# How do you solve (3x)/(x + 2)<5?

Jun 18, 2017

The solution is $x \in \left(- \infty , - 5\right) \cup \left(- 2 , + \infty\right)$

#### Explanation:

We cannot do crossing over, we rearrange the inequality

$\frac{3 x}{x + 2} < 5$

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

$\frac{3 x - 5 \left(x + 2\right)}{x + 2} < 0$

$\frac{3 x - 5 x - 10}{x + 2} < 0$

$\frac{- 2 x - 10}{x + 2} < 0$

$\frac{- 2 \left(x + 5\right)}{x + 2} < 0$

Let $f \left(x\right) = \frac{- 2 \left(x + 5\right)}{x + 2}$

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 a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 5$$\textcolor{w h i t e}{a a a a a a a}$$- 2$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$- \left(x + 5\right)$$\textcolor{w h i t e}{a a a a a a}$$+$$\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}$$-$

$\textcolor{w h i t e}{a a a a}$$\left(x + 2\right)$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\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}$$+$

$\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 a a a a a}$$-$$\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}$$-$

Therefore,

$f \left(x\right) < 0$, when $x \in \left(- \infty , - 5\right) \cup \left(- 2 , + \infty\right)$
graph{(3x)/(x+2)-5 [-36.53, 36.52, -18.28, 18.27]}