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Solution's forecast (Posted on 2017-10-24) Difficulty: 3 of 5
Prove that in the set of the equations

(i) x+y+z=a
(ii) 1/x+1/y+1/z=1/a

all possible solutions contain value a as an answer
for one of the uknowns.

See The Solution Submitted by Ady TZIDON    
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solution Comment 3 of 3 |
Here's a proof by contradiction that doesn't require a new variable.

Say none of (x,y,z)=a.  Then (x+y) = (a-z) <> 0

Recast the second equation to read (x+y)/xy + 1/z = 1/a.   Substitute for (x+y) and simplify giving az(a-z) + xy(a-z) = 0 and z=-xy/a.

Plug this value for z into the first equation and get rid of the fractions.  You're left with a^2 -ax -ay +xy = 0 which factors nicely as (a-x)(a-y) = 0 and the contradiction is reached.

  Posted by xdog on 2017-10-25 12:07:23
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