Find the ordered pair solutions for the system of equations. \[ \left\{\begin{array}{l}y=x^{2}+1 \\ y=x+1\end{array}\right. \] \( ([?], \square) \) and \( (\square], \square) \) Enter the smallest \( x \) first.

To solve the system of equations given by \[ \left\{\begin{array}{l}y=x^{2}+1 \\ y=x+1\end{array}\right. \] we can set the two equations equal to each other, since both are equal to \( y \): 1. \( x^2 + 1 = x + 1 \) Now, we can simplify this equation: 1. Subtract \( x + 1 \) from both sides: \[ x^2 + 1 - x - 1 = 0 \] This simplifies to: \[ x^2 - x = 0 \] 2. Factor the equation: \[ x(x - 1) = 0 \] 3. Solve for \( x \): - \( x = 0 \) - \( x = 1 \) Next, we will substitute these \( x \) values back into either equation to find the corresponding \( y \) values. For \( x = 0 \): \[ y = 0 + 1 = 1 \] So, we have the ordered pair \( (0, 1) \). For \( x = 1 \): \[ y = 1 + 1 = 2 \] So, we have the ordered pair \( (1, 2) \). The two ordered pairs that are solutions to the system are: \[ (0, 1) \quad \text{and} \quad (1, 2). \] Thus the final answer is: \[ ([0], [1]) \text{ and } ([1], [2]). \]