Answer
The inequality $\dfrac{x+4}{x-2}\leq1$ is valid for values only less than 2 (not including it) i.e. $(-\infty,2)$
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x+4}{x-2}\leq1$
$\dfrac{x+4}{x-2}-1\leq0$
$\dfrac{x+4}{x-2}-\dfrac{x-2}{x-2}\leq0$
$\dfrac{x+4-(x-2)}{x-2}\leq0$
$\dfrac{6}{x-2}\leq0$
Now, we find critical points by equating the numerator and denominator to zero:
$6=0$
$x-2=0$
There is only one critical point:
$x=2$
Next, we are going to take two values: one less than 2 and one more than 2 to test in the original equation and check if the inequality is true or not:
First test with a value less than 2:
$\dfrac{0+4}{0-2}\leq1$
$\dfrac{4}{-2}\leq1$
$-2\leq1 \rightarrow \text{ TRUE}$
Second test with a value more than 2:
$\dfrac{3+4}{3-2}\leq1$
$\dfrac{7}{1}\leq1$
$7\leq1 \rightarrow \text{ FALSE}$
These tests show that the inequality $\dfrac{x+4}{x-2}\leq1$ is valid for values only less than 2 (not including it) i.e. $(-\infty,2)$
