If the equations kx - 2y = 3 and 3x + y = 5 represent two intersecting lines at unique point, then the value of k is ……………. .

Answer

To find the value of \( k \) such that the lines represented by the equations \( kx - 2y = 3 \) and \( 3x + y = 5 \) intersect at a unique point, we will use the condition for two lines to be intersecting.

### Step-by-Step Solution:

1. **Write the equations in standard form**:
The equations are already in the standard form:
\[ kx - 2y - 3 = 0 \quad \text{(Equation 1)} \]
\[ 3x + y - 5 = 0 \quad \text{(Equation 2)} \]

2. **Identify coefficients**:
From the equations, we can identify the coefficients:
- For Equation 1:
- \( a_1 = k \)
- \( b_1 = -2 \)
- \( c_1 = -3 \)
- For Equation 2:
- \( a_2 = 3 \)
- \( b_2 = 1 \)
- \( c_2 = -5 \)

3. **Set up the condition for intersection**:
The lines will intersect at a unique point if the following condition holds:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
Substituting the values:
\[ \frac{k}{3} \neq \frac{-2}{1} \quad \text{and} \quad \frac{-2}{1} \neq \frac{-3}{-5} \]

4. **Calculate the ratios**:
- For \( \frac{-2}{1} \):
\[ -2 \text{ (which is } \frac{-2}{1} \text{)} \]
- For \( \frac{-3}{-5} \):
\[ \frac{-3}{-5} = \frac{3}{5} \]

5. **Set up inequalities**:
Now, we need to ensure:
\[ \frac{k}{3} \neq -2 \quad \text{and} \quad -2 \neq \frac{3}{5} \]

6. **Solve for \( k \)**:
From \( \frac{k}{3} \neq -2 \):
\[ k \neq -2 \times 3 = -6 \]

7. **Conclusion**:
Therefore, the value of \( k \) must not be equal to \( -6 \) for the lines to intersect at a unique point.

### Final Answer:
The value of \( k \) is \( k \neq -6 \).