y=2x-7 5x+3y+2z=2 x+4z=4 解下列三元一次方程组,要详细过程

2024-12-01 07:07:06
推荐回答(3个)
回答1:

y=2x-7 (1)
5x+3y+2z=2 (2)
x+4z=4 (3)
(1)代入(2)
5x+3(2x-7)+2z=2
11x+2z=23 (4)
(3)代入(4)
11(x+4z)-42z=23
44-42z=23
42z=21
z=0.5
x=2
y=-3

回答2:

y=2x-7 5x+3y+2z=2 x+4z=4

5x+3y+2z=2
5x+3(2x-7)+2z=2
5x+6x-21+2z=2

11x+2z=23
11x+44z=44
44z-2z=44-23
42z=21
z=1/2

x=4-4z=4-4*1/2=2

y=2x-7=2*2-7=-3

x=2
y=-3
z=1/2

回答3:

y=2x-7 5x+3y+2x=2 3x-4z=4 z=3/4x-1
5x+3(2x-7)+2(3/4x-1)=2
5x+6x-21+3/2x-2=2
25/2x=25
x=2
y=2*2-7=-3
z=3/4*2-1=1/2