Answer
The solution set of equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ is $\left\{ \left. 4 \right\} \right.$.
Work Step by Step
Consider the provided equation,
$\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$
Multiply both sides by ${{x}^{2}}+6x+8$
$\begin{align}
& \frac{2x}{{{x}^{2}}+6x+8}\cdot \left( {{x}^{2}}+6x+8 \right)+\frac{2}{x+2}\cdot \left( {{x}^{2}}+6x+8 \right)=\frac{x}{x+4}\cdot \left( {{x}^{2}}+6x+8 \right) \\
& 2x+\frac{2}{x+2}\cdot \left( {{x}^{2}}+6x+8 \right)=\frac{x}{x+4}\cdot \left( {{x}^{2}}+6x+8 \right) \\
\end{align}$
On factorization of ${{x}^{2}}+6x+8$ it becomes
$\begin{align}
& {{x}^{2}}+6x+8={{x}^{2}}+4x+2x+8 \\
& =x\left( x+4 \right)+2\left( x+4 \right) \\
& =\left( x+4 \right)\left( x+2 \right)
\end{align}$
Now replace ${{x}^{2}}+6x+8$ by $\left( x+4 \right)\left( x+2 \right)$
$\begin{align}
& 2x+\frac{2}{x+2}\cdot \left( x+4 \right)\left( x+2 \right)=\frac{x}{x+4}\cdot \left( x+4 \right)\left( x+2 \right) \\
& 2x+2\cdot \left( x+4 \right)=x\cdot \left( x+2 \right) \\
& 2x+2x+8={{x}^{2}}+2x \\
& 4x+8={{x}^{2}}+2x
\end{align}$
On further simplification
$\begin{align}
& 8={{x}^{2}}+2x-4x \\
& 8={{x}^{2}}-2x \\
& 0={{x}^{2}}-2x-8 \\
\end{align}$
It can be written as
${{x}^{2}}-2x-8=0$
Apply quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Here a is 1, b is $-2$ and c is $-8$
$\begin{align}
& x=\frac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)} \\
& =\frac{2\pm \sqrt{4+32}}{2} \\
& =\frac{2\pm \sqrt{36}}{2} \\
& =\frac{2\pm 6}{2}
\end{align}$
On further simplification
$\begin{align}
& x=\frac{2+6}{2} \\
& =\frac{8}{2} \\
& =4
\end{align}$
Or
$\begin{align}
& x=\frac{2-6}{2} \\
& =\frac{-4}{2} \\
& =-2
\end{align}$
Check:
Put $x=4$ in the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$
$\begin{matrix}
\frac{2\left( 4 \right)}{{{\left( 4 \right)}^{2}}+6\left( 4 \right)+8}+\frac{2}{4+2}\overset{?}{\mathop{=}}\,\frac{4}{4+4} \\
\frac{8}{16+24+8}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{4}{8} \\
\frac{8}{48}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\
\frac{1}{6}+\frac{2}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\
\end{matrix}$
On further simplification
$\begin{align}
& \frac{3}{6}\overset{?}{\mathop{=}}\,\frac{1}{2} \\
& \frac{1}{2}\overset{?}{\mathop{=}}\,\frac{1}{2} \\
\end{align}$
Which is true.
For $x=-2$
Put $x=-2$ in the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$
$\begin{align}
& \frac{2\left( -2 \right)}{{{\left( -2 \right)}^{2}}+6\left( -2 \right)+8}+\frac{2}{-2+2}\overset{?}{\mathop{=}}\,\frac{-2}{-2+4} \\
& \frac{-4}{4-12+8}+\frac{2}{0}\overset{?}{\mathop{=}}\,\frac{4}{8}
\end{align}$
Since the value of $\frac{2}{0}$ is undefined.
The solution does not check for $x=-2$
Therefore, the solution set of the equation $\frac{2x}{{{x}^{2}}+6x+8}+\frac{2}{x+2}=\frac{x}{x+4}$ is $\left\{ \left. 4 \right\} \right.$.
