# Chapter 7 - Algebraic Fractions - 7.6 - More Fractional Equations and Problem Solving - Problem Set 7.6 - Page 315: 38

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.

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