All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (1)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information