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$$ \begin{cases} \left(\frac{1+\sqrt{3}}{2}\right)^2x+\left(\frac{1+\sqrt{3}}{2}\right)y+1=0\\ \left(\frac{1-\sqrt{3}}{2}\right)^2x+\left(\frac{1-\sqrt{3}}{2}\right)y+1=0 \end{cases} $$

I tried setting the equations equal to each other and substituting variables but those methods just made the equation more conplicated.

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    $\begingroup$ Those radicals are just real numbers. You have ax + by + 1 = 0; cx + dy + 1 = 0 just solve those as normal. y = (-1 - cx)/d so ax + b(-1 - cx)/d + 1 = 0 so (a - bc/d)x = -1 + b/d so x = (-1+b/d)/(a-bc/d); y = (-1 - c( (-1+b/d)/(a-bc/d)))/d. $\endgroup$ – fleablood Dec 29 '16 at 19:11
  • $\begingroup$ I edited the original question. Can you please check that I didn't introduce a typo? It was easy to make a mistake when reading a rotated paged. $\endgroup$ – Michael Burr Dec 29 '16 at 19:16
  • $\begingroup$ Setting the equations equal to each other rarely results in something helpful when you have two variables. The problem is that that doesn't reduce the number of variables. $\endgroup$ – Michael Burr Dec 29 '16 at 19:16
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    $\begingroup$ You could use for example the fact that $\left(\frac{1+\sqrt{3}}{2}\right)\cdot\left(\frac{1-\sqrt{3}}{2}\right) = -\frac{1}{2}$ $\endgroup$ – Sil Dec 29 '16 at 19:19
  • $\begingroup$ Yes, no typos @Michael Burr $\endgroup$ – joko34 Dec 29 '16 at 20:02
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swetting $$a=\frac{1+\sqrt{3}}{2}$$ and $$b=\frac{1-\sqrt{3}}{2}$$ then we have to solve $$a^2x+ay=-1$$ $$b^2x+by=-1$$ from the first equation we get $$y=-\frac{1+a^2x}{a}$$ plugging this in the second equation we get $$a^2bx-b(1+a^2x)=-1$$ thus $$x=\frac{b-1}{ab^2-a^2b}$$ can you proceed?

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