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Given : 2х + 5у — z %3D27 x + y + z = 6 2y + 5z = -4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y + z = 6) 3y - 3z = 15 And we get : x + y + z = 6 2y + 5z = -4 Зу — 3z = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 6z 3D 30 - 6y + 15z = -12 -21z = 42 Then : z = -2 And we end up with: X +y + z =6 2y + 5z = 4 z= -2 By back-substituting: X = 5; y= 3 and z = -2

Given : 2х + 5у — z %3D27 x + y + z = 6 2y + 5z = -4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y + z = 6) 3y - 3z = 15 And we get : x + y + z = 6 2y + 5z = -4 Зу — 3z = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 6z 3D 30 - 6y + 15z = -12 -21z = 42 Then : z = -2 And we end up with: X +y + z =6 2y + 5z = 4 z= -2 By back-substituting: X = 5; y= 3 and z = -2

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Spot the mistakes or error in solving the linear system, and then say something about it.
Given : 2х + 5у — z %3D27 x + y + z = 6 2у + 52 %3D —4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y + z = 6) Зу — 32 %3D 15 And we get : x + y + z = 6 2y + 5z = -4 Зу — 3z = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 6z — 30 - 6y + 15z = -12 -21z = 42 Then : z = -2 And we end up with: X +y + z =6 2y + 5z = 4 z= -2 By back-substituting: X = 5; y= 3 and z = -2
Transcribed Image Text:Given : 2х + 5у — z %3D27 x + y + z = 6 2у + 52 %3D —4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y + z = 6) Зу — 32 %3D 15 And we get : x + y + z = 6 2y + 5z = -4 Зу — 3z = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 6z — 30 - 6y + 15z = -12 -21z = 42 Then : z = -2 And we end up with: X +y + z =6 2y + 5z = 4 z= -2 By back-substituting: X = 5; y= 3 and z = -2
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