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Algebra

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Let $x$ and $y$ be real numbers. If $x^2 + 3y^2 = 18$, then find the maximum value of $x + y$.

Jasmo Feb 2, 2025

1⁺⁰ Answers

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$x^2 + 3y^2 = 18\\ 3y^2=18-x^2\\ y=\sqrt{6-\frac{x^2}{3}}\\ x=\sqrt{18-3y^2}\\ \color{blue}x+y=\sqrt{18-3y^2}+\sqrt{6-\frac{x^2}{3}}$

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asinus Feb 3, 2025

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