Why '$3t \equiv 2 \pmod 5 \implies t \equiv 4 \pmod 5$' is true in resolving congruence?

Question

I want to resolve this system of equations

\begin{equation} \begin{cases} x & \equiv & 1 & \pmod 3\\ x & \equiv & 3 & \pmod 5\\ x & \equiv & 2 & \pmod 7\\ \end{cases} \end{equation}

I have solution for this, but there is one step, highlighted, which I don't understand

$$x \equiv 1 \pmod 3 \implies x = 3t + 1$$ $$3t + 1 \equiv 3 \pmod 5 \implies \color{red}{3t \equiv 2 \pmod 5 \implies t \equiv 4 \pmod 5} \implies t = 5u + 4 $$

and so on.

Answer 1

We have $$3t \equiv 2 \equiv 12 \pmod 5.$$

Recall the rule of division modulo $n$

If $\text{gcd}(a,n) = 1$ and $ab \equiv ac \pmod n$, then we have $b \equiv c \pmod n$.

Applying this with $a = 3$ and $n = 5$, we have $$3t \equiv 12 \pmod 5 \implies t \equiv 4 \pmod 5.$$