Question 1068229: Susan owns three food stands. One is in Tomsville, one is in Newbury Heights, and one is in Glen City. Each stand sells apples, cherries, and pears. The stands sell no other fruit. The table below gives the number of bags of each type of fruit sold at each location. (I tried to put an excel image of the table, but I can't figure out how to upload an image or excel sheet) The table is summarized below.
Number of Bags sold
Apples Cherries Pears
Tomsville STAND SOLD: 8 bags of apples, 3 bags of Cherries, and 4 bags of Pears,
Newbury Heights STAND SOLD: 8 bags of apples, 7 bags of Cherries, and 9 bags of Pears,
Glen City STAND SOLD: 2 bags of apples, 5 bags of Cherries, and 6 bags of Pears,
If the stand in Tomsville sold $54 worth of fruit, the stand in Newbury Heights sold $92 worth, and the stand in Glen City sold $52 worth, what is the cost for a bag of each type of fruit?
Cost for a bag of Apples: $ __________
Cost for a bag of Cherries $ __________
Cost for a bag of pears: $ __________
Appreciate an answer to this problem.
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! 8a+3c+4p = 54
8a+7c+9p = 92
2a+5c+6p = 52
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1) Rearrange:
2a+5c+6p = 52
8a+7c+9p = 92
8a+3c+4p = 54
2) Multiply all terms in the 1st equation times -4 and add to the second equation, and you get:
2a+5c+6p = 52
. -13c-15p = -116
8a+3c+4p = 54
3) Multiply the second equation times -4 and add to the third equation to get:
2a+5c+6p = 52
. -13c-15p = -116
. -17a-20p = -154
4) Multiply the 2nd equation times -17/13 and add the result to the third equation, and now we have:
2a+5c+6p = 52
. -13c-15p = -116
. . . -5/13p = -30/13
Solve for p:
-5/13p = -30/13; p = -30/13*-13/5 = -30/-5 = 6 So now we know that p = 6
-13c-15(6) = -116; -13c = -26; c = 2
8a+3c+4p = 54
8a+3(2)+4(6) = 54
8a = 24
a = 3
So there you have it:
apples: 3
cherries: 2
pears: 6
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