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: 32



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 $14$ on both sides of the equation to obtain: $14(2x-\frac{2x}{7}) = 14(\frac{x}{2}+\frac{17}{2}) \\28x - \frac{28x}{7} = \frac{14x}{2} +\frac{14(17)}{2} \\28x - 4x = 7x+7(17) \\24x = 7x+119$ Subtract $7x$ on both sides to obtain: $24x-7x=119 \\17x=119 \\\frac{17x}{17}=\frac{119}{17} \\x = 7$ Check: $\begin{array}{ccc} &2(7)-\frac{2(7)}{7} &= &\frac{7}{2}+\frac{17}{2} \\&14-\frac{14}{7} &= &\frac{24}{2} \\&14-2 &= &12 \\&12 &= &12\end{array}$
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