College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.2 - Real Numbers - P.2 Exercises - Page 15: 30

Answer

Answer: A. $\frac{1}{15}$ B. $\frac{35}{24}$

Work Step by Step

A.) $\frac{2}{3}-\frac{3}{5}$ To solve this we need to find the LCM of the denominator of both the fractions. I.e. LCM of 3 & 5 is 15. So now, Multiplay both fractions to make denominator as the LCM of both the denominators. $\frac{2\times5}{3\times5}-\frac{3\times3}{5\times3}$ = $\frac{10}{15}-\frac{9}{15}$ = $\frac{10-9}{15}$ = $\frac{1}{15}$ B.) $1+\frac{5}{8}-\frac{1}{6}$ = $\frac{1}{1}+\frac{5}{8}-\frac{1}{6}$ To solve this we need to find the LCM of the denominator of all the fractions. I.e. LCM of 1, 8 & 6 is 24. So now, Multiplay both fractions to make denominator as the LCM of both the denominators. $\frac{1\times24}{1\times24}+\frac{5\times3}{8\times3}-\frac{1\times4}{6\times4}$ = $\frac{24}{24}+\frac{15}{24}-\frac{4}{24}$ = $\frac{24+15-4}{24}$ = $\frac{35}{24}$
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