Question

If a:b=c:d, prove that: (i) xa+yb:xc+yd=b:d (ii) (6a+7b)(3c-4d)=(6c+7d)(3a-4b)

Answer 1

We have ,
a : b = c : d
Let a/b = c/d = k ( a constant )
Now ,
 LHS = xa + yb ÷ xc + yd
         = b( xa/b + y ) ÷ d( xc/d + y ) 
         = b(xk + y) ÷ d(xk + y) 
         = b/d = RHS

Given , 
a:b = c:d
a/b = c/d
ad = bc ------(1)
( 6a + 7b ) ( 3c - 4d )=( 6c + 7d ) ( 3a - 4b )
18ac - 24 ad + 21bc - 28bd =  18ac - 24bc + 21ad - 28bd
(- 24ad) + 21bc = (- 24bc) + 21ad
(- 24bc) + 21bc = (- 24bc) + 21bc    { from (1) }
(- 3bc) = (- 3bc)

Hence proved