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: 87


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

Work Step by Step

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