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

Mar 24, 2018

$x \ge \frac{10}{3}$

#### Explanation:

$\frac{x}{x - 5} \ge - 2$
$x \ge - 2 \left(x - 5\right)$
$x \ge - 2 x + 10$
$3 x \ge 10$
$x \ge \frac{10}{3}$

Mar 24, 2018

$x \ge \frac{10}{3}$

#### Explanation:

$x \ge - 2 \left(x - 5\right)$ multiply both sides by $x - 5$ to get rid of fraction
$x \ge - 2 x + 10$ distribute
$3 x \ge 10$ add 2x to both sides
$x \ge \frac{10}{3}$

Mar 24, 2018

$x \le \frac{10}{3} \mathmr{and} x > 5$

#### Explanation:

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

We wanna find the critical points of the inequality

$\frac{x}{x - 5} = - 2$

Cross multiply

$x = \left(- 2\right) \left(x\right) + \left(- 2\right) \left(- 5\right)$

$x = - 2 x + 10$

Add $2 x$ on both sides

$x + 2 x = - 2 x + 10 + 2 x$

$3 x = 10$

Then divide both sides by $3$

$\frac{\cancel{3} x}{\cancel{3}} = \frac{10}{3}$

$x = \frac{10}{3}$

Don't forget we were looking for the critical points:

Critical points

$x = \frac{10}{3}$ which makes both sides equal

$x = 5$

Ok now we can check the intervals in between critical points.

We have $x \le \frac{10}{3}$ which works in the original inequality

We also have $\frac{10}{3} \le x < 5$ which doesn't work in the original
inequality

And we have $x > 5$ which works in the original inequality

Thus,

$x \le \frac{10}{3} \mathmr{and} x > 5$