求解三元一次方程组 2X+3Y+4Z=1 5X-6Y+2Z=12 3X-3Y-14Z=-1

2025-05-16 06:56:56
推荐回答(2个)
回答1:

2(2X+3Y+4Z)+(5X-6Y+2Z)=2+12=14→9X+10Z=14
(2X+3Y+4Z)+(3X-3Y-14Z)=1-1=0→5X-10Z=0
(9X+10Z)+(5X-10Z)=14→14X=14 X=1
Z=1/2 ,Y=-1

回答2:

x=-3/5 y=-17/10 z=23/3