I am trying to solve a problem of breaking an amatuer cryptography.
The problem boils down to solving a combined system of linear and bilinear equations having $50$ unknowns.
For representational purposes, the equations look similar to the following, with $x$, $y$ and $z$ being the unknowns.
\begin{align} \begin{cases} 3x + 10y + 8z + 5xy &= 1470 \\ 2x + 10y + 3z + yz + xz &= 1210 \\ x + 5y + z + 3xy + 16xz &= 5540 \\ x + 3y + 8z + 12xy + 4yz &= 5110 \end{cases} \end{align}
The above system has the solution $x=10 , y=20, z=30$.
I want to know the method for solving these type of equations. Can gaussian elimination be applied on such a system ?
solving equation (1) for $$z$$ we obtain $$-5/8\,xy-3/8\,x-5/4\,y+{\frac{735}{4}}=0$$ (I) plugging this in the second equation we obtain $$-5/8\,{x}^{2}y-3/8\,{x}^{2}-7/2\,xy+{\frac {1477\,x}{8}}-5/8\,x{y}^{2} -5/4\,{y}^{2}+190\,y-{\frac{2635}{4}} =0$$ (II) pliugging (I) in the third equation $$-{\frac {141\,xy}{8}}-10\,{x}^{2}y-6\,{x}^{2}+{\frac {23525\,x}{8}}+{ \frac {15\,y}{4}}-{\frac{21425}{4}}=0$$ (III) plugging (I) in the fourth equation we get $$11/2\,xy-5/2\,x{y}^{2}-5\,{y}^{2}+728\,y-2\,x-3640=0$$ (IV) solving (IV) for $$y$$ we obtain $$y=-{\frac {48\,{x}^{2}-23525\,x+42850}{80\,{x}^{2}+141\,x-30}}$$ (V) plugging this in (III) factorizing and simplifying we get $$ \left( x-10 \right) \left( 13120\,{x}^{4}-5741148\,{x}^{3}-54229691 \,{x}^{2}+11341510\,x+205150100 \right) =0$$ the rest do it by yourself.