Answer
$\frac{11x^{2}-x-11}{(2x-1)(x+3)(3x+2)}; x \ne \frac{1}{2}, -3, -\frac{2}{3};$
Work Step by Step
$\frac{4x-1}{2x^{2}+5x-3} - \frac{x+3}{6x^{2}+x-2}$
Factors of $2x^{2}+5x-3$ are $(2x-1)(x+3)$ and
factors of $ 6x^{2}+x-2$ are $(2x-1)(3x+2)$
$=\frac{4x-1}{(2x-1)(x+3)} - \frac{x+3}{(2x-1)(3x+2)}$
$=\frac{(4x-1)(3x+2)-(x+3)(x+3)}{(2x-1)(x+3)(3x+2)}; x \ne \frac{1}{2}, -3, -\frac{2}{3};$
$=\frac{12x^{2}-3x+8x-2-[x^{2}+6x+9]}{(2x-1)(x+3)(3x+2)}; x \ne \frac{1}{2}, -3, -\frac{2}{3};$
$=\frac{12x^{2}-3x+8x-2-x^{2}-6x-9}{(2x-1)(x+3)(3x+2)}; x \ne \frac{1}{2}, -3, -\frac{2}{3};$
Combine like terms.
$=\frac{11x^{2}-x-11}{(2x-1)(x+3)(3x+2)}; x \ne \frac{1}{2}, -3, -\frac{2}{3};$
