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: 31

Answer

$5$

Work Step by Step

For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD of $10$ on both sides of the equation to obtain: $10(\frac{3x}{5}) - x = 10(\frac{x}{10}-\frac{5}{2}) \\\frac{30x}{5} - 10x = \frac{10x}{10} - \frac{50}{2} \\6x - 10x = x-25 \\-4x = x-25$ Add $4x$ and $25$ on both sides to obtain: $25=x+4x \\25=5x \\\frac{25}{5}=\frac{5x}{5} \\5 = x$ Check: $\begin{array}{ccc} &\frac{3(5)}{5} - 5&= &\frac{5}{10}-\frac{5}{2} \\&\frac{15}{5}-5 &= &\frac{1}{2}-\frac{5}{2} \\&3-5 &= &-\frac{4}{2} \\&-2 &= &-2\end{array}$
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