Answer
$x\leq-2\ x\geq\frac{3}{2}$
Interval: $(-\displaystyle \infty,-2]\cup[\frac{3}{2},\infty)$
Work Step by Step
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: $(-\displaystyle \infty,-2]\cup[\frac{3}{2},\infty)$
