Question

三元一次方程组球解1.X+Z=Y7Z=X+Y+2X+Y+Z=142.X:Y=3:2Y:Z=5:4X+Y+Z=66

Answer 1

1.因为Y=X+Z,所以14=X+Y+Z=2Y,所以Y=X+Z=7,因为7Z=X+Y+2=X+9=16-Z,所以Z=2,X=5

2.因为X:Y=3:2=15:10,Y:Z=5:4=10:8,所以X:Y:Z=15:10:8,设X=15t,Y=10t,Z=8t,所以X+Y+Z=33t=66,所以t=2,故X=30,Y=20,Z=16