# Solve Equation of three variables

I want to solve this system equations. I found 2 solutions, but it's was effort algebra with the second equation and take $z=0$ in (1), you can see 2 solutions with $x=-2,2$ and $y=1,-1$. $$(1) \ x^2+xyz+5z^3=4\\ (2) \ 3x^2-y^2+5xy=2$$

Are you know any algorithm, or method, or technique to find others solutions (?) I think that this system eq, have not other solutions in $\mathbb{R}$

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The system has infinite solutions. From the second equation, we get that $$y^2 - (5x)y + (2-3x^2) = 0 \implies y = \dfrac{5x \pm \sqrt{25x^2 - 4(2-3x^2)}}2$$ Hence, $$y = \dfrac{5x \pm \sqrt{37x^2 - 8}}2 \,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ Choose any $x \in \left(-\infty, - \sqrt{\dfrac8{37}} \right] \bigcup \left[\sqrt{\dfrac8{37}} , \infty \right)$. We can obtain $y$ from $(\star)$. Now after we get $x$ and $y$, solve the cubic equation in $z$ i.e. $$5z^3 + (xy)z + (x^2-4) = 0$$ which always has at-least one real solution.
@GerryMyerson But the OP mentions "I think that this system eq, have not other solutions in $\mathbb{R}$" –  user17762 Dec 1 '12 at 4:48