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

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

anonymous · Answer

@johnweldon1993 can you help??

anonymous · Answer

A x + y = –8
3x + y = –2

 B x + y = –8
5x + 3y = 14

C 7x + 7y = 28
3x + y = –2

D 7x + 7y = 28
5x + 3y = 10

johnweldon1993 · Answer

hmm okay....well in order to eliminate it in the first and second equations....we need to multiply that 2nd equation by 3

2x + 2y + z = 12

becomes

6x + 6y + 3z = 36

so now we have (the first 2 are)

x + y – 3z = –8
6x + 6y + 3z = 36

When we combine them we get

7x + 7y = 28

So now to eliminate the z in the third equation from the second equation we just combine them

2x + 2y + z = 12
3x + y – z = –2

After combining (and by that I mean adding them together) we have

5x + 3y = 10

so those 2 again are

7x + 7y = 28
5x + 3y = 10

Look familiar again? :)

anonymous · Answer

yess it does thankyou so much! :)

johnweldon1993 · Answer

No problem! Time to go eat just like primeralph lol :)

anonymous · Answer

lol okaay! idk what im gonna do now...