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: 25

Answer

$x=12$

Work Step by Step

For easier work, it is better to get rid of the fractions. This can be achieved by multiplying the LCD $6$ on both sides of the equation to obtain: $6(\frac{x}{3}) = 6(\frac{x}{2} - 2) \\\frac{6x}{3} = \frac{6x}{2} - 6(2) \\2x = 3x - 12$ Add $12$ on both sides and subtract $2x$ on both sides to obtain: $12=3x-2x \\12 = x$ Check: $\begin{array}{ccc} &\frac{12}{3} &= &\frac{12}{2} - 2 \\&4 &= &6-2 \\&4 &= &4 \end{array}$
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