# Question d7eb4

Dec 4, 2017

The solution is $x \in \left(- \infty , - \frac{2}{3}\right] \cup \left(2 , + \infty\right)$

#### Explanation:

Let $f \left(x\right) = \frac{3 x + 2}{2 x - 4}$

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}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- \frac{2}{3}$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaaa)2$\textcolor{w h i t e}{a a a a}$color(white)(aaaa)+oo

$\textcolor{w h i t e}{a a a a}$$3 x + 2$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a}$color(white)(aa)+$\textcolor{w h i t e}{a a a a}$color(white)(aaaaaaa)+

$\textcolor{w h i t e}{a a a a}$$2 x - 4$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$color(white)(aaaaa)-$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$color(white)(aa)+

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$0$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a a a}$color(white)(aa)+#

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(- \infty , - \frac{2}{3}\right] \cup \left(2 , + \infty\right)$

graph{(3x+2)/(2x-4) [-18.02, 18.03, -9.01, 9.01]}