.

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:

• \( (x, y) = (5, 2) \)
• \( (x, y) = (2, -1) \)