Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 10 - Section 10.2 - Systems of Linear Equations in Several Variables - 10.2 Exercises - Page 697: 27

Answer

$x=\frac{1}{4}$ $y=\frac{1}{2}$ $z=-\frac{1}{2}$

Work Step by Step

Express y from the first equation. $2y+4z=-1 \rightarrow y=-\frac{1}{2}-2z$ $-2x+y+2z=-1$ $4x-2y=0$ Substitute $ y=-\frac{1}{2}-2z$ into the second and the third equation. $-2x+(-\frac{1}{2}-2z)+2z=-1$ $4x-2(-\frac{1}{2}-2z)=0$ Simplify. $-2x=-\frac{1}{2}$ $4x+1+4z=0$ From $-2x=-\frac{1}{2}$ we can express $ x=\frac{1}{4}$ Substitute $ x=\frac{1}{4}$ into $4x+1+4z=0$ to get the value of z. $4(\frac{1}{4})+1+4z=0 \rightarrow z=-\frac{1}{2}$ Substitute $z=-\frac{1}{2}$ into $y=-\frac{1}{2}-2z$ to get the value of y. $y=-\frac{1}{2}-2(-\frac{1}{2}) \rightarrow y=\frac{1}{2}$
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