gaussian elimination

x+Y+Z=-2
X-y+5z+22
5x+y+z=-22

plug your equations in at

http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

to see all the steps of the solution.

Gaussian elimination is a method used to solve systems of linear equations. It involves transforming the system into echelon form, which simplifies the equations and leads to finding the solution.

To solve the system of equations you provided using Gaussian elimination, follow these steps:

Step 1: Write the augmented matrix:
Start by writing the coefficients of the variables and the constants on the right-hand side as an augmented matrix:

[1 1 1 | -2]
[1 -1 5 | 22]
[5 1 1 | -22]

Step 2: Perform row operations:
Perform row operations on the matrix to simplify it, while keeping the system of equations equivalent.

- R2 - R1 -> R2: Subtract the first row from the second row.
- -5R1 + R3 -> R3: Multiply the first row by -5 and add it to the third row.

The updated matrix becomes:

[1 1 1 | -2]
[0 -2 4 | 24]
[0 -4 -4 | 8]

Step 3: Continue with row operations:
Perform further row operations to simplify the matrix.

- -2R2 -> R2: Multiply the second row by -2.
- -4R2 + R3 -> R3: Multiply the second row by -4 and add it to the third row.

The updated matrix becomes:

[1 1 1 | -2]
[0 4 -8 | -48]
[0 0 12 | -40]

Step 4: Bring the matrix into row echelon form:
To bring the matrix into row echelon form, divide each row by its leading coefficient:

- (1/4)R2 -> R2: Divide the second row by 4.
- (1/12)R3 -> R3: Divide the third row by 12.

The updated matrix becomes:

[1 1 1 | -2]
[0 1 -2 | -12]
[0 0 1 | -10/3]

Step 5: Back-substitution:
Starting from the bottom row, substitute the values into the equations to solve for the variables.
We know that Z = -10/3.

Substitute Z into the second equation:
1 - 2Y = -12
-2Y = -13
Y = 13/2

Then substitute Y and Z into the first equation:
X + 13/2 - 10/3 = -2
(6X + 39 - 20)/6 = -2
6X + 19 = -12
6X = -31
X = -31/6

Therefore, the solution to the system of equations is X = -31/6, Y = 13/2, and Z = -10/3.