How do you solve and graph the absolute value inequality abs(4x + 1)≤ 5?

Apr 8, 2015

If $\left(4 x + 1\right)$ is positive or zero ($\rightarrow x \ge - \frac{1}{4}$)
and
$\left(4 x + 1\right) \le 5$
then
$x \le 1$

If $\left(4 x + 1\right)$ is negative ($\rightarrow x < - \frac{1}{4}$)
then $\left\mid 4 x + 1 \right\mid \le 5$
is equivalent to
$- 4 x - 1 \le 5$
then
$x \ge - \frac{3}{2}$

The solution graph is the vertical band $x \epsilon \left[- \frac{3}{2} , + 1\right]$