Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 12



Work Step by Step

Substituting the second equation ($y=x-2$) into the first one we get: $-x^2+2(x-2)^2=8\\-x^2+2(x^2-4x+4)=8\\x^2-8x+8=8\\x(x-8)=0$ Thus $x=0$ or $x=8$. If $x=0$, then $y=-2$, and if $x=8$, then $y=6$. Thus the solutions are: $(0,-2),(8,6)$
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