Solve for \( w \). \[ \frac{w-2}{w+1}+1=\frac{w-2}{w+2} \] If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

Solve the equation \( \frac{w-2}{w+1}+1=\frac{w-2}{w+2} \). Solve the equation by following steps: - step0: Solve for \(w\): \(\frac{w-2}{w+1}+1=\frac{w-2}{w+2}\) - step1: Find the domain: \(\frac{w-2}{w+1}+1=\frac{w-2}{w+2},w \in \left(-\infty,-2\right)\cup \left(-2,-1\right)\cup \left(-1,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{w-2}{w+1}+1\right)\left(w+1\right)\left(w+2\right)=\frac{w-2}{w+2}\times \left(w+1\right)\left(w+2\right)\) - step3: Simplify the equation: \(2w^{2}+3w-2=w\left(w+1\right)-2\left(w+1\right)\) - step4: Calculate: \(2w^{2}+3w-2=w^{2}+w-2w-2\) - step5: Simplify: \(2w^{2}+3w-2=w^{2}-w-2\) - step6: Cancel equal terms: \(2w^{2}+3w=w^{2}-w\) - step7: Move the expression to the left side: \(2w^{2}+3w-\left(w^{2}-w\right)=0\) - step8: Calculate: \(w^{2}+4w=0\) - step9: Factor the expression: \(w\left(w+4\right)=0\) - step10: Separate into possible cases: \(\begin{align}&w=0\\&w+4=0\end{align}\) - step11: Solve the equation: \(\begin{align}&w=0\\&w=-4\end{align}\) - step12: Check if the solution is in the defined range: \(\begin{align}&w=0\\&w=-4\end{align},w \in \left(-\infty,-2\right)\cup \left(-2,-1\right)\cup \left(-1,+\infty\right)\) - step13: Find the intersection: \(\begin{align}&w=0\\&w=-4\end{align}\) - step14: Rewrite: \(w_{1}=-4,w_{2}=0\) The solutions to the equation \( \frac{w-2}{w+1}+1=\frac{w-2}{w+2} \) are \( w=-4 \) and \( w=0 \).