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Math Help - [SOLVED] simultaneous equations

  1. #1
    Member i_zz_y_ill's Avatar
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    [SOLVED] simultaneous equations

    What is the most efficient way to solve this?

    x+y+z=1
    k(0.1^2x+0.01y+0.01z)=0.02
    k(0.01x+0.2^2y+0.01z)=0.03
    k(0.01x+0.01y+0.3^2z)=0.04

    for x,y,z and k?
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  2. #2
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    Using matrices and a computer.
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  3. #3
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    Quote Originally Posted by i_zz_y_ill View Post
    What is the most efficient way to solve this?

    x+y+z=1
    k(0.1^2x+0.01y+0.01z)=0.02
    k(0.01x+0.2^2y+0.01z)=0.03
    k(0.01x+0.01y+0.3^2z)=0.04

    for x,y,z and k?
    Dear i_zz_y_ill,

    x+y+z=1------------(1)
    k(0.1^2x+0.01y+0.01z)=0.02----------(2)
    k(0.01x+0.2^2y+0.01z)=0.03----------(3)
    k(0.01x+0.01y+0.3^2z)=0.04----------(4)

    First substract equation (4) and equation (2) ; you could find kz

    Then substract equation (3) and equation (2); you could find ky

    By substitution kz and ky to one of the equaitons (2),(3) or (4) you could find kx.

    Now x+y+z=1\Rightarrow{kx+ky+kz=k} you could find the value of k. Since you know the values of kx,ky and kz now you can find x,y and z

    Hope this will help you.
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  4. #4
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    The first thing I would do is multiply all except the first equation by 100!

    k(x+ y+ z)= 2

    k(x+ 4y+ z)= 3

    k(x+ y+ 9z)= 4

    Since the first equation was x+ y+ z= 1, k(x+ y+ z)= k= 2.

    Now, x+ 4y+ z= 3/2 and subtracting x+ y+ z= 1 from that
    3y= -1/2 so y= -1/6.

    x+ y+ 9z= 2 and subtracting x+ y+ z= 1 from that, 8z= 1 so z= 1/8.

    x- 1/6+ 1/8= x- 4/24+ 3/24= x- 1/24= 1 so x= 25/24.
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  5. #5
    Member i_zz_y_ill's Avatar
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    im sorry could you show me how exactly?
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