How many solutions this a multidimensional system have? How many solutions have this system of equations:
\begin{array}{l}
\ 3x^2+y+2xy^2-3 = 0 \\
\ x + 2yx = 0
\end{array}
The more general case is when we want to solve two nonlinear algebraic equations for two unknowns
\begin{array}{l}
\ {A_1}{x^2} + {B_1}xy + {C_1}{y^2} + {D_1}x + {E_1}y + {F_1} = 0 \\
\ {A_2}{x^2} + {B_2}xy + {C_2}{y^2} + {D_2}x + {E_2}y + {F_2} = 0
\end{array}
 A: Suppose that $x\neq 0$, then the second equation becomes $2y+1=0$ and then the first equation turns to be a quadratic in terms of $x$, and you will have two solutions for that $y$. Now, suppose that $x=0$, then we have another solution. 
In total, you left with three solution to that system of equations. 
A: In general an equation involving up to second degree i.e. up to $x^2$, $xy$ and/or $y^2$ will be a conic section - i.e. a parabola, circle, ellipse or hyperbola. Considering them geometrically these shapes can intersect at most four times.
Your equations however have a cubic term in the form of $xy^2$. In general a cubic could potentially lead to more solutions.
Your example however only has three solutions. From the second equation either $x=0$ or $y=-\frac{1}{2}$. If $x=0$ then the first equation is linear in $y$ with one solution. If $y=-\frac{1}{2}$ then the first equation is quadratic in $x$ with two solutions. So a total of three.
A: There are two roots of the second equation.
$x=0$ substituted into the first equation gives $ y=3.$
$y=-\frac12$ substituted into the first equation gives two solutions from the quadratic resulting in:
$$ x = \dfrac{-3 \pm \sqrt{82}}{2}  $$
So, in all 3 different roots and 3 solutions.
