答案:x=1,y=3,z=-1
由一式得x=y+z-1,设为四式
将四式代入二式和三式,得{3(y+z-1)+5y+7z=11,4(y+z-1)-y+2z=-1
即{8y+10z=14,3y+6z=3
由此可解出z=-1
将z=-1代入8y+10z=14,得y=3,再将y=3代入x=y+z-1,得出x=1
答案:x=1,y=3,z=-1
由一式得x=y+z-1,设为四式
将四式代入二式和三式,得{3(y+z-1)+5y+7z=11,4(y+z-1)-y+2z=-1
即{8y+10z=14,3y+6z=3
由此可解出z=-1
将z=-1代入8y+10z=14,得y=3,再将y=3代入x=y+z-1,得出x=1