Solve the system of linear equations

Question

2x+y-3z=4
4x+2z=10
-2x+3y-13z=-8

*im so lost (all these numbers are in one problem)

Answer 1

if you divide the 2nd by 2, all your equations start with 2x. That is good, since we could add or subtract them to eliminate the x

2x+y-3z=4
2x+z=10
-2x+3y-13z=-8

subtract 1st from the 2nd
-y + 4z = 6 , #4

add the 2nd and 3rd
3y - 12z = 2 , #5

#4 times 3---> -3y + 12z = 18
#5 as is -----> 3y - 12z = 2

add them: 0 = 20 , which is a contradiction
AHHH, there is no unique solution.

If you are studying equations of planes, you should categorize this as one of the special cases.

Answer 2

well, the easy way is to google Gauss Jordan reduction calculator.
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx

otherwise do it by elimination

First eliminate z from first two:

2x+y-3z = 4 times 2 = 4x+ 2y-6z = 8
4x+0y+2z=10 times 3 = 12x+0y+6z = 30
add the two
16 x + 2 y = 38
or
8 x + y = 19

Now eliminate z from the last 2
multiply second by 13 and last by 2
52 x + 0y + 26 z = 130
-4 x + 6 y -26 z = -16
------------------------- add the two
48 x + 6 y = 114
or
24 x + 3 y = 57
Now you have the two equations in x and y
8 x + 1 y = 19
24 x + 3 y = 57

multiply first by 3
24 x + 3 y = 57
Oh , my, system can not be solved, determinant is zero, two lines the same.

Answer 3

Your solution is illustrated by
http://www.google.com/search?q=intersection+of+3+planes&tbm=isch&imgil=VKeuvId73BKgcM%253A%253Bhttp%253A%252F%252Ft2.gstatic.com%252Fimages%253Fq%253Dtbn%253AANd9GcTP5OK2B8xVEFseHgCh5S25ao2K4pX-fClEpupY9ltfPRK6cZC8NA%253B332%253B265%253BL5ht1c6dmTcU5M%253Bhttp%25253A%25252F%25252Fstweb.peelschools.org%25252Fjfsweb%25252Fmga4u%25252Funit3%25252Flesson7.html&source=iu&usg=__Epsu58j8aAIOLuG__2toxIf0TSk%3D&sa=X&ei=SN6gU9wz2KPIBJaOgPAK&sqi=2&ved=0CCkQ9QEwAg&biw=1680&bih=936#imgdii=_

look at the 2nd picture, top left
Your planes intersect in pairs, each one giving us a parallel line as the intersection of the planes, two at a time.