Answer
The inequality is valid for values only between -2 and 9 (not including -2 since it is restricted by the denominator, but including 9) i.e. $(-2,9]$.
Work Step by Step
First, we are going to move everything to the left side to simplify:
$\dfrac{3x-5}{x+2}\leq2$
$\dfrac{3x-5}{x+2}-2\leq0$
$\dfrac{3x-5}{x+2}-\dfrac{2(x+2)}{x+2}\leq0$
$\dfrac{3x-5-(2x+4)}{x+2}\leq0$
$\dfrac{x-9}{x+2}\leq0$
Now, we find critical points by equating the numerator and denominator to zero:
$x-9=0$
$x+2=0$
There are two critical points:
$x_1+2=0\rightarrow x_1=-2$
$x-9=0\rightarrow x_2=9$
Next, we are going to take three values: one less than -2; one between -2 and 9; and one more than 9 to test in the original equation and check if the inequality is true or not:
First test with a value less than -2:
$\dfrac{3(-3)-5}{-3+2}\leq2$
$\dfrac{-9-5}{-1}\leq2$
$\dfrac{-14}{-1}\leq2$
$14\leq2 \rightarrow \text{ FALSE}$
Second test with a value between -2 and 9:
$\dfrac{3(-1)-5}{-1+2}\leq2$ $\dfrac{-3-5}{1}\leq2$
$\dfrac{-8}{1}\leq2$
$-8\leq2 \rightarrow \text{ TRUE}$
Third test with a value more than 9:
$\dfrac{3(10)-5}{10+2}\leq2$
$\dfrac{30-5}{12}\leq2$
$\dfrac{25}{12}\leq2$
$\dfrac{25}{12}\leq2 \rightarrow \text{ FALSE}$
These tests show that the inequality $\dfrac{3x-5}{x+2}\leq2$ is valid for values only between -2 and 9 (not including -2 since it is restricted by the denominator, but including 9) i.e. $(-2,9]$
