Help with solving these systems of equations!

Saaady

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Feb 1, 2016
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Stuck on the last 3 systems on my worksheet, if someone could give me more than just the answer but also teach me how, it would be much appreciated!
"Determine the nature of the intersection if it exists) between the following sets of planes. If it is a line, find the equation of it. If it is a point, determine it.
a.) x + y - z + 3 = 0
-4x + y + 4z - 7 = 0
-2x + 3y + 2z - 2 = 0


b.) 2x - 3y + 4z - 1 = 0

x - y - z + 1 = 0
-x + 2y -z + 2 = 0


c.) 2x - y + 2z + 1 = 0

-4x + 2y -4z -2 = 0
6x - 3y + 6x + 1 = 0
 
Stuck on the last 3 systems on my worksheet, if someone could give me more than just the answer but also teach me how, it would be much appreciated!
"Determine the nature of the intersection if it exists) between the following sets of planes. If it is a line, find the equation of it. If it is a point, determine it.
a.) x + y - z + 3 = 0
-4x + y + 4z - 7 = 0
-2x + 3y + 2z - 2 = 0


b.) 2x - 3y + 4z - 1 = 0

x - y - z + 1 = 0
-x + 2y -z + 2 = 0


c.) 2x - y + 2z + 1 = 0

-4x + 2y -4z -2 = 0
6x - 3y + 6x + 1 = 0
What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting
 
What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

Alright so I've managed to solve for the first two systems, however I'm stumped on the third one. I'll do my best to write it out here:


2x-y+2x =-1
-4x+2y-4z =2
6x-3y+6z =-1


I managed to bring this down to:


2x-y+2z =-1
0x+0y+0z = 0
6x-3y+6z =-1


This would imply that the system has infinitely many solution, however I kept going


2x-y+2z=1
0x+0y+0z =0
0x-0y-0z = 2


This final step is equal to a number, so that confuses me as that means there are no solutions.


Which one is it, thanks for any help!
 
Alright so I've managed to solve for the first two systems, however I'm stumped on the third one. I'll do my best to write it out here:


2x-y+2x =-1
-4x+2y-4z =2
6x-3y+6z =-1


I managed to bring this down to:


2x-y+2z =-1
0x+0y+0z = 0
6x-3y+6z =-1


This would imply that the system has infinitely many solution, however I kept going


2x-y+2z=1
0x+0y+0z =0
0x-0y-0z = 2


This final step is equal to a number, so that confuses me as that means there are no solutions.


Which one is it, thanks for any help!
Well I think you have the answer: "... that means there are no solutions." That 0x+0y+0z=0 just means that what ever satisfies (1) will satisfy (2)
 
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