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Thread: Equation system

  1. #1
    Member Jones's Avatar
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    Equation system

    Hi, i would like some help on how to solve the following equation system:

    \begin{array}{lcr}3x+10y+0.5k = 100 \\<br />
x+y+k = 100 \end{array}

    I know that you normally reduce down a system like this to only two variables

    For equations on the form ax+by=c there is a formula:
    (cu-bn,cv+an), n \in \mathbb{Z}

    But how do you solve this when you have to deal with floating point numbers?
    Last edited by mr fantastic; Oct 10th 2009 at 04:27 PM. Reason: Edited post title
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  2. #2
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    3x + 11y + 0.5k = 100 / *2

    6x + 22y + k = 200
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  3. #3
    Member Jones's Avatar
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    okay,

    So we have \begin{array}{lcr}6x + 20y + k = 200 \\<br />
x + y + k = 100 \end{array}

    If we subtract the second from the first, we end up with 5x+19y=100

    But how are we supposed to get the solution for k from this equation?
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  4. #4
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    Quote Originally Posted by Jones View Post
    Hi, i would like some help on how to solve the following equation system:

    \begin{array}{lcr}3x+10y+0.5k = 100 \\<br />
x+y+k = 100 \end{array}

    I know that you normally reduce down a system like this to only two variables

    For equations on the form ax+by=c there is a formula:
    (cu-bn,cv+an), n \in \mathbb{Z}

    But how do you solve this when you have to deal with floating point numbers?
    Your formula above only gives integer values of x and y. It is not just that a b and c must be integers.

    From 5x+ 19y= 100 you can get x= 20- (19/5)y.

    Putting that back into x+y+k= 100 gives 20- (19/5)y+ y+ k= 100 or -(14/5)y+ k= 80. Then k= (14/5)y+ 80.

    For any value of y, x= 20- (19/5)y and z= (14/5)y+ 80 so(20-(19/5)y, y, (14/5)y+ 80) is a solution.

    If you are only interested in integer values, it is easy to see that x and z will be integers if and only if y is a multiple of 5. If you let y= 5n, that becomes (20-19n, 5n, 14n+80).
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  5. #5
    Member Jones's Avatar
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    Thanks!

    So clearly n=1 is a solution that would work...
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