You have accurately solved the system of equations \( x = y + 3 \) and \( y^2 - x + 1 = 0 \). Here’s a concise summary of the steps you took and the final results:
Substitution: You substituted \( x \) from the first equation into the second equation, leading to: \[ y^2 - (y + 3) + 1 = 0 \]
Simplification: After simplifying, you obtained the quadratic equation: \[ y^2 - y - 2 = 0 \]
Factoring: You factored this equation as: \[ (y - 2)(y + 1) = 0 \]
Finding \( y \): Solving the factors gives: \[ y = 2 \quad \text{and} \quad y = -1 \]
Finding \( x \): Using these \( y \) values to find \( x \):
- For \( y = 2 \): \[ x = 5 \]
- For \( y = -1 \): \[ x = 2 \]
Results: You found the solution pairs:
- (5, 2)
- (2, -1)
The final solution set for the system of equations is indeed: \[ (5, 2) \quad \text{and} \quad (2, -1) \]
If you have any further questions or need clarification on specific parts, feel free to ask!