anonymous
  • anonymous
6.) Vera's piggybank contained 9.35 in quarters, dimes, and nickels. there were six more than two times as many nickels as quarters and four less dimes than quarters. how many of each kind of con was there in the bank?????? I don't need a answer. I need two equations..
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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jim_thompson5910
  • jim_thompson5910
thanks
anonymous
  • anonymous
your welcome
jim_thompson5910
  • jim_thompson5910
let q = number of quarters d = number of dimes n = number of nickels

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jim_thompson5910
  • jim_thompson5910
we are told that "there were six more than two times as many nickels as quarters ", so we can say that q = 2n + 6
jim_thompson5910
  • jim_thompson5910
there are " four less dimes than quarters.", so d = q - 4
jim_thompson5910
  • jim_thompson5910
there is $9.35 in the bank, so 0.25q + 0.10d + 0.05n = 9.35 25q + 10d + 5n = 935 .. multiply everything by 100
jim_thompson5910
  • jim_thompson5910
ok I made a typo, one sec
jim_thompson5910
  • jim_thompson5910
The equation q = 2n+6 should be n = 2q+6
jim_thompson5910
  • jim_thompson5910
25q + 10d + 5n = 935 25q + 10(q - 4) + 5n = 935 ... plug in d = q - 4 25q + 10q - 40 + 5n = 935 25q + 10q + 5n = 935+40 35q + 5n = 975 35q + 5(2q + 6) = 975 ... plug in n = 2q+6 35q + 10q + 30 = 975 45q + 30 = 975 45q = 975 - 30 45q = 945 q = 945/45 q = 21 So there are 21 quarters
jim_thompson5910
  • jim_thompson5910
n = 2q+6 n = 2*21+6 n = 42 + 6 n = 48 so there are 48 nickels
jim_thompson5910
  • jim_thompson5910
d = q - 4 d = 21 - 4 d = 17 and there are 17 dimes

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