Solving linear modular equations

Question

How does

$17r \equiv-2 \pmod {11} $

reduce to

$r\equiv-4 \pmod {11}$?

I know that $-2\equiv9 \pmod {11}$, but how do I simplify the $17$?

Answer 1

It's $6r$ cause $17r=11r+6r$ $$17r\equiv -2\pmod {11}\\6r\equiv-2\pmod{11}\\6r\equiv9\pmod{11}\\2r\equiv3\pmod{11}\\2r\equiv-8\pmod{11}\\r\equiv-4 \pmod{11}$$

Answer 2

First, $17\equiv 6\pmod {11}$, and since we have $6r$ it would be helpful to find a number that is an inverse of $6$ modulo $11$. A quick guess suggests that $6\cdot 2=12\equiv 1\pmod{11}$, and this means that $6^{-1}\equiv 2\pmod {11}$.

So we apply this as follows:

$$\begin{align}17r&\equiv -2\pmod{11}\\ \implies 6r&\equiv -2\\ \implies 6^{-1}\cdot 6r&\equiv 6^{-1}\cdot -2\\ \implies r\equiv 2\cdot -2&\equiv -4\pmod{11}\end{align}$$

Answer 3

${\rm mod}\ 11\!:\,\ \dfrac{-2}{17}\,\equiv\, \dfrac{20}{-5}\,\equiv\, -4$