College Algebra (10th Edition)

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

Chapter 8 - Review Exercises - Page 635: 59

Answer

See below.

Work Step by Step

If $x$ is the speed of the boat and $y$ is the speed of the river, then our equations are: $(x+y)2.5=100$ and $(x-y)3=100$. So $x+y=40$ and $x-y=\frac{100}{3}$. So $x=40-y$. then $x-y=(40-y)-y=\frac{100}{3}\\40-2y=\frac{100}{3}\\-2y=-\frac{20}{3}\\y=\frac{10}{3}$ Then $x=40-\frac{10}{3}=\frac{110}{3}$
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