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 - Chapter Review - Page 672: 36



Work Step by Step

$x+y-14=0\\y=14-x$ Thus plugging this into the first equation we get: $x^2+(14-x)^2-100=0\\2x^2+96-28x=0\\x^2-14x+48=0\\(x-6)(x-8)=0$ Thus $x=6$ or $x=8$. Thus if $x=6$, then $y=8$, and if $x=8$, then $y=6$. Thus the solutions are: $(6,8),(8,6)$
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