Answer
$\frac{3y}{4}$, $y\ne-\frac{2}{3}$
Work Step by Step
$\frac{6y+9y^2}{12y+8}$
The denominator must be different from 0.
$12y+8\ne0$
$12y\ne-8$
$y\ne\frac{-8}{12}$
$y\ne-\frac{2}{3}$
$\frac{6y+9y^2}{12y+8}=\frac{3y(2)+3y(3y)}{4(3y)+4(2)}=\frac{3y(2+3y)}{4(2+3y)}=\frac{3y}{4}$
