perplexus.info :: Just Math : Unraveling the Trio of Reciprocal Relations

Unraveling the Trio of Reciprocal Relations (Posted on 2024-07-06)

Find the real numbers x, y, z such that,

1/x + 1/(y+z) = 1/2
1/y + 1/(z+x) = 1/3
1/z + 1/(x+y) = 1/4

No Solution Yet Submitted by Danish Ahmed Khan

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Solution

| Comment 1 of 2

Take reciprocals and simplify:

xy/(x+y+z) + xz/(x+y+z) = 2

xy/(x+y+z) + yz/(x+y+z) = 3

xz/(x+y+z) + yz/(x+y+z) = 4

This in linear in terms of the three fractions, solve that to get:

xy/(x+y+z) = 1/2

xz/(x+y+z) = 3/2

yz/(x+y+z) = 5/2

Take ratios of these to get x/y = 3/5 and y/z = 1/3.

Then let x=3k, y=5k, and z=15k and substitute into an original equation to get k=23/30.

Then the solution is (x,y,z) = (23/10, 23/6, 23/2).

Posted by Brian Smith on 2024-07-06 21:32:39

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