The value of p, for which the equations `6x+py-5=0` and `3x+2y-8=0` have unique solution is :

Options

A. `p=4`
B. `pne4`
C. `p=-4`
D. `p ne-4`.

Answer

To find the value of \( p \) for which the equations \( 6x + py - 5 = 0 \) and \( 3x + 2y - 8 = 0 \) have a unique solution, we need to use the condition for a unique solution of a pair of linear equations.

### Step-by-Step Solution:

1. **Identify the coefficients from the equations:**
- The first equation is \( 6x + py - 5 = 0 \).
- Here, \( a_1 = 6 \), \( b_1 = p \), and \( c_1 = -5 \).
- The second equation is \( 3x + 2y - 8 = 0 \).
- Here, \( a_2 = 3 \), \( b_2 = 2 \), and \( c_2 = -8 \).

2. **Use the condition for a unique solution:**
- The condition for the equations to have a unique solution is given by:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
- Substituting the values we identified:
\[ \frac{6}{3} \neq \frac{p}{2} \]

3. **Simplify the left side:**
- Simplifying \( \frac{6}{3} \):
\[ 2 \neq \frac{p}{2} \]

4. **Cross-multiply to eliminate the fraction:**
- Cross-multiplying gives us:
\[ 2 \cdot 2 \neq p \]
- This simplifies to:
\[ 4 \neq p \]

5. **Conclusion:**
- Therefore, the value of \( p \) for which the equations have a unique solution is:
\[ p \neq 4 \]