### Instruction: Find $y$ if $\dfrac{y^2 - 9y + 8}{y-1} + \dfrac{3y^2 +16y-12 }{3y -2} = -3$.### Response: So we have $\dfrac{y^2 - 9y + 8}{y-1} + \dfrac{3y^2 +16y-12 }{3y -2} = -3$.
Right. We should simplify the left side.
Ok. So $\dfrac{y^2 - 9y + 8}{y-1} = \dfrac{(y-1)(y-8)}{y-1} = y-8$.
And $\dfrac{3y^2 +16y-12 }{3y -2} = \dfrac{(3y-2)(y+6)}{3y-2} = y+6$.
So the equation is now $y-8 + y+6 = -3$.
That simplifies to $2y-2=-3$.
So $2y=-1$.
And $y=\dfrac{-1}{2}$.