Find the value of k for which the system of equations `kx-4y =3, 7x -2y =5` has a unique solution.

Answer

To find the value of \( k \) for which the system of equations \( kx - 4y = 3 \) and \( 7x - 2y = 5 \) has a unique solution, we will use the condition for a unique solution in a system of linear equations.

### Step-by-Step Solution:

1. **Identify the coefficients**:
- From the first equation \( kx - 4y = 3 \):
- Coefficient of \( x \) (denoted as \( a_1 \)) = \( k \)
- Coefficient of \( y \) (denoted as \( b_1 \)) = \( -4 \)
- From the second equation \( 7x - 2y = 5 \):
- Coefficient of \( x \) (denoted as \( a_2 \)) = \( 7 \)
- Coefficient of \( y \) (denoted as \( b_2 \)) = \( -2 \)

2. **Apply the condition for a unique solution**:
- The condition for the system of equations to have a unique solution is:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
- Substituting the values:
\[ \frac{k}{7} \neq \frac{-4}{-2} \]

3. **Simplify the right side**:
- The right side simplifies to:
\[ \frac{-4}{-2} = 2 \]
- Therefore, the condition becomes:
\[ \frac{k}{7} \neq 2 \]

4. **Cross-multiply to solve for \( k \)**:
- Cross-multiplying gives:
\[ k \neq 2 \times 7 \]
- This simplifies to:
\[ k \neq 14 \]

5. **Conclusion**:
- The value of \( k \) for which the system of equations has a unique solution is:
\[ k \text{ must not equal } 14 \]