Answer
$(\pm2,\pm3)$
Work Step by Step
Adding the first and the second equation we get: $7x^2-28=0\\x^2=4\\x=\pm2$
$y^2=13-x^2=9$, thus $y=\pm3$
Thus the solutions are: $(\pm2,\pm3)$
Algebra 2 (1st Edition)
by Larson, Ron; Boswell, Laurie; Kanold, Timothy D.; Stiff, Lee
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 662: 22
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