# 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

• Whats the answer? Is it -2 – Rohan Shinde Aug 30 '18 at 17:03
• @Manthanein yeah, can you give me the hint though? – Pizzaroot Aug 30 '18 at 17:05
• You have obtained 3 linear equations in x,y ; y, z; and x, z. So try expressing y, z in terms of x. Now substitute the values if y, z in the first equation of all 3 original equations. Then factor out the x from obtained expression and since x is not equal to 0 then the other factor must be 0. So set that factor equal to 0 and continue with further algebra. You will automatically get the answer – Rohan Shinde Aug 30 '18 at 17:13

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.