We need to solve the following compound inequality: $\frac{7x}{2x+3} > -2$ and $\frac{3x}{2x+3} < 2$

1. Problem Description

We need to solve the following compound inequality:

$\frac{7x}{2x+3} > -2$ and $\frac{3x}{2x+3} < 2$

2. Solution Steps

Let's solve the first inequality:

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

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

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

$\frac{7x + 4x + 6}{2x+3} > 0$

$\frac{11x + 6}{2x+3} > 0$

The critical points are $x = -\frac{6}{11}$ and $x = -\frac{3}{2}$. We need to consider three intervals: $x < -\frac{3}{2}$, $-\frac{3}{2} < x < -\frac{6}{11}$, and $x > -\frac{6}{11}$.

If $x < -\frac{3}{2}$, both $11x+6$ and $2x+3$ are negative, so the fraction is positive.

If $-\frac{3}{2} < x < -\frac{6}{11}$, $11x+6$ is negative and $2x+3$ is positive, so the fraction is negative.

If $x > -\frac{6}{11}$, both $11x+6$ and $2x+3$ are positive, so the fraction is positive.

So, the solution for the first inequality is $x < -\frac{3}{2}$ or $x > -\frac{6}{11}$.

Now, let's solve the second inequality:

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

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

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

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

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

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

The critical points are $x = -6$ and $x = -\frac{3}{2}$. We need to consider three intervals: $x < -6$, $-6 < x < -\frac{3}{2}$, and $x > -\frac{3}{2}$.

If $x < -6$, both $x+6$ and $2x+3$ are negative, so the fraction is positive.

If $-6 < x < -\frac{3}{2}$, $x+6$ is positive and $2x+3$ is negative, so the fraction is negative.

If $x > -\frac{3}{2}$, both $x+6$ and $2x+3$ are positive, so the fraction is positive.

So, the solution for the second inequality is $x < -6$ or $x > -\frac{3}{2}$.

We need to find the intersection of the two solutions:

($x < -\frac{3}{2}$ or $x > -\frac{6}{11}$) and ($x < -6$ or $x > -\frac{3}{2}$)

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

3. Final Answer

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