Solve for positive solutions:
$a+b^2+2ab=9$
$b+c^2+2bc=47$
$c+a^2+2ac=16$
What I've done so far is add up all the equations to get: $a+b+c+(a+b+c)^2=72$, from which I got: $a+b+c=8$, but I still can't get the values of $a, b$, and $c$. I tried adding up the equations two at a time and got $(a+1)(b-7)=14$ but since the question's not asking for integer values, it doesn't quite help me that much.
With $$c=8-a-b$$ we get in equation (3): $$18a+2a^2-2ab+9-a^2-b=16-a^2$$(*) and in (2): $$b+(9+a^2-b)^2+2b(9+a^2-b)=47(**)$$ Now you can solve $(*)$ for $b$ and plug this in (**) $$b={\frac {2\,{a}^{2}+18\,a-7}{2\,a+1}}$$ Doing this we get after some algebra: $${\frac { \left( a-1 \right) \left( 3\,{a}^{3}-29\,{a}^{2}-236\,a+55 \right) }{ \left( 1+2\,a \right) ^{2}}} =0$$