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

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Discussion Overview

The discussion revolves around solving the equations involving real numbers \(x\), \(y\), and \(z\) given two specific conditions. Participants are tasked with finding the expression \(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\) based on these conditions.

Discussion Character

  • Homework-related, Mathematical reasoning

Main Points Raised

  • One participant presents the equations \(\dfrac{(x+y)(y+z)(z+x)}{xyz}=24\) and \(\dfrac{(x-2y)(y-2z)(z-2x)}{xyz}=10\) as the basis for the challenge.
  • Another participant acknowledges the challenge and expresses appreciation for the engagement.
  • A third participant provides feedback on a previous solution, noting a minor addition error while still commending the systematic approach taken.
  • A later reply reiterates the feedback about the addition error and adds a light-hearted comment about algebra.

Areas of Agreement / Disagreement

There is no consensus on the solution to the problem as participants are still discussing and refining their approaches. Some participants acknowledge errors in calculations without resolving the overall challenge.

Contextual Notes

Participants have not fully resolved the mathematical steps required to find the desired expression, and there are indications of potential calculation errors that may affect the outcome.

anemone
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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 $\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}$.
 
Last edited:
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Hi anemone,

Thanks for the challenge. :)

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,

\[\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=\frac{17}{6}\]
 
Last edited by a moderator:
Sudharaka said:
Hi anemone,

Thanks for the challenge. :)

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!

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

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

$\therefore \dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{61}{6}$
 
anemone said:
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!

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

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

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

Ah... silly algebra... :p
 

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