Question

If \( \frac{|x-3|}{x-3} > 0 \), then:

• A. \( x \in (-3, \infty) \)
• B. \( x \in (3, \infty) \)
• C. \( x \in (2, \infty) \)
• D. \( x \in (1, \infty) \)
• E. \( x \in (-1, \infty) \)

Hint:

The expression \( \frac{|a|}{a} \) gives the sign of \( a \). It simplifies many inequality problems instantly.

Answer 1

The Correct Option is B

Concept: The value of \( |a| \) is always non-negative. For the expression \( \frac{|a|}{a} \), the sign depends on \( a \): \[ \frac{|a|}{a} = \begin{cases} +1, & a > 0 \\ -1, & a < 0 \end{cases} \]

Step 1: Analyze the given expression.
\[ \frac{|x-3|}{x-3} \] Let \( a = x - 3 \). Then: \[ \frac{|x-3|}{x-3} = \frac{|a|}{a} \]

Step 2: Apply sign condition.
For the expression to be greater than zero: \[ \frac{|a|}{a} > 0 \Rightarrow a > 0 \] So, \[ x - 3 > 0 \]

Step 3: Solve inequality.
\[ x > 3 \]

Step 4: Final answer in interval form.
\[ \boxed{(3, \infty)} \]