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: 27



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: $6(20-\frac{x}{3}) = 6(\frac{x}{2}) \\120 - \frac{6x}{3} = \frac{6x}{2} \\120 - 2x = 3x$ Add $2x$ on both sides to obtain: $120=3x+2x \\120=5x \\\frac{120}{5} = \frac{5x}{5} \\24=x$ Check: $\begin{array}{ccc} &20 - \frac{24}{3} &= &\frac{24}{2} \\&20-86 &= &12 \\&12 &= &12 \end{array}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.