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: 11


$(3,0)$ or, $(-5,-4)$

Work Step by Step

Here, we have $x=2y+3$ Now, $y^2=(2y+3)^2-9$ This gives: $y^2+4y=0$ This yields : $y(y+4)=0$ when $y=0$ then we have $y=2(0)+3=3$ when $y=-4$ then we have $y=2(-4)+3=-5$ Hence, our answers are: $(3,0)$ or, $(-5,-4)$
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