Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 3 - Section 3.3 - Systems of Linear Equations in Three Variables - Exercise Set - Page 215: 15

Answer

The solution is $x=1 $, $ y=-5 $ and $z=-6 $

Work Step by Step

Given equations are $ x+y=-4 $ ... (1) $ y-z=1 $ ... (2) $ 2x+y+3z=-21 $ ... (3) Multiply equation (2) by $3 $. $ 3\cdot y - 3\cdot z= 3\cdot 1 $ $ 3y-3z=3 $ ... (4) Add equation (3) and (4). $ 2x+y+3z+3y-3z=-21+3 $ $ 2x+4y=-18 $ ... (5) Multiply equation (1) with $ -2$. $ (-2)\cdot x+(-2)\cdot y = -(-2)\cdot 4 $ $ -2x-2y=8 $ ... (6) add equation (5) and (6). $ 2x+4y-2x-2y=-18+8 $ $ 2y=-10 $ $ y=-5 $ Substitute into equation (1) $ x-5=-4 $ $ x=-4+5 $ $ x=1 $ Substitute the value of $y$ into equation (2). $ -5-z=1 $ $z=-5-1 $ $ z=-6 $
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