10(y+z)^2-29(y+z)+10
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+2y+3z=10,x-y+4z=10,x+3y+2z=2
x-y+4z=10,x+3y+2z=2,x+2y+3z=11.
x+y+z=4 2x+3y-z=6 3x+2y+2z=10
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
x+y−2z=52x−y+z=42x+y−3z=10
x/10 = y/8 =z/9 求x+2y+3z/y-5z
{5X+Y+Z①X+5Y+Z=-2②X+Y+5Z=10③
解一道三元一次方程x+y+z=13,y-2=z,100z+10y+x+99=100x+10z+y.
(x-2y+z)/9=(2x+y+3z)/10=-(3x+2y-4z)/3=1 连等,求x,y,z,