Question 904455: x y-2z=7, 2x-3y-2z=0, x-2y-3z=3 How to solve this linear equation by using: Addition method and row equivalent method?
Found 2 solutions by Fombitz, richwmiller:
Answer by Fombitz(32388)   (Show Source):
Answer by richwmiller(17219)  (Show Source):
You can put this solution on YOUR website! I am guessing that it should be +y in the first equation
The answers come out (23/3,22/3,10/-3) with -1y
and (1,2,-2) with +1y
x +y-2z=7,
2x-3y-2z=0,
x-2y-3z=3
1,1,-2,7
2,-3,-2,0
1,-2,-3,3
add down (-2) *row 1 to row 2
1,1,-2,7
0,-5,2,-14
1,-2,-3,3
add down (-1) *row 1 to row 3
1,1,-2,7
0,-5,2,-14
0,-3,-1,-4
divide row 2 by -5
1,1,-2,7
0,1,2/-5,-14/-5
0,-3,-1,-4
add down (3) *row 2 to row 3
1,1,-2,7
0,1,2/-5,14/5
0,0,11/-5,22/5
divide row 3 by 11/-5
1,1,-2,7
0,1,2/-5,14/5
0,0,1,-2
We now have the value for the last variable.
We will work our way up and get the other solutions.
add up (-2/-5) *row 3 to row 2
1,1,-2,7
0,1,0,2
0,0,1,-2
add up (2) *row 3 to row 1
1,1,0,3
0,1,0,2
0,0,1,-2
add up (-1) *row 2 to row 1
1,0,0,1
0,1,0,2
0,0,1,-2
final
1,0,0,1
0,1,0,2
0,0,1,-2
"1","2","-2"
(1,2,-2)
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