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I spent hours trying to solve this system of equations, they're two equations with two variables, and both are 2nd degree equations, I don't know, both equal zero, so each one equals the other, but I came to a step where I'm stuck.

The system is: \begin{cases} x^2+y^2-xy=21\\ x^2-8y^2+2xy = 0 \end{cases}

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    $\begingroup$ wolframalpha.com/input/… $\endgroup$ – Pawel Dec 17 '16 at 8:25
  • $\begingroup$ Perfect, thanks all, I don't know how did i forget the factoring solution, or just like Claude did. $\endgroup$ – Stephen Alexander Dec 17 '16 at 9:11
  • $\begingroup$ Please edit my solution if it requires $\endgroup$ – Blaise Thunderstorm Dec 18 '16 at 12:37
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Hint

Start setting $y=kx$ which makes the equations to be $$k^2 x^2-k x^2+x^2-21=0$$ $$-8 k^2 x^2+2 k x^2+x^2=0$$ From the second one, assuming $x\neq 0$ you end with $$8 k^2 -2 k+1=0$$ the roots of which being $k=-\frac 14$ and $k=\frac 12$. For each root, go back to the first equation which is simple.

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  • $\begingroup$ I looked into your method, it's quite nice, but, I can't use it anywhere right? I mean, a variable can hold anything, even 0, but I can't use this method in any equation unless they told me x is not equal to 0, right? Though, yours worked here. $\endgroup$ – Stephen Alexander Dec 17 '16 at 9:44
  • $\begingroup$ You don't need to be told x is not equal to 0. Just check it yourself. Assume x is 0. Then per the 1st equation y=sqrt(21) and per the 2nd equation y=0. CONTRADICTION!; hence we have proved that x is not 0. en.wikipedia.org/wiki/Proof_by_contradiction $\endgroup$ – Jose Antonio Reinstate Monica Dec 17 '16 at 13:03
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    $\begingroup$ @JoseAntonioDuraOlmos. I totally agree ! What was appealin (to me) was the homogeneous aspect of the equation. Cheers. $\endgroup$ – Claude Leibovici Dec 17 '16 at 13:34
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Take second eq as $8\cdot y^2-2\cdot x\cdot y-x^2=0$

Solve for $y$

$y=\frac{x}{2},\frac{-x}{4}$

Put in first equation

You would get

$x=\pm\sqrt{28},\pm4$ and $y=\pm\sqrt{7},\mp1$

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  • $\begingroup$ Very nice--this is simple, straightforward, the kind of solution anyone could do. But your final line is confusing and could use some editing in MathJax. Shall I do it for you? $\endgroup$ – Rory Daulton Dec 17 '16 at 19:17

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