# MATH

Solve the system of linear equations
2x+y-3z=4
4x+2z=10
-2x+3y-13z=-8

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

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3. 👁 147
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.

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posted by Reiny
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
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.

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posted by Damon
3. Your solution is illustrated by

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.

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posted by Reiny

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