# How do you solve | 2x + 1 | <| 3x - 2 | ?

Mar 13, 2016

x in ]-oo, 1/5 [ uu ] 3, +oo[

#### Explanation:

First find the zeros of each expression in the modules:

$2 x + 1 = 0 \implies x \equiv - \frac{1}{2}$

$3 x - 2 = 0 \implies x \equiv \frac{2}{3}$

Now we must divide the inequation in three parts:

1) When x>=2/3, both expression are larger than 0, therefore:

$| 2 x + 1 | < | 3 x - 2 |$

is equivalent to:

$2 x + 1 < 3 x - 2 \mathmr{and} x \ge \frac{2}{3}$

$- x < - 3 \mathmr{and} x \ge \frac{2}{3}$

$x > 3 \mathmr{and} x \ge \frac{2}{3}$

$x > 3$

2)if $- \frac{1}{2} \le x < \frac{2}{3}$:

3x-2 will be <=0, therefore $| 3 x - 2 | = - 3 x + 2$:

$2 x + 1 < - 3 x + 2 \mathmr{and} - \frac{1}{2} \le x \le \frac{2}{3}$:

$5 x < 1 \mathmr{and} - \frac{1}{2} \le x \le \frac{2}{3}$:

$x < \frac{1}{5} \mathmr{and} - \frac{1}{2} \le x \le \frac{2}{3}$:

$- \frac{1}{2} \le x < \frac{1}{5}$

3)if $x \le - \frac{1}{2}$

both expressions are negative so we must get the inverse for both ones:

$- 2 x - 1 < - 3 x + 2 \mathmr{and} x \le - \frac{1}{2}$

$x < 3 \mathmr{and} x < - \frac{1}{2}$

$x \le - \frac{1}{2}$

Now the solution is the union of the three solutions:

x in ]-oo, 1/2]uu[1/2, 1/5 [ uu ] 3, +oo[

x in ]-oo, 1/5 [ uu ] 3, +oo[