Answer
$(-4,1), \left(-\dfrac{5}{2},\dfrac{1}{4}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
y=(x+3)^2\\
x+2y=-2
\end{cases}$
We will use the substitution method. Solve Equation 1 for $y$ and substitute the expression of $y$ in Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
y=(x+3)^2\\
x+2(x+3)^2=-2
\end{cases}$
$x+2(x^2+6x+9)=-2$
$x+2x^2+12x+18=-2$
$2x^2+13x+18+2=0$
$2x^2+13x+20=0$
$2x^2+8x+5x+20=0$
$2x(x+4)+5(x+4)=0$
$(x+4)(2x+5)=0$
$x+4=0\Rightarrow x_1=-4$
$2x+5=0\Rightarrow x_2=-\dfrac{5}{2}$
Substitute each of the values of $x$ in the expression of $y$ to determine $y$:
$y=(x+3)^2$
$x_1=-4\Rightarrow y_1=(-4+3)^2=1$
$x_2=-\dfrac{5}{2}\Rightarrow y_2=\left(-\dfrac{5}{2}+3\right)^2=\dfrac{1}{4}$
The system's solutions are:
$(-4,1), \left(-\dfrac{5}{2},\dfrac{1}{4}\right)$
