由1),y=(4-2x)/(x+3)
由3),z=(33+x)/(x+3)
代入2）式：(4-2x)/(x+3)*(33+x)/(x+3)-(4-2x)/(x+3)+2(33+x)/(x+3)=5
去分母：(4-2x)(33+x)-(4-2x)(x+3)+2(33+x)(x+3)=5(x+3)^2
132-2x^2-62x-(12-2x^2-2x)+2(x^2+36x+99)=5(x^2+6x+9)
3x^2+18x-273=0
x^2+6x-91=0
(x+13)(x-7)=0
x=-13,7
x=-13时,y=(4-2x)/(x+3)=-3,z=(33+x)/(x+3)=-2
x=7时,y=(4-2x)/(x+3)=-1,z=(33+x)/(x+3)=4
因此原方程有两组解(-13,-3,-2),及(7,-1,4)