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: 16



Work Step by Step

Here, we have $2x=3y+4 \implies x=\dfrac{3y+4}{2}$ Now, $2(\dfrac{3y+4}{2})+6y=5y-4$ This gives: $4y=-8$ or, $y=-2$ Thus, $y=\dfrac{3(-2)+4}{2}=-1$ Solution is: $(x,y)=(-1,-2)$
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