Charles Davies.

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Font size 108 FIRST LESSONS IN ALGEBRA.

at 4 per cent. The interest which he received for the wholo
amounted to 4640 dollars. Required the two parts.

Ans. 64,000 and 36,000

15. At the close of an election, the successful candidate
of the unsuccessful candidate been also given to him, he
would have received three times as many as his competitor

wanting three thousand flvehundred : how many votes did

each receive ? „ {1st, 6500.

Ans. < '

( 2d, 5000.

16. A gentlemen bought a gold and a silver watch, and
a chain worth \$25. When he put the chain on the gold
watch, it was worth three and a half times more than the
silver watch ; but when he put the chain on the silver watch,
it was worth one half the gold watch and 1 5 dollars over :
what was the value of each watch ?

Gold watch \$80.

Ans.

Silver „ \$30.

17. There is a certain number expressed by two figures,
which figures are called digits. The sum of the digits is
11, and if 13 be added to the first digit the sum will be three
times the second : what is the number 1 Ans. 56.

retire ; there are then left two gentlemen to each lady.
After which, 45 gentlemen depart, when there are left 5
ladies to each gentleman : how many were there of each at
first? A 5 50 gentlemen.

19. A person wishes to dispose of his horse by lottery.
If he sells the tickets at \$2 each, he will lose \$30 on his
horse ; but if he sells them at \$3 each, he will receive \$30
more than his horse cost him. What is the value of the
horse and number of tickets 1 , ^ (Horse . . . \$150.

No. of tickets 60.

• {

EQUATIONS OF THE FIRST DEGREE. 109

20. A person purchases a lot of wheat at \$1 , and a lot of

rye at 75 cents per bushel, the whole costing him \$117,50.

He then sells \ of his wheat and \ of his rye at the same

rate, and realizes \$27,50. How much did he buy of each ?

. C 80bu. of wheat.

Ans. ■£

t 50bu. of rye.

Equations involving three or more unknown quantities.

77. Let us now consider the case of three equations
involving three unknown quantities.
Take the equations

5x—6y+ 4zz= 15,
7a:+4y — 3z=19,
2x-\- y+6*=46.

To eliminate z by means of the first two equations, mul-
tiply the first by 3 and the second by 4 ; then, since the
coefficients of z have contrary signs, add the two results
together. This gives a new equation :

43*— 2y = 121.

Multiplying the second equation by 2, a factor of the co-
efficient of z in the third equation, and adding them together,
we have

16*-j-9y=:84.

The question is then reduced to finding the values of x
and y, which will satisfy these new equations.

Now, if the first be multiplied by 9, the second by 2, and
the results be added together, we find

419#=1257, whence #==3.
10

110 FIRST LESSONS IN ALGEBRA.

We might, by means of the two equations involving a
and y, determine y in the same way we have determined x ;
but the value of y may be determined more simply, by ob-
serving that the last of these two equations becomes, by
substituting for x its value found above,

84—48
48 + 9y=84, whence y — — - — =4.

In the same manner the first of the three proposed equa-
tions becomes, by substituting the values of x and y,

24
15 — 24+4^=15, whence z=z — z=6.

Hence, to solve equations containing three or more un-
known quantities, we have the following

RULE.

I. To eliminate one of the unknown quantities, combine any
one of the equations with each of the others ; there will thus be
obtained a series of new equations containing one less unknown
quantity.

II. Eliminate another unknown quantity by combining one
of these new equations with the others.

III. Continue this series of operations until a single equa-
tion containing but one unknown quantity is obtained, from
which the value of this unknown quantity is easily found.
Then, by going back through the series of equations which have
been obtained, the values of the other unknown quantities may
be successively determined.

Quest. — 77. Give the general rule for solving equations involving
three or more unknown quantities 1 What is the first step 1 What the
second 1 What the third 1

EQUATIONS OF THE FIRST DEGREE. Ill

78. Remark. — It often happens that each of the pro-
posed equations does not contain all the unknown quantities.
In this case, with a little address, the elimination is very
quickly performed.

Take the four equations involving four unknown quantities :

(1.) 2x — 3y+2z=13. (3.) 4y+2z=U.

(2.) 4w— 2a=30. (4.) 5y+3w = 32.

By inspecting these equations, we see that the elimina-
tion of z in the two equations, (1) and (3), will give an
equation involving x and y ; and if we eliminate u in
the equations (2) and (4), we shall obtain a second equation,
involving x and y. These two last unknown quantities
may therefore be easily determined. In the first place, the
elimination of z in (1) and (3) gives

7y— 2x—\ ;

That of u in (2) and (4), gives

20y~{- 6a? = 38,

Multiplying the first of these equations by 3, and adding,

41y=41 ;
Whence y— 1.

