Help!! I've been working on this problem forever?

Question

I'm suppose to sole these systems of equations: (3 Variables)

x+9y+z=20
x+10-2z=18
3x+27y+2z=58

My teacher gave us the answers to make sure we were doing it right and the answer is { -18, 4, 2}

Answer 1

x + 9y + z = 20
x + 10y - 2z = 18
3x + 27y + 2z = 58

x + 10y - 2z = 18
3x + 27y + 2z = 58

the 2z's cancel out

4x + 37y = 76
37y = -4x + 76
y = (-4/37)x + (76/37)

x + 9((-4/37)x + (76/37)) + z = 20
x + (-36/37)x + (684/37) + z = 20
37x - 36x + 684 + 37z = 740
x + 37z = 56
37z = -x + 56
z = (-1/37)x + (56/37)

3x + 27((-4/37)x + (76/37)) + 2((-1/37)x + (56/37)) = 58
3x - (108/37)x + (2052/37) - (2/37)x + (112/37) = 58
111x - 108x + 2052 - 2x + 112 = 2146
x + 2164 = 2146
x = -18

y = (-4/37)(-18) + (76/37)
y = (72/37) + (76/37)
y = (72 + 76)/37
y = (148/37)
y = 4

z = (-1/37)(-18) + (56/37)
z = (18/37) + (56/37)
z = (18 + 56)/37
z = (74/37)
z = 2

x = -18
y = 4
z = 2

Answer 2

I think you've made a typo (or perhaps your teacher did). The second line should read:
x+10y-2z=18
That should help you get the values you need... otherwise read on:
I'm going to call the first line A, the second line B, and the third line C because it's easier than typing it all out a bunch of times.
The key to doing this problem is eliminating variables. The way to do this is to add the lines (or multiples of the lines) together. You can do this because it is equivalent to adding a number to both sides of an equation. So, if you start with A-B, you get:
3z-y=2. Solve for y and you get:
y=3z-2.
You've solved for y in terms of z. Fortuantely, your teacher (or whomever designed this problem) has a built in way of solving for z: Take 3A-C, which gives you:
z=2.
If you plug this back into your equation with just y and z you get:
y=4. If you plug these value back into any of the original equations you get
x=-18.

Answer 3

x = -18
y = 4
z = 2

-18+9(4)+2=20

3(-18)+27(4)+2(2)=58

These are two of them filled out but the middle one doesn't seem to work with those answers