\( \frac{w+3}{w+1}-1=\frac{w-5}{w-7} \) If there is more than one solution, separate them with commas. If there is no solution, click on "No solution".

Solve the equation \( \frac{w+3}{w+1}-1=\frac{w-5}{w-7} \). Solve the equation by following steps: - step0: Solve for \(w\): \(\frac{w+3}{w+1}-1=\frac{w-5}{w-7}\) - step1: Find the domain: \(\frac{w+3}{w+1}-1=\frac{w-5}{w-7},w \in \left(-\infty,-1\right)\cup \left(-1,7\right)\cup \left(7,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{w+3}{w+1}-1\right)\left(w+1\right)\left(w-7\right)=\frac{w-5}{w-7}\times \left(w+1\right)\left(w-7\right)\) - step3: Simplify the equation: \(2w-14=w\left(w+1\right)-5\left(w+1\right)\) - step4: Calculate: \(2w-14=w^{2}+w-5w-5\) - step5: Simplify: \(2w-14=w^{2}-4w-5\) - step6: Move the expression to the left side: \(2w-14-\left(w^{2}-4w-5\right)=0\) - step7: Subtract the terms: \(6w-9-w^{2}=0\) - step8: Factor the expression: \(-\left(3-w\right)^{2}=0\) - step9: Divide the terms: \(\left(3-w\right)^{2}=0\) - step10: Simplify the expression: \(3-w=0\) - step11: Move the constant to the right side: \(-w=0-3\) - step12: Remove 0: \(-w=-3\) - step13: Change the signs: \(w=3\) - step14: Check if the solution is in the defined range: \(w=3,w \in \left(-\infty,-1\right)\cup \left(-1,7\right)\cup \left(7,+\infty\right)\) - step15: Find the intersection: \(w=3\) The solution to the equation \( \frac{w+3}{w+1}-1=\frac{w-5}{w-7} \) is \( w=3 \).