Difference between revisions of "1966 AHSME Problems/Problem 17"
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== Solution == | == Solution == | ||
| − | + | Let <math>a=x^2</math> and <math>b=y^2</math>. | |
| + | We now have a system of 2 linear equations: | ||
| + | <math>a+4b=1\\4a+b=4</math> | ||
| + | Multiplying the first equation by 4 and then subtracting the second equation from the first one, we get: | ||
| + | <math>15b=0\\b=0</math> | ||
| + | Now, we can substitute b to solve for a: | ||
| + | <math>4a+0=4\\a=1</math> | ||
| + | Now note that <math>x^2=1 \rightarrow x=\pm 1</math> and <math>y^2=0\rightarrow y=0</math>, so the solutions are <math>(1,0), (-1, 0)</math>. There are two solutions, meaning that the answer is | ||
| + | <math>\fbox{C}</math> | ||
== See also == | == See also == | ||
Latest revision as of 10:35, 26 July 2021
Problem
The number of distinct points common to the curves 
and 
is:
Solution
Let 
and 
.
We now have a system of 2 linear equations:
Multiplying the first equation by 4 and then subtracting the second equation from the first one, we get:
Now, we can substitute b to solve for a:
Now note that 
and 
, so the solutions are 
. There are two solutions, meaning that the answer is
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 17 | |
| 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. 
