantity to
any power, raise the numeral coefficient, as in Arithinetic, to
Vie required power, and annex the powers of the given literal
quantities.
EXAMPLES.
1. Required the Sd power of 3ax^.
2. Required the 4th power of \y^.
3. Required the 3d power of — 2x.
4. Required the 4th power of — 3a:
6. Required the 2d power of 8a^i'.
6. Required the 3d power of b:^z.
Ans. 27a^x'^
Ans. If/
Ans. — 8ar*
Ans. Blx'^
Am. Gia'^b^.
Ans. 125x«^.
INVOLUTION. 143
7. Required the 3d power of Qcc'^f-x. . Ans. 2\^a}^tf^.
8. Required the 4tli power of 2am^ . Ans. 16a^6V.
(Art^ 73.) When the quantity to be raised to a power is
a fi'action, we must observe the rules for the multiplication
of fractions, and multiply numerators by numerators, and
denominators by denominators.*
Thus, the 2d power of - is — — -= —
Hence, to raise fractions to powers, we have the following
Rule. — Raise both numerator and denominator to the re-
quired power.
EXAMPLES.
Observe, that by the rules laid down for multiplication, the
even powers of minus quantities must be plus, and the odd
powers minus.
9.a%^ 4^4^12
1. Required the 2d power of -— ~. . . . Ans. ~p-j*
2. Required the 6th power of — --. . . . Ans. — — -•
* Suppose we were required to raise r to the fifth power, and did not
know whether the denominator was to be raised or not, we could decide
the point by means of an equation, as follows :
The fraction has soine value which we represent by a symbol, say P.
Then P=r' Now if we can find the true 5th power of P, it will be
the required 5th power of the fraction.
Clearing the equation of fractions, we have
bP=a
Taking the 5th power of both members gives
a>
By division, . . . . . P^—j-.
b"
This equation shows that to raise any fraction to any power, the
numerator and denominator must be raised to that power
14
ELEMENTARY ALGEBRA.
3.
Required the 6tli power of — • • •
Am. - ^ -.
4.
Required the 6\h. power of |a^5. . .
Am. 4>«
5.
Required the 2d power of -g. . . .
. 9
. . Ans. -.'
a*
6.
Required the 3d power of -^.
. . Ans. -r--'
7.
2y
Required the 4th power of — —. . .
16/
8.
Required the 3d power of ^. • •
(Art. 74.) The powers of compound quantities are raised
by the application of the rule for compound multiplication,
(Art. 14).
Let a-^b be raised to the 2d, 3d, 4th, &c., powers,
a +5
a-\-b
a +a6
ab-^b''
2d power or square, a^-^-^ab ~\-b^
a-\-b
a?-^S>a^b-\-ab^
a^b-\-^aJ)'-\-b^
3d power or cube, a?-{-2>d^b-{-?>aI^-\-b^
a-\-b
a^+2>a%-\-2,a^b''-\-ab''
a?b-\-Sa%^-\-^ab''-\-b''
The 4th power, . a^-{-^a^b^Q(rb'-^^ab^-\-h'
a-\-b
a^J[.^a%-\-Qa:'b''+^d'b^-\-ab''
a%-{-Aa?b^-\-Qa^b^-{-Aad'-^b''
The 6th power, . a^+5a:^b-{-\0a?b''-\'\0a^b^-\-bab'^-\-b^
INVOLUTION. 145
By inspecting the result of each product, we may arrive at
general principles, according to which any power of a bino-
mial may be expressed, without the labor of actual multipli-
cation. This theorem for abbreviating powers, and its general
application to both powers and roots, first shown by Sir Isaac
Newton, has given it the name of Newton's binomial, or the
binomial theorem.
Observations. — Observe the 5th power : a, being the first,
is called the leading term ; and h, the second, is called the
following term. The sum of the exponents of the two let-
ters in each and all of the terms amount to the index of
the power. In the 5th power, the sum of the exponents of a
and b is 5; in the 4th power it is 4; in the 10th power it
would be 10, &c. In the 2d power there are three terms ; in
the 3d power there are 4 terms ; in the 4th power there are
5 terms ; always one more term than the index of the power
denotes.
The 2d letter does not appear in the first term; the 1st
letter does not appear in the last term.
The highest power of the leading term is the index of the
given power, and the powers of that letter decrease by one
from term to term. The second letter appears in the 2d term,
and its exponent increases by one from term to term, aS the
exponent of the other letter decreases.
The 8th power of (a-\-b) is indicated thus, (a-{-6)^ Wlien
expanded, its literal part (according to the preceding obser-
vations) must commence with a^, and the sum of the expo-
nents of every term amount to 8, and they will stand thus
a\ a% a^h\ a%\ a'b\ a'b\ d'b\ ab\ b^
The coefiicients are not so obvious. However, we observe
that the coefficients of the first and last terms must be uniiy
The coefficients of the terms next to the first and last are
equal, and are the same as the index of the power. The
coefficients increase to the middle of the series, and then
decrease in the same manner, and it is manifested that there
13
146
ELEMENTARY ALGEBRA.
must be some law of connection between the exponents and
the coefficients.
By inspecting the 6th power of a-{-b, we find that the 2d
coefiicient is 5, and the 3d is 10.
5X4
10
The 3d coefiicient is the 2d, multiplied by the exponent of
the leading letter, and divided hy the exponent of the second
letter increased hy unity.
In the same manner, the fourth coefficient is the third, mul-
tiplied by the exponent of the leading letter, and divided by
the exponent of the second letter increased by unity, and so
on from coefiicient to coefficient.
The 4th coefficient is
The 6th is . . . .
The last is . . . .
Now let us expand .
For the 1st term write
For the 2d term write
8X7
2
28X6
10X3_
3
10X2_
4
5X1
10
5
1 understood.
For the 3d,
For the 4th,
For the 6th,
=28
3
56X6
^d'b
28a«52
lOa'b*
Now, as the exponents of a and b are equal, we have
arrived at the middle of the series, and of course to the high-
est coefficient. The coefficients now decrease in the reversa
order in which they increased.
INVOLUTION. 147
Hence, the expanded power is
Let the reader observe, that the exponent of b, mcreased
by unity, is always equal to the number of terms from the
beginning or from the left of the power. Thus, 5^ is in the
3d term, &c. Therefore in finding the coefficients, we may
divide by the number of terms already written, in place of
the exponents of the second term increased by unity.
If the binomial {a-\-b) becomes (a-j-l), that is, when b
becomes unity, the 8th power becomes,
a«+8a^-|-28a«+56a°+70a*+56a3H-28a24-8a+l.
• Any power of 1 is 1, and 1 as a factor never appears.
If a becomes 1, then the expanded power becomes,
H-8S+2862+56&'-l-706^+5655+285«-f-8i^+i«.
The manner of arriving at these results is to represent the
unit by a letter, and expand tlie simple literal terms, and after-
ward substitute their values in the result.
(Art. 74.) If we expand (a — b) in place of (a-|-i), the
exponents and coefficients will be precisely the same, but the
principles of multiplication of quantities afifected by different
signs will give the minus sign to the second and to levery
alternate term.
Thus, the 6th power of (a — b) is
a^—Qdb-[- 1 ba%^—ma%^-\- 1 5a'b'—6al^+b\
(Art. 75.) This method of readily expanding the powers
of a binomial quantity is one apphcation of the *' binomial theo-
rem,'' and it was thus by induction and by observations on
the result of particular cases that the theorem was established.
Its rigid demonstration is somewhat difficult, but its applica-
tion is simple and useful.
Its most general form may arise from expandmg {a-\-'b)^.
When w=3, we can readily expand it.
When 71=4, we can expand it.
148 ELEMENTARY ALGEBRA.
Wlien 71= any whole positive number, we can expand it.
Kow let us operate with n just as we would with a known
number, and we shall have
ji \
1
We know not where the series would terminate, until we
know the value of n. We are convinced of the truth of the
result, when n represents any positive whole number ; but let
n be negative or fractional, and we are not so sure of the
result.
The result would be true, however, whatever be the value
of n; but this requires demonstration, and a deeper investi-
gation than it would be proper to go into in a work like this.
When w is a fraction, the operation is extracting a root in
place of expanding a power.
But for the demonstration of the binomial theorem, and its
application to the extraction of roots, we refer the reader to
our University Edition of Algebra.
EXAMPLES.
1. Expand {x-^yf Ans. ar'+Sar'y-f Sa^-fy*.
2. Expand (y-f-^)^.
3. Required the third power of 3a;-|-2y.
We cannot well expand this by the binomial theorem,
because the terms are not simple literal quantities. But we
can assume Sx=a and 2y=b. Then
3a:+2y«a+i, and (a-t-i)3=a'+3«'i+3a6'-f *"
INVOLUTION. 149
Kow to return to the values of a and h, we have,
3ab^=:3 XSxX 4y^=36xy^
Hence, (3x+2yy=27u^+54x'2j-\-3Gxy'-\-Qf.
4. Required the 4th power of 2a^ — 3.
Let x=:2a^, y=o. Then expand (x — yy, and return the
values of x and y, and we shall find the result.
16a«— 96a«+216a^— 216a2+81.
5. Required the cube of (a-^b-\-c-\-d).
As we can operate in this summary manner only on bino-
mial quantities, we represent a-\-b by x, or assume x=a-]-b,
and y=c-\-d.
Then {xi-yy=a^^Sx'y+3xy^-\-f'
Returning the values of x and y, we have
(a+by+3(a-]-by{c+d)+3{a+bXc-{-dy+(c-\-dy.
Now we can expand by the binomial, these quantities con-
tained in parentheses.
6. Required the 4th power of 2a-\-3x.
Ans. 16a''-\-96a^x+216a^x^-{-216ax^-\-81x\
7. Expand (x'^-{-37fy.
Ans. a;^°+ 1 5a;«y2_|_9o^6y^270;2;y4.405a;y+2432/i°.
8. Expand {^a'+axy. Ans. 8a«+12a^a;+6aV+aV.
9. Expand (x—iy.
Ans. x^—6x^-\-15x'*—20x''-\-15x^—6x-\-l.
10. Expand (3a:— 5)^ Ans. 27x^—135x^-{-225x^—n5.
11. Expand (2a— 55)=^. Ans. Sa^—60a''b-{-150ab^—125bK
12. Expand {4a'b—2c^y.
Ans. 25Ga}^b*—512a%h^+3Ma^Pc'^—128(^bc^- 1 &<?,
ISO ELEMENTARY ALGEBKA.
EVOLUTION.
(Art. 76.) Evolution is tlie converse of involution.
Involution is the expanding of roots to powers. Evolutidte
is extracting the root when the power is given.
To find rules for operation, we must observe how powers
are formed, and then we shall be able to trace the alterations
hack. Thus, to square a, we double its exponent, which
makes a^, (Art. 71). The square of a^ is a"*, the cube of a^
is a°, &c. Take the 4th power of x, and we have x'^. The
wth power of ^^ is x^^, &c., &c.
Now, if mivltiplying exponents raises simple literal quanti-
ties to powers, dividing exponents must extract roots. Thus,
2.
the square root of a** is a^. The cube root of d^ must be a^.
The cube root of a must have its exponent, (1 understood)
divided by 3, which will make a^.
Therefore, roots are properly expressed hy frojctional exponents.
. The square root of a is d^, and the exponents, J , i, i, &c.,
indicate the third, fourth, and fifth roots. The 6th root
of or' is x^ ; hence, we perceive that the numerators of tht
exponent indicate the power of the quantity, and the denom-
nator the root of that power.
(Art. 77.) The square of ax is a^x^. We square both
factors, and so, for any other powers, we raise all the factors
to the required power. Conversely, then, we extract roots
by taking the required roots of all the factors. Thus the
cube root of 8a;^ is 2x.
The square root of 64a'* is obviously 8a^ and from these
examples we draw the following rule for the extraction of
roots of monomials.
EVOLUTION. 151
Rule . — Extract the root of the numeral coefficients, and
divide the exponent of each letter hy the index of the root.
EXAMPLES.
1 . What is the second root of SaVj/^ ? , . Ans. 2>ax^y^.
2. What is the third root of SaV ? . . . Ans. 9,a^y.
3. Whatis the/owr^A rootof 81aV2? . . Ans. 2,a^.
4. What is the fifth root of 32aVV^ ? • ^^«- 2a.tY-
For illustration, we will observe that this last example requires
us to find five equal factors, which, when multiplied together, will
produce 32 ; and^Zve equal factors, which, when multiplied together,
will produce a^,five equal factors, which, when multiplied together,
will produce x^^ , and ^re equal factors, which, when multiplied
together, will produce y^^.
Now, a* shows five equal factors, each equal to a ; therefore, a
is one of the factors required. In the same manner ar'" shows
ten equal factors, and the product of two of these, or x^ is one of
the five equal factors required. In the same manner y^ is another
of the equal factors ; and there is no trouble in finding any root
of any hteral monomial quantity; for all we have to do is to
divide its exponent (whatever it may be) by the index of the
proposed root. But when the factor is a numeral, like 32, we can
find the factor only by trial. Sometimes no such factor as the
one required exists ; in such cases we consider the number as a
letter with 1 understood for its exponent, and then divide such
exponent by the index of the root. For example, the fifth root
of 32 is (32)^; but this is only an indication of the root or fac-
tor, not an actual discovery of it. Take particular notice of the
following examples :
L 3.
6. What is the third root oi 1 a^x^ 1 . . Ans. I'^a^x.
The number 7 is here regarded as a letter.
6. What is the second root of 20aaj? Ans. ±(20)20^.
7. What is the fourth root of 16aV ?
Ans. — 2aa:', or 2aa;'.
8. What is the square root of 36aV ^ • • ^^^- i^ay^.
li^ ELEME^'TARY ALGEBRA.
(Art. 78.) The even roots of algebraic quantities may-
be taken with the double sign, as indicating either plus or
minus, for either quantity will give the same square, and we
may not know which of them produced the power, (Art 70).
For example, the square root of 16 may be either -[-4 or — 4,
for either of them, when multiplied by itself, will produce 16.
The cube root of a plus quantity is always plus, and the
cube root of a minus quantity is always minus. For -i-2a
cubed, gives -f-8a^, and — 2a cubed, gives -—8a'', and a may
represent any quantity whatever.
9. What is the fourth root of Sla^iV* ? Ans. itSaJV.
10. What is the third root of — 27a^V ? Ans. — 3a%.
U. What is the third root of I6a^? Ans. 2a(2a)'^.
In this example it is obvious that there are no three equal
factors, which, when multiplied together, will produce 16, and
there are no three equal factors expressed in entire quantities
L 1
that will produce a'*. Therefore, we must write (16)^ a^ for
the answer. But this is only indicating the operation, not
performing it, and we have no clearer idea of the result now
than at first. However, to see what can be done, we will
separate 16a!^ into the two factors (8a'')(2a). The first of
these is a complete third power, and the other is not ; but the
third root of the whole is the third root of the two factors
written together as a product; that is, 2a(2a)=*, and this is
all we can do to reduce or simplify it.
12. What is the second root of 20aV ? Ans. ±.2ax{5x)K
All the square factors in this are 4a?x^, the other fjictors
are 5x. We can take the second root of the square factors,
and of the others we cannot. In relation to them we can only
indicate the root. Therefore, the whole root is d^2ax(5x)'^.
13. What is the second root of 75 ? .
14. What is the second root of 9Qa^x ?
15. What is the third root of 32a^ ? .
16. What is the third root of 24a";J^ ?
17. What is the third root of 27a^ ? .
18. What is the third root of 19a ? .
EVOLUTION. 16»
Ans. ±5(3)^.
Ans. ±7a(2x)^.
Ans. 2a(4)^.
Ans. 2a{3x'^Y\
. Ans. 3a.
,Ans. (19a) 3.
(Art. 78.) By comparing examples 17 and 18, we perceive
that some monomials have such roots (or what is the same
thing), such equal factors as may be required, and some have
not. When no such factors exist, all we can do is to indicate
an operation, to be performed as example 18. So it is with
polynomials — some may have equal factors, and others not.
When equal factors do exist in any polynomial, they are com-
monly apparent to any one who has had a little experience in
raising roots to powers as explained in Articles 73, 74, and
75. For instance, any one can perceive that the polynomial
a'-\-2ab-\-P has two equal factors, each equal to (a-\-b) ; and
after a little more observation we can perceive that the poly-
nomial a^-\-3x^y-{-Sxi/-]ry^ has three equal factors, each equal
to (x-\-y), or perceive that (x-^ry) is its third root.
It is only regidar polynomials that have equal factors, and
it is only by observing how the powers are formed by multipli-
cation that we can determine
HOW TO EXTRACT ROOTS OF POLYNOMIALS.
On the supposition that we know that the square root of
the polynomial
a^-^2ab-\-h\ is (a-fi),
we propose to extract it out of the polynomial itself.
We know that a^, the first term, must have been formed by
the multiplication of a into itself, therefore, a must be part of
the root sought.
