by Guest » Sun Mar 31, 2013 7:25 am
Add up all the equations to get
[tex](2n-3)(x_1+x_2+\ldots+x_n) = \frac{n(n+1)}{2}[/tex]
Note that [tex]2n-3[/tex] is not 0 when [tex]n[/tex] is an integer, which means it is perfectly fine to write
[tex]2(x_1+x_2+\ldots+x_n) = \frac{n(n+1)}{2n-3}[/tex]
Subtract this from the i-th equation to get
[tex]-3x_i = i - \frac{n(n+1)}{2n-3}[/tex]
This must hold for all solutions, which means if there is a solution it is unique and given by
[tex]x_i = -\frac{i}{3} + \frac{n(n+1)}{3(2n-3)}[/tex]
All that remains is to show that this is indeed a solution, which we can easily do by showing it satisfies all the equations.
Hope this helped,
R. Baber.