Equation with sets So, I got this system of equations with sets (I need to find X):
$ \begin{cases} (A \cup X) \cup C = B \cap C  \\ X \setminus A = B^\complement \end{cases}$
I got, that 
$ \begin{cases} A \subseteq C \subseteq B \\ B^\complement \subseteq X \subseteq A \end{cases}$
But this means that X cannot exist like that.
So, am I right with my transformation & my conclusion?
 A: $C\subseteq\left(A\cup X\right)\cup C=B\cap C\subseteq C$. 
This allows
the conclusion that $A\cup X\subseteq C\subseteq B$.
Then $B^{c}=X\backslash A\subseteq B$. 
This can only be true if $B^c=\emptyset$.
It follows that $X\subseteq A\subseteq C\subseteq B$ and that $B$ is the universe you are working in.
There is nothing more 'to gain'.
A: $ \begin{cases} 
(A \cup X) \cup C = B \cap C \\ 
X \setminus A = B^\complement 
\end{cases}$
Substitution $L\cup M\leftarrow L\oplus M\oplus(L\cap M)$ gives 
$ \begin{cases} 
(A\oplus X\oplus(A\cap X)) \oplus C\oplus (A\oplus X\oplus(A\cap X))\cap C = B \cap C \\
X\cap A^\complement = B^\complement 
\end{cases}$
$ \begin{cases} 
A\oplus X\oplus(A\cap X)\oplus C\oplus (A\cap C)\oplus X\cap C\oplus(A\cap X\cap C)= B \cap C \\
X^\complement\cup A = B
\end{cases}$
Rewritten more algebraically 
$ \begin{cases} 
A+X+AX+C+AC+XC+AXC= BC \\
X^\complement+A+X^\complement A=B
\end{cases}$
Substitution of $B$ 
$A+X+AX+C+AC+XC+AXC= (X^\complement+A+X^\complement A)C=X^\complement C+AC+X^\complement AC$
Both sides multiplied with $X$
$XA+XX+XAX+XC+XAC+XXC+XAXC=XX^\complement C+XAC+XX^\complement AC$
and since $M+M=0$
$XA+X+AX+XC+XAC+XC+AXC=XAC\Leftrightarrow X=XAC\Leftrightarrow A\cap C\supseteq X $.
