# Solving System of Equations - precalc question

I need help to solve for the values of $x, y, z.$

$x+2y+z=6$

$2x-y+3z=-2$

$x+y-2z=0$

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Here they are: $x$, $y$ and $z$. –  azimut Mar 6 '13 at 17:11
There are many ways of solving this; are you familiar with matrix algebra? –  Alex Mar 6 '13 at 17:14
Hint: start with forming $-2E_1+E_2$ and $-E_1+E_3$ where $E_i$ is equation $i$ –  Maesumi Mar 6 '13 at 17:20
@Drew Hoehn: Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework - please tag it as such. Don't be afraid - people will still help. Also, some people on this site consider the use of imperatives (find, prove, explain etc.) rude. Consider editing you question –  Dennis Gulko Mar 6 '13 at 17:21
You can get $x$ in terms of $y$ and $z$ from one of the equations, and replace in the others. The two resulting equations are now in $y$ and $z$, use the same strategy to get a single equation in $z$. Then work backwards, with $z$ you can get the value for $y$, and $y$ and $z$ give you $x$. Solving such systems of linear equations is important enough than specialized techniques have been developed, as the other comments hint at. For simple cases this is enough. –  vonbrand Mar 6 '13 at 17:37

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 $x, y, z$.

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$x+2y+z=6$

$2x-y+3z=-2$

$x+y-2z=0$

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

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