BMO1 2012 3

3

Summing the three equations gives us:
x^2-4y+7+y^2-6z+14+z^2-2x-7=0
(x^2-2x)+(y^2-4y)+(z^2-6z)+14=0
(x-1)^2-1+(y-2)^2-4+(z-3)^2-9+14=0
(x-1)^2+(y-2)^2+(z-3)^2=0
Clearly (x-1)^2,(y-2)^2,(z-3)^2\geq0
\therefore we must have x=1,y=2,z=3 as only solution

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