We are asked to solve the inequality $|\frac{3x}{2x+3}| < 2$.

1. Problem Description

We are asked to solve the inequality $|\frac{3x}{2x+3}| < 2$.

2. Solution Steps

We have the inequality

$|\frac{3x}{2x+3}| < 2$.

This is equivalent to

$-2 < \frac{3x}{2x+3} < 2$.

We can split this into two inequalities:

$\frac{3x}{2x+3} < 2$ and $\frac{3x}{2x+3} > -2$.

First, consider $\frac{3x}{2x+3} < 2$.

$\frac{3x}{2x+3} - 2 < 0$

$\frac{3x - 2(2x+3)}{2x+3} < 0$

$\frac{3x - 4x - 6}{2x+3} < 0$

$\frac{-x-6}{2x+3} < 0$

$\frac{x+6}{2x+3} > 0$

We find the critical points $x = -6$ and $x = -\frac{3}{2}$.

We consider three intervals: $x < -6$, $-6 < x < -\frac{3}{2}$, and $x > -\frac{3}{2}$.

If $x < -6$, then $x+6 < 0$ and $2x+3 < 0$, so $\frac{x+6}{2x+3} > 0$.

If $-6 < x < -\frac{3}{2}$, then $x+6 > 0$ and $2x+3 < 0$, so $\frac{x+6}{2x+3} < 0$.

If $x > -\frac{3}{2}$, then $x+6 > 0$ and $2x+3 > 0$, so $\frac{x+6}{2x+3} > 0$.

Thus, $\frac{x+6}{2x+3} > 0$ when $x < -6$ or $x > -\frac{3}{2}$.

Now, consider $\frac{3x}{2x+3} > -2$.

$\frac{3x}{2x+3} + 2 > 0$

$\frac{3x + 2(2x+3)}{2x+3} > 0$

$\frac{3x + 4x + 6}{2x+3} > 0$

$\frac{7x+6}{2x+3} > 0$

We find the critical points $x = -\frac{6}{7}$ and $x = -\frac{3}{2}$.

We consider three intervals: $x < -\frac{3}{2}$, $-\frac{3}{2} < x < -\frac{6}{7}$, and $x > -\frac{6}{7}$.

If $x < -\frac{3}{2}$, then $7x+6 < 0$ and $2x+3 < 0$, so $\frac{7x+6}{2x+3} > 0$.

If $-\frac{3}{2} < x < -\frac{6}{7}$, then $7x+6 < 0$ and $2x+3 > 0$, so $\frac{7x+6}{2x+3} < 0$.

If $x > -\frac{6}{7}$, then $7x+6 > 0$ and $2x+3 > 0$, so $\frac{7x+6}{2x+3} > 0$.

Thus, $\frac{7x+6}{2x+3} > 0$ when $x < -\frac{3}{2}$ or $x > -\frac{6}{7}$.

We need both inequalities to be satisfied.

The first inequality is satisfied when $x < -6$ or $x > -\frac{3}{2}$.

The second inequality is satisfied when $x < -\frac{3}{2}$ or $x > -\frac{6}{7}$.

Therefore, we want the intersection of $(-\infty, -6) \cup (-\frac{3}{2}, \infty)$ and $(-\infty, -\frac{3}{2}) \cup (-\frac{6}{7}, \infty)$.

This is $(-\infty, -6) \cup (-\frac{6}{7}, \infty)$.

3. Final Answer

$x \in (-\infty, -6) \cup (-\frac{6}{7}, \infty)$