usually if it comes to worse, we can always just use the product of all denominators as the LCD, it will just be a CD but not LCD but as good, but usually we try to use the LCD.
in this case our denominators do not have a GCF, so the LCD will just be their product, so we'll use that.
![\bf \cfrac{3x}{x-1}+\cfrac{4}{7x}\implies \stackrel{\textit{using an LCD of (x-1)(7x)}}{\cfrac{(7x)3x~~+~~(x-1)4}{(x-1)(7x)}}\implies \cfrac{21x^2~~+~~(4x-4)}{(x-1)(7x)} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{21x^2+4x-4}{(x-1)(7x)}~\hfill](https://img.qammunity.org/2020/formulas/mathematics/high-school/2whx99pmukte4s5ykt7dj50z3j4l6vsy21.png)