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?
问题描述：
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?
答
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