SOLUTION: a Woman has 11 coins, consisting of nickels and dimes. she has three more dimes than nickels. how many of each type of coin does she have if the total value of the coins is 90 cent

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Question 273095: a Woman has 11 coins, consisting of nickels and dimes. she has three more dimes than nickels. how many of each type of coin does she have if the total value of the coins is 90 cents?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n = number of nickels
Let d = number of dimes
given:
n+%2B+d+=+11
d+=+n+%2B+3
5n+%2B+10d+=+90
By substitution:
5n+%2B+10%2A%28n+%2B+3%29+=+90
5n+%2B+10n+%2B+30+=+90
15n+=+90+-+30
15n+=+60
n+=+4
and
n+%2B+d+=+11
4+%2B+d+=+11
d+=+7
She has 7 dimes and 4 nickels
check:
5n+%2B+10d+=+90
5%2A4+%2B+10%2A7+=+90
20+%2B+70+=+90
90+=+90
OK