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: 10


$(-12, 1)$ or, $(-2,6)$

Work Step by Step

Here, we have $x=2y-14$ Now, $y(2y-14)=-12$ This gives: $y^2-7y+6=0$ This yields : $(y-1) (y-6)=0$ when $y=1$ then we have $y=2(1)-14=-12$ when $y=6$ then we have $y=2(6)-14=-2$ Hence, our answers are: $(-12, 1)$ or, $(-2,6)$
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