Solving Underdetermined Non-Linear Equation of Probabilities

Question

I'm trying to see if I can solve for g in the set of linear equations: $$ga+gb = m$$ $$ga+gb+2gab=n$$ where g, a, and b are all probabilities (i.e. between 0 and 1). I know that the problem is underdetermined, but I was wondering if through some clever algebra it is possible to get a unique solution for g or if there is a counterexample showing why it is not (i.e. two sets of g, a, and b that produce identical m and n). I tried squaring equation 1 and combining it with equation 2 to get (ga-gb) on one side of an equation, but that didn't work and I tried generating counterexamples from $$gab = g'a'b'$$ $$ga+gb = g'a'+g'b'$$ But I can't seem to either definitively prove that g is or isn't universally solvable. My intuition (because it is an underdetermined set of equations) leads me to believe that it isn't... Thanks so much!

Answer 1

Yes, there are several solution with the same $m$ and $n$. For example, taking $(g,a,b)=(\frac{1}{2},\frac{2}{3},0)$ and $(g,a,b)=(\frac{1}{3},1,0)$ the equations are both times satisfied with $m=n=\frac{1}{3}$.

For an example with $0<g,a,b<1$, take $(g,a,b)=(\frac{1}{3},\frac{1}{4},\frac{1}{5})$ and $(g,a,b)=(\frac{1}{4},\frac{9+\sqrt{21}}{30},\frac{9-\sqrt{21}}{30})$. Then both equations are satisfied with $$ (m,n)=\left(\frac{3}{20},\frac{11}{60}\right). $$