# How do you solve abs((x+1)/x)>2 using a sign chart?

Jul 9, 2018

The solution is $x \in \left(- \frac{1}{3} , 0\right) \cup \left(0 , 1\right)$

#### Explanation:

Start by solving

$\frac{x + 1}{x} = 0$

$\implies$, $x = - 1$ and $x \ne 0$

Let $g \left(x\right) = | \frac{x + 1}{x} |$

The sign chart is as follows

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a a}$$0$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$g \left(x\right)$$\textcolor{w h i t e}{a a a a}$$- \frac{x + 1}{x}$$\textcolor{w h i t e}{a a a a}$$\frac{x + 1}{x}$$\textcolor{w h i t e}{a a a a}$$\frac{x + 1}{x}$$\textcolor{w h i t e}{a a a a}$

In the interval ${I}_{1} = \left(- \infty , - 1\right)$

$- \frac{x + 1}{x} - 2 > 0$

$\implies$, $\frac{- x - 1 - 2 x}{x} > 0$

$\implies$, $\frac{- 3 x - 1}{x} > 0$

$\implies$, x>1/3

In the interval ${I}_{2} = \left(1 , + \infty\right)$

$\frac{x + 1}{x} - 2 > 0$

$\implies$, $\frac{x + 1 - 2 x}{x} > 0$

$\implies$, $\frac{1 - x}{x} > 0$

$\implies$, $x < 1$

Therefore ,

The solution is $x \in \left(- \frac{1}{3} , 0\right) \cup \left(0 , 1\right)$

graph{|(x+1)/x|-2 [-10, 10, -5, 5]}