Difference between revisions of "2002 AMC 10B Problems/Problem 12"
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== Solution == | == Solution == | ||
− | The domain over which we solve the equation is <math>\mathbb{R} | + | 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>. | ||
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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 | + | 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 4 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. |
+ | |||
+ | -Edited by XxHalo711 (typo within the solution) | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2002|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2002|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 16:59, 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.