Christmas Streak 19/88: Three Variables In Two Equations

$\cases{\sin\alpha+\sin\beta=1.5+\cos\gamma \\\\ \cos\alpha-\cos\beta=1-\sin\gamma}$

Square both sides to get

$\cases{\sin^2\alpha+\sin^2\beta+2\sin\alpha\sin\beta=2.25+3\cos\gamma+\cos^2\gamma \\\\ \cos^2\alpha+\cos^2\beta-2\cos\alpha\cos\beta=1-2\sin\gamma+\sin^2\gamma}$

Now add them sides by sides.

$(\sin^2\alpha+\cos^2\alpha)+(\sin^2\beta+\cos^2\beta)-(2\cos\alpha\cos\beta-2\sin\alpha\sin\beta)=3.25+3\cos\gamma-2\sin\gamma+(\sin^2\gamma+\cos^2\gamma)$

Move some stuffs around and we get

$(2\sin\gamma-3\cos\gamma)-(2\cos\alpha\cos\beta-2\sin\alpha\sin\beta)=2.25$

Finally, using the fact that $2\sin\gamma-3\cos\gamma=\sqrt{13}\sin\left(\gamma-\tan\left(\dfrac{3}{2}\right)\right)$ and $2\cos\alpha\cos\beta-2\sin\alpha\sin\beta=\cos(\alpha+\beta),$

$\sqrt{13}\sin\left(x-\tan^{-1}\left(\dfrac{3}{2}\right)\right)-2\cos(\alpha+\beta)=\boxed{2.25}.$