#### Answer

Pivots: 2, 1, 8; Solution: (2, 1, 1)

#### Work Step by Step

Start by converting your system of equations to an augmented matrix:
[2 3 1 | 8
4 7 5 | 20
0 -2 2 | 0]
Using Gaussian-elimination, create a triangular matrix:
$R_{2}$-(2$\times$$R_{1}$)=$R_{2'}$
[4 7 5 | 20]-[4 6 2 | 16] = [0 1 3 | 4]
Our new augmentred matrix is:
[2 3 1 | 8
0 1 3 | 4
0 -2 2 | 0]
$R_{3}$-((-2)$\times$$R_{2}$)=$R_{3'}$
[0 -2 2 | 0]-[0 -2 -6 | -8]=[0 0 8 | 8]
We now have a triangular matrix,
[2 3 1 | 8
0 1 3 | 4
0 0 8 | 8
We can convert this matrix back to a system of equations:
2x+3y+z=8
y+3z=4
8z=8
Now that we have a triangular system of equations, we can identify our pivot points (the first non-zero coefficient in the individual equations) as 2, 1, and 8.
We can now also solve this system using back-substitution:
8z=8
z=1
y+3(1)=4
y+3=4
y=1
2x+3(1)+(1)=8
2x+3+1=8
2x+4=8
2x=4
x=2
Our solution is: (2, 1, 1)