Question 223366
Solve each system by elimination.
1)
2x-y+z=-2  
x+3y-z=10  
x+0y+2z=-8  
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Let the 3rd equation be the 1st.
x+0y+2z=-8
2x-y+z=-2  
x+3y-z=10  
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2 times the 1st subtracted from the 2nd
Subtract the 1st from the 3rd
x+0y+2z=-8
0x-y-3z=14  
0x+3y-3z=18
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Add 3 times the 2nd to the 3rd
x+0y+2z=-8
0x-y-3z=14  
0x+0y-12z=60
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Solve the 3rd to get: z = -5
Substitute into the 2nd to get y = 1
Substitute those values into the 1st to get x = 2
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2)
x-2y+3z=12
-2x-y-2z=5
2x+2y-z=4
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2 times the 1st added to the 2nd
2 times the 1st subtracted from the 3rd
x-2y+3z=12
0x-5y+4z=29
0x+6y-7z=-20
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Add the 2nd to the 3rd
x-2y+3z=12
0x-5y+4z=29
0x+y-3z=9
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Add 5 times the 3rd to the 2nd:
x-2y+3z=12
0x+0y-11z=74
0x+y-3z=9
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Solve the 2nd to get: z = -74/11
Substitute into the 3rd to get y = -123/11
Substitute into the 1st to get x = 108/11
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Cheers,
Stan H.