Basic College Mathematics (10th Edition)

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

Chapter 9 - Basic Algebra - 9.7 Solving Equations with Several Steps - 9.7 Exercises - Page 676: 4

Answer

$k=2$

Work Step by Step

We are given the equation $6k+3=15$. In order to solve, we can first subtract 3 from both sides. $6k+3-3=15-3$ $6k=12$ Divide both sides by 6. $\frac{6k}{6}=\frac{12}{6}$ $k=2$
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