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!

$43$ 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]
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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.
