How do I simplify a modular congruence? If I have $$13x\equiv 15 \pmod{35}$$ or something like $$197x\equiv -1 \pmod{371}$$
How do I go about making the coefficient of the $13x$ and $197x$ $1$? Brute forcing this would take far too long. I know that you can use the Extended Euclidean Algorithm and Modular Inverses to solve, but I am unsure how.

Assuming you don't know how to use modular inverses (which is the way to do this and you should learn it) you can do it just be setting up equations
$13x \equiv 15 \pmod {35}$ means there exists an integer $m$ so that $13x =15 + 35m$ so $x = \frac{15 + 35m}{13}= 1 + 2m +\frac {2+9m}{13}$.
This means that $13|2+9m$ so $2+9m\equiv 0\pmod {13}$ and so $9m\equiv -2 \equiv 11\pmod {13}$.
That means there is an integer $n$ so that $9m = 11 + 13n$ and $m=\frac {11+13n}9=(n+1) + \frac {2+4n}9$.
That means $4n \equiv -2\equiv 7\pmod 9$.  Let $4n = 7 +9p$ so $n = \frac {7+9p}4 =(2p+1) + \frac{3+p}4$.
That means $p\equiv -3 \equiv 1\pmod 4$.
Let $p = 4z + 1$.
So $n = (2p+1) + \frac{3+p}4 = (2(4z+1)+1) + \frac {4z + 4}4 = 8z+3 + z + 1 =9z + 4$.
So $m =(n+1) + \frac {2+4n}9 = (9z+5) + \frac {2+4(9z+4)}9=(9z+5)+\frac{18+4\cdot 9z}9=(9z+5) + 2+4z = 13z +7$
So $x = 1 + 2m +\frac {2+9m}{13}= 1+2(13z +7) + \frac {2+9(13z+7)}{13}=(26z+15)+\frac {65 + 9\times 13z}{13}=(26z+15)+(5+9z)=20+35z$.
So $x \equiv 20 \pmod {35}$.
