Answer
$\displaystyle \frac{x(x-1)}{(x+5)(x-2)(x+3)}, \quad x\neq-5,2, -3$
Work Step by Step
Factor each denominator.
Factoring $x^{2}+bx+c$, we search for two factors of c (m and n) such that m+n=b. If they exist, $x^{2}+bx+c =(x+m)(x+n)$
$ x^{2}+3x-10=\quad$... we find factors $+5$ and $-2,$
$=(x+5)(x-2)$
$ x^{2}+x-6=\quad$... we find factors $+3$ and $-2,$
$=(x+3)(x-2)$
$LCD=(x+5)(x-2)(x+3)$
$\displaystyle \frac{3x}{x^{2}+3x-10}-\frac{2x}{x^{2}+x-6}=\frac{3x}{(x+5)(x-2)}-\frac{2x}{(x+3)(x-2)}$
$=\displaystyle \frac{3x}{(x+5)(x-2)}\times\frac{x+3}{x+3}-\frac{2x}{(x+3)(x-2)}\times\frac{x+5}{x+5}$
$=\displaystyle \frac{3x^{2}+9x}{(x+5)(x-2)(x+3)}-\frac{2x^{2}+10x}{(x+5)(x-2)(x+3)}$
$=\displaystyle \frac{3x^{2}+9x-2x^{2}-10x}{(x+5)(x-2)(x+3)}$
$=\displaystyle \frac{x^{2}-x}{(x+5)(x-2)(x+3)}$
$=\displaystyle \frac{x(x-1)}{(x+5)(x-2)(x+3)}, \quad x\neq-5,2, -3$