154 ELEMENTARY ALGEBRA.
The next term is 2aXb, that is tidce tlie root of the first term
into the second term of the root. Hence, if we divide the
second term of the square by twice the root of the first term,
we shall obtain b, the second term of the root, and as b must
be multiplied into itself to form a square, we add b to 2a, and
2a-{-b we call a divisor.
OPERATION.
a'-{-2ab-\-F(a-{-b
«2
2a-\-b)2ab-^b^
2ab+b^
"We take a for the first term of the root, and subtract its
square (a^) from the whole square. We then double a and
divide 2ab by 2a and w^e find b, which we place in both the
divisor and quotient. Then we multiply 2a-\-b by 5, and we
have 2ab-\-b^, to subtract from the tAVO remaining terms of
the square, and in this case nothing remains.
Again, let us take a-\-b-\-c, and square it. We shall find
its square to be
a'-}-2ab-{-b''-\-2ac-\-2bc-\-c'
a^+2ab-\-b''-{-2ac-\'2bc-{-c\a-\-b-\-c
2a-\-b 2ab+U'
2ab-^P
2a-r2b-^c 2ac-{-2bc-\-cr'
2ac-\-2bc-\-i^
By operating as before, we find the first two terms of the
root to be a-\-b, and a remainder of 2ac-\-2bc-\-(^. Double
the root already found, and we have 2a-\-2b for a partial
divisor. Divide the first term of the remainder 2ac by 2oy
and we have c for the third term of the root, which must be
added to 2a-\-2b to complete the divisor. Multiply the divisor
EVOLUTION. 155
by the last term of the root, and set the product under the
three terms last brought down, and we have no remainder.
Again, let us take a+^+c to square ; but before we square
it, let the single letter s=a-{-b.
Then we shall have s-\-c to square, which produces
To take the square root of this, we repeat the first opera-
tion, and thus the root of any quantity can be brought into
a binomial, and the rule for a binomial root will answer for a
root containing any number of terms by considering the root
already/ found, however great, as one term.
Hence, the following rule to extract the square root of a com-
pound quantity.
Rule . — Arrange the terms according to the powers of some
letter, beginning vnth the highest, and set the square root of the
first term in the quotient.
Subtract the square of the root thus found from the first tetTn,
and bring doion the next two tei-ms for a dividend.
Divide the first term of the dividend by double of the root
already found, and set the result both in the root and in the
divisor.
Multiply the divisor, thus completed, by the term of the root
last found, and subtract the product from the dividend, and so on.
EXAMPLES.
1 . What is the square root of
a^-\-4a'b-\-4b''—4a^-'8b-\-4(a^-\-2b—2
«4
2a'4"25 )4a'b-\-4P
4a^^4b^
2a^+4b—2 _4a2_86-f-4
1&6 ELEMENTARY ALGEBRA.
2. Wliat is the square root of 1 — 4b-\-4b^-{-2^ — 4^;y+?/2?
Ans. 1 — 26-ry.
3. What is the square root of 4x^—4x^-\'\Sx'^—6x-{-9 ?
Ans. 2x^ — ^ar-f-S.
4. What is the square root of 4x^—163^-\-24x^ — \6x-\-4 ?
Ans. 2aP — 4x-\-2.
5. What is the square root of IGx^ -f 24a^ + 892:= + 60a;
100? . A71S. 4x^-i-3x-hy0.
6. What is the square root of 4x'^^\6a^-^dx'^-{-16x-{-4 ?
Ans. 2x^ — 4x — 2
7. What is the square root of x^ -{-2iy-]-y^ -]-Gxz-\-Gi/z
+92:2? - jns. x+7/-^3z.
8. What is the square root of a^ — ab-\-lb^ ? Ans. a — ^b
a^ b^
9. What is the square root of -r^ — 2-[— j?
b^
. a b b a
Ans. or -.
0 a a b
2. J JL 2.
10. What is the square root of x"^ — 2x'^y^-\-y^ ?
. L 1 1 J.
Ans. x^ — y^ or y^ — x^.
(Art. 79.) Every square root will be equally a root if we
change the sign of all the terms. In the first example, for
instance, the root may be taken — a' — 2b-\-2, as well as
a^-\-2b — 2, for either one of these quantities, by squaring,
will produce the given square. Also, observe that every
square consisting of three terms only, has a binomial root.
Algebraic squares may be taken for formulas, correspond-
ing to numeral squares, and their roots may be extracted in
the same way, and by the same rule.
For example, a-{-b squared is a^-{-2ab-\-b^, and to apply
this to numerals, suppose a =40, and 5=7.
EVOLUTION. 157
Then the square of 40 is a'=1600
2(z5= 560
1?^ 49
Therefore, .... (47)2=2209
Now the necessary divisions of this square number, 2209,
are not visible, and the chief difficulty in discovering the root
is to make these separations.
The first observation to make is, that the square of 10 is
100, of 100 is 10000, and so on. Hence, the square root of
any square number less than 100, consists of one figure, and
of any square number over 100 and less than 10000, of two
figures, and . so on. Every two places in a power demanding
one place in its root.
Hence, to find the number of places or figures in a root,
we must separate the power into periods of two figures, begin-
ning at the unit's place. For example, let us require the
square root of 22 09. Here are two periods indicating two
places in the root, corresponding to tens and units. The
greatest square in 22 is 16, its root is 4, or 4 tens =40.
Hence, a =40.
22 09(40+7=47
a2=l6 00
2a+J=804-7=87 )6 09
6 09
Then 2a=80, which we use as a divisor for 609, and find
it is contained 7 times. The 7 is taken as the ■^lue of h, and
2a-{-b, the complete divisor, is 87, which, multiplied by 7, gives
the two last terms of the binomial square. 2ab-{-b^~560-\-49
=609, and the entire root, 40-|-7=47, is found.
Arithmetically, a may be taken as 4 in place of 40, and
1600 as 16, the place occupied by the 16 makes it 16
hundred, and the ciphers are superflous. Also, 2a may be
394
369
25 24
25 24
158 ELEMENTARY ALGEBRA.
considered 8 in place of 80, and 8 in 60 (not in 609) is
contained 7 times, &c.
If the square consists of more than two periods, treat it as
two, and obtain the two superior figures of the root, and when
obtained, bring down another period to the remainder, and
consider the root already obtained as one quantity, or one
figure.
For another example, let the square root of 399424 be
extracted.
39 94 24(632
36
123
1262
In this example, if we disregard the local value of the
figures, we have a=6, 2a=12, and 12 in 39, 3 times, which
gives 5=3. Afterward we suppose a=63, and 2a=:126,
126 in 252, 2 times, or the second value of 5=2. In the
same manner, we would repeat the formula of a binomial
square as many times as we have periods.
EXERCISES FOR PRACTICE.
1. What is the square root of 8836 ? . . . Ans. 94.
2. What is the square root of 106929? . . Ans, 327.
3. What is the square root of 4782969 ? . . A7is. 2187.
4. What is the square root of 43046721 ? . . Ans. Q5Q\.
5. What is the square root of 387420489 ? Ans. 19683.
When there are whole numbers and decimals, point off
periods both ways from the decimal point, and make the deci-
mal places even, by annexing ciphers when necessary, extend-
ing the decimal as far as desired. When there are decimals
only, commence pointing off from the decimal point.
(
EVOLUTION. 159
EXAMPLES.
1. What is the square root of 10.4976 ? . . Ans, 3.24.
2. What is the square root of 3271.4207? Ans. 57.19+.
3. What is the square root of 4795.25731 ?
Ans. 69.247+.
4. What is the square root of .0030 ? . . Ans. .06.
5. What is the square root of .00032754 ?
Ans. .01809+.
6. What is the square root of .00103041 ? Ans. .0321.
As the square of any quantity is the quantity multiplied
by itself, and the product of - by - (Art. 64) is j^; hence, to
take the square root of a fraction, we must extract the square
/ root of both numerator and depominator.
A fraction may be equal to a square, and the terms, as given,
not square numbers ; such may be reduced to square numbers.
EXAMPLES.
What is the square root of -^-^j ?
Observe tVi"— e I • Hence, the square root is f .
1 . What is the square root of /2V ? • • • • Ans. |.
2. What is the square root of i|| ? . . . . Ans. f .
3. What is the square root of f s|| ? . . . . Ans. f .
4. What is the square root of |||A ? . , . . . Ans. |.
When the given fractions cannot be reduced to square
terms, reduce the value to a decimal, and extract the root, as
in the last article.
TO EXTRACT THE CUBE ROOT OF COMPOUND
QUANTITIES.
(Art. 80.) We may extract the cube root in a similar
manner as the square root, by dissecting or retracing the
combination of terms in the formation of a binomial cube.
The cube of a-\-b is a^-\-'Sa'^b-\-2,alP-\-b'^ (Art. 67). Now,
to extract the root, it is evident we must take the root of the
160 ELEMENTARY ALGEBRA.
first term (a'), and the next term is 3a'^b. Three times the
square of the first letter or term of the root multiplied by tJie 9.d
term of the root.
Therefore, to find this second term of the root, we must
divide the second term of the power (3a^5) by three times
the square of the root already found (a).
Sa'b
When we can decide the value of b, we may obtain the
complete divisor for the remainder, after the cube of the first
term is subtracted, thus :
The remainder is . 3a^b-{-3a^b-\-b^
Take out the factor b, and 3a?-\-3ab-\-b^ is the complete
divisor for the remainder. But this divisor contains b, the
very term we wish to find by means of the divisor ; hence,
it must be found before the divisor can be completed. In
distinct algebraic quantities there can be no difficulty, as
the terms stand separate, and we find b by dividing simply
Sa^b by 3a^; but in numbers the terms are mingled together
and b can only be found by trial.
Again, the terms 3a?-\-3ab-\-b^ explain the common arith-
metical rule, as 3a^ stands in the place of hundreds, it corres-
ponds with the words : ** Multiply the square of the quotient
by 300," *♦ and the quotient by 30," (3a), &c.
By inspecting the various powers of a-\-b (Art. 73), we
draw the following general rule for the extraction of roots :
Rule . — Arrange the terms a/icording to the po^vers of some
letter; take the required root of the first term and place it in, the
quotient ; subtract its corresponding power from the first term,
and bring down the second term for a dividend.
Divide this term by tuice the root already found for the
SQUARE root, three times the square of it for the cube root, four
times the third power for the fourth root^ ^kc, and the qucHetU
EVOLUTION. 161
wUl be the next term of the root. Involve the whole of the root
ihvs found, to its proper power, which subtract from the given
quantity, and divide the first term of the remainder hy the same
divisor as before ; proceed in this manner till the whole root is
determined.
EXAMPLES.
1. What is the cube root of x^+6a^~403^-\-96x—64 ? -
x'-\-6aP—40x^-\-96x—64: (x^-\-2x—4.
x^
Divisor Sx'^) 6a:^= 1st remainder.
Divisor Sx* ) — 12a;'*=2d remainder.
a^-J^ex^—40x^+96x—6l
2. What is the cube root of 27a^4-108a=+144a+64 ?
Ans. 3a+4.
3. What is the cube root of c^—6a^x-h l^ax^—^a^ ?
Ans. a — 2x.
4. What is the cube root of x^—3a^-\-5sP—2x—l ?
Ans. x^ — X — 1.
6. What is the cube root of a^—6a''b-]-nah^--8b'''i
Ans. a — 2b.
3 1
6. What is the cube root of •'^+3.r+-+-3 ? ^^^ .1
X
7. Extract the fourth root of
a^+8a^+24a2+32a+16(a-f2
4a^) 8a^ <fec.
a^-l-8a''+24a2+32a+16
(Art. 81.) To apply this general rule to the extraction of
the cube root of numbers, we must first observe that the cube
of 10 is 1000, of 100 is 1000000, <fec.; ten times the root
producing 1000 times the power, or one cipher in the roa+
14
162 ELEMENTARY ALGEBRA.
producing 3 in the power; hence, any cube within 3 places
of figures can have only one in its root, any cube within 6
places can have only two places in its root, &c. Therefore,
we must divide off the given power into periods consisting of
three places, commencing at the unit. If the power contains
decimals, commence at the unit place, and count three places
each way, and the number of periods will indicate the number
of figures in the root.
EXAMPLES.
1. Required the cube root of 12812904.
12 812 904(234
a=2 a^z=z 8
Divisor 3^2= 12 )48
12167 = (23)^^
3(23)2=1587) 6459"(4
12 812 904=(234)«
Here, 12 is contained in 48, 4 times; but it must be
remembered that 12 is only atrial or partial divisor; when
completed it will exceed 12, and of course the next figlire of
the root cannot exceed 3.
The first figure in the root was 2. Then we assumed a=2.
Afterward we found the next figure must be 3. Then we
assumed a=23. To have found a succeeding figure, had
there been a remainder, we should have assumed a =234,
&c., and from it obtained a new partial divisor.
2. What is the cube root of 148877? .
3. What is the cube root of 571787 ? .
4. What is the cube root of 1367631 ?
6. What is the cube root of 2048383 ?
6. What is the cube root of 16581375 1
7. What is the cube root of 44361864 ?
8. What is the cube root of 100544625 ?
. . Am.
53.
Ans.
83.
, . Ans.
111.
. . Ans.
127.
. . Ans.
255.
. . Ans.
354.
. . Am.
465.
EVOLUTION. 163
(Art. 82.) The methods of dh'cct extraction of the cube
root of such numbers as have surd roots, are all too tedious
to be much used, and several eminent mathematicians have
given more brief and practical methods of approximation.
One of the most useful methods may be investigated as
follows :
Suppose a and a-\-c two cube roots, c being very small in
relation to a, a? and a^+Sa^c+Sac^-f-c^ are the cubes of the
supposed roots.
Now, if we double the first cube (a^), and add it to the
second, we shall have
If we double the second cube and add it to the first, we
shall have Sa^+Ga'c-f- 6(2(^4- Sc^
As c is a very small fraction compared to a, the terms con-
taining c^ and <? are very small in relation to the others ; and
the relation of these two sums will not be materially changed
by rejecting those terms containing c^ and c^, and the sums
will then be . . . . Sa^-f Sa^c
And ..... Za^-\-Qa?c
The ratio of these terms is the same as the ratio of a-j-c to
Or the ratio is 14— j—-
/•
But the ratio of the roots a to a+c, is 1 + -.
a
Observing again, that c is supposed to be very small in
relation to a, the fractional parts of the ratios — r— and - are
^ a-\-c a
both small, and very near in value to each other. Hence, we
have found an operation on two cubes which are near each
other in magnitude, that will give a proportion very near in
proportion to their roots ; and by knowing the root of one of
the cubes, by this ratio we can find the other.
164 ELEMENTARY ALGEBRA.
For example, let it be required to find the cube root of 28,
true to 4 or 5 places of decimals. As we wish to find the
cube root of 28, we may assume that 28 is a cube. 27 is a
cube near in value to 28, and the root of 27 we know to be 3.
Hence, a, in our investigation, corresponds to 3 in this
example, and c is unknown ; but the cube of a+c is 28, and
a^ is 27.
Then ... 27 28
2 2
64 66
Add ... 28 27
Sums . . . 82 : 83 : : 3 : a-\-c very nearly.
Or, (a-l-c)=Y/ =3.03658-1-, which is the cube root of
28, true to 6 places of decimals.
By the laws of proportion, which we hope more fully to
investigate in a subsequent part of his work, the above pro-
portion, . . . . 82 : 83 : : a : a+c,
may take this change, 82 : 1 : :a:c
Hence, c=j\; c being a correction to the known root, which,
being applied, will give the unknown or sought root.
From what precedes, we may draw the following rule for
finding approximate cube roots :
Rule . — Take the nearest rational cube to the given number y
or assume a root and cube it. Double this cube, and add the
number to it ; also double the number and add the assumed cube
to it. Then, by proportion, as the first sum is to the second,
so is the known root to the required root.
EXAMPLES.
1. What is the approximate cube root of 122 ?
Ans. 4.95967-f.
By the rule . 125X2=250 244
Add ... 122 125
372 : 369 : : 5 : Ans.
EVOLUTION. 165
2. What is tlie cube root of 10 ? . . Ans. 2.1 5U3-\-,
Assume 2.1 for the root, then 9.261 is its cube.
3. What is the approximate cube root of 720 ?
Ans. 8.9628+.
4. What is the approximate cube root of 345 ?
Ans. 7.013574-.
5. What is the approximate cube root of 520 ?
Ans. 8.04145-f.
6. What is the approximate cube root of 65?
Ans. 4.0207+.
7. What is the approximate cube root of 16 ?
The cube root of 8 is 2, and of 27 is 3; therefore the cube
root of 16 is between 2 and 3. Suppose it 2.5. The cube
of this root is 15.625, which shows that the cube root of 16
is a little more than 2.5, and by the rule
31.25 32
16 15.625
47.25 : 47.625 : : 2.5 : to the required root.
