If `(|x-3|)/(x-3)ge0`, then

Options

A. `x in[3,oo)`
B. `x in(3,oo)`
C. `x in(-oo,3)`
D. `x in(-oo,3)`

Answer

To solve the inequality \(\frac{|x-3|}{x-3} \geq 0\), we can analyze it step by step.

### Step 1: Understand the absolute value
The expression \(|x-3|\) can be defined in two cases based on the value of \(x\):
- Case 1: When \(x < 3\), \(|x-3| = -(x-3) = 3-x\).
- Case 2: When \(x \geq 3\), \(|x-3| = x-3\).

### Step 2: Analyze Case 1 (when \(x < 3\))
In this case, we have:
\[ \frac{|x-3|}{x-3} = \frac{3-x}{x-3} \]
This simplifies to:
\[ \frac{3-x}{x-3} = \frac{-(x-3)}{x-3} = -1 \]
Now we check the inequality:
\[ -1 \geq 0 \]
This is false, so this case does not satisfy the inequality.

### Step 3: Analyze Case 2 (when \(x \geq 3\))
In this case, we have:
\[ \frac{|x-3|}{x-3} = \frac{x-3}{x-3} \]
This simplifies to:
\[ 1 \]
Now we check the inequality:
\[ 1 \geq 0 \]
This is true, so this case satisfies the inequality.

### Step 4: Combine the results
From the analysis:
- Case 1 (when \(x < 3\)) does not satisfy the inequality.
- Case 2 (when \(x \geq 3\)) satisfies the inequality.

Thus, the solution set is:
\[ x \geq 3 \]
In interval notation, this is:
\[ [3, \infty) \]

### Final Answer:
The solution to the inequality \(\frac{|x-3|}{x-3} \geq 0\) is:
\[ x \in [3, \infty) \]