{20x+10y=z,5x+6y=1/3z,2x-3y=1
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
已知x::y:z=3:4:5,(1)求x+y分之z的值;(2)若x+y+z=6,求x,y,z.
{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2
解三元一次方程组:(1)x+2y+3z=11x-y+4z=10x+3y+2z=2(2)x:y=3:5y:z=5:6x+y
三元一次方程组数学题x+2y+2z=33x+y-2z=72x+3y-2z=10x-y=2z-x=3y+z=-1x-y-z
x/2=y/3=z/5 x+3y-z/x-3y+z
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
x+y+z=2 x-3y+2z=1 2x+2y+z+5 要具体步骤..
解方程一;{x+y-z=6,x-3y+2z=1,3x+2y-z=-7.②{3x+4z=7,2x+3y+z=9,5x-9y
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
x+2y=3 x+y+z=36 2x+y+z=15 2y=3z x-y=1 x+2y+z x-z=-1 2x+z-y=1