Answer
$-\frac{4y}{5}$, $y\ne\frac{1}{2}$
Work Step by Step
$\frac{4y-8y^2}{10y-5}$
The denominator must be different from 0.
$10y-5\ne0$
$10y\ne5$
$y\ne\frac{5}{10}$
$y\ne\frac{1}{2}$
$\frac{4y-8y^2}{10y-5}=\frac{-4y(-1)-4y(2y)}{5(2y)-5(1)}=\frac{-4y(2y-1)}{5(2y-1)}=-\frac{4y}{5}$
