College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.1 - Linear Equations - 1.1 Assess Your Understanding - Page 90: 36


The solution set is {$-\frac{5}{2}$}

Work Step by Step

$\frac{1}{2}-\frac{1}{3}p=\frac{4}{3}$ The LCD is 6 $6(\frac{1}{2}-\frac{1}{3}p)=\frac{4}{3}6$ Multiply both sides by 6. $6\cdot(\frac{1}{2})-6\cdot(\frac{1}{3}p)=\frac{4}{3}\cdot(6)$ Distribute and cancel out common factors. $3-2p=8$ Solve for p. $3-3-2p=8-3$ $-2p=5$ $\frac{-2}{-2}p=\frac{5}{-2}$ $p=-\frac{5}{2}$ The solution set is {$-\frac{5}{2}$}
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