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: 15


$(-1,2)$ or, $(4,-3)$

Work Step by Step

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