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To find the solution for the system of equations given: 1. \( y^2 - x = -1 \) 2. \( x = y + 3 \) First, substitute the expression for \( x \) from the second equation into the first equation: \[ y^2 - (y + 3) = -1 \] Simplify this equation: \[ y^2 - y - 3 = -1 \] Add 1 to both sides: \[ y^2 - y - 2 = 0 \] Now, factor the quadratic equation: \[ (y - 2)(y + 1) = 0 \] So, the solutions for \( y \) are: \[ y = 2 \quad \text{or} \quad y = -1 \] For \( y = 2 \): - Substitute back to find \( x \): \( x = 2 + 3 = 5 \) For \( y = -1 \): - Substitute back to find \( x \): \( x = -1 + 3 = 2 \) Therefore, the solutions are: - \( (x, y) = (5, 2) \) - \( (x, y) = (2, -1) \)