Difference between revisions of "2017 AMC 10B Problems/Problem 4"
m |
(→Solution) |
||
| Line 7: | Line 7: | ||
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{\textbf{(D)}\ 2}</math> | Substituting, we can convert the second equation into <math>\frac{x+3x}{3x-x}=\frac{4x}{2x}=\boxed{\textbf{(D)}\ 2}</math> | ||
| + | |||
| + | |||
| + | ==Solution(Cheap)== | ||
| + | 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> | ||
| + | Solution by sp1729 | ||
| + | |||
{{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 20:00, 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, 
Solution by sp1729
| 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. 
