If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and second equations, then the second and third equations.

x + y – 3z = –8
2x + 2y + z = 12
3x + y – z = –2

x + y = –8
3x + y = –2

x + y = –8
5x + 3y = 14

7x + 7y = 28
3x + y = –2

7x + 7y = 28
5x + 3y = 10

Question

If you were to use the elimination method to solve the following system, choose the new system of equations that would result after the variable z is eliminated in the first and second equations, then the second and third equations.

x + y – 3z = –8
2x + 2y + z = 12
3x + y – z = –2

x + y = –8
3x + y = –2

x + y = –8
5x + 3y = 14

7x + 7y = 28
3x + y = –2

7x + 7y = 28
5x + 3y = 10

mathstudent55 · Answer

To eliminate z using the first and second equations, multiply the second equation by 3 and add to the first one.
To eliminate z using the second and third equations, just add them together.

anonymous · Answer

x + y - 3z = - 8
2x + 2y +z = 12
3x + y - z = - 2

----------------
x + y - 3z = - 8
2x + 2y +z = 12 ---> (3)2x + 2y + z = 12
-----------------
x + y - 3z = - 8
6x + 6y + 3z = 36 (result of multiplying by 3)
-----------------
7x + 7y = 10 (result from equation 1 and 2)

2x + 2y + z = 12
3x + y - z = - 2
----------------
5x + 3y = 10 (result from equation 2 and 3)

ANSWER : 7x + 7y = 10
...............5x + 3y = 10

hero · Answer

That isn't even one of the answer choices

mathstudent55 · Answer

you mean 7x + 7y = 28
and          5x + 3y = 10
which is D

anonymous · Answer

my bad...it is D. Sorry for messing up.