# How do you solve |-6x + 3| ≥ 39?

Jan 17, 2017

$| - 6 x + 3 | \ge 39 \iff x \in \left(- \infty , - 6\right] \cup \left[7 , \infty\right)$

#### Explanation:

$\left(| x | = x \iff x \ge 0\right) \mathmr{and} \left(| x | = \left(- x\right) \iff x < 0\right)$

So, if $\left(- 6 x + 3\right) < 0$ then $| - 6 x + 3 | = 6 x - 3$

Then,

$6 x - 3 \ge 39$

$\iff$

$6 x \ge 42$

$\iff$

$x \ge 7$

Or, if $\left(- 6 x + 3\right) \ge 0$ then $- 6 x + 3 \ge 39$

$\iff$

$- 6 x \ge 36$

$\iff$

$x \le - 6$

Then $x \le - 6$ or $x \ge 7$

$\iff$ $x \in \left(- \infty , - 6\right] \cup \left[7 , \infty\right)$