very easy 的數學

2006-11-19 8:21 pm

解下列各方程 : (要有過程)
28y=2〔16ㄧ4(1ㄧy)〕
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回答 (6)

2006-11-19 8:26 pm

✔ 最佳答案

28y=2[16-4(1-y)]
28y=2[16-4+4y]
28y=24+8y
20y=24
y=6/5 (or 1.2)

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2006-11-19 9:32 pm

解下列各方程 : (要有過程)
28y=2〔16ㄧ4(1ㄧy)〕
28y=2(16-4+4y)
28y=2(12+4y)
28y=24+8y
28y - 8y=24
20y=24
y=24/20 = 6/5

參考: 自己

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2006-11-19 9:09 pm

28y = 2(16 - 4(1 - y)〕
28y = 2(12 + 4y)
7y = 6 + 2y
y = 6/5

我諗唔會有錯掛 -.-

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2006-11-19 8:43 pm

28y = 2〔16ㄧ4(1ㄧy)〕
28y = 2 (16 - 4 + 4y)
14y = 12 + 4y
10y = 12
y = 12/10
y = 6/5

一定是correct的!

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2006-11-19 8:34 pm

28y = 2(16 - 4(1 - y)〕
28y = 2(12 + 4y)
7y = 2(3 + y)
7y = 6 + 2y
6 = 5y
y = 6/5

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2006-11-19 8:33 pm

28y=2〔16ㄧ4(1ㄧy)〕
28y= 2(16ㄧ4+4y)
28y=32-8+8y
20y=24
y=24/20= 6/5

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