Answer
$ \displaystyle \frac{y(y+3)}{(y-2)(y-1)}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ = multiplying with $\displaystyle \frac{Q}{P}$
Rewrite the problem:
$ \displaystyle \frac{y^{2}+y}{y^{2}-4}\cdot\frac{y^{2}+5y+6}{y^{2}-1}=\qquad$ ... factor what we can
... $y^{2}+y=y(y+1)$
... $y^{2}-4$= difference of squares = $(y+2)(y-2)$
... $y^{2}+5y+6=$... factors of $+6$ whose sum is $+5$ ... are $+2$ and $+3$
$=(y+2)(y+3)$
...$y^{2}-1$= difference of squares $=(y+1)(y-1)$
Rewrite the problem:
=$ \displaystyle \frac{y(y+1)}{(y+2)(y-2)}\cdot\frac{(y+2)(y+3)}{(y+1)(y-1)}=\qquad$ ... reduce common factors
=$ \displaystyle \frac{y(1)}{(1)(y-2)}\cdot\frac{(1)(y+3)}{(1)(y-1)}=$
=$ \displaystyle \frac{y(y+3)}{(y-2)(y-1)}$
