Write the equations this way
4x + y = 7
-5x - y + z - w = 1
z + w = 7
Then put 1's in front of the letters
without coefficients showing, and +0
times the missing letters:
4x + 1y + 0z + 0w = 7
-5x - 1y + 1z - 1w = 1
0x + 0y + 1z + 1w = 7
Erase all the letters and
+ signs, move the minus
signs next to the coefficients
as negative signs, draw a line
where the equal signs are, and
put brackets around:
[ 4 1 0 0 | 7]
[-5 -1 1 -1 | 1]
[ 0 0 1 1 | 7]
Get a 0 where the -5 is by
adding 5 times the 1st row to
4 times the second row, and
restore the 1st row
5[ 4 1 0 0 | 7]
4[-5 -1 1 -1 | 1]
[ 0 0 1 1 | 7]
[ 4 1 0 0 | 7]
[ 0 1 4 -4 | 39]
[ 0 0 1 1 | 7]
Get a 0 where the 1 in row
1 is by adding -1 times row
2 to 1 times row 1, and
restore the 2nd row
1 [ 4 1 0 0 | 7]
-1 [ 0 1 4 -4 | 39]
[ 0 0 1 1 | 7]
[ 4 0 -4 4 |-32]
[ 0 1 4 -4 | 39]
[ 0 0 1 1 | 7]
Get a 0 where the -4 is in
the 1st row by multiplying
the 3rd row by 4 and adding
it to 1 times the 1st row,
and restore the 3rd row
1[ 4 0 -4 4 |-32]
[ 0 1 4 -4 | 39]
4[ 0 0 1 1 | 7]
[ 4 0 0 8 | -4]
[ 0 1 4 -4 | 39]
[ 0 0 1 1 | 7]
Get a 0 where the 4 is in
the 2nd row by multiplying
the 3rd row by -4 and
adding it to 1 times the
2nd row, and restore the
3rd row:
[ 4 0 0 8 | -4]
1[ 0 1 4 -4 | 39]
-4[ 0 0 1 1 | 7]
[ 4 0 0 8 | -4]
[ 0 1 0 -8 | 11]
[ 0 0 1 1 | 7]
Get a 1 where the first 4 is
in the 1st row by multiplying
the 1st row through by 1/4
[ 1 0 0 2 | -1]
[ 0 1 0 -8 | 11]
[ 0 0 1 1 | 7]
This is in row reduced
echelon form.
Put the letters, plus signs,
and equal signs back in:
1x + 0y + 0z + 2w = -1
0x + 1y + 0z - 8w = 11
0x + 0y + 1z + 1w = 7
Simplify to
x + 2w = -1
y - 8w = 11
z + w = 7
Eliminate the spaces (close it up:
x + 2w = -1
y - 8w = 11
z + w = 7
Solve for x, y and z
x = -1 - 2w
y = 11 + 8w
z = 7 - w
w can be any number, say c, then
x = -1 - 2c
y = 11 + 8c
z = 7 - c
w = c
So the solution is
(x, y, z, w) = (-1-c, 11+8c, 7-c, c)
It does turn out that w can be any number,
but as you see, z is not necessarily 7,
unless c happens to be chosen to be 0.
Edwin