SOLUTION: two men each travel 500 miles after leaving a certain place at the same time. The first man travels 10 miles per hour faster than the second man and hence arrives 2.5 hours earlier

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Question 414621: two men each travel 500 miles after leaving a certain place at the same time. The first man travels 10 miles per hour faster than the second man and hence arrives 2.5 hours earlier at the destination. How fast did each travel?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

first man travels 10 miles per hour faster than the second man

Let x and (x+10mph)represent the speed of the slower and faster man respectively

Question states***  NOte: D = r*t  OR   D/r = t

500/x - 500/(x+10)  = 2.5

Solving for x

500(x+10)-500x = 2.5x(x+10)

   2.5x^2 + 25x -5000 = 0

      x^2 + 10x -2000= 0

factoring

(x + 50)(x-40) = 0 Note:SUM of the inner product(50x) and the outer product(-40x) = 10x

(x + 50)= 0  |tossing out negative solution for time

(x - 40) = 0  x = 40mph, speed of slower man.  Faster man's speed is 50mph



CHECKING our Answer***

500mi/40mph - 500mi/50mph  = 12.5hr  - 10hr = 2.5hr