# How to solve complex absolute value inequalities like absx+ abs(x-2)< 5?

May 8, 2018

The solution is $x \in \left(- \frac{3}{2} , \frac{7}{2}\right)$

#### Explanation:

This is an inequality with absolute values.

$| x | + | x - 2 | < 5$

$| x | + | x - 2 | - 5 < 0$

Let $f \left(x\right) = | x | + | x - 2 | - 5$

$\left\{\begin{matrix}x \ge 0 \\ x - 2 \ge 0\end{matrix}\right.$, $\iff$, $\left\{\begin{matrix}x \ge 0 \\ x \ge 2\end{matrix}\right.$

Let's build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a a a}$$2$$\textcolor{w h i t e}{a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 2$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$| x |$$\textcolor{w h i t e}{a a a a a a a a a a}$$- x$$\textcolor{w h i t e}{a a a a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$x$

$\textcolor{w h i t e}{a a a a}$$| x - 2 |$$\textcolor{w h i t e}{a a a a a a}$$- x + 2$$\textcolor{w h i t e}{a a}$$- x + 2$$\textcolor{w h i t e}{a a a a}$$x - 2$

In the interval $\left(- \infty , 0\right)$

$- x - x + 2 - 5 < 0$, $\implies$, $2 x > - 3$, $\implies$, $x > - \frac{3}{2}$

$- \frac{3}{2} \in \left(- \infty , 0\right)$

In the interval $\left(0 , 2\right)$

$x - x + 2 - 5 < 0$, $\implies$, $0 - 3 < 0$, $\implies$, solution

In the interval $\left(2 , + \infty\right)$

$x + x - 2 - 5 < 0$, $\implies$, $2 x < 7$, $\implies$, $x < \frac{7}{2}$

$\frac{7}{2} \in \left(2 , + \infty\right)$

The solution is $x \in \left(- \frac{3}{2} , \frac{7}{2}\right)$

graph{|x|+|x-2|-5 [-10, 10, -5, 5]}