Looking for the correct method to solve a modulo congruence of the form,
$ax\equiv b(mod\ m)$
I know that the congruence is solvable if $(a,m)\ \vert \ b$. I'm just unsure of how to solve the congruence once I find that it IS solvable.
My working problem is,
$11x\equiv 21(mod\ 105) $
Thank you for the help! I'm so grateful for everyone's expertise.