be an edge from $S_j$ to $S_i$ with label $x^{-1}$. Our definition of $G$ does not necessarily preclude the existence of multiple edges between $S_i$ and $S_j$ (with different labels), or edges from $S_i$ to $S_i$. However, as part of the proof, we will show that this cannot happen, provided that we assume a suitable hypothesis on $a_1, a_2 {\text{ and }}a_3$.

First, we will show that $G$ captures almost all of the solutions to $a_1x_1 + a_2x_2 + a_3x_3 = 0$.

\[lem:ignore\_bad\_triples\] Say that a triple $S_i, S_j {\text{ and }}S_k$ is *good* if and only if the six relevant edges are present. For example, $S_i \rightarrow S_j$ has label $\frac{a_1}{a_2}$, $S_k \rightarrow S_j$ has label $\frac{a_3}{a_2}$, and so on. Say that a triple is *bad* otherwise.

Then, the total number of solutions to $a_1x_1 + a_2x_2 + a_3x_3 = 0$ among all of the bad triples is at most $\frac{{\varepsilon}}{4} |S|^2$.

There are six ways a triple $(S_i, S_j, S_k)$ can be bad. One such way is if there is no edge from $S_i$ to $S_j$ with label $\frac{a_1}{a_2}$.

Let us count the total number of solutions among triples for which the $\frac{a_1}{a_2}$ edge is missing. That is 