2002 AMC 10B Problem 12

Cross multiplying gives $(x - 1)(x - 6) = (x - 2)(x - k),$ which expands to $$\begin{aligned} x^2 - 7x + 6 &= x^2 - (2 + k)x \\ &\quad {}+ 2k. \end{aligned}$$ Cancelling $x^2$ leaves $(k - 5)x = 2k - 6.$ For $k$ equal to $1,$ $2,$ $3,$ or $4,$ this gives a valid value of $x$ that is neither excluded denominator value $2$ nor $6.$ For $k=5,$ the equation instead becomes $0\cdot x=4,$ which has no solution. Thus, the correct answer is E.