The problem shown is:

$\frac{1}{2} |x - 3| + 4 < 6$

Step-by-step solution:

1. Isolate the absolute value term: $\frac{1}{2} |x - 3| + 4 < 6$ Subtract 4 from both sides: $\frac{1}{2} |x - 3| < 2$

2. Eliminate the fraction: Multiply both sides of the inequality by 2 to get rid of the fraction: $|x - 3| < 4$

3. Solve the absolute value inequality: The inequality $|x - 3| < 4$ can be split into two cases: $-4 < x - 3 < 4$

4. Solve the two inequalities: Add 3 to all parts of the inequality: $-4 + 3 < x < 4 + 3$ $-1 < x < 7$

Final solution:

The solution is $-1 < x < 7$.

Graphing on the number line:

On the number line, this inequality corresponds to an open interval between -1 and 7, meaning the values of $x$ are between -1 and 7 but do not include -1 or 7.

Would you like further details or have any other questions?

Here are five related questions for further exploration:

1. How would the solution change if the inequality was $\frac{1}{2} |x - 3| + 4 \geq 6$?
2. What is the process for solving a similar inequality with a greater absolute value, such as $|x + 5|$?
3. How do you graph inequalities that include "or equal to" conditions on a number line?
4. How would the solution be affected if the absolute value term was multiplied by a different fraction?
5. What happens if you have multiple absolute value terms in an inequality?

Tip: Always isolate the absolute value expression first when solving absolute value inequalities to make the process more straightforward.