Evaluate $\frac{z-y}{z-x}$ for $x,\,y,\,z$ Real Numbers

[anemone]

Let $x,\,y,\,z$ be real numbers which satisfy the system below:

$x+\dfrac{1}{yz}=\dfrac{1}{5}$

$y+\dfrac{1}{xz}=-\dfrac{1}{15}$

$z+\dfrac{1}{xy}=\dfrac{1}{3}$

Evaluate $\dfrac{z-y}{z-x}$.

 

[kaliprasad]

Let $x,\,y,\,z$ be real numbers which satisfy the system below:

$x+\dfrac{1}{yz}=\dfrac{1}{5}$

$y+\dfrac{1}{xz}=-\dfrac{1}{15}$

$z+\dfrac{1}{xy}=\dfrac{1}{3}$

Evaluate $\dfrac{z-y}{z-x}$.

Spoiler

clearly x,y,z none of them is zero deviding 1st equation by x 2nd by y and 3rd by z we get

$1+\dfrac{1}{xyz}= \dfrac{1}{5x}=\dfrac{-1}{15y} = \dfrac{1}{3z}$
hence $5x=-15y=3z$
or $\dfrac{x}{z}= \dfrac{3}{5}$
$\dfrac{y}{z}= \dfrac{-1}{5}$
hence $\dfrac{z-y}{z-x}=\dfrac{1-\frac{y}{z}}{1-\frac{x}{z}}=\dfrac{1+\frac{1}{5}}{1-\frac{3}{5}}=3 $

 

[anemone]

Spoiler

clearly x,y,z none of them is zero deviding 1st equation by x 2nd by y and 3rd by z we get

$1+\dfrac{1}{xyz}= \dfrac{1}{5x}=\dfrac{1}{15y} = \dfrac{1}{3z}$
hence $5x=15y=3z$
or $\dfrac{x}{z}= \dfrac{3}{5}$
$\dfrac{y}{z}= \dfrac{1}{5}$
hence $\dfrac{z-y}{z-x}=\dfrac{1-\frac{y}{z}}{1-\frac{x}{z}}=\dfrac{1-\frac{1}{5}}{1-\frac{3}{5}}=2 $

Thanks kaliprasad for participating...but:

Spoiler

Please note that the RHS of the second equation has a minus sign..

 

[kaliprasad]

Thanks kaliprasad for participating...but:

Spoiler

Please note that the RHS of the second equation has a minus sign..

Thanks. I have done the correction based on the comment