# Chapter 7 - Algebraic Fractions - 7.6 - More Fractional Equations and Problem Solving - Problem Set 7.6 - Page 315: 37

Tom is going 17 mph, and Celia is going 20 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 Tom. Thus, we obtain: $\frac{60}{t-2} = \frac{ 85}{t} +3$ We create common denominators and solve for t: $\frac{60t}{t(t-2)} = \frac{85(t-2)}{t(t-2)} + \frac{3t(t-2)}{t(t-2)}$ Since the denominators are the same, we cancel them out and solve. (Recall, t cannot be negative.) $60t = 85t-170 + 3t^2 -6t \\ 3t^2 +19t - 170 = 0 \\ (t-5)(3t+4) = 0 \\t=5$ Thus, for Tom: $v = 85/5 = 17$ mph Since Celia was 3 mph faster, she went 20 mph.

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