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: 15

Answer

\[x=5\].

Work Step by Step

Consider the equation \[\frac{3}{x}=\frac{15}{25}\] The LCM is\[25x\] so multiply both the sides by \[25x\] \[\left( 25x \right)\times \frac{3}{x}=\left( 25x \right)\times \frac{15}{25}\] This gives, \[75=15x\] Or, \[15x=75\] Divide both the sides by 15 \[x=5\] Now, to check it, put \[x=5\]in the equation \[\begin{align} & \frac{3}{x}=\frac{15}{25} \\ & \frac{3}{5}=\frac{15}{25} \\ & \frac{3}{5}=\frac{3}{5} \\ \end{align}\] which is true. The required solution is\[x=5\].
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