Solve the following equations system: $$$\left\{ \begin{array} {rcl} 2x+2y+2z & = & 2 \\ 2x-2y+2z &=& 3\\ x+y-z &=& 1\end{array}\right.$$$
What happens if the following modification is carried out in the last equation? $$$\left\{ \begin{array} {rcl} 2x+2y+2z &=&2 \\ 2x-2y+2z &=& 3\\ x+y \ \fbox{+} \ z &=& 1\end{array}\right.$$$
Development:
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).
Solution:
$$x=\dfrac{5}{4}; \ y=-\dfrac{1}{4}; \ z=0$$
$$y=-x; \ z=1$$