## College Algebra (10th Edition)

$x\leq-2\ x\geq\frac{3}{2}$ Interval: $(-\displaystyle \infty,-2]\cup[\frac{3}{2},\infty)$
We solve: $|4x+1|\geq 7$ $4x+1\leq-7$ or $4x+1\geq 7$ $4x\leq-7-1$ or $4x\geq 7-1$ $4x\leq-8$ or $4x\geq 6$ $x\leq-2\ x\geq\frac{3}{2}$ Interval: $(-\displaystyle \infty,-2]\cup[\frac{3}{2},\infty)$