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To find the solution for the system of equations given:

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 this equation:

\[ y^2 - y - 3 = -1 \]

Add 1 to both sides:

\[ y^2 - y - 2 = 0 \]

Now, factor the quadratic equation:

\[ (y - 2)(y + 1) = 0 \]

So, the solutions for \( y \) are:

\[ y = 2 \quad \text{or} \quad y = -1 \]

For \( y = 2 \):

• Substitute back to find \( x \): \( x = 2 + 3 = 5 \)

For \( y = -1 \):

• Substitute back to find \( x \): \( x = -1 + 3 = 2 \)

Therefore, the solutions are:

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