= Eliminating a variable with a resultant
Consider the original pair of <polynomial equations>
$$
F(x,y)=y-x^2=0,\qquad G(x,y)=y+x^2-2=0.
$$
Treat them as <polynomials> in $y$ with <coefficients> in $\mathbb R[x]$. Their <Sylvester matrix> and <resultant> are
$$
S_y(F,G)=\begin{pmatrix}1&-x^2\\1&x^2-2\end{pmatrix},\qquad
R(x)=\operatorname{Res}_y(F,G)=2(x^2-1).
$$
Every common solution must therefore have $x=-1$ or $x=1$. Back-substitution into $F=0$ gives $y=x^2=1$, and both pairs also satisfy $G=0$. The complete real solution set is
$$
\boxed{\{(-1,1),(1,1)\}.}
$$
Here elimination and direct substitution agree: subtracting $F$ from $G$ gives $2x^2-2$, exactly the same condition.
\Image[/resultant-parabola-intersection-elimination.png]
{title=<Polynomial> elimination keeps the possible first coordinates of intersections}
{description=Original <parabola> example. Left: both equations vanish at the two marked intersections; dashed lines retain their $x$-coordinates on an auxiliary horizontal reference line at $y=-1/2$. Right: the <resultant> has zeros at those coordinates. Substitution recovers $y=1$ and verifies each full solution pair.}
In general, first solve the one-<variable> <resultant> equation, then recover and check the other coordinate in both original <polynomial equations>. A real coordinate candidate can correspond only to nonreal values of the eliminated <variable>. Also, vanishing <leading coefficients> require the separate check in <degree drop in a specialized resultant>. The general elimination-and-back-substitution method is described in the lecture's §3.1.<Algebraic Techniques and Semidefinite Optimization, MIT 6.972, Spring 2006/Lecture 6: Resultants and Discriminants>[Algebraic Techniques and Semidefinite Optimization, MIT 6.972, Spring 2006; MIT 6.972, Lecture 6, p. 6-3, §3.1 Polynomial equations: eliminate one variable, solve and back-substitute]{ref}
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