Difference between revisions of "2002 AMC 10B Problems/Problem 12"
(→Solution) |
(→Solution) |
||
| Line 8: | Line 8: | ||
== Solution == | == Solution == | ||
| − | The domain over which we solve the equation is <math>\mathbb{R} \ \{2,6\}</math>. | + | The domain over which we solve the equation is <math>\mathbb{R} \setminus \{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>. | We can now cross-multiply to get rid of the fractions, we get <math>(x-1)(x-6)=(x-k)(x-2)</math>. | ||
Revision as of 15:59, 7 August 2020
Problem
For which of the following values of 
does the equation 
have no solution for 
?
Solution
The domain over which we solve the equation is 
.
We can now cross-multiply to get rid of the fractions, we get 
.
Simplifying that, we get 
. Clearly for 
we get the equation 
which is never true. The answer is 
For other , one can solve for : 
, hence 
. We can easily verify that for none of the other 4 possible values of is this equal to 
or 
, hence there is a solution for in each of the other cases.
-Edited by XxHalo711 (typo within the solution)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
