#### Answer

(3,-2,-4)

#### 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 3 and the second equation by 1 and add to obtain:
$14x -8y = 58 $
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 -2 and the third equation by one and add to obtain:
$ -5x + 2y = -19 $
Plugging $x = 8/14y + 58/14$ into this equation, we obtain:
$-.857y = 1.71 \\ y= -2$
Now, we plug this value into one of the equations that only has x and y in them to find:
$ x = 3$
Finally, we plug the values of x and y into the first equation listed in the book to find:
$ z = -4$