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: 1


$(-3,5)$ or $(2,0)$

Work Step by Step

Here, we have $y=x^2-4$ Now, $x+(x^2-4)=2$ This gives: $x^2+x-6=0$ This yields : $(x+3)(x-2)=0$ when $x=-3$ then we have $y=(-3)^2-4=5$ and when $x=2$ then we have $y=(2)^2-4=0$ Hence, our answers are: $(-3,5)$ or $(2,0)$
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