College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 8 - Review Exercises - Page 636: 60

Answer

See below.

Work Step by Step

Let $x$ be Bruce's "speed", $y$ be Bryce's and $z$ be Marty's. Then our equations are: $(x+y)\frac{4}{3}=1$, $(z+y)\frac{8}{5}=1$, $(x+z)\frac{8}{3}=1$. So $(x+y)=\frac{3}{4}$, $(z+y)=\frac{5}{8}$, $(x+z)=\frac{3}{8}$. Thus $x=0.75-y$, so $(0.75-y)+z=0.375$, so $z-y=-0.375$, thus $z=-0.375+y$. So $(-0.375+y)+y=0.625$, thus $y=0.5$, so $z=0.125$ and $x=0.25$ Thus Bruce needs $\frac{1}{0.25}=4$ hours, Bryce needs $\frac{1}{0.5}=2$ hours and Marty needs $\frac{1}{0.125}=8$ hours
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