## College Algebra (10th Edition)

The solution is $[-6,1]$.
$|2u+5|\leq7$ Isolate the absolute value $-7\le2u+5\leq7$ Add 2 to each term $-7-5\le2u+5-5\leq7-5$ $-12\leq2u\leq2$ Divide each term by 3 $\frac{-12}{2}\leq\frac{2u}{2}\leq\frac{2}{2}$ $-6\leq{u}\leq1$ The solution is $[-6,1]$.