There are many ways to solve the problem. The conceptually simplest, but most tedious, is to test one by one the possibilities $x\equiv 0\pmod{5}$, $x\equiv 1\pmod{5}$, and so on up to $x\equiv 4\pmod{5}$. Quickly we find that $x\equiv 4\pmod{5}$. (This approach would become quite unpleasant if $5$ were replaced by $97$.)

It is simpler to use some algebra. So rewrite as $2x\equiv 3\pmod{5}$. Since $3\equiv 8\pmod{5}$, it is convenient to rewrite the congruence as $2x\equiv 8\pmod{5}$. Then since $2$ and $5$ are relatively prime, we can divide by $2$, getting $x\equiv 4\pmod{5}$.

A fancier version is to start from $2x\equiv 3\pmod{5}$. Now multiply both sides by $3$ (the modular inverse of $2$). We get $6x\equiv 9\pmod{5}$. But $6\equiv 1\pmod{5}$ and $9\equiv 4\pmod{5}$, so we conclude that $x\equiv 4\pmod{5}$.

Remark: We used congruence notation throughout, since it is very important to get accustomed to it. But $2x\equiv 3\pmod{5}$ means that $5$ divides $2x-3$. So we want to solve $2x-3=5k$, that is, $2x=3+5k$. So we want to find a $k$ such that $3+5k$ is divisible by $2$. It is clear that $k=1$ works, giving $2x=8$ so $x=4$. Any number congruent to $4$ modulo $5$ will also work, giving answer $x\equiv 4\pmod{5}$.

Mathematically, congruence modulo $n$ is an equivalence relation. We define:

$$a\equiv b \pmod n \iff n\mid (a - b)$$

Equivalently: When working in $\pmod n$, any number $a$ is congruent $\mod n$ to an integer $b$ if there exists an integer $k$ for which $\;nk = (a - b)$.

Now, let's compare the "discrepancies" in the equivalences you note (which are, in fact, all true):

$$13 \equiv \color{blue}{\bf 1} \pmod 4 \iff 4\mid (13-1) \iff 4\mid 12\; \checkmark$$ $$13\equiv \color{blue}{\bf 5} \pmod 4 \iff 4\mid (13-5) \iff 4\mid 8 \;\checkmark$$

• Note, indeed, that $\color{blue}{\bf 5 \equiv 1} \pmod 4$ since $4\mid(5 - 1) = 4$ $$ $$

$$9 \equiv \color{blue}{\bf 4} \pmod 5 \iff 5 \mid (9-4) \iff 5\mid 5 \;\checkmark$$ $$9 \equiv \color{blue}{\bf -1} \pmod 5 \iff 5 \mid (9 - (-1)) \iff 5\mid 10 \;\checkmark$$

• And again, note that $\color{blue}{\bf4 \equiv -1} \pmod 5$ since $5\mid(4-(-1)) = 5$ $$ $$

It is often customary to express equivalence modulo $n$, by choosing $b$ in $\;a \equiv b \pmod n\;$ to be such that $0 \leq b \lt n$. But this choice is simply a representative of all the numbers which belong to the same equivalence class, denoted $[b]$, $\pmod n$:

E.g. If $n = 4$, then one of the following holds: $$a \equiv b \pmod 4 \iff \begin{cases} a, b \in [0] = \{4k + 0\mid k\in \mathbb Z\} = \{\cdots, -8, -4, 0, 4, 8, 12,\cdots\} \\ \\ a, b \in [1] = \{4k + 1\mid k \in \mathbb Z\} = \{\cdots, -7, -3, 1, 5, 9, 13,\cdots\} \\ \\ a ,b \in [2] = \{4k + 2\mid k \in \mathbb Z\} = \{\cdots, -6, -2, 2, 6, 10, 14,\cdots\} \\ \\ a, b \in [3] = \{4k + 3\mid k \in \mathbb Z\} = \{\cdots, -5, -1, 3, 7, 11, 15, \cdots\} \end{cases} $$