Substituting this value in 7y—2x=l, we find

x=3.
Substituting for x its value in equation (2), it becomes

Au— 6 = 30:
Whence w = 9.

And substituting for y its value in equation (3), there
results z=5.

112

FIRST LESSONS IN ALGEBRA.

1. Given <

EXAMPLES.

x+ y+ z=29'
x+ 2y+ 3*= 62

> to find x, y and z.

Ans. x = 8, y = 9, z — \2.

2x+ 4y— 3z=:22~}

2. Given -i 4x— 2y + 5z— 18 *> to find oc, y and z.

I 6x+ 7y— z=z63]

Ans. x = 3, y=7, z=z4.

3. Given 4 — -x+— y+— z~\^ Y to find x, y and *.

Ans. x — 12, y—20, z=30.

4. Divide the number 90 into four sucli parts that the
first increased by 2, the second diminished by 2, the third
multiplied by 2, and the fourth divided by 2, shall be equal
to each other.

This question may be easily solved by introducing a new
unknown quantity.

Let x, y, z, and u, be the required parts, and desig-
nate by m the several equal quantities which arise from
the conditions. We shall then have

x+2=zm, y—2=zm, 2z=:m > ——m*

EQUATIONS OF THE FIRST DEGREE. 113

From which we find

m
x=m — 2, yrrm+2, z—— , u=2m.

Z

m

x-\-y-\-z-\-u=m-\-m-\ — — -\-2m=z4^m.

And since, by the conditions of the question, the first
member is equal to 90, we have

4^ = 90, or %m=z90;
hence m=20.

Having the value of m, we easily find the other values :
viz.

*=18, y=22, *=10, u = 40. >^

5. There are three ingots composed of different metals
mixed together. A pound of the first contains 7 ounces of
silver, 3 ounces of copper, and 6 of pewter. A pound of
the second contains 12 ounces of silver, 3 ounces of cop-
per, and 1 of pewter, A pound of the third contains 4
ounces of silver, 7 ounces of copper, and 5 of pewter. It
is required to find how much it will take of each of the
three ingots to form a fourth, which shall contain in a
pound, 8 ounces of silver, 3f of copper, and A\ of pewter.

Let x, y, and z represent the number of ounces which it

is necessary to take from the three ingots respectively, in

order to form a pound of the required ingot. Since there

are 7 ounces of silver in a pound, or 16 ounces, of the

first ingot, it follows that one ounce of it contains T 7 ^ of an

ounce of silver, and consequently in a number of ounces

7x
denoted by x f there is -— ounces of silver. In the same
16

10*

114 FIRST LESSONS IN ALGEBRA.

manner we would find that -—J- and — , express the num-

lb lb

ber of ounces of silver taken from the second and third, to

form the fourth ; but from the enunciation, one pound of this

fourth ingot contains 8 ounces of silver. We have, then,

for the first equation,

16 ' 16 16

or, making the denominators disappear,

7a:+12y + 4^=128.

As respects the copper, we should find

3x+3y+7z=z60,

and with reference to the pewter

6x+y+5z=68.

As the coefficients of y in these three equations, are
the most simple, it is most convenient to eliminate this un-
known quantity first.

Multiplying the second equation by 4, and subtracting
the first, we have

5^+24^=112.

Multiplying the third equation by 3, and subtracting the
second from the product,

15^+8^=144.

Multiplying this last equation by 3, and subtracting the
preceding one from the product, we obtain

40a?=:320,

whence a? =8.

EQUATIONS OF THE FIRST DEGREE. 115

Substitute this value for x in the equation

15,r+80=144 ;

it becomes 120 + 8^=144,

whence zz=i3.

Lastly, the two values x=8, z — 3, being substituted in
the equation

6a?+y+5.3r=68,

give 48+y+15 = 68,

whence y — 5-

Therefore, in order to form a pound of the fourth ingot,
we must take 8 ounces of the first, 5 ounces of the second,
and 3 of the third.

Verification.

