Answer
The required solution is $\left\{ \left( -3,0 \right),\left( 2,20 \right),\left( -2,4 \right) \right\}$
Work Step by Step
The given equations are,
$\begin{align}
& -4x+y=12\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& y={{x}^{3}}+3{{x}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
Solve the equation (I):
Put the value of $ y={{x}^{3}}+3{{x}^{2}}$ in equation (I)
$\begin{align}
& -4x+\left( {{x}^{3}}+3{{x}^{2}} \right)=12 \\
& {{x}^{3}}+3{{x}^{2}}-4x-12=0 \\
& {{x}^{2}}\left( x+3 \right)-4\left( x+4 \right)=0 \\
& \left( {{x}^{2}}-4 \right)\left( x+3 \right)=0
\end{align}$
On solving further:
$\begin{align}
& \left( x+3 \right)\left( x-2 \right)\left( x+2 \right)=0 \\
& x=-3,x=2,x=-2 \\
\end{align}$
Substitute the value of $ x=-3,x=2,x=-2$ into equation (II):
For,
$\begin{align}
& x=-3 \\
& y={{\left( -3 \right)}^{3}}+3\left( -{{3}^{2}} \right) \\
& =-27+27 \\
& =0
\end{align}$
For
$\begin{align}
& x=2 \\
& y={{\left( 2 \right)}^{3}}+3{{\left( 2 \right)}^{2}} \\
& =8+12 \\
& =20
\end{align}$
For
$\begin{align}
& x=-2 \\
& y={{\left( -2 \right)}^{3}}+3{{\left( -2 \right)}^{2}} \\
& =-8+12 \\
& =0
\end{align}$
Thus, the solution set of system is $\left\{ \left( -3,0 \right),\left( 2,20 \right),\left( -2,4 \right) \right\}$.
