Answer
$(-3,0), (2,20), (-2,4)$
Work Step by Step
We are given the system:
$\begin{cases}
y=x^3+3x^2\\
-4x+y=12
\end{cases}$
We will use the substitution method. Substitute the expression of $y$ from Equation 1 in Equation 2 to eliminate $y$ and determine $x$:
$-4x+x^3+3x^2=12$
$x^3+3x^2-4x-12=0$
$x^2(x+3)-4(x+3)=0$
$(x+3)(x^2-4)=0$
$(x+3)(x-2)(x+2)=0$
$x+3=0\Rightarrow x_1=-3$
$x-2=0\Rightarrow x_2=2$
$x+2=0\Rightarrow x_3=-2$
Substitute each of the values of $x$ in Equation 2 to determine $y$:
$-4x+y=12$
$x_1=-3\Rightarrow -4(-3)+y=12\Rightarrow 12+y=12\Rightarrow y_1=0$
$x_2=2\Rightarrow -4(2)+y=12\Rightarrow -8+y=12\Rightarrow y_2=20$
$x_3=-2\Rightarrow -4(-2)+y=12\Rightarrow 8+y=12\Rightarrow y_3=4$
The system's solutions are:
$(-3,0), (2,20), (-2,4)$
