The problem asks us to solve the inequality $|\frac{x-2}{x+3}| < 4$.

1. Problem Description

The problem asks us to solve the inequality $|\frac{x-2}{x+3}| < 4$.

2. Solution Steps

We can rewrite the inequality as:

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

This can be split into two inequalities:

$\frac{x-2}{x+3} < 4$ and $\frac{x-2}{x+3} > -4$

Let's solve the first inequality: $\frac{x-2}{x+3} < 4$.

Subtract 4 from both sides: $\frac{x-2}{x+3} - 4 < 0$.

Combine the terms on the left: $\frac{x-2 - 4(x+3)}{x+3} < 0$.

Simplify the numerator: $\frac{x-2 - 4x - 12}{x+3} < 0$, which gives $\frac{-3x - 14}{x+3} < 0$.

Multiply both sides by $-1$ and reverse the inequality sign: $\frac{3x + 14}{x+3} > 0$.

Now we analyze the sign of the expression. The critical points are $x = -3$ and $x = -\frac{14}{3}$.

We consider three intervals:

- $x < -\frac{14}{3}$: Both $3x + 14$ and $x+3$ are negative, so the fraction is positive.

- $-\frac{14}{3} < x < -3$: $3x + 14$ is positive and $x+3$ is negative, so the fraction is negative.

- $x > -3$: Both $3x + 14$ and $x+3$ are positive, so the fraction is positive.

Therefore, the solution to the first inequality is $x < -\frac{14}{3}$ or $x > -3$.

Now let's solve the second inequality: $\frac{x-2}{x+3} > -4$.

Add 4 to both sides: $\frac{x-2}{x+3} + 4 > 0$.

Combine the terms on the left: $\frac{x-2 + 4(x+3)}{x+3} > 0$.

Simplify the numerator: $\frac{x-2 + 4x + 12}{x+3} > 0$, which gives $\frac{5x + 10}{x+3} > 0$.

Simplify: $\frac{5(x+2)}{x+3} > 0$, so $\frac{x+2}{x+3} > 0$.

The critical points are $x = -2$ and $x = -3$.

We consider three intervals:

- $x < -3$: Both $x+2$ and $x+3$ are negative, so the fraction is positive.

- $-3 < x < -2$: $x+2$ is negative and $x+3$ is positive, so the fraction is negative.

- $x > -2$: Both $x+2$ and $x+3$ are positive, so the fraction is positive.

Therefore, the solution to the second inequality is $x < -3$ or $x > -2$.

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

The first solution is $x < -\frac{14}{3}$ or $x > -3$, and the second solution is $x < -3$ or $x > -2$.

Since $-\frac{14}{3} \approx -4.67$, the intersection is $x < -\frac{14}{3}$ or $x > -2$.

In interval notation, this is $(-\infty, -\frac{14}{3}) \cup (-2, \infty)$.

3. Final Answer

$(-\infty, -\frac{14}{3}) \cup (-2, \infty)$