Question 46398
This is a systems of equations because there are two linear conditions:
3candy + 1 gum = $1.75 [First condition]
2candy + 4 gum = $2.00 [Second condition]
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3C + 1G = 1.75 [Solve using any type of method elimination, substituion, etc.]
2C + 4G = 2.00 

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3C + 1G = 1.75 [Try the elimination method]
2C + 4G = 2.00 
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-2(3C + 1G = 1.75) [Multiply through with -2 to eliminate the "C" term 
+3(2C + 4G = 2.00) [Multiply through with 3 to eliminate the "C" term]
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-6C -2G = -3.50 [Multiply through with -2 to eliminate the "C" term]
+6C +12G = 6.00 [Multiply through with 3 to eliminate the "C" term]
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-6C -2G = -3.50  [Perform the indicated operations]
+6C +12G = 6.00 
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0C +10G = 2.50
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10G = 2.50  [Solve for "G"]
G = 2.50/10
Gum = .25 cents each 

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2C + 4(.25) = 2.00 [Plug-in G=.227 into either original equation]
2C + 1.00 = 2.00 [Solve for "C"]
2C + 1.00 - 1.00 = 2.00 - 1.00 [Simplify]
2C = 1.00
2C/2=1.00/2
Candy=0.50 cents each

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3C + 1G = 1.75 [Plug-in C=0.50 and G=.25 into the orginal equations]
3(0.50)+1(.25)=1.75[Plug-in C=0.50 and G=.25 into the orginal equations]
1.75=1.75 [Checks out]
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2C + 4G = 2.00 [Plug-in C=0.50 and G=.25 into the orginal equations]
2(0.50) + 4(.25) = 2.00 
1.00 + 1.00=2.00
2.00=2.00 [checks out]



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