Solving System of Equations: xy, yz, zx

[solakis1]

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]

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

Spoiler: my Solution

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

 

[solakis1]

nery good

 

[solakis1]

very good