At 1030AM two planes leave Houston one flying east at 560kmh

At 10:30A.M. two planes leave Houston one flying east at 560km/h and the other flying west at 640 km/h at what time will they be 2100 km apart

Solution

Let x=number of hours that will have elapsed when the planes are 2100km apart
d=rt
Now we are told that the distance flown by the slower plane (560x) plus the distance flown by the faster plane (640x) must equal 2100 km so our equation to solve is:
560x+640x=2100
1200x=2100
x=1.75 hrs=1h 45m
10:30 AM+(1h 45m)=12:15 PM
So at 12:15 PM, they will be 2100 miles apart
560(1.75)+640(1.75)=2100
980+1120=2100
2100=2100