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: 11

Answer

$(0,0,0)$

Work Step by Step

$x+5z = 0$ Equation $(1)$ $5x+y = 0$ Equation $(2)$ $y-3z = 0 $ Equation $(3)$ Equation $(2)$ $-$ Equation $(3)$ $5x+y -(y-3z) = 0 $ $5x+y-y+3z = 0$ $5x+3z = 0$ Equation $(4)$ Equation$(1) \times 5 $ $ - $ Equation $(4)$ $5(x+5z)-(5x+3z)=0$ $5x+25z -5x-3z =0$ $22z = 0$ $z = 0$ Substituting $z$ value in Equation$(1)$ $x+5z = 0$ $x+5(0) = 0$ $x = 0$ Substituting $x$ value in Equation$(2)$ $5x+y = 0$ $5(0)+y = 0$ $y = 0$ $x = 0$ $y = 0$ $z = 0$ Solution $(0,0,0)$
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.