Answer
$\left[-\frac{24}{7}, -2\right) \cup \left(0, 3 \right)$
Work Step by Step
1. Express the inequality in the form $f(x)<0, f(x)>0, f(x)\leq 0$, or $f(x)\geq 0,$ where $f$ is a Rational function.
$\displaystyle \frac{1}{x+2} +\frac{3}{x-3} \leq \frac{4}{x},$
$\displaystyle \frac{1}{x+2} +\frac{3}{x-3}- \frac{4}{x }\leq 0$,
$\displaystyle \frac{7x+24}{(x)(x+2)(x-3)}\leq 0$,
$f(x)=\displaystyle \frac{7x+24}{(x)(x+2)(x-3)}\leq 0$,
2.The cut points are:
$\displaystyle \frac{7x+24}{(x)(x+2)(x-3)}=0$
$x=-\frac{24}{7}$ or $x=-2$ or $x=0$ or $x=3$
3. Make a table or diagram: use the test values to make a table or diagram of the sign of each factor in each interval.
4. Test each interval's sign of $f(x)$ with a test value, $\begin{array}{llll} Intervals: & a=test.v. & f(a),signs & f(a) \leq 0 ? \\ & & \displaystyle \displaystyle \frac{7a+24}{(a)(a+2)(a-3)}& \\ (-\infty,-\frac{24}{7}) & -4 & \frac{(-)}{(-)(-)(-)}=(+) & F\\
(-\frac{24}{7}, -2) & -3 & \frac{(+)}{(-)(-)(-)}=(-) & T\\
(-2, 0) & -1 & \frac{(+)}{(-)(+)(-)}=(+) & F\\
(0, 3) & 1 & \frac{(+)}{(+)(+)(-)}=(-) & T\\
(3,\infty) & 10 & \frac{(+)}{(+)(+)(+)}=(+) & F\\
\end{array}$
5. Write the solution set, selecting the interval or intervals that satisfy the given inequality. If the inequality involves $\leq$ or $\geq$, include the boundary points.
Solution set: $\left[-\frac{24}{7}, -2\right) \cup \left(0, 3 \right)$
