Find the value(s) of p for the following pair of equations:
`3x- y-5=0 and 6x- 2y- p= 0`, if the lines represented by these equations are parallel.

Answer

To find the value(s) of \( p \) for which the lines represented by the equations \( 3x - y - 5 = 0 \) and \( 6x - 2y - p = 0 \) are parallel, we can follow these steps:

### Step 1: Identify the coefficients of the equations
From the first equation \( 3x - y - 5 = 0 \):
- Coefficient of \( x \) (denoted as \( a_1 \)) = 3
- Coefficient of \( y \) (denoted as \( b_1 \)) = -1
- Constant term (denoted as \( c_1 \)) = -5

From the second equation \( 6x - 2y - p = 0 \):
- Coefficient of \( x \) (denoted as \( a_2 \)) = 6
- Coefficient of \( y \) (denoted as \( b_2 \)) = -2
- Constant term (denoted as \( c_2 \)) = -p

### Step 2: Use the condition for parallel lines
For two lines to be parallel, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \quad \text{and} \quad \frac{a_1}{a_2} \neq \frac{c_1}{c_2} \]

### Step 3: Calculate the ratios
Calculate \( \frac{a_1}{a_2} \) and \( \frac{b_1}{b_2} \):
\[ \frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2} \]
\[ \frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2} \]

Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), we now need to ensure that:
\[ \frac{c_1}{c_2} \neq \frac{1}{2} \]

### Step 4: Set up the inequality
Substituting the values of \( c_1 \) and \( c_2 \):
\[ \frac{-5}{-p} \neq \frac{1}{2} \]
This simplifies to:
\[ \frac{5}{p} \neq \frac{1}{2} \]

### Step 5: Cross-multiply to find \( p \)
Cross-multiplying gives:
\[ 5 \cdot 2 \neq 1 \cdot p \]
\[ 10 \neq p \]

### Conclusion
Thus, the value of \( p \) must not be equal to 10 for the lines to be parallel.

### Final Answer
The lines represented by the equations are parallel if:
\[ p \neq 10 \]