Prove that in the set of the equations
all possible solutions contain value a as an answer
for one of the uknowns.
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