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Thread: Basic linear

  1. #1
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    Basic linear

    Hi!
    I´m having some issues solving this equation system.


    x_1+x_2+2x_3 = 3
    x_1-x_2+8x_3 = -1
    2x_1+3x_2+x_3 = 8

    Nomatter what I do I cant eliminate the 2 terms to get one of the unknows alone. No problem eliminating 1 term thought if you ad the first 2 terms x_2 is eliminated... great and now lets eliminate one more term....emm emmm can´t do it! whats wrong...?
    Any hints?
    Thanks in advance
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  2. #2
    Super Member craig's Avatar
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    Quote Originally Posted by Henryt999 View Post
    Hi!
    I´m having some issues solving this equation system.


    x_1+x_2+2x_3 = 3
    x_1-x_2+8x_3 = -1
    2x_1+3x_2+x_3 = 8

    Nomatter what I do I cant eliminate the 2 terms to get one of the unknows alone. No problem eliminating 1 term thought if you ad the first 2 terms x_2 is eliminated... great and now lets eliminate one more term....emm emmm can´t do it! whats wrong...?
    Any hints?
    Thanks in advance
    There are infinitely many solutions to this set of equations, you cannot get simple numeric values for x_1, x_2, x_3.

    The way I would do this is let one of the variables, (I'll use x_3), and let this equal any constant \lambda, so we have x_3 = \lambda

    Now try and solve your equations in terms of lamba( \lambda).

    You already know that x_3 = \lambda, your other 2 solutions will have the form x_n = a + b\lambda where a,b \in \mathbb{R}.

    Hope this helps.
    Last edited by craig; April 1st 2010 at 05:31 AM.
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  3. #3
    MHF Contributor

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    You can add the first two equations to get 2x_1+ 10x_3= 2 which is the same as x_1+ 5x_3= 1. If you multiply the first equation by 3 and subtract the third equation from it you also get x_1+ 5x_3= 1.

    Since those are the same equation, you are right- you cannot continue and get a single equation in one variable. And, as craig says, that means there are an infinite number of solutions.

    What you can do is this: solve x_1+ 5x_3= 1 for x_1 in terms of x_3: x_1= 1- 5x_3. Now put that back into the first equation, x_1+ x_2+ 2x_3= 1- 5x_3+ x_2+ 2x_3= 1- 3x_3+ x_2= 3 so x_2= 2+ 3x_3.

    We can now write the solution set as (x_1, x_2, x_3)= (1- 5x_3, 2+ 3x_3, x_3). We can also write it as (1, 2, 0)+ t(-5, 3, 1) where I have taken x_3 to be the parameter t. The solution set is a plane which, because the set of equations is not homogeneous, does not contain the origin.
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