Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 7 - Review: 53

Answer

The faster car travels at 30 mph, and the slower car travels at 20 mph.

Work Step by Step

Let $x$ mph be the speed of the faster car. Speed of the slower car = $x - 10$ Since both cars used the same time to travel 90 miles and 60 miles respectively, by $Speed = \frac{Distance}{Time}$, we have $\frac{90}{x} = \frac{60}{x - 10}$ $90(x - 10) = 60x$ $90x - 900 - 60x = 0$ $30x = 900$ $x = 30$ Therefore, the faster car travels at 30 mph, and the slower car travels at 30 - 10 = 20 mph.
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