Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 2.2 Junior Find the answer

Suppose that $x$ and $y$ are real numbers that satisfy the two equations $3x+2y=6$ and $9x^2+4y^2=468$ . What is the value of $xy$ ?

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since $3 x+2 y=6$ , then $(3 x+2 y)^{2}=6^{2}$ or $9 x^{2}+12 x y+4 y^{2}=36$ . Since $9 x^{2}+4 y^{2}=468$ , then $12 x y=\left(9 x^{2}+12 x y+4 y^{2}\right)-\left(9 x^{2}+4 y^{2}\right)=36-468=-432$ and so $x y=\frac{-432}{12}=-36$ .

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