We are asked to solve the inequality $\frac{2}{|x-3|-1} + \frac{3}{|x-3|+2} \le 0$.

1. Problem Description

We are asked to solve the inequality $\frac{2}{|x-3|-1} + \frac{3}{|x-3|+2} \le 0$.

2. Solution Steps

Let $y = |x-3|$. Then the inequality becomes

$\frac{2}{y-1} + \frac{3}{y+2} \le 0$.

To solve this inequality, we find a common denominator:

$\frac{2(y+2) + 3(y-1)}{(y-1)(y+2)} \le 0$

$\frac{2y+4+3y-3}{(y-1)(y+2)} \le 0$

$\frac{5y+1}{(y-1)(y+2)} \le 0$

The critical points are $y = -\frac{1}{5}$, $y = 1$, and $y = -2$. Since $y = |x-3| \ge 0$, we can ignore $y = -2$. Thus we analyze the sign of the expression $\frac{5y+1}{(y-1)(y+2)}$ for $y \ge 0$.

If $0 \le y < \frac{-1}{5}$, then $5y+1<0, y-1 < 0, y+2 > 0$, and $\frac{5y+1}{(y-1)(y+2)} > 0$. Impossible.

If $\frac{-1}{5} \le y < 1$, then $5y+1 \ge 0, y-1 < 0, y+2 > 0$, and $\frac{5y+1}{(y-1)(y+2)} \le 0$. Thus we have $-\frac{1}{5} \le y < 1$. Since $y \ge 0$, we have $0 \le y < 1$.

If $y = 1$, then the first term in the original inequality is undefined.

If $y > 1$, then $5y+1 > 0, y-1 > 0, y+2 > 0$, and $\frac{5y+1}{(y-1)(y+2)} > 0$.

Therefore, the only possible solution is $0 \le y < 1$. Substituting $y = |x-3|$, we have $0 \le |x-3| < 1$. This gives us

$-1 < x-3 < 1$.

Adding 3 to all sides, we get

$2 < x < 4$.

2. Final Answer

$2 < x < 4$