Answer
$\left\{-\dfrac{6}{5},4\right\}$
Work Step by Step
Multiplying both sides of the given equation, $
\dfrac{6}{x}+\dfrac{6}{x+2}=\dfrac{5}{2}
,$ by the $LCD=2x(x+2),$ then
\begin{align*}\require{cancel}
2x(x+2)\left(\dfrac{6}{x}+\dfrac{6}{x+2}\right)&=\left(\dfrac{5}{2}\right)2x(x+2)
\\\\
2(x+2)(6)+2x(6)&=5(x)(x+2)
\\
12x+24+12x&=5x^2+10x
\\
24x+24&=5x^2+10x
\\
0&=5x^2+10x-24x-24
\\
0&=5x^2-14x-24
\\
5x^2-14x-24&=0
.\end{align*}
Using the factoring of trinomials, the factored form of the equation above is
\begin{align*}
(5x+6)(x-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property), the solutions of the equation above are
\begin{align*}
\begin{array}{l|r}
5x+6=0 & x-4=0
\\
5x=-6 & x=4
\\\\
\dfrac{\cancel5x}{\cancel5}=-\dfrac{6}{5} &
\\\\
x=-\dfrac{6}{5} \end{array}
.\end{align*}
Since none of the solutions, when substituted in the original equation, result to a denominator of zero, both solutions are valid. Hence, the solutions of the equation $\dfrac{6}{x}+\dfrac{6}{x+2}=\dfrac{5}{2}$ are $\left\{-\dfrac{6}{5},4\right\}$.
