Alternative methods for solving a system of one linear one non linear simultaneous equations Take the equations $$x+y=5$$ $$x^2 + y^2 =13$$
The most basic method to solve this system is to first express the linear equation in terms of one of the variables and then sub that into the non-linear equation.
But I am curious if there are other methods to solve such a system ?

$\DeclareMathOperator{\lcm}{lcm}$
Compute the Gröbner basis of your system.  Let us start by writing this with zeroes on the right of the equals signs.  \begin{align*}
0 &= x+y-5  \\
0 &= x^2 + y^2 - 13  \text{.}  
\end{align*}
We pick a variable ordering.  Let us choose $x < y$.  (The given system is unchanged by the exchange of the variables $x$ and $y$, so we get the same computation, but with the variables swapped, if we choose the other ordering.)  We compute the first $s$-polynomial.  We need the LCM of the leading terms
$$  \lcm(x, x^2) = x^2  $$
and using this we get
\begin{align*}
0 &= \frac{x^2}{x}(x+y-5) - \frac{x^2}{x^2}(x^2 + y^2 - 13)  \\
  &= x^2 + xy - 5x -(x^2 + y^2 - 13)  \\
  &= xy - y^2 -5x + 13  \text{.}
\end{align*}
Now $\lcm(xy, x) = xy$ and
\begin{align*}
0 &= \frac{xy}{x}(x+y-5) - \frac{xy}{xy}(xy - y^2 -5x + 13)  \\
  &= xy + y^2 - 5y -(xy - y^2 -5x + 13)  \\
  &= 2y^2 +5x -5y -13
\end{align*}
and since we have a relation for $x$ and $y$ both of degree $1$, \begin{align*}
0 &= 2y^2 +5x - 5y - 13 -5(x+y-5)  \\
  &= 2y^2 +5x - 5y - 13 -5x -5y + 25  \\
  &= 2y^2 -10y + 12   \\
  &= 2(y^2 - 5y + 6)  \text{,}
\end{align*}
and since twice a thing is zero means the thing is zero, we have
$$  y^2 - 5y + 6 = 0  \text{.}  $$
Our collection of expressions which evaluate to zero is then (sorting by decreasing total degree, then according to the order we picked for the variables)
\begin{align*}
x^2 + y^2 - 13 &= 0  \\
xy - y^2 -5x + 13 &= 0  \\
y^2 - 5y + 6 &= 0  \\
x+y-5 &= 0  \text{.}
\end{align*}
Notice that in degree $2$ we slowly decreased the degree of the dependence on $x$ until we were left with a polynomial in $y$ alone.  Solving that polynomial, $y = 2$ or $y = 3$.  Then the collection becomes (by specializing the value of $y$ and appending a final equation for that value of $y$) either
\begin{align*}
x^2 - 9 &= 0  \\
-3x + 9 &= 0  \\
0 &= 0  \\
x-3 &= 0  \\
y -2 &= 0  \text{,}
\end{align*}
giving the solution $(x,y) = (3,2)$, or
\begin{align*}
x^2 - 4 &= 0  \\
-2x + 4 &= 0  \\
0 &= 0  \\
x-2 &= 0  \\
y - 3 &= 0  \text{,}
\end{align*}
giving the solution $(x,y) = (2,3)$.
