A man travels from Town X to Town Y at an average rate of 50

A man travels from Town X to Town Y at an average rate of 50 mph and returns at an average rate of 40 mph. He takes 1/2 hr longer than he would if he made the round trip at an average of 45 mph. What is the distance from Town X to Town Y?

Solution

t1 = d/50 + d/40 = 9d/200
t2 = 2d/45
t2 = t1 - 0.5
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Sub for t2 in 2nd eqn
t1-0.5 = 2d/45
t1 = 2d/45 + 0.5 = 9d/200
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2d/45 + 0.5 = 9d/200
200*2d + 4500 = 9d*45
400d - 405d = -4500
-5d = -4500
d = 900 miles