Question 323998: Solve the system of equations.
x+y–z+4=0
2x–3y–z=5
x+2y+2z=3
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! Solve the system of equations.
x+y–z+4=0
2x–3y–z=5
x+2y+2z=3
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x + y - z + 4 = 0
2x - 3y - z = 5
x + 2y + 2z = 3
in the 1st equation they got the + 4 on the left side of the equal sign, moving it to the other side and rewriting all 3 equations lining up the coefficients that go with the variables and the constants
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I like to use the elimination method which involves by a series of basic arithmetic operations breaking down the system of equations until all you are left with in each of the 3 equations is one variable equaling one constant.
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x + y - z = -4 (multiply 1st equation by 6)
2x - 3y - z = 5 (multiply 2nd equation by -3)
x + 2y + 2z = 3
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6x + 6y - 6z = -24
-6x + 9y + 3z = -15 (add 2nd equation to 1st equation)
x + 2y + 2z = 3 (multiply 3rd equation by 6)
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0x + 15y - 3z = -39
-6x + 9y + 3z = -15
6x + 12y + 12z = 18 (add 3rd equation to 2nd equation)
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0x + 15y - 3z = -39 (divide 1st equation by 3)
0x + 21y + 15z = 3 (divide 2nd equation by 3)
6x + 12y + 12z = 18 (move this equation to be the 1st equation)
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6x + 12y + 12z = 18
0x + 5y - z = -13 (subtract 2nd equation from 1st equation)
0x + 7y + 5z = 1
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6x + 7y + 13z = 31
0x + 5y - z = -13
0x + 7y + 5z = 1 (subtract 3rd equation from 1st equation)
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6x + 0y + 8z = 30
0x + 5y - z = -13 (multiply 2nd equation by 7)
0x + 7y + 5z = 1 (multiply 3rd equation by 5)
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6x + 0y + 8z = 30
0x + 35y - 7z = -91
0x + 35y + 25z = 5 (subtract 3rd equation from 2nd equation)
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6x + 0y + 8z = 30
0x + 0y - 32z = -96 (divide 2nd equation by 4)
0x + 35y + 25z = 5
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6x + 0y + 8z = 30
0x + 0y - 8z = -24 (add 2nd equation to the 1st equation)
0x + 35y + 25z = 5 (move 3rd equation to be the 2nd equation)
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6x + 0y + 0z = 6
0x + 35y + 25z = 5 (divide 2nd equation by 5)
0x + 0y - 8z = -24 (multiply 3rd equation by 5)
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6x + 0y + 0z = 6
0x + 7y + 5z = 1 (multiply 2nd equation by 8)
0x + 0y - 40z = -120
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6x + 0y + 0z = 6
0x + 56y + 40z = 8
0x + 0y - 40z = -120 (add 3rd equation to the 2nd equation)
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6x + 0y + 0z = 6 (divide 1st equation by 6)
0x + 56y + 0z = -112 (divide 2nd equation by 56)
0x + 0y - 40z = -120 (divide 3rd equation by -40)
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x + 0y + 0z = 1
0x + y + 0z = -2
0x + 0y + z = 3
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so x = 1, y = -2, z = 3
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check plugging in values in original equations:
(1) + (-2) - (3) = -4 --> 1 - 2 - 3 = -1 - 3 = -4
2(1) - 3(-2) - (3) = 5 --> 2 + 6 - 3 = 8 - 3 = 5
(1) + 2(-2) + 2(3) = 3 --> 1 - 4 + 6 = -3 + 6 = 3
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this method might not appeal to you but it is probably one of the easiest methods if not the easiest method, just do not get confused as to where you are in your sequence of operations
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