#### Answer

$\displaystyle \frac{1}{4}$

#### Work Step by Step

Step by step multiplication of rational expressions:
1. Factor completely what you can
2. Reduce (divide) numerators and denominators by common factors.
3. Multiply the remaining factors in the numerators and
multiply the remaining factors in the denominators. $(\displaystyle \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS})$
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Factor what we can:
... factor $ax^{2}+bx+c$ by searching for two factors of $ac$ whose sum is $b,$
... and, if they exist, rewrite $bx$ and factor in pairs.
$2y^{2}-9y+9= \qquad $...$(ac=18, b=-9)$, factors: $-6$ and $-3$
$=2y^{2}-6y-3y+9$
$= 2y(y-3)-3(y-3)=(y-3)(2y-3)$
$3y-y^{2}=-(y^{2}-3y)=-y(y-3)$
$8y-12=4(2y-3)$
The problem becomes
$...=\displaystyle \frac{2y\cdot[-(y-3)(2y-3)]}{-y(y-3)\cdot 4(2y-3)}\qquad$ ... divide out the common factors
$=\displaystyle \frac{\fbox{$-y$}\cdot\fbox{$(y-3)$}\fbox{$(2y-3)$}}{\fbox{$-y$}\fbox{$(y-3)$}\cdot 4\fbox{$(2y-3)$}}\qquad$
= $\displaystyle \frac{1}{4}$