Answer
The inequality is valid for x-values less than 3 and x-values more than 8 (not including them) i.e. $(-\infty,3)\cap (8,\infty)$
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x+2}{x-3}<2$
$\dfrac{x+2}{x-3}-2<0$
$\dfrac{x+2}{x-3}-\dfrac{2(x-3)}{x-3}<0$
$\dfrac{x+2-(2x-6)}{x-3}<0$
$\dfrac{-x+8}{x-3}\leq0$
Now, we find critical points by equating the numerator and denominator to zero:
$-x+8=0$
$x-3=0$
There are two critical points:
$x_1=3$
$x_2=8$
Next, we are going to take three values: one less than 3; one between 3 and 8; and one more than 8 to test in the original equation and check if the inequality is true or not:
First test with a value less than 3:
$\dfrac{2+2}{2-3}<2$
$\dfrac{4}{-1}<2$
$-4<2 \rightarrow \text{ TRUE}$
Second test with a value between 3 and 8:
$\dfrac{4+2}{4-3}<2$
$\dfrac{6}{1}<2$
$6<2 \rightarrow \text{ FALSE}$
Third test with a value more than 8:
$\dfrac{13+2}{13-3}<2$
$\dfrac{15}{10}<2$
$1.5<2 \rightarrow \text{ TRUE}$
These tests show that the inequality $\dfrac{x+2}{x-3}<2$ is valid for values less than 3 and values more than 8 (not including them) i.e. $(-\infty,3)\cap (8,\infty)$
