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The application of one variable linear equation -- the encounter problem of journey: a passenger car and a freight car leave from two cities 1000 km apart at the same time, and after 4.5 hours, the two cars are still 55 km apart. It is known that the speed of the passenger car is 1 / 3 faster than that of the freight car. How much is the speed of the passenger car and the freight car? (solved by one variable linear equation)
Let the speed of the freight car be X
Then: (x + 4x / 3) * 4.5 + 55 = 1000
The solution equation is: x = 90 * 4 / 3 = 120
The speed of passenger cars is 120 km / h, and that of freight cars is 90 km / h
A passenger car and a freight car leave from a and B cities at the same time. The passenger car travels 55 kilometers per hour and the freight car 45 kilometers per hour. After 5.5 hours, the two cars are still 80 kilometers apart. How far is the distance between a and B cities
After 5 hours, the two vehicles travel 5 * (45 + 55) = 500 km
So the distance between AB and ab is 500 + 80 = 580 km
I don't know if "5:5 hours" in the title means 5 hours
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