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Question 173846: Can you help me find a solution? I used a substitution method by it seem it does not work.
1.The Everton college store paid $1660 for an order of 44 calculators. The store paid $11 for each scientific calculator. The others, all graph calculators, cost the store $53 each. How many of each type of calculator was order?
The store ordered__scientific calculators and __graphing calculators.
2. Elimination method
0.3x-0.2y=4
0.2x+0.3y=-23/19
3. Substitution method
6x+7y=-22
-3x+y=20
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1.The Everton college store paid $1660 for an order of 44 calculators. The store paid $11 for each scientific calculator. The others, all graph calculators, cost the store $53 each. How many of each type of calculator was order?
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Value equation: 11s + 53g = 1660
Quantity equation: s + g = 44
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Substitution Method:
s = 44-g
Substitute into the 1st equation:
11(44-g) + 53g = 1660
484 - 11g + 53g = 1660
42g = 1176
g = 28 (# of graphing calculators purchased)
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Substitute into s = 44-g to get:
s = 44-28 = 16 (# of scientific calculators purchased)
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2. Elimination method
0.3x-0.2y=4
0.2x+0.3y=-23/19
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Modify the equations:
3x - 2y = 40
2x + 3y = -230/19
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Modify for elimination:
6x - 4y = 80
6x + 9y = -690/19
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Substract 1st from 2nd to get:
13y =(-690/19) - 80 = -116.3158
y = -8.9474
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Substitute into 3x-2y=40 to solve for x:
3x - 2(-8.9474) = 40
3x = 22.1053
x = 7.3684333...
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Cheers,
Stan H.
3. Substitution method
6x+7y=-22
-3x+y=20
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