x1+2x2+x3+x4=7

x1+2x2+2x3-x4=12

2x1+4x2 +6x4=4

1 2 1 1 7

1 2 2 -1 12

2 4 0 6 4

Fisrt we want to keep the first row the same, but we need to change the 2nd Row to get the first column 2nd row to be zero. We do this by (Row1-Row2).

1 2 1 1 7

(1-1) (2-2) (1-2) (1-(-1)) (7-12)

2 4 0 6 4

===> 1 2 1 1 7

0 0 -1 2 -5

2 4 0 6 4

Now we need to change the 3rd row;1st column to be a zero. Then we can have a pivot point in the first row. We then have to do this by (Row3 - 2*R1)

1 2 1 1 7

0 0 -1 2 -5

(2-(2)(1)) (4-(2)(2)) (0-(2)(1)) (6-(2)(1)) (4-(2)(7))

=====> 1 2 1 1 7

0 0 -1 2 -5

0 0 -2 4 -10

Now we need to change the -1 in the 2nd row to a +1. We do this by multiplying -1*Row2.

-1*Row2

====> 1 2 1 1 7

0 0 1 -2 5

0 0 -2 4 -10

Now we need to eliminate the -2 in row 3. We do this by (Row3 + 2 * Row2).

(Row3 + 2 * Row2)

====>

1 2 1 1 7

0 0 1 -2 5

(0+(2)(0)) (0+(2)(0)) (-2+(2)(1)) (4+(2)(-2)) (-10+(2)(5))

====> 1 2 1 1 7

0 0 1 -2 5

0 0 0 0 0

Now we make the #1 in the 1st Row;3 column a zero to make the 1 in the 2nd row; 3rd column a pivot point. We do this by (Row 1 - Row 2)

(Row 1 - Row 2)

=====>

(1-0) (2-0) (1-1) (1-(-1)) (7-5)

0 0 1 -2 5

0 0 0 0 0

======>

1 2 0 3 2

0 0 1 -2 5

0 0 0 0 0

And this is how I think you use Row Reduction Echelon Form , I tried my best to not leave any details out.