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: 4

Answer

Solution is $(1,2,5)$

Work Step by Step

$5x = 5$ Equation $(1)$ $2x+y = 4$ Equation $(2)$ $3x+y-4z = -15$ Equation $(3)$ From Equation $(1)$, we get $5x = 5$ $x = 1$ Substituting $x$ value in Equation $(2)$ $2x+y = 4$ $2(1)+y = 4$ $2+y = 4$ $y = 4-2$ $y = 2$ Substituting $x$ and $y$ values in Equation $(3)$ $3x+y-4z = -15$ $3(1)+2-4z = -15$ $3+2-4z = -15$ $5-4z = -15$ $-4z = -15-5$ $-4z = -20$ $z= 5$ $x = 1$ $y = 2$ $z = 5$ Solution is $(1,2,5)$
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