Solution
= Solution
For each crossing pair, delete one of its two edges. At most $t(G)$ distinct edges are deleted, and the graph left behind has a crossing-free drawing. The <planar graph edge bound> therefore gives
$$
e-t(G)\leq3n-6,
$$
so
$$
\boxed{t(G)\geq e-3n+6}.
$$
This is the linear estimate underlying the <crossing lemma>.