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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?

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Because you have three variables and two equations the answer set will be at least a line, so you could solve for the equation of that line. –  user45150 Jan 31 '13 at 7:13
    
Unfortunately I'm limited to the techniques taught in this section. Which is basically only sub and back-sub –  Ceelos Jan 31 '13 at 7:16
    
From the second $y=-2x$. Put it in the first to get $z=-x$. Hence, $(x,y,z)=(x,-2x,-x)=x(1,-2,-1)$ with any arbitrary $x$. –  Tapu Jan 31 '13 at 7:17

2 Answers 2

up vote 4 down vote accepted

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.

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ahhhhhh! I see what you were trying to illustrate and I do recall the teacher briefly discussing this at the end of lecture. Thank you very much! –  Ceelos Jan 31 '13 at 7:22
    
+1 for your nice illustration in $\mathbb R^3$ –  Babak S. Jan 31 '13 at 7:48

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.

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Nice "take" on the problem! + 1 –  amWhy Jan 31 '13 at 14:10

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