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Algebra2guy
Oct 9, 2010, 08:38 AM
w+2x+2y+z= -2
w+3x-2y-z= -6
-2w-x+3y+3z= 6
w+4x+y-2z= -14

Unknown008
Oct 9, 2010, 08:56 AM
You might want to use the Gaussian elimination method to make it easier?

If not, you can use the first and second equation to eliminate y and z directly.
Then use the first and third equation to get another equation in w and x.

Take those two equations and solve simultaneously.

Now, you have the value of w and x, you can solve any to equations simultaneously.

Post what you get! :)

Algebra2guy
Oct 9, 2010, 12:44 PM
Still don't get it. :(

Unknown008
Oct 10, 2010, 12:10 AM
Hm...

Take 1st and 2nd equations:

w+2x+2y+z= -2
w+3x-2y-z= -6

Eliminate 2y and z by adding both equations:

w+2x+2y+z + (w+3x-2y-z) = -2 + (-6)

This becomes:

2w + 5x + 0 + 0 = -8
2w + 5x = -8... (A)

I think I misread the second equation... but anyway, this only makes it a longer.
Take 3rd and 4th (I pre-multiplied the 4th by 3 and the 3rd by 2)

-4w - 2x + 6y + 6z = 12
3w + 12x + 3y - 6z = -42

Add;

-w + 10x +9y = -30... (B)

Take 2nd and 3rd (I pre-multiplied the 2nd by 3)

3w + 9x - 6y - 3z = -18
- 2w - x + 3y + 3z = 6

Add;

w + 8x - 3y + 0 = -12
w + 8x - 3y = -12... (C)

Multiply C by 3

3w + 24x - 9y = -36... (D)

Add D and B:

2w + 34x = -66... (E)

Now, take A and E together and solve simultaneously;
2w + 5x = -8
2w + 34x = -66

Subtract;

29x = -58
x = -2

Can you find the rest now? :)

Algebra2guy
Oct 10, 2010, 06:45 PM
w=1
x= -2
y= -1
z=3

Unknown008
Oct 10, 2010, 10:23 PM
Great! Yes, those are the values of w, x, y and z. Good job :)