Answer
$\left(-3, -\frac{1}{2}\right) \cup (2, \infty)$
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.
$\displaystyle \frac{x+2}{x+3}\lt\frac{x-1}{x-2}$,
$\displaystyle \frac{x+2}{x+3}-\frac{x-1}{x-2} \lt0$ ,
$\displaystyle \frac{-2x-1}{(x+3)(x-2)}\lt 0$ ,
$f(x)=\displaystyle \frac{-2x-1}{(x+3)(x-2)}\lt 0$,
2.The cut points are:
$\displaystyle \frac{-2x-1}{(x+3)(x-2)}=0$
$x=-3$ or $x=-\frac{1}{2}$ or $x=2$
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,
$\begin{array}{llll}
Intervals: & a=test.v. & f(a),signs & f(a) \lt 0 ? \\
& & \displaystyle \displaystyle \frac{-2a-1}{(a+3)(a-2)} & \\
(-\infty,-3) & -5 & \frac{(+)}{(-)(-)}=(+) & F\\
(-3, -\frac{1}{2}) & -2 & \frac{(+)}{(+)(-)}=(-) & T\\
(-\frac{1}{2}, 2) & 0 & \frac{(-)}{(+)(-)}=(+) & F\\
(2,\infty) & 10 & \frac{(-)}{(+)(+)}=(-) & T\\
\end{array}$
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: $\left(-3, -\frac{1}{2}\right) \cup (2, \infty)$
