What is the value of k such that the following pair of equations have infinitely many solutions?
x - 2 y = 3
-3 x + k y = -9

Options

A. `(-6)`
B. `(-3)`
C. 3
D. 6

Answer

To find the value of \( k \) such that the equations \( x - 2y = 3 \) and \( -3x + ky = -9 \) have infinitely many solutions, we need to use the condition for infinitely many solutions for a pair of linear equations. This condition states that the ratios of the coefficients of \( x \), \( y \), and the constants must be equal.

### Step-by-Step Solution:

1. **Identify Coefficients**:
From the first equation \( x - 2y = 3 \), we can rewrite it in the standard form \( ax + by + c = 0 \):
\[ 1x - 2y - 3 = 0 \]
Here, \( a_1 = 1 \), \( b_1 = -2 \), and \( c_1 = -3 \).

From the second equation \( -3x + ky = -9 \), we can also rewrite it:
\[ -3x + ky + 9 = 0 \]
Here, \( a_2 = -3 \), \( b_2 = k \), and \( c_2 = 9 \).

2. **Set Up the Ratios**:
For the equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Substituting the values we found:
\[ \frac{1}{-3} = \frac{-2}{k} = \frac{-3}{9} \]

3. **Simplify the Ratios**:
We can simplify \( \frac{-3}{9} \):
\[ \frac{-3}{9} = -\frac{1}{3} \]
So we have:
\[ \frac{1}{-3} = -\frac{1}{3} \]
This confirms that the first ratio is equal to the third ratio.

4. **Set the Second Ratio Equal**:
Now we set the first ratio equal to the second ratio:
\[ \frac{1}{-3} = \frac{-2}{k} \]

5. **Cross-Multiply**:
Cross-multiplying gives:
\[ 1 \cdot k = -2 \cdot (-3) \]
Simplifying this:
\[ k = 6 \]

### Conclusion:
Thus, the value of \( k \) such that the given pair of equations has infinitely many solutions is \( \boxed{6} \).