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: 14


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

Work Step by Step

Here, we have $y=\dfrac{4}{x}$ Now, $x^2+(\dfrac{4}{x})^2=8$ This gives: $x^4-8x^2+16=0$ This yields : $(x^2-4) (x^2-4)=0$ when $x=-2$ then we have $y=\dfrac{4}{-2}=-2$ when $x=2$ then we have $y=\dfrac{4}{2}=2$ Hence, our answers are: $(-2,-2)$, $(2,2)$
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