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: 12


$(-1,3)$ or, $(3, -5)$

Work Step by Step

Here, we have $y=-2x+1$ Now, $x^2+(-2x+1)^2=4$ This gives: $x^2-2x-3=0$ This yields : $(x+1)(x-3)=0$ when $x=-1$ then we have $y=-2(-1)+1=3$ when $x=3$ then we have $y=-2(3)+1=-5$ Hence, our answers are: $(-1,3)$ or, $(3, -5)$
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