Krdytkyu

I have the following system of four equations:

$x_1+x_2+x_3=K$

$Acdot x_3=y$

$Bcdot x_2=y$

$Ccdot x_1=y$

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:

$$begin{array}{cr}
x_1+x_2+x_3=K & (1)\
Acdot x_3=y & (2)\
Bcdot x_2=y & (3)\
Ccdot x_1=y & (4)end{array}$$

By multiplying (2)-(4), we obtain

$$begin{array}{cr}
Acdot B cdot C cdot x_3=B cdot C cdot y & (5)\
Acdot B cdot C cdot x_2=A cdot C cdot y & (6)\
Acdot B cdot C cdot x_1=A cdot B cdot y & (7)end{array}$$

Thus, by adding (5)-(7), we have

$$
(B cdot C + A cdot C + A cdot B)cdot y=Acdot B cdot C cdot (x_1 + x_2 + x_3) overset{(1)}{=} Acdot B cdot C cdot K quad (8)
$$

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.