# Solve $47x \equiv 4 \mod 17$ [duplicate]

I don't really understand how to solve the equation $$47x \equiv 4 \mod 17.$$ If someone could help, that would be much appreciated. Thanks!

As $$47\equiv-4\pmod{17},$$

$$47x\equiv4\pmod{17}\iff-4x\equiv4$$

As $$(4,17)=1,$$

$$-x\equiv1\iff x\equiv-1$$

$$\implies x\equiv-1+17$$

• Your last line is garbled... – TonyK Nov 26 '20 at 16:23
• @TonyK, Now sure if I have understood you correctly . – lab bhattacharjee Nov 26 '20 at 16:25
• Well you've fixed it now, haven't you? Initially it read $x\equiv -+17$. – TonyK Nov 26 '20 at 16:39