\( \frac{x}{x-1}-\frac{2}{x+2}=\frac{4 x-1}{x^{2}+x-2} \) What are all the solutions for the given equation?

To solve the equation \( \frac{x}{x-1}-\frac{2}{x+2}=\frac{4x-1}{x^{2}+x-2} \), we start by factoring the denominator on the right side, which is \( x^2 + x - 2 = (x-1)(x+2) \). This allows us to rewrite the equation as: \[ \frac{x}{x-1} - \frac{2}{x+2} = \frac{4x - 1}{(x-1)(x+2)} \] Next, we find a common denominator, which is \( (x-1)(x+2) \). Rewriting the left side, we get: \[ \frac{x(x+2) - 2(x-1)}{(x-1)(x+2)} = \frac{x^2 + 2x - 2x + 2}{(x-1)(x+2)} = \frac{x^2 + 2}{(x-1)(x+2)} \] Thus, the equation simplifies to: \[ \frac{x^2 + 2}{(x-1)(x+2)} = \frac{4x - 1}{(x-1)(x+2)} \] Since the denominators are the same, we can equate the numerators: \[ x^2 + 2 = 4x - 1 \] Rearranging this gives: \[ x^2 - 4x + 3 = 0 \] Factoring the quadratic: \[ (x - 1)(x - 3) = 0 \] Setting each factor to zero gives the potential solutions: \[ x = 1 \quad \text{and} \quad x = 3 \] However, we must check if any solutions cause the original equation to have undefined values. The denominators \( x - 1 \) and \( x + 2 \) cannot equal zero, hence \( x \neq 1 \) and \( x \neq -2 \). Thus, the solution \( x = 1 \) is not valid, leaving us with: \[ \boxed{3} \] as the only solution to the equation.