The easiest way which is widely known to calculate a modular inverse is finding the smallest $k\in\mathbb N$ such that the following expression is an integral integer:
$$\frac{1+k\cdot \varphi}{e}$$
If this is an integral integer, it is the inverse of $e$ modulo $\varphi$ and thus $d$ and you'll find it with at most $e$ tries.
The issue obviously know is that the run-time of this will grow linearly with the size of $e$, as opposed to the extended euclidean algorithm (EEA) which will run in time proportional to (a power of) the logarithm of $e$, which will especially be much faster than the above method when $e$ is chosen as large as the modulus. If you want to see examples and how-tos on the algorithm, Wikipedia is your friend as usual.