Help to solve $15x \equiv 20 \pmod{88}$ Solve $15x$ "congruent to" $20\mod 88$
So far I think I know $15\mod 88$ is $-41$ or if positive $47$`
 A: A general approach for problems of this type:
Notice that, if you can find an $y$ such that $15y \equiv 1 \pmod{88}$, then we can multiply both sides of this congruence by $y$ to yield $15xy \equiv 20y \pmod{88}$.  This becomes $x \equiv 20y \pmod{88}$.
So how do we find this $y$?  Notice that $\gcd(15, 88) = 1$.  Therefore, we can use the extended Euclidean algorithm to find the guaranteed $m, n \in \mathbb{Z}$ such that $15m + 88n = 1$.  Modding out by $88$ here yields $15m \equiv 1 \pmod{88}$, and so the $m$ we found is our $y$!
A: ${\rm mod}\ 88\!:\,\ x \equiv \dfrac{20}{15}\equiv \dfrac{4}3\equiv \dfrac{-84}3\equiv -28\equiv 60$
Beware $\ $ Modular fraction arithmetic is well-defined only for fractions with denominator coprime to the modulus, and we can only cancel factors coprime to the modulus (as we did for the factor $5$ above). See here for further discussion.
A: A trick to remember.
You are trying to solve the equation $15x - 20 = 88y$ for some $x,y$. Factorize five out, and you get that $5(3x-4) = 88y$. Now, since $5$ is prime and $5$ doesn
't divide $88$, it divides $y$. 
Now, it is a simple question of trial and error with multiples of $5$ (you don't have to check all values of $y\ $!). As it turns out, with $y=10$, $x=60$ works. 
