#### Answer

(0, -2, 3)

#### Work Step by Step

In order to solve systems of three linear equations, we multiply the first and second equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by 2 and the second equation by 3 and add to obtain:
$14x -y = 2$
We now multiply the first and third equation by values that will cancel out a variable when they are added. Thus, we multiply the first equation by one and the third equation by -3 and add to obtain:
$ -5x -11y = 22 $
Plugging $y = 14x -2 $ into this equation, we obtain:
$-159x +22 =22 \\ x=0$
Now, we plug this value into one of the equations that only has x and y in them to find:
$ y = -2$
Finally, we plug the values of x and y into the first equation listed in the book to find:
$ z = 3$