SOLUTION: Greetings. Anthony Vandorpe here I am working on a math word problem in the category of systems of linear equations. Here is the general form of the question they gave me:

Question 503548: Greetings.
Anthony Vandorpe here
I am working on a math word problem in the category of systems of linear equations.
Here is the general form of the question they gave me:
there are 2 runners, if they run towards each other, the time elapsed is 12 mins, and the distance is 2 miles presumably they meet at some point in that 2 mile distance).
If they start running in the same (at the same starting point?) direction, the faster runner will overtake the slower runner in 2 hours.
This may seem like a simple one to solve, in addition, once I get the 2 linear equations placed in standard form, I am to solve for the variables, and find I think, the running rate of the faster runner and the running rate of the slower runner. I just don't know how to set up this one. I would appreciate your help. Best regards, Anthony C. Vandorpe
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
there are 2 runners, if they run towards each other, the time elapsed is 12 mins, and the distance is 2 miles presumably they meet at some point in that 2 mile distance).
----
1st runner data:
distance = x miles ; time = 1/6 hr ; rate = distance/time = 6x mph
---
2nd runner data:
distance = 2-x miles ; time = 1/6 hr; rate = distance/time = 12-6x mph
-------------------------------------
If they start running in the same (at the same starting point?) direction, the faster runner will overtake the slower runner in 2 hours.
1st runner data:
rate = 6x mph ; time = 2 hrs ; distance = rt = 12x miles
---
2nd runner data:
rate = 12-6x mph ; time = 2 hrs ; distance = rt = 24-12x miles
----------------
Solve for "x":
distance = distance
12x = 24-12x
24x = 24
x = 1 mile
--------------------
You didn't say what the question was but
with the value of "x" you can figure out
the time and rate for each of the runners.
==============================================
Cheers,
Stan H.

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