Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.2 Applications and Modeling with Linear Equations - 1.2 Exercises - Page 102: 21

Answer

$2.7$ miles.

Work Step by Step

Since we have speed in miles/HOUR, convert the times from minutes to hours. $20$ min = $\displaystyle \frac{20}{60}= \displaystyle \frac{1}{3}$ hours, $45$ min =$\displaystyle \frac{45}{60}= \displaystyle \frac{3}{4}$ hours. Let the average speed on his bike be $x$ mph. The speed by car is then $(x+4.5)$ mph. Distance traveled in any of the two legs = (average speed)(time of travel). By car, distance = $(x+4.5)\displaystyle \cdot\frac{1}{3}$ By bike, distance = $x\displaystyle \cdot\frac{3}{4}.$ The two distances are equal, and we have the equation: $(x+4.5)\displaystyle \cdot\frac{1}{3}=x\cdot\frac{3}{4}\qquad$ ... multiply both sides by 12 (LCM of 3 and 4) $(x+4.5)\displaystyle \cdot\frac{1}{3}\cdot 12=x\cdot\frac{3}{4}\cdot 12$ $(x+4.5)\cdot 4=9x\qquad$ ... distribute and simplify $ 4x+18=9x\qquad$ ... subtract 4x $ 18=5x\qquad$ ... divide by 5 $x=\displaystyle \frac{18}{5}=3.6$ mph (by bike) By bike, distance = $x\displaystyle \cdot\frac{3}{4}=3.6\cdot\frac{3}{4}=2.7$ miles.
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