Basic College Mathematics (10th Edition)

Published by Pearson
ISBN 10: 0134467795
ISBN 13: 978-0-13446-779-5

Chapter 9 - Basic Algebra - Review Exercises - Page 698: 97



Work Step by Step

We are given the equation $4y-3=7y+12$. We can solve by first adding 3 to both sides. $4y-3+3=7y+12+3$ $4y=7y+15$ Subtract 7y from both sides. $4y-7y=7y-7y+15$ $-3y=15$ Divide both sides by -3. $y=-5$
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