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: 9



Work Step by Step

Substituting the second equation into the first one we get: $(-x)^2-x-6=0\\x^2-x-6=0\\(x+2)(x-3)=0$ Thus $x=-2$ or $x=3$. If $x=-2$, then $y=2$, and if $x=3$, then $y=-3$. Thus the solutions are: $(-2,2),(3,-3)$
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