Question

Solve the system using the elimination method.  −8x −4y+2z = −26, −2x+ 4y+2z = 10 and  −6x −8y− 5z = −41.

Answer 1

Elimination method :

Given equations are - 8 x -4 y + 2 z = - 26 -------> (1)

- 2 x + 4 y + 2 z = 10  --------> (2)

- 6 x - 8 y -  5 z = - 41 -------> (3)

Multiply each side by - 4 in eq ( 2)

We get the equation  8 x - 16 y -  8 z  = - 40---------> ( 4 )

To eliminate the x  value add  (1 )&(4).

 - 8 x - 4 y +  2 z = - 26

8 x - 16 y -  8 z  = - 40-

_______________

- 20 y- 6 z  =  - 66

- 10 y - 3 z  =  - 33  -----> (5)

Multiply each side by - 6 in eq (2 )

We get the equation 12 x - 24 y -12 z  = - 60 --------> (6)

Multiply each side by 2 in eq (3)

We get the equation - 12 x - 16 y - 10 z  = - 82 --------> ( 7 )

To eliminate the x  value add (6)&(7).

12 x - 24 y -12 z  = - 60

- 12 x - 16 y - 10 z  = - 82

_________________

- 40 y - 22 z = - 142

- 20 y - 11 z = - 71 --------> ( 8)

Multiply each side by 5- 10 in eq (5 ) 

We get the equation 20 y + 6 z = 66 -------> (9)

To eliminate the y  value add (8)&(9).

 - 20 y - 11 z = - 71

20 y + 6 z = 66

_____________

- 5 z = - 5

z = 1

Substitute the z value in eq ( 5 ).

- 10 y - 3 z  =  - 33

-10 y - 3(1) = - 33

- 10 y - 3 = - 33

- 10 y  = - 33 + 3

- 10 y  = - 30

y = 3 

Substitute the values z  & y values in eq ( 1 )

- 8 x - 4 ( 3)+ 2 (1) = - 26

- 8 x - 12 + 2 =  - 26

- 8 x - 10=  - 26

- 8 x = - 26 + 10

- 8 x = - 16

x = 2.

Solution of the system of  the equations is x = 2 , y  = 3 and  z = 1.