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

1. Problem Description

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

2. Solution Steps

We need to solve the inequality $|\frac{3x}{2x+3}| < 2$. This inequality is equivalent to $-2 < \frac{3x}{2x+3} < 2$. We will solve this compound inequality in two parts: $\frac{3x}{2x+3} < 2$ and $\frac{3x}{2x+3} > -2$. Also, we must have $2x+3 \ne 0$, so $x \ne -\frac{3}{2}$.

First, let's solve $\frac{3x}{2x+3} < 2$. Subtracting 2 from both sides, we get $\frac{3x}{2x+3} - 2 < 0$. Combining the terms, we have $\frac{3x - 2(2x+3)}{2x+3} < 0$, which simplifies to $\frac{3x - 4x - 6}{2x+3} < 0$, or $\frac{-x-6}{2x+3} < 0$. Multiplying by -1, we have $\frac{x+6}{2x+3} > 0$. The critical points are $x = -6$ and $x = -\frac{3}{2}$.

We test the intervals:

$x < -6$: Choose $x = -7$. Then $\frac{-7+6}{2(-7)+3} = \frac{-1}{-11} = \frac{1}{11} > 0$. This interval satisfies the inequality.

$-6 < x < -\frac{3}{2}$: Choose $x = -2$. Then $\frac{-2+6}{2(-2)+3} = \frac{4}{-1} = -4 < 0$. This interval does not satisfy the inequality.

$x > -\frac{3}{2}$: Choose $x = 0$. Then $\frac{0+6}{2(0)+3} = \frac{6}{3} = 2 > 0$. This interval satisfies the inequality.

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

Now, let's solve $\frac{3x}{2x+3} > -2$. Adding 2 to both sides, we get $\frac{3x}{2x+3} + 2 > 0$. Combining the terms, we have $\frac{3x + 2(2x+3)}{2x+3} > 0$, which simplifies to $\frac{3x + 4x + 6}{2x+3} > 0$, or $\frac{7x+6}{2x+3} > 0$. The critical points are $x = -\frac{6}{7}$ and $x = -\frac{3}{2}$.

We test the intervals:

$x < -\frac{3}{2}$: Choose $x = -2$. Then $\frac{7(-2)+6}{2(-2)+3} = \frac{-8}{-1} = 8 > 0$. This interval satisfies the inequality.

$-\frac{3}{2} < x < -\frac{6}{7}$: Choose $x = -1$. Then $\frac{7(-1)+6}{2(-1)+3} = \frac{-1}{1} = -1 < 0$. This interval does not satisfy the inequality.

$x > -\frac{6}{7}$: Choose $x = 0$. Then $\frac{7(0)+6}{2(0)+3} = \frac{6}{3} = 2 > 0$. This interval satisfies the inequality.

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

We need to find the intersection of the solutions to $\frac{3x}{2x+3} < 2$ and $\frac{3x}{2x+3} > -2$. The solution to the first inequality is $x < -6$ or $x > -\frac{3}{2}$. The solution to the second inequality is $x < -\frac{3}{2}$ or $x > -\frac{6}{7}$. The intersection of these solutions is $x < -6$ or $x > -\frac{6}{7}$.

3. Final Answer

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