Solve the following equation by factoring. \( \frac{7(x-2)}{x-3}+\frac{2}{x}=\frac{-2}{x(x-3)} \) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. 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 \( \frac{7(x-2)}{x-3}+\frac{2}{x}=\frac{-2}{x(x-3)} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{7\left(x-2\right)}{x-3}+\frac{2}{x}=\frac{-2}{x\left(x-3\right)}\) - step1: Find the domain: \(\frac{7\left(x-2\right)}{x-3}+\frac{2}{x}=\frac{-2}{x\left(x-3\right)},x \in \left(-\infty,0\right)\cup \left(0,3\right)\cup \left(3,+\infty\right)\) - step2: Rewrite the fraction: \(\frac{7\left(x-2\right)}{x-3}+\frac{2}{x}=-\frac{2}{x\left(x-3\right)}\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{7\left(x-2\right)}{x-3}+\frac{2}{x}\right)x\left(x-3\right)=-\frac{2}{x\left(x-3\right)}\times x\left(x-3\right)\) - step4: Simplify the equation: \(7x^{2}-12x-6=-2\) - step5: Move the expression to the left side: \(7x^{2}-12x-6-\left(-2\right)=0\) - step6: Subtract the numbers: \(7x^{2}-12x-4=0\) - step7: Factor the expression: \(\left(x-2\right)\left(7x+2\right)=0\) - step8: Separate into possible cases: \(\begin{align}&x-2=0\\&7x+2=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=2\\&x=-\frac{2}{7}\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=2\\&x=-\frac{2}{7}\end{align},x \in \left(-\infty,0\right)\cup \left(0,3\right)\cup \left(3,+\infty\right)\) - step11: Find the intersection: \(\begin{align}&x=2\\&x=-\frac{2}{7}\end{align}\) - step12: Rewrite: \(x_{1}=-\frac{2}{7},x_{2}=2\) The solutions to the equation \( \frac{7(x-2)}{x-3}+\frac{2}{x}=\frac{-2}{x(x-3)} \) are \( x = -\frac{2}{7} \) and \( x = 2 \). Therefore, the correct choice is: A. The solution set is \( \{-\frac{2}{7}, 2\} \).