## Algebra and Trigonometry 10th Edition

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