We need to solve the following inequality: $\frac{2}{|x-3|-1} + \frac{3}{|x-3|+2} \leq 0$

1. Problem Description

We need to solve the following inequality:

$\frac{2}{|x-3|-1} + \frac{3}{|x-3|+2} \leq 0$

2. Solution Steps

Let $u = |x-3|$. The inequality becomes:

$\frac{2}{u-1} + \frac{3}{u+2} \leq 0$

Combine the fractions:

$\frac{2(u+2) + 3(u-1)}{(u-1)(u+2)} \leq 0$

$\frac{2u+4 + 3u-3}{(u-1)(u+2)} \leq 0$

$\frac{5u+1}{(u-1)(u+2)} \leq 0$

The critical points are $u = -\frac{1}{5}$, $u = 1$, and $u = -2$. Since $u = |x-3| \geq 0$, we can ignore $u=-2$.

We consider the intervals $(-\infty, -\frac{1}{5}]$, $[-\frac{1}{5}, 1]$, and $[1, \infty)$. However, since $u \geq 0$, we only need to check the intervals $[0, -\frac{1}{5}]$, $[-\frac{1}{5}, 1]$ and $[1, \infty)$.

The interval $[0, -\frac{1}{5}]$ is an empty set.

So, we have intervals $[0, 1)$ and $(1, \infty)$.

Check the sign of $\frac{5u+1}{(u-1)(u+2)}$ on the intervals $[0, 1)$ and $(1, \infty)$:

If $0 \leq u < 1$, let $u=0$. Then $\frac{5(0)+1}{(0-1)(0+2)} = \frac{1}{(-1)(2)} = -\frac{1}{2} \leq 0$. This is true. Also, we have $5u+1 \geq 0$ and since we want $\frac{5u+1}{(u-1)(u+2)} \leq 0$, then $(u-1)(u+2) < 0$.

Since $u+2>0$, we need $u-1<0$ or $u<1$. Since $u \geq 0$, $0 \leq u < 1$ is valid.

If $u=1$, the expression is undefined.

If $u > 1$, let $u=2$. Then $\frac{5(2)+1}{(2-1)(2+2)} = \frac{11}{(1)(4)} = \frac{11}{4} > 0$.

If $u = -\frac{1}{5}$, $\frac{5u+1}{(u-1)(u+2)} = 0$.

So, we have $0 \leq u \leq -\frac{1}{5}$ or $0 \leq u < 1$. Since $|x-3| \geq 0$ for all real x.

$0 \leq |x-3| < 1$

$-1 < x-3 < 1$

$2 < x < 4$

$|x-3| = -\frac{1}{5}$

$x-3 = \frac{1}{5}$ or $x-3 = -\frac{1}{5}$

$x = \frac{16}{5}$ or $x = \frac{14}{5}$

$2 < x < 4$, $x= \frac{16}{5} = 3.2$, $x= \frac{14}{5} = 2.8$

Then $2 < x \leq \frac{14}{5} \cup \frac{16}{5} < x < 4$ or $(\frac{14}{5} \leq x < 4)$.

So the solution to the inequality is $2 < x < 4$.

$2 < x < 4$ is valid because if $x=3$, then $\frac{2}{0} + \frac{3}{5}$ is undefined. Thus, we have $x\neq 3$.

Thus, $2 < x < 4$.

$0 \leq |x-3| < 1$, so $-1 < x-3 < 1$, and thus $2 < x < 4$.

Or $5|x-3| + 1 = 0$

$|x-3| = -\frac{1}{5}$, but absolute value cannot be negative.

3. Final Answer

$2 < x < 4$