Algebra 2 (1st Edition)

Published by McDougal Littell
ISBN 10: 0618595414
ISBN 13: 978-0-61859-541-9

Chapter 3 Linear Systems and Matrices - 3.4 Solve Systems of Linear Equations in Three Variables - 3.4 Exercises - Skill Practice - Page 183: 33

Answer

$(2,6,-5)$ is a solution of the system.

Work Step by Step

Use elimination for the first and second equations: $x + y + z = 3$ $-x + 5y + z = 23$ ________________________ $6y+2z=26$ (1) Then continue to use elimination for the second and third equations: $3x - 4y + 2z=-28$ $-x + 5y + z = 23$ Multiply both sides of the first equation by $3$: $3x - 4y + 2z=-28$ $-3x + 15y + 3z = 69$ _______________________ $11y+5z=41$ (2) From (1) and (2): $6y+2z=26$ $11y+5z=41$ Multiply both sides of the first equation by $-5$ and the second equation by $2$: $-30y-10z=-130$ $22y+10z=82$ _________________________ $-8y=-48$ $y=6$ Solve for z: $11(6)+5z=41$ $5z=-25$ $z=-5$ Solve for x: $x+6+(-5)=3$ $x=2$ $(2,6,-5)$ is a solution of the system.
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