$18 x \equiv 43 \pmod {23 }$. Slightly similar to a question I submitted a few minutes ago, however this time my value for $x$ is $-602$, and I'm not sure how to write the general solution for $x$ .. help please?  
Help
 A: There is a unique solution to this equation modulo 23, so therefore, in general,
$$x\equiv 19\pmod{23}$$
A: First, let's simplify the equivalence:
$18x\equiv43$ mod 23
$18x\equiv20$ mod 23
Now, find the multiplicative inverse of 18 mod 23 which is 9 as $18 * 9 \equiv 162 \equiv 23*7 + 1 \equiv 1$ (This can be done by brute force if necessarily as 23 is a prime so the non-zero values do have multiplicative inverses with 1 as the identity.
$9*18x\equiv9*20$ mod 23
$x \equiv 180$ mod 23
$x \equiv 19$ mod 23 as 23*7+19 = 180
A: To get other solutions, the situation here is even simpler than my answer to your previous question: If
$$
x_1\equiv x_2\pmod{23}
$$
then
$$
23\mid18(x_1-x_2)
$$
and therefore
$$
18x_1\equiv18x_2\pmod{23}
$$
we don't even need to worry about $\gcd(18,23)$ for the preceding, though if we do consider it, we can get all solutions.
A: Try this chain (all mod $23$, which is prime)
$$18x\equiv 43$$
$$18x\equiv 66$$
$$3x\equiv 11$$
$(11+23=34, 34+23=57)$
$$3x\equiv 57$$
$$x\equiv 19$$
Knocking out prime factors of $2$ and $3$ in this way is easy, and systematic. I took the easy opportunity to knock out a factor of $6$ (could have started $18x \equiv 20$).
