Answer
The inequality is valid for values more than 4 (not including it since it is restricted by the denominator) i.e. $ (4,\infty)$
Work Step by Step
First, we are going to move everything to the left side and simplify:
$\dfrac{x+2}{x-4}\geq1$
$\dfrac{x+2}{x-4}-1\geq0$
$\dfrac{x+2}{x-4}-\dfrac{x-4}{x-4}\geq0$
$\dfrac{x+2-(x-4)}{x-4}\geq0$
$\dfrac{6}{x-4}\geq0$
Now, we find critical points by equating the numerator and denominator to zero:
$6=0$
$x-4=0$
There is only one critical point:
$x=4$
Next, we are going to take two values: one less than 4 and one more than 4 to test in the original equation and check if the inequality is true or not:
First test with a value less than 4:
$\dfrac{3+2}{3-4}\geq1$
$\dfrac{5}{-1}\geq1$
$-5\geq1 \rightarrow \text{ FALSE}$
Second test with a value more than 4:
$\dfrac{5+2}{5-4}\geq1$
$\dfrac{7}{1}\geq1$
$7\geq0 \rightarrow \text{ TRUEE}$
These tests show that the inequality $\dfrac{x+2}{x-4}\geq1$ is valid for values more than 4 (not including it since it is restricted by the denominator) i.e. $ (4,\infty)$
