For what value of ` p`, system of equations `2x + py = 8` and `x + y = 6` have no solution.

Answer

To determine the value of \( p \) for which the system of equations

\[ 2x + py = 8 \]
and
\[ x + y = 6 \]

has no solution, we need to use the condition for two linear equations to be inconsistent (i.e., have no solution). This condition is given by:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2} \]

where \( A_1, B_1, C_1 \) are the coefficients from the first equation and \( A_2, B_2, C_2 \) are the coefficients from the second equation.

### Step 1: Rewrite the equations in standard form
The first equation can be rewritten as:

\[ 2x + py - 8 = 0 \]

The second equation can be rewritten as:

\[ x + y - 6 = 0 \]

### Step 2: Identify coefficients
From the equations, we can identify the coefficients:

- For the first equation \( 2x + py - 8 = 0 \):
- \( A_1 = 2 \)
- \( B_1 = p \)
- \( C_1 = -8 \)

- For the second equation \( x + y - 6 = 0 \):
- \( A_2 = 1 \)
- \( B_2 = 1 \)
- \( C_2 = -6 \)

### Step 3: Set up the ratio condition
Now we set up the ratio condition:

\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2} \]

Substituting the values we identified:

\[ \frac{2}{1} = \frac{p}{1} \neq \frac{-8}{-6} \]

### Step 4: Simplify the ratios
This simplifies to:

\[ 2 = p \neq \frac{8}{6} \]

The fraction \( \frac{8}{6} \) simplifies to \( \frac{4}{3} \).

### Step 5: Conclusion
Thus, we have:

\[ p = 2 \quad \text{and} \quad 2 \neq \frac{4}{3} \]

This means that the value of \( p \) for which the system of equations has no solution is:

\[ \boxed{2} \]