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 662: 22



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: $(\pm2,\pm3)$
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