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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8daa39a-8c6c-4637-ab0b-0391e5544479%2F67fd3246-e539-4867-bf2e-14a49f6e233a%2Fu0mspwg_processed.jpeg&w=3840&q=75)
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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