若x+2y+3z=10,4x+2y+2z=15,则x+y+z=?

试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y) 已知,方程组:4x-3y-7z=0 x+2y=10z 则(x-y+z)÷(x+y+z)=—— x=y/z=z/3,x+y+z ＝12,求2x+3y+4z是多少, 若x-2y+4z=1,x+3y-7z=0,则y:z=? 若4x-3y-6z=0,x+2y-7z=0,求代数式5x*5x+2y*2y-z*z/2x*2x-3y*3y-10z*10 若{x+3y+10z=0 则 (x+y-z)/(x-y+z) x/2=y/3=z/5 x+3y-z/x-3y+z (4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai 若4x+3y+2z=15,x-2y+3z=10,则x+y+z=? 若x+2y+3z=10，4x+3y+2z=15，则x+y+z的值为（　　） 5,若x+2y+3z=10,4x+3y+2z=15,则x+y+z的值为（ ） x+2y+3z=10,x-y+4z=10,x+3y+2z=2