Question

Need Help ASAP!!!

Solve the system by elimination.

-2x+2y+3z=0
-2x-y+z=-3
2x+3y+3z=5

Answer 1

Steps:

isolate " x " for -2x + 2y + 3z = 0 ; x = - ( - 3z - 2y ) / 2

Subsititute : x = - ( - 3z - 2y ) / 2

- 2 * ( - ( - 3z - 2y ) / 2 ) - y + z = - 3

2 * ( - ( - 3z - 2y ) / 2 ) + 3y + 3z =5

isolate " z " for : - 2 ( - ( - 3z - 2y ) / 2 ) - y+ z = - 3 ; z = - ( - 3 + 3y / 2 )

subsititute z = - ( - 3 + 3y / 2 ) :

2* ( - ( -3* ( - ( - 3 +3y / 2 ) - 2y / 2 ) + 3y + 3 ( - ( - 3 + 3y / 2 ) = 5

isolate y for 2 ( -( - 3 ) - 3 + 3y / 2 ) - 2y / 2 ) + 3y + 3 ( - ( - 3 + 3y / 2 ) = 5

y = 1

for z = - ( - 3+ 3y ) / 2

subsititute y = 1

z = - ( - 3 + 3 * 1 ) / 2

z = 0

For x = - ( - 3z - 2y ) / 2

subsititute y = 1

subsititute z = 0

x = - ( - 3 *0 - 2 * 1 ) / 2

x = 1

x = 1 , y = 1 , z = 0

Answer 2

The answer to your question is:

x = 1

y = 1

z = 0

Explanation:

-2x + 2y + 3z = 0 (1)

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

2x + 3y + 3z = 5 (3)

Solve (1) and (2)

Multiply 2 by 2

-2x + 2y + 3z = 0

-4x -2y + 2z = -6

-6x + 5 z = -6 (4)

Solve (2) and (3)

Multiply 2 by 3

-6x - 3y + 3z = -9

2x + 3y + 3z = 5

-4x + 6z = -4 (5)

Solve (4) and (5)

Multiply (4) by 2 and (5) by -3

-12x + 10 z = -12

12x - 18z = 12

-6z = 0

z = 0

Then

-4x + 6(0) = -4

-4x = -4

x = -4/-4

x = 1

Finally

-2(1) - y + (0) = -3

-2 - y = -3

-y = -3 + 2

y = 1