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Simultaneous confusion

  1. Feb 20, 2004 #1
    how can the following pair of simultaneous eqns. be solved

    a1X^m +b1Y^n =c1
    a2X^m’+b2Y^n’=c2
    Please help:frown: [b(]
     
  2. jcsd
  3. Feb 20, 2004 #2

    NateTG

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    What do you mean by m' ? Is it supposed to be the derivative of m?
     
  4. Feb 20, 2004 #3
    No by m' I mean that the exponent of X in the first equation is not the same as the second equation.
    Here, Ive rewritten the equation:
    a1X^m +b1Y^n =c1
    a2X^p +b2Y^q =c2
     
  5. Feb 20, 2004 #4

    NateTG

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    So you have
    [tex]a_1x^{n_1}+b_1y^{m_1}=c_1[/tex]
    and
    [tex]a_2x^{n_2}+b_2y^{m_2}=c_2[/tex]
    Then
    [tex]x=(\frac{c_2-b_2y^{m_2}}{a_2})^{\frac{1}{n_2}}[/tex]
    Then you can substitute to get:
    [tex]a_1(\frac{c_2-b_2y^{m_2}}{a_2})^{\frac{n_1}{n_2}} + b_1y^{m_1}=c_1[/tex]
    which is an equation in a single variable. I'm not sure off the top of my head where to go from here, but that can definitely be solved numerically.
     
  6. Feb 20, 2004 #5
    I reached had reached so far, but I noticed that this is not an equation in one variable(duh!). But if the powers were really big it would become a hell of a job. Is there another way or a easier way to proceed from here.
     
  7. Feb 20, 2004 #6

    NateTG

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    Is there a particular problem that you're trying to solve?
     
  8. Feb 20, 2004 #7
    NO I'm thinking about this on a general basis.
     
  9. Feb 21, 2004 #8

    NateTG

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    Ok, so what are the variables, and what is known?
     
  10. Feb 21, 2004 #9
    a1X^m +b1Y^n =c1
    a2X^p +b2Y^q =c2

    That is the eqn.
    X,Y are the variables.
    Now please help me solve it
     
  11. Feb 21, 2004 #10

    matt grime

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    Sorry, but we can't help you. Not in general. You can rearrange all you want but, in general, you're going to end up with some high degree polynomial in one variable for which there are no general methods of solution (you can do up to degree 4 easily, degree 5 with a bit of ingenuity, but beyond that you're hoping for luck). I mean of course some nice exact algebraic solution. You could do it numerically and get approximate answers.

    You can get various constraints on the solutions (are they integers etc) but the methods would be ad hoc.

    Sorry, but that's what maths looks like in real life.
     
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