How can I determine the values of $x,y,z$ below?
$a,b,c$ - are given variables
$x,y,z$ - must be found $\begin{align*} ax&=S_1\\ by&=S_2\\ cz&=S_3\\ S_1-(y+z)&>0\\ S_2-(x+z)&>0\\ S_3-(x+y)&>0 \end{align*}$
How can I determine the values of $x,y,z$ below?
$a,b,c$ - are given variables
$x,y,z$ - must be found $\begin{align*} ax&=S_1\\ by&=S_2\\ cz&=S_3\\ S_1-(y+z)&>0\\ S_2-(x+z)&>0\\ S_3-(x+y)&>0 \end{align*}$
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 $z$.