47.25 : .375: : 2.5:. 01984
Assumed root . . 2.50000
Correction . . . .01984
Approximate root 2.51984
We give the last as an example to be followed in most
cases where the root is about midway between twointegi-al
numbers.
This rule may be used with advantage to extract the root
of perfect cubes, when the powers are very large.
EXAMPLE.
The number 22.069.810.125 is a cube ; required its root.
Dividing this cube into periods, we find that the root must
contam 4 figures, and the superior period is 22, and the cube
166 ELEMENTARY ALGEBRA.
root of 22 is near 3, and of course the whole root near 3000;
but it is less than 3000. Suppose it 2800, and cube this
number. The cube is 21952000000, which, being less than
the given number, shows that our assumed root is not large
enough.
To apply the rule, it will be sufficient to take six superior
figures of the given and assumed cubes. Then by the rule,
219520
2
220698
2
'\
439040
220698
441396
219520
659738
: 660916:: 2800
659738
659738
: 1178:: 2800
2800
942400
2336
659738)3298400(5
32986.90
Assumed root,
Correction, .
2800
6
True root, , 2805
The result of the last proportion is not exactly 5, as will
be seen by inspecting the work ; the slight imperfection arises
from the rule being approximate, not perfect.
When we have cubes, however, we can always decide the
unit figure by inspection, and, in the present example, the
unit figure in the cube being 5, the unit figure in the root
must be 5, as no other figui-e when cubed will give 5 in the
place of units.
[For several other abbreviations and expedients in extract'
ing cube root in numerals, see ** Robinson's Arithmetic.**]
EVOLUTION. 167
To obtain the 4tli root, we may extract the square root of
the square root. To obtain the 6th root, we may take the
square root first, and then the cube root of that quantity.
To extract odd roots of high powers in numeral quantities
is very tedious, and of no practical utility ; we, therefore, give
no examples.
(Art. 83.) It is sometimes necessary to multiply roots
together or to divide one by another, and we must, therefore,
find rules for such operations.
For instance, I wish to find the product of the square root
of 3 into the square root of 12, and I know not how to find
it, unless I first extract the square root of 3, and then of 12,
and multiply the two roots together. But this would require
a great amount of labor, and even then it would not be
done to accuracy, as no exact square roots of either 3 or 12
exist.
It is possible, however, that the product of these two roots
is the same as the square root of the product of 3 and 12,
that is, the square root of 36, which is 6; but how are we to
demonstrate whether this be true or not ?
In answer to this inquiry, we say let a and h represent any
two numbers then a- and h- will represent their square roots,
(Art. 76).
In Algebra we represent the product of any two quantities
by writing the quantities as factors with or without tlie
sign of multiplication between them. Thus, the product
L J. L 1
of X and y is x*y or xy, so the product of a^ into h"^ is a^'S-;
but the question is whether or not that this is the same as
{alf.
Now the product of these two roots must be some number.
Let that number be indicated by P. Then we shall have
this equation . . . P=:a^h'
168 ELEMENTARY ALGEBRA.
Square both members of this equation (and we square by
doubling the exponent of every factor, Art. 71), and we have
Kow, by considering ah as a single number, and extracting
the square root of both members, we have
This last equation answers the questioVy and we learn thai the
product of the roots is the same as the root of the product.
Hence, (3)^ X (12)2=(36)2=6.
What is the product of (2)2 by (8)^ Ans. 4.
What is the product of 5(5)2 by 3(8)2. . Ans. 15(40)2.
Hence, when we wish to multiply numbers together which
contain factors under the same root, we have the following
Rule . — Multiply the rational parts together for the rational
part of the product, and the radical parts together for the radical
part of the product.
EXAMPLES.
1. Multiply a(/^)2 by c(c?)2 Ans. ac{hdy.
2. Multiply 3(3)2 by 2(3)^. ...... Ans. 18.
3. Multiply 3(2)2 by 4(8)2 Ans. 48.
4. Multiply 2(14)^ by 3(4)3 Ans. Q{m)^.
But in 56 there is a cube factor 8, the other factor is 7,
therefore the last answer is 6(8)3(7)3=12(7)3.
5. Multiply 2(5)^ by 2(10)^ Ans. 20(2)i
(Art. 84.) The roots to be multiplied together may not be
the same — one may be a square root, the other a cube or some
other root. In such cases, is there any other mode of express-
ing the product except by a representation of the factors ?
EVOLUTION. 169
For example, what is the product of a* by 6*. Is there
another manner of expressing it than
As in the last article, the product must be some number
which we can represent by P.
Then F =:ah^
By squaring, . . . F^—ah^
By cubing, ... . . I^z=zc?l^
Taking the 6th root, . P =i{c^l^)^
In this manner we can find the product of other roots ; but
in a work like this it is not important to carry this subject to
any great length ; we give, however, the following
EXAMPLES.
1. What is the product of a^" by a^ ? . . . Ans. a*.
2. What is the product of (6)^ by (150)^? . Ans, 30.
3. What is the product of (^)^ by (|)^ ? . Ans. ^(3)*.
4. What is the product of (2)^ by (2)^? . Ans. (32)*.
^Art. 85.) As division is the converse of multiplication,
we may infer at once from Art. 83 that the quotient arising
from the division of one root by another, is the same as the
root of the quotient; but for greater clearness we had better
denote the quotient by a letter and use an equation.
For example, divide (8)2 by (2)^, the quotient is (4)* or
2; but to establish the principle of operation, let Q represent
the required quotient. Here, as in all examples of division,
the product of the divisor and quotient is equal to the divi-
dend. Therefore, in this case we must have
22 ^=8^
By squaring, . . . 2C*=8
15
170 ELEMENTARY ALGEBRA.
This equation shows that to obtain the true quotient, we
must divide one number hy the other regardless of the root, and
then vrrite tJie root over the quotient.
Thus, §=(4)2
EXAMPLES.
1. Divide (54)^ by (6)3 Ans. 92=3.
2. Divide 8(72)2 by 2(6)^. , . .* •, ., ^^5, 4(12)^.
3. Divide 3(10)2 by (15)2 ^^5. (6)2.
(15)2^=3(10)2
15 §2=9X10, or §=^=6
4. Divide 18 by 2(3)^ Ans. 3(3)^
5. Divide 6a by 3(a) 2 Ans. 2(a)"^
6. Divide (160)^ by (8)^ Ans. 2(5)^
7. Divide 9 by (27)t Ans. (3)2
8. Divide 1 by (j^)2 Ans. (3)2
9. Divide a by (a)2 Ans. (a)2
(Art. 86.) When the roots are different, t^^e proceed on
the same principle, which will be sufficient for every possible
case.
For example, divide 72 by 1'^', the quotient must be some
number which we can represent by Q, and from the equation
73^=72
Cubing, .... 7^5=72
Squaring, .... 7^§®=7^
Dividing by 7' . . (2^=7. Hence, Q=l^, Ans.
2. Divide {a^lPd^)^ hy d^ Ans. (ah)^.
SECTION IV.
EQUATIONS.
(Art. 87.) We have thus far been able to resolve only
simple equations, or equations of the first degree ; but many
problems and many philosophical investigations present equa-
tions of the second, third, and higher degrees, which may
demand a solution, as, for instance, the first example of Art.
86 incidentally demanded the solution of the equation
which is an equation of the sixth degree ; but it appears in so
simple a form that there is no mistaking the principle on which
its solution depends, and thus generally, When an unknown
quantity is involved to any power, we find the first power
(that is, the quantity itself) by extracting the corresponding
root of both members.
As in that equation . . ^=(7)^
In the same manner, if a;"=a, then x=a^.
The converse of these equations may often occur, that is,
the unknown quantity may appear under the form of a root,
as in the following equation :
Here, it is obvious that the value of x must be found by
cubing ; but if we cube the first member of the equation, we
must cube the second to preserve equality, (Ax. 9). That is
172 ELEMENTARY ALGEBRA.
(Art. 88.) From the foregoing observations, we draw the
following general rule of operation :
Rule . — To free a quantity from a power, extract the corves-
jponding root. To free it from a root, involve to the correspond-
ing power.
When the unknown quantity is connected to a known quan-
tity, and the whole number a power or root, the power or
root, as the case may be, is removed in the same manner as
before.
Thus (x-]rciy=cip
The value of x is found, by first taking the cube root of
both numbers and afterward transposing a.
Again .... (2a;-j-c)2'=a
Here, after squaring both members, we have
9.x-]rc=a?y a simple equation.
(Art. 89.) The equations that appear in Articles 87 and
88, and all other equations of like kind where the unknown
quantity is raised to a complete power, or is under some one
particular root, are called
PURE EQUATIONS.
Thus, the equation ax^=by or, which is the same thing,
«^=- is a pure equation, because the power of the unknown
quantity is complete; but the equation x^-^-hx^c is not a pure
equation, because it contains different powers of the unknown
quantity.
The equation {^x-\-a)^=.c is a pure equation ; but the equa-
tion sc^-\-x^c is an impure equation, because it contains no
complete power of the unknown quantity.
Again, x^-\-^ax-\-(i?=c-\-h is a pure equation, because the
first member is a complete power of (xAro), and {x-\-a) may
be represented by y, then y^:=^c-\'hy obviously a pure equation.
EQUATIONS. 173
There is no difficulty in resolving pure equations as we
have already seen, for all we have to do is to apply the rule
expressed in Art. 88; but impure equations in the higher
degrees, -present serious di^cuUies ; and even equations of the
second degree, when impure, compel us to complete the power
before we can solve the equations. Equations of the second
degree can be represented by a geometrical square ; and when
the equation is pure, the square corresponding to the first
member is complete, and when impure, it must be completed,
and the necessary operation is very appropriately called
COMPLETING THE SQUARE
But before we go into the investigation of completing a
square, we will give some examples to exercise the learner in
resolving pure equations.
EXAMPLES. •
1. Given ^4-\-(x—2)^==3, to find x. . » Ans. ar==27.
To remove the first radical sign, we square both members,
then .... 4-{-(a;— 2)2=9
Dropping 4 from both members, and then squaring, we
find a;— 2=25
2. Given x — ^x'^-\-6= — 2, to find x. . . Ans. x—^.
Transpose x for the purpose of having the quantity under
the radical stand alone as one member of the equation.
Thus, .... —JaP-hG^^—x
Now, by squaring, the radical sign will disappear ; but if
any other quantity were joined to this by + or — , the radical
could not disappear in the square.
The square is . . x'^-{-6=4 — ix-^-aP
3. Given x-\-JsP — 7=7, to find x. , , , Ans. ar=4.
4. Given Jx-jr^2=2-{-Jx, to ^nd x. . . Ans. x=4.
fit ELEMENTARY ALGEBRA.
t N. B. No rules can be given that will meet every case, for
the combination of quantities is too various. The pupil must
depend mainly on general principles and his own practical
experience.
5. Given 2-\-(3x)^=:j5x-\-4, to find x. . Am. a;=12.
/ 202;2 — 9 \2 J.
6. Given f — j =x^, to find x. . Ans. x= |.
x^
7. Given 3a;2— 29=--j-510, tofinda;. . Ans a:=14.
8. Given x-\-2='j4-\-xj64-\-x'^, to find x. Ans. x— 6.
9. Given x — ^Jx=Jx'^ — x, to find x, . . Ans. x=\^,
10. Given xJa?-\-a^=a? — x"^, to find x. Ans. x=aj^.
11. Given Jx— 32=16— Jx, to find x. . Ans. x=Sl.
For the sake of brevity, put a=l6, then the last equation
will be Jx — 2a=a — Jx. At the conclusion resubstitute the
value of a.
12. Given Jx — 16=8 — ^a:, to find a;. . . Ans. x=25)
13. Given a;^ — aa:^=a:, to find a:. . . . Ans. x=- .
1 — a
14. Givena;i-|-^/3+a;= — r=:, to find a:. . Ans. x= 1.
PROBLEMS PRODUCING PURE EQUATIONS.
(Art. 90.) In solving problems, it often depends on the
manner or means of notation we employ, whether the equation
comes out simple or complex, or whether it is a pure equation
or a common quadratic. For example,
1. Find two numbers, whose difference is 6, and their produa
40. Ans. 4 and 10.
If we represent the least number by x.
Then the greater number must be x -\-Q
Their product is x^-{-&x
But, by the problem this product is 40.
EQUATIONS. 175
Therefore, a:^-|-6^'=40; but as yet, we cannot solve this
equation, because it contains two separate powers of x; we
know not what to do with it. But in truth, the problem in
itself -is so simple we should be able to solve it, and must do
so by some artifice or other.
After some reflection, we conclude to
Put . X — 3 to represent the least number.
Then . a:-|-3 will represent the greatest number.
Product a^ — 9=40, a pure equation giving dr7 for the
value of X. (See i^t. 64).
The reason of taking the double sign is found in Art. 78.
If we take -{-1, then the least number is 4 and the greater
10. If we take — 7, the numbers are — 10 and — 4; but as
there are really.no such numbers as — 10 and — 4, we take
only +7 from the answer to x.
The solving of this problem shows us how to solve any
equation in the form of
a^-\-ax=b
Consider the first. member as the product of x, and .r-|-a,
the difference of theSe two factors is a.
Put . , . . . . a:=^y-- (1)
■ fc-\-a='y-\-^ (2)
Product, . . a;'+aa;=y2— --=5 ^3^
Hence, .... f=b+~ (4)
And y==±\f*+- (5)
■M
176 ELEMENTARY ALGEBRA.
The value af y, as determined in equation (5), put in equa-
tion (1), gives
(Art. 91). The student will perceive that Article 90 is a
digression, but one that should be pardoned. We now con-
tinue our problems in pure equations, and if any problem
does not produce such an equation, it will be because the
notation designed by the author has not been taken.
2. The sum of two numbers is 6, and the sum of their
cubes is 72. What are the numbers ? Ans. 2 and 4.
Let S-\-x= the greater.
And 3 — x=: the less.
3. Divide the number 56 into two such parts, that their
product shall be 640. Ans. 40 and 16.
Let 28+a;= the greater,
28 — x:= the less.
4. A and B distributed 1200 dollars each, among a certain
number of persons. A relieved 40 persons more than B, and
B gave to each individual 5 dollars more than A. How
many were relieved by A and B ? Ans. A, 120; B, 80.
Let a:-f-20= the number relieved by A.
And X — 20= the number relieved by B.
mi. 1200 , " 1200
Then .... -—-+5= — --
ar-|-20 ar— -20
T^•A'\^' 240 , ^ 240
Dividmg by 6, gives — ^ i ^
a?-[-20 a;— 20
To avoid numeral multiplication and division, put <i^=20,
and 5=240. Then the equation becomes
x-\-a X — a f
EQUATIONS. 177
6. Find a number, such that one-third of it multiplied by
one-fourth, shall produce 108. Ans. 36.
6. What number is that whose square plus 18 shall be
equal to half its square plus 301 ? Jins. 5.
7. What two numbers are those which are to each other as
5 to 6, and the difference of whose squares is 44 ?
Am. 10 and 12.
'Lei6x= the greater, and 5x= the less.
8. What two numbers are those which are to each other as
3 to 4, and the diflference of whose squares is 28 ?
Ans. 6 and 8.
9. What two numbers are those whose product is 144, and
the quotient of the greater by the less is 1 6 ?
Ans. 48 and 3.
10. The length of a lot of land is to its breadth as 9 to 5,
and it contained 405 square feet. Required the length and
breadth in feet. Ans. 27 and 1 5.
11. What two numbers are those whose difference is to the
greater as 2 to 9, and the difference of whose squares is 128 ?
A71S. 18 and 14.
12. Find two numbers in the proportion of ^ to |, the sum
of whose squares shall make 225 ? Ans. 9 and 12.
The thoughtful student will not use the fractional numbers
■i- and |; but he will use whole numbers in the same propor-
tion. Let this observation apply to other problems as well
as to this one.
13. There is a rectangular field, whose breadth is | of the
length. After laying out ^ of the whole ground for a garden,
it was found that there were left 625 square rods for mowing.
Required the length and breadth of the field.
Ans. Length, 30 rods ; breadth, 25.
14. Two men talking of their ages, one said that he was
94 years old. Then, replied the younger, the sum of your
178 ELEMENTARY ALGEBRA.
age and mine, multiplied by the difference between our ages,
will produce 8512. What is the age of the younger ?
A71S. 18 years.
15. A fisherman being asked how many fish he had caught,
replied, **If you add 14 to the number, the square root of
the sum, diminished by 8, will equal nothing." How many
had he caught ? Aiis. 50.
16. A merchant gains in trade a sum, to which 320 dollars
bears the same proportion as five times the sum does to 2500
dollars. What is the sum ? Ans. $400.
17. What number is that, the fourth part of whose square
being subtracted from 8, leaves a remainder equal to 4 ?
Ans. 4.
18. Find two numbers, such that the second power of the
greater, multiplied by the less, produces 448, and the second
power of the less, multiplied by the greater, gives 392.
