F.2數Simultaneous Equations 急!

2010-11-29 5:35 am
1))
4kg of tea and 6kg of coffee beans are sold for $240.1kg of tea and 8kg of coffee beans are sold for $242. Find the respective selling prices of tea and coffee beans per kg.


2)
Henry pays $1176 for 30 packs of mini size cookies and 40 packs of regular size cookies. Monica pays $768 for 40 packs ofmini size cookies and 20packs
of regular size cookies.Find the selling prices of each kind of cookies per pack.

Substitution Method/ Elimination Method
同埋要過程和 Let X be...y be....
THX

回答 (2)

2010-11-30 1:41 am
✔ 最佳答案
Let x be the selling prices of tea per kg,
Let y be the selling prices of coffee bean per kg:
4x+6y=240-------------1
x+8y=242---------------2

(2)x4-(1):
4x+32y-4x-6y=968-240
26y=728
y=28

Put y=28 into 2:
x=18

The selling prices of tea per kg is 18 dollars
the selling prices of coffee bean per kg is 28 dollars
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2010-11-29 17:48:36 補充:
Let x be the prices of mini size cookies,y be the prices of regular size cookies:

30x+40y=1176-------1
40x+20y=768------------2

2010-11-29 17:48:41 補充:
From2:
y=38.4-2x-----------3
Sub. 3 into 1:
30x+1536-80x=1176
-50x=-360
x=7.2
Put x=7.2 int0 3:
y=24
the prices of mini size cookies is $7.2 ,y be the prices of regular size cookies is 24 dollars
2010-11-29 6:06 am
1) Let x be tes and y be coffee beans.4x + 6y = 240 .................. (1)x + 8y = 242 .................. (2)(2) x 44x + 32y = 968 ................. (3)(3) - (1)26y = 728y = 24.16 [price of coffee beans]8y = 193.28x + 193.28 = 242x = 48.72 [price of tea]2) Let m be mini-size and r be regular size cookies30m + 40r = 1176 .............. (1)40m + 20r = 768 ................ (2)(2) x 280m + 40r = 1536 .............. (3)(30 - (1)50m = 360m = 7.20 [price of mini-sized cookies]In (2)28.8 + 20r = 76820r = 739.2r = 36.96 [price of regular-sized cookies]


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