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 364: 125


The provided statement makes sense.

Work Step by Step

It is easy to solve equations that do not contain fractions. If some equation contains fraction it is better to remove it as early as possible since in that way it makes the steps simpler and approaches to solution quickly. In the provided statement first subtract \[\frac{1}{5}\]from both sides of equation\[3x+\frac{1}{5}=\frac{1}{4}\]provides, \[\begin{align} & 3x+\frac{1}{5}=\frac{1}{4} \\ & 3x+\frac{1}{5}-\frac{1}{5}=\frac{1}{4}-\frac{1}{5} \\ & 3x=\frac{1}{4}-\frac{1}{5} \end{align}\] See still it needs at least two steps to remove fractions from the given equation. But if multiply both sides of equation by 20 gives; \[\begin{align} & 20\left( 3x+\frac{1}{5} \right)=20\left( \frac{1}{4} \right) \\ & \left( 20\times \left( 3x \right) \right)+\left( 20\times \frac{1}{5} \right)=\left( 20\times \frac{1}{4} \right) \\ & 60x+4=5 \end{align}\] Fractions are removed from the equation by multiplying both sides of the equation by the least common denominator 20in just two steps. Now, it is easier to solve the equation further. Therefore, the provided statement that, β€œAlthough I can solve \[3x+\frac{1}{5}=\frac{1}{4}\]by first subtracting \[\frac{1}{5}\] from both sides, I find it easier to begin by multiplying both sides by 20, the least common denominator.” makes sense. Hence, the provided statement makes sense.
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