Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 599: 16

The solutions of the system are $(1, -8)$ and $(2,-9)$.

$y=-x -7$ $y=x^2-4x-5$ Substitute $x +2$ for y: $-x -7=x^2-4x-5$ $x^2-3x+2=0$ $(x-1)(x-2)=0$ $x-1=0$ or $x-2=0$ $x=1$ or $x=2$ Find corresponding y-values. Use either original equation: $y=-x-7$ $y=-(1)-7$ or $y=-(2)-7$ $y=-8$ or $y=-9$ The solutions of the system are $(1, -8)$ and $(2,-9)$.