# How do you solve abs(5x+2)>=abs(3x-4)?

Mar 21, 2018

$x \le - 3$ and $x \ge \frac{1}{4}$

#### Explanation:

This inequality is equivalent to

$\sqrt{{\left(5 x + 2\right)}^{2}} \ge \sqrt{{\left(3 x - 4\right)}^{2}}$ now squaring both sides

${\left(5 x + 2\right)}^{2} \ge {\left(3 x - 4\right)}^{2}$ or

${\left(5 x + 2\right)}^{2} - {\left(3 x - 4\right)}^{2} \ge 0$ or

$\left(5 x + 2 - 3 x + 4\right) \left(5 x + 2 + 3 x - 4\right) \ge 0$ or

$\left(2 x + 6\right) \left(8 x - 2\right) \ge 0$ or

$\left(x + 3\right) \left(x - \frac{1}{4}\right) \ge 0$ or

$x \le - 3$ and $x \ge \frac{1}{4}$