2002 AMC 10B Problems/Problem 12

Problem

For which of the following values of $k$ does the equation $\frac{x-1}{x-2} = \frac{x-k}{x-6}$ have no solution for $x$ ?

$\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5$

Solution

The domain over which we solve the equation is $\mathbb{R} - \{2,6\}$ .

We can now cross-multiply to get rid of the fractions, we get $(x-1)(x-6)=(x-k)(x-2)$ .

Simplifying that, we get $7x-6 = (k+2)x - 2k$ . Clearly for $k=5$ we get the equation $-6=-10$ which is never true. The answer is $\boxed{\mathrm{ (E)}\ 5}$

For other $k$ , one can solve for $x$ : $x(5-k) = 6-2k$ , hence $x=\frac {6-2k}{5-k}$ . We can easily verify that for none of the other four possible values of $k$ is this equal to $2$ or $6$ , hence there is a solution for $x$ in each of the other cases.

-Edited by XxHalo711 (typo within the solution)

2002 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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2002 AMC 10B Problems/Problem 12

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