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Spot the mistakes/error in solving the linear system , and then say something about it. Given : 2x + 5y – z = 27 x + y + z = 6 2у + 52 -4 Eliminate x from 2nd & 3rd equation (by addition) 2а + 5у — 2 %3D 27 | 2 (x + y + z = 6) Зу — З2 — 15 | And we get : x + y + z = 6 2у + 52 -4 Зу — 32 = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 62 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 %3D

Spot the mistakes/error in solving the linear system , and then say something about it. Given : 2x + 5y – z = 27 x + y + z = 6 2у + 52 -4 Eliminate x from 2nd & 3rd equation (by addition) 2а + 5у — 2 %3D 27 | 2 (x + y + z = 6) Зу — З2 — 15 | And we get : x + y + z = 6 2у + 52 -4 Зу — 32 = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 62 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 %3D

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Spot the mistakes/error in solving the linear system , and then say something about it. Given : 2а + 5y — 2 3D 27 x + y + z = 6 2у + 52 -4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y+ z = 6) Зу — З2 — 15 | And we get : x + y + z = 6 2у + 52 -4 Зу — 32 = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 62 30 бу + 152: –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 %3D
Transcribed Image Text:Spot the mistakes/error in solving the linear system , and then say something about it. Given : 2а + 5y — 2 3D 27 x + y + z = 6 2у + 52 -4 Eliminate x from 2nd & 3rd equation (by addition) 2x + 5y – z = 27 2 (x + y+ z = 6) Зу — З2 — 15 | And we get : x + y + z = 6 2у + 52 -4 Зу — 32 = 15 Eliminate y from 3rd and 4th equation (by subtraction) бу — 62 30 бу + 152: –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 %3D
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