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Math Help - System of three equations and four variables

  1. #1
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    System of three equations and four variables

    How to solve this system?
    System of three equations and four variables-333.jpg
    Thanks a bunch!
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    Re: System of three equations and four variables

    ONE question: are you serious?

    Well, I'd attack that (if attackable!) this way:
    you have terms that contain :
    x1x2; let a = that term
    x1x3; let b...
    x1x4; let c...
    x2x3; let d...
    x2x4; let e...
    x3x4; let f...

    So the equations can be rewritten:
    sqrt(c) - sqrt(b) = sqrt(f) [1]
    sqrt(c) - sqrt(a) = sqrt(e) [2]
    sqrt(b) - sqrt(a) = sqrt(d) [3]

    Put those 3 equations through the elimination process...and see what happens...
    Last edited by Wilmer; July 18th 2012 at 04:46 PM.
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    Re: System of three equations and four variables

    Quote Originally Posted by Wilmer View Post
    ONE question: are you serious?
    Why not? Is this problem too easy?

    Quote Originally Posted by Wilmer View Post
    So the equations can be rewritten:
    sqrt(c) - sqrt(b) = sqrt(f) [1]
    sqrt(c) - sqrt(a) = sqrt(e) [2]
    sqrt(b) - sqrt(a) = sqrt(d) [3]
    Put those 3 equations through the elimination process...and see what happens...
    Three equations (square after elimination process) with 6 variables (instead of 4 initial variables) and new roots (after being squared) gives nothing good
    Any new ideas?
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    Re: System of three equations and four variables

    That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the x_i's are ordered.
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    Re: System of three equations and four variables

    Quote Originally Posted by richard1234 View Post
    That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the x_i's are ordered.
    agrees... any new ideas?
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    Re: System of three equations and four variables

    Quote Originally Posted by studenttt View Post
    Why not? Is this problem too easy?
    No; the other way: it appears impossible, with 4 unknowns and 3 equations.
    Do you have any background...like, where does it come from?
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    Re: System of three equations and four variables

    Quote Originally Posted by studenttt View Post
    Three equations (square after elimination process) with 6 variables (instead of 4 initial variables) and new roots (after being squared) gives nothing good
    Any new ideas?
    Nope. And it'll only get worse...
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    Re: System of three equations and four variables

    Quote Originally Posted by Wilmer View Post
    Do you have any background...like, where does it come from?
    No background... teacher took it from his book (wrapped, so I do not know what book it was).
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    Re: System of three equations and four variables

    Don't give up! I'm playing with it now; getting interesting stuff!
    I'll let you know later...

    Have a look here; that's why I say "interesting":
    solve sqrt(a^2-b^2)-sqrt(c^2-d^2)-sqrt(e^2-f^2)=0 for a - Wolfram|Alpha
    Last edited by Wilmer; July 19th 2012 at 06:01 AM.
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  11. #11
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    Re: System of three equations and four variables

    Any new ideas?
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  12. #12
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    Re: System of three equations and four variables

    Not from me!

    I make that a "lost cause": 3 equations, 4 variables; you need minimum of 4 equations.

    Also, almost impossible to manipulate even without the square roots.

    And all of the "math heavyweights" at this site have (wisely!) stayed away from this mess...
    Last edited by Wilmer; July 24th 2012 at 07:15 PM.
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    Re: System of three equations and four variables

    You could set all the x_i's equal to get a solution, but you have four variables/three equations.
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    Re: System of three equations and four variables

    Or one way is to set each term equal to a square, and get:
    c^2 = a^2 + b^2
    c^2 = d^2 + e^2
    b^2 = d^2 + f^2
    ...in other words, that's 3 right triangles, possibly:
    1073^2 = 448^2 + 975^2
    1073^2 = 495^2 + 952^2
    975^2 = 495^2 + 840^2

    Then, using my c^2 (as example):
    x1^2 x4^2 / 4 - (x1 - x4)^2 = 1073^4

    But that also turns out as ugly as the 7 witches of MacBeth!
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  15. #15
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    Re: System of three equations and four variables

    Quote Originally Posted by Wilmer View Post
    I make that a "lost cause": 3 equations, 4 variables; you need minimum of 4 equations.
    Do you think we really need minimum of 4 equations? I know a lot of examples when it is possible to solve a system of eqns where a number of eqns is less than a number of variables. E.g.:
    1) One eq and 2 vars: (x-2)^2+(y-3)^2=0. Answer: x=2, y=3
    2) System of two eqns and three variables:
    (x-2)^2+(y-3)^2+(z-4)^2=0
    x+y+z=9
    Answer: x=2, y=3, z=4.

    Sometimes (a case when a number of eqns is less than a number of variables) we just have more than one solution. E.g.: system of two eqns and three variables:
    x+y+z=5
    x+y-z=3
    Answer: (x,y,z) may take the form (x,4-x,1) so we get many roots.
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