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標題:

一d消元同代入法既問題

發問:

1.Solve the simultaneous equations a. 3a-2b=18 a+b=21 b.using the result of (a), solve the simultaneous equations: 3/x-2/y=18 1/x+1/y=21 plz help!!

• 幫我翻譯一篇報導....20點(急急急!!!!!)
• help me to do this mathematics
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• 幫我翻譯一篇報導....20點(急急急!!!!!)
• help me to do this mathematics
• 20分,thanks@1@

 

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最佳解答:

A) 3a - 2b = 18 - 1 a + b = 21 - 2 2*3, 3a + 3b = 63 - 3 1-3, (3a - 2b) - (3a + 3b) = 18 - 63 3a - 2b - 3a - 3b = -45 -5b = -45 5b = 45 b = 9 Substitute b = 9 into 2 a + b = 21 a + 9 = 21 a = 12 B) By using the result of A, a = 1/x 12 = 1/x x = 1/12 b = 1/y 9 = 1/y y = 1/9 記住,, 佢b題個條式會誤導你,, 但係你唔洗理佢,, 總之你一睇落去,, 就睇得出a = 1/x , b = 1/y 咁你就用返上面個兩條式,, 否則就會錯架啦,, 2007-03-21 20:34:58 補充： 寫漏左,,1, 2 ,3式個d記住加返括號

其他解答:

1.Solve the simultaneous equations a. 3a-2b=18-----------[1] a+b=21 a = 21 - b------------[2] sub[2] into [1] 3(21 - b) - 2b = 18 63 - 3b - 2b = 18 45 = 5b b = 9 a = 12 b.using the result of (a), solve the simultaneous equations: 3/x-2/y=18-------------[3] 1/x+1/y=21-----------------[4] now a = 1/x,b = 1/y x = 1/a and y = 1/b x = 1/12 and y = 1/9|||||3a-2b=18.....(1) a+b=21....(2) From (2), a+b=21 a=21-b....(3) Sub (3) into (1) ,3a-2b=18 3(21-b)=18 63-b=18 -b=-45 b=45