# System of linear equation question: Intersection of three equations

1. Mar 10, 2009

### zeion

1. The problem statement, all variables and given/known data

Solve the system of the linear equations and interpret your solution geometrically:
2x + y + 2z - 4 = 0 [1]
x - y - z - 2 = 0 [2]
x + 2y -6z - 12 = 0 [3]

3. The attempt at a solution

I've tried to eliminate the y variable:
[1] + [2]
3x + z - 6 = 0 [4]

[1]x(-2) + [3]
-3x -10z + 4 = 0 [5]

Now solve for z
[4] + [5]
-9z - 2 = 0
z = -2/9

Is this correct so far?
I'm not sure what to do now, do I do this whole process again and solve for another variable? Or can I sub the z into [4] or [5]?

2. Mar 10, 2009

### Staff: Mentor

Yes, substitute your z value into equations 4 and 5, to solve for x. Now you know z and x, so substitute them into any of your first 3 equations.

To check, make sure that all three of your starting equations are true statements when you replace x, y, and z with the values you have found. If all three equations are satisified, you're golden.

3. Mar 10, 2009

### zeion

Sub z into [4]
3x + (-2/9) - 6 = 0
x = 56/27

Sub z into [5]
-3x -10(-2/9) + 4
x = 56/27

So x = 56/27
Now when I sub in the values x = 56/27 and z = -2/9 into [1] and [2] I get y = 8/27,
but when I sub it into [3] I get 116/27.. Did I do something wrong?

4. Mar 11, 2009

### Staff: Mentor

Yes.
That + 4 should be -4.
I get z = -10/9

5. Mar 11, 2009

### zeion

I don't really get it.

[1] x (-2) = (-2)(2x + y + z - 4) = -4x -2y -4z +16
+ [3]
-4x -2y - 4z +16 + x +2y -6z -12 = 0
-3x -10z + 4 = 0

16 - 12 = 4

6. Mar 11, 2009

### zeion

I've just tried to eliminate x first instead of y.. and I get totally different values for z -_-

7. Mar 11, 2009

### Staff: Mentor

-2 * -4 = 8, not 16

8. Mar 11, 2009

### zeion

Oh geez. I must be blind. I have the correct answer now, thanks for your help.