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: \[ (5, 2) \quad \text{and} \quad (2, -1) \]