# How do you solve (2x)/(x-5)>=3?

Aug 4, 2018

The solution is $x \in \left(5 , 15\right]$

#### Explanation:

You cannot do crossing over in inequalities

$\frac{2 x}{x - 5} \ge 3$

$\iff$, $\frac{2 x}{x - 5} - 3 \ge 0$

$\iff$, $\frac{2 x - 3 \left(x - 5\right)}{x - 5} \ge 0$

$\iff$, $\frac{2 x - 3 x + 15}{x - 5} \ge 0$

$\iff$, $\frac{- x + 15}{x - 5} \ge 0$

Let $f \left(x\right) = \frac{- x + 15}{x - 5}$

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 a a}$$5$$\textcolor{w h i t e}{a a a a a a a}$$15$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x - 5$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$- x + 15$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a a a}$color(white)(aa)+$\textcolor{w h i t e}{a 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 a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a}$$+$$\textcolor{w h i t e}{a a a}$$0$$\textcolor{w h i t e}{a a}$$-$

Therefore,

$f \left(x\right) \ge 0$ when $x \in \left(5 , 15\right]$

graph{(2x)/(x-5)-3 [-25.56, 32.17, -14.65, 14.22]}