FRACTIONS.] Ex. 3. px+q B A L G E c BRA 533 (x a) (x b) (x c) x a x b x c gives px + q = A.(x b)(x c) + B(# a) (x c) and C(x-a)(x-b), = A(a -b)(a- c), &c., px + q pa+q 1 (x a) (x b) (x c) (a b)(a c) x a pb + q 1 pc + q 1 (6 a) (6 c) a; - 6 (c a) (c - 6) x c a+b b + c Ex. 4. Find the sum of c + a (6 c) (c a) (c -a) (a b) Let gives = s; and write in alphabetical order; it a s b- s _ c - s (a-b)(a-c) + (b-a)(b^c) + (c-a)(a-b) i.e. (Ex. 3), the A, B, C of the resolved fraction, x-a _ A B C (x a)(x &) (x c) x-a x-6 x-c and since x - s = A(x - b) (x - c) + ~B(x - a) (x - c) + C(x - a) (z - I), the sum required, being the coefficient of y?, is equal to 0. The reader will easily extend this process to other cases, as, for instance, to prove _ bed _ cda _ (a-b)(a-c)(a-d) + (b-c)(b-d)(b-a) + _ dab _ abc _ 1 (c-d)(c-a)(c-b) + (d-a)(d-b)(d-c) = 34. PJROB. VII. To Multiply Fractions. Rule. Multiply the numerators of the fractions for the numerator of the product, and the denominators for the denominator of the product. The demonstration follows at once from the definition of a fraction given in Art. 26 ; thus since T x b = a f ^xd = c ) we have rX&x-iXcZ = ac, i.e., by the commutative law u r x -j x bd = ac . b a But ac , j bd xbd=at a c ac 35. PROB. VIII. To Divide Fractions. Rule. Multiply the dividend by the reciprocal of the divisor, the product will be the quotient required. This rule requires no demonstration. Examples in Multiplication and Division of Fractions. Ex. 1. Multiply ?-- by -^ r , . Since ?-- = ^-% o a J a?-l z o a ao .*. the product is a* a* a ab a 3 -& 2 ab~b Ex. 2. Multiply /- 2 3 * +2 by *+**+!. 3 z 3 +2x 2 +2x + l * x 2 -5x + 4 Because the numerator of the first fraction, and the denominator of the second both become 0, when 1 is written for x, each is divisible by x - 1 (Art. 20). In the same way the denominator of the first fraction, and the numerator of the second are both divisible by x + l. Hence, x 3 -3x+2 x 2 + 2x+l x 3 -3x+2 x 2 + x-2 x+l x 3 + 2x 2 -x-2 x-4 "x 2 + z + l x 3 -3x 2 -3x-4 Ex. 3. Divide - - by vr - -= o a J o 2 a 2 The quotient is ab Ex. 4. Reduce 1
2bc )
" to factorials. -a 2 a 2 -(5-c) 2 c-g) c. 5. Eeduce 1 - ( ^-7-7 i - ( a8 t, 5 " t ~. c2 .,r- ) = o + . to factorials. -b} (c + d-a Miscellaneous Examples in Fractions. 11 x Ex. 1. Find the value of 2abc when x = , ao + ac + oc first, there results ab-ax bx ab (x-a) (6c-cx) Writing down every term with a; x x x -+T+- abc = 0. a(x-b) b(x-a) c(x-a)(x-b) (x-a)(x-V) Ex. 2. Find the value of - H -; + , when x-3a x-36 x + 2c - + - = - and x = 2(a + b - c). abc Restore symmetry by writing - c for c; the numerator of the sum is (a; - 3b) (x - 3c) + (x- 3a) (x - 3c) + (x- 3a) (x - 36) = 3 {a; 2 - 2(a + b + c)x + 3(ab + ac + be)}. But x = 2(a + b + c), whence the first and second terms make up : and - + - + - = 0, is the third term divided by abc, abc J . . the sum required is 0. Ex. 3. Given that (a 2 + be) (b 2 + ac) (c 2 + ab) + (a 2 - be) (b 2 - ac) (c 2 - ab) = 0, when multiplied out and reduced, may be written a 3 + b 3 + c s + abc = 0, prove that (a 2 + be) (b 2 + ac) (c 2 + ab) - (a 2 - be) (b 2 - ac} (c 2 - ab) = 0, may be reduced to ~7 + TT + ~3+"T" = 0. The latter given equality. sj3 /jo / ft hf* by dividing it by a 2 bc x b 2 ac x c-ab, becomes ab ab bc a 2 / ao . bc a? be which is identical with the first given equality, but with -, r, -, written in place of a, 5, c. The result therefore of
d G