Ans The numbers are 8 and 7.
Let x=i the greater, ?/= the less.
Then a:2y:=448 (1)
And 3^2^392 (2)
The product of (1) and (2) is
iry=(448)(392) (3)
This equation indicates that there may be cube factors in
448 and in 392. Therefore, we will try to find them by
dividing by 8, the least cube number above unity.
Thus, .... 8)448 8)392
8)56 7)49
Hence, . . ay=(8«8«7)(8-7'7)=83»73 (4)
Or, . . . ay=8«7 (5)
Divide (1) by (5), and we have x==S. •
19. A man purchased a field, whose length was to its
breadth as 8 to 5. The number of dollars paid per acre was
EQUATIONS. 179
equal to the number of rods in the length of the field : and
the number of dollars given for the whole, was equal to 13
times the number of rods round the field. Required the
length and breadth of the field.
Ans. Length, 104; breadth, 65 rods.
Let 8a;= the length, and 5x=: the breadth,
40ar' a^
Then . .— — =— =number of acres.
160 4
And . —XSx=2x^= the whole number of dollars.
4
Again, Sx-{-5x=:lSx= half round the field.
And 13a;*2»13= thirteen times round, which is equal to
the dollars paid.
20. There is a stack of hay, whose length is to its breadth
as 5 to 4, and whose highth is to its breadth as 7 to 8. It is
worth as many cents per cubic foot as it is feet in breadth ;
and the whole is worth at that rate 224 times as many cents
as there are square feet in the bottom. Required the dimen-
sions of the stack.
Ans. Length, 20; breadth, 16; and highth 14 feet.
7x
Let 5x= the length, 4x= the breadth, and -^= highth.
Then [ 5x*4x* — J 4x= cost in cents.
Again, 5x'4x= square feet on the bottom;
Hence, . 224* 5a?' 4a:= cost in cents;
Ix
Therefore, ^x'4x'—*4x=29.4'bx*4x
2
By striking out equal factors, we have
7a;'2a:=224
21. It is required to divide the number of 14 into two such
parts, that the quotient of the greater divided by the less,
180 ELEMENTARY ALGEBRA.
may be to the quotient of the less divided by the greater, as
16:9. Ans. The parts are 8 and 6.
Let x= the greater part. Then 14 — x= the less.
/J* 1 4 X
Per question, — : : : 16 : 9.
1 4 — X X
Multiply extremes and means, and — = — ^^
Clearing of fractions, we have 9a;^=16(14 — xy.
By evolution, . 3x=4(l4 — ^a;)=4'14 — Ax.
By transposition, 7aJ=4*14.
By division . . rr=4»2=8, the greater part.
We solved the last four examples for the purpose of strongly
recommending the factor system, which has not been prac-
ticed or appreciated half as much as its merits deserve.
(Art. 92.) We will now return to equations in the form of
x^-}-ax=b, for if we know how to solve them, we need not be
so particular in our notation as we have been in the last
Article.
As we have before remarked, all these equations can be
resolved by considering the first member as the product of
two factors, one of which is x, the other (x-{-a), and their
difference is a. Then if we put y — - for one factor, and y+-
for the other, we shall have y^ — 7=^' ^^^ y^=^-{--
That is, if to b, or to its equal, x^-jrax, we add — , the
square of - (the square of half the coefficient of the first power
of a;), we shall have y, that is, some square. Hence, x'^-\-a3t
H — is a square, and we have the equation
EQUATIONS. 181
Now, the first member is a square mform; but whether it
is a numerical square or not, depends on C^+j) being a
numerical square ; but whatever it is, we have, by extracting
the square root of both members
■=-%^4'
.-><
We have found this formula before, in Art. 90.
(Art. 93.) An equation in the form of x^ — ax=bf its first
member may be considered the product of the two factors, x
and {x — a), and these two factors diflfer by a.
Let y^l=x (1)
And y — ~x — a (2)
At
Multiply (1) and (2), and we have
y^ — --=a;(a; — a) =5
Hence, 2^=^+1' (3)
Equation (3) shows that whether we have an equation in
the form of x^-\'ax-=.h, or of o^—ax=.h; that is, whether ax
is either -plus or minus ^ we make the first member a square by
the addition of precisely the same quantity — , which is the
square of half the coefficient of x.
In other words, . . . x^-{-ax-\-—
4
And ... ... x^ — ^'^'7
The square root of the first is
The square root of the second is
are bpth squares.
See Art. 78.
182
ELEMENTARY ALGEBRA.
Hence, to complete the square of the first member of any
equation in the form of x^-}-ax=b, or of x^ — ax=b, or more
generally when the exponent of the unknown quantity in the
first term is double of that in the other, we have the following
Rule . — Add the sqttare of half the coefficient of the lowest
jyower of the unknown quantity.
EXAMPLES.
Complete the squares in the followmg equations :
N. B. We add a quantity to the first member to complete
its square, we add the same quantity to the second member
to preserve equality.
1. ar»-f 4a;=96. .
. . Ans.
a^+4a;-l-4=96+4.
2. ar^— 4a;=46. .
. . Ans.
a;2_4a;-|_4==49.
3. ^2_7^^8. . .
. . Ans.
ar^— 7a:-{-V=8+V.
4. a;'*— 2a;2=24. .
. . Ans.
a:4_2:c24-l =24-1-1.
5. a;^"— 4ic"=a. .
. . Ans.
a:2«_4^n_|_4_a-|-4.
6. a^4-6a;=16. .
. . \Ans.
ar^4-6a;+9=25.
7. ar*— 15a;=— 54.
. . Ans.
a?—\bx^-^^=^-\^—b^.
8. ar'— |ar=ip. .
, . Ans.
^=-|^+i=4^+i.
9. x'-^x=\. . .
. . Ans.
^^-f+rV.^i+t'^.
10. . ^.=-. . .
. . Ans.
"^ 'f'^Ab'^-'d'^'ib-'
Find the values of x in each of the ten preceding equations.
First by extracting the square root of both members.
. . . Therefore, a:=8 or —12.
1. a;4-2=dbl0.
2. a;— 2=ir7.
a;=5 or — 9.
EQUATIONS.
183
3. .1;— 1=±|. . .
4. x^—l=dz5. . .
5. a;™— 2=±(a+4)
Therefore, . .
6. x-{-S=diz5. .
7. a;— V =±f .
8. a:— i=±V.
9. o:-/^
Therefore, ir=8 or
a;=(6)2 orV— 4.
a { c , a? ^
ar=">/2-l-7a4-4 or (2— 7a-|-4)'*
. . Therefore, a;=2 or — 8.
. . " ar=9 or 6.
. . " a;=7 or — V.
. . " x=\ or
1
6*
I
2 . T
26 I dT^b'' I
The preceding ten equations are all prepared for completing
the square ; that is, the highest power of the unknown quan-
tity stands first, and is positive.
It is necessary that it should be positive, because we must
take its square root, and there is no square root to a negative
quantity. Therefore, in reducing an equation preparatory to
completing its square, if the highest power comes out mi7iits,
make it plus, by changing all the signs to both members of
the equation.
* '• Example. Find the values of x from the following equation
36
2
x-\-t
' Mtfltiplying by 2, and afterward dropping 2 from both
'• • '■■ • 72
•memJbei^ we have . .
-x=Q —
a; 4-2
•/ Clearing of fractions,
' • —f—2x== Bx+ 1 6—72
Tran^osmg; 8x, uniting 16 and — 72, and afterward chang-
VRg all- {he signs, we have
ar'4-10a;=66. Hence, x=4, or —14
184 ELEMENTARY ALGEBRA.
2. Find the values of x from the equation 3x^-\-2x — 9=76.
Ans. x=b or — 5|.
X 2,X X
3. Find a; from 2:^4--=-- r+H- Ans. x=\ or —2^.
Zoo
x^
4. Find x from - — 30-\-x=2x — 22.
Ans. ar=8 or — 4.
Cu X
5. Find a; from -• — --j-7i=8i. . Ans. a:=l|^ or — |.
^ o
/^ 2a;
6. Find a; from — — 16=— — 14f. Ans. x=3 or — ^.
/p a;-4-3
7. Find x from ——=-—;—. . ^W5. a;=12 or —2.
2
8. Find a; from a; — 1+ :=0. • Ans. x=3 or 2.
a: — 4
9. Find x from J~^= ;f. . Ans. x=36 or 12.
20 X — 6
g/j. J x-\~ 1
10. Find X from --=— -r— . . Ans. x=4 or — 4.
a; — 1 2a;+7
(Art. 94.) For a more definite understanding of quadra-
tics, we will solve and strictly examine the following equation :
a;2-f4a;=60
Completing the square, then
ar^+4a;+4=64
Extracting the square root of both members, *,•
And a:-}-2=±8 »■
The reason of taking the double sign to 8 has been ^eiVeral
times explained. '^
If we take +8, then . «=6 V
If . . —8, then . a:=— 10
That is, either +6, or — 10 will verify the equation.
For 6'+4»6=60 *;; -'
Also, . . (—10)2— 4'10=60 • * *•
If ar=6, . . . . ar— 6=0 (1) * "
If ic=— 10 . . . ar4-10=0 (2)
EQUATIONS. 185
If we multiply equations (1) and (2) together, we sliall
have as follows :
X — 6
a^—6x
10a;— 60
a^-^4x—60=0
As the two factors are in value equal to 0, the product of
the two must, of course, equal 0, and we have the equation
as above. Transpose — 60, and we have x-\-4x=60, the
onginal equation.
Thus we perceive, that a quadratic equation may be con-
sidered as the product of two simple equations, and the
values of x in the simple equations are said to be roots of the
quadratic, and this view of the subject gives the rationale of
the unknown quantity having two values.
This example shows us how to form an equation when
we have the two roots ; that is, gives us the following
Rule . — Connect each root with a contrary sign to an un-
known quantity. Take the product of the two binomial factors
thus fanned for the first member of the equation sought, and 0
for the other member.
EXAMPLES.
1. Find the equation which has 3 and — 2 for its roots.
Ans. x^ — x — 6=0.
2. Find the equation which has 5 and — 9 for its roots.
Ans. aP-{-4x — 45=0.
3. Find the equation which has 7 and — 7 for its roots.
Ans. a;2— 49=0.
4. Find the equation which has 8 and — 12 for its roots.
Ans. x^-\-4x—9e=0.
16
166 ELEMENTARY ALGEBRA.
Let the pupil observe that this last equation is equation 1
in Art. 93, and he can take the roots of those ten equations
and deduce the equations again if desirable.
5. Find the equation which has a and h for its roots.
Ans. x^ — [a-\-b)x-{-ah=^0.
If one of the roots is negative, suppose — a, the equation is
then ar^+(a — h)x — ah=0
If h is negative and a positive, the equation is
^^-{-{b — a)x — ab=0
If both roots are negative, then the equation is
x''^{a-^h)x^ab=0
]Sro^, let the pupil observe that the exponent of the highest
power of the unknown quantity is 2; and there are two roots.
The coefficient of the first power of the unknown quantity is the
algebraic sum of the two roots, with their signs changed; and
the absolute term, independent of the unknown quantity, is the
product of the roots (the sign conforming to the rules of
multiplication).
From these observations we can instantly form the equation
when the two roots are given, without the formality of going
through the multiplication ; for example.
Find the equation which has 7 and — 9 for its roots.
7— 9 =—2 ; changed +2, 7(— 9)=— 63
Hence, x'^-i-2x — 63=0 is the equation.
(Art. 95.) When a quadratic equation is formed, or found,
or given, we may consider it as the product of two binomial
factors, and those factors may be obvious to one who fully
understands the subject, or any one can find them who can
resolve the equation.
In giving examples in factoring (Art. 26), we omitted
trinomial quantities of the second degree. The reason of
EQUATIONS. , 187
that omission must now be perfectly comprehended by the
careful student. We now return to that subject, and require
the factors composing the expression
Put the expression equal to zero, and resolve the quadratic,
and we shall find the roots to be — 2 and — 3. Therefore,
the sought factors are (^+2) and (x-\-3).
Find the factors composing each of the following express-
ions. Each expression must be taken as a quadratic equation
presented for solution :
1. aP—x—20=zO Ans. (x—5)(x+4).
2. a^— 7a4-12. ........ Ans. (a— 3)(a— 4).
3. a^— 7a— 8 Ans. (a—S)(a-{-l).
4. x'—x—SO Ans. {x—6){x-\-5).
5. x^-\-7x—\8 Ans. (x-h9)(x—2).
6. x^-\-2ax-^a^ Ans. (x-\-a)(x-\-a).
7. x^ — 2ax-\-a^ Ans. (x — a)(x — a).
8. a^ — a^ Ans. (x-\-a)(x — a).
9. x^ — 2a;+4 Ans. (x — r)(x — r).
In this last expression r=\±J — 3, and it not being a
numerical quantity, the roots are said to be imaginary/.
(Art. 96). When a quadratic equation is reduced to the
form of x^-\-ax=b, to complete the square of the first mem-
ber we take the half of the coefficient of x to square, there-
fore, it will be more convenient to represent that coefficient by
2a in place of a, and as it may have the negative as well as
the positive sign, and as b can be negative as well as positive ;
therefore, for a representation of every variety of quadratic
equations, we have the four general forms.
x^-\-2ax=b (1)
a^^2ax=b (2)
x' — 2ax=—b (3)
3^'^2ax=:—b (4)
188 ELEMENTARY ALGEBRA.
A solution of each of these equations gives for the values
of a; as follows :
x=—a±:Jb +a2 (1)
x=^-\-a±Jb +a' (2)
x=-^a^J^—b (3)
x=^—a±ijc^^ (4)
The quantity x has no conceivable value in equations (3)
and (4) when applied to any problem in which h has a greater
numerical value than a?, for the solution requires the square
root of Ja^ — b, a negative quantity ; and there being no square
roots to such quantities, we have no conception of any value
to X, and, of course, we call the value imaginary.
After we reduce an equation to one of the preceding forms,
the solution is only substituting particular values for a and b;
but in many cases it is more easy to resolve the equation as
an original one, than to refer and substitute from the formula.
(Art. 97.) We may meet with many quadratic equations
that would be very inconvenient to reduce to the form of
a?-{-2a^=b; for when reduced to that form, 2a and b may
both be troublesome fractions.
Such equations may better be left in the form of
aoi^-\-bx=^c
An equation in which the known quantities, a, b, and c, are
all whole numbers, and prime to each other.
"VVe now desire to find some method of making the first
member of this equation a square, without making fractions.
We therefore cannot divide by a, because b is not divisible by
a, the two letters being prime to each other by hypothesis.
But the first term of a binomial square is always a square ;
therefore, if we desire the first member of our equation to be
converted into a binomial square, we must render the first
term a square, and we can accomplish this by multiplying
every term by a.
EQUATIONS. iP
The equation then becomes
d^oi^-\-hax=ca
Put y=ax
Then y^A^hy^ca
Complete the square, by the preceding rule, and we have
We are sure the first member is a square ; but one of the
terms is fractional, a condition we wished to avoid ; but the
denominator of the fraction is 4, a squarcy and a square mul-
tiplied by a square produces a square.
Therefore, multiply by 4, and we have the equation
An equation in which the first member is a binomial square,
and not fractional.
If we return the values of y and y^ this last equation
becomes 4a'a:^-{-4a5a:-f-5^=4ac+5^
Compare this with the primitive equation
"We multiplied this equation first by a, then by 4, and in
addition to this, we find H^ on both sides of the rectified equa-
tion, h being the coefficient of the first power of the unknown
quantity. From this it is obvious, that to convert the express-
ion ax^-^-hx into a binomial square, we may use the following
Rule . — Multiply hy four times the coefficient of x^, and add
the square of the coefficient of x.
To preserve equality, both sides of an equation must be
multiplied by the same quantity, and the same addition must
be made to both sides. We operate on the first member of
an equation to make it a square; we operate on the second
member to preserve equality.
yfb ELEMENTARY ALGEBRA.
EXAMPLES.
1. Given 6a;^+4a;=^4, to find the values of x.
By the rule, we multiply 4 times 6, and add to both mem-
bers 4K That is,
4«5V+80a;+16=4080+16
By extracting square root, we have
2«5a:+4=±64, a:=6 or — 6f
By extracting the square root of the first member, the
second term always disappears; it is, therefore, not necessary
to compute it, and for that reason we may simply represent it
by a letter, as in the following example :
2. Given 7x^ — 20a?=32, to find the values of x.
Multiply by 4 times 7 and add 20^.
Then . . 4«7V-— ^+400=896+400
Square root, . 2- 7a; — 20= ±36; hence, a;=2 or — |.
3. Given 2x^ — 5a:=117, to find the values of x.
Ans. x=9 or — 6^.
4. Given 3aP — 5a; =28, to find the values of x.
Ans. x=4 or — J.
6. Given 3a;^ — a;=70, to find the values of x.
Ans. x=5 or — y.
6. Given 6a;^+4a;=273, to find the values of x.
Ans. x=7 or — 7|.
