Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 1 - Section 1.4 - Solving Linear Equations - Exercise Set - Page 50: 28



Work Step by Step

For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD $30$ on both sides of the equation to obtain: $30(\frac{x}{5}-\frac{1}{2}) = 30(\frac{x}{6}) \\\frac{30x}{5} - \frac{30}{2} = \frac{30x}{6} \\6x - 15 = 5x$ Add $15$ and subtract $5x$ on both sides to obtain: $6x-5x=15 \\x=15$ Check: $\begin{array}{ccc} &\frac{15}{5} - \frac{1}{2} &= &\frac{15}{6} \\&3-\frac{1}{2} &= &\frac{5}{2} \\&\frac{6}{2}-\frac{1}{2} &= &\frac{5}{2} \\&\frac{5}{2} &= &\frac{5}{2}\end{array}$
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