#### Answer

True

#### 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:
$x -5y = -20$
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 one and add to obtain:
$ 3x + 3y = 12 $
Plugging $x = 5y -20$ into this equation, we obtain:
$18y - 60 = 12 \\ y =4$
Now, we plug this value into one of the equations that only has x and y in them to find:
$ x = 0$
Finally, we plug the values of x and y into the first equation listed in the book to find:
$ z = 8$