# How do you solve 3\leq \frac { 2- x } { x + 2}?

Aug 28, 2017

The solution is $x \in \left(- 2 , - 1\right]$

#### Explanation:

We rewrite the inequality as

$3 \le \frac{2 - x}{x + 2}$

$3 - \frac{2 - x}{x + 2} \le 0$

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

$\frac{3 x + 6 - 2 + x}{x + 2} \le 0$

$\frac{4 x + 4}{x + 2} \le 0$

$\frac{4 \left(x + 1\right)}{x + 2} \le 0$

Let $f \left(x\right) = \frac{4 \left(x + 1\right)}{x + 2}$

We can build the 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}$$- 2$$\textcolor{w h i t e}{a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a}$$+ \infty$

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

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

Therefore,

$f \left(x\right) \le 0$ when $x \in \left(- 2 , - 1\right]$

graph{3-(2-x)/(x+2) [-28.87, 28.86, -14.43, 14.44]}