Answer
$\dfrac{-2k-19}{(k+3)(k-2)}$
Work Step by Step
Using the $LCD=
(k+3)(k-2)
,$ the given expression, $
\dfrac{2}{k+3}-\dfrac{5}{k-2}
$, is equivalent to
\begin{align*}
&
\dfrac{2}{k+3}\cdot\dfrac{k-2}{k-2}-\dfrac{5}{k-2}\cdot\dfrac{k+3}{k+3}
\\\\&=
\dfrac{2k-4}{(k+3)(k-2)}-\dfrac{5k+15}{(k+3)(k-2)}
\\\\&=
\dfrac{2k-4-4k-15}{(k+3)(k-2)}
\\\\&=
\dfrac{-2k-19}{(k+3)(k-2)}
.\end{align*}
Hence, the expression $
\dfrac{2}{k+3}-\dfrac{5}{k-2}
$ simplifies to $
\dfrac{-2k-19}{(k+3)(k-2)}
$.
