# How do you solve 3abs(4x-5)>=12?

May 23, 2015

Given $3 | 4 x - 5 | \ge 12$

divide both sides by $3$ to get:

$| 4 x - 5 | \ge 4$

Expanding the modulus, this means

$4 x - 5 \le - 4$ or $4 x - 5 \ge 4$

Add $5$ to both sides of both of these to get:

$4 x \le 1$ or $4 x \ge 9$

Divide both sides of both of these by $4$ to get:

$x \le \frac{1}{4}$ or $x \ge \frac{9}{4}$

In general, you can perform any of the following operations on an inequality and preserve its truth...

(1) Add or subtract the same value on both sides.
(2) Multiply or divide by the same positive value on both sides.
(3) Multiply or divide by the same negative value on both sides and reverse the inequality ( $<$ becomes $>$, $\ge$ becomes $\le$, etc.).