Solve for x . If there is more than one solution, separate the \( \frac{8}{x-9}+\frac{x}{x+5}=\frac{-13 x+13}{x^{2}-4 x-45} \) \( x=\square \) Question Help: \( \square \) Video \( \square \) Message instructor

Solve the equation \( \frac{8}{x-9}+\frac{x}{x+5}=\frac{-13x+13}{x^{2}-4x-45} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{8}{x-9}+\frac{x}{x+5}=\frac{-13x+13}{x^{2}-4x-45}\) - step1: Find the domain: \(\frac{8}{x-9}+\frac{x}{x+5}=\frac{-13x+13}{x^{2}-4x-45},x \in \left(-\infty,-5\right)\cup \left(-5,9\right)\cup \left(9,+\infty\right)\) - step2: Multiply both sides of the equation by LCD: \(\left(\frac{8}{x-9}+\frac{x}{x+5}\right)\left(x-9\right)\left(x+5\right)=\frac{-13x+13}{x^{2}-4x-45}\times \left(x-9\right)\left(x+5\right)\) - step3: Simplify the equation: \(-x+40+x^{2}=-13x+13\) - step4: Move the expression to the left side: \(-x+40+x^{2}-\left(-13x+13\right)=0\) - step5: Calculate: \(12x+27+x^{2}=0\) - step6: Factor the expression: \(\left(3+x\right)\left(9+x\right)=0\) - step7: Separate into possible cases: \(\begin{align}&3+x=0\\&9+x=0\end{align}\) - step8: Solve the equation: \(\begin{align}&x=-3\\&x=-9\end{align}\) - step9: Check if the solution is in the defined range: \(\begin{align}&x=-3\\&x=-9\end{align},x \in \left(-\infty,-5\right)\cup \left(-5,9\right)\cup \left(9,+\infty\right)\) - step10: Find the intersection: \(\begin{align}&x=-3\\&x=-9\end{align}\) - step11: Rewrite: \(x_{1}=-9,x_{2}=-3\) The solutions to the equation \( \frac{8}{x-9}+\frac{x}{x+5}=\frac{-13x+13}{x^{2}-4x-45} \) are \( x=-9 \) and \( x=-3 \).