## Intermediate Algebra (6th Edition)

The solution is $(-\infty,-3]\cap(-\infty,0]$
$4x+2\le-10$ and $2x\le0$ Solve both inequalities separately: $4x+2\le-10$ Take $2$ to the right side: $4x\le-10-2$ $4x\le-12$ Take $4$ to divide the right side: $x\le-\dfrac{12}{4}$ $x\le-3$ The solution is $(-\infty,-3]$ $2x\le0$ Take $2$ to divide the right side: $x\le\dfrac{0}{2}$ $x\le0$ The solution is $(-\infty,0]$ Since the compound inequality has the word "and", the solution is the intersection of the two solution sets found. The solution is $(-\infty,-3]\cap(-\infty,0]$