For what value of k does the equation 6(x-1) + 9 = kx + 7 have no solution?

Simplify the left side: $6(x - 1) + 9 = 6x - 6 + 9 = 6x + 3$, so the equation is $6x + 3 = kx + 7$. It has no solution when the $x$-coefficients match but the constants differ. With $k = 6$, the equation becomes $6x + 3 = 6x + 7$, or $3 = 7$, which is false for every $x$. So $k = 6$, choice D. Choice A (7) is the constant on the right side. Choice B (9) is the constant added on the left before distributing. Choice C ($-6$) has the wrong sign and gives the solution $x = \frac{1}{3}$.