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: 38



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 $30$ on both sides of the equation to obtain: $30(\frac{3x}{5} - \frac{x-3}{2}) = 30(\frac{x+2}{3}) \\\frac{90x}{5} - \frac{30(x-3)}{2} =\frac{30(x+2)}{3} \\18x - 15(x-3) = 10(x+2) \\18x - 15x +45 = 10x+20 \\3x+45 =10x+20$ Subtract $3x$ and $20$ on both sides to obtain: $45-20=10x-3x \\25=7x \\\frac{25}{7} = \frac{7x}{7} \\\frac{25}{7}=x$ Check: $\begin{array}{ccc} &\dfrac{3(\frac{25}{7})}{5} - \dfrac{\frac{25}{7}-3}{2} &= &\dfrac{\frac{25}{7}+2}{3} \\&\dfrac{(\frac{75}{7})}{5} - \dfrac{\frac{25}{7}-\frac{21}{7}}{2} &= &\dfrac{\frac{25}{7}+\frac{14}{7}}{3} \\&\frac{75}{7(5)} - \dfrac{\frac{4}{7}}{2} &= &\dfrac{\frac{39}{7}}{3} \\&\frac{15}{7} - \frac{4}{7(2)}&= &\frac{39}{7(3)} \\&\frac{15}{7} - \frac{2}{7} &= &\frac{13}{7} \\&\frac{13}{7} &= &\frac{13}{7}\end{array}$
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