Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 4 - Section 4.2 - Solving Systems of Linear Equations in Three Variables - Exercise Set - Page 220: 26

Answer

$$(1/3,0,-2)$$

Work Step by Step

We add the first two equations to find: $$ 6x-4y=2$$ We add this to the third equation to obtain: $$ 3z=-6$$ $$z=-2$$ Next, we find x. We substitute z into the first two equations, and then we subtract the second from the first to find: $$-2y-4=-4$$ This gives: $$y=0$$ Lastly, we find x, substituting y and z into the original first equation: $$3x-3(0)+(-2)=-1$$ $$x=1/3$$ Thus, the solution is: $$(1/3,0,-2)$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.