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We are given the equation:

\[
|2x - 11| = \frac{1}{2}x + k
\]

and the condition that it has exactly two distinct solutions. Our goal is to find the range of possible values of \( k \).

### Step-by-step Analysis:

For the absolute value equation \( |2x - 11| = \frac{1}{2}x + k \) to have exactly two distinct solutions, the equation must be structured in a way that guarantees two solutions.

#### 1. Break the equation into two cases:
The absolute value function can be rewritten as two separate equations, depending on whether the expression inside the absolute value is positive or negative.

- **Case 1:** \( 2x - 11 = \frac{1}{2}x + k \)
- **Case 2:** \( -(2x - 11) = \frac{1}{2}x + k \)

#### 2. Solve each case separately:

##### Case 1: \( 2x - 11 = \frac{1}{2}x + k \)

Multiply both sides by 2 to eliminate the fraction:

\[
4x - 22 = x + 2k
\]

Now, rearrange to get:

\[
3x = 2k + 22 \quad \Rightarrow \quad x = \frac{2k + 22}{3}
\]

##### Case 2: \( -(2x - 11) = \frac{1}{2}x + k \)

Simplify the left-hand side:

\[
-2x + 11 = \frac{1}{2}x + k
\]

Multiply both sides by 2:

\[
-4x + 22 = x + 2k
\]

Now, rearrange to get:

\[
-5x = 2k - 22 \quad \Rightarrow \quad x = \frac{22 - 2k}{5}
\]

#### 3. Analyze conditions for two distinct solutions:

For the system to have **exactly two distinct solutions**, the two solutions found above must be different from each other. Thus, we require that:

\[
\frac{2k + 22}{3} \neq \frac{22 - 2k}{5}
\]

#### 4. Solve for the inequality:

Cross-multiply to eliminate the fractions:

\[
5(2k + 22) \neq 3(22 - 2k)
\]

Expand both sides:

\[
10k + 110 \neq 66 - 6k
\]

Now, simplify and solve for \( k \):

\[
10k + 6k \neq 66 - 110 \quad \Rightarrow \quad 16k \neq -44
\]

Finally, divide by 16:

\[
k \neq -\frac{44}{16} \quad \Rightarrow \quad k \neq -\frac{11}{4}
\]

#### 5. Conclusion:

The equation will have exactly two distinct solutions for all values of \( k \), except \( k = -\frac{11}{4} \).

### Range of \( k \):
The range of possible values for \( k \) is all real numbers except \( k \neq -\frac{11}{4} \).

---

Would you like any further details or clarifications?

Here are some related questions for practice:

1. How would the solution change if there were no distinct solutions?
2. How does the method change if the equation had more than two solutions?
3. What is the geometric interpretation of absolute value equations with respect to number of solutions?
4. How can we graph the two cases of this equation to visualize the solutions?
5. What happens if we change the coefficient of \( x \) in the absolute value term?

**Tip:** Always check for extraneous solutions when working with absolute value equations, as the squaring process can sometimes introduce invalid solutions.

The equation |2x - 11| = 1/2x + k has exactly two distinct solutions. Find the range of possible values of k.

This solution explains how to solve the absolute value equation |2x - 11| = 1/2x + k to find the range of possible values for k that result in exactly two distinct solutions. Explore step-by-step calculations and insights into absolute value and linear equation concepts. Suitable for Grades 10-12.

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