SOLUTION:
Solve:
1.
2x+y+2z=11,
3x+2y+2z=8,
X+4y+3z=0
Algebra.Com
Question 888225:
Solve:
1.
2x+y+2z=11,
3x+2y+2z=8,
X+4y+3z=0
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
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2x+y+2z=11,
3x+2y+2z=8,
x+4y+3z=0
BTW X is not the same as x.
2,1,2,11
3,2,2,8
1,4,3,0
divide row 1 by 2/1
1,1/2,1,11/2
3,2,2,8
1,4,3,0
add down (-3/1) *row 1 to row 2
1,1/2,1,11/2
0,1/2,-1,-17/2
1,4,3,0
add down (-1/1) *row 1 to row 3
1,1/2,1,11/2
0,1/2,-1,-17/2
0,7/2,2,-11/2
divide row 2 by 1/2
1,1/2,1,11/2
0,1,-2,-17
0,7/2,2,-11/2
add down (-7/2) *row 2 to row 3
1,1/2,1,11/2
0,1,-2,-17
0,0,9,54
divide row 3 by 9/1
1,1/2,1,11/2
0,1,-2,-17
0,0,1,6
We now have the value for the last variable.
We will work our way up and get the other solutions.
add up (2/1) *row 3 to row 2
1,1/2,1,11/2
0,1,0,-5
0,0,1,6
add up (-1/1) *row 3 to row 1
1,1/2,0,-1/2
0,1,0,-5
0,0,1,6
add up (-1/2) *row 2 to row 1
1,0,0,2
0,1,0,-5
0,0,1,6
final
1,0,0,2
0,1,0,-5
0,0,1,6
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