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: 6



Work Step by Step

Here, we have $y=x^2+4x+5$ Now, $x^2+4x+5=x^2+2x-1$ This gives: $2x=-6$ This yields : $x=-3$ when $x=-3$ then we have $y=(-3)^2+4(-3)+5=2$ Hence, our answers are: $(-3,2)$
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