Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 1 - Section 1.8 - Exponents and Order of Operations - Exercise Set - Page 102: 120

Answer

$\frac{1}{6}$ is indeed a solution of this equation.

Work Step by Step

Let's plug in $\frac{1}{6}$ for $x$ to see if the equation holds true. If it does, then $\frac{1}{6}$ is a solution of the equation: $$-\frac{1}{2} = \frac{1}{6} - \frac{2}{3}$$ To work with fractions, we need to make sure they have the same denominator. We must find the lowest common denominator for all three fractions. $6$ seems to be the common denominator for all three fractions. We need to multiply each fraction by the lowest common denominator to get rid of the fractions: $$6(-\frac{1}{2}) = 6(\frac{1}{6}) - 6(\frac{2}{3})$$ Cancel out common factors: $$-3 = 1 - 4$$ Subtract: $$-3 = -3$$ Because the two sides of the equation equal one another, $\frac{1}{6}$ is indeed a solution of this equation.
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