Answer
{$(-3,4), (0,1)$}
Work Step by Step
Since, we have first equation: $y=x^2+2x+1$ and second equation: $x+y=1$
From the first equation, we have
$-x+1=x^2+2x+1$
or, $x^2+3x=0$
or, $x(x+3)=0$
Thus, $x=${$-3,0$}
From the second equation, we have
$-3+y=1$
or, $y=4$
Now, $0+y=1 \implies y=1$
Hence, the solution set: {$(-3,4), (0,1)$}
