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: 1. \( y = x^{2} + 1 \) 2. \( y = x + 1 \) we can set the two equations equal to each other since they both equal \( y \): \[ x^{2} + 1 = x + 1 \] Now, we'll simplify this equation: 1. Subtract \( x + 1 \) from both sides: \[ x^{2} + 1 - x - 1 = 0 \] \[ x^{2} - x = 0 \] 2. Factor the left-hand side: \[ x(x - 1) = 0 \] This gives us two potential solutions for \( x \): 1. \( x = 0 \) 2. \( x = 1 \) Now, we'll find the corresponding \( y \) values for both \( x \) values using either of the original equations. Let's use \( y = x + 1 \). For \( x = 0 \): \[ y = 0 + 1 = 1 \] So, the ordered pair is \( (0, 1) \). For \( x = 1 \): \[ y = 1 + 1 = 2 \] So, the ordered pair is \( (1, 2) \). Thus, the solutions for the system of equations are: \( (0, 1) \) and \( (1, 2) \).