I am thinking how to solve the four simultaneous equations for $x,y,z,u$ where $\beta, c, a, b$ are arbitrary real numbers.
$$\frac{a-x}{b} - z - u = 0,\\ x - min(\beta u, c y) =0,\\ x -y = 0,\\ x-1 = 0$$
From the last two equations we have $x=1, y=1$ but we have to find values of $z, u$ as well.
In the second equation, we will have $1 - min(\beta u, c) = 0$ but then after that how we proceed?
We consider two cases, suppose $min(\beta u, c) = \beta u$, then $u = \frac{1}{\beta}$. Next we consider the other case, $min(\beta u, c) = c$, then we get $c = 1$ but $c $ is an arbitrary real number..
If $u = \frac{1}{\beta}$, then from the first equation we have $z = \frac{a-1}{b} - \frac{1}{\beta}$. So $(x,y,z,u) = (1,1,\frac{a-1}{b} - \frac{1}{\beta}, \frac{1}{\beta})$.
For the second case, we had $c = 1$ and thus from the first equation, we have $z+u = \frac{a-1}{b}$..this can have infinite solutions
So I am puzzled which one should I consider