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

Locate the zeros of each expression in the modules.
${\displaystyle 2x+1=0\Rightarrow x\equiv -\frac{1}{2}}$
${\displaystyle 3x-2=0\Rightarrow x\equiv \frac{2}{3}}$
The equation must now be divided into three parts:
1) When $x\ge 2/3$, both expression are larger than 0, hence:
${\displaystyle |2x+1|<|3x-2|}$
is equivalent to:
${\displaystyle 2x+1<3x-2\text{and}x\ge \frac{2}{3}}$
${\displaystyle -x<-3\text{and}x\ge \frac{2}{3}}$
${\displaystyle x>3\text{and}x\ge \frac{2}{3}}$
${\displaystyle x>3}$
2)if ${\displaystyle -\frac{1}{2}\le x<\frac{2}{3}}$:
3x-2 will be <=0, therefore ${\displaystyle |3x-2|=-3x+2}$:
${\displaystyle 2x+1<-3x+2\text{and}-\frac{1}{2}\le x\le \frac{2}{3}}$:
${\displaystyle 5x<1\text{and}-\frac{1}{2}\le x\le \frac{2}{3}}$:
${\displaystyle x<\frac{1}{5}\text{and}-\frac{1}{2}\le x\le \frac{2}{3}}$:
${\displaystyle -\frac{1}{2}\le x<\frac{1}{5}}$
3)if ${\displaystyle x\le -\frac{1}{2}}$
Given that both statements are negative, we should find their inverses.
${\displaystyle -2x-1<-3x+2\text{and}x\le -\frac{1}{2}}$
${\displaystyle x<3\text{and}x<-\frac{1}{2}}$
${\displaystyle x\le -\frac{1}{2}}$
Now the solution is the union of the three solutions:
${\displaystyle x\in ]-\infty ,\frac{1}{2}]\cup [\frac{1}{2},\frac{1}{5}[\cup ]3,+\infty [}$
${\displaystyle x\in ]-\infty ,\frac{1}{5}[\cup ]3,+\infty [}$