The number of values of `k` for which the system of equations:
`kx+(3k+2)y=4k`
`(3k-1)x+(9k+1)y=4(k+1)` has no solution, are

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To determine the number of values of \( k \) for which the given system of equations has no solution, we will use the condition for inconsistency in a system of linear equations. The equations are:

1. \( kx + (3k + 2)y = 4k \)
2. \( (3k - 1)x + (9k + 1)y = 4(k + 1) \)

### Step 1: Identify coefficients
From the equations, we can identify the coefficients:
- For the first equation:
- \( a_1 = k \)
- \( b_1 = 3k + 2 \)
- \( c_1 = 4k \)

- For the second equation:
- \( a_2 = 3k - 1 \)
- \( b_2 = 9k + 1 \)
- \( c_2 = 4(k + 1) = 4k + 4 \)

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

### Step 3: Set up the first ratio
Calculating \( \frac{a_1}{a_2} \):
\[ \frac{a_1}{a_2} = \frac{k}{3k - 1} \]

Calculating \( \frac{b_1}{b_2} \):
\[ \frac{b_1}{b_2} = \frac{3k + 2}{9k + 1} \]

Setting these two ratios equal:
\[ \frac{k}{3k - 1} = \frac{3k + 2}{9k + 1} \]

### Step 4: Cross-multiply
Cross-multiplying gives:
\[ k(9k + 1) = (3k + 2)(3k - 1) \]

Expanding both sides:
\[ 9k^2 + k = (9k^2 - 3k + 6k - 2) = 9k^2 + 3k - 2 \]

### Step 5: Simplify the equation
Subtract \( 9k^2 \) from both sides:
\[ k = 3k - 2 \]

Rearranging gives:
\[ k - 3k = -2 \implies -2k = -2 \implies k = 1 \]

### Step 6: Check the second condition
Now we need to check the second condition:
\[ \frac{c_1}{c_2} = \frac{4k}{4k + 4} \]
We need to ensure:
\[ \frac{3k + 2}{9k + 1} \neq \frac{4k}{4k + 4} \]

Calculating \( \frac{c_1}{c_2} \):
\[ \frac{c_1}{c_2} = \frac{4k}{4(k + 1)} = \frac{k}{k + 1} \]

### Step 7: Set up the inequality
We need:
\[ \frac{3k + 2}{9k + 1} \neq \frac{k}{k + 1} \]

Cross-multiplying gives:
\[ (3k + 2)(k + 1) \neq (9k + 1)k \]

Expanding both sides:
\[ 3k^2 + 3k + 2k + 2 \neq 9k^2 + k \]
\[ 3k^2 + 5k + 2 \neq 9k^2 + k \]

Rearranging gives:
\[ 0 \neq 6k^2 - 4k - 2 \]

### Step 8: Factor the quadratic
Factoring \( 6k^2 - 4k - 2 \):
\[ 3k^2 - 2k - 1 = 0 \]

Using the quadratic formula:
\[ k = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ k = \frac{2 \pm \sqrt{4 + 12}}{6} = \frac{2 \pm 4}{6} \]
This gives:
\[ k = 1 \quad \text{or} \quad k = -\frac{1}{3} \]

### Conclusion
Thus, the values of \( k \) for which the system has no solution are \( k = 1 \) and \( k = -\frac{1}{3} \). Therefore, the number of values of \( k \) for which the system has no solution is:

**Answer: 2**