Operations on congruence equations?

Question

I have to do back substitution for my homework, and I have to modify x ≡ 1 (mod 5) to x=5t+1, which I understand. What I don't understand is when I put this into the next equation which becomes 5t + 1 ≡ 2 (mod 6), how this transforms to t ≡ 5 (mod 6). My book just says this as a fact, and doesn't say at all how to solve for t in 5t + 1 ≡ 2 (mod 6). How is this answer arrived upon?

Answer 1

$5t+1\equiv 2\pmod 6\implies 5t\equiv 1\pmod 6\implies 25t=t\equiv 5\pmod 6$, as $5^2\equiv 1\pmod 6$.

We multiply by the inverse of 5 on either side in the final step, which happens to be 5.

Answer 2

maybe worthwhile to remark that because $(5,6)=1$ and $5 \lt 6$ the ring isomorphism $\mathbb{Z}_5 \oplus \mathbb{Z}_6 \cong \mathbb{Z}_{30}$ can be given an explicit form: $$ (a,b) \Rightarrow a +5b $$ this gives the multiplication: $$ (a,b) \star (c,d) = (ac,ad+bc-bd) $$ (where components are computed modulo $5$ and $6$ respectively)

in the opposite direction: $$ c \Rightarrow \left( c- 5 \lfloor \frac{c}{5} \rfloor, \lfloor \frac{c}{5} \rfloor \right) $$