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Mock AIME 3 2006-2007 Problems/Problem 1: Difference between revisions

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Created page with '== Problem == <math>x</math> and <math>y</math> are real numbers such that <math>x^2 + 3y = 9</math>, and <math>2y^2 - 6x = 18</math>. If the 4 solutions to the system are <math>…'
 
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*[[Mock AIME 3 2006-2007/Problem 2 | Next Problem]]
*[[Mock AIME 3 2006-2007 Problems/Problem 2 | Next Problem]]


*[[Mock AIME 3 2006-2007]]
*[[Mock AIME 3 2006-2007]]

Latest revision as of 16:52, 11 March 2025

Problem

$x$ and $y$ are real numbers such that $x^2 + 3y = 9$, and $2y^2 - 6x = 18$. If the 4 solutions to the system are $(x_1,y_1),(x_2,y_2),(x_3,y_3),$ and $(x_4,y_4)$, then find $(y_1 - x_1) + (y_2 - x_2) + (y_3 - x_3) + (y_4 - x_4)$.

Solution

Substitute $y = \frac{9 - x^2}{3}$ into the second to get the quartic polynomial: \[x^4 - 18x^2 - 27x = 0.\]

From Vieta's formulae, $\sum x = 0$.

Now create another quadratic equation in terms of $y$ by making the substitution for $x = \frac{9 - y^2}{3}$ and the polynomial \[y^4 - 6y^2 + 27y = 0\] results.

Thus, $\sum y = 0$.

Finally, the answer is $\sum y - \sum x = 0 - 0 = \boxed{000}$.

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