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

(2002 AMC 10B Problem 12)
 
Line 1: Line 1:
12. For which of the following values of <math>k</math> does the equation <math>\frac{x-1}{x-2} = \frac{x-k}{x-6}</math> have no solution for <math>x</math>?
+
== Problem ==
 +
 
 +
For which of the following values of <math>k</math> does the equation <math>\frac{x-1}{x-2} = \frac{x-k}{x-6}</math> have no solution for <math>x</math>?
  
 
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math>
 
<math>\textbf{(A) } 1\qquad \textbf{(B) } 2\qquad \textbf{(C) } 3\qquad \textbf{(D) } 4\qquad \textbf{(E) } 5</math>
 +
 +
 +
== Solution ==
 +
 +
The domain over which we solve the equation is <math>\mathbb{R} - \{2,6\}</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=\boxed{5}</math> we get the equation <math>-6=-10</math> which is never true.
 +
 +
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.
 +
 +
== See Also ==
 +
{{AMC10 box|year=2002|ab=B|num-b=11|num-a=13}}

Revision as of 07:30, 2 February 2009

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 $k=\boxed{5}$ we get the equation $-6=-10$ which is never true.

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 (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS