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


Choices:



A.x + y = –8
3x + y = –2

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

C.x + y = –8
5x + 3y = 14

D.7x + 7y = 28
3x + y = –2

Answer 1

to eliminate from the 1st and 2nd you have to multiply the 2nd by 3. Then you have
x+y-3z=-8
6x+6y+3z=36 and the result is 7x+7y=28
to eliminate from the 2nd and 3rd you leave it as it is and the result is 5x+3y=10
Therefore the answer is B

Answer 2

ok thank u :)

Answer 3

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