Answer
The solution is, $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$
Work Step by Step
The given equations are,
${{x}^{3}}+y=0$ (I)
And
$2{{x}^{2}}-y=0$ (II)
For solving this equation use the addition method:
Add equation (i) and (ii) to eliminate y:
$\begin{align}
& {{x}^{3}}+y=0 \\
& 2\underline{{{x}^{2}}-y=0} \\
& {{x}^{3}}+2{{x}^{2}}=0 \\
\end{align}$
${{x}^{2}}\left( x+2 \right)=0$ (III)
${{x}^{2}}=0$ or $ x+2=0$
$ x=0$ or $ x=-2$
If $ x=0$ then substitute $ x=0$ in equation (I):
$\begin{align}
& {{\left( 0 \right)}^{3}}+y=0 \\
& y=0
\end{align}$
If $ x=-2$ then substitute $ x=-2$ in equation (I):
$\begin{align}
& {{\left( -2 \right)}^{3}}+y=0 \\
& y=8
\end{align}$
Therefore, the solution set is $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$
Hence, the solution set of the system is, $\left\{ \left( 0,0 \right),\left( -2,8 \right) \right\}$.
