The problem is to solve the following equation for $x$: $\frac{2x-1}{x^2-1} = \frac{2}{x+2}$

1. Problem Description

The problem is to solve the following equation for $x$:

$\frac{2x-1}{x^2-1} = \frac{2}{x+2}$

2. Solution Steps

First, we factor the denominator on the left side of the equation.

$x^2 - 1 = (x-1)(x+1)$

So the equation becomes:

$\frac{2x-1}{(x-1)(x+1)} = \frac{2}{x+2}$

Next, we cross-multiply:

$(2x-1)(x+2) = 2(x-1)(x+1)$

Expanding both sides gives:

$2x^2 + 4x - x - 2 = 2(x^2 - 1)$

$2x^2 + 3x - 2 = 2x^2 - 2$

Subtract $2x^2$ from both sides:

$3x - 2 = -2$

Add 2 to both sides:

$3x = 0$

Divide both sides by 3:

$x = 0$

Now we need to check if this solution is valid. We need to make sure that the denominators in the original equation are not zero.

For the left side, we have $x^2 - 1$. If $x=0$, $x^2 - 1 = 0^2 - 1 = -1$, which is not zero.

For the right side, we have $x+2$. If $x=0$, $x+2 = 0+2 = 2$, which is not zero.

Thus, $x=0$ is a valid solution.

3. Final Answer

$x = 0$