<s>[INST] 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.
 [/INST] As others have noted, all the usual methods of solving systems of linear equations (such as Gaussian elimination) in the field of real numbers work just as well in the finite field of integers modulo 2, also known as $GF(2)$.
In this field, addition corresponds to the XOR operation, while multiplication corresponds to AND (as it does in the reals, if the operands are restricted to $0$ and $1$).  As both $0$ and $1$ are their own additive inverses in $GF(2)$ (since $0 \oplus 0 = 1 \oplus 1 = 0$), subtraction is also equivalent to XOR, while division is trivial (dividing by $1$ does nothing, dividing by $0$ is undefined).

So, let's try solving your example equations.  Since martini already did Gaussian elimination, let me do Gauss–Jordan:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
1x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
The first coefficient of the first equation is already $1$, so we just subtract (XOR) that equation from the second and the third:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
Now the second coefficient of the second equation is $0$, so we need to choose some later row that has a $1$ for that coefficient and swap those rows:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 1y & \oplus & 1z & \oplus & 0w & = & 1 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 1z & \oplus & 1w & = & 1 \\
\end{array}$$
... and then subtract that row from all the others with a non-zero second coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\
\end{array}$$
Then we subtract the third row from those with a non-zero third coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 1w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 1w & = & 0 \\
0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\
\end{array}$$
... and finally the fourth row from those with a non-zero fourth coefficient:
$$\begin{array}{rcrcrcrcr}
1x & \oplus & 0y & \oplus & 0z & \oplus & 0w & = & 0 \\
0x & \oplus & 1y & \oplus & 0z & \oplus & 0w & = & 1 \\
0x & \oplus & 0y & \oplus & 1z & \oplus & 0w & = & 0 \\
0x & \oplus & 0y & \oplus & 0z & \oplus & 1w & = & 0 \\
\end{array}$$
And now we can read  out the result: $x = 0$, $y = 1$, $z = 0$, $w = 0$.
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