# Prerequisites - P.5 - Rational Expressions - Exercises - Page 48: 23

$\frac{3y}{4}$, $y\ne-\frac{2}{3}$

#### Work Step by Step

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

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