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}} | ||
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Revision as of 21:49, 16 February 2017
Problem
Supposed that 
and 
are nonzero real numbers such that 
. What is the value of 
?
Solution
Rearranging, we find 
, or 
Substituting, we can convert the second equation into 
Solution(Cheap)
Substituting each and with 
, we see that the given equation holds true, as 
. Thus, 
| 2017 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
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| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
