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

1. Problem Description

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

2. Solution Steps

The inequality $|\frac{3x}{2x+3}| < 2$ can be rewritten as:

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

This is equivalent to the system of inequalities:

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

First, let's solve $\frac{3x}{2x+3} < 2$:

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

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

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

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

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

We analyze the sign of $\frac{x+6}{2x+3}$. The critical points are $x = -6$ 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 $\frac{x+6}{2x+3} > 0$.

Therefore, the solution to $\frac{x+6}{2x+3} > 0$ is $x < -6$ or $x > -\frac{3}{2}$.

Next, let's solve $\frac{3x}{2x+3} > -2$:

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

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

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

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

We analyze the sign of $\frac{7x+6}{2x+3}$. The critical points are $x = -\frac{6}{7}$ and $x = -\frac{3}{2}$.

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

Therefore, the solution to $\frac{7x+6}{2x+3} > 0$ is $x < -\frac{3}{2}$ or $x > -\frac{6}{7}$.

Now we need to find the intersection of the two solutions:

$(x < -6$ or $x > -\frac{3}{2})$ and $(x < -\frac{3}{2}$ or $x > -\frac{6}{7})$

The intersection is $x < -6$ or $x > -\frac{6}{7}$.

3. Final Answer

$x < -6$ or $x > -\frac{6}{7}$

In interval notation: $(-\infty, -6) \cup (-\frac{6}{7}, \infty)$