Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 586: 33

The student didn't see that the determinant is zero (and thus there are no intersecting points between the two equations).

$x^2+2x+4=y$ $y=x+1$ $x^2+2x+4=x+1$ $x^2+2x+4-x-1=x+1-x-1$ $x^2+x+3 = 0$ $b^2-4ac$ $1^2-4*1*3$ $1-12=-11$ Since the determinant is negative, then there are no real-number solutions.