Results 1 to 9 of 9

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

  1. #1
    Newbie
    Joined
    Jan 2006
    Posts
    4

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Are you familiar with Matrices and RREF? You can use Matrices to simplify this process a lot.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2006
    Posts
    4

    what?

    no i havent what are those?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    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?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2006
    Posts
    4

    problem

    uh no i do not. o and this is not for school so i havent exactly learned it yet.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2006
    Posts
    4

    meaning?

    what exactly does that mean?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    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
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Challenging Problem
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 22nd 2009, 11:18 AM
  2. Challenging Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: August 15th 2009, 09:53 AM
  3. Challenging problem
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: July 13th 2009, 05:49 AM
  4. challenging sequences problem
    Posted in the Pre-Calculus Forum
    Replies: 7
    Last Post: February 26th 2009, 04:41 PM
  5. challenging probability problem
    Posted in the Statistics Forum
    Replies: 1
    Last Post: September 20th 2007, 03:49 AM

Search Tags


/mathhelpforum @mathhelpforum