Answer
$\frac{2x^2-9x}{(x-3)(x+4)(x-4)}$
Work Step by Step
Step 1. Factor each denominator, we have $x^2+x-12=(x-3)(x+4)$ and $x^2-16=(x+4)(x-4)$
Step 2. The least common denominator can be identified as $(x-3)(x+4)(x-4)$
Step 3. Convert each expression to have the least common denominator then perform operations:
$\frac{3x(x-4)}{(x-3)(x+4)(x-4)}-\frac{x(x-3)}{(x-3)(x+4)(x-4)}=\frac{2x^2-9x}{(x-3)(x+4)(x-4)}$ where $x\ne 3, \pm 4$
