8. \( \begin{aligned} \frac{y}{y+4}-\frac{32}{y^{2}-16} & =\square \\ & =\square \\ & =\square \\ & =\square\end{aligned} \)

To solve the equation \( \frac{y}{y+4}-\frac{32}{y^{2}-16} = \square \), we should start by simplifying the left-hand side. First, notice that \( y^2 - 16 \) can be factored as \( (y-4)(y+4) \). Rewrite the equation: \[ \frac{y}{y+4} - \frac{32}{(y-4)(y+4)} \] To subtract these fractions, we need a common denominator, which is \( (y-4)(y+4) \): \[ \frac{y(y-4) - 32}{(y-4)(y+4)} \] Now simplify the numerator: \[ \frac{y^2 - 4y - 32}{(y-4)(y+4)} \] Next, factor the numerator \( y^2 - 4y - 32 \): To do this, we need factors of \(-32\) that add up to \(-4\). Those factors are \(4\) and \(-8\): \[ y^2 - 8y + 4y - 32 = (y - 8)(y + 4) \] Thus, we can rewrite our fraction: \[ \frac{(y - 8)(y + 4)}{(y - 4)(y + 4)} \] Now, we notice that we can cancel the \( (y + 4) \) terms (as long as \(y \neq -4\)): \[ \frac{y - 8}{y - 4} \] Thus, the simplified expression is: \[ \frac{y - 8}{y - 4} \] You can now express your final answer as: \[ \frac{y - 8}{y - 4} = \square \] Or if you have further values for \(y\), you can substitute those in for additional output in the form \(=\square\).