How to solve these $ 2x + 4y + 3x^{2} + 4xy =0$ and $ 4x + 8y + 2x^{2} + 4y^{3}$ = $0 $

Question

I need to solve these two equations .

$ 2x + 4y + 3x^{2} + 4xy =0$

$ 4x + 8y + 2x^{2} + 4y^{3}$ = $0 $

I have added them , subtracted them . Nothing is helping here . Can anyone give hints ? Thanks

Answer 1

It is not so hard. Just eleiminate $y$ from the first equation; this gives $$y=\frac{-3 x^2-2 x}{4 (x+1)}$$ Plug it in the second equation and simplify as much as you can; you should arrive to $$(-27 x^3-22 x^2+28 x+24)\frac{x^3}{16 (x+1)^3}=0$$ So, beside the trivial solution $x=0$, remain the roots of the cubic $$-27 x^3-22 x^2+28 x+24=0$$ Use Cardano to show that there is only one real root.

By inspection of $$f(x)=-27 x^3-22 x^2+28 x+24$$ $f(0)=24$, $f(1)=3$, $f(2)=-224$; so the solution is very close to $x=1$. To approximate it, use Newton starting at $x=1$; the first iterates are $\frac{100}{97}$, $\frac{24311038}{23603689}$ which is the solution for more than six significant figures.

If you finish the steps for solving a cubic, you should find that the exact solution is $$x=\frac{1}{81} \left(\sqrt[3]{150704-1296 \sqrt{1113}}+2 \sqrt[3]{ 18838+162 \sqrt{1113}}-22\right)$$

Answer 2

Rearrange the first equation to get $y$ as a function of $x$. Plug that into the second equation.
I think that, after rearrangement, you should get a polynomial in $x$ of degree 6.