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Math Help - three variables

  1. #1
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    three variables

    What I am doing wrong here?

    Also does it matter that there are only two variables in the third line? All the examples I have looked at for solving three equations with three variables had three variables in each line.

    2x + y + z = 21
    2y + x + z = 20
    2x + 2y = 22

    define x

    x + y = 11
    x = y -11

    substitute into line 1 and 2:

    2(11 - y) + y + z = 21
    22 - 2y + y + z = 21
    22 + z = 21 + y
    z = y - 1

    2y + (y-11) + z = 20
    3y - 11 + z = 20
    3y + z = 31

    substitute again:

    3y + (y-1) = 31
    4y = 32
    y = 8


    thanks!
    Last edited by aquajam; December 17th 2009 at 04:53 PM.
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  2. #2
    Super Member bigwave's Avatar
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    i would suggest elimanating the z in eq1 and eq2
    and factor out 2 in the eq3

    this will make it very easy

    you should get

    x=6
    y=5
    z=4
    Last edited by bigwave; December 17th 2009 at 05:18 PM. Reason: more info
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  3. #3
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    Sorry, I don't follow. You would need to explain it a bit further.

    Also, I would really like to know why my method doesn't work, and if the lack of three variables in the third line is an issue.


    thanks
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  4. #4
    Super Member bigwave's Avatar
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    your method probably works but substituting in equations into equations gets very difficult and error prone, you want reduce the equations down to very simple terms

    so the first 2 equations are

    2x+y+z=21
    2y+x+z=20 (multiply thru with -1 and re oder the terms)

    so now we have

    2x+y+z=21
    -x-2y-z=-20

    add these and we have

    x - y = 1

    now take eq3 factor out 2 you have

    x + y = 11

    now add these equations together and we have

    x=6

    y and z are very easy to get from here

    did you see the magic??
    Last edited by bigwave; December 17th 2009 at 05:33 PM. Reason: spelling
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  5. #5
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    Yes I see.

    But is that method only possible because of the third line in this case?

    What about something like:

    2x + y + z = 72
    2y + x + z = 64
    2z + x + y = 68

    adding lines 1 and 2

    x - y = 8

    then I would have to follow my original method of substitution, no?

    Sorry, I'm just trying to figure out how the proper method for doing these.
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  6. #6
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    Thumbs up

    Quote Originally Posted by aquajam View Post
    define x

    x + y = 11
    x = y -11
    it should be x = 11 - y
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  7. #7
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    ^^
    That's what I thought at first, and I tried what you suggested. I must have made a mistake somewhere, I'll try again.
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  8. #8
    Super Member bigwave's Avatar
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    not necessarily
    with 3 variables in 3 equations
    try to eliminate one of variables first
    it doesn't mater which one just pick the easy one first
    often you can eliminate 2 variables in one sweep
    again try avoid plugging in an equation into an equation with simultaneous equations
    there are times when there is no choice.. but don't see it here.

    the next simultaneous equation can be very easy by just eliminating the z

    not questioning your method.... but it takes more steps
    my view anyway.
    Last edited by bigwave; December 17th 2009 at 06:04 PM. Reason: spelling
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  9. #9
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    Okay, I see my mistake in the first one.

    I had I should have used (11-y) in the second substitution. Had it right in the first one.

    I also see your point about being eliminate variables more simply. It's just all the tutorials I've looked at, suggested the longer way.
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