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: 89


The solution set is \[\left\{ \frac{7}{2} \right\}\].

Work Step by Step

The equation is\[\frac{x-2}{5}=\frac{3}{10}\]. Apply the cross-product principle in the equation. \[\begin{align} & \frac{x-2}{5}=\frac{3}{10} \\ & 10\left( x-2 \right)=3\times 5 \\ & 10x-20=15 \\ \end{align}\] Add20to both sides of the equal sign. \[\begin{align} & 10x-20+20=15+20 \\ & 10x=35 \\ \end{align}\] Divided by \[10\], both sides of the equal sign. \[\begin{align} & \frac{10x}{10}=\frac{35}{10} \\ & x=\frac{7}{2} \\ \end{align}\] Check the proposed solution. Substitute \[\frac{7}{2}\] for x in the original equation \[\frac{x-2}{5}=\frac{3}{10}\]. \[\begin{align} & \frac{\frac{7}{2}-2}{5}=\frac{3}{10} \\ & \frac{\frac{7-4}{2}}{5}=\frac{3}{10} \\ & \frac{3}{2\times 5}=\frac{3}{10} \\ & \frac{3}{10}=\frac{3}{10} \\ \end{align}\] This true statement \[\frac{3}{10}=\frac{3}{10}\] verifies that the solution set is \[\left\{ \frac{7}{2} \right\}\]. Thus, the solution set is \[\left\{ \frac{7}{2} \right\}\].
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