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

First, we subtract 2 from both sides of the inequality to get:

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

Next, we find a common denominator and combine the terms:

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

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

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

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

$\frac{x+6}{2x+3} > 0$ (Multiply both sides by -1 and flip the inequality sign).

Now, we find the critical points by setting the numerator and denominator equal to zero:

$x+6 = 0 \Rightarrow x = -6$

$2x+3 = 0 \Rightarrow x = -\frac{3}{2}$

We analyze the intervals determined by the critical points $x = -6$ and $x = -\frac{3}{2}$:

1. $x < -6$: Choose $x = -7$. Then $\frac{-7+6}{2(-7)+3} = \frac{-1}{-14+3} = \frac{-1}{-11} = \frac{1}{11} > 0$. So the inequality holds for $x < -6$.

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

3. $x > -\frac{3}{2}$: Choose $x = 0$. Then $\frac{0+6}{2(0)+3} = \frac{6}{3} = 2 > 0$. So the inequality holds for $x > -\frac{3}{2}$.

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

3. Final Answer

$x < -6$ or $x > -\frac{3}{2}$