# How do you solve 1/3 |1/2x-2| >= 2?

Jul 18, 2018

The solution is $x \in \left(- \infty , - 8\right] \cup \left[16 , + \infty\right)$

#### Explanation:

The point to consider is

$\frac{1}{2} x - 2 = 0$

$\implies$, $\frac{1}{2} x = 2$

$\implies$, $x = 4$

There are $2$ intervals to consider

${I}_{1} = \left(- \infty , 4\right)$ and ${I}_{2} = \left(4 , + \infty\right)$

In the first interval

$\frac{1}{3} \left(- \frac{1}{2} x + 2\right) - 2 \ge 0$

$- \frac{1}{6} x + \frac{2}{3} - 2 \ge 0$

$\frac{1}{6} x \le - \frac{4}{3}$

$x \le - 8$

This solution $\in {I}_{1}$

In the second interval

$\frac{1}{3} \left(\frac{1}{2} x - 2\right) - 2 \ge 0$

$\frac{1}{6} x - \frac{2}{3} - 2 \ge 0$

$\frac{1}{6} x \ge \frac{8}{3}$

$x \ge 16$

This solution $\in {I}_{2}$

The solution is $x \in \left(- \infty , - 8\right] \cup \left[16 , + \infty\right)$

graph{1/3|1/2x-2|-2 [-13.97, 18.06, -6.42, 9.6]}