c. $$ \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} $$

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Solve the equation $$ \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} $$.
Solve the equation by following steps:
- step0: Solve for $$x$$:
  $$\frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2}$$
- step1: Find the domain:
  $$\frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2},x \in \left(-\infty,-2ight)\cup \left(-2,2ight)\cup \left(2,+\inftyight)$$
- step2: Rewrite the fraction:
  $$\frac{x+6}{x^{2}-4}-\frac{2}{x-2}=-\frac{1}{x+2}$$
- step3: Multiply both sides of the equation by LCD:
  $$\left(\frac{x+6}{x^{2}-4}-\frac{2}{x-2}ight)\left(x+2ight)\left(x-2ight)=-\frac{1}{x+2}	imes \left(x+2ight)\left(x-2ight)$$
- step4: Simplify the equation:
  $$-x+2=-x+2$$
- step5: The statement is true:
  $$x \in \mathbb{R}$$
- step6: Check if the solution is in the defined range:
  $$x \in \mathbb{R},x \in \left(-\infty,-2ight)\cup \left(-2,2ight)\cup \left(2,+\inftyight)$$
- step7: Find the intersection:
  $$x \in \left(-\infty,-2ight)\cup \left(-2,2ight)\cup \left(2,+\inftyight)$$
The solution to the equation $$ \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} $$ is $$ x \in (-\infty,-2) \cup (-2,2) \cup (2,+\infty) $$.
The solution to the equation is all real numbers except $$ x = -2 $$ and $$ x = 2 $$.

c. \( \frac{x+6}{x^{2}-4}-\frac{2}{x-2}=\frac{-1}{x+2} \)

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