# How can I determine these 3 variables?

How can I determine the values of $x,y,z$ below?

$a,b,c$ - are given variables

$x,y,z$ - must be found \begin{align*} ax&=S_1\\ by&=S_2\\ cz&=S_3\\ S_1-(y+z)&>0\\ S_2-(x+z)&>0\\ S_3-(x+y)&>0 \end{align*}

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Well, no they are not the same. Look carefully and you will see that every S is bigger than the variables of the other two S. –  Rma Jun 19 '12 at 10:37
I'm Sorry, it was my mistake. Now it's correct. –  Rma Jun 19 '12 at 10:54
I don't understand the question. What are $S_1$, $S_2$, $S_3$? If they are known, the answer is obviously $x=S_1/a$, and so forth, and the final inequalities are just so much noise – they may or may not hold. And if they are unknown, the answer is not unique, if one exists. –  Harald Hanche-Olsen Jun 19 '12 at 11:37
I understand, yes the $S_1 ,S_2 ,S_3$ are just notations. But if an x,y,z exists, how to determine the interval in which it is located. –  Rma Jun 19 '12 at 12:03

First of all, if the $S_i$ are known, you are done. If the $S_i$ are just placeholders we might as well get rid of them and obtain\begin{align*} ax&>(y+z)\\ by&>(x+z)\\ cz&>(x+y) \end{align*}

This system might or might not have solutions. This pretty much depends on $a,b,c$. If for example $a=b=c=2$, then summing the three inequalities gives $2(x+y+z)>2(x+y+z)$, if $a=b=c=3$ any triple $x=y=z>0$ is a solution.

However if we have a solution $(x,y,z)$, then all positive scalar multiples of this solution are also solutions. Hence we can reduce the question to the three cases $z=0$, $z=1$ and $z=-1$. Thereby we have reduced the problem to a 2-dimensional one.

In each of these three cases one is left with three linear inequalities in $x$ and $y$. Each of them can be expressed geometrically as a line and solutions lie on one distinct side of these lines (depending on the inequality sign). Thus the lines cuts out the set of solutions. In other words, we can now choose $x$ arbitrarily and obtain either the empty set or a well defined interval for the solution of $y$.

Edit: Concerning your example $a=2$, $b=3$, $c=7$, I really advise you to try it yourself. My answer provides all tools for the general solution and plugging in some numbers is the best way to understand it. But to give you a start: if $(x,y,z)$ is a solution with $z>1$, then $(x/z,y/z)$ lies in the triangle here. You will get such a solution for each positive $z$.

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And if i have 3 variables I need to draw a 3 dimensional graph ? –  Rma Jun 19 '12 at 12:04
No, that's the whole point of the first step. –  Simon Markett Jun 19 '12 at 12:25
I don't know what to say. If you want you can try it with $x=2,y=3,z=7$. From what my teacher said, any combination has a interval within which you can find values for $a,b,c$ –  Rma Jun 19 '12 at 12:33
You did say in your question, that a,b,c are given and x,y,z must be found! That's the opposite from what you wrote now! –  Simon Markett Jun 19 '12 at 12:40
Sorry, i meant to say $a=2,b=3,c=7$ –  Rma Jun 20 '12 at 10:29