$$ \begin{cases} \left(\frac{1+\sqrt{3}}{2}\right)^2x+\left(\frac{1+\sqrt{3}}{2}\right)y+1=0\\ \left(\frac{1-\sqrt{3}}{2}\right)^2x+\left(\frac{1-\sqrt{3}}{2}\right)y+1=0 \end{cases} $$
I tried setting the equations equal to each other and substituting variables but those methods just made the equation more conplicated.
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?