Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 2 - Section 2.1 - Linear Equations in One Variable - Exercise Set - Page 56: 87



Work Step by Step

We are given the equation $\frac{7}{11}x+9=\frac{3}{11}x-14$ and the equation $\frac{7}{11}x=\frac{3}{11}x+K$ In order to find the value of K that makes both equations equivalent, we must move all constant terms in the first equation to the right side. For $\frac{7}{11}x+9=\frac{3}{11}x-14$, subtract 9 from both sides. $\frac{7}{11}x=\frac{3}{11}x-5$. Therefore, both equations will be equivalent when $K=-23$.
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