Difference between revisions of "2017 AMC 10B Problems/Problem 4"
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==Solution== | ==Solution== | ||
Rearranging, we find <math>3x+y=-2x+6y</math>, or <math>5x=5y\implies x=y</math> | Rearranging, we find <math>3x+y=-2x+6y</math>, or <math>5x=5y\implies x=y</math> | ||
| − | Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{ | + | Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math> |
{{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}} | {{AMC10 box|year=2017|ab=B|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:29, 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 
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
