\( \frac{7(x-3)}{x-4}+\frac{5}{x}=\frac{-5}{x(x-4)} \) Solve the following equation by factoring. A. The solution set is \( \{ \). (Type an integer or a simplified fraction. Use a comma to separate answers as needed.) B. There is no solution.

Solve the equation \( \frac{7(x-3)}{x-4}+\frac{5}{x}=\frac{-5}{x(x-4)} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{7\left(x-3\right)}{x-4}+\frac{5}{x}=\frac{-5}{x\left(x-4\right)}\) - step1: Find the domain: \(\frac{7\left(x-3\right)}{x-4}+\frac{5}{x}=\frac{-5}{x\left(x-4\right)},x \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)\) - step2: Rewrite the fraction: \(\frac{7\left(x-3\right)}{x-4}+\frac{5}{x}=-\frac{5}{x\left(x-4\right)}\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{7\left(x-3\right)}{x-4}+\frac{5}{x}\right)x\left(x-4\right)=-\frac{5}{x\left(x-4\right)}\times x\left(x-4\right)\) - step4: Simplify the equation: \(7x^{2}-16x-20=-5\) - step5: Move the expression to the left side: \(7x^{2}-16x-20-\left(-5\right)=0\) - step6: Subtract the numbers: \(7x^{2}-16x-15=0\) - step7: Factor the expression: \(\left(x-3\right)\left(7x+5\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-3=0\\&7x+5=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=3\\&x=-\frac{5}{7}\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=3\\&x=-\frac{5}{7}\end{align},x \in \left(-\infty,0\right)\cup \left(0,4\right)\cup \left(4,+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=3\\&x=-\frac{5}{7}\end{align}\) - step12: Rewrite: \(x_{1}=-\frac{5}{7},x_{2}=3\) The solution to the equation \( \frac{7(x-3)}{x-4}+\frac{5}{x}=\frac{-5}{x(x-4)} \) is \( x = -\frac{5}{7} \) or \( x = 3 \). Therefore, the solution set is \( \{ -\frac{5}{7}, 3 \} \).