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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

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  • $\begingroup$ Whats the answer? Is it -2 $\endgroup$ – Rohan Shinde Aug 30 '18 at 17:03
  • $\begingroup$ @Manthanein yeah, can you give me the hint though? $\endgroup$ – Pizzaroot Aug 30 '18 at 17:05
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    $\begingroup$ 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 $\endgroup$ – Rohan Shinde Aug 30 '18 at 17:13
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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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