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

 Solution's forecast (Posted on 2017-10-24)
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.)
 Sollution | Comment 1 of 3

Let z=a+k, k<>0, k<>-y.  Then x+y+a+k=a, or x= -y �"k

Substitute in the second equation

1/(-y-k) + 1/y + 1/(a+k)=1/a

Multiply through by y

y/(-y-k) + 1 +y/(a+k) = y/a

Multiply through by  (-y-k)

y + (-y-k) +y(-y-k)/(a+k)=y*(-y-k)/a ;

Continue simplifying and you finally get

-a^2k-ak^2=-yka-yk^2

Or y=a

Therefore if z<>a, then y must =a, and x must = -z  QED

If k=-y

Then x + y + a - y = a, or, x=0.  In that case the second equation is undefined anyway.

Edited on October 24, 2017, 5:07 pm

Edited on October 24, 2017, 5:07 pm
 Posted by Kenny M on 2017-10-24 17:06:56

 Search: Search body:
Forums (0)