Can some explain how $2x \equiv 3 \pmod 5$ can be reduced to $x \equiv 4 \pmod 5$?

Question

Given $2x \equiv 3 \pmod 5$ then

1. $2x \equiv 3 \pmod 5$

2. $3(2x) \equiv 3(3) \pmod 5$, which somehow changes to

3. $x \equiv 9 \pmod 5$

and finally even more confusing is its reduction to

$$x \equiv 4 \pmod 5$$

4??? What happened to the $9$ and $3$?

We are solving these as an equivlance relation

equivalnece realtion defined as it partitions ℤ into equivalence classes where [] denotes the equivalence class that contains integer .

Answer 1

Note

$$\begin{equation}\begin{aligned} 3(2x) \equiv 3(3) \pmod 5 \\ 6x \equiv 9 \pmod 5 \\ x + 5x \equiv 9 \pmod 5 \\ x \equiv 9 \pmod 5 \end{aligned}\end{equation}\tag{1}\label{eq1A}$$

This is because $5x \equiv 0 \pmod 5$. Similarly, you have

$$\begin{equation}\begin{aligned} x \equiv 9 \pmod 5 \\ x \equiv 5 + 4 \pmod 5 \\ x \equiv 4 \pmod 5 \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

since $5 \equiv 0 \pmod 5$. Note, in general, $a \equiv b \pmod c \iff a - b \equiv 0 \pmod c$, i.e., $a - b = kc$ for some integer $k$. For \eqref{eq1A}, you have $5x \equiv 0 \pmod 5$ since $5x - 0 = (x)5$, and for \eqref{eq2A}, you have $5 - 0 = (1)5$.

Considering the equivalence class of $[4]$ modulo $5$, note all values, $b$, in this equivalence class are such that $b \equiv 4 \pmod 5 \iff b = 5k + 4$ for some integer $k$. In your specific case, you get $b = 9$ where $k = 1$ since $9 = 5(1) + 4$. In general, $[4] = \{\ldots, -11, -6, -1, 4, 9, 14, \ldots\}$ as $k$ goes through the integers from negative infinity to positive infinity.

Answer 2

Okay, so in step two, you have $6x \equiv 9$ (mod 5).

Then, it also works out that $6x-5x \equiv 9-5$ (mod 5)

(Because you're just subtracting /adding multiples of $5$, it doesn't change the equivalence relation)

Answer 3

In modular arithmetic you can reduce constants any time you wish. Since $3 \times 2 = 6 \equiv 1 \pmod{5}$ and $3 \times 3 = 9 \equiv 4 \pmod{5}$ you see that $x \equiv1 \pmod{5}$ when you multiply both sides of the original congruence by $3$. The reason you choose to multiply by $3$ is precisely because it's the multiplicative inverse of $2$: $3 \times 2 = 6 \equiv 1 \pmod{5}$.

Answer 4

The inverse of $2$ mod $5$ is $3$, because $3\cdot 2=6\cong1\pmod5$. Use this fact to solve for $x$, by multiplying both sides by $2^{-1}=3$, just as you would an ordinary algebraic equation.

Get $2^{-1}\cdot2x\cong2^{-1}\cdot3\pmod5\implies x\cong3\cdot3\pmod5\implies x\cong9\pmod5\implies x\cong4\pmod5$.