Results 1 to 3 of 3
Like Tree1Thanks
  • 1 Post By jakncoke

Math Help - Solving for a system of equations

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    New York
    Posts
    30

    Solving for a system of equations

    The problem is as follows:

    Sometimes we may need to solve a system of equations that isn't linear, but wants to be. With this in mind, solve the following system of nonlinear equations for x, y, and z.

    xz + yz + xy = -1
    xz - yz + 2xy = 5
    xz - 2yz + 2xy = 7


    I put the equations into matrices and got so far as xz = 20, -yz = -5, -xy = -16. But now I don't know where to go... any help would be much appreciated!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,607
    Thanks
    591

    Re: Solving for a system of equations

    Hey widenerl194.

    Hint: Try getting y = f(x,z), z = g(x,y) and then substitute both to get a function of h(x) = 0. (Example x = 20/z).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member jakncoke's Avatar
    Joined
    May 2010
    Posts
    388
    Thanks
    80

    Re: Solving for a system of equations

    Note that if you use the first equation to get xz = -1 - yz - xy

    plug it into eqn 2 and 3.

    to get (2) -1 - yz - xy - yx + 2xy = 5 or (2) -2yz + xy = 6
    and again for (3) -1 -yz - xy - 2yz + 2xy = 7 or (3) -3yz + xy = 8
    use yz=p and xy=q to solve the linear system
     \begin{bmatrix}-2 & 1 \\ -3 & 1 \end{bmatrix} x = \begin{bmatrix} 6 \\ 8 \end{bmatrix}
    solving it you get p = -2, q = 2, or yz = -2 and xy = 2.
    Plugging this information into any 3 of your original equations gives you xz = -1

    so you got to solve 3 equations
    (1)xz = -1
    (2)yz=-2
    (3)xy = 2

    taking the first one  z = \frac{-1}{x} plug into (2) to get  y \frac{-1}{x} = -2
    taking 3,  x = \frac{2}{y}
    so  y \frac{1}{\frac{2}{y}} = 2 or the solution to y^2 = 4 which is  y = +2, -2

    now working backwords we get x = +1 or -1, z = +1 or -1.
    Now since xz = -1, set x = -1, z = 1 or x = 1 and z = -1
    and since yz = -2, since z = 1, y = -2 or z = -1 and z = 2

    so the solution set is (x,y,z) where x=+1 or -1, y = +2 or -2 and z = +1 or -1 with the only restriction that the sign of x,z are reverse (e.g if x = 1, then z = -1) and the sign of y and z are reverse (e.g if z = -1, then y = 2)

    The solution set is (x,y,z) =  \{(-1,-2,1), (1,2,-1) \}
    Last edited by jakncoke; February 12th 2013 at 07:57 PM.
    Thanks from widenerl194
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Solving for a system of equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: December 1st 2012, 07:59 PM
  2. solving system of equations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: March 21st 2012, 02:05 AM
  3. Help with solving system of equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 13th 2012, 05:35 PM
  4. Solving system of equations
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: February 15th 2010, 02:19 PM
  5. Solving the system of equations
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 25th 2009, 06:50 PM

Search Tags


/mathhelpforum @mathhelpforum