Question

((y-3)/(7y-4))-((y-3)/(y-4))

Answer 1

=(y^2-3y-4y+12-7y^2+21y+4y-12)/(7y^2-4y-28y+16=(-6y^2+18y)/(7y^2-32y+16)

Answer 2

$frac{y-3}{7y-4}-frac{y-3}{y-4}=frac{(y-3)(y-4)-(y-3)(7y-4)}{(7y-4)(y-4)}=$

$frac{(y-3)(y-4)-(y-3)(7y-4)}{(7y-4)(y-4)}=frac{(y-3)(y-4-7y+4)}{(7y-4)(y-4)}=$

$frac{(y-3)(-6y)}{(7y^2-32y+16)}=frac{-6y^2+18y}{7y^2-32y+16}$