I have an additional question if someone would please help. If so, please explain how this. I simply do not understand how the numerator changes in this case.

$\displaystyle

\frac{1(x-1)+1(x+2)}{(x+2)(x+3)(x-1)}=\frac{2x+1}{(x+2)(x+3)(x-1)}

$

ExpressionSOLVED

$\displaystyle \frac{3}{2x^2+4x-48}+\frac{7}{6x-24}+\frac{1}{4x^2+20x-24}$

=

$\displaystyle \frac{3}{2(x-4)(x+6)}+\frac{7}{6(x-4)}-\frac{1}{4(x+6)(x-1)}$

The author then lists the correct LCD as $\displaystyle 12(x-4)(x+6)(x-1)$. I agree with everything here except the coefficient of 12 should it not be

LCD =$\displaystyle 2 * 6 * 4 (x-4)(x+6)(x-1)$ , where here I took the coeffecient factored out from each denominators expression and multiplied them by each other giving us a LCD of $\displaystyle 48(x-4)(x+6)(x-1)$