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To solve the system of equations: 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 and solve for \( y \): \[ y^2 - y - 3 = -1 \] \[ y^2 - y - 3 + 1 = 0 \] \[ y^2 - y - 2 = 0 \] Factor the quadratic equation: \[ (y - 2)(y + 1) = 0 \] Set each factor equal to zero: 1. \( y - 2 = 0 \) → \( y = 2 \) 2. \( y + 1 = 0 \) → \( y = -1 \) Now find the corresponding \( x \) values using \( x = y + 3 \): - For \( y = 2 \), \( x = 2 + 3 = 5 \) - For \( y = -1 \), \( x = -1 + 3 = 2 \) The solutions are: - \( (x, y) = (5, 2) \) - \( (x, y) = (2, -1) \)