2(x+3y)+(x+y+z)=5.8和3(x+3y)+(x+y+z)=6.3组成方程组,问x+y+z=多少?

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2(x+3y)+(x+y+z)=5.8和3(x+3y)+(x+y+z)=6.3组成方程组,问x+y+z=多少?
2(x+3y)+(x+y+z)=5.8和3(x+3y)+(x+y+z)=6.3组成方程组,问x+y+z=多少?

2(x+3y)+(x+y+z)=5.8和3(x+3y)+(x+y+z)=6.3组成方程组,问x+y+z=多少?
2(x+3y)+(x+y+z)=5.8 (1)
3(x+3y)+(x+y+z)=6.3 (2)
(1)×3-(2)×2
3(x+y+z)-2(x+y+z)=5.8×3-6.3×2
所以x+y+z=4.8

2(x+3y)+(x+y+z)=5.8和3(x+3y)+(x+y+z)=6.3
设X+3y=Q x+y+z=W
由上式可得：
2Q+W=5.8
3Q+W=6.3
Q=0.5
W=4.8 即得：x+y+z=4.8

这么简单
设a=x+3y
b=x+y+z得
2a+b=5.8
3a+b=6.3
=>b=4.8=>x+y+z=4.8

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