#### Answer

no solution

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
To solve the given equation, $
\dfrac{2}{x+2}+\dfrac{1}{x+4}=\dfrac{4}{x^2+6x+8}
,$ factor all expressions that can be factored. Then multiply both sides by the $LCD$ and isolate the variable. Finally, do checking and ensure that any denominator does not become $0.$
$\bf{\text{Solution Details:}}$
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the factored form of the equation above is
\begin{array}{l}\require{cancel}
\dfrac{2}{x+2}+\dfrac{1}{x+4}=\dfrac{4}{(x+2)(x+4)}
.\end{array}
The $LCD$ of the denominators, $x+2,x+4,$ and $(x+2)(x+4)$ is $(x+2)(x+4)$ since it is the lowest expression which can be divided exactly by all the given denominators.
Multiplying both sides by the $LCD=
(x+2)(x+4)
,$ the given expression is equivalent to
\begin{array}{l}\require{cancel}
(x+2)(x+4)\left( \dfrac{2}{x+2}+\dfrac{1}{x+4} \right)=\left( \dfrac{4}{(x+2)(x+4)} \right) (x+2)(x+4)
\\\\
(x+4)(2)+(x+2)(1)=1(4)
\\\\
2x+8+x+2=4
\\\\
2x+x=4-8-2
\\\\
3x=-6
\\\\
x=-\dfrac{6}{3}
\\\\
x=-2
.\end{array}
If $x=-2,$ the part of the given equation, $
\dfrac{2}{x+2}
,$ becomes $
\dfrac{2}{0}
,$ which is undefined. Hence, there is $\text{
no solution
.}$