Question 438675: Please help me solve this equation:
Betty has $8.60 in nickels, dimes, and quarters. She has 6 fewer quarters than nickels and 3 more dimes than twice the number of nickels. How many of each coin does she have?
Answer by oberobic(2304)   (Show Source):
You can put this solution on YOUR website! With coin problems you have to keep track of the counts and the values of the coins.
n = number of nickels
d = number of dimes
q = number of quarters
5n = value of the nickels in cents
10d = value of the dimes in cents
25q = value of the quarters in cents
Note that problem setups that have dollars and cents have to be converted to cents.
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"Betty has $8.60" is a value statement.
5n + 10d + 25q = 860 cents
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She has "6 fewer quarters than nickels"
q = n-6
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"3 more dimes than twice the number of nickels"
d = 2n + 3
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Now we can look back at the value equation to see what we can substitute.
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5n + 10d + 25q = 860
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Substitute q=n-6 and d= 2n+3, which would make the equation entirely in terms of n.
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5n + 10(2n+3) + 25(n-6) = 860
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5n + 20n + 30 + 25n - 150 = 860
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Collect like terms
50n - 120 = 860
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Add 120 to both sides
50n = 980
Divide both sides by 50
n = 19.8
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This presents us with an unsolvable problem because you cannot have a fractional number of nickels.
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That having been said, if she were to have had $8.80 instead of $8.60, then we might be able to solve it.
50n - 120 = 880
50n = 1000
n = 20
d = 2n+3
d = 43
q = n-6
q = 14
Checking to see if these total $8.80
5n=5(20) = $1
10d = 10(43) = $4.30
25q = 25(14) = $3.50
$1 + $4.30 + $3.50 = $8.80
True.
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But if $8.60 is in the presented problem, then it is unsolvable with "real" coins.
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Done.
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