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system of equations

solakis

Active member
Dec 9, 2012
352
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
 

kaliprasad

Well-known member
Mar 31, 2013
1,331
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


$y= \frac{2abc}{ac - ab + bc}$
 

solakis

Active member
Dec 9, 2012
352
nery good
 

solakis

Active member
Dec 9, 2012
352