We are given the following equation and asked to solve for $x$: $$ \frac{3}{x-m} - \frac{2}{x+m} = \frac{3x-7m}{x^2 - m^2} $$

1. Problem Description

We are given the following equation and asked to solve for $x$:

$$\frac{3}{x-m} - \frac{2}{x+m} = \frac{3x-7m}{x^2 - m^2}$$

2. Solution Steps

First, we note that $x^2 - m^2 = (x-m)(x+m)$. Thus, the common denominator for the left-hand side is also $x^2 - m^2$.

We rewrite the left-hand side with the common denominator:

$$\frac{3(x+m)}{(x-m)(x+m)} - \frac{2(x-m)}{(x+m)(x-m)} = \frac{3x+3m - (2x - 2m)}{x^2 - m^2}$$

$$\frac{3x + 3m - 2x + 2m}{x^2 - m^2} = \frac{x + 5m}{x^2 - m^2}$$

Now, we have the equation:

$$\frac{x + 5m}{x^2 - m^2} = \frac{3x - 7m}{x^2 - m^2}$$

If $x^2 - m^2 \neq 0$, then we can multiply both sides by $x^2 - m^2$, giving us:

$$x + 5m = 3x - 7m$$

Now we solve for $x$:

$$12m = 2x$$

$$x = 6m$$

Since we multiplied by $x^2 - m^2$, we need to check if $x=6m$ makes the denominator $0$.

$x^2 - m^2 = (6m)^2 - m^2 = 36m^2 - m^2 = 35m^2$.

If $m=0$, then $x=0$, but the original equation becomes $\frac{3}{x} - \frac{2}{x} = \frac{3x}{x^2}$, which simplifies to $\frac{1}{x} = \frac{3}{x}$. This implies $1=3$, which is false. Therefore, $m \neq 0$.

Thus, if $m \neq 0$, then $x = 6m$ is a solution.

3. Final Answer

$x = 6m$