In $\mathbb Z/p\mathbb Z$, is $(a+b\cdot r)$ a random value for fixed $a,b$ and random $r$? Let $p$ be a prime number.
Consider two fixed values $a,b\in\mathbb Z/p\mathbb Z$, where $b\neq0$, and a uniformly random value $r\leftarrow \mathbb Z/p\mathbb Z$.

Is $v=a+b\cdot r$ a uniformly random value in $\mathbb Z/p\mathbb Z$?
Yes. For fixed $a\in\mathbb Z/n\mathbb Z$ and $b\in(\mathbb Z/n\mathbb Z)^\ast$ — note that $b$ must be invertible modulo $n$, which need not necessarily be a prime (but if it is, invertibility is equivalent to $b\neq0$), the map
$$ f\colon\;\mathbb Z/n\mathbb Z\to\mathbb Z/n\mathbb Z,\; r \mapsto a+br $$
is a bijection, hence it preserves uniform distribution. That is: If $R\colon\;\Omega\to\mathbb Z/n\mathbb Z$ is a uniformly distributed random variable, then $f(R)\colon\Omega\to\mathbb Z/n\mathbb Z$ is as well.

The direct argument is: If $\forall y\in\mathbb Z/n\mathbb Z.\;\Pr[R=y]=1/n$, then for any $x\in\mathbb Z/n\mathbb Z$,
$$ \Pr[f(R)=x] = \Pr[a+bR=x] = \Pr[R=b^{-1}(x-a)] = 1/n \text, $$
therefore $f(R)$ is uniformly distributed on $\mathbb Z/n\mathbb Z$.