<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] Other answers have given standard, elementary methods of solving simultaneous equations. Here are a few other ones that can be more long-winded and excessive, but work nonetheless.

Method $1$: (multiplicity of $y$)

Let $y=kx$ for some $k\in\Bbb R$. Then $$3x+2y=36\implies x(2k+3)=36\implies x=\frac{36}{2k+3}\\5x+4y=64\implies x(4k+5)=64\implies x=\frac{64}{4k+5}$$ so $$36(4k+5)=64(2k+3)\implies (144-128)k=(192-180)\implies k=\frac34.$$ Now $$x=\frac{64}{4k+5}=\frac{64}{4\cdot\frac34+5}=8\implies y=kx=\frac34\cdot8=6.\quad\square$$ 


Method $2$: (use this if you really like quadratic equations :P)

How about we try squaring the equations? We get $$3x+2y=36\implies 9x^2+12xy+4y^2=1296\\5x+4y=64\implies 25x^2+40xy+16y^2=4096$$ Multiplying the first equation by $10$ and the second by $3$ yields $$90x^2+120xy+40y^2=12960\\75x^2+120xy+48y^2=12288$$ and subtracting gives us $$15x^2-8y^2=672$$ which is a hyperbola. Notice that subtracting the two linear equations gives you $2x+2y=28\implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!

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