Find the solution of binary xor operator equation

Question

I am working in binary xor operator $\mathbb Z_2$. I have to resolve my problem such as $$\begin {cases} x_1+x_2+x_3=1\\ x_1+x_2=0\\ x_1+x_3=1\\ \end {cases}$$

Could you suggest to me any method to find the solution. I hear that gaussian elimination is one option. How about the other solution? Solution for this equation is $$x_1=0,x_2=0,x_3=1$$ For more complex equation $$\begin {cases} x_1+x_2=0\\ x_2+x_3=1\\ x_3+x_4=1\\ \end {cases}$$ In this case we can see we have only 3 equation but we must find 4 variances. The solution is $$x_1=0;x_2=0; x_3=1, x_4=0$$

Answer 1

Substitution works fine. Plug your second into the first to get $x_3=1$ and so on. Another ways is that the second gives you $x_1=x_2$ and you can plug that in to eliminate one of the variables.

Answer 2

Since you ask for any method, I will add that making an exhaustive search for a solution is perfectly viable here. For the first exercise, you only have $2^3=8$ cases to try and in the second $2^4=16$.

It may not be as elegant or give as much insight and understanding, but sometimes it's the most practical way to go.