Answer
$\frac{(y-3)}{(y+3)(y+1)}$.
Work Step by Step
The given expression is
$=\frac{y}{y^2+5y+6}-\frac{2}{y^2+3y+2}$
Factor $y^2+5y+6$
Rewrite the middle term $5y$ as $3y+2y$.
$=y^2+3y+2y+6$
Group terms.
$=(y^2+3y)+(2y+6)$
Factor each term.
$=y(y+3)+2(y+3)$
Factor out $(y+3)$.
$=(y+3)(y+2)$
Factor $y^2+3y+2$
Rewrite the middle term $3y$ as $2y+y$.
$=y^2+2y+y+2$
Group terms.
$=(y^2+2y)+(y+2)$
Factor each term.
$=y(y+2)+1(y+2)$
Factor out $(y+2)$.
$=(y+2)(y+1)$
Substitute the factor into the given expression.
$=\frac{y}{(y+3)(y+2)}-\frac{2}{(y+2)(y+1)}$
LCM $=$ greatest power of all the prime factors.
LCM $=(y+3)(y+2)(y+1)$
$=\frac{(y+1)}{(y+1)}\cdot \frac{y}{(y+3)(y+2)}-\frac{(y+3)}{(y+3)}\cdot \frac{2}{(y+2)(y+1)}$
Simplify.
$= \frac{y(y+1)}{(y+3)(y+2)(y+1)}-\frac{2(y+3)}{(y+3)(y+2)(y+1)}$
Add both numerators because both denominators are equal.
$= \frac{y(y+1)-2(y+3)}{(y+3)(y+2)(y+1)}$
Use the distributive property in the numerator.
$= \frac{y^2+y-2y-6}{(y+3)(y+2)(y+1)}$
$= \frac{y^2-y-6}{(y+3)(y+2)(y+1)}$.
Factor $y^2-y-6$.
Rewrite the middle term $-y$ as $-3y+2y$.
$=y^2-3y+2y-6$
Group terms.
$=(y^2-3y)+(2y-6)$
Factor each term.
$=y(y-3)+2(y-3)$
Factor out $(y-3)$.
$=(y-3)(y+2)$
Substitute the factor into the fraction.
$= \frac{(y-3)(y+2)}{(y+3)(y+2)(y+1)}$.
Cancel common terms.
$= \frac{(y-3)}{(y+3)(y+1)}$.
