<s>[INST] How to solve $100x +19 =0 \pmod{23}$ How to solve $100x +19 =0 \pmod{23}$, which is $100x=-19 \pmod{23}$ ?
In general I want to know how to solve $ax=b \pmod{c}$.
 [/INST] $100x +19 =0 \pmod{23}$
$100x  ≡ -19 \pmod{23}$
$92x+8x=4-23\pmod{23}$
$=>8x≡4 \pmod{23}$ dividing both sides by $23$ and taking resdiues,
So,  $2x≡1 \pmod{23}$  dividing both sides by $4$ which is possible as $(4,23)=1,$
Now $\frac{23}{2}=11+\frac{1}{2}$  So, $23-2\cdot 11=1$
(Please refer to this  and this, for the theorem used.)
So, $2x≡23-2\cdot 11 \pmod{23}=>2x≡-22\pmod{23}$
$x≡-11\pmod{23}$ dividing both sides by $2$ as $(2,23)=1$
$x≡-11\pmod{23}≡12\pmod{23}$
To solve the general problem, $ax≡b\pmod{c}$,
$d=(a,b)$ must divide $c$ to admit any solution(according to this or this).
If we divide either side by $d$, so that $Ax≡B\pmod{C}$  where $\frac{a}{A}=\frac{b}{B}=\frac{c}{C}=d$ and (A,B)=1.
Using convergent property of continued fraction, we can get integers $r,s$ such that $rA-sB=±1$
For example, $\frac{17}{13}=1+\frac{4}{13}=1+\frac{1}{\frac{13}{4}}=1+\frac{1}{3+\frac{1}{4}}$
So, the convergent here is $1+\frac{1}{3}=\frac{4}{3}$
So. $17\cdot 3 -  13\cdot 4$ must be $±1$ , and is $-1$.
Put $rA-sB$ with proper sign in place of $(1)$,  $Ax≡B(1)\pmod{C}$
One example can be found here.
Please refer to this for the details.
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