Solving System of Equations: xy, yz, zx

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SUMMARY

The forum discussion focuses on solving the system of equations defined by the relationships $\dfrac{xy}{x+y}=a$, $\dfrac{yz}{y+z}=b$, and $\dfrac{zx}{z+x}=c$, where a, b, and c are non-zero constants. Participants emphasize the importance of understanding the properties of symmetric functions and the application of algebraic manipulation techniques to derive solutions. The discussion highlights the necessity of substituting variables and simplifying expressions to find values for x, y, and z effectively.

PREREQUISITES
  • Understanding of algebraic manipulation techniques
  • Familiarity with symmetric functions in mathematics
  • Knowledge of systems of equations
  • Basic skills in substitution methods
NEXT STEPS
  • Explore advanced techniques in solving nonlinear systems of equations
  • Learn about symmetric polynomials and their applications
  • Investigate the use of graphical methods for visualizing solutions
  • Study numerical methods for approximating solutions to complex equations
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Mathematicians, students studying algebra, and anyone interested in solving complex systems of equations.

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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