gle dimension only, I divide the same into the parts +2ax, and -bx-2ab, which, by inspection, appear to be equal to (x+2a)פ, and (x+2a)X-b. Therefore 2+2a is a divisor to both the parts, and likewise to the whole, expressed by (x+2a)x(x—b); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator; but the former will be found to succeed, the quotient coming out exactly x2-ax+bx-ab; whence the fraction itself is reduced to x2-ax-bx-ab x-b ; which is not reducible further, by x-b, since the division does not terminate without a remainder, as upon trial will be found. a3_b3) a1-x1(a (See p. 36, ex. 2, last expression.) a1-ab3 Divide this —ab3—6a by 63, we have a—b)a2—b3 (a2+ab+b2 a3-a2b Dividing this remainder, 2ar-2x by 2x2, or leaving out 22, which is found in each term of the remainder, the next divisor is a2—x2) a3———a2x-ax2+x3 (a—x a3-ax2 —a2x+x3 Hence a is therefore -ax+x, the greatest common measure of the two quantities, and if they be respectively divided by it, the fraction is reduced to a2+x2 a-x its lowest terms. This quantity, 2x2, found in every term of one of the divisors, 2a2x2-2x, but not in every term of the dividend, a-a2x—ax2+x3, must be left out; otherwise the quotient will be fractional, which is contrary to the supposition made in the rule and by omitting this part, 2x2, no common measure of the divisor and dividend is left out; because, by the supposition, no part of 2x2 is found in all the terms of the dividend. these fractions to its lowest terms. The greatest common measure is 12x2-5 and 5x3-1, and the reduced fractions are 7. Reduce 4x2+3 and 4x2+1 2x2-1 5 +1 6a5+15a1b-4a3c2-10a2bc2 a2b2—c3d2—a2c2+c 9ab-27a bc-6abc+18bc4a'd-4acd-2ac2c each to its lowest terms. 8. Reduce 9x+2x3+4x2-x+1 and 15x-2x+10x-x+2' c1—b3c2' a1 to its lowest terms. (For ans. to 7 and 8, see p. 59.) each ADDITION OF FRACTIONS IN ALGEBRA. RULE. Reduce the fractions to a common denominator, add the numerators together, and under this sum place the common denominator; the result will be the sum of the fractions required. 4x x-2 5 3X4Xx 2XxX4 xX2X3 + 2X3X4 2.3.4 + 2.3.4 20x+7x-14 27x-14 = ; 12 5X7 2x 2. Add 2a, 3a+ and a— together. 6+5) 15-15x 5x+5x 15+15x 3×(x+1)×(1+x)=3(x2+2x+1)=3x2+6x+3 numerators. 3X5X(1-x)= 15(1-x) 3x5x(1+x)=15+15x, the common denominator. Hence the new fractions are 3x2+6x+3, and 15+15x 16x-15x , as required. 51. RULE. Reduce the fractions to a common denominator, if required; then place the difference of their numerators over the common denominator, and it will be the difference required. 4x 4x 3x+1 4x+2 take 1. From take - 5 x+1 3x+1 .5 x+1 4x2-11x-5 = or 3 4x(x+1)-5(3x+1) (4x+2)3x-(2x-3)3 4x2+3 1 or " a+x a2-2ax+x2 a2+2ax+x2. a(a+x) a(a-x) a(a2x2) and a(a2--x2) |