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: 28

Answer

$x=\frac{1}{3}$ $y=-\frac{1}{3}$ $z=\frac{2}{3}$

Work Step by Step

Express y from the first equation. $y-z=-1 \rightarrow y=-1+z $ $6x+2y+z=2$ $-x-y-3z=-2$ Substitute $ y=-1+z$ into the second and the third equation. $6x+2(-1+z)+z=2$ $-x-(-1+z)-3z=-2$ Simplify. $6x+3z=4$ $-x-4z=-3$ Multiply the second equation by 6. $-x-4z=-3 \rightarrow -6x-24z=-18$ Add the equations to eliminate x. $-21z=-14$ $z=\frac{2}{3}$ Substitute for z and solve for y and x. $y=-1+z \rightarrow y=-\frac{1}{3}$ $6x+3z=4 \rightarrow x=\frac{1}{3}$
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