Answer
$\left\{\dfrac{4}{5},3\right\}$
Work Step by Step
Multiplying both sides by the $LCD=
3x(x-2)
,$ the given equation, $
\dfrac{2}{x}+\dfrac{1}{x-2}=\dfrac{5}{3}
,$ is equivalent to
\begin{align*}
3x(x-2)\left(\dfrac{2}{x}+\dfrac{1}{x-2}\right)&=\left(\dfrac{5}{3}\right)3x(x-2)
\\\\
3(x-2)(2)+3x(1)&=5(x)(x-2)
\\
6x-12+3x&=5x^2-10x
\\
0&=5x^2+(-10x-6x-3x)+12
\\
0&=5x^2-19x+12
\\
5x^2-19x+12&=0
.\end{align*}
Using factoring of trinomials, the equation above is equivalent to
\begin{align*}
(x-3)(5x-4)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x-3=0 & 5x-4=0
\\
x=3 & 5x=4
\\\\
& x=\dfrac{4}{5}
.\end{array}
Hence, the solution set of the equation $
\dfrac{2}{x}+\dfrac{1}{x-2}=\dfrac{5}{3}
$ is $\left\{\dfrac{4}{5},3\right\}$.
