How do you solve x/(x+3)>=2?

Dec 27, 2016

The answer is x in [-6, -3[

Explanation:

We cannot do crossing over.

We rewrite the inequality as $\frac{x}{x + 3} - 2 \ge 0$

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

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

Let $f \left(x\right) = - \frac{x + 6}{x + 3}$

We can make a time chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- 6$$\textcolor{w h i t e}{a a a a a}$$- 3$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$- x - 6$$\textcolor{w h i t e}{a a a a}$$+$$\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 a a}$$x + 3$$\textcolor{w h i t e}{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}$∣∣$\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 a}$$-$$\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}$$-$

Therefore,

$f \left(x\right) \ge 0$, when x in [-6, -3[
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