<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] If you prefer using parametric form or your equations are already in parametric form, this is how you can proceed:
We know that $(0,18)$ is a solution to $3x+2y=36$ and $(0,16)$ is a solution to $5x + 4y = 64$. Therefore the equations in parametric form become:
$${ \begin{pmatrix} 0 \\ 18 \\  \end{pmatrix} } + t_1 { \begin{pmatrix} 2 \\ -3 \\  \end{pmatrix} }  \tag{3}$$
$$\pmatrix { \begin{matrix} 0 \\ 16 \\  \end{matrix} } + t_2 { \begin{pmatrix} 4 \\ -5 \\  \end{pmatrix} }  \tag{4}  $$ 
Equate the $x$ and $y$ coordinates:
$$0+2t_1 = 0+4t_2 \Rightarrow t_1 = 2t_2 \tag{5}$$
$$18-3t_1 = 16-5t_2 \Rightarrow 18-6t_2 = 16-5t_2 \text{ (using (5) }) \Rightarrow2 = t_2 \tag{6}$$
and now substitute back into $(4)$:
$$\pmatrix { \begin{matrix} 0 \\ 16 \\  \end{matrix} } + 2 { \begin{pmatrix} 4 \\ -5 \\  \end{pmatrix} }   =\pmatrix { \begin{matrix} 8 \\ 6 \\  \end{matrix} }  $$ 
i.e $x=8, y=6$.
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