There are two ways to do this by matrices.  
Augmented matrix (Gauss-Jordan) and inverse
method.  I picked the 1st method:



We abbreviate that system with this augmented matrix:



We want 0's where the 2 and 5 are.
and 1's where the 3 (upper left corner) 
and -4 are.

We'll get the 0's first.

To get a 0 where the 5 is,
multiply row 1 by -5 and 
row 2 by 3 and replace row 2.

Here's the work:      -15 -10 | -15
                       15 -12 |  81
                     ---------------
                        0 -22 |  66

Since it turns out that that all those numbers
can be divided through by -22, we will do that
too, getting 

                        0   1 |  -3 

and replace row 2 by that:



Then we get a 0 where the 2 is by multiplying the
second row by -2 and adding it to the first row:

Here's the work:        3   2 |   3
                        0  -2 |   6
                     ---------------
                        3   0 |   9

Since it turns out that that all those numbers
can be divided through by 3, we will do that
too, getting 

                        1   0 |   3


and replace row 1 by that:



This augmented matrix is the abbreviation for 
this system:



or x = 3
   y = -3

Edwin