<s>[INST] Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:

$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$

I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation eliminates $4y$ to solve strictly for $x$; i.e. $6x - 5x = 72 - 64 \Rightarrow x = 8$. Substituting $x=8$ into $(2)$ reveals that $y=6$.
I could also subtract $(1)$ from $(2)$ and divide by $2$, yielding $x+y=14$. Let $$\begin{align}3x+3y - y &= 36 \tag{1a}\\ 5x + 5y - y &= 64\tag{1b}\end{align}$$ then expand brackets, and it follows that $42 - y = 36$ and $70 - y = 64$, thus revealing $y=6$ and so $x = 14 - 6 = 8$.
I can even use matrices!
$(1)$ and $(2)$ could be written in matrix form:
$$\begin{align}\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}\begin{bmatrix} x \\ y\end{bmatrix}&=\begin{bmatrix}36 \\ 64\end{bmatrix}\tag3 \\ \begin{bmatrix} x \\ y\end{bmatrix} &= {\begin{bmatrix} 3 &2 \\ 5 &4\end{bmatrix}}^{-1}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}4 &-2 \\ -5 &3\end{bmatrix}\begin{bmatrix}36 \\ 64\end{bmatrix} \\ &=\frac12\begin{bmatrix} 16 \\ 12\end{bmatrix} \\ &= \begin{bmatrix} 8 \\ 6\end{bmatrix} \\ \\ \therefore x&=8 \\ \therefore y&= 6\end{align}$$

Question
Are there any other methods to solve for both $x$ and $y$?
 [/INST] Construct the Groebner basis of your system, with the variables ordered $x$, $y$: 
$$  \mathrm{GB}(\{3x+2y-36, 5x+4y-64\}) = \{y-6, x-8\}  $$
and read out the solution.  (If we reverse the variable order, we get the same basis, but in reversed order.)  Under the hood, this is performing Gaussian elimination for this problem.  However, Groebner bases are not restricted to linear systems, so can be used to construct solution sets for systems of polynomials in several variables.

Perform lattice reduction on the lattice generated by $(3,2,-36)$ and $(5,4,-64)$.  A sequence of reductions (similar to the Euclidean algorithm for GCDs): \begin{align*}
(5,4,-64) - (3,2,-36) &= (2,2,-28)  \\
(3,2,-36) - (2,2,-28) &= (1,0,-8)  \tag{1} \\
(2,2,-28) - 2(1,0,-8) &= (0,2,-12)  \tag{2} \\
\end{align*}
From (1), we have $x=8$.  From (2), $2y = 12$, so $y = 6$.  (Generally, there can be quite a bit more "creativity" required to get the needed zeroes in the lattice vector components.  One implementation of the LLL algorithm, terminates with the shorter vectors $\{(-1,2,4), (-2,2,4)\}$, but we would continue to manipulate these to get the desired zeroes.)
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