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Difference between revisions of "2002 AMC 10B Problems/Problem 12"

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We can now cross-multiply to get rid of the fractions, we get <math>(x-1)(x-6)=(x-k)(x-2)</math>.
 
We can now cross-multiply to get rid of the fractions, we get <math>(x-1)(x-6)=(x-k)(x-2)</math>.
  
−
Simplifying that, we get <math>7x-6 = (k+2)x - 2k</math>. Clearly for <math>k=5</math> we get the equation <math>-6=-10</math> which is never true. The answer is <math>\boxed{\mathrm{ (E)}\ 5}</math>
+
Simplifying that, we get <math>7x-6 = (k+2)x + 2k</math>. Clearly for <math>k=5</math> we get the equation <math>-6=10</math> which is never true. The answer is <math>\boxed{\mathrm{ (E)}\ 5}</math>
  
 
For other <math>k</math>, one can solve for <math>x</math>: <math>x(5-k) = 6-2k</math>, hence <math>x=\frac {6-2k}{5-k}</math>. We can easily verify that for none of the other four possible values of <math>k</math> is this equal to <math>2</math> or <math>6</math>, hence there is a solution for <math>x</math> in each of the other cases.
 
For other <math>k</math>, one can solve for <math>x</math>: <math>x(5-k) = 6-2k</math>, hence <math>x=\frac {6-2k}{5-k}</math>. We can easily verify that for none of the other four possible values of <math>k</math> is this equal to <math>2</math> or <math>6</math>, hence there is a solution for <math>x</math> in each of the other cases.

Revision as of 21:03, 23 September 2015

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.

See Also

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
All AMC 10 Problems and Solutions

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