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. infinite
B. 1
C. 2
D. 3

Explanation

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

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

### Step 1: Identify the slopes of the equations
For the system of equations to have no solution, the lines represented by these equations must be parallel. This occurs when their slopes are equal but their intercepts are different.

The slope of a line in the form \( Ax + By = C \) is given by \( -\frac{A}{B} \).

**For Equation 1:**
- \( A = k \)
- \( B = 3k + 2 \)

The slope \( m_1 \) is:
\[ m_1 = -\frac{k}{3k + 2} \]

**For Equation 2:**
- \( A = 3k - 1 \)
- \( B = 9k + 1 \)

The slope \( m_2 \) is:
\[ m_2 = -\frac{3k - 1}{9k + 1} \]

### Step 2: Set the slopes equal to each other
For the lines to be parallel, we set \( m_1 = m_2 \):
\[ -\frac{k}{3k + 2} = -\frac{3k - 1}{9k + 1} \]

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

### Step 4: Expand both sides
Expanding both sides:
\[ 9k^2 + k = (3k)(3k) + (3k)(2) - (1)(3k) - (1)(2) \]
\[ 9k^2 + k = 9k^2 + 6k - 3k - 2 \]
\[ 9k^2 + k = 9k^2 + 3k - 2 \]

### Step 5: Simplify the equation
Subtract \( 9k^2 \) from both sides:
\[ k = 3k - 2 \]
Rearranging gives:
\[ k - 3k = -2 \]
\[ -2k = -2 \]
\[ k = 1 \]

### Step 6: Conclusion
The value \( k = 1 \) is the only value for which the lines are parallel. Since we are asked for the number of values of \( k \) for which the system has no solution, the answer is:

\[ \text{The number of values of } k \text{ is } 1. \]