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Math Help - Basic linear

  1. #1
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    Basic linear

    Hi!
    Im 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 cant 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!
    Im 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 cant 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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