Elementary Algebra

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

Chapter 5 - Exponents and Polynomials - Chapters 1-5 Cumulative Review Problem Set: 48

Answer

$n=\frac{9}{14}$

Work Step by Step

We will solve the equation by first finding the least common denominator on the left side of the equation and then cross multiplying the resultant expression: $\frac{3n+1}{5}+\frac{n-2}{3}=\frac{2}{15}$ $\frac{3(3n+1)+5(n-2)}{15}=\frac{2}{15}$ $\frac{9n+3+5n-10}{15}=\frac{2}{15}$ $\frac{9n+5n-10+3}{15}=\frac{2}{15}$ $\frac{14n-7}{15}=\frac{2}{15}$ $15(14n-7)=2(15)$ $14n-7=2$ $14n=2+7$ $14n=9$ $n=\frac{9}{14}$
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