Is this method allowed ?

$$beginpmatrix3&2&36\5&4&64 endpmatrix sim beginpmatrix1& 2/3&12\5&4&64 endpmatrix sim beginpmatrix1&2/3&12\0&2/3&4 endpmatrix sim beginpmatrix1&0&8\0&2/3&4 endpmatrix sim beginpmatrix1&0&8\0&1&6 endpmatrix$$

Which yields $x=8$ and $y=6$

The first step is $R_1 to R_1 times frac13$

The second step is $R_2 to R_2 - 5R_1$

The third step is $R_1 to R_1 -R_2$

The fourth step is $R_2 to R_2times frac32$

Here $R_i$ denotes the $i$ -th row.

How about using Cramer's Rule? Define $Delta_x=left[beginmatrix36 & 2 \ 64 & 4endmatrixright]$, $Delta_y=left[beginmatrix3 & 36\ 5 & 64endmatrixright]$
and $Delta_0=left[beginmatrix3 & 2\ 5 &4endmatrixright]$.

Now computation is trivial as you have: $x=dfracdetDelta_xdetDelta_0$ and $y=dfracdetDelta_ydetDelta_0$.

By false position:

Assume $x=10,y=3$, which fulfills the first equation, and let $x=10+x',y=3+y'$. Now, after simplification

$$3x'+2y'=0,\5x'+4y'=2.$$

We easily eliminate $y'$ (using $4y'=-6x'$) and get

$$-x'=2.$$

Though this method is not essentially different from, say elimination, it can be useful for by-hand computation as it yields smaller terms.

Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.

Any method you can come up with will in the end amount to Cramer's rule, which gives explicit formulas for the solution. Except special cases, the solution of a system is unique, so that you will always be computing the ratio of those determinants.

Anyway, it turns out that by organizing the computation in certain ways, you can reduce the number of arithmetic operations to be performed. For $2times2$ systems,
the different variants make little difference in this respect. Things become more interesting for $ntimes n$ systems.

Direct application of Cramer is by far the worse, as it takes a number of operations proportional to $(n+1)!$, which is huge. Even for $3times3$ systems, it should be avoided. The best method to date is Gaussian elimination (you eliminate one unknown at a time by forming linear combinations of the equations and turn the system to a triangular form). The total workload is proportional to $n^3$ operations.

The steps of standard Gaussian elimination:

$$begincasesax+by=c,\dx+ey=f.endcases$$

Subtract the first times $dfrac da$ from the second,

$$begincasesax+by=c,\0x+left(e-bdfrac daright)y=f-cdfrac da.endcases$$

Solve for $y$,

$$begincasesax+by=c,\y=dfracf-cdfrac dae-bdfrac da.endcases$$

Solve for $x$,

$$begincasesx=dfracc-bdfracf-cdfrac dae-bdfrac daa,\y=dfracf-cdfrac dae-bdfrac da.endcases$$

So written, the formulas are a little scary, but when you use intermediate variables, the complexity vanishes:

$$d'=frac da,e'=e-bd',f'=f-cd'to y=fracf'e', x=fracc-bya.$$

Anyway, for a $2times2$ system, this is worse than Cramer !

$$begincasesx=dfracce-bfDelta,\y=dfracaf-cdDeltaendcases$$ where $Delta=ae-bd$.

For large systems, say $100times100$ and up, very different methods are used. They work by computing approximate solutions and improving them iteratively until the inaccuracy becomes acceptable. Quite often such systems are sparse (many coefficients are zero), and this is exploited to reduce the number of operations. (The direct methods are inappropriate as they will break the sparseness property.)

$$beginalign3x+2y&=36 tag1\ 5x+4y&=64tag2endalign$$

From $(1)$, $x=frac36-2y3$, substitute in $(2)$ and you'll get $5(frac36-2y3)+4y=64 implies y=6$ and then you can get that $x=24/3=8$

Another Method
From $(1)$, $x=frac36-2y3$

From $(2)$, $x=frac64-4y5$

But $x=x implies frac36-2y3=frac64-4y5$ do cross multiplication and you'll get $5(36-2y)=3(64-4y) implies y=6$ and substitute to get $x=8$

Fixed Point Iteration

This is not efficient but it's another valid way to solve the system. Treat the system as a matrix equation and rearrange to get $beginbmatrix x\ yendbmatrix$ on the left hand side.

Define
$fbeginbmatrix x\ yendbmatrix=beginbmatrix (36-2y)/3 \ (64-5x)/4endbmatrix$

Start with an intial guess of $beginbmatrix x\ yendbmatrix=beginbmatrix 0\ 0endbmatrix$

The result is $fbeginbmatrix 0\ 0endbmatrix=beginbmatrix 12\ 16endbmatrix$

Now plug that back into f

The result is $fbeginbmatrix 12\ 6endbmatrix=beginbmatrix 4/3\ 1endbmatrix$

Keep plugging the result back in. After 100 iterations you have:

$beginbmatrix 7.9991\ 5.9993endbmatrix$

Here is a graph of the progression of the iteration:

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 $kinBbb R$. Then $$3x+2y=36implies x(2k+3)=36implies x=frac362k+3\5x+4y=64implies x(4k+5)=64implies x=frac644k+5$$ so $$36(4k+5)=64(2k+3)implies (144-128)k=(192-180)implies k=frac34.$$ Now $$x=frac644k+5=frac644cdotfrac34+5=8implies y=kx=frac34cdot8=6.quadsquare$$

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

How about we try squaring the equations? We get $$3x+2y=36implies 9x^2+12xy+4y^2=1296\5x+4y=64implies 25x^2+40xy+16y^2=4096$$ 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 $2x+2y=28implies y=14-x$ so you have the nice quadratic $$15x^2-8(14-x)^2=672.$$ Enjoy!

It is clear that:

• 
  $x=10$, $y=3$ is an integer solution of $(1)$.


• 
  $x=12$, $y=1$ is an integer solution of $(2)$.

Then, from the theory of Linear Diophantine equations:

• Any integer solution of $(1)$ has the form $x_1=10+2t$, $y_1=3-3t$ with $t$ integer.


• Any integer solution of $(2)$ has the form $x_2=12+4t$, $y_2=1-5t$ with $t$ integer.

Then, the system has an integer solution $(x_0,y_0)$ if and only if there exists an integer $t$ such that

$$10+2t=x_0=12+4tqquadtextandqquad 3-3t=y_0=1-5t.$$

Solving for $t$ we see that there exists an integer $t$ satisfying both equations, which is $t=-1$. Thus the system has the integer solution
$$x_0=12+4(-1)=8,; y_0=1-5(-1)=6.$$

Note that we can pick any pair of integer solutions to start with. And the method will give the solution provided that the solution is integer, which is often not the case.