<Question>
For how many values of $x$ is the expression $\frac{x-5}{x^2-9}$  not defined?
</Question>

<think>
<Plan_1>
Okay, so I need to figure out for how many values of x the expression (x - 5)/(x² - 9) is not defined. Let me start by recalling when a rational expression like this isn't defined. I think it's when the denominator is zero because division by zero is undefined. So, the main task here is to find the values of x that make the denominator zero and then check if those values also make the numerator zero. If they do, then those points might be removable discontinuities, but the expression would still be considered undefined at those points because the denominator is zero.

First, let's look at the denominator: x² - 9. I need to find when this equals zero.
So, solving x² - 9 = 0. That factors into (x - 3)(x + 3) = 0, right? Because it's a difference of squares. So, setting each factor equal to zero gives x - 3 = 0 => x = 3 and x + 3 = 0 => x = -3. Therefore, the denominator is zero when x is 3 or -3. That means the expression is not defined at x = 3 and x = -3.

Now, I should check the numerator at these points to see if there's a common factor. The numerator is (x - 5).

<prompt_1.1> Check the numerator at x = 3 and x = -3 to see if there are any common factors with the denominator. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let's plug in x = 3 into the numerator: 3 - 5 = -2, which is not zero. Similarly, x = -3 in the numerator gives -3 - 5 = -8, which is also not zero. So, the numerator doesn't cancel out the denominator at either of these points. That means both x = 3 and x = -3 are vertical asymptotes or points where the function is not defined.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Therefore, there are two values where the expression is not defined.

Wait, but let me make sure I didn't miss anything. The question is asking for the number of values, not their specific values.

<prompt_2.1> Confirm the number of values where the expression is not defined, considering the possibility of common factors. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
So, even though the expression could be simplified if there were common factors, in this case, there aren't any. The numerator is (x - 5) and the denominator factors into (x - 3)(x + 3). Since (x - 5) doesn't share any factors with the denominator, there's no simplification possible. Therefore, both x = 3 and x = -3 must be excluded from the domain of the expression.

So, the expression is not defined at two points: 3 and -3.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Hence, the number of values is 2.

Wait a second, but let me double-check.

<prompt_3.1> Double-check the solution by re-solving the problem to ensure no mistakes were made. </prompt_3.1>

Is there any other value where the expression could be undefined? The numerator is x - 5, which is defined for all real numbers, so the only potential issue is with the denominator.

<prompt_3.2> Consider if there are any other values where the expression could be undefined, given the numerator and denominator. </prompt_3.2>

Alternatively, maybe I can think about the domain of the function. The function is a rational function, and its domain is all real numbers except those that make the denominator zero. Since denominator is zero at x = 3 and x = -3, those two are excluded from the domain. So, the expression is not defined at those two points, hence two values.

Alternatively, maybe someone might think that x =5 is a point where the expression is undefined, but plugging x=5 into the denominator: 5² -9 =25 -9=16, which is not zero. So, denominator is 16, numerator is 0.

<prompt_3.3> Check if x = 5 is a point where the expression is undefined and confirm the final answer. </prompt_3.3>
</Plan_3>

<Execution_3>
<execution_3.1>
The denominator is x² - 9. Setting that equal to zero gives x² = 9, so x = ±3. Yep, that's two solutions. And the numerator is x - 5, which doesn't factor into those. So, both x = 3 and x = -3 make the denominator zero, and since they don't make the numerator zero, those are the points of discontinuity. Therefore, the expression is not defined at two values of x.
</execution_3.1>

<execution_3.2>
Therefore, the only values where the expression is not defined are the roots of the denominator, which are 3 and -3. Therefore, the answer should be 2.

But let me just verify once more. If I substitute x = 3 into the denominator, 3² - 9 = 9 - 9 = 0. Numerator is 3 - 5 = -2. So, the expression becomes -2/0, which is undefined. Similarly, x = -3: denominator is (-3)^2 -9 = 9 -9 =0, numerator is -3 -5 = -8, so -8/0, undefined. So yes, both points make the expression undefined.
</execution_3.2>

<execution_3.3>
So, the expression would be 0/16=0, which is defined. So x=5 is fine. So, x=5 is not an issue.

Therefore, the answer should be 2. There are two values where the expression is not defined: 3 and -3.
</execution_3.3>
</Execution_3>

<Final_answer>
Therefore, there are $\boxed{2}$ values of $x$ for which the expression $\frac{x-5}{x^2-9}$ is not defined.
</Final_answer>
</think>