A system of linear equations won't have a solution if the two equations in the system contradict each other.  Multiplying the first equation by 2, we get the equations
\begin{align*}
12x + 8y &= 14, \\
Kx + 8y &= 7.
\end{align*}If $K = 12,$ then the two equations contradict each other.  Otherwise, we can subtract them to get $(K - 12) x = -7.$  We can solve this equation to find $x,$ then substitute to find $y.$

So the value of $K$ for which the system has no solution is $K = \boxed{12}.$
Suppose $K$ is a constant, such that no pair of real numbers $(x,y)$ satisfies this system of equations: $$6x + 4y = 7$$ $$Kx + 8y = 7$$

What is the value of $K$?
{'answer': '12', 'subject': 'Algebra', 'level': 4, 'unique_id': 'test/algebra/1884.json'}