\(S_1\)
\(\begin{eqnarray*} x + y &=& 0\\ x + y^2 &=& 0 \end{eqnarray*}\)
Solving the first equation for \(y\) in terms of \(x\) yields \( y = - x\).
Substituting into the second equation and simplifying yields \(x + x^2 = 0\).
Factoring the expression on the left yields
\(x ( 1 + x ) = 0\)
So, either \(x = 0\) and \(y = 0\), or \(x = -1\) and \(y = 1\).
This also means that the graphs of the two equations intersect at the points \((0,0)\) and \((-1,1)\) in the plane.