Solve for \( y \) \[ \frac{4}{y+4}=\frac{8}{2 y+8}-3 \]

Solve the equation \( \frac{4}{y+4}=\frac{8}{2y+8}-3 \). Solve the equation by following steps: - step0: Solve for \(y\): \(\frac{4}{y+4}=\frac{8}{2y+8}-3\) - step1: Find the domain: \(\frac{4}{y+4}=\frac{8}{2y+8}-3,y\neq -4\) - step2: Divide the terms: \(\frac{4}{y+4}=\frac{4}{y+4}-3\) - step3: Multiply both sides of the equation by LCD: \(\frac{4}{y+4}\times \left(y+4\right)=\left(\frac{4}{y+4}-3\right)\left(y+4\right)\) - step4: Simplify the equation: \(4=-8-3y\) - step5: Swap the sides: \(-8-3y=4\) - step6: Move the constant to the right side: \(-3y=4+8\) - step7: Add the numbers: \(-3y=12\) - step8: Change the signs: \(3y=-12\) - step9: Divide both sides: \(\frac{3y}{3}=\frac{-12}{3}\) - step10: Divide the numbers: \(y=-4\) - step11: Check if the solution is in the defined range: \(y=-4,y\neq -4\) - step12: Find the intersection: \(y \in \varnothing \) The solution to the equation \( \frac{4}{y+4}=\frac{8}{2y+8}-3 \) is that there are no real solutions for \( y \).