System of equations in $\mathbb{F}_p$ I have:$$\begin{matrix}X&+&Y&+&Z&=&3,\\& &Y&+&Z&=&2,\\&-&Y&+&Z&=&0.\end{matrix}$$
Let furthermore $p$ be prime. How can I solve this system with respect to $\mathbb{F}_p$?
 A: Continuing from the comments. If we are not in a field of characteristic $2$, we can solve the system of linear equations in the usual manner and get $X = Y = Z = 1$ as the unique solution.
Thus, that is the solution in any $\Bbb F_p$ if $p > 2$.

For the case $p = 2$, we see that the equations become
\begin{align}
X + Y + Z &= 1,\\
Y + Z &= 0,\\
Y + Z &= 0.
\end{align}
(Recall that $Y = -Y$ in char $2$.)
Now we see that we actually have a redundancy in the system. (If you're familiar with linear algebra, we see that the coefficients matrix is not full rank and thus, we won't get a unique solution.)
Substituting $Y+Z = 0$ in the first equation gives us that $X = 1$. Thus, we are only left with $Y + Z = 0$ which has two solutions. Thus, the complete set of solutions is
$$(X, Y, Z) \in \{(1, 1, 1), (1, 0, 0)\}.$$
A: In the first equation, replace $Y+Z$ with $2$ (by the 2nd equation). You get $X=1$. Add the 2nd and the third equation side-by-side, you get $2Z=2$. Now $if p\neq 2$ this yields $Z=1$ and then $Y=1$. Otherwise, if $p=2$, you have two solutions for $Y,Z$: either $Y=Z=1$ or $Z=Y=0$.
