## Elementary Algebra

Published by Cengage Learning

# Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set - Page 468: 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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