#### Answer

$(-2, 3, 5)$

#### 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:
$8x +y = -13$
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:
$ 13x +11y =7 $
Plugging $x =-1/8y -13/8$ into this equation, we obtain:
$9.375y -21.125 = 7 \\ y =3 $
Now, we plug this value into one of the equations that only has x and y in it to find:
$ x = -2$
Finally, we plug the values of x and y into the first equation listed in the book to find:
$ z = 5$