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Math Help - Tell me where i went wrong (systems of equations in 3 variables)

  1. #1
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    Tell me where i went wrong (systems of equations in 3 variables)

    ive tried it a bunch of times ending up with the same thing, the only thing left is to have someone point out where i screwed up. The answer is (9,9,6) but i want to understand how to get there

    Ok so heres the problem

    1) 6x-9y-z=-33
    2)x-3y+4x=6
    3)5x+y+z=60

    I start with 1 and 3 to form equation 4

    1) 6x-9y-z=-33
    3)5x+y+z=60
    ------------------
    4)11x -8y = 27 (cancel z, add)

    Then I use equations 1 and 2 to form equation 5

    1) 6x-9y-z=-33
    2)x-3y+4x=6
    ------------------
    5) 25x - 39y = -126 (I multiply eq1 by 4)

    4) 11x -8y = 27
    5) 25x - 39y = -126
    ------------------------
    -429x = -1053
    200x = 1008
    ------------------
    -229x = -45

    x=-45/-229

    I dont know if im just making a bone headed mistake over and over, or if theres something im straight up forgetting. but this is as far as i get, knowing that x must be 9.
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  2. #2
    Senior Member yeKciM's Avatar
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    hmmm... i don't see why did you do that ? from the start... i assume you have 3 equations ? express from 1st or third "z" and just put in another equations and solve them for x,y ...


    \begin{matrix}<br />
 6x&-9y  &-z  &=33 \\ <br />
 5x&-3y  &  &=6 \\ <br />
 5x& +y & +z & =60<br />
\end{matrix}

    from third ....

     z = 60-5x-y

    and put in first


    \begin{matrix}<br />
 6x&-9y  &-(60-5x-y)  &=33 \\ <br />
 5x&-3y  &  &=6 \\ <br />
\end{matrix}

    and you'll get

     6x-9y-60+5x+y = 33
     5x-3y = 6

    can you continue from here ?



    P.S. x =9 , y = 9 and z=6 are not the solutions .... x = -33/7 ... y = -69/7 ... z= 654/7 are the solutions of the system of equations
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  3. #3
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    your way seems much simpler, however my teacher wants me to use the elimination method for this, because the book we have says to.
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  4. #4
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by NecroWinter View Post
    your way seems much simpler, however my teacher wants me to use the elimination method for this, because the book we have says to.
    okay ...
    than add first to third equation, so you lose "z" don't do anything to the second ...

    \begin{matrix}<br />
6x&-9y &-z &=33 \\ <br />
5x&-3y & &=6 \\ <br />
5x& +y & +z & =60<br />
\end{matrix}

    second ::
     5x-3y = 6

    first + third ::
     11x-8y = 93

    can you go from here ?
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  5. #5
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    yes i can, this is where i go from there (the 33 is negative )

    4)11x -8y = 27 (cancel z, add)

    Then I use equations 1 and 2 to form equation 5

    1) 6x-9y-z=-33
    2)x-3y+4x=6
    ------------------
    5) 25x - 39y = -126 (I multiply eq1 by 4)

    4) 11x -8y = 27
    5) 25x - 39y = -126
    ------------------------
    -429x = -1053
    200x = 1008
    ------------------
    -229x = -45

    x=-45/-229
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  6. #6
    Senior Member yeKciM's Avatar
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    wait, hmmmm.... let's go like this ....

     6x - 9y - z = -33 .......... (1)
     x - 3y + 4x = 6 .......... (2)
     5x + y + z = 60 .......... (3)

    add 1 to 3

    11x -8y = 27 .......... (1+ 3)
    5x - 3y = 6.......... (2)

    multiply (1+3) with (-3) and (2) with (8) ....

    -33x +24y = -81 .......... (1+ 3)
    40x - 24y = 48.......... (2)

    now add them ....

    \displaystyle  7x = -33 \Rightarrow x = - \frac  {33}{7}

    now you have "x" ... and go back to the equation (2) ... and put x in

     x - 3y + 4x = 6 .......... (2)
    \displaystyle  5 \cdot(- \frac{33}{7})- 3y  = 6 .......... (2)

    so now multiply this all with 7 so you have ...

    \displaystyle -165-21y = 42 \Rightarrow -21y = 207 \Rightarrow y = -\frac {69}{7}

    so you have and "x" and "y" now put it in the first (strarting equations) or third one and you'll get "z"



    P.S. I'm again saying that your "answer" that you have isn't correct if you don't trust me put those "solutions" (9,9,6) to the second equation (for example) and see if you have correct answer perhaps it's just typo but in system of equations
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  7. #7
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    Quote Originally Posted by NecroWinter View Post
    ive tried it a bunch of times ending up with the same thing, the only thing left is to have someone point out where i screwed up. The answer is (9,9,6) but i want to understand how to get there

    Ok so heres the problem

    1) 6x-9y-z=-33
    2)x-3y+4x=6
    3)5x+y+z=60

    I start with 1 and 3 to form equation 4

    1) 6x-9y-z=-33
    3)5x+y+z=60
    ------------------
    4)11x -8y = 27 (cancel z, add)
    Okay, that is correct.

    Then I use equations 1 and 2 to form equation 5

    1) 6x-9y-z=-33
    2)x-3y+4x=6
    ------------------
    5) 25x - 39y = -126 (I multiply eq1 by 4)
    I would have said "add 4 times equation 1 to equation 2" but this is correct.


    4) 11x -8y = 27
    5) 25x - 39y = -126
    ------------------------
    -429x = -1053
    200x = 1008
    A little too much "short hand"! I presume you mean that you multiplied equation 4 by -39 to get -429x+ 312y= -1053 and multiplied equation 5 by 8 to get 200x- 312y= 1008 - but 8 times -126 is -1008.

    It should be 200x- 312y= -1008.
    -----------------
    -229x = -45

    x=-45/-229
    When you add -429x+ 312y= -1053 and 200x- 312y= -1008 you get -229x= -2061 so x= -2061/-229= 9.

    I dont know if im just making a bone headed mistake over and over, or if theres something im straight up forgetting. but this is as far as i get, knowing that x must be 9.
    Yep, a bone headed mistake! You dropped a negative sign.
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