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I have a system $$ 5x^2 - 5y^2 - 3x + 9y = 0, $$ $$ 5x^3 + 5y^3 - 13xy - 15x^2 - y^2 = 0. $$

How to solve it?

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Dirty Solution: $$5x^2-5y^2-3x+9y =0 \\ \rightarrow x^2-y^2-3/5x+9/5x =0 \\ \rightarrow x^2-3/5x + 9/100 - (y^2-9/5y+81/100) = -72/100 \\ \rightarrow (x-3/10)^2 - (y-9/10)^2 = -72/100 \\ \rightarrow x = \pm \sqrt{(y-9/10)^2 - 72/100} + 3/10$$ Plug into the second question. –  Ben Feb 25 '13 at 22:42
    
$(0,0)$ works. Mostly i would draw some careful pictures in the $x-y$ plane to see what is reasonable. The first one is a hyperbola, not centered at the origin, but one branch passes through it. I imagine $y=-x$ is an asymptote for the second one, not sure. –  Will Jagy Feb 26 '13 at 0:17

1 Answer 1

up vote 2 down vote accepted

First, you can observe that if $y=0$, then $x=0$, which gives out a solution.

Let us assume $y\neq 0$ now.

Take $\dfrac{x}{y}=\lambda$.

Then the first equation is:

$$(5\lambda^2-5)y^2+(9-3\lambda)y=0$$ which is $$(5\lambda^2-5)y+(9-3\lambda)=0$$

So $$y = \dfrac{3\lambda-9}{5\lambda^2-5}$$

The second one is: $$(5\lambda^3+5)y^3-(15\lambda^2+13\lambda+1)y^2=0$$ which is $$(5\lambda^3+5)y-(15\lambda^2+13\lambda+1)=0$$

Thus $$y = \dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$,

According to these, we have $$\dfrac{3\lambda-9}{5\lambda^2-5}=\dfrac{15\lambda^2+13\lambda+1}{5\lambda^3+5}$$

which is equal to:

$$6\lambda^4+11\lambda^3-7\lambda^2-8\lambda+4=0$$ Factorize it: $$(\lambda+1)(\lambda+2)(2\lambda-1)(3\lambda-2)=0$$

plug in $y = \dfrac{3\lambda-9}{5\lambda^2-5}$.

And $\lambda\neq -1$.

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