Solve for \( y \). \[ -\frac{7}{y+1}=-6-\frac{2}{y-2} \] 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 \( -\frac{7}{y+1}=-6-\frac{2}{y-2} \). Solve the equation by following steps: - step0: Solve for \(y\): \(-\frac{7}{y+1}=-6-\frac{2}{y-2}\) - step1: Find the domain: \(-\frac{7}{y+1}=-6-\frac{2}{y-2},y \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(-\frac{7}{y+1}\times \left(y+1\right)\left(y-2\right)=\left(-6-\frac{2}{y-2}\right)\left(y+1\right)\left(y-2\right)\) - step3: Simplify the equation: \(-7y+14=-6y^{2}+4y+10\) - step4: Move the expression to the left side: \(-7y+14-\left(-6y^{2}+4y+10\right)=0\) - step5: Calculate: \(-11y+4+6y^{2}=0\) - step6: Factor the expression: \(\left(1-2y\right)\left(4-3y\right)=0\) - step7: Separate into possible cases: \(\begin{align}&1-2y=0\\&4-3y=0\end{align}\) - step8: Solve the equation: \(\begin{align}&y=\frac{1}{2}\\&y=\frac{4}{3}\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&y=\frac{1}{2}\\&y=\frac{4}{3}\end{align},y \in \left(-\infty,-1\right)\cup \left(-1,2\right)\cup \left(2,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&y=\frac{1}{2}\\&y=\frac{4}{3}\end{align}\) - step11: Rewrite: \(y_{1}=\frac{1}{2},y_{2}=\frac{4}{3}\) The solutions to the equation \( -\frac{7}{y+1}=-6-\frac{2}{y-2} \) are \( y = \frac{1}{2} \) and \( y = \frac{4}{3} \).