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February 1st, 2016, 07:25 AM  #1 
Newbie Joined: Feb 2016 From: Canada Posts: 1 Thanks: 0  Help with solving these systems of equations?
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 Last edited by skipjack; February 1st, 2016 at 10:30 AM. 
February 1st, 2016, 09:34 AM  #2  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Quote:
x+ y z= 3 4x+ y+ 4z= 7 2x+ 3y+ 2z= 2. Obviously whoever gave you this problem expects you to know how to do that. But there are many different ways to do that and I not know which you have learned. I would notice, for example, that the first two equations both have "+ y" so that subtracting one equation from the other eliminates "y". And subtracting 3 times either of those from the third equation also eliminates y so you have two equations with x and z only. Quote:
 
February 1st, 2016, 10:54 AM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
(a). Adding double the first equation to the second gives 2x + 3y + 2z  1 = 0, which is incompatible with the third equation. Hence the three planes have no point in common. (b). Subtracting the third equation from the second gives 2x  3y  1 = 0. Subtracting this from the first equation gives 4z = 0, so z = 0. Adding the second and third equations and replacing z with zero gives y + 3 = 0, so y = 3. It's now easy to show that x = 4. Hence the three planes intersect in the single point (x, y, z) = (4, 3, 0). (c). The first two equations are equivalent. The variable z doesn't occur in the third equation  is that a typing error? 

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