If there be 7 ounces of silver in 16 ounces of the first
ingot, in 8 ounces of it, there should be a number of ounces
of silver expressed by

7x8
. 16 '
In like manner,

12x5 4x3

and

16 16

will express the quantity of silver contained in 5 ounces of
the second ingot, and 3 ounces of the third.
Now, we have

7X8 12x5 , 4X3 128 -

16 16 ' 16 16 T '

therefore, a pound of the fourth ingot contains 8 ounces of
silver, as required by the enunciation. The same condi-
tions may be verified relative to the copper and pewter.

116 FIRST LESSONS IN ALGEBRA.

6. A's age is double B's, and B's is triple 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, and C's =14.

7. A person bought a chaise, horse, and harness, for
£60 ; the horse came to twice the price of the harness,
and the chaise to twice the price of the horse and harness.
What did he give for each ?

, =£13 6s. Sd. for the horse.
Ans. ? £ 6 13s. 4d. for the harness.
v j£40 for the chaise.

8. To divide the number 36 into three such parts that
•| of the first, ^ of the second, and -J of the third, may be
all equal to each other. Ans. 8, 12, and 16.

9. If A and B together can do a piece of work in 8 days,
A and C together in 9 days, and B and C in ten days ; how
many days would it take each to perform the same work
alone ? " Ans, A 14f|, B 17|f, C 23 T 7 T .

10. Three persons, A, B, and C, begin to play together,
having among them all \$600. At the end of the first game
A has won one-half of B's money, which, added to his own,
makes double the amount B had at first. In the second
game, A loses and B wins just as much as C had at the
beginning, when A leaves off with exactly what he had at
first. How much had each at the beginning ?

Ans. A \$300, B \$200, C \$100.

11. Three persons, A, B, and C, together possess \$3640.
If B gives A \$400 of his money, then A will have \$320
more than B ; but if B takes \$140 of C's money, then B
and C will have equal sums. How much has each ?

Ans. A \$800, B \$1280, C \$1560.

12. Three persons have a bill to pay, which neither
alone is able to discharge. A says to B, " Give me the
4th of your money, and then I can pay the bill." B says
to C, " Give me the 8th of yours, and I can pay it. But

EQUATIONS OF THE FIRST DEGREE. 117

C says to A, " You must give me the half of yours before
I can pay it, as I have but \$8." What was the amount of
their bill, and how much money had A and B ?

. j Amount of the bill, \$13.
( A had \$10, andB \$12.

13. A person possessed a certain capital, which he placed
out at a certain interest. Another person, who possessed
10000 dollars more than the first, and who put out his capi-
by 800 dollars. A third person, who possessed 15000 dol-
lars more than the first, putting out his capital 2 per cent,
Required the capitals of the three persons, and the rates of
interest.

Ans ( Sums at interest, \$30000, 40000, 45000.

* Rates of interest, 4 5 6 pr. ct.

14. A widow receives an estate of \$15000 from her de-
ceased husband, with directions to divide it among two sons
and three daughters, so that each son may receive twice as
much as each daughter, and she herself to receive \$1000
more than all the children together. What was her share,
and what the share of each child ?

/■ The widow's share, \$8000.

Ans. 1 Each son's, 2000.

v Each daughter's, 1000.

15. A certain sum of money is to be divided between
three persons, A, B, and C. A is to receive \$3000 less
than half of it, B \$1000 less than one third part, and C to
receive \$800 more than the fourth part of the whole. What
is the sum to be divided, and what does each receive ?

f Sum, \$38400.

B „ 11800.

LC „ 10400

Ans. <

118 FIRST LESSONS IN ALGEBRA.

CHAPTER IV.

Of Powers.

79. If a quantity be multiplied several times by itself,
the product is called the power of the quantity. Thus,

a — a is the root, or first power of a.

axa = a 2 is the square, or second power of a.

aXaXa — a 3 is the cube, or third power of a.

aXflXflXfl=fl 4 is the fourth power of a.

axaxaxax a=za 5 is the fifth power of a.

In every power there are three things to be considered •

1st. The quantity which is multiplied by itself, and which
is called the root or the first power.

2nd. The small figure which is placed at the right, and
a little above the letter. This figure is called the exponent
of the power, and shows how many times the letter enters
as a factor.

3rd. The power itself, which is the final product, or
result of the multiplications.

Quest. — 79. If a quantity be continually multiplied by itself, what is
the product called 1 How many things are to be considered in every
power 1 What are they 1

OP POWERS. 110

For example, if we suppose #=3, we have

a = 3 the root, or 1st power of 3.

a 2__32__3 X 3_ 9 the sec ond power of 3.

a 3 — 3 3 — 3 X3x3= 27 the third power of 3.

