Answer
The inequality is valid for values less than -2 and values more than -1 (including them) i.e. $(-\infty,-2]\cap [-1,\infty)$.
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x-1}{x+2}\geq-2$
$\dfrac{x-1}{x+2}+2\geq0$
$\dfrac{x-1}{x+2}+\dfrac{2(x+2)}{x+2}\geq0$
$\dfrac{x-1+2x+4}{x+2}\geq0$
$\dfrac{3x+3}{x+2}\geq0$
Now, we find critical points by equating the numerator and denominator to zero:
$3x+3=0$
$x+2=0$
There are two critical points:
$x_1+2=0\rightarrow x_1=-2$
$3x_2+3=0\rightarrow x_2=-\frac{3}{3}=-1$
Next, we are going to take three values: one less than -2; one between -2 and -1; and one more than -1 to test in the original equation and check if the inequality is true or not:
First test with a value less than -2:
$\dfrac{-3-1}{-3+2}\geq-2$
$\dfrac{-4}{-1}\geq-2$
$4\geq-2 \rightarrow \text{ TRUE}$
Second test with a value between -2 and -1:
$\dfrac{-1.5-1}{-1.5+2}\geq-2$
$\dfrac{-2.5}{0.5}\geq-2$
$-5\geq-2 \rightarrow \text{ FALSE}$
Third test with a value more than -1:
$\dfrac{0-1}{0+2}\geq-2$
$\dfrac{-1}{2}\geq-2$
$0.5\geq-2 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x-1}{x+2}\geq-2$ is valid for values less than -2 and values more than -1 (including them) i.e. $(-\infty,-2]\cap [-1,\infty)$
