modular arithmetic help 
*

*$3t_1 \equiv 1 \pmod 5$

*$t_1 \equiv 2 \pmod 5$


how can we derive line 2 from line 1? 
 A: Because $5$ is prime, we know that the modular multiplicative inverse exists for the co-prime equivalence classes; that is $\forall a:\gcd(a, 5)=1, \exists b: ab\equiv 1 \bmod 5 $
So $a^{-1}$ means something and is unique to equivalence. In particular, $3^{-1} \equiv 2 \bmod 5 $ because $3\times 2 =6 \equiv 1 \bmod 5$.
From line 1, all calculated $\bmod 5$: 
$$\begin{align}
2\times 3t_1 &\equiv 2\times 1  \\
\implies 6t_1&\equiv 2 \\
\implies 5t_1 + t_1 &\equiv 2 \\
\implies t_1 &\equiv 2
\end{align}$$
A: You have to know the inverse of $3$ modulo $5$. The general way to have the inverse is from a Bézout's relation between the number and the modulus, from the extended Euclidean algorthm, but here it is obvious:
$$2\cdot 3 -5=1, \quad\text{whence}\quad3^{-1}=2\mod 5.$$
A: $3t_1 = 5k + 1 = 6k + (1-k) \implies 3 \mid (1-k) \implies 1-k = 3m \implies k = 1- 3m\implies t_1 = \dfrac{6k+(1-k)}{3}= 2k+ \dfrac{1-k}{3} = 2k+m = 2(1-3m) + m = -5m+2 = 2\pmod 5$ as claimed.
A: $$3t_1 \equiv 1 \pmod 5\iff3t_1=1+5m\iff 6t_1=5t_1+t_1=2+5(2m)$$ Hence $$t_1\equiv 2\pmod 5 $$
