I did it a different way, but got the same answer; If c + b = 2 and c + a = 3, then a = b + 1; (b + 1) + b = 1, 2b + 1 = 1, b = 0; c + 0 = 2, c = 2; {a, b, c} = {1, 0, 2}.

Or use elimination to eliminate b from the first two equations, then add the resulting equation to the third equation to get c. Once you have b and c, its easy to get a. Change the first to b+a=1, then invert to -b-a=-1. Add this to b+c=2, you get c-a=1. Add this to third equation to get 2c=4 Solve for c= 2. After that its just a matter of plugging in numbers in the original equations: If b+c=2 and c=2 then b=0, and if b=0 and a+b=1, then a=1.

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a+b+c=b+c+a . Возможно 1+2+3 тоже , равно шести 6 .