SOLUTION: At 11am two cars started traveling toward each other from towns 315kilometers apart. They passed each other at 2:30pm. If the rate of the faster car exceeded the rate of the slo

Let r = the rate of the slower car.

So r+6 = the rate of the faster car

So their approach rate is the sum of their rates = r+r+6 = 2r+6
From 11AM to 2:30PM is 3.5 hours.

So their original distance apart went from 315 km to 0 km distance 
apart in 3.5 hours, so

using Distance = Rate∙Time, with Distance = 315 km, 
Approach Rate = 2r+6, and Time = 3.5 hours

315 = (2r+6)(3.5)
 
Solve that and get r = 42 km/hr for the slower car, and
r+6 = 42+6 = 48 km/hr for the faster car.

Edwin