Isolate 2 variables in 2 formulas I am developing a robotic arm and I need to know the angles of $2$ parts of this arm (bars) and I have $2$ variables with $2$ formulas and I tried my best and I still cant isolate those $2$ variables.
In the formulas below, $[A,B,C,D]$ are constants, and $[x,y]$ are variables in degrees (not radians):


*

*$A\cdot\cos{x}-B=A\cdot\sin{y-90}$

*$C\cdot\cos{Y-90}--A\cdot\sin{X}=D$


How can I isolate $x$ and $y$?
I tried my best and I ended up with a big $\arccos$ with $\sin$ inside which had also another $\cos$ inside which I could not work with it anymore and keep isolating.
EDIT:
Thank you for the person who edited this question and made if more readable, I am not used to MATHEXCHANGE sintax. 
I also would like to complement this question saying that I really need to isolate X and Y cause I am using an Arduino microcontoller which is not powerful and not good for an iterative process to solve this equations.
FINAL EDIT:
A REALLY NICE GUY HELPED ME WITH THIS in this thread -> How to get alfa and beta from this image and there he solved this problem with a very creative alternative.
 A: If the angles are in degrees, you are in luck!
Remember, $$\cos(y-90^\circ) = \sin(y)\\
\sin(y-90^\circ)=-\cos(y)$$
So, your equations become
$$A\cos x + C\sin y = B\\
C\sin y - A\sin x = D$$
You can now use $\sin^2 x + \cos^2 x$ and then first extract $\sin y$ (and out of that, $y$).
A: simplifying the system you will get
$$a\cos(x)=b-a\cos(y)$$
and 
$$a\sin(x)=c\sin(y)-d$$
thus you will get
$$a^2=(b-\cos(y))^2+(c\sin(y)-d)^2$$
can you prceed?
using the $$tan$$ half angle formulas we get
$${a}^{2}-{b}^{2}+2\,{\frac {b \left( 1- \left( \tan \left( y/2 \right) 
 \right) ^{2} \right) }{1+ \left( \tan \left( y/2 \right)  \right) ^{2
}}}-{\frac { \left( 1- \left( \tan \left( y/2 \right)  \right) ^{2}
 \right) ^{2}}{ \left( 1+ \left( \tan \left( y/2 \right)  \right) ^{2}
 \right) ^{2}}}-4\,{\frac {{c}^{2} \left( \tan \left( y/2 \right) 
 \right) ^{2}}{ \left( 1+ \left( \tan \left( y/2 \right)  \right) ^{2}
 \right) ^{2}}}+4\,{\frac {c\tan \left( y/2 \right) d}{1+ \left( \tan
 \left( y/2 \right)  \right) ^{2}}}-{d}^{2}
=0$$
