解三元一次方程(要有步骤):(1)2x+2y+z=4 ① 2x+y+2z=7 ② x+2y+2z=-6 ③
2个回答

1) 2x+2y+z=4 ①

2x+y+2z=7 ②

x+2y+2z=-6 ③

②-①:z-y=3 z=3+y

代入①、③:

2x+3y=1 ④

x+4y=-12 ⑤

④-2⑤:-5y=25 y=-5 x=-12-4y=8 z=-2

2)x+y+z =80 ①

x-y =6 ②

x+y-7z =0 ③

①-③:8z=80 z=10

②+③ 2x=76 x=38

由②:y=x-6=32

3)5x+4y+z=0 ①

3x+y-4z=1 ②

x+y+z=-2 ③

①-③:4x+3y=2

②+4③:7x+5y=-7

仿上解得:x= -31 y=42 z= -13

4) 设这个三位数形如xyz

则依题可列:

100*x+10*y+z=27*(x+y+z)

x+z=y+1

100*z+10*y+x=100*x+10*y+z+99

解之得:x =2 y =4 z =3

即这个三位数是:243.