SOLUTION: In a collection of nickels and dimes the number of dimes is 3 more than twice the number of nickels. If the value of the collection is $1.80, how many coins are there?

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Question 104837: In a collection of nickels and dimes the number of dimes is 3 more than twice the number of nickels. If the value of the collection is $1.80, how many coins are there?
Answer by solver91311(24713) About Me  (Show Source):
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n = number of nickels
d = number of dimes
value of one nickel is 5 cents, so the value of n nickels is 5n cents
value of one dime is 10 cents, so the value of d dimes is 10d cents
The number of dimes (d) is (=) 3 more (+3) than twice the number of nickels (2n), so

d+=+2n+%2B+3

The value of the collection, that is the value of the dimes plus the value of the nickels is $1.80 or 180 cents, so

5n+%2B+10d+=+180

Substitute the expression for d from the first equation into the second equation:
cartoon%285n+%2B+10%2Ared%28d%29+=+180%2C5n+%2B+10%2Ared%28%282n%2B3%29%29=180%29

Simplify and solve:
5n+%2B+10%282n%2B3%29=180
5n%2B20n%2B30=180
25n=150
n=6

d=2%286%29%2B3=15

So there are d%2Bn=15%2B6=21 coins.

Check:
10%2A15=150
5%2A6=30
and 150%2B30=180