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