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-   -   Solving Systems of Equations in Three Variables (https://www.askmehelpdesk.com/showthread.php?t=403343)

  • Oct 6, 2009, 03:30 PM
    N_Rivera
    Solving Systems of Equations in Three Variables
    Ok so I don't know I can do them but sometimes I get one variable wrong but the other two are correct, maybe I'm mixing up the equations or numbers?


    7x+5y+z=0
    -x+3y=2z=-16
    x-6y-z=-18



    3x-5y+z=9
    x-3y-2z=-8
    5x-6y+3z=15


    4x-3y+2z=12
    x+y-z=3
    -2-2y+2z=5

    -x-3y+z=54
    4x+2y-3z=-32
    2y+8z=78
  • Oct 6, 2009, 04:07 PM
    sGt HarDKorE

    You have to solve for a variable first. Tip: Solve for the easiest one. For example in your first one, I'd sovle for z and use this equation.

    7x+5y+z=0
    -7x -7x
    --------------------
    5y+z=-7x
    -5y -5y
    ----------------------
    z=-7x-5y

    Now plug that into the other equations:

    -x+3y=2z=-16
    x-6y-z=-18

    The way to know if your answers are correct are try plugging them in once you have found all the variables.

    Edit: I noticed in your first problem that theirs 2 equation signs in "-x+3y=2z=-16." Assuming it is not a typo, you could just do the following to find z.

    2z=-16
    z=-8
  • Oct 9, 2009, 01:02 AM
    morgaine300

    Can you show us the work you've done? It would help to straighten out the issue you're having.
  • Oct 9, 2009, 09:04 AM
    Perito
    Quote:

    I don't know I can do them but sometimes I get one variable wrong but the other two are correct, maybe I'm mixing up the equations or numbers?

    7x+5y+z=0
    -x+3y=2z=-16
    x-6y-z=-18

    It sounds like you do know how to do them, but you make mistakes. Nearly everyone does that. That's why you substitute back into all of the equations once you find a solution.

    Have you learned Cramer's Rule yet? For me, that simplifies the type of equation you've got -- but you can even make mistakes with that.

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