7. Given 2a;^+3a;=65, to find the values of a;.
An^. x=5 or — 6|.
8. Given 33:^+53;= 42, to find the values of ar.
Ans. x=3 or — 4'.
9. Given Sa;'— .7a;+ 1 6 =f 1 8 1 . to find a;.
Ans. a;=5 or — 4J.
EQUATIONS. i^l
10. Given 10a^—8;r+8=320, to find ir.
Ans. x=G or — 5}.
11. Given 3x^+2x=4, to find ar. . Ans. a;=— i-ii^ls.
12. Given 5a;2+7a;=7, to find a;. Ans. x— — ^^-^.f^J^.
240 216
13. Given |-to= :-, to find a:. . Ans.x=.1b.
X X — 15
QUESTIONS
GIVING RISE TO QUADRATIC EQUATIONS.
1. If four times the square of a certain number be dimin-
ished by twice the number it "will leave a remainder of 30.
What is the number ? Ans. 3.
N. B. The number 3 is the only number that will answer
the required conditions — the algebraic expression — f will
also answer the conditions ; but the expression is not a num-
ber in any arithmetical sense.
2. A person purchased a number of horses for 240 dollars.
If he had obtained 3 more for the same money, each horse
would have cost him 4 dollars less. Required the number of
horses. Ans. 12.
3. A grazier bought as many sheep as cost him 240 dol-
lars, after reserving 15 out of the number, he sold the remain-
der for 216 dollars, and gained 40 cents a head on the
number sold. How many sheep did he purchase 1 Ans. 15.
(See equation 13 just passed over).
4. A company dining at a house of entertainment, had to
pay 3 dollars and 50 cents ; but before the bill was presented
two of them went away ; in consequence of which, those who
remained, had to pay each 20 cents more than if all had been
present. How many persons dined ? Ans. 7.
5. There is a certain number, which being subtracted from
22, and the remainder multiplied by the number, the product
will be 117. What is the number ? Ans. 13 or 9.
192 ELEMENTARY ALGEBRA.
6. In a certain number of hours a man traveled 36 miles,
but if he had traveled one mile more per hour, he would have
taken 3 hours less than he did to perform his journey. How
many miles did he travel per hour ? Ans. 3 miles.
7. A man being asked how much money he had in his
purse, answered, that the square root of the number taken
from half the number would give a remainder of 180 dollars.
How much money had he ? Ans. $400.
8. Divide 100 into two such parts, that the sum of their
square roots may be 14. Ans. 64 and 36.
9. Divide the number 14 into two such parts, that the sum
of the squares of those parts shall be 100. Ans. 8 and 6.
10. Divide the number a into two such parts, that the sum
of the squares of those parts shall be b.
Ans. ^{a±ij^b—a?).
11. It is required to divide the number 24 into two such
parts, that their product may be equal to 35 times their
difference. Ans. 10 and 14.
12. The sum of two numbers is 8, and the sum of their
cubes 152. What are the numbers ? Ans. 3 and 5.
Let 4 — x= the less number.
And 4+a;= the greater number.
Then put a=4, cube, &c.
(Art. 98.) In the preceding examples wc have only con-
sidered the resulting equation after all the other unknown
quantities have been eliminated.
In solving a problem, however, the operator may use one,
two, three, or more unknown quantities, and operate as in sim-
ple equations, and in eliminating one quantity after another,
there will result a final equation, which may be of the first,
second, Hiird, or higher degree, according to the conditions of
the problem and the tact of the operator in taking hold of the
matter.
EQUATIONS. 193
(Art. 99.) When two independent equations are drawn
from a problem, if one of them is quadratic, the other must
be simple, or the resulting equation cannot be brought down
to the second degree, except in rare cases, where the two
equations are homogeneous or are symmetrical.
(Art. 100.) Two equations essentially quadratic, involving
two unknown quantities, depend for their solution on a result-
ing equation of the fourth degree.
(Art. 101.) No words will cover every case of similar or
symmetrical equations ; but as a general thing, in similar
equations we may change x to y, and y to a; without changing
the form of the equations or falsifying them.
Thus, a:+y=^a
And icy=b
Are similar equations, and x-^y=a
Are both similar and symmetrical.
When equations are similar, we can reduce them without
completing the square, as the learner will discover by the fol-
lowing examples :
1. Given \ , !• to find ar and y.
( xy=h S
Squaring the first, . . x^-\-9.xyAry'^=-a^ (1)
Four times the second, . - Axy =45 (2)
Subtracting (2) from (1) ^— i^H^y^Z^^HJj ^gj
Square root of (3), , . . . x — y—dczja? — 46 (4)
But a;+y=a (5)
Add (4) to (5), . . . ^. . 2a;=a±Va2^45 (6)
Take (4) from (5) .... Zy=azpja''—4b (7)
Dividing (6) and (7) by 2 and we have the values of x
and y, and by reason of the double sign each letter has two
valueff. When the quantity ^0^—45 is a ccjmplete secoad
17
ELEMENTARY ALGEBRA.
power, the values of x and y will be rational, otherwise they
will contain surds.
2. Given \ ^, T ^9~f }J\ Mo find x and y.
( a?-\-f=h (2) )
Squaring (1), . . x^-\-9.3cy-]ry'^-=o^ (3)
Subtracting (2), .... 2xy==a?—h (4)
Taking (4) from (2), a?—'2xy-\-y'^=9,h—a? (5)
Square root of (5), . . x — y=ziztj2b — a^ (6)
Adding (6) and (1), . . . 2x=a^j2b-~a^
Taking (6) from (1), . . . 2y=a:=f.j20—a^
3. Given ■! , , ^ /«{ f to find x and y.
I a^—f=b (2) )
Divide (2) by (1), and x—y=-. (3)
Equation (1) and (3) will give the values sought.
4. Given i ^ , o , \^{ !• to find a; and y.
( ^-{-y^=b (2) j
.Cube(l), . ar'-f3a:V+3a^2^_y3^^3 ^3^
From (3) take (2), and Sx^y-^Sxy=a^-^ (4)
Or, Sxy(x-\-y)=a^-b
Divide (4) by (1), and . . 3xy=a^—- (5)
Equations (1) and (5) combined, will make an example the
same in forni as example 1, and may be solved in the same
manner.
6. Given •< , ~_ \J. y to find a? and y.
The equation resulting from these cannot be reduced to the
second degree, and we mention the fact to save the time of
the operator from making useless trials.
EQUATIONS
Divide (2) by (1), then we have
1
>
x'-\-Ty\-f=
0
~ a
(3)
Squaring (1), . . a?—2xy-\-y^=
=a'
(4)
Taking (4) from (3), . . 3ary=
b
=
a
a^
(5)
Dividing (5) by (3), . . xy=
h
~3a"
a"
3
Adding (6) and (3), x'-^^ciyVy^^
_45
"3a"
a^
3
I9S
(6)
(7)
Square root of (7), . . a;+y=d=^r — (8)
Combine (1 ) and (8) for the values of x and y.
To exercise in these general principles j we give the following
numeral examples :
7. Given -< „ ?• to find x and y. Ans. i „ ^
( xy=35) (y=7 or 6.
8. Given ■< f- to find x and y. Ans. \
i xy= 42) ^ iy=7
« ^. (a;+y=1125) , ^ , , . (^
9. Given -< „ „ , , ^ >- to fii^d a; and y. Ans. ■{
-« ^. (^ — y= 4) ^ , , . far=5 or — 1,
10. Given ■{ „ , ^^^v tofindicandy. -<4ws. •<
(a:3 — yS_-^24) ^ (y=l or — 5.
cjPJrf=l9(x-{-y)l
i x—y=3 j
or — 6.
563,
562.
11. Givenf-^^-^^(^+2/)^,^fi^^^^^y.
12. Given'
a; y 6
1 1 __31
x=5 or — 2,
y=2 or — 5.
^ to find X and y. Ans. \
xz=2 or 3,
3 or 2.
To solve this, put, -= F, and -=Q, then we have an
example in the form of example 2.
196 ELEMENTARY ALGEBRA.
13. Given i , ' >• to find a; and y. Ans. •! „ '
(Art. 102.) . Equations are homogeneous, when the sum of
the exponents of the unknown quantities is the same in every
term.
Thus, 2x^— X2j=e ) , .. •.
, , ^ y are homogeneous equations, because
the sum of the exponents of x and y is the same in every
term ; that is, 2.
Such equations may always be resolved by putting one
unknown quantity equal to the other multiplied by a new
unknown factor. '
To solve these equations, put x=vi/.
Substituting this value of x in the two equations, and they
become .... 2i;y — vy^==6 (1)
And 2f-\-3vf==S (2)
^™-(') 2^=2^ (')
From (2) ..... . f=^^ (4)
Equating (3) and (4), clearing of fractions and reducing,
we have .... 8v'^ — 13?;= 6 (5)
Thics, from every pair of homogeneous equations, we may
have a resulting equation of the second degree in reference to the
new factor introduced.
Solving equation (6), we find z;=2 or — f .
Taking 2 for the value, a:=2y, and from equation (4)
o
y^= — 7--=l. Hence, y=d=l, which gives ar=2, or — 2.
The equations . ic^+y^ — x — ^y=78
xy-\-x-\-y=2>^y are both quadratic,
and, therefore, by Art. 100, they will produce a resulting
equation of the fourth degree ; but they are also similar and
symmetrical, and for this reason, it is possible to bring out a
EQUATIONS. 197
resulting quadratic, but no general rule of operation can be
laid down, and the operator must depend mainly on his own
acquired tact and skill.
To resolve these equations, we double the second, and add
it to the first, we then have
The first member of this equation is obviously the same as
(a;-{-y)2+(rc+y) = 156
For the purpose of simplification, put (a;-{-y)=s.
Then, s^+5=i. 1 56, a quadratic equation in relation to s, and
a solution gives s, ora;+y=12. This equation taken from
the second of the primitive equations gives xy=9.7, and from
these last two equations, x=9 or 3, and y=3 or 9.
There are a great variety of circumstances that may come
in aid, or deter the solution of equations ; but it is not proper
to notice them in an elementary work like this. For a more
full development of these particulars in equations, see Robin-
son's Algebra, University Edition.
(Art. 103.) It is not essential that the unknown quantity
should be in involved literally to its first and second powers ;
it is only essential that the index of one power should be
double that of the other.* In such cases, the equations can be
resolved as quadratics. For example, x^ — 4x^=z621 is an
impure equation of the sixth degree, yet with a view to its
solution, it may be called a quadratic. For we can assume
* From this and the following article we perceive that the term quad-
ratic equations, is far more proper and comprehensive than equations of
the second degree.
"We speak of this because it has been suggested to us, that the modern
rules of science required the systematic use of the term equations of
the first, second, third, &c., degrees. The author of this work is modern
in all his views, and is an advocate for modern improvements ; but it
must be improvements, not merely varieties, or changes in technicalities.
19S ELEMENTARY ALGEBRA.
y=si?; then y^=x^, and the equation becomes i/^ — 43/=621,
a quadratic in relation to y, giving y='21, or — 23.
Therefore, .' . . ar'=27 or —23
And .... ar=3 or V— 23
There are other values of x; but it would be improper to
seek for them now; such inquiries belong to the higher order
of equations.*
For another example, take a? — x^=56, to find the values
of X.
Here we perceive one exponent of x is double that of the
other; it is therefore essentially a quadratic.
Such cases can be made clear by assuming the lowest
power of the unkno'svn quantity equal to any single letter.
In the present case, assume y=x'^ ; then y^=x^, and the equa-
tion becomes , . . y^ — y=56
A solution gives y=8, or — 7, and by returning to the
3. 3. i
assumption, y=x^, we find X" = S, or x-=2, or x=4.
(Art. 104.) When a compound quantity appears under
different powers or fractional exponents, one exponent being
double of the other, we may put the quantity equal to a single
letter, and make its quadratic form apparent and simple. For
example, suppose the values of x were required in the equation
2x''-]-3x+9—5j2x^-\-3x-\-9=G
Assume J2x^-jr^x-i-9=y
Then by involution, . . 2x^ -\-3x-{-9 =y- (A)
And the equation becomes . y^ — 5y=6 (jB)
Which equation gives y=6 or — 1. These values of y,
substituted for y in equation (A), give
2r^-l- 3a; -1-9=36
Or, 2ar»-f3a;-|-9=l
From the first of these we find . . x=3 or — 4^.
* See Algebra, University Edition.
EQUATIONS. 199
From the last, we find x=l( — 3±:J — 55), imaginary
quantities.
We give a few examples to fix the principles explained in
Articles 103 and 104.
1. Given x-\-3-\-2(x-{-3)'^=35, to find one value of x.
Ans. x=22.
2. Given (f+2yy-\-4(i/^+2y)=96, to find one value of y.
Ans. y=2.
3. Given 10+a; — (10+a;)2 = 12, to find one value of x.
Ans. x=Q,
4. Given ( --Hy J+ [ -+y ) =30 to find y.
Ans. y=3 or 2, or — 3±73.
5. Given (a;-l-12)^+(a:-i-12)^=6, to find the values of x.
Ans. x—4 or 69.
6. Given (a;4-«)^ + 25(a;+a)*=35^ to find the values of ar.
Ans. ir=Z»^ — a or 81^'* — a.
It is very seldom that problems produce such compound
equations as the last six, or indeed never will unless expressly
designed so to do. The following is one :
1. A poulterer going to market to buy turkeys, met with
four flocks. In the second, were 6 more than 3 times the
square root of double the number in the first. The third
contained 3 times as many as in the first and second ; and the
fourth contained 6 more than the square of one-third the num-
ber in the third; and the Avhole number was 1938. How
many were in each flock? Ans. 18, 24, 126, 1770.
Let .... 2x^= the number in the first.
Then . . 6ar-|-6= the number in the second,
3(2x^-\-6x-\-6)— the number in the third,
(2x^-\-6x-\-6y-{-6= the number in the fourth.
Assume 2x^-\-6x-\-6==y. Then the whole sum is
y^-|-4y+6=1938
200 ELEMENTARY ALGEBRA.
Subtracting 2 from both members, and extracting square
root, we have . . . 2/-|-2=44
We do not take the minus sign, for minus cannot apply to
this problem.
From the assumed equation, we have
2. If a certain number be increased by 3, and the square
root of the sum taken and added to the number, the sum will
be 17. What is the number ? Ans. 13.
3. The square of a certain number, and 1 1 times the num-
ber makes 80. What is the number ? Ans. 5.
4. Find two numbers, sucli that the less may be to the
greater as the gi-eater is to 12, and that the sum of their
squares may be 45. Ans. 3 and 6.
6. "What two numbers are those, whose difference is 3, and
the difference of their cubes 189 ? Ans. 3 and 6.
6. What two numbers are those, whose sum is 5, and the
sum of their cubes 35 ? Ans. 2 and 3.
7. A merchant has a piece of broadcloth and a piece of
eilk. The number of yards in both is 110; and if the square
of the number of yards of silk be subtracted from 80 times
the number of yards of broadcloth, the difference will be 400,
How many yards are there in each piece ?
Ans. 60 of silk; 50 of broadcloth.
8. A is 4 years older than B ; and the sum of the squares
of their ages is 976. What are their ages ?
Ans. A's age, 24 years; B's, 20 years.
9. Divide the number 10 into two such parts, that the
square of 4 times the less part, may be 112 more than the
square of 2 times the greater. Ans. 4 and 6.
10. Find two numbers, such that the sum of their squares
may be 89, and their sum multiplied by the greater, may
produce 104. A7is. 5 and 8
EQUATIONS. 201
11. What number is that, which, being divided by the
product of its two digits, the quotient is 5]; but when 9 is
subtracted from it, there remains a number having the same
digits inverted ? Ans. 32.
12. Divide 20 into three parts, such that the continual
product of all three may be 270, and that the difference of
the first and second may be 2 less than the difference of the
second and third. Ans. 5, 6, and 9.
13. A regiment of soldiers, consisting of 1066, formed into
two squares, one of which has four men more in a side than
the other. What number of men are in a side of each of the
squares? Ans. 21 and 25.
14. The plate of a lookingglass is 18 inches by 12, and is
to be framed with a frame of equal width, whose area is to
be equal to that of the glass. Required the width of the
frame. Ans. 3 inches.
15. A square courtyard has a rectangular gravel walk
round it. The side of the court wants two yards of being-
six times the width of the gravel walk, and the number of
square yards in the walk exceeds the number of yards in the
periphery of the court by 164. Required the area of the
court. Ans. 256 yards.
16. A and B start at the same time to travel 150 miles; A
travels 3 miles an hour faster than B, and finishes his jour-
ney 8i hours before him ; at what rate per hour did each
travel ? Ans. 9 and 6 miles per hour.
17. A company at a tavern had 1 dollar and 75 cents to
pay ; but before the bill was paid two of them went away,
when those who remained had each 10 cents more to pay;
how many were in the company at first ? Ans. 7.
