Mock AIME 3 2006-2007 Problems/Problem 1

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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Mock AIME 3 2006-2007

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Mock AIME 3 2006-2007 Problems/Problem 1

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