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

Options

A. `k = 3`
B. `k = 4`
C. `k = 5`
D. none of the above

Answer

To determine the value of \( k \) for which the equations \( 4x + ky = 4 \) 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. **Write the equations in standard form**:
The given equations are:
\[ 4x + ky - 4 = 0 \quad (1) \]
\[ 3x + 2y - 6 = 0 \quad (2) \]

2. **Identify coefficients**:
From equation (1), we have:
- \( a = 4 \)
- \( b = k \)
- \( c = -4 \)

From equation (2), we have:
- \( m = 3 \)
- \( n = 2 \)
- \( p = -6 \)

3. **Set up the ratios**:
For the equations to have infinitely many solutions, the following must hold:
\[ \frac{a}{m} = \frac{b}{n} = \frac{c}{p} \]
This translates to:
\[ \frac{4}{3} = \frac{k}{2} = \frac{-4}{-6} \]

4. **Simplify the ratios**:
We first simplify \( \frac{-4}{-6} \):
\[ \frac{-4}{-6} = \frac{2}{3} \]

Now we have:
\[ \frac{4}{3} \neq \frac{2}{3} \]
This shows that the first and last ratios are not equal.

5. **Set up the equation for \( k \)**:
Now we can set up the equation for \( k \):
\[ \frac{4}{3} = \frac{k}{2} \]

6. **Cross-multiply to solve for \( k \)**:
Cross-multiplying gives:
\[ 4 \cdot 2 = 3k \]
\[ 8 = 3k \]
\[ k = \frac{8}{3} \]

7. **Conclusion**:
The value of \( k \) for which the equations have infinitely many solutions is:
\[ k = \frac{8}{3} \]