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.
Here's a proof by contradiction that doesn't require a new variable.
Say none of (x,y,z)=a. Then (x+y) = (az) <> 0
Recast the second equation to read (x+y)/xy + 1/z = 1/a. Substitute for (x+y) and simplify giving az(az) + xy(az) = 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 (ax)(ay) = 0 and the contradiction is reached.

Posted by xdog
on 20171025 12:07:23 