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Math Help - algebra system

  1. #1
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    algebra system

    I have to write the following in terms of b1, b2, b3

    x2 + x3 = b1
    x1 + x3 = b2
    x1 + x2 = b3


    ex: x1 = only b's
    x2 = only b's
    x3 = only b's

    I can't figure it out?? Thanks for any advice.
    Last edited by jzellt; October 28th 2008 at 12:26 AM.
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  2. #2
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    Quote Originally Posted by jzellt View Post
    I have to write the following in terms of b1, b2, b3

    x2 + x3 = b1
    x1 + x3 = b2
    x1 + x2 = b3


    ex: x1 = only b's
    x2 = only b's
    x3 = only b's

    I can't figure it out?? Thanks for any advice.
    This is how I'd go about it, be aware that it's a bit cumbersome and I'm sure someone else will be able to show you another way of going about it.

    label x2 + x3 = b1 as equation 1
    x1 + x3 = b2 as equation 2
    x1 + x2 = b3 as equation 3.

    You can start with any equation, I'll start with 1. rewrite in terms of an x value. I'll use x3.

    x3 = b1 - x2

    sub into equation two, x1 + b1 - x2 = b2

    rewrite interms of an x, I'll use x1.

    x1 = b2 - b1 + x2

    sub into equation 3.

    b2 - b1 + x2 + x2 = b3

    rewrite interms of x2

    x2 = (1/2)(b3 - b2 + b1)

    That's one of your answers, now sub that back into equation 1.

    (1/2)(b3 - b2 + b1) + x3 = b1

    rewrite for x3:

    x3 = b1 - (1/2)(b3 - b2 + b1)

    That's the second of your answers, finally sub this value into equation two.

    x1 + b1 - (1/2)(b3 - b2 + b1) = b2

    rewrite for x1

    x1 = b2 - b1 + (1/2)(b3 - b2 + b1)

    Hope that works out right!
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  3. #3
    MHF Contributor

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    Another way to do this, perhaps slightly simpler, is to subtract the equations.

    You have x2 + x3 = b1, x1 + x3 = b2, and x1 + x2 = b3. Subtracting the first equation from the second eliminates x3: x1- x2= b1- b2. The third equation does not have x3 and just adding it to this new equation eliminates x2: 2x1= b1-b2+b3. x1= (1/2)(b1-b2+b3). You can now put that value of x1 back into the third equation and solve for x2. Once you have both x1 and x2, putting those expression into either of the first two equations gives and equation for x3.
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