Solve for \( y \). \( 4+\frac{1}{y+1}=\frac{5}{y+4} \) If there is more than one solution, separate them with commas, If there is no solution, dick on "No solution". \( y=\square \)

Solve the equation \( 4+\frac{1}{y+1}=\frac{5}{y+4} \). Solve the equation by following steps: - step0: Solve for \(y\): \(4+\frac{1}{y+1}=\frac{5}{y+4}\) - step1: Find the domain: \(4+\frac{1}{y+1}=\frac{5}{y+4},y \in \left(-\infty,-4\right)\cup \left(-4,-1\right)\cup \left(-1,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(4+\frac{1}{y+1}\right)\left(y+1\right)\left(y+4\right)=\frac{5}{y+4}\times \left(y+1\right)\left(y+4\right)\) - step3: Simplify the equation: \(4y^{2}+21y+20=5y+5\) - step4: Move the expression to the left side: \(4y^{2}+21y+20-\left(5y+5\right)=0\) - step5: Calculate: \(4y^{2}+16y+15=0\) - step6: Factor the expression: \(\left(2y+3\right)\left(2y+5\right)=0\) - step7: Separate into possible cases: \(\begin{align}&2y+3=0\\&2y+5=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=-\frac{3}{2}\\&y=-\frac{5}{2}\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&y=-\frac{3}{2}\\&y=-\frac{5}{2}\end{align},y \in \left(-\infty,-4\right)\cup \left(-4,-1\right)\cup \left(-1,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&y=-\frac{3}{2}\\&y=-\frac{5}{2}\end{align}\) - step11: Rewrite: \(y_{1}=-\frac{5}{2},y_{2}=-\frac{3}{2}\) The solutions to the equation \(4+\frac{1}{y+1}=\frac{5}{y+4}\) are \(y=-\frac{5}{2}\) and \(y=-\frac{3}{2}\). Therefore, the solutions are \(y=-2.5\) and \(y=-1.5\).