The system of equations given is:

$1) \ 6 + 7r = pw$ $2) \ 7r - 5w = 5w + 11$

To solve for $p$, we must determine the condition under which the system has no solution. For a system of two linear equations to have no solution, the lines represented by the equations must be parallel. In terms of algebra, this happens when the two equations are inconsistent — their slopes (or ratios of coefficients) are the same, but their constants differ.

Step 1: Simplify the second equation

Rearrange the second equation to get: $7r - 5w = 5w + 11$ Subtract $5w$ from both sides: $7r - 10w = 11$ Thus, the second equation becomes: $3) \ 7r - 10w = 11$

Step 2: Rewrite the first equation in a similar form

The first equation is: $6 + 7r = pw$ Subtract 6 from both sides: $7r = pw - 6$ Rearrange this: $4) \ 7r - pw = -6$

Step 3: Set up the condition for no solution

For the system to have no solution, the coefficients of $r$ and $w$ in both equations must be proportional, but the constants must differ.

From the two equations: $7r - 10w = 11$ $7r - pw = -6$

We compare the coefficients of $r$ and $w$. The ratio of the coefficients of $w$ must be equal: $\frac{-10}{-p} = \frac{11}{-6}$ Cross-multiplying: $-10 \times (-6) = -p \times 11$ $60 = -11p$ Thus, $p = -\frac{60}{11}$.

Final Answer:

$p = -\frac{60}{11}$

Do you want more details or have any questions?

Here are five related questions:

1. What is the condition for two linear equations to have infinitely many solutions?
2. How can you determine if two lines are perpendicular from their equations?
3. What is the general solution method for a system of linear equations with two variables?
4. How would the solution change if the system had exactly one solution?
5. Can a system of linear equations have more than two solutions?

Tip: When solving systems of linear equations, always check the relationships between the coefficients before concluding no solution, one solution, or infinitely many solutions.