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>. | ||
| − | == | + | == Also see == |
{{AMC12 box|year=2000|num-b=19|num-a=21}} | {{AMC12 box|year=2000|num-b=19|num-a=21}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 13:17, 21 January 2019
Problem
If 
and 
are positive numbers satisfying
![\[x + 1/y = 4,\qquad y + 1/z = 1, \qquad \text{and} \qquad z + 1/x = 7/3\]](http://latex.artofproblemsolving.com/8/0/9/809268c34d341e924e9b6500bbdc5de487d5445d.png)
Then what is the value of 
?
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.
![\[4 = x + \frac{1}{y} = x + \frac{1}{1 - \frac{1}{z}} = x + \frac{1}{1-\frac{1}{7/3-1/x}} = x + \frac{7x-3}{4x-3}\]](http://latex.artofproblemsolving.com/9/f/0/9f0d43b0925f112e173739d093abd90c9a6ecf41.png)
Multiplying out the denominator and simplification yields 
, so 
. Substituting leads to 
, and the product of these three variables is 
.
Also see
| 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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
