The problem is NP-hard over any field not of characteristic 2. We show this by reduction from an NP-complete NAE-3-SAT problem of checking satisfiability of a boolean formula $\land_{i} \mathrm{NAE}(z_{i, 1}, z_{i, 2}, z_{i, 3})$, where $z_{i, j}$ are variables $x_{\cdot}$ or their negations, and $\mathrm{NAE}(\ldots)$ is true iff not all of its arguments have the same value.
The reduction is as follows: introduce variables $x_i$ subject to $x_i = \pm 1$, where we interpret values $1$ and $-1$ as $\top$ and $\bot$, respectively. For each $\mathrm{NAE}(\ldots)$ we impose a condition $s_a x_a + s_b x_b + s_c x_c = \pm 1$, where $x_a, x_b, x_c$ are variables in arguments of NAE, and each of the signs $s_a, s_b, s_c$ is $\pm 1$ depending on whether the corresponding variable was taken with negation. It is easy to check that the imposed conditions are equivalent to the original formula, hence the reduction is proper (and of course, polynomial). The reduction works both for machines with precise and approximate real registers (as long as it handles numbers with $O(1 / n)$ relative precision). It also works for any field $F$ not of characteristic 2 if we take $1 \in F$ to be any element of order at least 3, or zero.
In fields of characteristic 2 the given equations (treated as $a_i \cdot x = \pm b_i$) are equivalent to $a_i x = b_i$, which can be solved with any standard LES solving algorithm (for instance, Gaussian elimination).
