Solving System of Equations - precalc question I need help to solve for the values of $x, y, z.$
$x+2y+z=6$
$2x-y+3z=-2 $
$x+y-2z=0$
 A: $x+2y+z=6$
$2x-y+3z=-2 $
$x+y-2z=0$
From first equation we get 
$[1].....x=6-2y-z$ 
put that $x$ in two last equations we get
$$2(6-2y-z)-y+3z=-2 $$
$$6-2y-z+y-2z=0$$
or
$$14-5y+z=0 $$
$$6-y-3z=0$$
now from first equation we get
$[2].....z=5y-14$
replacing that $z$ in second equation we get
$$6-y-3(5y-14)=0\iff6-y-15y+42=0\iff-16y=-48\iff y=3$$
put this value on [2] we get
$z=5\cdot3-14=1$
now I think you can continue on your own 
A: Assuming this is being asked outside of the context of linear algebra, or prior to being introduced to matrices, I'll model the process of approaching problems like this in a systematic way:
We have the system of three linear equations in three unknowns: 
$$x+2y+z=6\tag{1}$$
$$2x-y+3z=-2 \tag{2}$$
$$x+y-2z=0\tag{3}$$
Note that if we subtract equation $(3)$ from equation $(1)$, we get:
$y + 3z = 6\tag{i}$
And if we subract $2 \times$ equation $(3)$ from equation $(2)$, we get
$-3y + 7z =  - 2\tag{ii}$
Now, we can multiply equation $(i)$ by $3$ and then add it to equation $(ii)$ to eliminate $y$ and solve for $z$:
$(3i + ii) \iff \quad 16z = 16 \implies z = 1\tag{z=1}$
Now, we go back to, say $(i)$ and use the fact that $z = 1$ to solve for $y$:
$y + 3z = 6 \implies y + 3\cdot 1 = 6 \iff y = 3\tag{y = 3}$.
Now go back to any of the original equations $(1), (2), \text{ or}\;(3)$ to solve for $\,x\,$ knowing $\;z = 1$ and knowing $\;y = 3$.
Then you will have found the value of each of $x, y, z$.
