How one can deduce that $tx≡2t[mod(z)]$?

Question

Let $x,y,z,t$ four positive integers. If $$x≡2[mod(y)]$$ and $$z=ty$$ Then how one can deduce that $$tx≡2t[mod(z)]$$

Answer 1

$x \equiv 2 [mod(y)]$ is equivalent to say that there exists an integer $n$ such that $x = 2 + ny$. Multiplying this last equation by $t$, you get $tx = 2t + n(ty) = 2t + nz$ or, equivalently, $tx \equiv 2t[mod(z)]$.

Answer 2

$$ x=2 (\mod y) $$ means that there exists a $k\in \mathbb Z$ such that $x=2+k*y$.

That means we can write $tx$ as $2t+k*t*y$, and then it should be obvious.

Answer 3

$$\eqalign{ x\equiv2\pmod y &\Rightarrow x-2=ky\quad\hbox{for some $k\in\Bbb Z$}\cr &\Rightarrow tx-2t=kty\quad\hbox{for some $k\in\Bbb Z$}\cr &\Rightarrow tx-2t=kz\quad\hbox{for some $k\in\Bbb Z$}\cr &\Rightarrow tx\equiv 2t\pmod z\ .\cr}$$