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 661: 10



Work Step by Step

Substituting the second equation into the first one we get: $x^2+(2x-10)^2-25=0\\x^2+4x^2-40x+100-25=0\\5x^2-40x+75=0\\x^2-8x+15=0\\(x-3)(x-5)=0$ Thus $x=5$ or $x=3$. If $x=5$, then $y=0$, and if $x=3$, then $y=-4$. Thus the solutions are: $(5,0),(3,-4)$
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