When I am faced with the following:
$15 \equiv x$ (mod 10)
Ignore the simplicity of this. What I am interested is is the following in any way connected to the solution for $x$:
$\frac{15}{5} \equiv \frac{x}{5}$ (mod 2)
The congruence was divided by $5$. Does solving this for $x$ in any way connect to the first congruence?
$x \equiv ca \pmod{\!cn}\,\Rightarrow\, c\mid x,\ $ so $\,\ c(x/c)\equiv ca \pmod{\!cn} \iff x/c \equiv a \pmod{\! n}$
because $\,cn\mid cX-ca\iff n\mid X-a,\,$ since $\ \dfrac{cX-ca}{cn} = \dfrac{X-a}n\,$
Remark $ $ For the general case (and its fractional viewpoint) see this answer.