Solve the system: 2x+y-2z=-1; 3x-3y-z=5; x-2y+3z=6

Question

2x + y - 2z = -1, 3x - 3y - z = 5 and x 2y + 3z = 6.

Answer 1

Elimination method :

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

3 x - 3 y  -  z = 5  ----> (2)

x - 2 y + 3 z = 6 -----> (3)

Multiply each side by - 2  in eq ( 2)

We get the equation - 6 x + 6 y + 2 z  = - 10  ---------> ( 4 )

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

2 x + y - 2 z = - 1

- 6 x  + 6 y  + 2 z  =  -10

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- 4 x + 7 y = - 11 ---> ( 5 )

Multiply each side by 3  in eq ( 2)

We get the equation 9 x - 9 y - 3z  =  15 --------> ( 6 )

To eliminate the z  value add (6)&(3).

9 x - 9 y - 3z  =  15

x - 2 y + 3 z = 6

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10 x -11y = 21   ---> (7)

Multiply each side by  10  in eq (5 ) 

We get the equation - 40 x + 70 y   = - 110   -------> ( 8 )

Multiply each side by  4 in eq (7 ) 

We get the equation  40 x -44 y   = 84   -------> ( 9 )

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

- 40 x + 70 y  = - 110

40 x - 44 y   = 84

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26 y  = - 26

y = -1

Substitute the y value in eq ( 5 ).

- 4 x + 7 ( - 1 ) = -11

- 4x = -11 + 7

- 4x  =  - 4

x  =  1

Substitute the values x & y valuesin eq ( 1 )

2 ( 1 ) + ( - 1 ) - 2z = - 1

2 -1 -  2 z  = -1

2 = 2 z

1 = z

Therefore  x = 1, y = -1 , z = 1.