Difference between revisions of "2017 AMC 10B Problems/Problem 4"
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Substituting each <math>x</math> and <math>y</math> with <math>1</math>, we see that the given equation holds true, as <math>\frac{3(1)+1}{1-3(1)} = -2</math>. Thus, <math>\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}</math> | Substituting each <math>x</math> and <math>y</math> with <math>1</math>, we see that the given equation holds true, as <math>\frac{3(1)+1}{1-3(1)} = -2</math>. Thus, <math>\frac{x+3y}{3x-y}=\boxed{\textbf{(D)}\ 2}</math> | ||
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{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}} | ||
+ | {{AMC12 box|year=2017|ab=B|before=First Problem|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:01, 6 February 2018
Contents
[hide]Problem
Supposed that and are nonzero real numbers such that . What is the value of ?
Solutions
Solution 1
Rearranging, we find , or . Substituting, we can convert the second equation into .
Solution 2
Substituting each and with , we see that the given equation holds true, as . Thus,
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 10 Problems and Solutions |
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by First Problem |
Followed by Problem 4 |
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.