Answer
$x=\dfrac{4}{3}$
Work Step by Step
The values of $x$ in the equation $
\dfrac{2}{x+3}=\dfrac{1}{x^2-9}-\dfrac{1}{x-3}
$ are
\begin{array}{l}
\dfrac{2}{x+3}=\dfrac{1}{(x+3)(x-3)}-\dfrac{1}{x-3}
\\\\
(x+3)(x-3)\left( \dfrac{2}{x+3} \right) =\left( \dfrac{1}{(x+3)(x-3)}-\dfrac{1}{x-3} \right)(x+3)(x-3)
\\\\
(x-3)(2)=1(1)-1(x+3)
\\\\
2x-6=1-x-3
\\\\
2x+x=1-3+6
\\\\
3x=4
\\\\
x=\dfrac{4}{3}
.\end{array}
