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The given inequality is:

\[
\left| \frac{1}{x} - 3 \right| > 6
\]

To solve this, we will follow the steps for solving an absolute value inequality.

### Step 1: Break the absolute value inequality into two cases.
The general rule for absolute value inequalities \( |A| > B \) is:

\[
A > B \quad \text{or} \quad A < -B
\]

Here, \( A = \frac{1}{x} - 3 \) and \( B = 6 \).

So, the two cases are:

#### Case 1: 
\[
\frac{1}{x} - 3 > 6
\]

#### Case 2: 
\[
\frac{1}{x} - 3 < -6
\]

### Step 2: Solve each case.

#### Case 1: \( \frac{1}{x} - 3 > 6 \)

Add 3 to both sides:

\[
\frac{1}{x} > 9
\]

Now, take the reciprocal of both sides (note: taking the reciprocal reverses the inequality sign):

\[
x < \frac{1}{9}
\]

#### Case 2: \( \frac{1}{x} - 3 < -6 \)

Add 3 to both sides:

\[
\frac{1}{x} < -3
\]

Now, take the reciprocal of both sides (again, reverse the inequality sign):

\[
x > -\frac{1}{3}
\]

### Step 3: Combine the solutions.

The combined solution is:

\[
x < \frac{1}{9} \quad \text{or} \quad x > -\frac{1}{3}
\]

### Step 4: Exclude zero.
Since \( x = 0 \) makes the expression undefined (division by zero), we exclude \( x = 0 \).

Thus, the solution in interval notation is:

\[
(-\infty, -\frac{1}{3}) \cup (0, \frac{1}{9})
\]

Would you like more details or explanations on any part of this?

Here are some related questions for practice:

1. How would the solution change if the inequality were \( \left| \frac{1}{x} - 3 \right| \leq 6 \)?
2. What happens if you multiply or divide by a negative number in inequalities?
3. How do you handle inequalities when there are two absolute value expressions?
4. Can you graph the solution of \( (-\infty, -\frac{1}{3}) \cup (0, \frac{1}{9}) \)?
5. How do you solve rational inequalities like \( \frac{1}{x} + 4 < 5 \)?

**Tip:** When working with absolute values, always remember to split the equation or inequality into two cases and solve each one separately.

Learn to solve the absolute value inequality |1/x - 3| > 6 with a detailed, step-by-step breakdown. This solution covers algebraic concepts such as absolute value inequalities and rational inequalities. Suitable for students in Grades 9-11.

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