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: 22


$(-1,-2)$, $(-1,2)$, $(1,-2)$, $(1,2)$

Work Step by Step

Here, we have $-2(2x^2-y^2)=(-2)(-2) \implies -4x^2+2y^2=4$ Now, $(3x^2-2y^2) +(-4x^2+2y^2) =-5+4$ This gives: $-x^2=-1 \implies x^2=1$ when $x=1$ then we have $2(1)^2-y^2=2 \implies y^2=4$ This gives: $y=2, -2$ when $x=1$ then we have $2(-1)^2-y^2=2 \implies y^2=4$ This gives: $y=2, -2$ Hence, our answers are: $(-1,-2)$, $(-1,2)$, $(1,-2)$, $(1,2)$
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