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: 3

Answer

$(-2,5,1)$

Work Step by Step

$x+y = 3$ Equation $(1)$ $2y=10$ Equation $(2)$ $3x+2y-3z = 1$ Equation $(3)$ From Equation $(2)$, we get $2y=10$ $y = 5$ Substituting $y$ value in Equation $(1)$ $x+y = 3$ $x+5 = 3$ $x = 3-5$ $x = -2$ Substituting $x$ and $y$ values in Equation $(3)$ $3x+2y-3z = 1$ $3(-2)+2(5)-3z = 1$ $-6+10-3z = 1$ $4-3z = 1$ $-3z = 1-4$ $-3z = -3$ $z = 1$ $x = -2$ $y=5$ $z= 1$ Solution is $(-2,5,1)$
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