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Solving for y

  1. Oct 18, 2005 #1
    I'm trying to solve the following equation for y
    [tex] 2y^2 + xy = x^2 + 3 [/tex]
    So far I've gotten it down to [tex] y = \frac{x^2 + 3}{2y = x} [/tex]
    Or I've tried [tex] (2y-x)(y+x) = 3 [/tex]
    But I'm stuck at that. Any advice would be appreciated.
     
  2. jcsd
  3. Oct 18, 2005 #2
    Well for this sort of thing, I think we need to treat y as a variable and x as a constant. This leaves us with a quadratic expression in y (with x^2 + 3 as a constant), which means that we can complete the square and make y the subject of the equation...

    Final tip: In your final answer, there should be a "plus-minus" symbol somewhere...

    All the best!
     
  4. Oct 19, 2005 #3

    TD

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    You won't be able to get a unique solution for y, since the equation is quadratic in y. If you see that, you can just solve it like any other quadratic equation, using the abc-formula. The only difference is that there won't only be numerical coefficients, but also x's but that's no problem.
     
  5. Oct 19, 2005 #4
    So it would be
    [tex] y = \frac{-x \pm \sqrt{9x^2 + 24} }{4} [/tex]
     
  6. Oct 19, 2005 #5

    HallsofIvy

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    Yes, that's exactly it.
     
  7. Oct 19, 2005 #6
    Much obliged!
     
  8. Oct 19, 2005 #7

    TD

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    Exactly, as you see: you can use it with variables as well :smile:
     
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