I've been solving various quadratic equation for some time now, and I found 2 equations that I am unable to solve:

$x^2-y^2=5 $


This is one of the equations. The problem I have is if I exchange $y$ with and expression with $x$ I will always get $xy$, and I just can't get rid of $xy$ and I can't express $xy$ by anything else.

  • $\begingroup$ What's your question? Do you just want the answers, or are you looking for help with the solving process? For $x^2-y^2=5$, did you get as far as $y^2 = x^2 - 5$? I don't see where $xy$ comes in. $\endgroup$ – LarsH Apr 1 '12 at 9:55
  • $\begingroup$ I don't want the answer. Of course I've expressed $y^2$ as $x^2-5$. If i use that in second equation i will always get $xy$ $\endgroup$ – Transcendental Apr 1 '12 at 9:58
  • $\begingroup$ If you want to solve these two equations simultaneously, for both $x$ and $y$, you should probably edit your question. Both @Didier and I misunderstood your goal as wanting to solve each equation for $y$, separately. $\endgroup$ – LarsH Apr 1 '12 at 10:12

I understand that you want to solve these equations simultaneously?

First take the equation $$\frac{1}{x}+\frac{1}{y}=−\frac{1}{6}$$

We'll rearrange this to find $y$: $$\frac{1}{y}=−\frac{1}{6}-\frac{1}{x}$$

Now put the RHS into one fraction: $$\frac{1}{y}=-\frac{6+x}{6x}$$ Then reciprocate both sides: $$y=-\frac{6x}{6+x}$$

Now, you can substitute this into the first equation (or, another method that you prefer). Be careful though - this may result in a quartic equation. Consider writing the first equation as $(x-y)(x+y) = 5$ before substituting.

Hopefully it should be clear what to do from here.

  • 2
    $\begingroup$ Good insight into what the OP (may have) meant, and good pedagogy (anticipating the pitfalls). $\endgroup$ – LarsH Apr 1 '12 at 10:08
  • $\begingroup$ Thanks: your comment discussion helped to confirm this for me. $\endgroup$ – Ronald Apr 1 '12 at 10:10
  • $\begingroup$ Ooooh, so that's what he means by "If i use that in second equation". $\endgroup$ – LarsH Apr 1 '12 at 10:13
  • $\begingroup$ It's a shame that :) is too few characters for a valid comment. ;) $\endgroup$ – Ronald Apr 1 '12 at 10:18

$$y=\pm\sqrt{x^2-5}\qquad\text{and}\qquad y=-6x/(6+x)$$


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