How to solve | 2 x + 1 | &lt; | 3 x - 2 | ?

Question

How to solve             |      2      x      +      1      |        &amp;lt;          |      3      x      -      2      |      ?

Jaime Lutz · Accepted Answer

Locate the zeros of each expression in the modules.

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

3 x - 2 = 0 \Rightarrow x \equiv \frac{2}{3}

The equation must now be divided into three parts:
1) When

x \geq 2 / 3

, both expression are larger than 0, hence:

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

is equivalent to:

2 x + 1 < 3 x - 2 and x \geq \frac{2}{3}

- x < - 3 and x \geq \frac{2}{3}

x > 3 and x \geq \frac{2}{3}

x > 3

2)if

- \frac{1}{2} \leq x < \frac{2}{3}

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

| 3 x - 2 | = - 3 x + 2

:

2 x + 1 < - 3 x + 2 and - \frac{1}{2} \leq x \leq \frac{2}{3}

:

5 x < 1 and - \frac{1}{2} \leq x \leq \frac{2}{3}

:

x < \frac{1}{5} and - \frac{1}{2} \leq x \leq \frac{2}{3}

:

- \frac{1}{2} \leq x < \frac{1}{5}

3)if

x \leq - \frac{1}{2}

Given that both statements are negative, we should find their inverses.

- 2 x - 1 < - 3 x + 2 and x \leq - \frac{1}{2}

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

x \leq - \frac{1}{2}

Now the solution is the union of the three solutions:

x \in] - \infty, \frac{1}{2}] \cup [\frac{1}{2}, \frac{1}{5} [\cup] 3, + \infty [

x \in] - \infty, \frac{1}{5} [\cup] 3, + \infty [

How to solve |2x+1|<|3x-2|?