Question

1. $\frac{6}{2x - 8}+\frac{x}{x^{2}-16}$
2. $\frac{x + 5}{x^{2}-4}-\frac{x + 1}{x - 2}$
3. $\frac{-8x - 6}{x^{2}-x - 6}-\frac{x + 3}{3 - x}$

Explanation:

Step1: Simplify denominators

For $\frac{6}{2x - 8}+\frac{x}{x^{2}-16}$, factor $2x - 8=2(x - 4)$ and $x^{2}-16=(x + 4)(x - 4)$. The common - denominator is $2(x + 4)(x - 4)$.
$\frac{6}{2(x - 4)}+\frac{x}{(x + 4)(x - 4)}=\frac{6(x + 4)}{2(x + 4)(x - 4)}+\frac{2x}{2(x + 4)(x - 4)}=\frac{6x+24 + 2x}{2(x + 4)(x - 4)}=\frac{8x + 24}{2(x + 4)(x - 4)}=\frac{4(x + 3)}{(x + 4)(x - 4)}$

Step2: Simplify another rational - expression

For $\frac{x + 5}{x^{2}-4}-\frac{x + 1}{x - 2}$, factor $x^{2}-4=(x + 2)(x - 2)$. The common - denominator is $(x + 2)(x - 2)$.
$\frac{x + 5}{(x + 2)(x - 2)}-\frac{(x + 1)(x + 2)}{(x + 2)(x - 2)}=\frac{x + 5-(x^{2}+3x + 2)}{(x + 2)(x - 2)}=\frac{x + 5 - x^{2}-3x - 2}{(x + 2)(x - 2)}=\frac{-x^{2}-2x + 3}{(x + 2)(x - 2)}$

Step3: Simplify the third rational - expression

For $\frac{-8x - 6}{x^{2}-x - 6}-\frac{x + 3}{3 - x}$, factor $x^{2}-x - 6=(x - 3)(x+2)$ and rewrite $\frac{x + 3}{3 - x}=-\frac{x + 3}{x - 3}$. The common - denominator is $(x - 3)(x + 2)$.
$\frac{-8x - 6}{(x - 3)(x + 2)}+\frac{x + 3}{x - 3}=\frac{-8x - 6+(x + 3)(x + 2)}{(x - 3)(x + 2)}=\frac{-8x - 6+x^{2}+5x + 6}{(x - 3)(x + 2)}=\frac{x^{2}-3x}{(x - 3)(x + 2)}=\frac{x(x - 3)}{(x - 3)(x + 2)}=\frac{x}{x + 2}$

Answer:

The simplifications are correct as shown in the step - by - step process. The results $\frac{4(x + 3)}{(x + 4)(x - 4)}$, $\frac{-x^{2}-2x + 3}{(x + 2)(x - 2)}$, and $\frac{x}{x + 2}$ are accurate for the given rational expressions.