What is the best way to solve modular arithmetic equations such as $9x \equiv 33 \pmod{43}$? What is the best way to solve equations like the following:
$9x \equiv 33 \pmod{43}$
The only way I know would be to try all multiples of $43$ and $9$ and compare until I get $33$ for the remainder.
Is there a more efficient way ? 
Help would be greatly appreciated!
 A: How would we solve it in $\mathbb{R}$?  Divide both sides by $9$ of course—or, in other words, multiply both sides by the multiplicative inverse of $9$.  This setting is no different.  
The challenge is knowing the multiplicative inverse of $9$ in $\mathbb{Z}_{43}$.  What is key$^\dagger$ is that $\gcd(9,43)=1$, which guarantees integers $n$ and $m$ such that $9n + 43m = 1$.  Modding out by $43$, we see that  $9n \equiv 1 \pmod{43}$.  Thus, multiplying both sides of $9x \equiv 33 \pmod{43}$ by $n$ gives us $x$.
The integers $n$ and $m$ can be found by using the extended Euclidean algorithm.

$^\dagger$ This coprimality condition is if-and-only-if.  An integer $x$ will not have a multiplicative inverse $(\text{mod} \ n)$ if $\gcd(x,n) \neq 1$.
A: $\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{9}{x}\equiv\mathrm{33}\left({mod}\mathrm{43}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
$\mathrm{3}{x}\equiv\mathrm{11}\left({mod}\mathrm{43}\right)\equiv\mathrm{54}\left({mod}\mathrm{43}\right) \\ $
$\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\equiv\mathrm{18}\left({mod}\mathrm{43}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $
A: $43$ is prime.  And even if it weren't, $\gcd(9,43)= 1$.
That means we know that $9^{-1}_{43}$ exists.  That is there is a number so that $9\cdot 9^{-1}_{43} \equiv 1 \pmod {43}$.  And we could calculate  (or guess) that $9^{-1}_{43}\equiv 24$ (because $9\cdot 24\equiv 1 \pmod {43}$) and we can solve $9x \equiv 33\pmod {43} \implies 24\cdot 9 x\equiv x \equiv 24\cdot 33\pmod {43}$.
But we don't have to.  If $\gcd(9,43) =1$ and if $9^{-1}$ exists, then in this case we are allows to know that division is acceptable, and we are, in this case, allowed to divide both sides by $3$.
$9x \equiv 33 \pmod {43}$
$3x\equiv 11 \pmod {43}$  (we wouldn't be allowed to do this if $\gcd(3,43)\ne 1$)
Now $3$ is such a small number that we know that one these:  $11, 11 + 43, 11+2*43$ is divisible by $3$.
As it turns out $3x \equiv 11 \equiv 11 + 43 \equiv 54\pmod {43}$.
And we can divide both sides by $3$ to get
$x \equiv 18\pmod {43}$.[1]
..........
Alternatively we can do baby steps.
$9x \equiv 33 \pmod {43} \implies$ there is a $k$ so that
$9x = 33 + 43k$.  Divide by $3$ and we get
$3x = 11 + 43\frac k3$.  That will be an integer so long as we choose a $k$ that is divisible by $3$.  Relable $m= \frac k3$
$3x = 11 + 43m$
$3x =11 + 43 + 43(m-1)$
$3x = 54 +43(m-1)$
$x = 18 + 43\frac {m-1}3$ which is an integer so long as we select $m-1$ so that $m-1$ is divisible by $3$.  (For example the easiest example is $m-1=0$, $m=1$, $k = 3$ an $9*18 = 33 + 3\cdot 43$.  But we don't need to find the actually $k$)
......
[1] Notice our earlier answer $24\cdot 33\equiv 24\cdot 32 + 24\equiv 48\cdot 16 + 24 \equiv 5\cdot 16 + 24\equiv 80 + 24 \equiv 86 + 18 \equiv 18\pmod {43}$.  So our answers are consistent.
A: As the OP observes,
$$9x \equiv 33 \pmod{43} \text{ iff } 9x - 33 = 43k$$
but this leaves us with a considerable amount of calculating as we start increasing $x$ from $2$ on up.
But by modding out by $9$  on the right side equation, we can eliminate the $x$ variable:
$$\tag 13 \equiv 7k \pmod{9}$$
So we can solve the starting problem if we can solve $\text{(1)}$, but this one can be done by inspection - we get $k = 3$ as the minimal positive integer choice for $k$, and thus
$$ 9x - 33 = 43 \times 3 \text{ iff } x = 18$$
