What they refer to as a "worked example" is just a verification of the fact that the multiplicative inverse of $17$ modulo $\phi(3233) = 3120$ is $2753$.
In order to find the multiplicative inverse of $17$ modulo $3120$, suppose $17x \equiv 1 \bmod 3120$. Then $3120 \mid (17x - 1)$ so $17x - 1 = 3120y$ for some $y \in \mathbb{Z}$. Rearranging, we obtain the equation $17x - 3120y = 1$. Now we use the Euclidean algorithm to find the greatest common divisor of $17$ and $3120$:
\begin{align*}
3120 &= 183\times 17 + 9\\
17 &= 1\times 9 + 8\\
9 &= 1\times 8 + 1.
\end{align*}
So the greatest common divisor of $17$ and $3120$ is $1$; note, if it weren't $1$, then the equation $17x - 3120y = 1$ would have no solutions. Using back substitution, we have
\begin{align*}
1 &= 9 - 1\times 8\\
&= 9 - 1\times(17 - 1\times 9)\\
&= 2\times 9 - 1\times 17\\
&= 2\times(3120 - 183\times 17) - 1\times 17\\
&= 2\times 3120 - 367\times 17.
\end{align*}
Therefore $x = -367$, $y = 2$ is a solution to $17x - 3120y = 1$. So $(-367)\times 17 \equiv 1 \bmod 3120$. Note that for any $t \in \mathbb{Z}$, $(-367 + 3120t)\times 17 \equiv -367\times 17 \bmod 3120 \equiv 1\bmod 3120$. Taking $t = 1$, we see that $-367 + 3120\times 1 = 2753$.