Question

Can any system be solved using the multiplication method?

Answer 1

I assume that you are only asking about linear systems.
I also need to assume that, like me, you count discovering that a system has no solutions (is inconsistent) counts as "solving the system"
Third, I assume that the number of variables equals the number of equations.
Finally, I assume that the "multiplication method" is the same as the method I was taught to call the "addition/subtraction method".

With these four assumptions, the answer is,

Yes. Any linear system of `n` equations in `n` variables can be solved by this method:

`ax+by=c`
`dx+ey=f`

Mutiply the first equation by `d` and the second by `-a` to get:

`adx+bdy=cd`
`-adx-aey=-af`

Add to get:

`bdy-aey = cd-af`

`(bd-ae)y = cd-af`

So, as long as we don't have `0` on the left and non-zero on tlhe right (which would indicate an inconsistent system), and we also don't have `0=0` (infinitely many solutions) we get:

`y = (cd-af)/(bd-ae)`

Using the number we get for `y`, we can go back and find `x`.