Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 2 - Real Numbers - 2.2 - Addition and Subtraction of Rational Numbers - Problem Set 2.2: 89

Answer

$\frac{20}{9}n$

Work Step by Step

We first apply the distributive property. Then, we simplify the resultant expression: $n+\frac{4}{3}n-\frac{1}{9}n$ =$n(1+\frac{4}{3}-\frac{1}{9})$ =$n(\frac{1(9)+4(3)-1(1)}{9})$ =$n(\frac{9+12-1}{9})$ =$n(\frac{21-1}{9})$ =$n(\frac{20}{9})$ =$\frac{20}{9}n$
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