Solve the systems of equations (this example is also shown in our video lesson)
^$$\left\{\begin{matrix} x+2y-z=4\\ 2x+y+z=-2\\ x+2y+z=2 \end{matrix}\right.$$
First we add the first and second equation to make an equation with two variables, second we subtract the third equation from the second in order to get another equation with two variables. Now we have a system of two equations with two variables:
$$\left \{\begin{matrix} 3x+3y & = & 2\\ x-y & = & -4\\ \end{matrix}\right.$$
We then multiply the second equation with 3 on both sides and add that to the first equation:
$$\\ 6x=-10\\ \\ x=\frac{-10}{6}$$
We plug this value into the $3x+3y=2$ equation in order to determine our $y$-value:
$$\begin{array}{lcl} \\ 3\cdot \frac{-10}{6}+3y&=&2\\ \\ -5+3y&=&2\\ 3y&=&7\\ \\ y&=&\frac{7}{3}\\ \end{array}$$
Last we plug our x- and y-value into any equation in first system in order to determine our z-value:
$$\begin{array}{lcl} x+2y-z&=&4\\ \frac{-10}{6}+2\cdot \frac{7}{3}+z&=&2\\ 3+z&=&2\\ z&=&-1\\ \end{array}$$
But I'm confused, why do we choose only one equation to determine the value of $z$ maybe we could get different values for $z$ in other equations and therefore the first $system$ of equations have no solutions in the set of real numbers.