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 - Quiz for Lessons 9.6-9.7 - Page 664: 10



Work Step by Step

Subtracting the first equation from the second one we get: $3y^2-3y-6=0\\y^2-y-2=0\\(y-2)(y+1)=0$ Thus $y=2$ or $y=-1$ $x^2=4+y$, thus if $y=2$, then $x=\pm\sqrt6$, and if $y=-1$, then $x=x=\pm\sqrt3$. Thus the solutions are: $(x=\pm\sqrt6,2),(x=\pm\sqrt3,-1)$
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