Answer
$\left(-3, 0 \right) \cup \left(2, \frac{9}{2}\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.
$\displaystyle \frac{1}{x-2} +\frac{2}{x+3} \geq \frac{3}{x},$
$\displaystyle \frac{1}{x-2} +\frac{2}{x+3}- \frac{3}{x }\geq 0$,
$\displaystyle \frac{-4x+18}{(x)(x-2)(x+3)}\geq 0$,
$f(x)=\displaystyle \frac{-4x+18}{(x)(x-2)(x+3)}\geq 0$,
2.The cut points are:
$\displaystyle \frac{-4x+18}{(x)(x-2)(x+3)}=0$
$x=-3$ or $x=0$ or $x=2$ or $x=\frac{9}{2}$
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,
$\begin{array}{llll} Intervals: & a=test.v. & f(a),signs & f(a) \geq 0 ? \\ & & \displaystyle \displaystyle \frac{-4a+18}{(a)(a-2)(a+3)} & \\ (-\infty,-3) & -5 & \frac{(+)}{(-)(-)(-)}=(-) & F
\\ (-3, 0) & -1 & \frac{(+)}{(-)(-)(+)}=(+) & T
\\ (0, 2) & 1 & \frac{(+)}{(+)(-)(+)}=(-) & F
\\ (2, \frac{9}{2}) & 3 & \frac{(+)}{(+)(+)(+)}=(+) & T
\\ (\frac{9}{2},\infty) & 10 & \frac{(-)}{(+)(+)(+)}=(-) & F
\\ \end{array}$
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(-3, 0 \right) \cup \left(2, \frac{9}{2}\right] $
