Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 7 - Section 7.1 - Systems of Linear Equations in Two Variables - Exercise Set - Page 818: 15



Work Step by Step

Here, we have $-3y=8-4x \implies y=-\dfrac{8-4x}{3}$ Now, $3x+4(-\dfrac{8-4x}{3})=x+3(-\dfrac{8-4x}{3})+14$ This gives: $\dfrac{10x}{3} =\dfrac{50}{3}$ or, $x=5$ Thus, $y=-\dfrac{8-4(5)}{3}=4$ Solution is: $(x,y)=(5,4)$
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