<s>[INST] Finding the value of $\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42} $ given a system of three equations Let $a, b, c, x, y, z$ be real numbers that satisfy the three equations
$$ 13x+by+cz=0 $$
$$ ax+23y+cz=0 $$
$$ ax+by+42z=0 $$
Suppose that $ a\neq13 $ and $x\neq0$. What is the value of $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42} $$

I tried
$$ (13-a)x+(b-23)y=0 $$
$$ (23-b)y+(c-42)z=0 $$
$$ (13-a)x+(c-42)z=0 $$

$$ (a-13)x=(b-23)y=(c-42)z $$
But I don't know how to continue further
Maybe
$$ \frac{1}{(a-13)x}=\frac{1}{(b-23)y}=\frac{1}{(c-42)z} $$
But any hint will be appreciated
 [/INST] Your current progress is excellent.
Note that $$\frac{1}{(a-13)x}=\frac{1}{(b-23)y}=\frac{1}{(c-42)z}$$ means that $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{13x}{(a-13)x}+\frac{23y}{(b-23)y}+\frac{42z}{(c-42)z}$$ or that $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{13x+23y+42z}{(a-13)x}$$ due to the equalities.
Adding the three equations given gives $$13x+23y+42z=-2(ax+by+cz)=-2(ax-13x)=-2x(a-13)$$ from the first equation hence $$\boxed{\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{-2(a-13)x}{(a-13)x}=-2}$$
P.S. The values of $13,23,42$ are totally arbitrary. This works for any triplet of non-zero integers.
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