Question 198645
Joe has a collection of nickels and dimes that is worth $6.80. If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10 How many dimes does he have? 
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Let n = nickels
and d = dimes
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From:"If the number of dimes were doubled and the number of nickels increased by 6, the value of the coins would be $11.10"
.05(n+6) + .10(2d) = 11.10
.05n+ .30 + .20d = 11.10
.05n + .20d = 10.80
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From: "Joe has a collection of nickels and dimes that is worth $6.80"
.05n + .10d = 6.80
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Our two equations:
.05n + .20d = 10.80
.05n + .10d = 6.80
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Multiplying the bottom equation by -1 and adding to the first:
 .05n + .20d = 10.80
-.05n - .10d = -6.80
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        .10d = 4.00
           d = 40 dimes
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Plug the above into the following equation and solve for n:
.05n + .10d = 6.80
.05n + .10(40) = 6.80
.05n + 4.00 = 6.80
.05n = 2.80
n = 56 nickels