### Instruction: Find $y$ if $\dfrac{y^2 - 9y + 8}{y-1} + \dfrac{3y^2 +16y-12 }{3y -2} = -3$.### Response: Hey, let's simplify this equation.
Ok, let's do the first part: $\dfrac{y^2-9y+8}{y-1}$.
I see. We can factor the top to get $(y-1)(y-8)$.
And that simplifies to $\dfrac{(y-1)(y-8)}{y-1}$.
Right, and then we cancel the $y-1$ terms to get $y-8$.
Now, let's simplify the second part: $\dfrac{3y^2+16y-12}{3y-2}$.
Ok. We can factor the top to get $(3y-2)(y+6)$.
And that simplifies to $\dfrac{(3y-2)(y+6)}{3y-2}$.
Right, and then we cancel the $3y-2$ terms to get $y+6$.
Now, we can go back to the original equation and substitute in those values.
So we get $y-8+y+6=-3$.
And if we simplify that, we get $2y-2=-3$.
Right. So, $2y=-1$.
And finally, $y=\dfrac{-1}{2}$.