18. A set out from C, toward D, and traveled 7 miles a
day. After he had gone 32 miles, B set out from D toward
C, and went every day j\ of the whole journey; and after he
202 ELEMENTARY ALGEBRA.
had traveled as many days as he went miles in a day, he met
A. Required the distance from C to D.
Ans. 76 or 152 miles; both numbers will answer the con-
dition.
19. A farmer received 24 dollars for a certain quantity of
wheat, and an equal sum at a price 25 cents less by the
bushel for a quantity of barley, which exceeded the quantity
of wheat by 16 bushels. How many bushels were there of
each ? Ans. 32 bushels of wheat, and 48 of barley.
20. A laborer dug two trenches, one of which was 6 yards
longer than the other, for 17 pounds, 16 shillings, and the
digging of each of them cost as many shillings per yard as
there were yards in its length. What was the length of each ?
Ans. 10 and 16 yards.
21. A and B set out from two towns which were distant
from each other 247 miles, and traveled the direct road till
they met. A went 9 miles a day, and the number of days at
the end of which they met, was greater, by 3, than the num-
ber of miles which B went in a day. How many miles did
each travel? Ans. A, 117, and B 130 miles.
22. The fore wheels of a carriage make 6 revolutions more
than the hind wheels, in going 120 yards ; but if the circum-
ference of each wheel be increased 1 yard, they will make only
4 revolutions more than the hind wheels, in the same distance ;
required the circumference of each wheel.
Ans. 4 and 5 yards.
23. There are two numbers whose product is 120. If 2
be added to the lesser, and 3 subtracted from the greater, the
product of the sum and remainder will also be 120. What
are the numbers ? Ans. 15 and 8.
24. There are two numbers, the sum of whose squares
exceeds twice their product, by 4, and the difference of tlieir
squares exceeds half their product, by 4; required the
numbers. A7is. 6 and 8.
EQUATIONS. 203
25. "What two numbers are those, which being both mul-
tiplied by 27, the first product is a square, and the second
the root of that square ; but being both multiplied by 3, the
first product is a cube, and the second the root of that cube ?
Ans. 243 and 3.
26. A man bought a horse, which he sold, after some time,
for 24 dollars. At this sale he loses as much per cent, upon
the price of his purchase as the horse cost him. What did
he pay for the horse ?
Ans. He paid f 60 or $40 ; the problem does not decide
which sum.
27. What two numbers are those whose product is equal to
the difference of theii squares ; and the greater number is to
the less as 3 to 2 ? Ans. No such numbers exist.
28. What two numbers are those, the double of whose
product is less than the sum of their squares by 9, and half
their product is less than the difference of their squares by 9 ?
Ans, The numbers are 9 and 12.
Will the student show that examples 24 and 28 are essen-
tially the same.
SECTION V.
ARITHMETICAL PROGRESSION.
(Art. 105.) A series of numbers or quantities, increasing
or decreasing by the same diflference, from term to term, is
called arithmetical progression.
Thus, 2, 4, 6, 8, 10, 12, &c., is an increasing or ascending
arithmetical series, having a common diflference of 2; and 20,
17, 14, 11, 8, &c., is a decreasing series, whose common dif-
ference is 3.
"We can more readily investigate the properties of an arith-
metical series from literal than from numeral terms. Thus,
let a represent the first term of a series, and d the common
diflference. Then
a, (a-^d), (a-\-2d), (a-{-3d), (a-{-4d), &c.,
represents an ascending series ; and
a, (a-^), (a— 2d), (a— 3d), (a—id), <fec.,
represents a descending series.
Observe that the coeflScient of d in any term, is equal to
the number of the preceding term.
The first term exists without the common diflference. All
other terms consist of the first term and the common diflfer*
ence multiplied by one less than the number of terms.
ARITHMETICAL PROGRESSION. 205
Thus, if the first term of an arithmetical series is a, and d
the common difference, the tenth term would be expressed by
The 1 7th term by . a-\-\&d
The 53d term by . a-\-52d
The wth term by . a-\-{n — \)d
When the series is decreasing, the sign to the term contain-
ing d will be minus, the 20th term, for example, would be
a— 19g?
The wth term . . . a — {n — 1 )d
We add a few examples to exercise the pupil in finding any
term of a series, when the first term, a, and the common dif-
ference, c?, are given.
1. Whena=2 and c?=3, what is the 10th term?
Ans. 29.
2. When a= 3 and fl?=2, what is the 12th term?
Ans. 25.
3. When a=7 and d=\0, what is the 21st term ?
Ans. 207.
4. When a=l and d—\y what is the 100th term ?
Ans. 50|,
6. When a=3 and <f=i, what is the 100th term ?
Ans. 36.
6. When a=0 and c?=|, what is the 89th term ?
Ans. 1 1 .
7. When a=Q and c^=— i, what is the 20th term ?
Ans. —31.
8. When a=30 and c?= — 3, what is the 31st term ?
Ans. — 60.
Wherever the series is supposed to terminate, is the last
term, and if such term be designated by Z, and the number
of tenns by n, the last term must bfe a+(rj — \)d, or
ELEMENTARY ALGEBRA.
a — [n — '\)d, according as the series may be ascending or
descending, which we draw from inspection.
Hence, .... L=adz(n — l)d (A)
(Art. 106.) It is manifest, that the sum of the terms will
be the same, in whatever order they are written.
Take, for instance, the series . 3, 5, 7, 9,11,
And the same inverted, . . .11, 9, 7, 5, 3.
The sums of the terms will be 14, 14, 14, 14, 14.
Take . .a a-\- d, a-]-2d, a-\-2d, a-\-4d,
Inverted, . . a-\-4d, a-\-3d, a-\-2d, a-\- d, a
Sums, . . 2a+4f^, 2a-{-4d, 2a-\-4d, 2a-{-4d, 2a-\-4d,
Here we discover the important property, that, in arithme-
tical progression, the sum of the extremes is equal to the sum of
any other two terms equally distant from the extremes. Also,
that twice the sum of any series is equal to the extremes, or first
and last term repeated as many times as the series contains terms.
Hence, if S represents the sum of a series, and n the num-
ber of terms, a the first term, and L the last term, we shall
have 2S=n{a-^L)
Or, S^lia-^L) {B)
The two equations (.^4) and [B) contain five quantities, a,
d, L, %, and S; any three of them being given, the other two
can be determined.
Two independent equations are sufficient to determine two
unknown quantities (Art. 45), and it is immaterial which two
are unknown, if the other three are given.
By examining the two equations they will become familiar.
X=a+(«-l)/ {A)
ARITHMETICAL PROGRESSION. 207
Equation (A) and (B) furnish all the rules given in Arith-
metics in relation to arithmetical progression.
For instance, the rule to find the last term of any arithme-
tical series, is equation (A) put in words, thus :
Rule . — Multiply the common difference by the nuniber of
terms less one, and to the product add the first term.
A rule for finding the sum of any series, we draw from
equation (^), thus :
Rule . — Multiply the sum of the extremes hy half the number
of terms.
EXAMPLES.
1. The first term of an arithmetical series is 5, the last
term 92, and the number of terms 30. What is the sum of
the terms? Ans. 1465.
2. The first term of an arithmetical series is 2, the number
of terms 10, and the last term 30. What is the sum of the
terms? Ans. 160.
3. The first term of an arithmetical series is 5, the common
difierence 3, and the number of terms 30. What is the last
term? Ans. 92.
4. The first term of an arithmetical series is 7, the last
term 207, and the number of terms 21. What is the sum of
the terms ? Ans. 2247.
5. The first term of an arithmetical series is 6, the last
term — 3^, and the number of terms 20. What is the sum
of the terms ? Ans. 25.
The two equations [A) and [B) cover the whole subject
of arithmetical progression, when any three of the five quan-
tities are given ; for there would be two unknown quantities,
and we have two equations, which are sufficient to find them ;
we, therefore, give the following miscellaneous examples :
Use the equations vMhout modification or change, by putting
in the given values just as they stand, and afterward reduce
them as numeral equations.
208 ELEMENTARY ALGEBRA.
EXAMPLES.
1. The sum of an arithmetical series is 1455, the first term
5, and the number of terms 30. What is the common
difference? A7is. 3.
Here, ^=1455, a=5, n=SO. Z and cf are sought.
Equation (B), 1455=(5+Z)15. Reduced, Z=92.
Equation (A), 92=5-\-29d. Reduced, d=S, Am.
2. The sum of an arithmetical series is 567, the first term
7, and the common difference 2. What is the number of
terms? Ans. 21.
Here, >S'=567, a=7, d=2. L and n are sought.
Equation {A), Z=7-\-2n — 2=5+ 2w
Equation (J5), 667=(7+5-h2w)|=6w+»=»
Or, n^+6n+9=576
w+3=24, or «=21, Ans.
3. Find seven arithmetical means between 1 and 49.
Observe that the series must consist of 9 terms.
Hence, a=l, Z=49, n=9.
Ans. 7, 13, 19, 25, 31, 37, 43.
4. The first term of an arithmetical series is 1, the sum of
the terms 280, the number of terms 32. What is the com-
mon difference, and the last term ? Ans. d=^, Z=16^.
5. Insert three arithmetical means between i and ^.
Ans. The means are f , /^, |{.
6. Insert five arithmetical means between 5 and 1 5.
Ans. The means ars 6|, 81, 10, llf, 13^.
7. Suppose 100 balls be placed in a straight line, at the
distance of a yard from each other; how far must a person
travel to bring them one by one to a box placed at the distance
of a yard from the first ball ?
Ans. 6 miles and 1300 yards.
ARITHMETICAL PROGRESSION. 209
8. A speculator bought 47 house lots in a certain village,
giving 10 dollars for the first, 30 dollars for the second, 50
dollars for the third, and so on. What did he pay for the
whole 47 ? Ans. $22090.
9. In gathering up a, certain number of balls, placed on the
ground in a straight line, at the distance of 2 yards from each
other, the first being placed 2 yards from the box in which
they were deposited, a man, starting from the box, traveled
11 miles and 840 yards. How many balls were there ?
Ans. 100.
10. How many strokes do the clocks of Venice, which go
on to 24 o'clock, strike in a day ? Ans. 300.
11. In a descending arithmetical series the first term is 730,
the common difi'erence 2, and the last term 2. What is the
number of terms ? Ans. 365.
12. The sum of the terms of an arithmetical series is 280,
the first term 1 , and the number of terms 32. What is the
common difference ? Ans. -J.
13. The sum of the terms of an arithmetical series is 950,
thie common difference 3, and the number of terms 25. What
is the first term ? Ans. 2.
14. What is the sum of n terms of the series 1 , 2, 3, 4, 5,
&C.1 Ans. S=:-(l-{-n)
2^ /
15. Suppose a man owes 1000 dollars, what sum shall he
pay daily so as to cancel the debt, principal and interest, at
the end of a year, reckoning it at 6 per cent, simple interest ?
Divide 1000 dollars by 365, and call the quotient a. This
would be the sum he must pay daily, provided there were no
interest to be paid.
Cast the interest on a for one day, at 6 per cent, and call
this interest i.
18
210 ELEMENTARY ALGEBRA.
Then the first day he must pay a-\-i
The second day, .... a-f-Si
The third day, a-{-3i; and so, on in arith-
metical progression.
The last day he must pay . a-{-365i
Altogether, he must pay . I I c
365.
Or, he must pay daily, . . a4-183i= the answer.
(Art. 107.) Bodies falling near the surface of the earth, sind
unresisted by the atmosphere, fall in the first second of time
16Jj feet, and increase the distance which they fall 2(16^^)
feet every second. Hence, lGj\ feet may be considered the
first term of an arithmetical series, and 2(\6j\) the common
difference. We call IGJ^- feet ff, the symbol for gravity.
Then ff is the first term of an arithmetical series, and 2g the
common difi*erence. Hence, g, 3g, bg, Ig^ 9^, &c., are the
spaces corresponding to 1, 2, 3, 4, &c., seconds.
These facts being admitted, show a formula for the fall of
a body in 10 seconds, and for its fall the last second of the
ten,
From (A) . . . L^g-^-^'Stg
From(J5) . . . S=b(2g-{-\^g)
Hence, its fall during the last second of the ten is 19^, and
the whole space fallen through is 100^, which is the square of
the seconds multiplied by the force of gravity, and this is the
general rule in Astronomy.
But this manner of arriving at the result is not recommended,
except as an exercise in progression.
PROBLEMS IN ARITHMETICAL PROGRESSION
TO WHICH THE FRKCEDINO FORMULAS, {A) AND (B),
DO NOT IMMEDIATELY APPLY.
(Art. 108.) When three quantities are in arithmetical pro-
gression, it is evident that the middle one must be the exact
inean of the three, otherwise, it would not be arithmetical
ARITHMETICAL PROGRESSION. 211
progression ; therefore the sum of the extremes must be double
that of the mean.
Take, for example, any three consecutive terms of a series,
as a-i-2c?, a-\-^d, a-\-4d
and we perceive, by inspection, that the sum of the extremes
is double that of the mean.
When there are four terms, the sum of the extremes is
equal to the sum of the means, by (Art. 106).
To facilitate the solution of problems, when three terms are
in question, let them be represented by (x — y), x, (x-{-y),
y being the common difference.
When four numbers are in question, let them be repre-
sented by (a:— 3y), {x—y), {xArv), {x^^y)y ^y being
the common difference.
So in general for any other number, assume such terms
that th^ common difference will disappear by addition,
EXAMPLES.
1. Three numbers are in arithmetical progression, the
product of the first and second is 15, and of the first and third
is 21. What are the numbers ? Ans. 3, 5, and 7.
2. There are four numbers in arithmetical progression, the
sum of the two means is 25, and the second, multiplied by
the common difference is 50. What are the numbers ?
Ans. 5, 10, 15, and 20.
3. There are four numbers in arithmetical progression, the
product of the first and third is 5, and of the second and
fourth is 21. What are the numbers 1 Ans. 1, 3, 5, and 7.
4. There are five numbers in arithmetical progression, the
sum of these numbers is 65, and the sum of their squares
1005. What are the numbers? Ans. 5, 9, 13, 17, and 21.
Let x^= the middle term, arid y the common difference.
212 ELEMENTARY ALGEBRA.
Then x — 2y, x — y, x, x-\-y, x-\-2y, will represent the
numbers, and their sum will be 5x=65, or a;=13. Also, the
sum of their squares will be
5ar^+10y2=l005, or a^^-f 23/2= 201.
5. The sum of three numbers in arithmetical progression
is 15, and their continued product is 105. What are the
numbers? " Ans. 3, 5, and 7.
6. There are three numbers in arithmetical progression,
their sum is 18, and the sum of their squares 158. What
are those numbers ? ' Ans. 1, 6, and 1 1.
7. Find three numbers in arithmetical progression, such
that the sum of their squares may be 56, and the sum arising
by adding together once the first and twice the second, and
thrice the third, may amount to 28. Aiis. 2, 4, 6.
8. Find three numbers having equal differences, so that
their sum may be 12, and the sum of their fourth powers
962. Ans. 3, 4, 5.
9. Find three numbers having equal differences, and such
that the square of the least added to the product of the two
greater, may make 28, but the square of the greatest added
to the product of the two less, may make 44. Ans. 2, 4, 6.
10. Find three numbers in arithmetical progression, such
that their sum shall be 1 5, and the sum of their squares 93.
Ans. 2, 5, and 8.
11. Find three numbers in arithmetical progression, such
that the sum of the first and third shall be 8, and the sum of
the squares of the second and third shall be 52.
Ans. 2, 4, and 6.
12. Find four numbers in arithmetical progression, such
that the sura of the first and fourth shall be 13, and the dif-
ference of the squares of the two means sliall be 39.
Ans. 2, 5, 8, and 11.
13. Find seven numbers in arithmetical progression, such
ARITHMETICAL PROGRESSION.
that the sum of the first and sixth shall he 14, and the pro-
duct of the third and fifth shall be 60.
Ans. 2, 4, 6, 8, 10, 12, and 14.
15. Find five numbers in arithmetical progression, such
that their sum shall be 25, and their continued product 945.
Ans. 1, 3, 5, 7, and 9.
16. Find four numbers in arithmetical progression, such
that the dijQference of the squares of the first and second shall
be 12, and the diflference of the squares of the third and fourth
shall be 28. Ans. 2, 4, 6, and 8.
GEOMETRICAL PROGRESSION.
(Art. 109.) When a series of numbers or quantities in-
crease or decrease by a constant multiplier from term to term,
the numbers or quantities are said to be in geometrical pro-
ffression, and the constant multiplier is called the ratio.
Thus, let a be the first term of the progression, and r the
ratio, then a, ar, ar^, ar^, ar*, &c., will represent the series.
If r is greater than 1 , the series will be ascending; if less
than 1 , the series will be descending, and if r= 1 , every terra
of the series will be the same in value.
For example, 2, 6, 18, 54, 162, &c., is a geometrical series
in which the first term a is 2, and the ratio is 3.
