Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.4 - Systems of Nonlinear Equations in Two Variables - Exercise Set - Page 850: 16


$(-2,-1)$ or, $(-6,3)$

Work Step by Step

Here, we have $x=-y-3$ Now, $(-y-3)^2+2y^2=12y+18$ This gives: $3y^2-6y-9=0 $ This yields : $(y+1) (y-3)=0$ when $y=-1$ then we have $x=-(-1)-3=-2$ when $y=3$ then we have $x=-(3)-3=-6$ Hence, our answers are: $(-2,-1)$ or, $(-6,3)$
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