# How do you solve abs(2/3 k - 1)>=2?

May 21, 2018

The solutions are $k \in \left(- \infty , - \frac{3}{2}\right] \cup \left[\frac{9}{2} , + \infty\right)$

#### Explanation:

This is an inequality with absolute values.

$| \frac{2}{3} k - 1 | \ge 2$

The solutions are obtained as follows :

$\left\{\begin{matrix}\frac{2}{3} k - 1 \ge 2 \\ - \frac{2}{3} k + 1 \ge 2\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}\frac{2}{3} k \ge 3 \\ \frac{2}{3} k \le - 1\end{matrix}\right.$

$\iff$, $\left\{\begin{matrix}k \ge \frac{9}{2} \\ k \le - \frac{3}{2}\end{matrix}\right.$

The solutions are

$k \in \left(- \infty , - \frac{3}{2}\right] \cup \left[\frac{9}{2} , + \infty\right)$

graph{|2/3x-1|-2 [-6.22, 9.58, -3.73, 4.17]}