2|x-4|+3 = kx + 5 find the value of k if there is 2 solutions and no solutions

Let's solve the equation \(2|x - 4| + 3 = kx + 5\) for two cases: 

1. When there are exactly **two solutions**.
2. When there are **no solutions**.

---

### Step 1: Analyze the equation structure

We need to solve the equation \(2|x - 4| + 3 = kx + 5\). 

Recall that the **absolute value equation** \( |x - 4| \) will split into two cases:  
- **Case 1**: \( x - 4 \geq 0 \implies |x - 4| = x - 4 \)
- **Case 2**: \( x - 4 < 0 \implies |x - 4| = -(x - 4) = -x + 4 \)

Thus, the original equation splits into **two linear equations**:

1. **If \(x \geq 4\)**:  
   \(2(x - 4) + 3 = kx + 5\)

2. **If \(x < 4\)**:  
   \(2(-x + 4) + 3 = kx + 5\)

---

### Step 2: Solve the two cases

#### Case 1: \(x \geq 4\)
The equation becomes:  
\[
2(x - 4) + 3 = kx + 5
\]
Simplify:  
\[
2x - 8 + 3 = kx + 5 \implies 2x - 5 = kx + 5
\]
Rearrange:  
\[
2x - kx = 10 \implies (2 - k)x = 10
\]
Thus:  
\[
x = \frac{10}{2 - k} \quad \text{(Solution 1, valid for \(x \geq 4\))}
\]

---

#### Case 2: \(x < 4\)
The equation becomes:  
\[
2(-x + 4) + 3 = kx + 5
\]
Simplify:  
\[
-2x + 8 + 3 = kx + 5 \implies -2x + 11 = kx + 5
\]
Rearrange:  
\[
-2x - kx = -6 \implies -(2 + k)x = -6 \implies (2 + k)x = 6
\]
Thus:  
\[
x = \frac{6}{2 + k} \quad \text{(Solution 2, valid for \(x < 4\))}
\]

---

### Step 3: Conditions for Two Solutions or No Solutions

#### **Two Solutions Case**
For the equation to have two solutions, both solutions need to exist **in their valid domains**:

1. \(x = \frac{10}{2 - k} \geq 4\)
2. \(x = \frac{6}{2 + k} < 4\)

We need to solve both inequalities to find the valid range of \(k\).

1. \( \frac{10}{2 - k} \geq 4 \)  
   \[
   10 \geq 4(2 - k) \implies 10 \geq 8 - 4k \implies 4k \geq -2 \implies k \geq -\frac{1}{2}
   \]

2. \( \frac{6}{2 + k} < 4 \)  
   \[
   6 < 4(2 + k) \implies 6 < 8 + 4k \implies 4k > -2 \implies k > -\frac{1}{2}
   \]

Thus, for two solutions to exist, \(k > -\frac{1}{2}\).

---

#### **No Solutions Case**
For no solutions to exist, the two cases must yield contradictions. One way this can happen is if the solutions do not lie in their respective domains.

This will happen if:
1. \(x = \frac{10}{2 - k} < 4\)
2. \(x = \frac{6}{2 + k} \geq 4\)

We can solve these inequalities to find the condition for \(k\).

1. \( \frac{10}{2 - k} < 4 \)  
   \[
   10 < 4(2 - k) \implies 10 < 8 - 4k \implies 4k < -2 \implies k < -\frac{1}{2}
   \]

2. \( \frac{6}{2 + k} \geq 4 \)  
   \[
   6 \geq 4(2 + k) \implies 6 \geq 8 + 4k \implies 4k \leq -2 \implies k \leq -\frac{1}{2}
   \]

Thus, for no solutions, \(k \leq -\frac{1}{2}\).

---

### Step 4: Final Answer

- **For two solutions**: \(k > -\frac{1}{2}\).
- **For no solutions**: \(k \leq -\frac{1}{2}\).

---

Do you have any questions or want more details? Here are some related questions:  

1. How do we solve inequalities involving absolute value expressions?  
2. What happens if \(k = -\frac{1}{2}\) exactly—how many solutions will there be?  
3. How does changing the slope \(k\) affect the number of solutions of the equation?  
4. Can you explain how absolute values split equations into two cases?  
5. How can we graph the solutions to better visualize the answer?

**Tip:** Always carefully manage the domain restrictions when dealing with absolute value equations.

2|x - 4| + 3 = kx + 5 find the value of k if there are 2 solutions and no solutions

Solve the equation 2|x - 4| + 3 = kx + 5 for values of k that result in two or no solutions. This problem explores absolute value properties, linear equations, and inequalities, and is suitable for high school students in Grades 9-12.

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