6 + 7 r = p w 7 r - 5 w = 5 w +

The correct answer is $10$. Solving by substitution, the given system of equations, where $p$ is a constant, can be written so that the left-hand side of each equation is equal to $7r$. Subtracting $6$ from each side of the first equation in the given system, $6+7r=pw$, yields $7r=pw-6$. Adding $5w$ to each side of the second equation in the given system, $7r-5w=5w+11$, yields $7r=10w+11$. Since the left-hand side of each equation is equal to $7r$, setting the the right-hand side of the equations equal to each other yields $pw-6=10w+11$. A linear equation in one variable, $w$, has no solution if and only if the equation is false; that is, when there's no value of $w$ that produces a true statement. For the equation $pw-6=10w+11$, there's no value of $w$ that produces a true statement when $pw=10w$. Therefore, for the equation $pw-6=10w+11$, there's no value of $w$ that produces a true statement when the value of $p$ is $10$. It follows that in the given system of equations, the system has no solution when the value of $p$ is $10$.