Elementary Algebra

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

Chapter 7 - Algebraic Fractions - 7.6 - More Fractional Equations and Problem Solving - Problem Set 7.6: 38

Answer

The express train is going 70 mph, and the freight train is going 50 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 the express train. Thus, we obtain: $\frac{300}{t+2} = \frac{ 280}{t} -20 $ We create common denominators and solve for t: $ \frac{300t}{t(t+2)} = \frac{280(t+2)}{t(t+2)} + \frac{-20t(t+2)}{t(t+2)}$ Since the denominators are the same, we cancel them out and solve. (Recall, t cannot be negative.) $300t = 280t +560 -20t^2 -40t \\ 20t^2 +60t -560 = 0 \\ 20(t+7)(t-4) \\ t = 4$ Thus, the express train went: $ v = 280/4 = 70$ Since the freight train was 20 mph slower, it went 50 mph.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.