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Math Help - A system of equations.

  1. #1
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    A system of equations.

    Dear Mathematical community,

    (x^2)yz+3y-z = 31
    (x^2)+(y^2)-z = 11
    3xy+zy = 24


    I've been working with this problem for the last week or so and it seems every way I approach the problem it becomes over complicated. I've tried every last elimination and substitution method I know along with using inverse matrices. So any help, or constructive comments would be much appreciated.

    Thanks,
    Max Thompson

    P.S. I know the real integer solutions to this problem, the real issue is I don't know how to algebraically arrive at said solutions.
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  2. #2
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    Quote Originally Posted by iammax27 View Post
    Dear Mathematical community,

    (x^2)yz+3y-z = 31
    (x^2)+(y^2)-z = 11
    3xy+zy = 24


    I've been working with this problem for the last week or so and it seems every way I approach the problem it becomes over complicated. I've tried every last elimination and substitution method I know along with using inverse matrices. So any help, or constructive comments would be much appreciated.

    Thanks,
    Max Thompson

    P.S. I know the real integer solutions to this problem, the real issue is I don't know how to algebraically arrive at said solutions.
    The substitution method might be easier than the elimination method in this case because you can easily write equation 2 in terms of z, whereas making the variables have the same coefficients could be tiresome.
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  3. #3
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    OK, seeing as this was quite difficult, I've had a go of it.

    x^2yz + 3y - z = 31
    x^2 + y^2 - z = 11
    3xy + yz = 24.

    From equation 2 we can see that

    z = x^2 + y^2 - 11.


    Substituting into the other equations:

    x^2y(x^2 + y^2 - 11) + 3y - (x^2 + y^2- 11) = 31
    3xy + y(x^2 + y^2 - 11) = 24.


    Working on the second of these equations:

    y(3x + x^2 + y^2 - 11) = 24

    x^2 + 3x + y^2 - 11 = \frac{24}{y}

    x^2 + 3x - 11 = \frac{24}{y} - y^2

    x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2 - 11 = \frac{24}{y} - y^2

    \left(x + \frac{3}{2}\right)^2 - \frac{53}{4} = \frac{24}{y} - y^2

    \left(x + \frac{3}{2}\right)^2 = \frac{53}{4} + \frac{24}{y} - y^2

    x + \frac{3}{2} = \pm \sqrt{\frac{53}{4} + \frac{24}{y} - y^2}

    x = -\frac{3}{2} \pm \sqrt{\frac{53}{4} + \frac{24}{y} - y^2}.


    You will now have to substitute this into

    x^2y(x^2 + y^2 - 11) + 3y - (x^2 + y^2- 11) = 31

    and solve for y.
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  4. #4
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    (@ Prove it) I also took substitution as my first approach due to the enormity of work I would have to do to simplify using elimination, however, with the substitution I found myself arriving at two distinctive end results. One, I almost got everything in terms of x but found myself with a polynomial expression to the degree of 7 with 2 extra z's. Then the other way I found myself dealing with a giant complex fraction with x and z still present in the equation (same as before, 2 z's). I showed this to my physics professor and he's still a work in progress. Along with my Math professor, whom I sat with for an upmost of 2 hours trying to solve this problem.

    So if anyone can actually work through it and show me their work I would love to see, and discuss said work, however, I know most people don't have lots of free time to spend doing arbitrary mathematics.

    Still any thoughts or comments are much appreciated.

    (Sorry, I wrote this response before I saw that you had worked it out, but thanks I'll take A look at your answer and get back to you)
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  5. #5
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    @ prove it

    Thanks for the help, that was fantastic. I think the problem was that I sometimes become too tunnel visioned that I miss to do things like completing the square. Still thanks for the help.
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