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Math Help - system of equations

  1. #1
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    system of equations

    I have to try and solve this system of equations but don't know where to start

    4u + v = 2y + 2z = j + k
    2u - 2v = 2y - 2z = j - 2k
    2u + 2v = 4y - z = j - k
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  2. #2
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    Re: system of equations

    Quote Originally Posted by moriman View Post
    I have to try and solve this system of equations but don't know where to start

    4u + v = 2y + 2z = j + k
    2u - 2v = 2y - 2z = j - 2k
    2u + 2v = 4y - z = j - k
    That is is 6 equations in 6 variables.

    First separate the equations:
    (1) 4u+ v= 2y+ 2z
    (2) 2u- 2v= 2y- 2z
    (3) 2u+ 2v= 4y- z

    (4) 2y+ 2z= j+ k
    (5) 2y- 2z= j- 2k
    (6) 4y- z= j- k

    Add 2 times (1) to (2) to get 10u= 6y+ 2z or (7) 5u= 3y+ z
    Add (2) and (3) to get (8) 6u= 6y- 3z .

    Multiply (7) by 6 to get 30u= 18y+ 6z and multiply (8) by 5 to get 30u= 30y- 3z.
    That gives 18y+ 6z= 30y- 3z which reduces to 12y= 9z or 4y= 3z. Replacing y by (3/4) z, 30u= (45/2)z- 3z= (39/2)z so that u= (39/30)z= (13/10)z. Replace u and y by those in (1) and solve for v as multiple of z. You can now replace y in (4), (5), and (6) by (3/4)z to get three equations in z, j, and k. Solve those three equations for the three values, then use that value of z to find y, u, and v.
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  3. #3
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    Re: system of equations

    thanks.

    Here what i've done

    (1) 4u + v= 2y + 2z
    (2) 2u - 2v= 2y - 2z
    (3) 2u + 2v= 4y - z
    (4) 2y + 2z= j + k
    (5) 2y - 2z= j - 2k
    (6) 4y - z= j - k

    2*(1)+(2) 10u = 6y + 2z ____ 5u = 3y + z (7)

    Add (2) and (3) to get (8) 6u= 6y- 3z .
    Should that actually be

    (2)+(3) 4u= 6y - 3z (8)

    4*(7) 20u = 12y + 4z (9)
    5*(8) 20u = 30y - 15z (10)

    so 12y + 4z = 30y - 15z
    19y = 18z
    y = (18/19)z

    substituting this in (10) 20u = (540/19)z - 15z
    u = (51/76)z

    Replace u and y by those in (1)and solve for v as multiple of z.
    substituting u and y in (1) v = (23/19)z

    You can now replace y in (4), (5), and (6)by (3/4)z to get three equations in z, j, and k.
    y = (18/19)z

    (4) 2y + 2z= j+ k _______ j + k = (74/19)z
    (5) 2y - 2z= j- 2k ______ j - 2k = -(2/19)z
    (6) 4y - z= j- k ________ j - k = (53/19)z

    (4)+(6) j = (127/38)z
    substituting j in (5) k = 131/76 z

    Solve those three equations for the three values, then use that value of z to find y, u, and v.
    What values of y, u and v am I solving for?
    I already have
    y = (18/19)z
    u = (51/76)z
    v = (23/19)z

    I know the values for j, k, u, v and y that I have are only correct in (1) but don't know what I've done wrong.

    [edit]
    My original problem with this system was that I thought I only had 3 equations no matter what I did with them. i.e.

    (1) 4u + v= 2y + 2z
    (4) 2y + 2z= j + k

    isn't this just one equation, telling me that 4u + v = j + k ?

    Thanks for your help
    Last edited by moriman; September 26th 2011 at 02:01 AM.
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  4. #4
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    Re: system of equations

    Where d'heck do these equations come from?!

    (1) 4u+ v= 2y+ 2z
    (2) 2u- 2v= 2y- 2z
    (3) 2u+ 2v= 4y- z

    An integer solution for these 3: u=15, v = 14, y=19, z=18

    As far as j and k goes, these seem to have been thrown in there in a "hope for the best" manner:
    in other words, there' s something wrong with them....but I can't make it out exactly...

    Are you SURE there is a solution?
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  5. #5
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    Re: system of equations

    The last 3 equations are a dependant system. (As are the first 3)

    Solutions for these three:

    2y + 2z = j + k (1)
    2y - 2z = j - 2k (2)
    4y - z = j - k (3)

    (1)-(2) 4z=3k (4)

    2*(1)-(3) 5z=j+3k (5)

    substituting z=(3/4)k in (5) gives j=(3/4)k so j=z

    substituting these back into (1) y=(1/8)k

    so one of the infinite solutions would be y=1 z=6 j=6 k=8
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  6. #6
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    Re: system of equations

    I think I've finally figured that it can't be figured

    4u + v = 2y + 2z = j + k
    2u - 2v = 2y - 2z = j - 2k
    2u + 2v = 4y - z = j - k

    Set 1

    4u + v = 2y + 2z
    2u - 2v = 2y - 2z
    2u + 2v = 4y - z

    valid when...

    ..... v = (14/15)u
    ..... y = (19/15)u
    ..... z = (6/5)u

    Set 2

    2y + 2z = j + k
    2y - 2z = j - 2k
    4y - z = j - k

    valid when...

    ..... y = (1/8)k
    ..... z = (3/4)k
    ..... j = (3/4)k

    Set 3

    4u + v = j + k
    2u - 2v = j - 2k
    2u + 2v = j - k

    valid when...

    ..... u = (9/8)k
    ..... v = (1/4)k
    ..... j = (15/4)k

    I think this means the 3 sets cannot be correct simultaneously since we have

    ..... j = (3/4)k
    ..... j = (15/4)k

    and also

    ..... v = (14/15)u where 14*(9/8)k <> 15*(1/4)k

    Am I correct that there is no solution (apart from the obvious zero solutions)?

    Thanks
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  7. #7
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    Re: system of equations

    Quote Originally Posted by moriman View Post
    I think I've finally figured that it can't be figured
    ...............
    Am I correct that there is no solution (apart from the obvious zero solutions)?
    Agree.
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