title: 有理数x，y，z满足方程组x=6
desc: 有理数x，y，z满足 方程组：x=6-3y
                            x+3y-2xy+2z^2=0
   求2^2y+z的值     
answer: 解：
  x=6-3y => x+3y=6
  把x+3y=6代入x+3y-2xy+2z^2=0得：
  6-2xy+2z^2=0
  把x=6-3y代入6-2xy+2z^2=0得：
  6-2(6-3y)y+2z^2=0
  3-(6-3y)y+z^2=0
  3-6y+3y^2+z^2=0
  3(1-2y+y^2)+z^2=0
  3(1-y)^2+z^2=0
  则1-y=0,z=0,即y=1,z=0
  则2^2y+z=2^2+0=4
  