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- Thread starter solakis
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- Mar 31, 2013

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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

$\frac{1}{y} + \frac{1}{x} = \frac{1}{a}\dots(1)$

$\frac{1}{y} + \frac{1}{z} = \frac{1}{b}\dots(2)$

$\frac{1}{z} + \frac{1}{x} = \frac{1}{c}\dots(3)$

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

$y= \frac{2abc}{ac - ab + bc}$

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very good