Answer
$\frac{34}{(3x-4)(2x+3)}$
Work Step by Step
By inspection, we see that the LCM of the denominators is $(2x-1)(3x+1)$. Therefore, we create a common denominator using this LCM and obtain:
$\frac{6}{3x-4}-\frac{4}{2x+3}$
=$\frac{6(2x+3)-4(3x-4)}{(3x-4)(2x+3)}$
=$\frac{12x+18-12x+16}{(3x-4)(2x+3)}$
=$\frac{0x+34}{(3x-4)(2x+3)}$
=$\frac{34}{(3x-4)(2x+3)}$
