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/(-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
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Posted by Kenny M
on 2017-10-24 17:06:56 |