If 3x + 2y = 11 & kx + 4y = 22 are coincide lines find value of k.

Options

A. 5
B. 6
C. 0
D. `-6`

Answer

To solve the problem, we need to find the value of \( k \) such that the lines represented by the equations \( 3x + 2y = 11 \) and \( kx + 4y = 22 \) are coincident.

### Step-by-Step Solution:

1. **Identify the coefficients of the equations:**
- For the first equation \( 3x + 2y - 11 = 0 \):
- \( a_1 = 3 \)
- \( b_1 = 2 \)
- \( c_1 = -11 \)

- For the second equation \( kx + 4y - 22 = 0 \):
- \( a_2 = k \)
- \( b_2 = 4 \)
- \( c_2 = -22 \)

2. **Set up the condition for coincident lines:**
The lines are coincident if the ratios of the coefficients are equal:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

3. **Write the ratios:**
\[ \frac{3}{k} = \frac{2}{4} = \frac{-11}{-22} \]

4. **Simplify the ratios:**
- The ratio \( \frac{2}{4} \) simplifies to \( \frac{1}{2} \).
- The ratio \( \frac{-11}{-22} \) simplifies to \( \frac{1}{2} \).

Thus, we have:
\[ \frac{3}{k} = \frac{1}{2} \]

5. **Cross-multiply to solve for \( k \):**
\[ 3 \cdot 2 = 1 \cdot k \]
\[ 6 = k \]

6. **Conclusion:**
The value of \( k \) is \( 6 \).

### Final Answer:
\[ k = 6 \]