2002 AMC 10B Problem 12

The equation is defined for $x \ne 2, 6$. On that domain, cross-multiply:
$$(x-1)(x-6) = (x-k)(x-2).$$
Expanding, $x^2 - 7x + 6 = x^2 - (k+2)x + 2k$. The $x^2$ terms cancel, leaving a linear equation:
$$(k+2)x - 7x = 2k - 6 \quad\Longrightarrow\quad (k-5)\,x = 2k - 6.$$

If $k = 5$ the left side is $0$ while the right side is $4$, so there is no $x$ at all.

For the other choices the coefficient $k - 5$ is nonzero and $x = \frac{2k-6}{k-5}$ is a candidate: $k = 1, 2, 3, 4$ give $x = 1, \frac23, 0, -2$. None of these is $2$ or $6$, so each is a genuine solution and those equations are solvable.

The answer is $\boxed{\textbf{(E)}\ 5}$.

After clearing denominators, a rational equation of this shape becomes linear, and a linear equation $ax = b$ fails to have a solution only when $a = 0$ and $b \ne 0$. The second check, that the solution does not hit an excluded value, is what separates a careful solver from a lucky one; here it happens to be clean, but it can bite.

Rewrite each side: $\frac{x-1}{x-2} = 1 + \frac{1}{x-2}$ and, for $k = 5$, $\frac{x-5}{x-6} = 1 + \frac{1}{x-6}$. Equality would require $x - 2 = x - 6$, which is impossible.

• Forgetting that x = 2 and x = 6 are excluded values and not checking whether the solution lands on them for the other choices.
• A sign slip in $-7x = -(k+2)x + \ldots$ that makes the coefficient vanish at $k = -9$ or $k = 3$ instead of $k = 5$.