For what value(s) of` k, 2x – ky = 4 and 3x + 2y = 6 `has infinitely many solutions?

Options

A. `k = (4)/(3)`
B. `k =- (4)/(3)`
C. `k = (3)/(4)`
D. `k = - (3)/(4)`

Answer

To determine the value(s) of \( k \) for which the equations \( 2x - ky = 4 \) and \( 3x + 2y = 6 \) have infinitely many solutions, we can use the condition for two linear equations to have infinitely many solutions. This condition states that the ratios of the coefficients of \( x \), \( y \), and the constant terms must be equal.

### Step-by-Step Solution:

1. **Rewrite the equations in standard form**:
- The first equation: \( 2x - ky - 4 = 0 \) can be rewritten as \( 2x - ky + (-4) = 0 \).
- The second equation: \( 3x + 2y - 6 = 0 \) can be rewritten as \( 3x + 2y + (-6) = 0 \).

2. **Identify coefficients**:
- From the first equation \( 2x - ky - 4 = 0 \), we have:
- \( a = 2 \)
- \( b = -k \)
- \( c = -4 \)
- From the second equation \( 3x + 2y - 6 = 0 \), we have:
- \( l = 3 \)
- \( m = 2 \)
- \( n = -6 \)

3. **Set up the ratio condition for infinitely many solutions**:
- According to the condition for infinitely many solutions:
\[ \frac{a}{l} = \frac{b}{m} = \frac{c}{n} \]
- Substituting the values we identified:
\[ \frac{2}{3} = \frac{-k}{2} = \frac{-4}{-6} \]

4. **Simplify the ratios**:
- The third ratio simplifies to:
\[ \frac{-4}{-6} = \frac{2}{3} \]
- Thus, we have:
\[ \frac{2}{3} = \frac{-k}{2} \]

5. **Cross-multiply to solve for \( k \)**:
- Cross-multiplying gives:
\[ 2 \cdot 2 = 3 \cdot (-k) \]
- This simplifies to:
\[ 4 = -3k \]

6. **Solve for \( k \)**:
- Dividing both sides by -3:
\[ k = -\frac{4}{3} \]

### Final Answer:
The value of \( k \) for which the equations have infinitely many solutions is:
\[ k = -\frac{4}{3} \]