Note that $25 | 15 \cdot 25$, so $15 \cdot 25 \equiv 0 \pmod{25}$. On the other hand, we see that
$$1 \equiv -15 \cdot 25 - 34 \cdot 11 \equiv 0 + (-34) \cdot 11 \pmod{25}$$
Finally,
$$-34 \equiv -34 + 50 \equiv 16 \pmod{25}$$
so we can rewrite the above as
$$1 \equiv 16 \cdot 11 \pmod{25}$$
and $16$ is the desired inverse.
Note that by direct computation,
$$16 \cdot 11 = 176 = 7 \cdot 25 + 1$$
Alternatively, we can use the Euclidean algorithm and find that
\begin{align*}
25 &= 2 \cdot 11 + 3 \\
11 &= 3 \cdot 3 + 2 \\
3 &= 1 \cdot 2 + 1
\end{align*}
Thus,
\begin{align*}
1 &= 3 - 1 \cdot 2 \\
&= 3 - 1 \cdot (11 - 3 \cdot 3) = 4 \cdot 3 - 1 \cdot 11 \\
&= 4 \cdot (25 - 2 \cdot 11) - 1 \cdot 11 = 4 \cdot 25 - 9 \cdot 11
\end{align*}
So $1 \equiv -9 \cdot 11 \equiv 16 \cdot 11 \pmod{25}$ as before.
If you want the multiplicative inverse of $2$ mod $7$, then you want to find an integer $n$ such that $2n = 7k + 1$, where $k$ is a nonnegative integer. Try $k = 1$, because that's the easiest thing to do. Then $2n= 8$, and $n = 4$.
Best Answer
This is a coincidence, and this is how it happened:
Because of the miscalculation, we now found the inverse of 43 modulo 660. We found that $1=43\cdot307-660\cdot20$. But coincidently, we have $660\cdot20=13200=600\cdot22$, so $1=43\cdot307-600\cdot22$. This means that we also have found the inverse of 43 modulo 600.
Edit: We can do the same trick for any divisor of 13200. Take 825 for example. We get $1=43\cdot307-825\cdot16$, so we also have found the inverse of 43 modulo 825.