Linear systems of three equations with three unknowns

First of all, the third equation is placed at the top, and the variable $x$ is eliminated from the second equation. $$\left\{ \begin{array} {rcl} x+y-z &=& 1 \\ 2x+2y+2z & = & 2 \\ 2x-2y+2z &=& 3 \end{array}\right.$$ $$E2' = E2 - 2\cdot E1$$ $$\left\{ \begin{array} {rcl} x+y-z &=& 1 \\ 4z & = & 0 \\ 2x-2y+2z &=& 3 \end{array}\right.$$

It is possible to observe,also, that the variable y has been eliminated from the second equation. So $z=0$ and we have a system of two equations and two variables, which it is possible to solve easily by replacement: $$\left\{ \begin{array} {rcl} x+y &=& 1 \\ 2x-2y &=& 3 \end{array}\right. \\ 2(1-y)-2y=3 \Rightarrow 2-4y=3 \Rightarrow y=-\dfrac{1}{4} \\ x=\dfrac{5}{4} $$

And so, $$x=\dfrac{5}{4}; \ y=-\dfrac{1}{4}; \ z=0$$

The change of sign in the third equation shows that equations 1 and 3 are proportional, that is, they contain the same information. $$\left\{ \begin{array} {rcl} 2x+2y+2z &=&2 \\ 2x-2y+2z &=& 3\\ x+y \ \fbox{+} \ z &=& 1\end{array}\right.$$ $$\left\{ \begin{array} {rcl} 2x-2y+2z &=&3 \\ x+y+z &=& 1\end{array}\right.$$ $$E1'=E1-2\cdot E2$$ $$\left\{ \begin{array} {rcl} z&=&1 \\ x+y+z &=& 1\end{array}\right.$$ $$x+y+1=1 \Rightarrow y=-x $$

Thus we have more variables than equations, so it will not be possible to find a simple solution. The degree of freedom that there is results in the solution being a straight line (intersection of two planes).