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Isolating variable confounded by cursed reciprocals

  1. Apr 28, 2014 #1

    Atomised

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    1. The problem statement, all variables and given/known data

    Given

    1) xy = 1

    2) x(t^2)-y = (t^3)-1/t

    Express x in t, y in t



    2. The attempt at a solution


    x/t = [(tx+1)(td-1)] / [(t^2)-1] but I still can't separate t and x, driving me mad it is.

    Also subtracted 1) from 2) to obtain

    (t^2)x - (t^3) +1/t -1/x = 0

    but no progress from there.
     
  2. jcsd
  3. Apr 28, 2014 #2

    AlephZero

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    That's a good start. You can get rid of the fractions by multiplying everything by tx.

    Then, if you know t and you want to find x, what sort of equation do you have?
     
  4. Apr 28, 2014 #3

    Atomised

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    (x - t) (x t^3 + 1) = 0, therefore x = t, and (1) implies y = 1/t.

    Solved thank you very much. I am self studying and this forum is incredibly helpful.
     
  5. Apr 28, 2014 #4

    Ray Vickson

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    If your second equation is
    [tex] t^2 x - y = t^3 - \frac{1}{t}[/tex]
    then, substituting y = 1/x from the first equation gives you
    [tex] t^2 x - \frac{1}{x} = t^3 - \frac{1}{t}[/tex]
    so if you divide through by t you find that the 'variable' ##z = xt## obeys
    [tex] z - \frac{1}{z} = t^2 - \frac{1}{t^2}[/tex]
    There are two solutions for ##z##.
     
  6. Apr 28, 2014 #5

    AlephZero

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    There are two solutions to that quadratic equation. not one. if AB = 0, either A = 0, or B = 0.
     
  7. Apr 29, 2014 #6

    Atomised

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    Full Solution

    [itex]x_1=t\\ y_1=1/x\\ x_2=-1/t^3\\ y_2=-t^3[/itex]
     
  8. Apr 29, 2014 #7

    Atomised

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    It is not immediately apparent to me what the advantage of making this substitution is. Is it that it makes apparent that $$z=t^2$$ and therefore $$x=t$$

    Or could it be that it is suggesting the difference of two squares identity, which I cannot see how to use?




    .
     
  9. Apr 29, 2014 #8

    Ray Vickson

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    You can do it however you want; it all amounts in the end to finding the two solutions of a quadratic equation. However, the form ##z - 1/z = t^2 - 1/t^2## has nice "symmetry" that allows one to spot the two solutions right away, with almost no work. The solutions are ##z = t^2## and ##z = -1/t^2##. The point is that both sides of this equation involve differences of terms that are reciprocals of each other, and that makes the solution easy to write down.

    I cannot figure out why you refuse to acknowledge that there are two solutions.
     
  10. Apr 29, 2014 #9

    Atomised

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    Fully acknowledge two solutions - it was hare brained of me to omit in earlier post. I will experiment with substitutions more from now on.
     
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