Answer
$x=\frac{\sqrt 3}{2}x'-\frac{1}{2}y'$ and $y=\frac{1}{2}x'+\frac{\sqrt 3}{2}y'$
Work Step by Step
1. Based on the given equation, we have $A=11, B=10\sqrt 3, C=1$, the rotation angle satisfies $cot(2\theta)=\frac{A-C}{B}=\frac{11-1}{10\sqrt 3}=\frac{\sqrt 3}{3}$, thus $2\theta=\frac{\pi}{3}$ and $\theta=\frac{\pi}{6}$.
2. The formulas for the rotation are $x=x'cos\theta-y'sin\theta=\frac{\sqrt 3}{2}x'-\frac{1}{2}y'$ and $y=x'sin\theta+y'cos\theta=\frac{1}{2}x'+\frac{\sqrt 3}{2}y'$
