Find the value(s) of p for the following pair of equations:
`2x+ 3y-5=0 and px- 6y-8=0`, if the pair of equations has a unique solution.

Answer

To find the value(s) of \( p \) for the given pair of equations such that they have a unique solution, we start with the equations:

1. \( 2x + 3y - 5 = 0 \)
2. \( px - 6y - 8 = 0 \)

### Step 1: Identify coefficients
We need to identify the coefficients from both equations in the standard form \( ax + by + c = 0 \).

From the first equation:
- \( a_1 = 2 \)
- \( b_1 = 3 \)
- \( c_1 = -5 \)

From the second equation:
- \( a_2 = p \)
- \( b_2 = -6 \)
- \( c_2 = -8 \)

### Step 2: Condition for unique solution
For the pair of equations to have a unique solution, the condition is:
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]

### Step 3: Substitute coefficients into the condition
Substituting the values we identified:
\[ \frac{2}{p} \neq \frac{3}{-6} \]
This simplifies to:
\[ \frac{2}{p} \neq -\frac{1}{2} \]

### Step 4: Cross-multiply to eliminate the fractions
Cross-multiplying gives:
\[ 2 \cdot (-2) \neq 1 \cdot p \]
This simplifies to:
\[ -4 \neq p \]

### Step 5: Conclusion
Thus, the value of \( p \) must not be equal to \(-4\) for the pair of equations to have a unique solution.

**Final Answer:**
The value of \( p \) must not be equal to \(-4\).

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