{3x+2y+z=13 {x+y+z=7 {2x+3y-z=12

问题描述:

{3x+2y+z=13 {x+y+z=7 {2x+3y-z=12

带入消元 z=2x+3y-12带入前2式 ==>
1)式:3x+2y+2x+3y-12=13 ==> 5x+5Y=25 ==> x+ y = 5
2)式:x+y+ 2x + 3y -12 = 7 ==> 3x+4y =19
由上2个 ==> x=1 y=4 ==> z =2