How can I go ahead solving this equation ?
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$2x = \lambda(10x + 6y)$
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$2y = \lambda(10y + 6x)$
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$5x^2 + 6xy + 5y^2 = 1$
My attempt was dividing number $1$ by number two. Eliminating the Lagrange multiplier, and then solving for $x$ and $y$. But that did not work. Do you have any advice or any simpler methods for solving this system of equation?
Best Answer
I would usually solve for $\lambda$ and equate the two values of $\lambda$ which gives $$\frac{2x}{10x+6y}=\frac{2y}{10y+6x}(=\lambda)$$ $$x(5y+3x)=y(5x+3y)$$ $$5xy+3x^2=5xy+3y^2$$ $$x^2=y^2$$ $$y=\pm x$$ Hence the last equation is $$10x^2\pm6x^2=1$$ $$4x^2,16x^2=1$$ $$x^2=\frac14,\frac1{16}$$ $$x=\pm\frac12,\pm\frac14$$ Hence solutions are $$(\frac12,\frac12),(\frac12,-\frac12),(\frac14,\frac14),(\frac14,-\frac14),(-\frac12,\frac12),(-\frac12,-\frac12),(-\frac14,\frac14),(-\frac14,-\frac14),$$