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: 19


$(-3,-2)$, $(-3,2)$,$(3,-2)$, $(3,2)$

Work Step by Step

Here, we have $(x^2+y^2)+(x^2-y^2)=13+5$ Now, $x^2+y^2+x^2-y^2=18$ This gives: $2x^2=18$ This yields : $x^2=9 \implies x=3,-3$ when $x=3$ then we have $9+y^2=13 \implies y^2=4$ and $y=2,-2$ when $x=-3$ then we have $(-3)^2+y^2=13 \implies y^2=4$ and $y=2,-2$ Hence, our answers are: $(-3,-2)$, $(-3,2)$,$(3,-2)$, $(3,2)$
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