Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>. | Multiplying out the denominator and simplification yields <math>4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0</math>, so <math>x = \frac{3}{2}</math>. Substituting leads to <math>y = \frac{2}{5}, z = \frac{5}{3}</math>, and the product of these three variables is <math>1</math>. | ||
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+ | == Video Solution by OmegaLearn == | ||
+ | https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s | ||
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+ | ~ pi_is_3.14 | ||
== Video Solution == | == Video Solution == |
Latest revision as of 06:34, 4 November 2022
Contents
[hide]Problem
If and are positive numbers satisfying
Then what is the value of ?
Solution 1
We multiply all given expressions to get: Adding all the given expressions gives that We subtract from to get that . Hence, by inspection, . ~AopsUser101
Solution 2
We have a system of three equations and three variables, so we can apply repeated substitution.
Multiplying out the denominator and simplification yields , so . Substituting leads to , and the product of these three variables is .
Video Solution by OmegaLearn
https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s
~ pi_is_3.14
Video Solution
See Also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.