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Math Help - Math 20 3 variable systems

  1. #1
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    Math 20 3 variable systems

    I've tried this question so many times and I can't seem to get the right answer. Any takers??

    Solve this system using substitution or elimination:
    d+e-2f=-5
    3d-e-4f=11
    2d+4e+f=4

    Here is what I did:
    3(d+e-2f=-5)
    3d-e-4f=11
    3d+3f-6f=-15
    -3d-e-4f=11
    4e-2f=-26

    2(d+e-2f=-5)
    4e-2f=-26
    2d+2e-4f= -10
    -2d+4e+f=4
    -2e-5f=-14

    2(-2e-5f=-14)
    4e-2f=-26
    -4e-10f=-28
    +4e-2f=-26
    -12f=-54
    f=-9/2

    Then:
    4e-2f=-26
    4e-2(-9/2)=-26
    4e + 9 = -26
    4e = -35
    e= -35/4

    Then:
    d+e-2f = -5
    d+ (-35/4) -2(-9/2) = -5
    d-35/4 + 9 = -5
    d= -91/4

    I have no idea where I went wrong. The answer is(33/4,-17/4,9/2)
    So i got one variable right, but messed up on the others... ugh
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  2. #2
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    Be careful with signs!!!!!

    You obviously understand the process.
    The line -12f = -54 implies f = +9/2.
    Substitute +9/2 as you did with -9/2, and the remaining variables can be found.
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  3. #3
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    oh my gosh.
    how did i miss that??
    thanks
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  4. #4
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    Quote Originally Posted by t-dot View Post
    Solve this system using substitution or elimination:
    d+e-2f=-5 [1]
    3d-e-4f=11 [2]
    2d+4e+f=4 [3]
    You seem to go to a lot of unnecessary work.

    Can be much simpler; like:

    Add [1] + [2] to get 4d - 6f = 6 ; 2d - 3f = 3 [4]

    Multiply [2] by 4 : 12d - 4e - 16f = 44
    Add to [3] to get 14d - 15f = 48 [5]

    Multiply [4] by -7 : -14d + 21f = -21
    Add to [5] to get 6f = 27 ; f = 9/2

    Now get d and e.
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