Answer
$\displaystyle \frac{3y^{2}+8y-5}{(y+1)(y-4)}$
Work Step by Step
We rewrite the factor $y-4$ as $-(4-y)$
in the denominator of the first term:
$\displaystyle \frac{(y-3)(y+2)}{(y+1)(y-4)}=\frac{(y-3)(y+2)}{-(y+1)(4-y)}=-\frac{(y-3)(y+2)}{(y+1)(4-y)}$
We now have
$=-\left( \displaystyle \frac{(y-3)(y+2)}{(y+1)(4-y)}+\frac{(y+2)(y+3)}{(y+1)(4-y)}+\frac{(y+5)(y-1)}{(y+1)(4-y)}\right)$
To add/subtract rational expressions with the same denominator,
add/subtract numerators and place the sum/difference over the common denominator.
$=-\left( \displaystyle \frac{(y-3)(y+2)+(y+2)(y+3)+(y+5)(y-1)}{(y+1)(4-y)} \right)$
$=-\left( \displaystyle \frac{(y+2)[y-3+y+3]+[y^{2}-y+5y-5]}{(y+1)(4-y)} \right)$
$=-\left( \displaystyle \frac{(y+2)[2y]+y^{2}-y+5y-5}{(y+1)(4-y)} \right)$
$=-\left( \displaystyle \frac{2y^{2}+4y+y^{2}-y+5y-5}{(y+1)(4-y)} \right)$
$=-\left( \displaystyle \frac{3y^{2}+8y-5}{(y+1)(4-y)} \right)$
$... $for $ax^{2}+bx+c$ we search for factors of $ac$ whose sum is $b$
... if sucessful, rewrite $bx $ and factor in pairs.
$... $here, factors of $3\cdot(-5)=-15$ ... whose sum is $8 ..$. do not exist
... so the numerator is left as it is.
Apply the minus back into the factor of the denominator...
= $\displaystyle \frac{3y^{2}+8y-5}{(y+1)(y-4)}$
