x+y/2=z+x/3+y=z/4,x+2y+3z=60,求x,y,z的值初一三元一次方程
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x+y/2=z+x/3+y=z/4,x+2y+3z=60,求x,y,z的值初一三元一次方程
x+y/2=z+x/3+y=z/4,x+2y+3z=60,求x,y,z的值
初一三元一次方程
x+y/2=z+x/3+y=z/4,x+2y+3z=60,求x,y,z的值初一三元一次方程
设X+Y/2=Y+Z/3=Z+X/4=K,
则有
X+Y=2K,(1)
Y+Z=3K,(2)
Z+X=4K,(3)
(1)+(2)+(3)得
2(X+Y+Z)=9K,
所以X+Y+Z=4.5K,(4)
(4)-(1)得
Z=2.5K,
(4)-(2)得
X=1.5K,
(4)-(3)得
Y=0.5K,
因为X+2Y+3Z=60,
所以1.5K+2*0.5K+3*2.5K=60,
所以3K+2K+15K=120,
所以20K=120,
所以K=6,
所以X=1.5K=9,
Y=0.5K=3,
Z=2.5K=15.
x+y/2=z+x/3=y+z/4 x+y+z=27
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