Difference between revisions of "2000 AMC 12 Problems/Problem 20"
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<cmath>x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3</cmath> | <cmath>x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3</cmath> | ||
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+ | Then what is the value of <math>xyz</math> ? | ||
<math>\text {(A)}\ 2/3 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 4/3 \qquad \text {(D)}\ 2 \qquad \text {(E)}\ 7/3</math> | <math>\text {(A)}\ 2/3 \qquad \text {(B)}\ 1 \qquad \text {(C)}\ 4/3 \qquad \text {(D)}\ 2 \qquad \text {(E)}\ 7/3</math> | ||
__TOC__ | __TOC__ | ||
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === |
Revision as of 10:24, 30 March 2008
Problem
If and are positive numbers satisfying
Then what is the value of ?
Contents
[hide]Solution
Solution 1
Multiplying all three expressions together,
Thus
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 .
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 |