Answer
The solution set is $(-\infty, -5]$.
Work Step by Step
Multiply 4 to both sides to have:
$\\4(x+2) \le 4\cdot \frac{1}{4}(x-7)
\\4x+8 \le 1(x-7)
\\4x+8 \le x-7$
Subtract 8 to both sides to have:
$\\4x+8-8 \le x-7-8
\\4x \le x -15$
Subtract $x$ to both sides to have:
$\\4x-x \le x-15-x
\\3x \le -15$
Divide 3 to both sides to have:
$\\x \le -\frac{15}{3}
\\x \le -5$
