For what value of k does the equation 5(x-2)+12 = kx+6 have no solution?

Distribute on the left: $5(x-2)+12 = 5x - 10 + 12 = 5x + 2$, so the equation is $5x + 2 = kx + 6$. A linear equation has no solution when both sides have the same $x$-coefficient but different constants. Setting $k = 5$ gives $5x + 2 = 5x + 6$, which simplifies to $2 = 6$, impossible. So $k = 5$, choice A. Choice B ($k=6$) matches the constant term, not the coefficient, and gives one solution. Choice C ($-5$) is a sign slip and gives $x = -0.4$. Choice D ($k=2$) mixes up the constant on the left and also gives one solution.