# Solve system of equations with square roots and fractions

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

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

• 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. – fleablood Dec 29 '16 at 19:11
• 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. – Michael Burr Dec 29 '16 at 19:16
• 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. – Michael Burr Dec 29 '16 at 19:16
• You could use for example the fact that $\left(\frac{1+\sqrt{3}}{2}\right)\cdot\left(\frac{1-\sqrt{3}}{2}\right) = -\frac{1}{2}$ – Sil Dec 29 '16 at 19:19
• Yes, no typos @Michael Burr – joko34 Dec 29 '16 at 20:02

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?