Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - Chapter Summary, Review, and Test - Review Exercises - Page 403: 12

Answer

\[x=2\]

Work Step by Step

Consider the equation \[\frac{2x}{3}=\frac{x}{6}+\frac{1}{1}\] The least common denominator is 6 so multiply both the sides by 6 as follows: \[\begin{align} & \frac{2x}{3}\times 6=\left( \frac{x}{6}+\frac{1}{1} \right)\times 6 \\ & 4x=x+6 \end{align}\] Simplifying \[\begin{align} & 3x=6 \\ & x=2 \end{align}\] Now, to check it, put \[x=2\]in the equation \[\begin{align} & \frac{2x}{3}=\frac{x}{6}+1 \\ & \frac{2\times 2}{3}=\frac{2}{6}+1 \\ & \frac{4}{3}=\frac{1}{3}+1 \\ & \frac{4}{3}=\frac{4}{3} \end{align}\] which is true. Hence, the solution of the given equation is\[x=2\].
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