If 2cos theta -sin theta =x …

Given, 2 cos$\theta$ - sin$\theta$ $= x$  and
cos$\theta$ - 3 sin$\theta$ $= y$
Put the values of x and y in $2x^2 + y^2 − 2xy$ (LHS) , we get 
= 2(2 cos$\theta$ − sin$\theta$)² + (cos$\theta$ − 3 sin$\theta$)² − 2(2 cos$\theta$ − sin$\theta$)(cos$\theta$ − 3 sin$\theta$)
= 2(4cos² $\theta$− 4cos$\theta$ sin$\theta$ + sin²$\theta$) + (cos² $\theta$− 6cos$\theta$ sin$\theta$ + 9sin² $\theta$) − 2(2cos²  $\theta$− 7cos$\theta$ sin$\theta$ + 3sin² $\theta$)
= 8cos²$\theta$ − 8cos$\theta$ sin$\theta$ + 2sin²$\theta$ + cos²$\theta$ − 6cos$\theta$ sin$\theta$ + 9sin²$\theta$ − 2(2cos²  $\theta$− 7cos$\theta$ sin$\theta$ + 3sin² $\theta$)
= 8cos²$\theta$ − 8cos$\theta$ sin$\theta$ + 2sin²$\theta$ + cos²$\theta$ − 6cos$\theta$ sin$\theta$ + 9sin²$\theta$ − 4cos²$\theta$ + 14cos$\theta$ sin$\theta$ − 6sin²$\theta$
= 5cos²$\theta$ + 5sin²$\theta$
= 5(cos²$\theta$ + sin²$\theta$)
$=5$
= RHS