Difference between revisions of "2013 AMC 12A Problems/Problem 8"
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Cross multiply in either equation, giving us <math>xy=2</math>. | Cross multiply in either equation, giving us <math>xy=2</math>. | ||
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+ | <math>\boxed{\textbf{(D) }{2}}</math> | ||
==Solution 2== | ==Solution 2== | ||
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<math>xy = 2</math> | <math>xy = 2</math> | ||
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+ | == Video Solution == | ||
+ | https://youtu.be/ba6w1OhXqOQ?t=1129 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/CCjcMVtkVaQ | ||
+ | ~sugar_rush | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}} | {{AMC12 box|year=2013|ab=A|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:15, 15 January 2021
Problem
Given that and are distinct nonzero real numbers such that , what is ?
Solution 1
Since , we may assume that and/or, equivalently, .
Cross multiply in either equation, giving us .
Solution 2
Since
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1129
~ pi_is_3.14
Video Solution
https://youtu.be/CCjcMVtkVaQ ~sugar_rush
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.