Let r_{n} = rate or speed of the the train northbound

r_{s }= rate or speed of the the train southbound

d_{n} = distance travelled of the the train northbound

d_{s }= distance travelled of the the train southbound

Since the northbound train travels 18 miles per hour faster than the southbound train, we can represent it this way:

r_{n} = 18 + r_{s }

To solve this problem, we can use the common formula d = rt (distance = rate multiplied by the time travelled). Therefore the distance travelled by the train northbound after t = 2 hours is:

d_{n} = 2 r_{n }or_{ }d_{n} = 2 (18 + r_{s})

Furthermore, the distance travelled by the train southbound after t = 2 hours is:

d_{s} = 2 r_{s }

If we add these two distances together they should add up to 244 miles according to the problem.

2 (18 + r_{s}) + 2 r_{s }= 244

36 + 2 r_{s} + 2 r_{s }= 244

36 + 4 r_{s }= 244

r_{s }= **52 mph**

r_{n} = 18 + r_{s }= 18 + 52 = **70 mph**