Find distribution of $Y=aX + Z$ given $X=x$ 
Suppose $X$ and $Z$ are random variables with densities $f_X$ and $f_Z$ respectively. Define $Y = aX + Z$. Find the conditional distribution $f_Y(y|x)$.

If we are given that $X=x$, then essentially we have $Y = ax + Z$. So $f_Y(y|x)$ should be the density of $ax + Z$. The density of $ax + Z$ is just $f_Z(y-ax)$. So
$$f_Y(y|x) = f_Z(y-ax)$$
But I am wondering is possible to do it in a perhaps more rigorous way? For example, I considered something like
$$\mathbb{P}(Y\le y | X=x) = \mathbb{P}(aX + Z \le y | X=x)$$
, but then don't know how to proceed, partly because I think $\mathbb{P}(X=x)$ is zero, and also how do I manipulate $\mathbb{P}( \{aX + Z \le y \}\cap \{X =x \})$?
 A: You are not told that $X$ and $Z$ are independent, but you are correct that $\{\omega:Z\leq y-aX, X=x\}=\{\omega:Z\leq y-ax, X=x\}$
So directly: $f_{Y\mid X}(y\mid x) ~{= f_{aX+Z\mid X}(y\mid x) \\= \begin{vmatrix}\dfrac{\partial~(y-ax)}{\partial~y\hspace{7ex}}\end{vmatrix}f_{Z\mid X}(y-ax\mid x)\\= f_{Z\mid X}(y-ax\mid x)}$

Using the CDF:  ${F_{Y\mid X=x}(y) ~{= \mathsf P(Y\leq y\mid X=x) \\ = \mathsf P(Z\leq y-aX\mid X=x)\\ = \mathsf P(Z\leq y-ax\mid X=x) }\\~\\ f_{Y\mid X}(y\mid x) ~{= \begin{vmatrix}\dfrac{\partial~ \mathsf P(Z\leq y-ax\mid X=x)}{\partial~y\hspace{22ex}}\end{vmatrix} \\ = \text{as above} \\ = f_{Z\mid X}(y-ax\mid x) } }$

And of course if $X$ and $Z$ actually are independent, then $f_{Y\mid X}(y\mid x) = f_Z(y-ax)$.
A: If they were to be independent, then you can use convolution to get the density of Y and then divide by density of X.  In other words,
$f_{Y/X}(y/X=x) = \dfrac{\int_{-\infty}^{\infty}f_{Z}(y-ax).f_{X}(x)dx}{\int_{-\infty}^{\infty}f_{X}(x)dx}$
Am I correct @Graham Kemp?
