Answer
$ \displaystyle \frac{y-5}{2}$
Work Step by Step
Dividing with $\displaystyle \frac{P}{Q}$ = multiplying with $\displaystyle \frac{Q}{P}$
Rewrite the problem:
$ \displaystyle \frac{y^{2}-25}{2y-2}\cdot\frac{y^{2}+4y-5}{y^{2}+10y+25}=\qquad$ ... factor what we can
...$y^{2}-25$= difference of squares = $(y+5)(y-5)$
....$2y-2=2(y-1)$
...$y^{2}+4y-5=$... factors of $-5$ whose sum is $+4$ ... are $+5$ and $-1$
$=(y+5)(y-1)$
...$y^{2}+10y+25=(y)^{2}+2(y)(5)+(5)^{2}=(y+5)^{2}$
Rewrite the problem:
=$ \displaystyle \frac{(y+5)(y-5)}{2(y-1)}\cdot\frac{(y+5)(y-1)}{(y+5)(y+5)}=\qquad$ ... reduce common factors
=$ \displaystyle \frac{(1)(y-5)}{2(y-1)}\cdot\frac{(1)(y-1)}{(1)(1)}=$
=$ \displaystyle \frac{(y-5)(1)}{2(1)}$
=$ \displaystyle \frac{y-5}{2}$
