How do you solve abs(3x+12)>=42?

Dec 15, 2016

The solutions are $x \ge 10$ and $x \le - 18$

Explanation:

We have 2 equations to solve

$3 x + 12 \ge 42$ and $- 3 x - 12 \ge 42$

$3 x > \left(42 - 12\right)$ and $3 x \le \left(- 12 - 42\right)$

$3 x \ge 30$ and $3 x \le - 54$

$x \ge 10$ and $x \le - 18$