Answer
$x= \frac{\sqrt 5}{5}(2x'-y')$ and $y= \frac{\sqrt 5}{5}(x'+2y')$
Work Step by Step
Step 1. From the given equation, we have $A=1, B=4, C=-2$. Thus we have $cot2\theta=\frac{A-C}{B}=\frac{1+2}{4}=\frac{3}{4}$. Thus $2\theta$ is in Quadrant I and $cos2\theta=\frac{3}{\sqrt {3^2+4^2}}=\frac{3}{5}$ (form a right triangle with sides $3,4,5$ here).
Step 2. We can find values of $sin\theta$ and $\cos\theta$ as $sin\theta=\sqrt {\frac{1-cos2\theta}{2}}=\sqrt {\frac{1-(\frac{3}{5})}{2}}=\frac{\sqrt 5}{5}$ and $cos\theta=\sqrt {1-sin^2\theta}=\frac{2\sqrt 5}{5}$
Step 3. Using the transformation formula $x=x'cos\theta-y'sin\theta$ and $y=x'sin\theta+y'cos\theta$, we have
$x=x'(\frac{2\sqrt 5}{5})-y'(\frac{\sqrt 5}{5})=\frac{\sqrt 5}{5}(2x'-y')$
$y=x'(\frac{\sqrt 5}{5})+y'(\frac{2\sqrt 5}{5})=\frac{\sqrt 5}{5}(x'+2y')$
