I have the following system of four equations:
$x_1+x_2+x_3=K$
$A\cdot x_3=y$
$B\cdot x_2=y$
$C\cdot x_1=y$
From these four equations I would like to derive a function for y of A,B,C and K as such:
$f(A,B,C,K)=y$
How would I go about solving this problem?
Let me firstly enumerate your equalities:
$$\begin{array}{cr} x_1+x_2+x_3=K & (1)\\ A\cdot x_3=y & (2)\\ B\cdot x_2=y & (3)\\ C\cdot x_1=y & (4)\end{array}$$
By multiplying (2)-(4), we obtain
$$\begin{array}{cr} A\cdot B \cdot C \cdot x_3=B \cdot C \cdot y & (5)\\ A\cdot B \cdot C \cdot x_2=A \cdot C \cdot y & (6)\\ A\cdot B \cdot C \cdot x_1=A \cdot B \cdot y & (7)\end{array}$$
Thus, by adding (5)-(7), we have
$$ (B \cdot C + A \cdot C + A \cdot B)\cdot y=A\cdot B \cdot C \cdot (x_1 + x_2 + x_3) \overset{(1)}{=} A\cdot B \cdot C \cdot K \quad (8) $$
Now, we distinguish between two cases:
Case 1: $BC + AC + AB \neq 0$. Then we can rearrange (8) to $$ y = \frac{ABCK}{BC + AC + AB} $$ Case 2: $BC + AC + AB = 0$. Then by (8) it follows that $ABCK = 0$, so one of $A,B,C,K$ has to be zero. If $A,B$ or $C$ are zero, then (2), (3) or (4) imply $y=0$. Otherwise, we can assume $A,B,C \neq 0, K=0$.
But I have to admit: at this point I'm stuck as well. I tried to rearrange (2)-(4): $$\begin{array}{cr} x_3=\frac{y}{A} & (9)\\ x_3=\frac{y}{B} & (10)\\ x_3=\frac{y}{C} & (11)\end{array}$$ So, by (1) and $K=0$: $\left(\frac{1}{A}+\frac{1}{B}+\frac{1}{C}\right)y = 0$
Sadly, because $ABC \neq 0$, if we divide the case-2-equation we get $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} = 0$. So at this point I don't know how to continue.
Note: This answers assumes we are working over a field.