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: 9


$(\dfrac{-3}{2}, -4)$ or, $(2,3)$

Work Step by Step

Here, we have $y=2x-1$ Now, $x(2x-1)=6$ This gives: $2x^2-x-6=0$ This yields : $(2x+3) (x-2)=0$ when $x=\dfrac{-3}{2}$ then we have $y=2(\dfrac{-3}{2})-1=-4$ when $x=2$ then we have $y=2(2)-1=3$ Hence, our answers are: $(\dfrac{-3}{2}, -4)$ or, $(2,3)$
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