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Thread: Basic linear

  1. #1
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    Basic linear

    Hi!
    Im having some issues solving this equation system.


    $\displaystyle x_1+x_2+2x_3 = 3$
    $\displaystyle x_1-x_2+8x_3 = -1$
    $\displaystyle 2x_1+3x_2+x_3 = 8$

    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.


    $\displaystyle x_1+x_2+2x_3 = 3$
    $\displaystyle x_1-x_2+8x_3 = -1$
    $\displaystyle 2x_1+3x_2+x_3 = 8$

    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 $\displaystyle x_1, x_2, x_3$.

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

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

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

    Hope this helps.
    Last edited by craig; Apr 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 $\displaystyle 2x_1+ 10x_3= 2$ which is the same as $\displaystyle x_1+ 5x_3= 1$. If you multiply the first equation by 3 and subtract the third equation from it you also get $\displaystyle x_1+ 5x_3= 1$.

    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 $\displaystyle x_1+ 5x_3= 1$ for $\displaystyle x_1$ in terms of $\displaystyle x_3$: $\displaystyle x_1= 1- 5x_3$. Now put that back into the first equation, $\displaystyle x_1+ x_2+ 2x_3= 1- 5x_3+ x_2+ 2x_3= 1- 3x_3+ x_2= 3$ so $\displaystyle x_2= 2+ 3x_3$.

    We can now write the solution set as $\displaystyle (x_1, x_2, x_3)= (1- 5x_3, 2+ 3x_3, x_3)$. We can also write it as $\displaystyle (1, 2, 0)+ t(-5, 3, 1)$ where I have taken $\displaystyle x_3$ 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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