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: 24



Work Step by Step

Substituting the first equation into the second one we get: $10y-6=-2\\10y=4\\y=0.4$ If $y=0.4$, then $x^2=4$, thus $x=\pm2$. Thus the solutions are: $(2,0.4),(-2,0.2)$
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