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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 $ax + b = c\pmod{n}.$

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    $7j+\color{blue}6\equiv4\pmod5\iff 7j\equiv4-\color{blue}6=-2\pmod5\iff 7j\equiv3\pmod5$ because $-2\equiv3\pmod5$ – J. W. Tanner Nov 03 '21 at 01:16
  • Thank you that was very helpful! That makes a lot more sense now. I seen that 4 - 6 = -2 but I wasn't sure what to do next. Where does the 2 in 2j≡3mod5 come from? is it just the absolute value from -2? – null Nov 03 '21 at 01:24
  • $7j\equiv2j\pmod5$ because $7\equiv2\pmod5$; glad to help – J. W. Tanner Nov 03 '21 at 01:31
  • Awesome! I very much appreciate the help. If you want to post as an answer I'll accept. – null Nov 03 '21 at 01:34
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    I would like to post an answer, but I'm afraid that someone will say that this question is similar to a previous one, and then I could be suspended for answering a duplicate. Anyways, welcome to Mathematics Stack Exchange. – J. W. Tanner Nov 03 '21 at 01:40
  • They add $-6$ to both sides fo the congruence using the Congruence Sum Rule in the linked dupe, i.e. $, A\equiv B,\ {-}6\equiv -6\Rightarrow A-6\equiv B-6.\ $ Congruences are generalized equations, and like equations they are preserved by adding (or multiplying) both sides by equal numbers (says the Sum & Product Rules). – Bill Dubuque Nov 03 '21 at 01:42
  • SImilarly $\ 7\equiv 2\Rightarrow 7j\equiv 2j\ $ by the Congruence Product Rule. Thus by induction, the congruence rules imply that we can replace arguments of sums and products (but not expts!) by any congruent argument and we will obtain a congruent result - as explained here – Bill Dubuque Nov 03 '21 at 01:52
  • Typo alert: "by equal numbers" should be "by congruent numbers" in the 2nd last comment. – Bill Dubuque Nov 03 '21 at 07:57

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