«*== 3 4 =3 X3 X3x3= 81 the fourth power of 3.

c 5 = 3 5 = 3x 3x3x3x3=243 the fifth power of 3.

In these expressions, 3 is the root, 1, 2, 3, 4 and 5 are
the exponents, and 3, 9, 27, 81 and 243 are the powers.

To raise monomials to any poiver.

80. Let it be required to raise the monomial 2a 3 b 2 to
the fourth power. We have

{2a 3 b 2 y=z2a 3 b 2 X 2a 3 b 2 X 2a 3 b 2 X 2a 3 b 2 ,

which merely expresses that the fourth power is equal to
the product which arises from writing the quantity four
times as a factor. By the rules for multiplication, this pro-
duct becomes

(2a 3 b 2 Y=2*a 3 + 3 + 3 + 3 b 2 + 2 + 2 + 2 = 2*a 12 b B ',

from which we see,

1st. That the coefficient 2 must be raised to the 4th
power ; and,

v 2nd. That the exponent of each letter must be multiplied
by 4, the exponent of the power.

As the same reasoning would apply to every example,
we have, for the raising of monomials to any power, the
following

120 • FIRST LESSONS IN ALGEBRA.

RULE

I. Raise the coefficient to the required power.

II. Multiply the exponent of each letter by the exponent of
the power.

EXAMPLES.

1 . What is the square of 3a 2 y 3 1 Ans. 9<2 4 y 6 .

2. What is the cube of 6a 5 y 2 x ? Ans. 2l6a l5 y G x 3 .

3. W T hat is the fourth power of 2a 3 y 3 b 5 1

Ans. I6a 12 y l2 b 20 .

4. What is the square of a 2 b h y 3 ? Ans. a*b 10 y G .

5. What is the seventh power of a 2 bcd 3 ?

Ans. a li b 1 c 1 d 2 ^.

6. W r hat is the sixth power of a 2 b 3 c 2 d 1 Ans. a l2 b 18 c 12 d 6 .

7. What is the square and cube of —2a 2 b 2 ?

Square. Cube.

—2a 2 b 2 —2a 2 b 2

- 2a 2 b 2 -2a 2 b 2

+ 4a 4 6 4 +4a 4 & 4

-2a 2 b 2

By observing the way in which the powers are formed,
we may conclude,

1st. When the root is positive, all the powers will be positive.
2nd. When the root is negative, all the even powers will be
positive and all the odd powers negative.

Quest. — 80. What is a monomial 1 Give the rule for raising a
monomial to any power. When the root is positive, how will the powers
be ! When the root is negative, how will the powers be 1

OF POWERS. 121

8. What is the square of —2a±b 5 ? Ans. 4a 8 b 10 .

9. What is the cube of —5a 5 y 2 c? A?is. — I25a 15 y e c 3 .

10. What is the eighth power of — a 3 xy 2 1

Ans. -\-a 24 x 8 y 16 .

11. What is the seventh power of — a 2 yx 2 ?

Ans. — a l4: y' r x u >

12. What is the sixth power of 2ab 6 y 5 ?

Ans. 64a 6 b 36 y 30 .

13. What is the ninth power of — cdx 2 y 3 1

Ans. — c 9 d Q x 18 y 2 \

14. What is the sixth power of —3ab 2 d ?

Ans. 729 a 6 b 12 d 6 .

15. What is the square of — 10 a 2 b 2 c 3 1 Ans. lOOaWc 6 .

16. What is the cube of — 9a 6 b 5 d 3 f 2 ?

Ans. — 729a 18 b 15 d 9 / 6 .

17. What is the fourth power of — 4a 5 b 3 c i d 5 1

Ans. 256a 20 b l2 c 16 d 20 .

18. What is the cube of —4a 2 b 2 c 3 d 1

Ans. — 64a*b 6 c 9 d 3 .

19. What is the fifth power of 2a 3 b 2 xy ?

Ans. 32a l5 b 10 x 5 y 5 .

20. What is the square of 20xHjW ? Ans. 400x 8 y 8 c 10 .

21. What is the fourth power of 3a 2 b 2 c 3 ?

Ans. 8la 8 b 8 c™.

22. What is the fifth power of —c 2 d 3 x 2 y 2 ?

Ans. -c 10 rf 15 * 10 y 10 .

23. What is the sixth power of —ac 2 df\

Ans. a 6 c l2 d 6 f 6 -

24. What is the fourth power of — 2a 2 c 2 d 3 1

Ans. I6a 8 c 8 d u .
11

122 FIRST LESSONS IN ALGEBRA.

To raise Polynomials to any power.

