Answer
$\underline{\left( -3,5 \right]}$
Work Step by Step
Rewrite the given inequality as a “related” function,
$\begin{align}
& f\left( x \right)=0 \\
& \frac{x-5}{x+3}=0
\end{align}$
Solve the above rational equation by multiplying both sides by $x+3$.
$\begin{align}
& \left( x+3 \right)\left( \frac{x-5}{x+3} \right)=0\cdot \left( x+3 \right) \\
& x-5=0 \\
& x=5
\end{align}$
Since $x=5$ is a solution of the related equation, use 5 while dividing the number line into intervals. Since the rational expression is undefined for $x=-3$ use $-3$ also.
Select one test value from each interval and determine the sign of $f\left( x \right)$ over that interval.
For interval A:
$\begin{align}
& f\left( -5 \right)=\frac{-5-5}{-5+3} \\
& =\frac{-10}{-2} \\
& =5 \\
& \not{\le }0
\end{align}$
So interval A is not part of the solution set.
For interval B:
$\begin{align}
& f\left( 0 \right)=\frac{0-5}{0+3} \\
& =\frac{-5}{3} \\
& \le 0
\end{align}$
So interval B is part of the solution set.
For interval C:
$\begin{align}
& f\left( 7 \right)=\frac{7-5}{7+3} \\
& =\frac{2}{10} \\
& =\frac{1}{5} \\
& \not{\le }0
\end{align}$
So interval C is not part of the solution set.
Indicate the sign of $f\left( x \right)$ on the number line.
Look for the solution of $f\left( x \right)=\frac{x-5}{x+3}\le 0$
So the rational function is negative in the interval B and $-3$ is not a solution.
Therefore the solution set of the original inequality is $\left( -3,5 \right]$.
