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 - Page 315: 48

Answer

Pipe A takes 4 hours, and Pipe B takes 12 hours.

Work Step by Step

Let t be the time it takes Pipe A. Thus: $ 1/ 3 = \frac{1}{t} + \frac{1}{t+8} $ We create common denominators to obtain: $ .33 = \frac{t+t+8}{t(t+8)} \\ .33t^2 + 2.67t = 2t +8 \\ .33t^2 +.67t -8 = 0 \\ t^2 +2t -24 = 0 \\(t+6)(t-4) =0$ Since t can't be negative, we see that it takes Pipe A 4 hours. This means it takes Pipe B 4+8=12 hours.
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.