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3 Solve the following system of linear equations in 3 variables. Enter your solution as: (x,y,z)=L 3 - 2x + y + 4z = 5 x + 3y - z = 2 4x + y - 6z = 11 Show Your Work

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The given system of equations is expressed as

- 2x + y + 4z = 5 equation 1

x + 3y - z = 2 equation 2

4x + y - 6z = 11 equation 3

From equation 2, z = x + 3y - 2

Substituting z = x + 3y - 2 into equation 1 and 3, it becomes

For equation 1,

- 2x + y + 4(x + 3y - 2) = 5

- 2x + y + 4x + 12y - 8 = 5

- 2x + 4x + y + 12y = 5 + 8

2x + 13y = 13 equation 4

For equation 3,

4x + y - 6(x + 3y - 2) = 11

4x + y - 6x - 18y + 12 = 11

4x - 6x + y - 18y + 12 = 11

- 2x - 17y = 11 - 12

- 2x - 17y = - 1 equation 5

We would eliminate x from equation 4 and 5 by adding both equations. It becomes

2x + - 2x + 13y + - 17y = 13 + - 1

2x - 2x + 13y - 17y = 13 - 1

- 4y = 12

y = 12/- 4

y = -3

Substituting y = - 3 into equation 5, it becomes

- 2x - 17*-3 = -1

- 2x + 51 = - 1

- 2x = - 1 - 51

- 2x = - 52

x = - 52/- 2

x = 26

Substituting x = 26, y = - 3 into z = x + 3y - 2, it becomes

z = 26 + 3*-3 - 2

z = 26 - 9 - 2

z = 26 - 11

z = 15

The solution is

[26, - 3, 15)

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