Answer
$\dfrac{y-1}{y+1}; y \ne -1, -\frac{2}{3}$
Work Step by Step
Factor the numerator and the denominator completely to obtain:
$=\dfrac{(3y+2)(y-1)}{(3y+2)(y+1)}$
Cancel the common factors to obtain:
$\require{cancel}
\\=\dfrac{\cancel{(3y+2)}(y-1)}{\cancel{(3y+2)}(y+1)}
\\=\dfrac{y-1}{y+1}; y \ne -1, -\frac{2}{3}$
($y$ cannot be $-1$ and $-\frac{2}{3}$ because they will make the original expression undefined).
