Answer
$(-\infty, -\frac{1}{2})\cup[6,\infty)$
Work Step by Step
Step 1. Rewrite the inequality as $\frac{x+7}{2x+1}-1\leq0\longrightarrow \frac{x+7-2x-1}{2x+1}\leq0\longrightarrow \frac{-x+6}{2x+1}\leq0\longrightarrow \frac{x-6}{2x+1}\geq0$
Step 2. Identify the boundary points $x=-\frac{1}{2}, 6$ and separate the number line into intervals $(-\infty, -\frac{1}{2})$, $(-\frac{1}{2},6)$, and $(6,\infty)$
Step 3. Use test points $x=-1,0,7$ to get the signs of the left side of the inequality as $+,-,+$
Step 4. Based on the signs and consider the boundary points for the equal sign, we have the solution as $(-\infty, -\frac{1}{2})\cup[6,\infty)$
