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如果题目是:
3(x-3)+10(三分之二+y)=13
22(x-3)+5(三分之二+y)=27
解:设x - 3 = X,2/3 + y = Y 则原方程组可化为:
3X + 10Y = 13 (1)
22X + 5Y = 27 (2)
(2) x 2 得: 44X +10Y = 54 (3)
(3) - (1) 得: 41X = 41 X = 1
把X = 1 代入(1)得:Y = 1
∴X = 1
Y = 1
∴x = 4
y = 1/3
3(x-3)+10(三分之二+y)=13
22(x-3)+5(三分之二+y)=27
解:设x - 3 = X,2/3 + y = Y 则原方程组可化为:
3X + 10Y = 13 (1)
22X + 5Y = 27 (2)
(2) x 2 得: 44X +10Y = 54 (3)
(3) - (1) 得: 41X = 41 X = 1
把X = 1 代入(1)得:Y = 1
∴X = 1
Y = 1
∴x = 4
y = 1/3
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