MHB Solve x/y+y/z+z/x for Real Numbers with 24,10

[anemone]

Let $x,\,y,\,z$ be real numbers satisfying

$\dfrac{(x+y)(y+z)(z+x)}{xyz}=24$

$\dfrac{(x-2y)(y-2z)(z-2x)}{xyz}=10$

Find $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$.

 

[Sudharaka]

Hi anemone,

Thanks for the challenge. :)

Spoiler

Simplifying the first equation we get,

\[\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}=22\]

Simplifying the second equation we get,

\[\frac{2x-z}{y}+\frac{2y-x}{z}+\frac{2z-y}{x}=\frac{17}{2}\]

Adding these two equations we get,

\[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{17}{6}\]

 

[anemone]

Hi anemone,

Thanks for the challenge. :)

Spoiler

Simplifying the first equation we get,

\[\frac{$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$}{z}+\frac{x+z}{y}+\frac{y+z}{x}=22\]

Simplifying the second equation we get,

\[\frac{2x-z}{y}+\frac{2y-x}{z}+\frac{2z-y}{x}=\frac{17}{2}\]

Adding these two equations we get,

\[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{17}{6}\]

Hi Sudharaka, :)

Thanks for participating. You've used a systematic approach to solve for this challenge. Unfortunately you've made a minor addition error, but it is a good solution nevertheless. Well done, Sud!

Spoiler

$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$

 

[Sudharaka]

Hi Sudharaka, :)

Thanks for participating. You've used a systematic approach to solve for this challenge. Unfortunately you've made a minor addition error, but it is a good solution nevertheless. Well done, Sud!

Spoiler

$\dfrac{x+y}{z}+\dfrac{x+z}{y}+\dfrac{y+z}{x}+
\dfrac{2x-z}{y}+\dfrac{2y-x}{z}+\dfrac{2z-y}{x}=22+\dfrac{17}{2}$

$\rightarrow 3\left(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\right)=\dfrac{61}{2}$

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$

Ah... silly algebra... :p