{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)
~
(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)
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