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: 27

Answer

$(12, 6, 4)$

Work Step by Step

$\frac{3}{4}x-\frac{1}{3}y+\frac{1}{2}z=9$ Equation $(1)$ $\frac{1}{6}x+\frac{1}{3}y-\frac{1}{2}z=2$ Equation $(2)$ $\frac{1}{2}x-y+\frac{1}{2}z=2$ Equation $(3)$ Equation $(1)$ + Equation $(2)$ $(\frac{3}{4}x-\frac{1}{3}y+\frac{1}{2}z) + (\frac{1}{6}x+\frac{1}{3}y-\frac{1}{2}z) = 9+2$ $\frac{11}{12}x=11$ $x=12$ Equation $(2)$ + Equation $(3)$ $(\frac{1}{6}x+\frac{1}{3}y-\frac{1}{2}z) + (\frac{1}{2}x-y+\frac{1}{2}z)=2+2$ $\frac{4}{6}x-\frac{2}{3}y=4$ Equation $(4)$ Substitute $x=12$ into Equation $(4)$ to get $y$. $(\frac{4}{6}\times12)-\frac{2}{3}y=4$ $8-\frac{2}{3}y=4$ $-\frac{2}{3}y=-4$ $y=6$ Substitute known values for $x$ and $y$ into Equation $(3)$ to get $z$. $(\frac{1}{2}\times12) - 6 + \frac{1}{2}z=2$ $6-6+\frac{1}{2}z=2$ $\frac{1}{2}z=2$ $z=4$ $(12, 6, 4)$ satisfies all of the given equations. Solution is $(12, 6, 4)$.
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