The series 9, 3, 1, i, i, ^\, <fec., is also a geometrical series
in which the first term a is 9, and the multiplier, the ratio.
The series 3, 3, 3, 3, <fec., is also a geometrical series in
which the first term a, is 3, and the multiplier, the ratio, is 1.
Geometry compares magnitudes, and inquires how many
times one magnitude is greater than another, and thus, in the
814 ELEMENTARY ALGEBRA.
series, 2, 4, 8, 16, <fec., 4 is two times 2, 8 is 2 times 4, &,c.-
hence, numbers so compared, and a regular series thus
obtained, is called a geometrical series.
(Art. 110.) In any given series we may find the ratio, by
dividing any term by its preceding term.
(Art. 111.) Taking the general series a, ar, ar^y ar^, ar"^,
&c., under inspection, we find that the^r*^ power of r is a
factor in the second term, the second power of r in the third
term, the third power of r in the fourth term, and thus, uni-
versally, the power of the ratio in any term, is one less than
the number of the term.
The first term is a factor in every term. Hence, the 10th
term of this general series is ar^. The 1 7th term would be
ar^^. The 25th term would be ar^S and, in general, the wth
term would be ar'^~\
Therefore, if n represent the number of terms in any
series, and L the last term, then
Z=ar™-» (1)
Wherever a series is supposed to terminate, is the last term,
and equation ( 1 ) is a general representation of it ; and if we
multiply that equation by r, we shall have
rL=ar^
(Art. 112.) Let S represent the sum of any geometrical
series, then we have
S=s:a-\-a.r'\-ar^'^ai*, &c., to ar*"^
Multiply this equation by r, and we have
rSs=ar-\-ar'-\-a7^, <fec., to ar"~'-}-ar"
Subtracting the upper equation from the lower, and observ-
ing that rL=ar^
Then . . . (r--l)S=rL—a
Therefore, . . . 5« (2)
GEOMETRICAL PROGRESSION. 215
As the equations (1) and (2) are fundamental, and cover
the whole subject of geometrical progression, let them be
brought together for critical inspection.
Here we perceive five quantities, a, r, n, Z, and S^ and any
three of them being given in any problem, the other two can
be determined from the equations,
L^ar-^-"- (1)
8="^ (2)
"These two equations furnish the rules given for the opera-
tions in common arithmetic.
Thus, in almost every Arithmetic, the rules for finding the
ifest term of any arithmetical series is expressed in the follow-
ing words :
R tj L E . — Raise the ratio to a power one less than the number
of terms, and multiply that number by the first term.
This rule is simply equation ( 1 ) put in words.
Equation (2) gives the following rule for the sum of a
series.
Rule . — Multiply the last term by the ratio, and from the
product subtract the first term, and divide the remairider by the
ratio less one.
GENERAL EXAMPLES IN GEOMETRICAL
PROGRESSION.
1. What is the ratio of the series 2, 6, 18, 54, &c.?
Ans. 3.
2. What is the ratio of the series 5, 20, 80, &c.? Ans. 4.
3. What is the ratio of the series |-, |-, 2-\, &c.l Ans. ?j,
4. What is the ratio of the series y^.^ ^i_^ toV o» ^^•
Ans. yV-
5. What is the ratio of the series f, 1, ^, &c.? Ans. |.
216 ELEMENTARY ALGEBRA.
6. What is the ratio of the series 8, 20, 60, &c.? Ans. ^
7. What is the ratio of the series -, — , — , &c.? Ans. —
8. What is the ratio of the series a, — h, -\ — &c.?
Ans. •
a
9. What is the 11th term of the series 1, 2, 4, &c.?
Ans. 1024.
10. What is the 9th term of the series 5, 20, 80, &c.?
Ans. 327680.
11. What is the 8th term of the series 2, 6, 18, &c.?
Ans. 4374.
12. What is the 6th term of the series 1, |,f^,(fec.
Ans. /^V*-
13. What is the sum of 8 terms of the series 2, 6, 18, &c.?
Ans. 6560.
,^. „ rZ—a 3*4374—2 ^^^^
(2) S= -= =6560
^ ^ r— 1 2
14. What is the sum of 10 terms of the series 4, 12, 36,
&c.? Ans, 118096.
15. What is the sum of 9 terms of the series 5, 20, 80,
&c.? Ans. 436905.
16. What is the sum of 5 terms of the series 3, 4|-, 63,
&c.? ^ Ans. 39y9^.
17. What is the sum of 10 terms of the series 1, |, ^,
&c.? Ans. VVoW-
18. A man purchased a house, giving 1 dollar for the first
door, 2 dollars for the second, 4 dollars for the third, and so
on, there being 1 0 doors. What did the house cost him ?
Ans. ^1023.
(Art. 113.) By equation (2), and the rule subsequently
given, we perceive that the sum of a series depends on the
first and last terms and the ratio, and not on the number of
terms; and whether the terms be many or few, there is no
GEOMETRICAL PROGRESSION. 217
variation in the rule. Hence, we may require the sum of any
descending series, as 1, ^, j, |, &c., to infinity, provided we
determine the last term. Now, we perceive the magnitude of
the terms decrease as the series advances; the hundredth
term would be extremely small, the thousandth term would
be very much less, and the infinite term nothing ; not too small
to be noted, as some tell us, but absolutely vMhing.
Hence, in any decreasing series, when the number of terms
is conceived to be infinite, the last term, L, becomes 0, and
equation (2) becomes
— a
By change of signs . s=-
This gives the following nile for the sum of a decreasing
infinite series :
Rule . — Divide the first term hy the difference hetvxen unity
and the ratio.
EXAMPLES.
1. Find the value of 1, f , t\, &c., to infinity.
a=r\f r=f Ans. 4.
/ 2. Find the exact value of the series 2, 1, ^, &c., to infinity.
/ Ans. 4.
t 3. Find the exact value of the series 6, 4, &c., to infinity.
Ans. 18.
4. Find the exact value of the decunal .3333, &c., to
infinity. Ans. \.
This may be expressed thus : fV+rf 7' <^c- Hence, a=^,
6. Find the value of .323232, <fec., to infinity. m
«=t¥5 » «»'=«T5¥5iri therefore, r«xiv -4nir.(|f.
19
218 ELEMENTARY ALGEBRA, r
6. Find the value of .777, &c., to infinity. . . Am. |.
7. Find the sum of the infinite series 1 + -^H — ^H — 6+, &c.
Ans. - — r-
x^ — 1
8. Find the sum of the infinite geometrical progression
a — h-\ -A — ^ — , (fee, in which the ratio is .
a a^ a? a ^2
Am. —r-
a-\-o
(Art. 114.) When three numbers are in geometrical pro-
gression, the product of the extremes is equal to the square
of the mean.
This principle is. obvious from the general series
a, ar, ar^, ar^, ar*, ar^, &c.
Taking any three consecutive terms anywhere along the
series, we observe, that the product of the extremes is equal to
the square of the mean.
That is, if the three terms taken, are a, ar, ar^,
a^r^=(ary
If ar^, ar^, ar"^ are the three terms,
a7^Xa7^=(a7^y
Hence, to find a geometrical mean between two numbers, we
must multiply them together, and take the square root. ^
we take four consecutive terms, the product of the extremes will
he equal to the product of the means.
(Art. 115.) This last property belongs equally to geome-
trical proportion, as well a^ to a geometrical series, and the
learner must be careful not to confound proportion with a
series.
a: ar: :h:hr, is a geometrical proportion, not a continued
series. The ratio is the same in the two couplets, but the
magnitudes, a and h, to which the ratio is applied, may be
very diflferent.
GEOMETRICAL PROGRESSION. 219
We may suppose a : ar two consecutive terms of one series,
and b : br any two consecutive terms of another series having
the same ratio as the first series, and, being brought together,
they form a geometrical proportion. Hence, the equality of the
ratio constitutes proportion.
EXAMPLES.
1. Find the geometrical mean between 2 and 8. Ans. 4.
(Art. 114) ^2X8=4
2. Find the geometrical mean between 3 and 12.
Ans. 6.
3. Find the geometrical mean between 5 and 80.
Ans. 20.
4. Find the geometrical mean between a and b.
Ans. {aby,
5. Find the geometrical mean between \ and 9. Ans. |,
6. Find the geometrical mean between 3a and 27a.
Ans. 9a.
7. Find the geometrical mean between 1 and 9. Ans. 3.
8. Find the geometrical mean between 2 and 3. Ans. JQ,
9. Find two geometrical means between 4 and 256.
N. B. When the two means are found, the series will con-
sist oi four terms, and 4 will be the first term and 256 will be
the last term.
Comparing this with the general series,
a, ar, ar^, aT^y we have
a=:4 and ar'=256
Hence, . . . r^=64 or r=4
Therefore, 16 and 64 are the means required.
10. Find three geometrical means between 1 and 16.
220 ELEMENTARY ALGEBRA.
Here, the first term of tlie series is 1, the last term IG, and
the number of terms 5, because three terms are required, and
two are ah'eady given.
Now, by equation (1), L^^ar'^"^
That is, .... ar"-^=16
But as a=l, and »=5, this equation is
Hence, . . • . • r=2
Therefore, the means required are 2, 4, and 8.
We may obtain the ratio when the first and last terms are
given, by the following formula : ^
\ a
1 1 . The first and last terms of a geometrical series are
2 and 162, and the number of terms 5 ; required the ratio.
Ans. 3
12. The first term of a geometrical series is 28, the last
term 17500, and the number of terms 6; what is the ratio?
Ans. 5.
PROBLEMS THAT INVOLVE THE PRINCI-
PLES OF GEOMETRICAL PROGRESSION.
(Art. 116.) When we wish to express three unknown
quantities in geometrical progression, we may represent them
by X, Jxy, y, or by x^y ry, y^, or by Xy icy, xy'^y for either of
these correspond with Art. 114; that is, the product of the
extremes is equal to the square of the mean.
When we wish to express four unknown quantities in geo-
metrical progression, we may express them by ar, ary, ay*, ary*,
or by P, X, y, Q.
The object of this last notation, is to reduce P and Q to
terms expressed by x and y, thus :
GEOMETRICAL PROGRESSION. 221
Taking the first three terms only, we shall have
Or, P=- . , .
y
Taking the last three terms only, we shall have
Therefore, four quantities in geometrical progression may
be expressed by x and y only^ and the terms stand symmetri-
cally thus :
x" y"
7.' ^' y^ t:
y X
In a similar manner, we might express more terms by x
and y only, and have them stand symmetrically, if it were
proper to extend this subject in a work as elementary as this.
1. Three numbers are in geometrical progression, the sum
of the first and second is 90, and the sum of the second and
third is 1 80. What are the numbers ?
Am. 30, 60, and 120.
Represent the numbers by a;, xy, and xy^.
2. The sum of three numbers in geometrical progression
is 7, and the sum of their squares is 21. "What are the
numbers ? Ans. 1 , 2, 4.
This problem furnishes the following equations :
a:+»Jx2j-\ry=7 (1)
x'+xy+y'^n (2)
From (1) , . . . X'\-y—a—J^ (3)
From (2) .... ^-\-f^^a—xy (4)
Squaring (3), . x^-{-^xy-\-y'^=c?-^'iLaJxy\-xy (5)
Subtracting (4) from (5), ^xy=a^-^Za—^aJxy-\-2xy (6)
222 ELEMENTARY ALGEBRA.
Dropping 9.xy from botli members, dividing by a, and
transposing, we have . . ^Jxy=^a — 3
That is, V^=2 (7)
This value of Jxy put in equation (3), gives
3.4.^=5 (8)
From equations (7) and (8), we find x and y, as taught in
Art. 101.
3. The sum of the first and third of four numbers in geo-
metrical progression is 20, and the sum of the second and
fourth is 60. What are the numbers ? Ans. 2, 6, 1 8, 54.
4. Divide the number 210 into three parts, so that the last
shall exceed the first by 90, and the. parts be in geometrical
progression. Ans. 30, 60, and 120.
5. The sum of four numbers in geometrical progression is
30; and the last term divided by the sum of the mean terms
is li. What are the numbers? Ans. 2, 4, 8, and 16.
6. The sum of the first and third of four numbers in geo-
metrical progression is 148, and the sum of the second and
fourth is 888. What are the numbers ?
Ans. 4, 24, 144, and 864.
7. The continued product of three numbers in geometrical
progression is 216, and the sum of the squares of the ex-
tremes is 328. What are the numbers ? Ans. 2, 6, 18.
8. The sum of three numbers in geometrical progression
is 13, and the sum of the extremes being multiplied by the
mean, the product is 30. What are the numbers ?
Ans. 1, 3, and 9.
9. There are three numbers in geometrical progression
whose product is 64, and the sum of their cubes is 584.
What are the numbers ? Ans. 2, 4, and 8.
GEOMETRICAL PROGRESSION. 223
Let ar^, xy, y^ represent the three numbers.
Then, x'if^Q\ (1)
(^■\-j^y^-\-y^=bZ^ (2)
. . ar^^ . . =64
Also, . .
Add, . .
And . .
Square root
a:«4-2a^2/2+2/^=648=324*2 (3)
. . ^■^f=\zji (4)
From (2) subtract three times (1), and we have
JB«— 2a^/-f2^=392=196'2 (5)
Square root, . . . y^ — x^=\^J^ (6)
We give the minus sign to s^, because y must be greater
than X from the position it occupies in our notation, and o^ — y^
or y^ — 01^ y when squared, will produce the same power.
Subtracting (6) from (4), and
2ar^=4<y2
Or, Q?=2j^
Squaring, .... a;^=8
Cube root, .... ar*=2, Ans.
10. There are three numbers in geometrical progression,
the sum of the first and last is 52, and the square of the
mean is 100. What are the numbers? Ans. 2, 10, 50.
11. There are three numbers in geometrical progression,
their sum is 31, and the sum of the squares of the first and
last is 626. What are the numbers ? Ans. 1, 5, 25.
12. It is required to find three numbers in geometrical pro-
gression, such that their sum shall be 14, and the sum of
their squares 84. Ans. 2, 4, and 8.
13. There are four numbers in geometrical progression, the
second of which is less than the fourth by 24; and the sum
of the extremes is to the sum of the means, as 7 to 3. What
are- the numbers ? An^. 1, 3, 9, and 27.
224 ELEMENTARY ALGEBRA.
14. The sum of four numbers in geometrical progression
is equal to the common ratio +1, and the first term is j\.
What are the numbers ? Ans. j\, j\, j%, 3.7.
PROPOETION.
(Art. 117.) Two magnitudes of the same kind can be
compared with each other, and the numerical relation
between them determined. The manner of determining this
relation, is to divide one by the other, and the quotient is
called the ratio betweenthe two magnitudes. When two quan-
tities have the same ratio as two other quantities, the four
quantities may constitute a proportion.
Therefore, proportion is the equality of ratios.
Proportion is written in two ways,
Thus, aib: :c:d
Or thus, . . . . a:h = c:d
The last is the modern method, and means that the ratio
of a to 5 is equal to the ratio of c to d.
If a is taken as the unit of measure between a and h, then
- IS the numerical ratio between these two magnitudes.
If c is taken for the unit of measure between c and d, then
d.
- is the numerical ratio between these two magnitudes.
c ^
The magnitudes a and h may be very different in kind from
those of c and d; for instance, a and b may be bushels of
wheat, and c and d sums of money.
This manner of comparing magnitudes, by taking one of
them as a whole (regardless of other units) is called
PROPORTION. 225
geometrical proportion, and if there are more than two magni-
tudes having the same ratio, the magnitudes are said to be in
geometrical progression.
Two magnitudes compared by raiio are called a couplet.
Thus, a : 5 is a couplet, and c.dis another couplet.
The first magnitude of a couplet is called the antecedent^
the second the conseqtteni.
A ratio can exist between two magnitudes ; but a proportion
requires four — tivo antecedents and two consequents having the
same ratio.
Thus, if . . , . a: h=c : d
Then -=- by the def. of proportion,
0) c
All operations in proportion rest on this fundamental equa-
tion ; and to prove a principle or an operation true, we directly,
or remotely compare the principle or the operation to this
equation, and if we find a correspondence, the principle or
the operation is true — otherwise, false.
PROPOSITION I.
In every proportion, the product of the extremes is equal to the
product of the means.
Let a: b=c : d represent any proportion,
b d
Then, -=- must be a true equation.
a c ^
Multiply both members of this equation by ac, and as the
product of equal factors are equal (Ax. 3),
Therefore, . . . . cb=ad
That is, the product of c and b, the means, is equal to the
product of a and d, the extremes.
S c H o L I u M. — Divide both members of this equation by a,
Then -=ti
a
226 ELEMENTARY ALGEBRA.
This equation shows, that the fourth term of any proportion
may be found from the first three, by the following
Rule . — Multiply the second and third terms of the proper'
tion together, and divide that product hy the first term.
This is a part of the well known rule of three, in Arithmetic.
