Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 7 - Section 7.4 - Adding and Subtracting Radical Expressions - 7.4 Exercises: 46

Answer

$\dfrac{14\sqrt{3}}{9}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given radical expression, $ \dfrac{4\sqrt{3}}{3}+\dfrac{2\sqrt{3}}{9} ,$ change the terms to similar fractions by using the $LCD.$ $\bf{\text{Solution Details:}}$ To simplify the expression above, change the expressions to similar fractions (same denominator) by using the $LCD$. The $LCD$ of the denominators $ 3 $ and $ 9 $ is $ 9 $ since it is the lowest number that can be exactly divided by the denominators. Multiplying the terms by an expression equal to $1$ that will make the denominator equal to the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{4\sqrt{3}}{3}\cdot\dfrac{3}{3}+\dfrac{2\sqrt{3}}{9} \\\\= \dfrac{12\sqrt{3}}{9}+\dfrac{2\sqrt{3}}{9} \\\\= \dfrac{12\sqrt{3}+2\sqrt{3}}{9} \\\\= \dfrac{14\sqrt{3}}{9} .\end{array}
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