#### Answer

(5, 2, -1)

#### 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 1 and add to obtain:
$x + 5y =15$
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 3 and the third equation by 1 and add to obtain:
$ 5x - 3y = 19 $
Plugging $x = -5y +15$ into this equation, we obtain:
$-28y + 75 = 19 \\ y =2 $
Now, we plug this value into one of the equations that only has x and y in them to find:
$ x = 5$
Finally, we plug the values of x and y into the first equation listed in the book to find:
$ z = -1$