# Can any system be solved using the multiplication method?

May 14, 2015

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:

$a x + b y = c$
$\mathrm{dx} + e y = f$

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

$a \mathrm{dx} + b \mathrm{dy} = c d$
$- a \mathrm{dx} - a e y = - a f$

$b \mathrm{dy} - a e y = c d - a f$
$\left(b d - a e\right) y = c d - a f$
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:
$y = \frac{c d - a f}{b d - a e}$
Using the number we get for $y$, we can go back and find $x$.