Basic College Mathematics (10th Edition)

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

Chapter 9 - Basic Algebra - Mixed Review Exercises - Page 700: 21



Work Step by Step

We are given the equation $3(k-6)=6-12$. In order to solve, we must first use the distributive property to simplify the left side. $3(k-6)=3(k)+3(-6)=3k-18=6-12$ Combine like terms on the right side. $3k-18=6-12=-6$ Add 18 to both sides. $3k-18+18=-6+18$ $3k=12$ Divide both sides by 3. $\frac{3k}{3}=\frac{12}{3}$ Therefore, $k=4$.
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