[tex]\bf \cfrac{5}{3x-12}+\cfrac{3x+1}{x^2-x-12}-\cfrac{2}{3}\implies \cfrac{5}{3(x-4)}+\cfrac{3x+1}{(x-4)(x+3)}-\cfrac{2}{3}[/tex]
now, let's take a peek at the denominators, so we have "3" and "x-4" repeated, so we'll use them only once in our LCD, so using those repeated factors once, our LCD boils down to 3(x-4)(x+3).
[tex]\bf \stackrel{\textit{using an LCD of }3(x-4)(x+3)}{\cfrac{(x+3)5~~~~+~~~~(3)(3x+1)~~~ ~-~~~~[(x-4)(x+3)]2}{3(x-4)(x+3)}} \\\\\\ \cfrac{5x+15~~+~~9x+3~~-~~[x^2-x-12]2}{3(x-4)(x+3)} \\\\\\ \cfrac{5x+15~~+~~9x+3~~-~~(2x^2-2x-24)}{3(x-4)(x+3)} \\\\\\ \cfrac{5x+15~~+~~9x+3~~-~~2x^2+2x+24}{3(x-4)(x+3)} \\\\\\ \cfrac{42+16x-2x^2}{3(x-4)(x+3)}\implies \cfrac{2(21+8x-x^2)}{3(x-4)(x+3)}[/tex]