# How do you solve abs(2x+9)-2x>=0?

Feb 24, 2017

$\left\mid 2 x + 9 \right\mid - 2 x \ge 0$ is true for all $x \in \mathbb{R}$

#### Explanation:

$\left\mid 2 x + 9 \right\mid - 2 x \ge 0 \to \left\mid 2 x + 9 \right\mid - \left(2 x + 9\right) + 9 \ge 0$

then, for $x \ne - \frac{9}{2}$

$1 - \frac{2 x + 9}{\left\mid 2 x + 9 \right\mid} + \frac{9}{\left\mid 2 x + 9 \right\mid} \ge 0$

Calling $\frac{9}{\left\mid 2 x + 9 \right\mid} = \epsilon \ge 0$ we have

$1 \pm 1 + \epsilon \ge 0$. This is always true so

$\left\mid 2 x + 9 \right\mid - 2 x \ge 0$ is true for all $x \in \mathbb{R}$