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: 18


$(\dfrac{1}{5},\dfrac{18}{5})$ or, $(1,2)$

Work Step by Step

Here, we have $y=4-2x$ Now, $(x+1)^2+[(4-2x)^2-2]^2=4$ This gives: $5x^2-6x+1=0$ This yields : $(5x-1) (x-1)=0$ when $x=\dfrac{1}{5}$ then we have $y=4-2(\dfrac{1}{5})=\dfrac{18}{5}$ when $x=1$ then we have $y=4-2(1)=2$ Hence, our answers are: $(\dfrac{1}{5},\dfrac{18}{5})$ or, $(1,2)$
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