8 1 . The power of a polynomial, like that of a mono-
mial, is obtained by multiplying the quantity continually by
itself. Thus, to find the fifth power of the binomial a~\-b t
we have

a -\- b 1st power.

a + b

a 2 -\~ ab

+ ab+b 2

a 2 -\-2ab -\-b 2 2nd power.

a + b
a 3 +2a 2 b + ab 2

+ a 2 b-\- 2ab 2 + b 3
a 3 +3a 2 b+ 3ab 2 + b 3 . . . 3rd power.

a + b

a±+3a 3 b + 3a 2 b 2 + ab 3

+ a 3 b+ 3a 2 b 2 -\-3ab 3 + jjg
« 4 +4a 3 5+ 6a 2 b 2 + 4ab 3 + ¥ 4th power.
« + b
a 5 +4a*b+ 6a 3 b 2 + 4a 2 W-\- ab*

+ a*b-\- 4a 3 b 2 + 6a 2 b 3 + 4ab±+b 5
a 5 -\-5^b-{-l0a 3 b 2 -\-l0a 2 b 3 -^5a¥-\-b 5 Ans.

Remark. — 82. It will be observed that the number of
multiplications is always 1 less than the units in the expo-

Quest. — 81. How is the power of a polynomial obtained.

OF POWERS

123

nent of the power. Thus, if the exponent is 1, no multipli-
cation is necessary. If it is 2, we multiply once ; if it is
3, twice ; if 4, three times, &c. The powers of polyno-
mials may be expressed by means of the exponent. Thus,
to express that a-^b is to be raised to the 5th power, we write

(a+bY,

which expresses the fifth power of a-\-b.

2. Find the 5th power of the binomial a— b.

a — b

a 2 — ah

— ab +b 2

a 2 —2ab -\-b 2 2nd power.

a — b

a?—2a 2 b +

ab 2

— a 2 b+

2ab 2 — b 3

a 3 —3a 2 b+

3ab 2 —b 3 ... 3rd power.

a — b

a±-3a 3 b-i-

3a 2 b 2 — ab 3

— a 3 b-\-

3a 2 b 2 - 3ab 3 + &

a±—4a 3 b+

6a 2 b 2 — 4ab 3 + b± 4th power.

a — b

a s—±a±b +

6a 3 b 2 — 4a 2 b 3 + ab*

— a*b+

4a 3 b 2 — 6a 2 b 3 +4ab*—b 5

a 5 —5a*b-\-l0a 3 b 2 —10a 2 b 3 +5a¥—b 5 Ans.

Quest. — 82. How does the number of multiplications compare with
the exponent of the power 1 If the exponent is 4, how many multipli-
cations 1

124 FIRST LESSONS IN ALGEBRA.

3. What is the square of 5a — 2c -\-d.

5a — 2c + d

5a — 2c + d

— Wac+ 4c 2 — 2cd

25a 2 - 20ac+l0ad-\-4c 2 —4cd+d 2 Ans.

4. Find the 4th power of the binomial 3a— 2b.

3a — 2b 1st power.

Sa — 2b

9a 2 — 6ab

— 6ab + 4b 2

9a 2 — I2ab+4b 2 2nd power.

3a — 2b

27 a 3 — 36a 2 b+12ab 2

— I8a 2 b+24ab 2 — 8b 3
27a 3 — 54a 2 b+36ab 2 — 8b 3 . . 3rd power.

3a — 2b

8la*-l62a 3 b+l08a 2 b 2 -24ab 3

- 54a 3 b-t-l08a 2 b 2 —72ab 3 +l6b*
8la*—2l6a 3 b+216a 2 b 2 — 96ab 3 +l6b± Ans,

5. What is the square of the binomial a -{-11

Ans. a 2 +2a+l.

6. What is the square of the binomial a — 11

Ans. a 2 —2a-\-l.

7. What is the cube of 9a — 3b ?

Ans. 729a 3 — 729a 2 b-\-243ab 2 — 27b 3 .

8. What is the third power of a— 1 1

Ans f a 3 —3a 2 +3a — \.

OF POWERS. 125

9. What is the 4th power of x— y 1

Arts, # 4 — 4x 3 y-}-6x 2 y 2 — 4xy 3 -{-y*.

10. What is the cube of the trinomial (x+y+z!

Ans. x 3 + 3x 2 y + 3x 2 z+ 3xy 2 + 3xz 2 + 3y 2 z+ 3yz 2 + 6xyz
-\-y 3 -\-z 3 .

