Results 1 to 12 of 12

Math Help - Finding solution (X, Y, Z) that satisfies EQW

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    22

    Finding solution (X, Y, Z) that satisfies EQW

    Hey i've been strugling with this question for quite some time now, almost long enough for me to give up. Is there ANYONE out there who can solve it?

    Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW

    Xy = Z X Y
    XZ = Y X Z
    YZ = X Y Z
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by Mathmen View Post
    Hey i've been strugling with this question for quite some time now, almost long enough for me to give up. Is there ANYONE out there who can solve it?

    Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW

    Xy = Z X Y
    XZ = Y X Z
    YZ = X Y Z
    Use the first equation to get Z in terms of X and Y, then plug that Z into the other two equations, those two will then become 2 equations in 2 unknowns and will be solveable for X and Y. Once you have X and Y, you can use them in the first equation to solve for Z.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    22
    Y2X + XY + Y2 + Y = X
    X2Y + X2 +XY +X = y


    I know but this is what i ended up with:

    Y2X + XY + Y2 + Y (X2Y + X2 + XY + X) = Z -(Y2X + XY + Y2 + Y) -(X2Y + X2 +XY +X)
    Does anyone know if i'm on the right track. I'm not really sure how to simplify this =/
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,660
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Mathmen View Post
    Find all the solutions (X, y, Z) of Real Numbers which satisfy the following EQW
    Xy = Z – X – Y
    XZ = Y – X – Z
    YZ = X – Y – Z
    Notice that (0,0,0) is a solution.

    If Z=0~\&~XY\not=0\, \Rightarrow \,\left\{ \begin{array}{l} x - y = 0 \\  y - x = 0 \\  \end{array} \right.\, \Rightarrow \,x = y\, \Rightarrow \,x^2  =  - 2x.
    We can rule that out.

    Suppose that XYZ \not=0.
    \left\{ \begin{array}{l}<br />
 XY = Z - X - Y \\ <br />
 XZ = Y - X - Z \\ <br />
 YZ = X - Y - Z \\ <br />
 \end{array} \right.\, \Rightarrow \,XY + XZ =  - 2X\, \Rightarrow \,Y + Z =  - 2
    Now carry on.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,367
    Thanks
    42
    You should have obtained (I prefer little variables)

     <br />
z = xy + x + y<br />

    from which substituting into the remaining two equations

     <br />
x(xy + x + 2y + 2) = 0,<br />
     <br />
y(xy + 2x + y + 2) = 0,<br />

    Now consider the four cases x,y = 0, x not 0, y not zero, both x,y not zero.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2009
    Posts
    22
    Quote Originally Posted by danny arrigo View Post
    You should have obtained (I prefer little variables)

     <br />
z = xy + x + y<br />

    from which substituting into the remaining two equations

     <br />
x(xy + x + 2y + 2) = 0,<br />
     <br />
y(xy + 2x + y + 2) = 0,<br />

    Now consider the four cases x,y = 0, x not 0, y not zero, both x,y not zero.
    The thing is, it is exactly from here i'm really not sure of what i'm doing or what algrebraic rules work. Anyone who could help me get it? hope it is not too much to ask

    Regards Mathmen
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,660
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Mathmen View Post
    The thing is, it is exactly from here i'm really not sure of what i'm doing or what algrebraic rules work. Anyone who could help me get it? hope it is not too much to ask
    Have you looked into my approach?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    May 2009
    Posts
    22
    Quote Originally Posted by Plato View Post
    Have you looked into my approach?
    Yes i analysed your approach and it did make sense but do i plug in the statements in the equation and from there work out what x.y, and z is?

    if it is possible could you show me where your approach leads to, because now i have just been getting different solutions can't really figure out which way is the right way
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,660
    Thanks
    1616
    Awards
    1
    From the point I said "carry on", you can do the same steps and get:
    \left\{ \begin{array}{l}<br />
 X + Y =  - 2 \\ <br />
 X + Z =  - 2 \\ <br />
 \end{array} \right..

    What system of three is easy to solve.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    May 2009
    Posts
    22
    Quote Originally Posted by Plato View Post
    From the point I said "carry on", you can do the same steps and get:
    \left\{ \begin{array}{l}<br />
 X + Y =  - 2 \\ <br />
 X + Z =  - 2 \\ <br />
 \end{array} \right..

    What system of three is easy to solve.
    Is this as simplified as the equation can get or can you find out what one unknown is equal to?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,660
    Thanks
    1616
    Awards
    1
    Quote Originally Posted by Mathmen View Post
    Is this as simplified as the equation can get or can you find out what one unknown is equal to?
    Can you solve a system of three equations in three unknowns?
    It cannot be any easier than that!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Newbie
    Joined
    May 2009
    Posts
    22
    Quote Originally Posted by Plato View Post
    Can you solve a system of three equations in three unknowns?
    It cannot be any easier than that!
    If that is the answer i guess i finally understood it Thanks a lot for all your help Plato and the others, you sure did help me.

    Best regards Mathmen
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show a solution satisfies the DE
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: March 28th 2011, 01:59 PM
  2. Finding a Non-linear Map That Satisfies af(x) = f(ax)
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: November 14th 2010, 01:20 AM
  3. Finding a constant that satisfies a limit
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: November 8th 2010, 12:34 PM
  4. Finding the solution of Un and Vn
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: March 21st 2010, 05:43 PM
  5. Finding the general solution from a given particular solution.
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: October 7th 2009, 01:44 AM

Search Tags


/mathhelpforum @mathhelpforum