{x+y-z=6,x-y+2z=-7,3x+2y+z=7解三元一次方程,
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{x+y-z=6,x-y+2z=-7,3x+2y+z=7解三元一次方程,
x+y-z=6 (1)
x-y+2z=-7 (2)
3x+2y+z=7 (3)
分析:利用加减消去法,将未知数逐一消去
(1) + (2):2x + z = -1 (3) (消去y)
(1)*2:2x + 2y - 2z = 12 (4)
(3) - (4):x + 3z = -5 (5) (消去y)
(5)*2:2x + 6z = -10 (6)
(6) - (3):5z = -9 (消去x)
z = -9/5
将 z = -9/5 代入 (3)式:得 2x + (-9/5) = -1
2x = 4/5
x = 2/5
将 x = 2/5,z = -9/5,代入 (1)式:得 2/5 + y - (-9/5) = 6
y = 19/5
所以答案为 x = 2/5,y = 19/5,z = -9/5 (经验算,准确无误)
x-y+2z=-7 (2)
3x+2y+z=7 (3)
分析:利用加减消去法,将未知数逐一消去
(1) + (2):2x + z = -1 (3) (消去y)
(1)*2:2x + 2y - 2z = 12 (4)
(3) - (4):x + 3z = -5 (5) (消去y)
(5)*2:2x + 6z = -10 (6)
(6) - (3):5z = -9 (消去x)
z = -9/5
将 z = -9/5 代入 (3)式:得 2x + (-9/5) = -1
2x = 4/5
x = 2/5
将 x = 2/5,z = -9/5,代入 (1)式:得 2/5 + y - (-9/5) = 6
y = 19/5
所以答案为 x = 2/5,y = 19/5,z = -9/5 (经验算,准确无误)
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