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2000 AMC 12 Problems/Problem 20: Difference between revisions

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If <math>x,y,</math> and <math>z</math> are positive numbers satisfying
If <math>x,y,</math> and <math>z</math> are positive numbers satisfying


<cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = 7/3</cmath>
<cmath>x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}</cmath>


Then what is the value of <math>xyz</math> ?
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)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}</math>


__TOC__
== Solution 1 ==
 
== Solution ==
=== Solution 1 ===
We multiply all given expressions to get:
We multiply all given expressions to get:
<cmath>(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</cmath>
<cmath>(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}</cmath>
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~AopsUser101
~AopsUser101


=== Solution 2 ===
== Solution 2 ==
We have a system of three equations and three variables, so we can apply repeated substitution.  
We have a system of three equations and three variables, so we can apply repeated substitution.  


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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 ==
== Video Solution by Power Solve==
https://www.youtube.com/watch?v=ZiJ29GzhyUY
==Video Solution(Fast, quick, and learn more about Algebraic Manipulations!)==
https://youtu.be/rP8_-36n1ps
~MK
 
== Video Solution by OmegaLearn ==
https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s
 
~ pi_is_3.14
 
== Video Solution ==
https://youtu.be/ph8o017pw_o
 
== See Also ==
{{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}}

Latest revision as of 18:00, 25 September 2025

Problem

If $x,y,$ and $z$ are positive numbers satisfying

\[x + \frac{1}{y} = 4,\qquad y + \frac{1}{z} = 1, \qquad \text{and} \qquad z + \frac{1}{x} = \frac{7}{3}\]

Then what is the value of $xyz$ ?

$\text {(A)}\ \frac{2}{3} \qquad \text {(B)}\ 1 \qquad \text {(C)}\ \frac{4}{3} \qquad \text {(D)}\ 2 \qquad \text {(E)}\ \frac{7}{3}$

Solution 1

We multiply all given expressions to get: \[(1)xyz + x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} + \frac{1}{xyz} = \frac{28}{3}\] Adding all the given expressions gives that \[(2) x + y + z + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 + \frac{7}{3} + 1 = \frac{22}{3}\] We subtract $(2)$ from $(1)$ to get that $xyz + \frac{1}{xyz} = 2$. Hence, by inspection, $\boxed{xyz = 1 \rightarrow B}$. \[\] ~AopsUser101

Solution 2

We have a system of three equations and three variables, so we can apply repeated substitution.

\[4 = x + \frac{1}{y} = x + \frac{1}{1 - \frac{1}{z}} = x + \frac{1}{1-\frac{1}{7/3-1/x}} = x + \frac{7x-3}{4x-3}\]

Multiplying out the denominator and simplification yields $4(4x-3) = x(4x-3) + 7x - 3 \Longrightarrow (2x-3)^2 = 0$, so $x = \frac{3}{2}$. Substituting leads to $y = \frac{2}{5}, z = \frac{5}{3}$, and the product of these three variables is $1$.

Video Solution by Power Solve

https://www.youtube.com/watch?v=ZiJ29GzhyUY

Video Solution(Fast, quick, and learn more about Algebraic Manipulations!)

https://youtu.be/rP8_-36n1ps ~MK

Video Solution by OmegaLearn

https://www.youtube.com/watch?v=SpSuqWY01SE&t=374s

~ pi_is_3.14

Video Solution

https://youtu.be/ph8o017pw_o

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

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