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I am looking for stationary points for a function and ended up with this system: of equations

$2x^3y+2y=0$
$3x^2y^2-12x^2=0$

I have never encountered a system like this where the variables are multiplied to each other but i imagine that it's dangerous to divide with a variable since you can't be sure that it isn't $0$.

So how would you go about solving this?

2 Answers 2

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Since

$$ (2x^3+2)y=0$$

then either $x=-1$ or $y=0$.

If $x=-1$ the second equation $3x^2y^2-12x^2=0 $gives $3y^2-12=0$ so $y=\pm2$ so we have two solutions $(-1,2),\,(-1,-2)$.

On the other hand if $y=0$ then the second equation gives $0-12x^2=0$ so then $x=0$. So we have a third solution $(0,0)$.

Solutions: $(-1,2)\,(-1,-2)\,(0,0)$.

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Hint:

$2yx^3+2y$=$2y(x^3+1)$

$3x^2y^2-12x^2=3x^2(y^2-4)=3x^2(y-2)(y+2)$

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    So in situations like this you factor and try to find the zeros of the functions? Would y1=0, y2=2, y3=-2 and x1=0, x2=-1 in this case? I'm not sure i understand. – 2017-01-31