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First of all, I won't have the opportunity to ask him because this was last task in our last reading and now we have semester vacation and preparing for exam, that's why I need ask you here and I hope you can tell me what my mistake is.

So task was:

Determine the solution set of the following system of equation in $\mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3}$

$$\text{I: }x_{1}+2x_{2}+x_{4}=2$$

$$\text{II: }2x_{1}+x_{2}+x_{3}+2x_{4}=0$$

$$\text{III: }2x_{2}+x_{3}+x_{4}=0$$

$$\text{IV: }x_{1}=1$$

Here is what I did:

Insert $x_{1}$ in all equations:

$$\text{I: }2x_{2}+x_{4}=1$$

$$\text{II: }x_{2}+x_{3}+2x_{4}=-2$$

$$\text{III: }2x_{2}+x_{3}+x_{4}=0$$

Now $\text{I}-\text{III}: -x_{3} = 1 \Leftrightarrow x_{3} = -1$

Insert this in $\text{III}:2x_{2}-1+x_{4} \Leftrightarrow 2x_{2}+x_{4}=1$

Now we see that $\text{I}$ equals $\text{III}$ which means we can take one out, so we have $2$ equations to work with now and these are $\text{I}$ and $\text{II}$.

$\text{I}: x_{4}=1-2x_{2}$

Insert this in $\text{II}: x_{2}-1+2(1-2x_{2})=-2 \Leftrightarrow x_{2}-1+2-4x_{2}=-2 \Leftrightarrow -3x_{2}+1=-2 \Leftrightarrow -3x_{2}=-3 \Leftrightarrow x_{2}=1$

Insert all in $\text{I}: x_{4}=1-2 \cdot 1 = -1$

Thus, $x_{1}=1, x_{2}=1, x_{3}=-1, x_{4}=-1$

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    Your I-III should be I + III. – 2017-02-16

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In one of your steps you divide by $3$, which in $Z_3$, which probably means $\mathbb{Z}$ modulo $3$, is zero. Concretely, when you pass from $-3x_2=-3$ to $x_2=1$.

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    Understood, thank you :) I hope with all the information given here, I can correct it myself. Good I can simply insert and test my solutions, I can imagine there will be several. – 2017-02-16
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To give more of an overall picture of what you have done:

You have solved the system over $\mathbb Q$ and found a unique integer solution. This solution is of course also a solution over $\mathbb Z_3$, but it does not necessarily mean that this solution is also unique over $\mathbb Z_3$. And in fact it turns out that there are more solutions over $\mathbb Z_3$, since the equation $-3x_2=-3$ means that $x_2$ can be arbitrary when you work over $\mathbb Z_3$.

You certainly have done some correct calculations until you hit the step $-3x_2=-3$. Though I can understand the teacher to give you zero points (and not something like 50% of the points), since your solution suggests that you have not understood at all what to take care of when working over a finite field $\mathbb Z_p$. To make sure you understand it in the future, it can be very helpful to give you zero points.

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Does -3 exist in $Z_3$? No. Can you substitute something which is unsure to exist? No. So, you are solving this as you had to solve it in $Z \times Z \times Z \times Z$, simply in $Z_3$ it is not $-1$, it is 2,...

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    Oh I think now I see... Will re try that with your hint! – 2017-02-16
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    Well, $\;3,\,-3\;$ **certainly** exist in $\;\Bbb Z_3\;$ ...! They just are the same as zero and thus one cannot divide by zero... – 2017-02-16
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    @DonAntonio I meant standard representation of $Z_3$, but the point is the same. I can take a triangle, a square and a hexagon and declare that triangle + triangle is triangle, triangle + square is square, triangle + hexagon is hexagon, square + hexagon is triangle, square + square is hexagon, hexagon + hexagon is square, addition is commutative, and voila, I got $Z_3$ with addition. – 2017-02-18