two cars start 100 miles apart the faster goes 90 mph the sl

two cars start 100 miles apart. the faster goes 90 mph. the slower goes 60 mph. how far from the starting location of the faster car will they meet? how much time will have passed?
i tried d= rt
distance r(t)
100= (90+60)t
100/150=150/150(t)
.667=t
60 miles =d??

Solution

Distance(d) equals rate (r) times time(t) or d=rt; r=d/t and t=d/r
Distance travelled by faster car=rt=90t
Distance travelled by slower car=rt=60t
Now when the above two distances add up to 100 miles the two cars will have met, so:
60t+90t=100
150t=100 or
t=2/3 hour
So distance travelled by faster car =90*(2/3)=60 mi
Distance travelled by slower car =60*(2/3)=40 mi