Difference between revisions of "2000 AMC 12 Problems/Problem 11"
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− | <math>\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b}{a-b} =2 \Rightarrow \text{(E)}</math>. | + | <math>\frac {a}{b} + \frac {b}{a} - ab = \frac{a^2 + b^2}{ab} - (a - b) = \frac{a^2 + b^2}{a-b} - \frac{(a-b)^2}{(a-b)} = \frac{2ab}{a-b} = \frac{2(a-b)}{a-b} =2 \Rightarrow \text{(E)}</math>. |
Another way is to solve the equation for <math>b,</math> giving <math>b = \frac{a}{a+1};</math> then substituting this into the expression and simplifying gives the answer of <math>2.</math> | Another way is to solve the equation for <math>b,</math> giving <math>b = \frac{a}{a+1};</math> then substituting this into the expression and simplifying gives the answer of <math>2.</math> |
Revision as of 15:03, 7 May 2017
- The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.
Contents
Problem
Two non-zero real numbers, and
satisfy
. Which of the following is a possible value of
?
Solution
.
Another way is to solve the equation for giving
then substituting this into the expression and simplifying gives the answer of
Solution 2
This simplifies to . The two integer solutions to this are
and
. The problem states than
and
are non-zero, so we consider the case of
. So, we end up with
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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.