解三元一次方程组:x(y-z)=27①,y(x-z)=35②,z(x+y)=28③.(求正整数解)
问题描述:
解三元一次方程组:x(y-z)=27①,y(x-z)=35②,z(x+y)=28③.(求正整数解)
答
①+②+③,得x(y-z)+y(x-z)+z(x+y)==90,化简为2xy==90,变形xy=45①-②+③,得x(y-z)-y(x-z)+z(x+y)==20,化简为2yz=20,变形yz=10-①+②+③,得-x(y-z)+y(x-z)+z(x+y)==36,化简为2xz==36,变形xz=18此时,方程组变形为{xy=4...