Question 54056
{{{abc = a+b+c}}}
{{{abc - c = a+b}}}
{{{c(ab - 1) = a+b}}}
{{{c = (a+b)/(ab - 1)}}}

I will choose values for a and b arbitrarily. Suppose a = 6 and b = 2:

{{{c = 8/11}}}
{{{6+2+8/11 = 6*2*8/11}}}

Therefore, we can choose any value for two of the values, and find a value for which this equation is true.

Similarly with all the other equations, as long as you don't restrict them to the naturals, and algebra isn't capable of that.