how to solve system of linear equations of XOR operation? how can i solve this set of equations ? to get values of $x,y,z,w$ ?
$$\begin{aligned} 1=x \oplus y \oplus z \end{aligned}$$
$$\begin{aligned}1=x \oplus y \oplus w \end{aligned}$$
$$\begin{aligned}0=x \oplus w \oplus z \end{aligned}$$
$$\begin{aligned}1=w \oplus y \oplus z \end{aligned}$$
this is not a real example, the variables don't have to make sense, i just want to know the method.

The other answers are fine, but you can use even more elementary (if somewhat ad hoc) methods, just as you might with an ordinary system of linear equations over $\Bbb R$. You have this system:
$$\left\{\begin{align*}
1&=x\oplus y\oplus z\\
1&=x\oplus y\oplus w\\
0&=x\oplus w\oplus z\\
1&=w\oplus y\oplus z
\end{align*}\right.$$
Add the first two equations: 
$$\begin{align*}
(x\oplus y\oplus z)\oplus(x\oplus y\oplus w)&=(z\oplus w)\oplus\Big((x\oplus y)\oplus(x\oplus y)\Big)\\
&=z\oplus w\oplus 0\\
&=z\oplus w\;,
\end{align*}$$
and $1\oplus 1=0$, so you get $z+w=0$. Substitue this into the last two equations to get $0=x\oplus 0=x$ and $1=y\oplus 0=y$. Now you know that $x=0$ and $y=1$ so $x\oplus y=1$. Substituting this into the first two equations, we find that $1=1\oplus z$ and $1=1\oplus w$. Add $1$ to both sides to get $0=z$ and $0=w$. The solution is therefore $x=0,y=1,z=0,w=0$.
