Why is $5t +1\equiv 2 \pmod 6$ equivalent to $t \equiv 5 \pmod 6$?

Question

I feel like I am missing something really obvious here but I can't explain why these two are equivalent. I understand that $5t \equiv 1 \pmod 6$ but I can't make the next step.

Answer 1

If you were solving the equation $5t = 1$ you'd multiply both sides by $5^{-1} = \frac{1}{5}$ which is the number which satisfies $5^{-1}.5 = 5.5^{-1} = 1$. Then $t$ would be isolated on the left hand side.

When solving such an equation modulo $m$, the approach is the same. The interpretation of $5^{-1}$ is the same (it undoes multiplication by $5$) but what $5^{-1}$ is equal to is different. In this case, $5^{-1}$ is the unique element of $\{0, 1, \dots, m-1\}$ such that $5.5^{-1} = 5^{-1}.5 \equiv 1 \pmod m$ - such an element exists if and only if $5$ and $m$ have no common factors (i.e. they are coprime).

To solve the equation $5t \equiv 1 \pmod 6$, you need to find what $5^{-1}$ is, keeping in mind that $5^{-1} \in \{0, 1, \dots, 5\}$.

Answer 2

Starting with $$5t \equiv 1 \pmod 6,$$ if we add $5+t$ both sides, we obtain $$6t+5 \equiv 6+t \pmod 6,$$ or equivalently $$t \equiv 5 \pmod 6.$$