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. `oo`
B. 1
C. 2
D. 3

Answer

To determine the number of values of \( k \) for which the system of equations has no solution, we start with the given equations:

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

### Step 1: Identify the coefficients
From the equations, we can identify the coefficients:
- For the first equation:
- \( a_1 = k + 1 \)
- \( b_1 = 8 \)
- \( c_1 = 4k \)

- For the second equation:
- \( a_2 = k \)
- \( b_2 = k + 3 \)
- \( c_2 = 3k - 1 \)

### Step 2: Apply the condition for no solution
For the system of equations to have no solution, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

### Step 3: Set up the first ratio
We calculate the first ratio:
\[ \frac{a_1}{a_2} = \frac{k+1}{k} \]
\[ \frac{b_1}{b_2} = \frac{8}{k+3} \]

Setting these two ratios equal gives us:
\[ \frac{k+1}{k} = \frac{8}{k+3} \]

### Step 4: Cross-multiply and simplify
Cross-multiplying, we get:
\[ (k + 1)(k + 3) = 8k \]
Expanding the left side:
\[ k^2 + 3k + k + 3 = 8k \]
\[ k^2 + 4k + 3 = 8k \]
Rearranging gives:
\[ k^2 - 4k + 3 = 0 \]

### Step 5: Factor the quadratic equation
Now we factor the quadratic:
\[ (k - 3)(k - 1) = 0 \]
Thus, the solutions are:
\[ k = 1 \quad \text{or} \quad k = 3 \]

### Step 6: Check the second condition
Next, we need to check the second condition:
\[ \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

Calculating \( c_1 \) and \( c_2 \):
- \( c_1 = 4k \)
- \( c_2 = 3k - 1 \)

Thus:
\[ \frac{c_1}{c_2} = \frac{4k}{3k - 1} \]

### Step 7: Check for \( k = 1 \)
Substituting \( k = 1 \):
\[ \frac{c_1}{c_2} = \frac{4(1)}{3(1) - 1} = \frac{4}{2} = 2 \]
Now, substituting into \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{1 + 1}{1} = 2 \]
Since \( \frac{c_1}{c_2} = \frac{a_1}{a_2} \), \( k = 1 \) does not satisfy the condition for no solution.

### Step 8: Check for \( k = 3 \)
Substituting \( k = 3 \):
\[ \frac{c_1}{c_2} = \frac{4(3)}{3(3) - 1} = \frac{12}{8} = \frac{3}{2} \]
Now, substituting into \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{3 + 1}{3} = \frac{4}{3} \]
Since \( \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \), \( k = 3 \) satisfies the condition for no solution.

### Conclusion
The only value of \( k \) for which the system of equations has no solution is \( k = 3 \).

Thus, the number of values of \( k \) for which the system has no solution is **1**.