Algebra 1: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281140
ISBN 13: 978-0-13328-114-9

Chapter 9 - Quadratic Functions and Equations - Chapter Test - Page 607: 22


The solutions of the system are $(3,13)$ and $(-4,6)$.

Work Step by Step

$ y=x+10$ $y=x^2+2x-2$ Substitute $x+10$ for y $x+10=x^2+2x-2$ $x^2+x-12=0$ $(x-3)(x+4)=0$ $x-3=0$ or $x+4=0$ $x=3$ or $x=-4$ Find corresponding y-values. Use either original equation $ y=x+10$ $y=3+10$ or $ y=(-4)+10$ $y=13$ or $y=6$ The solutions of the system are $(3,13)$ and $(-4,6)$.
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