Spot the mistakes or error in solving the linear system, and then say something about it. Given :2х + 5у — z %3D27x + y + z = 62у + 52 %3D —4Eliminate x from 2nd & 3rd equation (by addition)2x + 5y – z = 272 (x + y + z = 6)Зу — 32 %3D 15And we get :x + y + z = 62y + 5z = -4Зу — 3z= 15Eliminate y from 3rd and 4th equation (by subtraction)бу — 6z — 30-6y + 15z = -12-21z =42Then : z = -2And we end up with:X +y + z =62y + 5z = 4z= -2By back-substituting:X = 5; y= 3 and z = -2

Given : System of linear equation : 2x + 5y - z = 27 x + y + z = 6 2y + 5z = -4

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 for College Students

10th Edition

ISBN:9781285195780

Author:Jerome E. Kaufmann, Karen L. Schwitters

Publisher:Jerome E. Kaufmann, Karen L. Schwitters

Chapter13: Conic Sections

Section13.1: Circles

Problem 48PS

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