solve a system of linear equations with three variables:

Question

x+y-3z=1

2x-y+z=9

3x+y-4z=8

Answer 1

x+y-3z = 1------> (1)

Multiple to each side by negitive 2,3.

-2x-2y+6z = -2-------> (2)

-3x-3y+9z = -3------> (3)

2x-y+z = 9 ------> (4)

3x+y-4z = 8 -------> (5)

To elminate x value add equations (4) and (2).

2x-y+z = 9

-2x-2y+6z = -2

____________

-3y+7z = 7-----> (6)

To elminate x value add equations (5) and (3).

-3x-3y+9z = -3

3x+y-4z = 8

_____________

-2y+5z = 5 -------> (7)

Multiple to each side of eqution(7) by negitive 3.

6y-15z = -15 -------> (8)

Multiple to each side of eqution (6) by3.

-6y+14z = 14----> (9)

To eliminate y value, add (8)and (9).

6y-15z = -15

-6y+14z = 14

___________

-z = -1

⇒ z = 1

-2y+5*1 = 5

-2y = 5-5

-2y = 0

⇒ y = 0

Substitute z,y in (5).

3x+0-4*1 = 8

3x-4 = 8

Add 4 to each side.

3x-4+4 = 8+4

3x = 12

Divide by 3 to each side.

3x/3 = 12/3

⇒ x = 4

Solution x = 4, y = 0, z = 1.