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Algebra Difficulty 2.6 Junior Find the answer Canada

Suppose that $x$ and $y$ are real numbers that satisfy the two
equations $\begin{align*} 3x+2y &= 6\\ 9x^2+4y^2&= 468\end{align*}$ What is the value of $xy$ ?

Pick one

Solution

Since $3x + 2y = 6$ , then
$(3x + 2y)^2 = 6^2$ or $9x^2 + 12xy + 4y^2 = 36$ .

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

(With some additional work, we can find that the solutions to the system
of equations are $(x,y) = (-4,9)$ and
$(x,y) = (6,-6)$ .)

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.