3. Add 3a+26-5, a+b-c, and 6a-2c+3, together. 4. Add 2a+3b4c-9, and 5a-3b+2c-10, together. 5. Add x2+ax2+bx+2, and x2+cx2+dx-1, together.

SUBTRACTION.

39. SUBTRACTION in Algebra is the method of finding the dif ference between two algebraical quantities, and connecting those quantities together with their proper signs.

RULE 1. Set those quantities from which the subtraction is to be made in one line, and the quantities to be subtracted in a line below them, observing to place like quantities under each other when they occur.

RULE 2. Subtraction in Algebra is performed by simply chang ing the signs (+ into and into) of the lower line of quantities to be subtracted, and then adding or connecting them as in addition, and the result will be the difference required.

The answer or remainder, +5a

Here the quantity to be subtracted is—3a+2y.

+9a + 3y 9a3y4 3; we therefore change its signs according to the 2d rule, and it becomes +3a2y+3, and this added to the other quantity gives the remainder +9a + 3y — 4.

Ex. 15. -6a+13x-4b-12c 9a+ 4x+4b- 5c

6xy-3(xy)-4xy+12a (ax+6) 3x2y+3√(xy) 7xy 4a (ax+6). 3a+ 9x-8b- 7c 3x2y—6√(xy) —11xy3+ 8a√(ax+6) Ex. 18. From the sum of 4ax—150+4x+, 5x2+3ax+10x2, and 90-2ax-12/(x), take the sum of 2ax-80+7x2, 7x2-8ax -70, and 30—4/(x)—2x2+4a2x2.

4ax-150+ 4x2
3ax+5x2+10x

-2ax 90-12x

and 2ax-80+7x2
-Sax-70-7x

4a2x2+30-4x-2x2

Then,

5ax— 60+5x2+2x

-6ax—120+3x+5x2+4a2x2

-6ax-120+5x2+3x2+4a2x2

Ans. 11ax+60-√x—4a2x2.

EXAMPLES WITH LITERAL COEFFICIENTS.

NOTE. When the quantities that are to be subtracted have lit

eral coefficients, the opera

tion may be performed by

ax-b

CX

ax2+bx cx2-cx

placing the coefficients, with (a–c)x+d—b (a—c) x2+(b+c)x their proper signs, between brackets, as in addition, and then subjoining the common quantity, the same as in the margin. Ex. 19. From pxy + qxz-rz2+s

Take mxy-pqxz—nz2+a

Remainder, (pm)xy+(1+p) qxz+(n-r) 22+s-a.

Ex. 20. From a(x—y)}+bxy+c(a+x)2

Take (xy)-bxy+ (a+c) (a+x)2.

Remainder, (a-1) (x—y)1+2bxy—a(a+x)2.

EXAMPLES FOR PRACTICE.

1. Required the difference of a+b and

2

-b

Answer, b.

" 2 2. From 3x-2a-b+7, take 8-3b+a+4x. 3. From 3a+b+c-2d, take b-8c+2d-8. 4. From 5ab+262-c+bc-b, take b2-2ab+bc. 5. From ax-ba+cx-5d, take bx2+ex-12d.

MULTIPLICATION.

40. Multiplication in Algebra is the method of finding the product of two or more indeterminate algebraic quantities, and is generally divided into three cases.

CASE I. When both factors are simple quantities.

RULE. Multiply the coefficients of the two quantities together, and annex to the result (or product) all the letters in both factors, which will give the whole product required. If the signs of the factors be like, that is, both or both the sign of the product is +; but if they are unlike, or one of them and the other+, the sign of the product is: and this is commonly expressed by saying, like signs give plus, and unlike signs minus.

CASE II. When one factor is a compound and the other a simple. RULE. Multiply each or every term of the compound factor separately by the simple factor, and to each product prefix its proper sign, and the result will be the whole product.

-70x+14ax 36xy-3axy+18xy-62xy/b8/bx-2ab

CASE III. When both factors are compound quantities. RULE. Multiply every part of the multiplicand by each part of the multiplier, placing the products one after the other, with their proper signs; then add the several products together, as in common multiplication.

Ex. 1.

Multiply a+ Ex. 2. a+b
by a+b

a-b

a2+ab

-ab-b2

a2

Ex. 3. a2+ab+b2

a-b

When we have two or more quantities to multiply together, it is indifferent which two we begin with; for the products will always be the same, as will appear from the following example.

Let it be proposed to find the value or product of the four following factors, viz:

(I.)

(II.) ·

(a+b) (a2+ab+b2)

1st. Multiply the factors I. and II.

a2+ab+b2

a+b

a3+a2b+ab2
+a2b+ab2+b3

+2ab+2ab+b3

(IV.)

and (a2 ab+b2)

(III.)

(a-b)

a2-ab+b2

Next the factors III. and IV.

a-b

a3_abab

-a2bab2-b3

a3 2a3b+2ab2 b3

It remains now to multiply the first product I. II. by the second product III. IV.

2d. Change the order of the question; that is, multiply the factors I. and III., then II, and IV. together.

It will be proper to illustrate this example by a numerical application. Suppose a 3, and b-2, we shall have a+b=5, and a-b-1; further, a2 9, ab-6, and 6-4, therefore a2+ab+b2= 19, and a2+ab+b7: so that the product required is that of 5×19X1X7=665. Now a 729, and 6o-64; consequently the product is ao—bo—665, as we have already seen.