The number of values of k, for which the system of eauations:
`(k+1)x+8y=4k`
`kx+(k+3)y=3k-1`
has no solution is,

Options

A. infinite
B. 1
C. 2
D. 3

Answer

To find the number of values of \( k \) for which the system of equations has no solution, we need to analyze the given equations:

1. \( (k+1)x + 8y = 4k \)
2. \( kx + (k+3)y = 3k - 1 \)

### Step 1: Identify coefficients
From the equations, we can identify:
- For the first equation:
- Coefficient of \( x \) (denoted as \( a_1 \)) = \( k + 1 \)
- Coefficient of \( y \) (denoted as \( b_1 \)) = \( 8 \)
- Constant term (denoted as \( c_1 \)) = \( 4k \)

- For the second equation:
- Coefficient of \( x \) (denoted as \( a_2 \)) = \( k \)
- Coefficient of \( y \) (denoted as \( b_2 \)) = \( k + 3 \)
- Constant term (denoted as \( c_2 \)) = \( 3k - 1 \)

### Step 2: Apply the condition for no solution
The system of equations will have no solution if:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{but} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

This translates to:
\[ \frac{k + 1}{k} = \frac{8}{k + 3} \quad \text{and} \quad \frac{4k}{3k - 1} \neq \frac{k + 1}{k} \]

### Step 3: Solve the first equation
Cross-multiplying gives:
\[ (k + 1)(k + 3) = 8k \]
Expanding this:
\[ k^2 + 3k + k + 3 = 8k \]
\[ k^2 + 4k + 3 - 8k = 0 \]
\[ k^2 - 4k + 3 = 0 \]

### Step 4: Factor the quadratic equation
Factoring the equation:
\[ (k - 3)(k - 1) = 0 \]
Thus, we have:
\[ k = 3 \quad \text{or} \quad k = 1 \]

### Step 5: Solve the second condition
Next, we check the second condition:
\[ \frac{4k}{3k - 1} \neq \frac{k + 1}{k} \]
Cross-multiplying gives:
\[ 4k^2 \neq (k + 1)(3k - 1) \]
Expanding the right-hand side:
\[ 4k^2 \neq 3k^2 + k - 3k + 1 \]
This simplifies to:
\[ 4k^2 \neq 3k^2 - 2k + 1 \]
Rearranging gives:
\[ k^2 + 2k - 1 \neq 0 \]

### Step 6: Solve the quadratic inequality
To find the values of \( k \) that do not satisfy this equation, we can find the roots:
\[ k = \frac{-2 \pm \sqrt{(2)^2 + 4(1)(1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = -1 \pm \sqrt{2} \]

### Step 7: Determine the valid values of \( k \)
The roots are \( k = -1 + \sqrt{2} \) and \( k = -1 - \sqrt{2} \). We need to check the intervals around these roots to find where \( k^2 + 2k - 1 \neq 0 \).

### Conclusion
The values \( k = 1 \) and \( k = 3 \) do not equal \( -1 + \sqrt{2} \) or \( -1 - \sqrt{2} \). Therefore, the only values of \( k \) for which the system has no solution are \( k = 1 \) and \( k = 3 \).

Thus, the number of values of \( k \) for which the system of equations has no solution is **2**.