Question 54438
Using matrices is simply using elementary row operations to get 1x=? and 1y=? so the original equation in matrix form becomes the co-efficients of x and y and the constant term:
2 -3  4
5 4   33
Ideally, we want a matrix that is(see below) where ? is the answer for each variable.
1 0  ?
0 1  ?

start by dividing the top row by 2 to get a 1 in the first column of the first row
1  (-3/2)   2
5   4       33

Then get a zero in the first column of the second row by taking row 2 and subtracting 5 times row 1: (note: only row two changes in this step)

1      (-3/2)             2    =    1  (-3/2)    2
0   (8/2)-((-3)(5)/2)    33    =    0  (23/2)    23


Next get a one in the second column of the second row by multiplying the row by (2/23):

1  (-3/2)   2
0   1       2

Now to get a zero in the 2nd column of the first row, take row 1 and add (3/2) times the second row (because the term we're looking to get rid of is negative)
Note: in the first column (3/2) times 0 is still 0 so 1-0 will still be one.
Note: multiplying the second row is only for applying it to row 1, after the operation has been completed, the second row goes back to its original form.

1      (-3/2)      2     =  1    0    5
0(3/2) 1(3/2)  2(3/2)    =  0    1    2

Now putting that back into linear algebra form 
1x +0y = 5   x=5
0x +1y = 2   y=2

Check this by substituting the x and y values into the original equation:
2x -3y =4         2(5)-3(2) =4    10-6=4
5x +4y =33        5(5)+4(2) =33   25+8=33