solve the system of linear equations

Question

 -x-y+3=-1,  -x+y-3=3 and  x-y-z=5.

Answer 1

Elimination method :

The system of linear equations are

- x - y + 3 = - 1      → (1)

- x + y - 3 = 3        → (2)

    x - y - z = 5       → (3)

Since the coefficient of the x - term in two equations i.e, (1) & (2) is same, i can eliminate these terms by subtracting the equations.

Write the equations in column form and subtract to eliminate variable x.

    - x - y + 3 = - 1

(-)  - x + y - 3 = 3

 _____________________

- 2y + 6 = - 4

- 2y = - 4 - 6 = - 10

⇒ y = 10/2 = 5.

Substitute the value of y = 5 in equation (1) and solve for x.

- x - 5 + 3 = - 1

- x - 2 = - 1

- x = - 1 + 2 = 1

⇒ x = - 1.

Substitute the values of x = - 1 and y = 5 in equqtion (3) and solve for z.

(- 1) - 5 - z = 5

- 1 - 5 - z = 5

- 6 - z = 5

z = - 6 - 5

⇒ z = - 11.

Solution of the system is (x, y, z ) = (- 1, 5, - 11).

Answer 2

Elimination method :

Given equations are - x - y +3 = - 1 ---> (1)

- x + y - 3 = 3 ------> (2)

x -  y -  z = 5 -----> (3)

To eliminate the y  value add  (1 )&(2).

- x - y + 3 = - 1

- x + y - 3 = 3 

________________

- 2 x  = 2

x = - 1

To eliminate the x and y  values add  (2)&(3)

- x + y - 3 = 3

x -  y -  z = 5

__________

- 3  -  z  =  8

-  z  = 8 + 3

-  z  = 11

z = - 11  

Therefore z = - 11.

Substitute the  x  & z values in eq ( 1 )

x -  y -  z = 5

- 1  -  y - (- 11 ) = 5

- 1 - y + 11 =  5

- y + 10  = 5

- y  =  5 - 10

- y = - 5

y = 5.

Solution of the system of  the equations is x = - 1 , y = 5   and  z = - 11.