Question

Annette drove to shawnee in 4 hours and drove back in 3 hours. what were her speeds if her speed coming back was 11 miles per hour greater than her speed going

Answer 1

The distance covered in both trips is the same.
If x mph is the going speed, then coming back speed = (x+11) mph

Distance = speed * time

Therefore,
4*x = 3*(x+11) => 4x = 3x+33 => 4x-3x = 33 => x = 33 mph, and x+11 = 44 mph

The going speed is 33 mph while coming back speed is 44 mph