Decide what values of the variable cannot possibly be solutions for the equation. Do not solve. \( \frac{1}{x-1}+\frac{1}{x+2}=\frac{1}{x^{2}+x-2} \)

To determine the values of the variable \( x \) that cannot possibly be solutions for the equation \[ \frac{1}{x-1}+\frac{1}{x+2}=\frac{1}{x^{2}+x-2}, \] we need to identify the values that would make any of the denominators equal to zero, as division by zero is undefined. 1. **Identify the denominators:** - The left side has two denominators: \( x - 1 \) and \( x + 2 \). - The right side has the denominator \( x^2 + x - 2 \). 2. **Set each denominator to zero and solve for \( x \):** - For \( x - 1 = 0 \): \[ x = 1 \] - For \( x + 2 = 0 \): \[ x = -2 \] - For \( x^2 + x - 2 = 0 \): We can factor this quadratic equation: \[ x^2 + x - 2 = (x - 1)(x + 2) = 0 \] This gives us the same solutions: \[ x = 1 \quad \text{and} \quad x = -2 \] 3. **Conclusion:** The values of \( x \) that cannot possibly be solutions for the equation are: \[ x = 1 \quad \text{and} \quad x = -2. \]