Precalculus (6th Edition) Blitzer

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

Chapter 9 - Cumulative Review Exercises - Page 1038: 8



Work Step by Step

Step 1. Multiply 2 to the first equation and multiply 3 to the second equation: $\begin{cases}6x+8y=4\\6x+15y=-3 \end{cases} $ Step 2. Taking the difference between the two, we have $7y=-7$; thus $y=-1$ Step 3. Back-substitute $y=-1$ in either equation; we have $3x-4=2$ and thus $x=2$ Step 4. The solution is $(2,-1)$
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