X = y + 3, y²- x + 1 = 0.

To solve the system of equations given by \( x = y + 3 \) and \( y^2 - x + 1 = 0 \), we can follow these steps: 1. **Substitute** \( x \) from the first equation into the second equation. Given: \[ x = y + 3 \] Substitute \( x \) into the second equation: \[ y^2 - (y + 3) + 1 = 0 \] 2. **Simplify the equation**: \[ y^2 - y - 3 + 1 = 0 \] \[ y^2 - y - 2 = 0 \] 3. **Factor the quadratic equation**: We can factor the quadratic as: \[ (y - 2)(y + 1) = 0 \] 4. **Solve for \( y \)**: This gives us two solutions: \[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \] \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] 5. **Find the corresponding \( x \) values**: Now we need to find the \( x \) values corresponding to each \( y \) value using \( x = y + 3 \): - For \( y = 2 \): \[ x = 2 + 3 = 5 \] - For \( y = -1 \): \[ x = -1 + 3 = 2 \] 6. **Results**: The solutions to the system of equations are: - \( (x, y) = (5, 2) \) - \( (x, y) = (2, -1) \) Thus, the final solution set is: \[ (5, 2) \quad \text{and} \quad (2, -1) \]