Answer
$x= \frac{1}{2}(\sqrt 3x'-y')$ and $y= \frac{1}{2}(x'+\sqrt 3y')$
Work Step by Step
Step 1. From the given equation, we have $A=11, B=10\sqrt 3, C=1$. Thus we have $cot2\theta=\frac{A-C}{B}=\frac{11-1}{10\sqrt 3}=\frac{\sqrt 3}{3}$, which gives $2\theta=\frac{\pi}{3}$ and the rotation angle $\theta=\frac{\pi}{6}=30^\circ$
Step 2. Using the transformation formula $x=x'cos\theta-y'sin\theta$ and $y=x'sin\theta+y'cos\theta$, we have
$x=x'cos30^\circ-y'sin30^\circ=\frac{1}{2}(\sqrt 3x'-y')$ $y=x'sin30^\circ+y'cos30^\circ=\frac{1}{2}(x'+\sqrt 3y')$
