三元一次方程组x+y+z=111y:x=3:2y:z=5:4的解是.
2019-05-07
优质解答
∵y:x=3:2,y:z=5:4,
∴y:x:z=15:10:12,
设x=10k,y=15k,z=12k,
∵x+y+z=111,
∴10k+15k+12k=111,
∴k=3,
∴x=30,y=45,z=36,
故答案为:.
∵y:x=3:2,y:z=5:4,
∴y:x:z=15:10:12,
设x=10k,y=15k,z=12k,
∵x+y+z=111,
∴10k+15k+12k=111,
∴k=3,
∴x=30,y=45,z=36,
故答案为:.