SOLUTION: If Noelle has 3 more nickels than dimes and they have a combined value of 90 cents, how many of each coin does she have?

Question 975764: If Noelle has 3 more nickels than dimes and they have a combined value of 90 cents, how many of each coin does she have?
Found 2 solutions by algebrahouse.com, Alan3354:
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
d = number of dimes
d + 3 = number of nickels {Noelle has 3 more nickels than dimes}

0.1d + 0.05(d + 3) = 0.9 {value of coin times number of coins equals total value}
0.1d + 0.05d + 0.15 = 0.9 {used distributive property}
0.15d + 0.15 = 0.9 {combined like terms}
0.15d = 0.75 {subtracted 0.15 from each side}
d = 5 {divided each side by 0.15}
d + 3 = 8 {substituted 5, in for d, into d + 3}

5 dimes and 8 nickels

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Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
If Noelle has 3 more nickels than dimes and they have a combined value of 90 cents, how many of each coin does she have?
=================
Without the 3 extra nickels she has pairs of dimes & nickels worth 75 cents.
75/15 = 5 pairs
--> 5 dimes & 8 nickels

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