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: 20


$(-1,0)$, $(1,0)$

Work Step by Step

Here, we have $(4x^2-y^2)+(4x^2+y^2)=4+4$ Now, $4x^2-y^2+4x^2+y^2=8$ This gives: $8x^2=8$ This yields : $x^2=1 \implies x=1,-1$ when $x=1$ then we have $4 (1)^2+y^2=4 \implies y^2=0$ and $y=0$ when $x=-1$ then we have $4 (-1)^2+y^2=4 \implies y^2=0$ and $y=0$ Hence, our answers are: $(-1,0)$, $(1,0)$
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