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We know that, for example,

$2x \equiv 3 \mod 4$

has no solutions since

$2\mid 2x, 2\mid 4,$ however 2 does not divide 3.


So my question is, how does one get from

$2x \equiv 3 \mod 5$

to

$x \equiv 4 \mod 5$?

In other words, could someone please explain the steps involved going from $2x = 3 + 5k$ to $x = 4 + 5k$, where $k\ \epsilon \ \mathbb{Z}$.

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Well, there can be many ways. For example, double the congruence $2x\equiv 3\pmod 5$: $\ 4x\equiv 6\pmod 5$, but because $4\equiv -1$ and $6\equiv 1$, this yields $$-x\equiv 1\pmod 5$$ which gives $x\equiv -1\ (\equiv 4) \pmod 5$.

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