First week in Linear Algebra, need some help on this simple problem If you have 
$5x + 2y + z = 0$ 
$2x + y      = 0$  and you're asked to solve using back-substitution how would you go about doing it?  
Initially I thought just simply the following: 
$x + \frac{2}{5} y + \frac{1}{5} z = 0$  (divide by $5$) 
$2x + y            = 0$  
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$ 
$\displaystyle 2(x + \frac{2}{5} y + \frac{1}{5} z) + \frac{1}{5} y - \frac{2}{5} z = 0$ (sub method I was taught in class) 
But after simplifying I realized I basically just made the problem worse because I ended with: 
$\displaystyle x + \frac{2}{5} y + \frac{1}{5} z = 0$ 
    $\displaystyle \frac{1}{5} y + \frac{2}{5} z = 0$ 
So I pretty much still have 3 unknowns. Any suggestions or hints?
 A: I will write up what I wrote above into a solution:
Note that we begin with 
$$5x + 2y + z = 0 $$
$$2x + y = 0$$
Subtracting twice the second from the first we get $x+z=0$, therefore, $z=-x$. Therefore, we get solutions of the form $y=-2x$ and $z=-x$ where $x$ can be anything. This means that all solutions are on the line $$(x,-2x,-x)$$ and anything on this line is a solution. 
Note since that there are less equations than variables there cannot be a unique solution, which is why we get a line of solutions.
A: Of course the accepted answer is completing one, but look at this one. Maybe it looks fine either. We have: $$
 \left\{
        \begin{array}{ll}
            5x+2y+z=0  \\
            2x+y=0 & 
        \end{array}
    \right.$$
Now multiply the second equation by $-2$:
$$ \left\{
        \begin{array}{l1}
            5x+2y+z=0  \\
            -4x-2y=0 & 
        \end{array}
    \right.$$ Add two equations above to eliminate $y$: $$(5x-4x)+z=0$$ so $x=-z$ Now, put $x=-z$ into the first equation: $$5x+2y-x=0$$ so $4x+2y=0$ or $y=-2x$. This is what you see in another answer as: $(x,-2x,-x)$ which is a 3D line.
