ALGEBRA prob HELP!!!

[cobrown]

I need to solve for x, y AND z...this one is killin me...
22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14

 

[daon]

There is more than one way to solve this problem, is there any specific method you need to use?

 

[cobrown]

I HAVE BEEN using the elimination method for the other ones....

 

[tkhunny]

I need to solve for x, y AND z...this one is killin me...
22X + 5Y + 7Z = 12
10X + 3Y + 2Z = 5
9X + 2Y + 12Z = 14

Develop an algorithm lacking in panic. You don't have to get '1' or '0' with every move. Show a little patience and try to stay in the integers until you absolutely HAVE to do otherwise. DON'T try to do too much in one pass.

1) 22X + 5Y + 7Z = 12
2) 10X + 3Y + 2Z = 5
3) 9X + 2Y + 12Z = 14

Keep 2
2) - 3) Replaces 3)

1) 22X + 5Y + 7Z = 12
2) 10X + 3Y + 2Z = 5
3) X + Y - 10Z = -9

1) - Twice *2) Replaces 1)

1) 2X - 1Y + 3Z = 2
2) 10X + 3Y + 2Z = 5
3) X + Y - 10Z = -9

The coefficients are lots smaller and I haven't seen a fraction. That is good. Mix it up a bit.

Keep 3
2) less 5 times 1) Replaces 2

1) 2X - 1Y + 3Z = 2
2) 8Y -13z = -5
3) X + Y - 10Z = -9

1) less twice 3) Replaces 1)

1) - 3Y + 23Z = 20
2) 8Y -13z = -5
3) X + Y - 10Z = -9

I have the first column cleared out.
I still haven't seen a fraction.
I haven't used any values worse than the ones I started with.

What is 2) + three * 1)?