Difference between revisions of "1981 AHSME Problems/Problem 6"
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Latest revision as of 13:35, 28 June 2025
Problem
If 
then 
equals
Solution
We can cross multiply both sides of the equation and simplify.
![\[x(y^2 + 2y - 2) = (x-1)(y^2 + 2y - 1)\]](http://latex.artofproblemsolving.com/a/6/a/a6ade39e0e30d561d34f1f7a0424ed8739d21096.png)
![\[xy^2 + 2xy - 2x = xy^2 + 2xy - x - y^2 - 2y + 1\]](http://latex.artofproblemsolving.com/1/b/1/1b18b6e7b030feff8a95e7a5c02757e52089beab.png)
![\[-x = -y^2 - 2y + 1\]](http://latex.artofproblemsolving.com/5/3/7/5373f3f1b1cd03df689803c0bacc077d10c8bac5.png)
![\[x = y^2 + 2y - 1\]](http://latex.artofproblemsolving.com/6/9/3/693f25182ec1ff5724d70fe8c231898895706590.png)
The answer is 
-edited by coolmath34
See also
| 1981 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
