# What is the solution of the following system?: x + 9y + z = -12, x + y - 9z = 7, -12x + 4y + z = -4

Mar 24, 2018

$x = - \frac{151}{1016} , y = - \frac{1233}{1016} , z = - \frac{118}{127}$

#### Explanation:

We want to solve

{: (color(white)(aaa)x + 9y + z = -12), (color(white)(aaaaaa)x + y - 9z = 7) ,(-12x + 4y + z = -4):}}

We start by putting the system in echelon form using Gaussian elimination

1) Add $- 1$ lots of the 1st equation to the second

{: (color(white)(aaaaa)x + 9y + z = -12), (color(white)(aaaaaaa)-8y - 10z = 19) ,(color(white)(aa)-12x + 4y + z = -4):}}

2) Add $12$ lots of equation 1 to equation three

{: (color(white)(a)x + 9y + z = -12), (color(white)(aaa)-8y - 10z = 19) ,(112y + 13z= -148):}}

3) Add 14 lots of equation 2 to equation three

{: (x + 9y + z = -12), (color(white)(aa)-8y - 10z = 19) ,( color(white)(aaaaa)-127z= 118):}}

We have the system in echelon form, so we now back-substitute.

$z = - \frac{118}{127}$

$y = - \frac{1}{8} \left(19 + 10 \left(- \frac{118}{127}\right)\right) = - \frac{1233}{1016}$

$x = \left(- 12 - 9 \left(- \frac{1233}{1016}\right) - \left(- \frac{118}{127}\right)\right) = - \frac{151}{1016}$