If the system of pair of linear equations `kx + 4y = 2, 7x + 3y = 1` has a unique solution, then the value of k is :

Options

A. k = 6
B. `k ne 6`
C. `k ne (3)/(2)`
D. `k ne (2)/(3)`

Answer

To determine the value of \( k \) for which the system of equations \( kx + 4y = 2 \) and \( 7x + 3y = 1 \) has a unique solution, we need to analyze the conditions for a unique solution in a system of linear equations.

### Step 1: Identify the coefficients
The given equations can be rewritten in the standard form:
1. \( kx + 4y = 2 \) (Equation 1)
2. \( 7x + 3y = 1 \) (Equation 2)

From these equations, we can identify the coefficients:
- For Equation 1: \( a_1 = k \), \( b_1 = 4 \), \( c_1 = 2 \)
- For Equation 2: \( a_2 = 7 \), \( b_2 = 3 \), \( c_2 = 1 \)

### Step 2: Use the condition for a unique solution
A system of linear equations has a unique solution if the ratio of the coefficients of \( x \) and \( y \) are not equal:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Substituting the values of \( a_1, a_2, b_1, \) and \( b_2 \):
\[ \frac{k}{7} \neq \frac{4}{3} \]

### Step 3: Cross-multiply to eliminate the fractions
Cross-multiplying gives us:
\[ 3k \neq 28 \]

### Step 4: Solve for \( k \)
To find the value of \( k \) that we must avoid, we can solve the equation:
\[ 3k = 28 \]
Dividing both sides by 3:
\[ k = \frac{28}{3} \]

### Step 5: Conclusion
For the system of equations to have a unique solution, \( k \) must not equal \( \frac{28}{3} \). Therefore, the value of \( k \) can be any real number except \( \frac{28}{3} \).

### Final Answer
The value of \( k \) is:
\[ k \in \mathbb{R} \setminus \left\{ \frac{28}{3} \right\} \]