One method is simply the Euclidean algorithm:
\begin{align*}
31 &= 4(7) + 3\\\
7 &= 2(3) + 1.
\end{align*}
So $ 1 = 7 - 2(3) = 7 - 2(31 - 4(7)) = 9(7) - 2(31)$. Viewing the equation $1 = 9(7) -2(31)$ modulo $31$ gives $ 1 \equiv 9(7)\pmod{31}$, so the multiplicative inverse of $7$ modulo $31$ is $9$. This works in any situation where you want to find the multiplicative inverse of $a$ modulo $m$, provided of course that such a thing exists (i.e., $\gcd(a,m) = 1$). The Euclidean Algorithm gives you a constructive way of finding $r$ and $s$ such that $ar+ms = \gcd(a,m)$, but if you manage to find $r$ and $s$ some other way, that will do it too. As soon as you have $ar+ms=1$, that means that $r$ is the modular inverse of $a$ modulo $m$, since the equation immediately yields $ar\equiv 1 \pmod{m}$.
Another method is to play with fractions Gauss's method:
$$\frac{1}{7} = \frac{1\times 5}{7\times 5} = \frac{5}{35} = \frac{5}{4} = \frac{5\times 8}{4\times 8} = \frac{40}{32} = \frac{9}{1}.$$
Here, you reduce modulo $31$ where appropriate, and the only thing to be careful of is that you should only multiply and divide by things relatively prime to the modulus. Here, since $31$ is prime, this is easy. At each step, I just multiplied by the smallest number that would yield a reduction; so first I multiplied by $5$ because that's the smallest multiple of $7$ that is larger than $32$, and later I multiplied by $8$ because it was the smallest multiple of $4$ that is larger than $32$. Added: As Bill notes, the method may fail for composite moduli.
Both of the above methods work for general modulus, not just for a prime modulus (though Method 2 may fail in that situation); of course, you can only find multiplicative inverses if the number is relatively prime to the modulus.
Update. Yes, your method for general linear congruences is the standard one. We have a very straightforward method for solving congruences of the form $$ax \equiv b\pmod{m},$$
namely, it has solutions if and only if $\gcd(a,m)|b$, in which case it has exactly $\gcd(a,m)$ solutions modulo $m$.
To solve such equations, you first consider the case with $\gcd(a,m)=1$, in which case $ax\equiv b\pmod{m}$ is solved either by finding the multiplicative inverse of $a$ modulo $m$, or as I did in method $2$ above looking at $\frac{b}{a}$.
Once you know how to solve them in the case where $\gcd(a,m)=1$, you can take the general case of $\gcd(a,m) = d$, and from
$$ax\equiv b\pmod{m}$$
go to
$$\frac{a}{d}x \equiv \frac{b}{d}\pmod{\frac{m}{d}},$$
to get the unique solution $\mathbf{x}_0$. Once you have that unique solution, you get all solutions to the original congruence by considering
$$\mathbf{x}_0,\quad \mathbf{x}_0 + \frac{m}{d},\quad \mathbf{x}_0 + \frac{2m}{d},\quad\ldots, \mathbf{x}_0 + \frac{(d-1)m}{d}.$$
If $1-n$ is a modular inverse, then so is $(1-n) + kn^2$ for any integer k. In your case, to get the inverse between $0$ and $n^2$, you can take $k=1$ to get the inverse as $n^2-n+1$.
Best Answer
You have to write $$ 1 = 240x+17y $$ so $$ 240x\equiv 1\pmod{17} $$
The Euclidean algorithm applied to $240$ and $17$ gives \begin{align} \color{red}{240} &= \color{red}{17}\cdot 14 + \color{red}{2} \\ \color{red}{17} &= \color{red}{2}\cdot 8 + \color{red}{1} \end{align} The successive remainders are colored red. Now start from the top: $$ \color{red}{2}=\color{red}{240}-\color{red}{17}\cdot 14 $$ Go one line down: $$ \color{red}{1}=\color{red}{17}-\color{red}{2}\cdot 8 $$ Substitute the value you have for $\color{red}{2}$: $$ \color{red}{1}=\color{red}{17}-(\color{red}{240}-\color{red}{17}\cdot 14)\cdot 8 =\color{red}{17}\cdot(1+132)-\color{red}{240}\cdot 8 =\color{red}{240}\cdot(-8)+\color{red}{17}\cdot 133 $$ Thus you can take $x=-8$, or else $x=9$ since $-8\equiv 9\pmod{17}$. Indeed $$ 240\cdot9=2160=17\cdot127+1 $$ or $$ 240\cdot 9\equiv 1\pmod{17}. $$
Just remember operating on the “red numbers” as if they were letters. Express each remainder in terms of the previous ones and substitute in the equations below the first. Only terms with $\color{red}{240}$ and $\color{red}{17}$ multiplied by integers will remain.