To solve the inequality $\frac{x-2}{x-9} \leq 0$, we can first find the critical points where the expression is equal to zero or undefined.
The expression is undefined when the denominator is equal to zero, so we have $x - 9 = 0 \Rightarrow x = 9$. This is a critical point.
The expression is equal to zero when the numerator is equal to zero, so we have $x - 2 = 0 \Rightarrow x = 2$. This is another critical point.
Now, let's create a number line with the critical points at 2 and 9:
---o-----------o---
Using test points, we can determine the sign of the expression in each interval:
In the interval $(-\infty, 2)$, we can choose $x = 0$ as a test point. $\frac{0-2}{0-9} = \frac{-2}{-9} = \frac{2}{9} > 0$
In the interval $(2, 9)$, we can choose $x = 5$ as a test point. $\frac{5-2}{5-9} = \frac{3}{-4} = -\frac{3}{4} < 0$
In the interval $(9, \infty)$, we can choose $x = 10$ as a test point. $\frac{10-2}{10-9} = \frac{8}{1} = 8 > 0$
Therefore, the solution to the inequality $\frac{x-2}{x-9} \leq 0$ is $x \in (2, 9)$.
To solve the inequality $\frac{x-2}{x-9} \leq 0$, we can first find the critical points where the expression is equal to zero or undefined.
The expression is undefined when the denominator is equal to zero, so we have $x - 9 = 0 \Rightarrow x = 9$. This is a critical point.
The expression is equal to zero when the numerator is equal to zero, so we have $x - 2 = 0 \Rightarrow x = 2$. This is another critical point.
Now, let's create a number line with the critical points at 2 and 9:
---o-----------o---
Using test points, we can determine the sign of the expression in each interval:
In the interval $(-\infty, 2)$, we can choose $x = 0$ as a test point. $\frac{0-2}{0-9} = \frac{-2}{-9} = \frac{2}{9} > 0$
In the interval $(2, 9)$, we can choose $x = 5$ as a test point. $\frac{5-2}{5-9} = \frac{3}{-4} = -\frac{3}{4} < 0$
In the interval $(9, \infty)$, we can choose $x = 10$ as a test point. $\frac{10-2}{10-9} = \frac{8}{1} = 8 > 0$
Therefore, the solution to the inequality $\frac{x-2}{x-9} \leq 0$ is $x \in (2, 9)$.