Elementary Algebra

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

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set: 30

Answer

Two possibilities: Jay went 45 mph and Jean went 48 mph OR Jay went 7.5 mph and Jean went 10.5 mph

Work Step by Step

Recall that velocity is equal to change in distance over change in time. We call t the time it takes Jay. Thus, we obtain: $\frac{336}{t+2} = \frac{225}{t} + 3 $ We create common denominators and solve for t: $ \frac{336t}{t(t+2)} = \frac{225(t+2)}{t(t+2)} + \frac{3t(t+2)}{t(t+2)}$ Since the denominators are the same, we cancel them out and solve. (Recall, t cannot be negative.) $ 0 = (t-30)(t-5)$ Since both values yield positive results, we must consider both: $ v_{Jay} = 225 / 30 = 7.5 $ $ v_{Jean} = 7.5 + 3 = 10.5$ OR $v_{Jay} = 225/5 = 45$ $ v_{Jean} = 45 + 3 = 48$
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