For what value of k, the two lines 2x - 4y = 6 and 3x - 6y = k are parallel forever?

Options

A. `k = -9` only
B. `k = 9`
C. All real values except 9
D. none of these

Answer

To determine the value of \( k \) for which the lines \( 2x - 4y = 6 \) and \( 3x - 6y = k \) are parallel, we can follow these steps:

### Step 1: Write the equations in standard form
The given equations are:
1. \( 2x - 4y = 6 \)
2. \( 3x - 6y = k \)

We can rearrange both equations into the standard form \( Ax + By + C = 0 \).

For the first equation:
\[ 2x - 4y - 6 = 0 \quad \text{(let's denote this as Equation 1)} \]

For the second equation:
\[ 3x - 6y - k = 0 \quad \text{(let's denote this as Equation 2)} \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:
- For Equation 1: \( a_1 = 2 \), \( b_1 = -4 \), \( c_1 = -6 \)
- For Equation 2: \( a_2 = 3 \), \( b_2 = -6 \), \( c_2 = -k \)

### Step 3: Use the condition for parallel lines
Two lines are parallel if the ratios of their coefficients are equal, but the ratio of the constant terms is not equal. This can be expressed as:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{c_1}{c_2} \neq \frac{a_1}{a_2} \]

### Step 4: Calculate the ratios
Calculating the ratios:
\[ \frac{a_1}{a_2} = \frac{2}{3} \]
\[ \frac{b_1}{b_2} = \frac{-4}{-6} = \frac{2}{3} \]

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), we can proceed to check the constant term ratio.

### Step 5: Set up the inequality for the constant terms
Now we need to ensure that:
\[ \frac{c_1}{c_2} \neq \frac{2}{3} \]
Substituting the values:
\[ \frac{-6}{-k} \neq \frac{2}{3} \]
This simplifies to:
\[ \frac{6}{k} \neq \frac{2}{3} \]

### Step 6: Solve the inequality
Cross-multiplying gives:
\[ 6 \cdot 3 \neq 2 \cdot k \]
\[ 18 \neq 2k \]
Dividing both sides by 2:
\[ 9 \neq k \]

### Conclusion
Thus, the lines \( 2x - 4y = 6 \) and \( 3x - 6y = k \) are parallel for all values of \( k \) except \( k = 9 \).

### Final Answer
The value of \( k \) for which the lines are parallel forever is:
\[ k \in \mathbb{R} \text{ except } k = 9 \]