perplexus.info :: Just Math : In pairs

In pairs (Posted on 2019-07-09)

xy/(x+y)=3
yz/(y+z)=4
zx/(z+x)=5
xyz/(xy+yz+zx)=?

No Solution Yet Submitted by Danish Ahmed Khan

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solution

| Comment 1 of 3

Invert each equation to get an easily solved system expressed in reciprocals of x,y,z.

x=120/17

y=120/23

z=120/7

xyz/(xy+yz+zx)=1/(1/x+1/y+1/z)=120/47.

Posted by xdog on 2019-07-09 17:39:39

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