Let $R(x)=\frac{x-6}{x-8}$. Solve for $R(x) \geq 0$

Step 1 ：Find the critical points of the function \(R(x)\). The critical points are the values of \(x\) that make \(R(x) = 0\) or undefined. The function \(R(x)\) is equal to zero when the numerator is zero. So, we solve the equation \(x - 6 = 0\) to get \(x = 6\). The function \(R(x)\) is undefined when the denominator is zero. So, we solve the equation \(x - 8 = 0\) to get \(x = 8\). So, the critical points are \(x = 6\) and \(x = 8\).

Step 2 ：Test the intervals determined by the critical points. We choose a test point in each interval and evaluate the sign of \(R(x)\) at that point. For \(x < 6\), we can choose \(x = 0\). Then \(R(0) = \frac{0 - 6}{0 - 8} = \frac{6}{8} = 0.75 > 0\). For \(6 < x < 8\), we can choose \(x = 7\). Then \(R(7) = \frac{7 - 6}{7 - 8} = -1 < 0\). For \(x > 8\), we can choose \(x = 10\). Then \(R(10) = \frac{10 - 6}{10 - 8} = 2 > 0\).

Step 3 ：From the above, we see that \(R(x) \geq 0\) when \(x \leq 6\) or \(x \geq 8\). So, the solution to the inequality is \(x \in (-\infty, 6] \cup [8, \infty)\).

Step 4 ：Check our solution by substituting a few values from each interval into the inequality. For example, if we choose \(x = 5\) and \(x = 9\), we find that \(R(5) = 1 > 0\) and \(R(9) = 3 > 0\), which confirms that our solution is correct.

Step 5 ：\(\boxed{x \in (-\infty, 6] \cup [8, \infty)}\)