The value of k, for which the system of equations `kx- 3y+6= 0, 4x- 6y+15=0` represent parallel lines, is _____

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the value of \( k \) for which the system of equations \( kx - 3y + 6 = 0 \) and \( 4x - 6y + 15 = 0 \) represent parallel lines, we will use the condition for parallel lines in the context of linear equations.

### Step-by-Step Solution:

1. **Identify the Coefficients**:
The given equations can be rewritten in the standard form \( ax + by + c = 0 \):
- For the first equation \( kx - 3y + 6 = 0 \):
- \( a_1 = k \)
- \( b_1 = -3 \)
- \( c_1 = 6 \)
- For the second equation \( 4x - 6y + 15 = 0 \):
- \( a_2 = 4 \)
- \( b_2 = -6 \)
- \( c_2 = 15 \)

2. **Apply the Condition for Parallel Lines**:
The condition for two lines to be parallel is given by:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]
Substituting the values we have:
\[ \frac{k}{4} = \frac{-3}{-6} \]

3. **Simplify the Right Side**:
Simplifying \( \frac{-3}{-6} \):
\[ \frac{-3}{-6} = \frac{3}{6} = \frac{1}{2} \]
Thus, we have:
\[ \frac{k}{4} = \frac{1}{2} \]

4. **Cross Multiply to Solve for \( k \)**:
Cross multiplying gives:
\[ k \cdot 2 = 1 \cdot 4 \]
This simplifies to:
\[ 2k = 4 \]

5. **Solve for \( k \)**:
Dividing both sides by 2:
\[ k = \frac{4}{2} = 2 \]

6. **Check the Condition for Non-Equality**:
Now we need to ensure that the lines are not coincident:
\[ \frac{c_1}{c_2} = \frac{6}{15} = \frac{2}{5} \]
Since \( \frac{1}{2} \neq \frac{2}{5} \), the lines are indeed parallel.

### Final Answer:
The value of \( k \) for which the system of equations represents parallel lines is \( \boxed{2} \).