# How do you solve the system x-6y+4z=-12, x+y-4z=12, and 2x+2y+5z=-15?

##### 1 Answer
Nov 10, 2016

Please see the explanation for steps leading to the answer:

$x = 0 , y = 0 , \mathmr{and} z = - 3$

#### Explanation:

Write the 3 equations as an augmented matrix:

[ (1,-6,4,|,-12), (1,1,-4,|,12), (2,2,5,|,-15) ]

Subtract row 1 from row 2:

[ (1,-6,4,|,-12), (0,7,-8,|,24), (2,2,5,|,-15) ]

Multiply row 1 by -2 and add to row 3:

[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,14,-3,|,9) ]

Multiply row 2 by -2 and add to row 3:

[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,0,13,|,-39) ]

Divide row 3 by 13:

[ (1,-6,4,|,-12), (0,7,-8,|,24), (0,0,1,|,-3) ]

Multiply row 3 by 8 and add to row 2:

[ (1,-6,4,|,-12), (0,7,0,|,0), (0,0,1,|,-3) ]

Multiply row 3 by -4 and add to row 1:

[ (1,-6,0,|,0), (0,7,0,|,0), (0,0,1,|,-3) ]

Divide row 2 by 7:

[ (1,-6,0,|,0), (0,1,0,|,0), (0,0,1,|,-3) ]

Multiply row 2 by 6 and add to row 1:

[ (1,0,0,|,0), (0,1,0,|,0), (0,0,1,|,-3) ]

$x = 0 , y = 0 , \mathmr{and} z = - 3$