Question

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)

Answer 1

①+②+③,得x(y-z)+y(x-z)+z(x+y)==90,化简为2xy==90,变形xy=45

①-②+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=45,yz=10,xz=18}

④*⑤*⑥,得xy*yz*xz=8100,化简为x²y²z²=8100,开平方得xyz=±90

题目要求求整数解,所以xyz取90,xyz=90⑦

⑦÷④,得z=2

⑦÷⑤,得x=9

⑦÷⑥,得y=5

解得{x=9,y=5,z=2}