What is the solution of \( \frac{-8}{2 y-8}=\frac{5}{y+4}-\frac{7 y+8}{y^{2}-16} ? \)

To solve the equation \[ \frac{-8}{2y-8}=\frac{5}{y+4}-\frac{7y+8}{y^{2}-16}, \] first, notice that the denominator \(y^2 - 16\) factors as \((y - 4)(y + 4)\). Rewrite the equation accordingly, ensuring that all fractions have a common denominator. Then, rewriting gives: \[ \frac{-8}{2(y-4)} = \frac{5(y-4) - (7y+8)}{(y+4)(y-4)}. \] Now, substitute \(2(y-4)\) to simplify the expression on the left side: \[ \frac{-8}{2(y-4)} = \frac{-4}{y-4}. \] Next, multiply both sides by \(2(y - 4)(y + 4)\) to eliminate the fractions: \[ -8(y + 4) = 2[5(y - 4) - (7y + 8)]. \] Expanding both sides gives: \[ -8y - 32 = 10y - 20 - 14y - 16. \] Combining like terms on the right: \[ -8y - 32 = -4y - 36. \] Now, isolate \(y\) by adding \(4y\) to both sides: \[ -4y - 32 = -36. \] Then, add \(32\) to both sides: \[ -4y = -4. \] Finally, divide by \(-4\): \[ y = 1. \] Thus, the solution is \[ \boxed{1}. \] Don't forget to check if \(y = 1\) makes any denominators zero! Here, \(2y - 8 = 2(1) - 8 = -6\) and \(y + 4 = 1 + 4 = 5\) neither of which are zero, so \(y = 1\) is valid.