You need to use gaussian elimination, hence, you need to convert the original form of the system to a triangular form.

You need to eliminate the variable x from the first and second equations and then, from the first and the third equations such that:

`{(4x+8y-z=10|*(-3)), (3x-8y+9z=14|*4):}` `=gt{(-12x-24y+3z=-30),(12x-32y+36z=56):}`

Adding the equations yields:

`-56y + 39z = 26`

Considering the first and the third equations yields:

`{(4x+8y-z=10|*(7)), (7x+6y+5z=0|*(-4)):}` `=>{(28x+56y-7z=70),(-28x-24y-20z=0):}`

Adding the equations yields:

`32y - 27z = 70`

You need to use the equations -`56y + 39z = 26`  and `32y - 27z = 70`  to eliminate y such that:

`{(-56y + 39z = 26|*4),(32y - 27z = 70|*7):}=gt` `{(-224y + 156z = 104),(224y - 189z = 490):}`

`-33z = 594 => z = -18`

Substituting -18 for z in equation `32y - 27z = 70`  yields:

`32y - 27(-18)= 70 => 32y =-416 => y = -13`

Substituting -18 for z and -13 for y in equation `4x+8y-z=10`  yields:

`4x-104+18=10 => 4x=96 => x = 24`

Hence, evaluating the solution to the given system, using gaussian elimination, yields `x = 24 , y = -13 , z = -18` .

Write the three equations as an augmented matrix:

4  8 -1 |  10

3 -8  9 |  14

7  6  5 |   0

The result of using gaussian elimination on this matrix will be an augmented matrix in echelon form:

1  a  b |  c        or   1  a  b |  c      

0  1  d |  e              0  0  1 |  d          

0  0  1 |  f               0  0  0 |  e

First multiply row 1 by 1/4:

1   2  -1/4 | 5/2

3  -8   9    | 14

7   6   5    |  0

Now take 3*row 1 from row 2 and 7*row 1 from row 3:

1   2  -1/4   |   5/2

0 -14  39/4  |  13/2

0 -8    27/4  |  -35/2

Now multiply row 2 by -1/14:

1   2  -1/4    |   5/2

0   1  -39/56 |  -13/28

0  -8   27/4  |   -35/2

Now add 8*row 2 to row 3:

1   2  -1/4      |   5/2

0   1  -39/56   |  -13/28

0   0   33/28   |  -297/14

Finally multiply row 3 by 28/33

1   2  -1/4    |    5/2

0   1  -39/56 |  -13/28

0   0   1       |  -18

Gauss-jordan elimination will result in the identity matrix on the left of the augmented matrix.

Add 39/56*row 3 to row 2 and 1/4*row 3 to row 1:

1   2   0   |  -2

0   1   0   |  -13

0   0   1   |  -18

Finally take 2*row 2 from row 1:

1   0   0   |   24

0   1   0   |  -13

0   0   1   |  -18

Therefore the solution to the system of equations is

`x = 24`,  `y = -13`   and `z = -18`