Difference between revisions of "1966 AHSME Problems/Problem 15"
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== Solution == | == Solution == | ||
| − | + | From <math>x-y>x</math>, we get that <math>-y>0\implies y<0</math>. | |
| + | From <math>x+y<y</math>, we get that <math>x<0</math>. | ||
| + | So, our final answer is <math>\fbox{D}</math>. | ||
== See also == | == See also == | ||
Latest revision as of 21:11, 14 January 2018
Problem
If 
and 
, then
Solution
From , we get that 
.
From , we get that 
.
So, our final answer is 
.
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
