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Math Help - a very challenging problem! please help(3 variables)

  1. #1
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    a very challenging problem! please help(3 variables)

    ok these are the equations...
    8k - 5n = 8r
    3u + 2o = 189
    5d + 2c = 6m
    5L + 4m = 23b
    3r - m = 3u
    3k - 4i = 2i
    3n - 3o = c
    o - c = 4d
    8i + 9r = 9n
    10m + 7d = 3n
    4d + 4e = 6b
    4y - 6u = 2m
    6b + 3c = 14m

    now i need to find out what each variable really is...any help?
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  2. #2
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    Are you familiar with Matrices and RREF? You can use Matrices to simplify this process a lot.
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  3. #3
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    what?

    no i havent what are those?
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  4. #4
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    Alright. Well first of all, if your teacher won't let you use any method possible than this is pointless. If you can, this involves making a Matrix that has N rows, where N is the number of variables and N+1 columns. Do you have a calculator that can perform Matrix operations?
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  5. #5
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    problem

    uh no i do not. o and this is not for school so i havent exactly learned it yet.
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  6. #6
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    Alright, well this is going to require a ridiculous amount of substitution then. Find a way to substitute two of the three variables into one, solve and repeat. It'll take a while, but you can do it.
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  7. #7
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    meaning?

    what exactly does that mean?
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  8. #8
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    That means, for example, you have:
    x+3y=2
    2x+4y=1
    Then solve for the first equation:
    x=2-3y
    Thus substitute that into 2nd equation:
    2(2-3y)+4y=1
    Now solve for y
    One you have y you can solve for x.
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by jackiekka
    ok these are the equations...
    8k - 5n = 8r
    3u + 2o = 189
    5d + 2c = 6m
    5L + 4m = 23b
    3r - m = 3u
    3k - 4i = 2i
    3n - 3o = c
    o - c = 4d
    8i + 9r = 9n
    10m + 7d = 3n
    4d + 4e = 6b
    4y - 6u = 2m
    6b + 3c = 14m

    now i need to find out what each variable really is...any help?
    This same question has already been asked. The discussion for that
    instantiation can be found here:

    http://www.mathhelpforum.com/math-he...ead.php?t=1552

    RonL
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