Let $\mathbb{Z}_2 = \{0,1\}$ be the integers mod $2$. Let $A \in \mathbb{Z}_2^{m \times n}$ and $u,v \in \mathbb{Z}_2^m$. Consider the problem of determining whether there is a vector $x \in \mathbb{Z}_2^n$ such that $Ax = u \vee Ax = v.$ One can reduce this problem to the problem of determining whether there is a solution to this system: \begin{pmatrix} 0 & \ldots & 0 & 1 & 1 \\ A_{1,1} & \ldots & A_{1,n} & u_1 & v_1 \\ \vdots & \ddots & \vdots & \vdots & \vdots \\ A_{m,1} & \ldots & A_{m,n} & u_m & v_m \\ \end{pmatrix} \begin{pmatrix} x_1 \\ \vdots \\ x_n \\ y_1 \\ y_2\end{pmatrix} = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix} since each line gives $A_i \cdot x = y_1 u_i + y_2 v_i$ and $(y_1, y_2) \in \{(0,1), (1,0)\}$.
I wonder if there is such a strategy to reduce $Ax = u \vee Ax = v$ to $By = w$ over $\mathbb{Z}_3$.