# How do solve (2x)/(x-2)<=3 algebraically?

Aug 6, 2018

Multiply both sides by $x - 2$

$2 x \le 3 \left(x - 2\right)$

Distribute the 3

$2 x \le 3 x - 6$

Move $x$ terms to one side

$- x \le - 6$

Divide out negative one from both sides

$x \ge 6$

*When dividing or multiplying by a negative, you have to switch the greater or less than sign.

Aug 7, 2018

$x \in \left(- 00 , 2\right) \cup \left[6 , \infty\right)$

#### Explanation:

$\text{subtract 3 from both sides}$

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

$\text{combine the left side as a single fraction}$

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

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

$\text{find the critical values of numerator/denominator}$

$6 - x = 0 \Rightarrow x = 6 \leftarrow \textcolor{b l u e}{\text{is a zero}}$

$x - 2 = 0 \Rightarrow x = 2$

$\text{these values divide the domain into 3 intervals}$

$\left(- \infty , 2\right) \cup \left(2 , 6\right] \cup \left[6 , \infty\right)$

$\text{select a value for x as a "color(red)"test point in each interval}$

$x = 1 \to \frac{6 - 1}{1 - 2} = - 5 < 0 \leftarrow \textcolor{b l u e}{\text{valid}}$

$x = 3 \to \frac{6 - 3}{3 - 2} = 3 > 0 \leftarrow \textcolor{b l u e}{\text{not valid}}$

$x = 10 \to \frac{6 - 10}{10 - 2} = - \frac{1}{2} < 0 \leftarrow \textcolor{b l u e}{\text{valid}}$

$x \in \left(- \infty , 2\right) \cup \left[6 , \infty\right)$
graph{(2x)/(x-2)-3 [-10, 10, -5, 5]}