How to prove that $b\_n=\frac{7n-19}{3n+7}$ converges to $\frac73$?

In other words, we want to prove $\forall\epsilon>0, \exists N\in \mathbb{R}$ such that $n>N\implies |\frac{7n-19}{3n+7}-\frac73|<\epsilon$.

My scratch: $|\frac{7n-19}{3n+7}-\frac73|=\frac{106}{3(3n+7)}=\frac{106}{9n+21}\lt\epsilon$, then $n>\frac{106-21\epsilon}{9}$, so we can pick $N=\frac{106-21\epsilon}{9}$.

Formal: Let $\epsilon>0$, let $N=\frac{106-21\epsilon}{9}$, then $n>\frac{106-21\epsilon}{9}$, then $\frac{106}{3(3n+7)}\lt \epsilon$, then $|\frac{7n-19}{3n+7}-\frac73|\lt\epsilon.$

However, the answer gives that we should pick $N=\max\{1,\frac{106-21\epsilon}{9}\}$. Why is the $1$ significant?