Answer
$\dfrac{y+5}{2(y+1)}; y\ne-1, 5$
Work Step by Step
Factor the numerator and the denominator completely to obtain:
$=\dfrac{(y-5)(y+5)}{2(y^2-4y-5)}
\\=\dfrac{(y-5)(y+5)}{2(y-5)(y+1)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{\cancel{(y-5)}(y+5)}{2\cancel{(y-5)}(y+1)}
\\=\dfrac{y+5}{2(y+1)}; y\ne-1, 5$
($y$ cannot be $-1$ and $5$ because they will make the original expression undefined.)
