\(\begin{cases}y^{2}-x=-1\\ x=y+3\end{cases}\)

To solve the system of equations: \[ \begin{cases} y^{2} - x = -1 \\ x = y + 3 \end{cases} \] we can substitute the second equation into the first equation. 1. From the second equation, we have \(x = y + 3\). 2. Substitute \(x\) in the first equation: \[ y^{2} - (y + 3) = -1 \] 3. Simplify the equation: \[ y^{2} - y - 3 + 1 = 0 \] \[ y^{2} - y - 2 = 0 \] 4. Now, we can factor this quadratic equation: \[ (y - 2)(y + 1) = 0 \] 5. Setting each factor to zero gives us the solutions for \(y\): \[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \] \[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \] 6. Now, we will substitute these \(y\) values back into the equation \(x = y + 3\) to find the corresponding \(x\) values. For \(y = 2\): \[ x = 2 + 3 = 5 \] For \(y = -1\): \[ x = -1 + 3 = 2 \] 7. Therefore, the solutions to the system of equations are: \[ (5, 2) \quad \text{and} \quad (2, -1) \] The final result is: \[ \boxed{(5, 2) \text{ and } (2, -1)} \]