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: 27


no solution.

Work Step by Step

Adding twice the first and negative one times the second equation we get: $2x^2-20-x-8=0\\2x^2-x-28=0\\(2x+7)(x-4)=0$ Thus $x=-3.5$ or $x=4$ Plugging this into the first equation we get: if $x=-3.5$: $12.25+2y^2-10=0$ but this has no solution. if $x=4$: $16+2y^2-10=0$, but this has no solution Thus there are no solutions.
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