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: 13



Work Step by Step

Substituting the second equation into the first one we get: $6x^2+3(-x+2)^2=12\\6x^2+3(x^2-4x+4)=12\\9x^2-12x+12=12\\3x(3x-4)=0$ Thus $x=0$ or $x=\frac{4}{3}$. If $x=0$, then $y=2$, and if $x=\frac{4}{3}$, then $y=\frac{2}{3}$. Thus the solutions are: $(0,2),(\frac{4}{3},\frac{2}{3})$
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