## Algebra: A Combined Approach (4th Edition)

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.