The number of values of `k` for which the lines `(k+1)x+8y=4ka n dk x+(k+3)y=3k-1` are coincident is __________

Explanation

To find the number of values of \( k \) for which the lines \( (k+1)x + 8y = 4k \) and \( kx + (k+3)y = 3k - 1 \) are coincident, we will use the condition for two lines to be coincident. The lines are coincident if the ratios of the coefficients of \( x \), \( y \), and the constant terms are equal.

### Step 1: Identify the coefficients
From the first line \( (k+1)x + 8y = 4k \):
- Coefficient of \( x \) (let's call it \( a_1 \)) = \( k + 1 \)
- Coefficient of \( y \) (let's call it \( b_1 \)) = \( 8 \)
- Constant term (let's call it \( c_1 \)) = \( 4k \)

From the second line \( kx + (k+3)y = 3k - 1 \):
- Coefficient of \( x \) (let's call it \( a_2 \)) = \( k \)
- Coefficient of \( y \) (let's call it \( b_2 \)) = \( k + 3 \)
- Constant term (let's call it \( c_2 \)) = \( 3k - 1 \)

### Step 2: Set up the equations based on the condition for coincident lines
The lines are coincident if:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

This gives us the following equations:
1. \( \frac{k + 1}{k} = \frac{8}{k + 3} \)
2. \( \frac{8}{k + 3} = \frac{4k}{3k - 1} \)

### Step 3: Solve the first equation
From the first equation:
\[ \frac{k + 1}{k} = \frac{8}{k + 3} \]
Cross-multiplying gives:
\[ (k + 1)(k + 3) = 8k \]
Expanding:
\[ k^2 + 4k + 3 = 8k \]
Rearranging:
\[ k^2 - 4k + 3 = 0 \]
Factoring:
\[ (k - 3)(k - 1) = 0 \]
Thus, \( k = 3 \) or \( k = 1 \).

### Step 4: Solve the second equation
From the second equation:
\[ \frac{8}{k + 3} = \frac{4k}{3k - 1} \]
Cross-multiplying gives:
\[ 8(3k - 1) = 4k(k + 3) \]
Expanding:
\[ 24k - 8 = 4k^2 + 12k \]
Rearranging:
\[ 4k^2 - 12k + 8 = 0 \]
Dividing by 4:
\[ k^2 - 3k + 2 = 0 \]
Factoring:
\[ (k - 2)(k - 1) = 0 \]
Thus, \( k = 2 \) or \( k = 1 \).

### Step 5: Identify common roots
The common root from both equations is \( k = 1 \).

### Conclusion
The number of values of \( k \) for which the lines are coincident is \( 1 \).

### Final Answer
The number of values of \( k \) is \( \boxed{1} \).

---