MATH
posted by Anonymous on .
Solve the system of linear equations
2x+y3z=4
4x+2z=10
2x+3y13z=8
*im so lost (all these numbers are in one problem)

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+y3z=4
2x+z=10
2x+3y13z=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. 
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+y3z = 4 times 2 = 4x+ 2y6z = 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. 
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%253AANd9GcTP5OK2B8xVEFseHgCh5S25ao2K4pXfClEpupY9ltfPRK6cZC8NA%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.