11. What is the cube of the trinomial 2a 2 — 4ab-\-3b 2 1
Ans. 8a 6 —48a 5 b + 132a*b 2 —208a 3 b 3 +l98a 2 b±—l08ab 5

+ 27b 6 .

To raise a Fraction to any Power.

83. The power of a fraction is obtained by multiplying
the fraction by itself ; that is, by multiplying the numerator
by the numerator, and the denominator by the denominator.

Thus, the cube of — , which is written
b

/ a \ 3 a a a a 3

\TJ = T x T x T = T 3 '

is found by cubing the numerator and denominator sepa-
rately.

d Q

2. What is the square of the fraction r-r — ?

b-\-c

We have

/a — c \ 2 _ (a— c) 2 _ a 2 — 2ac-\-c 2
\b+e) ~ (b+c) 2 ~~¥+2b^+^ US '

3. What is the cube of ~f^- 1 Ans. -^~~.

3bc 27o 3 c s

Quest.— 83. How do you find the power of a fraction 1
11*

126 FIRST LESSONS IN ALGEBRA.

db 2 C

4. What is the fourth power of — — ; ?
1 2# 2 y 2

Ans.

I6x 8 y 8 '

5. What is the cube of

a? 3 — 3a? 2 y+3j?3/ 2 — y 3

# 3 +3£ 2 y-f-3#3/ 2 -f-?/ 3

2## ^

6. What is the fourth power of ? .Arcs.

Aay 1 6y 4

7. What is the fifth power of ? Arcs.

8. What is the square of

IQyz 32y 5 z 5

ax—y ?

by — x

a 2 x 2 — 2axy-\-y 2

b 2 y 2 —2bxy-\-x 2

9. What is the cube of — -— - — ?

x-\-2y

8a 3 — 36a 2 b-\-54ab 2 —27b 3
x 3 + 6x 2 y + 1 2xy 2 + 8y 3

Binomial Theorem.

8 4. The method which has been explained of raising a
polynomial to any power, is somewhat tedious, and hence
other methods, less difficult, have been anxiously sought
for. The most simple which has yet been discovered, is
the one invented by Sir Isaac Newton, called the Binomial
Theorem.

Quest. — 84. What is the object of the Binomial Theorem 1 Who
discovered this theorem 1

BINOMIAL THEOREM. 127

85. In raising a quantity to any power, it is plain that
there are four things to be considered : —

1st. The number of terms of the power.
2nd. The signs of the terms.
3rd. The exponents of the letters.
4th. The coefficients of the terms.

Of the Terms.

86. If we take the two examples of Article 81, which
we there wrought out in full ; we have

(a+b) 5 = a 5 +5a*b-\-l0a 3 b 2 +10a 2 b 3 +5a¥+b 5 ;

(a — b) 5 = a 5 — 5a*b+10aW — 10a 2 b 3 + 5ab* - V i .

By examining the several multiplications, in Art. 8 1 , we
shall observe that the second power of a binomial contains
three terms, the third power four, the fourth power five, the
fifth power six, &c ; and hence we may conclude — That
the number of terms in any power of a binomial, is one greater
than the exponent of the power.

Of the Signs of the Terms.

87. It is evident that when both terms of the given bi-
nomial are plus, all the terms of the power will be plus.

2nd. If the second term of the binomial is negative, then
all the odd terms, counted from the left, will be positive, and
all the even terms negative.

Quest. — 85. In raising a quantity to any power, how many things
are to be considered'? What are they] — 86. How many terms are
there in any power of a binomial 1 If the exponent is 3, how many
terms'? If it is 4, how many terms'? If 51 &c. — 87. If both terms
of the binomial are positive, how are the terms of the power 1 If the
second term is negative, how are the signs of the terms 1

128 FIRST LESSONS IN ALGEBRA.

Of the Exponents.

88. The letter which occupies the first place in a bino-
letter in the binomials a-\-b, a—b.

1st. It is evident that the exponent of the leading letter
in the first term will be the same as the exponent of the
power ; and that this exponent will diminish by unity in
each term to the right, until we reach the last term, which
does not contain the leading letter.

2nd. The exponent of the second letter is 1 in the second
term, and increases by unity in each term to the right,
until we reach the last term, in which the exponent is the
same as that of the given power.

3rd. The sum of the exponents of the two letters, in any
term, is equal to the exponent of the given power. This
last remark will enable us to verify any result obtained by

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