Answer
$\frac{2y}{3y+7}$.
Work Step by Step
The given expression is
$=\frac{\frac{2y}{y^2+4y+3}}{\frac{1}{y+3}+\frac{2}{y+1}}$
Factor $y^2+4y+3$.
Rewrite the middle term $4y$ as $3y+1y$
$=y^2+3y+1y+3$
$=(y^2+3y)+(1y+3)$
$=y(y+3)+1(y+3)$
$=(y+3)(y+1)$ plug into the given expression.
$=\frac{\frac{2y}{(y+3)(y+1)}}{\frac{1}{y+3}+\frac{2}{y+1}}$
Multiply the numerator and the denominator by $(y+3)(y+1)$.
$=\frac{(y+3)(y+1)}{(y+3)(y+1)}\cdot \frac{\frac{2y}{(y+3)(y+1)}}{\frac{1}{y+3}+\frac{2}{y+1}}$
Use the distributive property.
$=\frac{(y+3)(y+1)\cdot\frac{2y}{(y+3)(y+1)}}{(y+3)(y+1)\cdot\frac{1}{y+3}+(y+3)(y+1)\cdot\frac{2}{y+1}}$
Simplify.
$=\frac{2y}{(y+1)+2(y+3)}$
$=\frac{2y}{y+1+2y+6}$
$=\frac{2y}{3y+7}$.
