Answer
The solution set of equation $\frac{2}{x-3}-\frac{4}{x+3}=\frac{8}{x{}^{2}-9}$ is $\left\{ \left. 5 \right\} \right.$.
Work Step by Step
Consider the provided equation:
$\frac{2}{x-3}-\frac{4}{x+3}=\frac{8}{x{}^{2}-9}$
Multiply both sides by ${{x}^{2}}-9$ and use the difference of the square as follows:
$\begin{align}
& \frac{2}{x-3}\cdot \left( x{}^{2}-9 \right)-\frac{4}{x+3}\cdot \left( x{}^{2}-9 \right)=\frac{8}{x{}^{2}-9}\cdot \left( x{}^{2}-9 \right) \\
& \frac{2}{x-3}\cdot \left( x+3 \right)\left( x-3 \right)-\frac{4}{x+3}\cdot \left( x+3 \right)\left( x-3 \right)=\frac{8}{x{}^{2}-9}\cdot \left( x{}^{2}-9 \right) \\
& 2\cdot \left( x+3 \right)-4\cdot \left( x-3 \right)=8 \\
& 2x+6-4x+12=8
\end{align}$
Take the terms containing x on the left hand side and constant terms on the right side
$\begin{align}
& 2x+6-4x+12=8 \\
& 2x-4x=8-6-12 \\
& -2x=-10
\end{align}$
Divide both sides by $-2$
$\begin{align}
& \frac{-2x}{-2}=\frac{-10}{-2} \\
& x=5
\end{align}$
Check:
For $x=5$
Put $x=5$ in $\frac{2}{x-3}-\frac{4}{x+3}=\frac{8}{x{}^{2}-9}$
$\begin{align}
& \frac{2}{5-3}-\frac{4}{5+3}\overset{?}{\mathop{=}}\,\frac{8}{25-9} \\
& \frac{2}{2}-\frac{4}{8}\overset{?}{\mathop{=}}\,\frac{8}{16} \\
& 1-\frac{1}{2}\overset{?}{\mathop{=}}\,\frac{1}{2} \\
& \frac{1}{2}=\frac{1}{2}
\end{align}$
Which is true.
Therefore, the solution set of the equation $\frac{2}{x-3}-\frac{4}{x+3}=\frac{8}{x{}^{2}-9}$ is $\left\{ \left. 5 \right\} \right.$.
