Answer
$\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$
Work Step by Step
Step 1:
Factor each denominator
For $x^{2}+bx+c$, we search for factors of c whose sum is b:
$ x^{2}-7x+6 = \quad$ ... -6 and -1...
$=(x-6)(x-1)$
$ x^{2}-2x-24 = \quad$ ... $-6$ and $+4$...
$=(x-6)(x+4)$
Step 2: The LCM is the product of each of these factors raised to a power equal to the greatest number of times that the factor occurs in the polynomials.
LCM = $(x-6)(x-1)(x+4)$
Step 3:
Write each rational expression using the LCM as the denominator.
$\displaystyle \frac{x}{ (x-6)(x-1)}-\frac{x}{ (x-6)(x+4)}$=
$= \displaystyle \frac{x(x+4)}{ (x-6)(x-1)(x+4)}-\frac{x(x-1)}{ (x-6)(x+4)(x-1)}$=
$=\displaystyle \frac{x^{2}+4x-(x^{2}-x)}{ (x-6)(x-1)(x+4)}$
$=\displaystyle \frac{5x}{ (x-6)(x-1)(x+4)}$
