Afreight train completes its journey of 150 miles 1 hour earlier if its original speed is increased by 5 miles/hour. what is the trains original sped?

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.

The answer is 63.!