Chapter 9 - Quadratic Functions and Equations - 9-8 Systems of Linear and Quadratic Equations - Practice and Problem-Solving Exercises - Page 599: 18

The solutions of the system are $(8, 42)$ and $(-2,2)$.

$y=4x+10$ $y=x^2-2x-6$ Substitute $4x+10$ for y: $4x+10=x^2-2x-6$ $x^2-6x-16=0$ $(x-8)(x+2)=0$ $x-8=0$ or $x+2=0$ $x=8$ or $x=-2$ Find corresponding y-values. Use either original equation $y=4x+10$ $y=4(8)+10$ or $y=4(-2)+10$ $y=42$ or $y=2$ The solutions of the system are $(8, 42)$ and $(-2,2)$.