Solve for \( x \). If there is more than one solution, separate the solutions with a comma. \( \frac{2}{x+4}+\frac{x}{x+6}=\frac{-3 x-8}{x^{2}+10 x+24} \) \( x=\square \) Next Question \( \gg \)

Solve the equation \( \frac{2}{x+4}+\frac{x}{x+6}=\frac{-3x-8}{x^{2}+10x+24} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\frac{2}{x+4}+\frac{x}{x+6}=\frac{-3x-8}{x^{2}+10x+24}\) - step1: Find the domain: \(\frac{2}{x+4}+\frac{x}{x+6}=\frac{-3x-8}{x^{2}+10x+24},x \in \left(-\infty,-6\right)\cup \left(-6,-4\right)\cup \left(-4,+\infty\right)\) - step2: Rewrite the fraction: \(\frac{2}{x+4}+\frac{x}{x+6}=-\frac{3x+8}{x^{2}+10x+24}\) - step3: Multiply both sides of the equation by LCD: \(\left(\frac{2}{x+4}+\frac{x}{x+6}\right)\left(x+4\right)\left(x+6\right)=-\frac{3x+8}{x^{2}+10x+24}\times \left(x+4\right)\left(x+6\right)\) - step4: Simplify the equation: \(6x+12+x^{2}=-3x-8\) - step5: Move the expression to the left side: \(6x+12+x^{2}-\left(-3x-8\right)=0\) - step6: Calculate: \(9x+20+x^{2}=0\) - step7: Factor the expression: \(\left(4+x\right)\left(5+x\right)=0\) - step8: Separate into possible cases: \(\begin{align}&4+x=0\\&5+x=0\end{align}\) - step9: Solve the equation: \(\begin{align}&x=-4\\&x=-5\end{align}\) - step10: Check if the solution is in the defined range: \(\begin{align}&x=-4\\&x=-5\end{align},x \in \left(-\infty,-6\right)\cup \left(-6,-4\right)\cup \left(-4,+\infty\right)\) - step11: Find the intersection: \(x=-5\) The solution to the equation \( \frac{2}{x+4}+\frac{x}{x+6}=\frac{-3x-8}{x^{2}+10x+24} \) is \( x = -5 \).