Answer
$(-\infty,3)\cup[7,\infty)$.
Work Step by Step
Step 1. For $\frac{x+1}{x-3}-2\le0 \Longrightarrow \frac{-x+7}{x-3}\le0 $, identify boundary points as $x=3,7$.
Step 2. Form intervals $(-\infty,3),(3,7],[7,\infty)$.
Step 3. Choose test values for each interval $x=0,4,8$.
Step 4. Test the inequality to get results: $True,\ False,\ True$
Step 5. We have the solution $(-\infty,3)\cup[7,\infty)$.
