Let the train's original speed be s. Recall that distance = (speed)(time). Here, 150 miles = s(time), or 150 = s*t. If its original speed is increased by 5 mph, then the time required tomake the trip is 1 hour less than before;
distance = speed times time, so
150 mi = (s+5)(t-1). Let's eliminate t and solve for s:
Since 150 = s*t, t = 150/s. Subbing 150/s for t in the 2nd equation, we get:
150 s
150 mi = (s+5)(150/s - 1), or 150 = (s+5)(------- - ---- )
s s
Mult. both sides by s to elim. the fraction(s):
150s = (s+5)(150 - s)
Then 150s = 150s - s^2 + 750 - 5s, or
0 = -s^2 - 5s + 750
or 0 = s^2 + 5s - 750
Thus, 0 = (s-25)(s+30), and the roots are s=25 and-30. Only a positive original speed makes sense, so the answer is s = 25 mph.