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algebra question

Topic closed. 6 replies. Last post 12 years ago by johnph77.

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Canada
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October 25, 2003
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Posted: May 16, 2005, 10:38 pm - IP Logged

Hi, I'm working on something of a system for pick 3.  However I need help in calculating by using algebra. I am not familiar with algebra, so I was wondering how to get the correct answer.  If someone can give me the correct answer, It will be greatly appreciated.  This ties in with with a system I'm working with( which if successful I'll gladly share with everyone). So here goes:

A - 1437 = B   

B - 402 = C

C = ( A - (1437 + 402 )

So the obvious question is what are the values for A B & C.

Like I said this ties into a system, which I hope to work! Good luck

    Blackie's avatar - Norfolk 20Sunrise%20Nov%2016.jpg
    Norfolk , Va
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    Posted: May 16, 2005, 10:58 pm - IP Logged

    Don't know if it will help but you can look in algebra if you scroll down and see if it will help.

    http://www.newton.dep.anl.gov/askasci/math98.htm<http://www.newton.dep.anl.gov/askasci/math98.htm>

     

     Good Luck,

    Blackie.                           

      Littleoldlady's avatar - basket
      Clarksville
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      Posted: May 16, 2005, 10:59 pm - IP Logged

      Flaw, you have C equaling 2 different values.  You have not defined A very clear..the way u have it set up, the value for A would be B+1437...this looks like some type of circle..what is the original equation?

      If you know your number is going to hit, have patience and then KILL IT!

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        Canada
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        Posted: May 16, 2005, 11:29 pm - IP Logged

        Sorry for any confusion, but like I said I'm not familiar with algebra.

         

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          Canada
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          Posted: May 16, 2005, 11:34 pm - IP Logged

          The  correct numbers are A = 2459 B= 1022 C = 620 .  How do I arrive at the correct numbers for A B and C next time? These numbers are based on calculations for previous draws.

            RJOh's avatar - chipmunk
            mid-Ohio
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            Posted: May 17, 2005, 12:28 am - IP Logged

            If those numbers were calculated, you can calculate the new numbers the same way.  If they were observed then you won't know what the new ones are until you observe them.  Some one in the math section may be able to tell you what you need to calculate the results you are looking for.  Good luck to you.

            RJOh

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              johnph77's avatar - avatar
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              Posted: May 17, 2005, 2:24 am - IP Logged



              The  correct numbers are A = 2459 B= 1022 C = 620 .  How do I arrive at the correct numbers for A B and C next time? These numbers are based on calculations for previous draws.



              You could have assigned any value to any one of your variables (A, B or C) and, as long as the difference between the variables remained constant, the equations would be correct. As an example, let's say C=0. Solving the formulas for the remaining variables you would have B=402 (B=C+402) and A=1839 (B=A-1437). So the question here wouldn't be what the correct numbers were, but how were the formulas derived? What determined the value of the constants 1437 and 402?

              gl

              j

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