# Difference between revisions of "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)

## See Also

 2002 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 11 Followed byProblem 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 All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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