Solving System of Equations: xy, yz, zx

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Discussion Overview

The discussion revolves around solving a system of equations involving three variables, x, y, and z, expressed in terms of parameters a, b, and c. The equations are presented in a specific fractional form, and the participants engage in a dialogue about the problem's formulation.

Discussion Character

  • Homework-related

Main Points Raised

  • Participants present the same system of equations multiple times, indicating a focus on the problem itself.
  • Some participants express approval of the problem, with comments such as "nery good" and "very good," though no further elaboration on the equations or solutions is provided.

Areas of Agreement / Disagreement

There is no substantive disagreement or agreement on the solution or approach, as the discussion primarily consists of repeated problem statements and expressions of approval.

Contextual Notes

The discussion lacks detailed exploration of methods or solutions, and no assumptions or mathematical steps are explicitly addressed.

Who May Find This Useful

Individuals interested in solving systems of equations or those seeking examples of mathematical problems may find this discussion relevant.

solakis1
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Solve the following systemof equations:

$\dfrac{xy}{x+y}=a$

$\dfrac{yz}{y+z}=b$

$\dfrac{zx}{z+x}=c$

where a,b,c are not zero
 
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solakis said:
Solve the following systemof equations:

$\dfrac{xy}{x+y}=a$

$\dfrac{yz}{y+z}=b$

$\dfrac{zx}{z+x}=c$

where a,b,c are not zero
Inverting the 3 we get$\frac{1}{y} + \frac{1}{x} = \frac{1}{a}\dots(1)$$\frac{1}{y} + \frac{1}{z} = \frac{1}{b}\dots(2)$$\frac{1}{z} + \frac{1}{x} = \frac{1}{c}\dots(3)$Add the 3 to get$2(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$Or $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{2}(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$Subtracting (1) from above we get$\frac{1}{z} = \frac{1}{2}(\frac{1}{b} + \frac{1}{c}- \frac{1}{a})$Or $\frac{1}{z} = \frac{1}{2}\frac{ac + ab - bc}{abc}$Or $z= \frac{2abc}{ac + ab - bc}$ Similarly $x= \frac{2abc}{ab + bc - ac}$And $y= \frac{2abc}{ac - ab + bc}$
 
nery good
 
solakis said:
very good
 

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