Question

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Answer 1

I would use the elimination method. Eliminating z

equation (1) x 4 - equation (2)

8x + 4y + 4z = -28
- x - 3y + 4z = -14
-- --------------------
7x + 7y = -14 (4)

and then equation (1) x 3 + equation (3)

6x + 3x + 3z = -21
x - 2y - 3z = -11
------------------
7x + y = -32 (5)

now you have 2 equations in 2 unknowns

7x + 7y = -14 (4)
7x + y = -32 (5)

then equation (4) - equation (5) to eliminate x

6y = 18

y = 3

substitute into equation (5)

7x + 3 = -32

7x = -35

x = -5

substitute into equation 1 to find z

2(-5) + 3 + z = - 7

- 7 + z = -7

z = 0

then the solution is

x = -5, y = 3 and z = 0

Answer 2

i'm so lost...

Answer 3

There are several variables in this system that have a coefficient of 1.

Let's solve the 3rd equation for x:
x = 2y + 3z - 11

Substituting into the 1st and 2nd equations gives

2(2y+3z-11) + y + z = -7
2y + 3z - 11 - 3y + 4z = -14

Multiplying out all brackets in the 1st equation yields

4y + 6z - 22 + y + z = -7
2y + 3z - 11 - 3y + 4z = -14.

Simplifying both equations yields

5y + 7z = 15
-y + 7z = -3

Now, one variable has the same coefficient in both equations (namely z), so this set of 2-by-2 equations can be solved by elimination: Subtracting the 2nd equation from the 1st yields

6y = 18, so y = 3.
get it?

Answer 4

these are examples can you relate them to the problem you are
completing?

Answer 5

I understand now, I was just a little confused before.

Answer 6

okay good(: