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

This is equivalent to $-2 < \frac{3x}{2x+3} < 2$.

We have two inequalities to solve:

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

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

Let's solve the first inequality:

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

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

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

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

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

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

The critical points are $x = -6$ and $x = -\frac{3}{2}$.

We consider the intervals $(-\infty, -6)$, $(-6, -\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$.

If $x<-6$, say $x=-7$, $\frac{-7+6}{2(-7)+3} = \frac{-1}{-11} = \frac{1}{11} > 0$.

If $-6 < x < -\frac{3}{2}$, say $x = -2$, $\frac{-2+6}{2(-2)+3} = \frac{4}{-1} = -4 < 0$.

If $x > -\frac{3}{2}$, say $x=0$, $\frac{0+6}{2(0)+3} = \frac{6}{3} = 2 > 0$.

So the solution to the first inequality is $x < -6$ or $x > -\frac{3}{2}$.

Now let's solve the second inequality:

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

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

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

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

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

The critical points are $x = -\frac{6}{7}$ and $x = -\frac{3}{2}$.

We consider the intervals $(-\infty, -\frac{3}{2})$, $(-\frac{3}{2}, -\frac{6}{7})$ and $(-\frac{6}{7}, \infty)$.

If $x < -\frac{3}{2}$, say $x = -2$, $\frac{7(-2)+6}{2(-2)+3} = \frac{-8}{-1} = 8 > 0$.

If $-\frac{3}{2} < x < -\frac{6}{7}$, say $x = -1$, $\frac{7(-1)+6}{2(-1)+3} = \frac{-1}{1} = -1 < 0$.

If $x > -\frac{6}{7}$, say $x = 0$, $\frac{7(0)+6}{2(0)+3} = \frac{6}{3} = 2 > 0$.

So the solution to the second inequality 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 $x > -\frac{6}{7}$.

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

3. Final Answer

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