# Solving congruences involving addition for CRT

I'm trying to follow this example on solving systems of congruences to understand CRT, however I do not understand how the author solves the following congruence:

$$7j + 6 \equiv 4 \pmod 5 \\ j \equiv 4 \pmod 5$$

If I were to attempt to solve this congruence, I would try $$7j + 6 \equiv 4 \pmod 5 \\ 7j \equiv -2 \pmod 5$$ but I do not know if subtracting 6 from both sides is valid.

Assuming it is, I would just guess and check numbers until I found a $j$ such that $5|(7j + 2)$.

So how do I solve this congruence (ideally, without guessing and checking)?

We have $$7j+6\equiv 4\bmod 5,$$ subtracting $6$ gives $$7j\equiv -2\equiv 3\bmod 5,$$ that is $$2j\equiv3\bmod 5.$$ Now we must find a number $b$ such that $2b\equiv 1\bmod 5$ in order to isolate $j$. It can be checked that $b=3$ does the trick. So, multiplying the congruence through by $3$ gives $$j\equiv 9\equiv 4\bmod 5.$$
• Just to make sure I understand, you go from $7j \equiv -2 \pmod 5$ to $7j \equiv 3 \pmod 5$ by adding 5 to the right-hand side, which you can do because that's like adding 0 when dealing with regular equations? – Charles German Jun 2 '18 at 1:42
• indeed - in general; $a\equiv b\bmod n$ and $a\equiv b+kn\bmod n$ are the same. in this case, these two statements are equivalent: "a number has a remainder of $-2$ when divided by $5$" and "a number has a remainder of $3$ when divided by $5$. – thesmallprint Jun 2 '18 at 1:55