# How do you solve abs(( 2 - 5x) / 4 ) >= 2/3?

May 26, 2015

One way to approach this kind of problem is to split it into two cases:
one when the argument of the absolute value is $< 0$
and the other when the argument of the absolute value is $\ge 0$

Given $\left\mid \frac{2 - 5 x}{4} \right\mid \ge \frac{2}{3}$

Case 1: $\frac{2 - 5 x}{4} < 0$
Note that this implies $x > \frac{2}{5}$

$\left\mid \frac{2 - 5 x}{4} \right\mid \ge \frac{2}{3}$
becomes equivalent to
$\frac{5 x - 2}{4} \ge \frac{2}{3}$

$5 x - 2 \ge \frac{8}{3}$

$5 x \ge \frac{14}{3}$

$x \ge \frac{14}{15}$
Note that if $x \ge \frac{14}{15}$ the restriction $x > \frac{2}{5}$ becomes irrelevant.

Case 2: $\frac{2 - 5 x}{4} \ge 0$
Note this implies $x \le \frac{2}{5}$

$\left\mid \frac{2 - 5 x}{4} \right\mid \ge \frac{2}{3}$
becomes equivalent to
$\frac{2 - 5 x}{4} \ge \frac{2}{3}$

$2 - 5 x \ge \frac{8}{3}$

$2 \ge \frac{8}{3} + 5 x$

$- \frac{2}{3} \ge 5 x$

$- \frac{10}{3} \ge x$
Again, note that $x \le - \frac{10}{3}$ implies $x \le \frac{2}{5}$ so the restriction that $x \le \frac{2}{5}$ is irrelevant.

Combining Case 1 and Case 2:
$\left\mid \frac{2 - 5 x}{4} \right\mid \ge \frac{2}{3}$
if
$x \ge \frac{14}{15}$ or $x \le - \frac{10}{3}$