Thinking Mathematically (6th Edition)

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

Chapter 6 - Algebra: Equations and Inequalities - 6.2 Linear Equations in One Variable and Proportions - Exercise Set 6.2 - Page 363: 92

Answer

the solution set is \[\left\{ 0 \right\}\].

Work Step by Step

The equation is \[\frac{x}{2}+\frac{2x}{3}+3=x+3\]. Subtract \[3\]from both sides of the equal sign. \[\begin{align} & \frac{x}{2}+\frac{2x}{3}+3-3=x+3-3 \\ & \frac{x}{2}+\frac{2x}{3}=x \\ & \frac{3x}{6}+\frac{4x}{6}=x \\ & \frac{7x}{6}=x \\ \end{align}\] Multiply by \[6\] both sides of the equal sign. \[7x=6x\] Subtract \[6x\]from both sides of the equal sign. \[\begin{align} & 7x-6x=6x-6x \\ & x=0 \\ \end{align}\] Check the proposed solution.Substitute 0 for x in the original equation \[\frac{x}{2}+\frac{2x}{3}+3=x+3\] \[\begin{align} & \frac{0}{2}+\frac{2\times 0}{3}+3=0+3 \\ & 0+0+3=3 \\ & 3=3 \\ \end{align}\] This true statement \[3=3\] verifies that the solution set is \[\left\{ 0 \right\}\]. Thus, the solution set is \[\left\{ 0 \right\}\].
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