Question 374821
Let the speed of the slower car = s
then the speed of the faster car = s + 12 
so within one hour the distance between them is 
d = s+12 + s
d = 2s + 12
In 6 hours the distance would be 6d = 6(2s+12)
so....
6(2s+12 ) = 648
divide both sides by  6
2s+12 = 648/6
2s+12 = 108
subtract 12 on both sides
2s = 108-12
2s = 96
divide both sides by 2
s = 48
so the speed of the slower car = 48 mph
and the speed of the faster car = 48 + 12 = 60 mph