Answer
$\displaystyle \frac{y^{2}-9y-8}{(y-2)(y-3)}$
Work Step by Step
The third term can be rewritten as
$\displaystyle \frac{(y+5)(2y+1)}{[-(y-3)][-(y-2)]}=\frac{(y+5)(2y+1)}{(y-3)(y-2)}$
So, we have
$=\displaystyle \frac{(y+1)(2y-1)}{(y-2)(y-3)}+\frac{(y+2)(y-1)}{(y-2)(y-3)}-\frac{(y+5)(2y+1)}{(y-3)(y-2)}$
To add/subtract rational expressions with the same denominator,
add/subtract numerators and place the sum/difference over the common denominator.
$= \displaystyle \frac{(y+1)(2y-1)+(y+2)(y-1)-(y+5)(2y+1)}{(y-2)(y-3)}$
$= \displaystyle \frac{(2y^{2}-y+2y-1)+(y^{2}-y+2y-2)-(2y^{2}+y+10y+5)}{(y-2)(y-3)}$
$= \displaystyle \frac{2y^{2}+y-1+y^{2}-y+2y-2-2y^{2}-11y-5}{(y-2)(y-3)}$
= $\displaystyle \frac{y^{2}-9y-8}{(y-2)(y-3)}$
... factoring $x^{2}+bx+c,$ we search for factors of $c$ whose sum is $b$
... no integer factors of $-8$ whose sum is $-9$ exist
... the numerator can not be factored, we leave it as it is.
