Can anyone help me to understand what to do with the subtraction or addition on the LHS of a congruence? I found this question -
Solving congruences involving addition for CRT
But I don't understand how the answer goes from $$7j+6\equiv 4 \pmod{5}$$ to $$7j\equiv −2\equiv 3 \pmod{5}$$ from subtracting 6, and then something happens and things become familiar where $2j\equiv 3\pmod{5}$.
What rules are being applied here? What is being subtracted, and why does the $4 \pmod{5}$ turn to $3\pmod{5}$?
I could solve $2j\equiv 3\pmod{5}$, but I don't know how we got there from $7j+6\equiv 4 \pmod{5}.$
I just don't know what to do about $b$ given $ax + b = c\pmod{n}.$
4 - 6 = -2
but I wasn't sure what to do next. Where does the2
in2j≡3mod5
come from? is it just the absolute value from-2
? – null Nov 03 '21 at 01:24