You can
put this solution on YOUR website!lets call s the rate of travel of the slow car and f the rate of travel of the fast car.
we know that s=1/2f+13 and we know that the rate multiplied by the time which is 6s+6f=420
so plugging the value for s in the first equation into the second we get 6(1/2f+13)+6f=420
3f+78+6f=420 distribute
9f=342 combined like terms
f=38 so the fast car travels at 38 mph
s=1/2(38)+13=32 so the slow car travels and 32 mph.
the fast car has to be faster than the slow car otherwise the slow car is the fast car........wow