general solution sum of two uniform random variables aY+bX=Z? is there a general solution to that? I have seen simple examples for Y+X=Z but I was wondering how this would be with rescaling? 
 A: Define $Y'=aY$ and $X'=bX$, find their distribution, and you are back to the problem you know how to solve: $Y'+X'=Z$ (convolution).
A: If we have a variable $X\sim U(0,1)$ and multiply it by $a$, then $aX\sim U(0,a)$.
Assume that we're dealing with independent continuous uniform on $(0,a)$ and $(0,b)$ respectively (with $a<b$)
(This assumption is not restrictive since we can obtain the general case from this easily.)
Then the joint density is $\frac{1}{ab} I_{(0,a)}\times I_{(0,b)}$.
Since the bivariate density is constant where it's non-zero, we can just draw it "looking from above" by marking the boundary of that non-zero region.

... and so by elementary geometric argument (along the lines of (i) recognize that density increases linearly as the sum, $z$ goes from $0$ to $a$, stays constant until $b$ and then decreases linearly to $a+b$, and (ii) that the height in the middle section must be $1/b$ to get unit area, then (iii) the equations of the three non-zero sections follow immediately by inspection), the density of the convolution is 
$f(z) = \begin{cases} 0 & z\leq 0\\
z/ab & 0<z<a \\
1/b & a\leq z<b \\
(a+b-z)/ab & b\leq z<a+b \\
0 & z\geq a+b
\end{cases}$
[While formal integration will obviously work, it's somewhat quicker - for me at least - to proceed by something like the above reasoning, where one simply draws the density and then writes the result down immediately.]

The general case:
Imagine instead we had independent $U(c,a+c)$ and $U(d,d+b)$. Then the above density would simply be shifted right by $c+d$.