PROPOSITION II.
Conversely. ]f the product of two quantities is equal to the
product of two others, then two of them may he taken for the
means, and the other two for the extremes of a proportion.
Let ch=ad
Divide both members of this equation by any one of the
four factors, say c, then we have
ad
c
Divide this last equation by another of the factors, say a.
Then -=-
a c
This is the fundamental equation for proportion, and gives
a : b=c :d
Now, as the principle is established, we may proceed more
summarily, and take the two factors in one member for the
extremes,
Thus, .... a: = :d
To fill up the means, we must take the factor which has
the same name as a to stand before the equality, and the
other factor to stand after the equahty will be of the same
name as d, and the proportion will be complete.
If the quantities are all numerals, it is immaterial which
factor stands first in the means.
Thus, a : b=c : d ) are proportions equally true in numeri-
Or, . a : c=6 :d ) cal values.
PROPORTION. 227
Scholium . — A proportion and an equation may be re-
garded as but a different form for the same expression, and
every equation may be put into a proportion. For example,
Wliat proportion is equivalent to the following equation ?
(xy=a{a-\-b)
Ans. . . . ; a;:a=a-l-J:y,
Or, (cy: a=^a-{-h : 1
Or, a : rr=y : {a-\-h)
What proportion is equivalent to the equation
cd
a
Thus, we might give examples without end.
Ans, . , x\ { — |- 1 j =a : 1
PROPOSITION III.
If three quantities are in continued proportion, the product
of tlie extremes is equal to the square of the mean.
If a: b=b : c
From proposition 1, ac=bb=h'^
That is, if proposition 1 is true, the truth of this proposition
follows as an inevitable consequence.
If ac=b^, then b=Jac, which shows that
The mean proportional between two quantities is found by
extracting the square root of their product.
PROPOSITION IV.
If four quantities are in proportion, they will be in proportion
by INVERSION, that is, the second will be to the first, as the fourth
to the third.
Let a'.b=:^c:d
228 ELEMENTARY ALGEBRA.
Then, by the definition of ratio and proportion, we have
a c
h . a
Divide 1 by - and the quotient is -.
d . c
Divide 1 by - and the quotient is -.
But equals divided by equals must produce equal quotient*
(Ax. 4).
Therefore, . .
a c
Or, ....
. b:a= die
In numbers if .
. 3:5=12:20
Then ....
. 5:3=20:12
PROPOSITION V.
Magnitudes which are proportional to the same proportionals,
are proportional to^each other.
If a:b=F:Q
a:b=F: Q )
c:d=P:Q J
. , , ^ ^ , Then a : b=c : d.
And
b Q
From the first proportion, -=p
a Jr
By the second, . . . -=5
C Jr
Therefore, .... -=- (Ax. 1).
a c ^
Or, a\b=^c\d
PROPOSITION VI.
If four magnitudes be in proportion^ they must be in propor-
tion by COMPOSITION ; that is, the first will be to the sum of the
first and second, as the third will be to the sum of the third and
fourth ; and the first is to the difference between the first and
second, as the third is to the difference between the third and fourth.
PROPORTION. 229
On the supposition that . a : 5=c : d
We are to prove that a : a-\-h=^c : c-j-rf
T. , . . h d
From the supposition, . -=-
Add each member of this equation to unity, and then we
have 1+-=1+-
a c
Reducing these mixed quantities to improper fractions,
Ajid a+6^c+d
a c
That is, . . . . a: a-|-5=c : c-\-d
Subtracting each member of the original equation from
unity, and we have
a c
a — b c — d
ur, . . • • •
. =
a c
Therefore, . . .
a : a — 5=c : c — d
Scholium . — This composition may be carried to almost
ly extent, as we see
by the following investigation :
Take Ae equation.
b_d
a c
Multiply both members by m, then
mb^md
a c
Add each member of this equation to n, ,.
Then, ....
, nib , md
. w+ — =wi —
a c
By reduction, . .
na-\-mb iic-{-md
a c
Hence, ... a
•.na'\-mb=sQ'.nG-^md
230 ELEMENTARY ALGEBRA.
PROPOSITION VII.
If four quantities he in proportion, the sum of the two quan-
iities which form the first couplet is ip their difference, as the sum
of the two quantities which form the second couplet is to their
difference.
On the supposition that • • a : 5=c : d
We are to prove that a-\-h : a — 5=c+a? ; c — d
From proposition 6, . . a\a-\-h^=c\c-\-d (1)
Also, a\a — h=c\c^d (2)
■n /H\ a-\-h c-\-d
From(l), __L-=-_L_
^ ' a c
Dividing both members of this equation by {a-\-h), and
multiplying both members by c, we have
c c-\-d
a a+d
Operating in the same manner with (2), we shall find
c c — d
a a — b
c-\-d c- w - . . .
Therefore, — r7= 1 (Ax. 1).
a+o a — 0 ^ ^
Whence, .... a-]rh\c-\-d=za — b:c — d
Or, a — h : c — d=a-]rh : c-{-d
PROPOSITION VIII.
If four quantities he in proportion, either couplet may he mul-
tiplied or divided hy any number whatever, and the quantities
will still he in proportion.
Let ........ a : h=c : d
Then, -=-
PROPORTION. . 231
Multiplying both numerator and denominator of either of
these fractions by any number, n.
Then, — =-
na c
b 7id
Also, -= —
a nc
That is, , . . , na: nb=^c : d
Also, a\ h=nc : nd
Here, n may represent any number whatever ; and if it
represents a whole number, as 3, 7, 8, &c., then the couplet
is multiplied. If n represent a fraction, as i, i, |, &c., then
the couplet is divided.
PROPOSITION IX.
If four quantities be in proportion, the antecedents may be
multiplied by any numJyer, and they will still be in proportion;
also, the consequents may be multiplied by any number, and the
four quantities will still be in proportion.
Let a: b=.c : d
Then . *=^
a c
Multiplying this equation by m, then
mb md
a c
Therefore, . ' . . a: 7rib=c -.md
Divide both members of the original equation by m.
Then, ±=-^
ma mc
Hence, . . . . ma: b=m^ : d
232 ELEMENTARY ALGEBRA,
PROPOSITION X.
If four magnitudes he in 2}roportionf like powers or roots of
the same will be in proportion.
Let ...... a:h=c'.d
Then, *=^
a c
Raise both members of this equation to any power denoted
by n.
Then, — =—
Hence," .... a»:5"=c":c?»
By extracting any root of the primitive equations, which
may be designated by -, we have
1
1 "~"
1
a«
c"
1
1
I 1
:c^ : (f
Hence, . . .
PROPOSITION XI.
If four quantities he in proportion^ also four others, the
^product or quotient of the two, term ly term, will still form
proportions.
If a:h=c'.d
And, x'.y=^m:n
Then we are to prove that
ax : hy=cm : dn
. . abed
And, -:-=- :-
X y m n
From the first proportion we have
U. (,)
a c ^ ^
From the second, . . -=- (2)
X m ^ ^
PROPORTION. 233
By multiplying these two equations together, term by term,
ire find
ly nd
ax mc
That is, . . . . ax: bi/=mc : nd
Apply proposition 1 to the two given proportions, and we
have ad=bc
And, nx=.my
Dividing one of these equations by the other, we have
\x) Kni \y)\m)
By the reverse application of proposition 1 , we have
ah c d
X ' y m ' n
PROPOSITION XII.
If any number of proportionals have the same ratio, any one
vf the antecedents will be its consequent, and as the sum of all the
antecedents to the sum of all the consequeMs.
Let a : b= a : h
Also, . . . . . a : 5= c : d
a:b=m:n
(fec.=&c.
Then we are to prove that
a:b=(a-\-c-\-m):(b-^d-\-n)
From the first prop., ab=ab
From the second, . ad=cb
From the third, . . an=nb
By addition, a(b-{-d-\-n)=b(a-\-c-]-m)
By prop. 1, . . . a:b=(a-\rc-^m):(b-{-d-{-n)
The following examples are intended to illustrate the prac-
tical utility of the foregoing propositions :
20
234
ELEMENTARY ALGEBRA
EXAMPLES.
1. Find two numbers, the greater of which is to the less as
their sum to 42, and the greater to the less as their difference
is to 6.
Let . x= the greater and y= the less.
mi 1 Tx- ( x:y=x-\-y:42
Then, by conditions i „
•' ( x: y=-x — y : 6
(Prop. 6), . . ic+y:42=a; — y\^
Changing means, x-\-y : x — ?/=42 : 6
(Prop. 9), . . . 2a;:2y=48:36
(Prop. 8), . . . a; : y= 4 : 3
With these proportions of x and y, we return to the original
conditions ; applying proposition 5, and we have
4 : ^—x-\-y : 42
4 : 3=a;— y : 6
From the first, . . a;-}-y=56
From the second, . x — ^2/= 8
Hence, ic=32, y=24
2. Divide the number 14 into two such parts, that the quo-
tient of the greater, divided by the less, shall be to the less,
divided by the greater, as 100 to 16.
Let . a:= the greater, and y= the less part,
Then, -:^=100:16
y X
And, a;+2/=14
(Prop. 8), ... a^:y2=:i00:16
(Prop. 10), . . . ar:y=10:4
Hence, 9,x=5y
But, x-{-y=14
Therefore, a:=10, y=4
3. Find three numbers in geometrical progression whose
sum is 13, and the sum of the extremes is to the double of
the mean as 10 to 6.
Let a:, xy, and xy^ represent the numbers,
PROPORTION.
Then, by the conditions, x-\rX2/-\-xy^=\3
And, xf-\-x : 2xy=10 : e
(Prop. 8), 2/2+l:2y=10:6
(Prop. 7), (y2+2y+l):(y'— 2y+l)=16:4
(Prop. 10), y+l:y— 1=4:2
(Prop. 8), 2y:2=6:2
Hence, y=3, x=l
4. The product of two numbers is 35, and the difference
of their cubes is to the cube of their difference as 109 to 4.
What are the numbers ? Ans. 7 and 5.
Let X and y represent the numbers. Then, by the given
conditions, xy=^35
And, a^—y'^:(x—yy=\09:4
Divide the first couplet by (x — y) (Prop. 8). Then we
have .... x^-{-xy-\-f : (x—i/y=\09 : 4
Expanding (x — yy, and then making appHcation of (Prop.
7), we have .... Sxy:(x — yY=105:4
But, 3ary=105
Therefore, (x — yy=4:
And, . X — y=2
5. What two numbers are those, whose difference is to
their sum as 2 to 9, and whose sum is to their product as 1 8
to 77? Am. 11 and 7.
6. Two numbers have such a relation to each other, that
if 4 be added to each, they will be in proportion as 3 to 4;
and if 4 be subtracted from each, they will be to each other
as 1 to 4. What are the numbers ? Ans. 5 and 8.
7. Divide the number 16 into two such parts that their
product shall be to the sum of their squares as 15 to 34.
Ans. 10 and 6,
8. There are two numbers whose product is 320; and the
difference of their cubes, is to the cube of their difference, as
61 to 1. What are the numbers? Ans. 20 and 16.
236 ELEMENTARY ALGEBRA.
CONCLUSION.
We conclude this volume by giving a general investigation
of the rules in Arithmetic for the computation of interest, and
the adjustment of accounts in fellowship.
Interest is a percentage paid for the tise of money for a spe-
cified time.
On this single definition, all the rules of computation are
founded. The unit for time is commonly one year.
Let r represent the interest corresponding to unity of prin-
cipal remaining at interest for unity of time.
Then, as a double capital would demand double interest
for the same time, a treble capital, treble interest, and so on.
Therefore, if P represents any principal or capital, we have
the following proportion :
Trin. Int. Prin. Int.
1 : r=F : rP
The last term of this proportion shows that to find the in-
terest of any principal for one year, we must multiply that
principal by the decimal rate per cent.*
For double the length of time, the interest must be double,
for treble the length of time, the interest must be treble ; and
so on.
Now, let t represent the length of time that any principal,
P, remains at interest, r being the rate per cent., and / the
aggregate interest, then we shall have this general equation.
Prt=I (1)
This gives the following universal rule for computing
interest.
Rule . — Multij^ly the principal hy the decimal rate per cent.,
and that product by tltc time,
* Rate per cent is but another definition for the interest of unity of prin-
cipal for unity of time.
PROPORTION. 237
If we consider that the principal and interest added together
must give the amount, and if we put A to represent the
amount, then we shall have
Fri-hF=A (2)
Equations (1) and (2) embrace all the conditions in relation
to interest, and furnish all the rules for computations.
For instance, equation (1) gives
Equation (2) gives . P= j—
That is, when any problem requires the finding of the
principal, observe the following rules :
R u L E 1 . — Divide the interest ly the ^product of the rate and
time.
R u L E 2 . — Divide the amount ly the product of the rate
and tivMy increased ly unity.
Equation ( 1 ) gives . ^—-p
That is, to find the time, we have the following rule :
Rule . — Divide the whole interest hy the interest for one
year.
Equation (1) gives . ^='p>
To find the rate per cent., take the following rule :
Rule . — Divide the interest hy the product of the principal
and time.
FELLOWSHIP.
Two men united capital to engage in a certain enterprise,
the first put in a dollars, the second I dollars, and they gained
g dollars. Give a rule for the equitable division of this gain.
ELEMENTARY ALGEBRA.
Let X represent the portion belonging to tliat one which paid
in a dollars, and y the portion of the other.
Then, .... x+y=ff
But their portions of the gain should be in just the same
proportion as their capital paid in,
That is, ... . a? : y=a : b, or hx=ay
Multiply the first equation by a, then
ax-\-ay=agy or ax-{-hx=ag
Or, .=^, y^ ^
'a-\-b* ^ a^b
Hence, we have the following rule to find each man's share.
Rule . — Multiply the gain by each man's stock, and divide
the product by the whole capital invested.
Again, suppose three persons. A, B, and C, enter into
partnership, and furnish capital in proportion to a, b, and c,
and they gain a sum, g, what is each man's share of it ?
Let . x= A's share, y= B's share, and ^=C's share,
Then, . . . x-jry-^z=g
(1)
And, x:y=a:b,
also,
y:z=a:c
From the first propor., y=-
(2)
ex
From the second, . . «= —
a
(3)
These values of y and 2, put in equation (1), give
bx ex
x-i h— =0', or ax-\-ox-]rcx=:.ag
a a
Or ^=-4^^ (4)
a-\-b-\-c ^ '
This value of x put in (2) gives y, and in (3) gives z.
FELLOWSHIP. 239
Here, again, we find that each man's share of the gain is
equal to the whole gain multiplied hy his particular portion oj
Ike stock, and that product divided hy the whole stock invested.
The same results would be obtained in relation to any nuni-
ber of partners. Observe, that g can be of any value, posi-
tive, negative, or zero. When it is zero, each numerator is
zero ; and, thus, x, y, and z becopae zero, as they ought
in that case. When g is negative, it denotes loss, and losses
must be shared in the same proportion.
It is not necessary that a, h, and c should designate the
actual stock of each partner if they represent their due pro-
portional parts.
In taking up a book on common Arithmetic, we find the
following rule for fellowship :
As the whole amount of stock or labor
Is to each Trumps portion.
So is the whole property, loss or gain^
To each man's share of it.
These four lines express either one of the equations, (4),
(5), or (6); for, by resolving (4), for example, into a propor-
tion, we have . . (a-{-b-jrc) :a=g:x
Thus, we perceive that this, like most other arithmetical
rules, is the result of algebraic investigation.
Let us now consider the case in which time is an element,
and for the sake of clearness we will suppose an example.
Two men, A and B, hired a pasture, for which they agreed
to pay g dollars. A ptU in a coivs 3 weeks, B put in b cows
for 6 weeks; what shall each pay ?
Consider that, a cows for three weeks would consume as
much as 2>a cows for one week. Also, h cows for five weeks
would consume as much as bh cows for one week. Thus, we
reduce all action to some unit of time. To be more general, we
will consider 3 and 5 as t and t' , any number of weeks or
days whatever, then the action will be at and W, and it is
240 ELEMENTARY ALGEBRA.
evident fhat the partners must pay in proportion to this action,
or in this case, to the consumption of the cows.
Now, let x=: what A must pay, and y= what B must pay.
Then, .... a;-|-y=:^ (1)
And, .... x:y—at:hi!
Hence, y=(--Ja; (2)
This value of y put in equation (1), gives
x-{—x=g, or {ai^bt:)x={at)g (3)
Cll
Equation (4) will furnish the following proportion:
[at-\-ht') : ai=g : a
Equation (5), at-{-bt' : ht'r=g \y
Taking up a work on Arithmetic, I found the following rule
for computing results in compound fellowship.
Rule . — Multiply the active agents hy the time each was in
action. Then by proportion.
As the sum of the products
' Is to each 2mrticular product,
So is the whole gain or loss
To each man's share of it.
Now, it is evident that the words of this rule were dictated
by the preceding proportions.
THE END.
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