# Chapter 11 - Additional Topics - 11.2 - 3 x 3 Systems of Equations - Problem Set 11.2 - Page 484: 9

(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: $z = -4$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.