{x+y-z=6,x-y+2z=-7,3x+2y+z=7解三元一次方程,过程详细点
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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
解: (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
解: (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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