Question 154010: Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
According to my calculations the answer should be 88 dimes. I came to this because originally there should have been 44 dimes=$4.40 and 25 nickels=$1.25 which would make the original equation of $5.65. So if I had 88 dimes=$8.80 and 33 nickels=$1.65 this totals $10.45.
I am pretty sure my answer is correct but I am at a loss as to how to show the formula in algebraic equation.
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Joe has a collection of nickels and dimes that is worth $5.65. If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?
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Write an equation for each statement:
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"Joe has a collection of nickels and dimes that is worth $5.65."
.05n + .10d = 5.65
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" If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. "
.05(n+8) + .10(2d) = 10.45
simplify
.05n + .4 + .20d = 10.45
.05n + .20d = 10.45 - .40
.05n + .20d = 10.05
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Now subtract the 1st equation from the above equation and eliminate n, find d
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