Answer:
[tex]\frac{-2x+11}{(x-4)(x+1)}[/tex]
Step-by-step explanation:
I don't think we can factor this so we'll have to multiply to make the denominators the same
[tex]\frac{3(x+1)}{(x^2-3x-4)(x+1)}-\frac{2(x^2-3x-4)}{(x+1)(x^2-3x-4)}\\\\\frac{3x+3-(2x^2-6x-8)}{(x^2-3x-4)(x+1)}=\frac{-2x^2+9x+11}{(x^2-3x-4)(x+1)}\\-2x^2+9x+11=(x+1)(-2x+11)\\\\x^2-3x-4=(x+1)(x-4)\\\frac{(x+1)(-2x+11)}{(x+1)(x-4)(x+1)}=\frac{-2x+11}{(x-4)(x+1)}[/tex]
