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.
Let z=a+k, k<>0, k<>y. Then x+y+a+k=a, or x= y �"k
Substitute in the second equation
1/(yk) + 1/y + 1/(a+k)=1/a
Multiply through by y
y/(yk) + 1 +y/(a+k) = y/a
Multiply through by (yk)
y + (yk) +y(yk)/(a+k)=y*(yk)/a ;
Continue simplifying and you finally get
a^2kak^2=ykayk^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 20171024 17:06:56 