SOLUTION: Juan has 15 pennies and 8 more nickles than dimes. He has 51 coins in all. How many dimes does he have?
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Question 12177: Juan has 15 pennies and 8 more nickles than dimes. He has 51 coins in all. How many dimes does he have?
Answer by bonster(299) (Show Source): You can put this solution on YOUR website!
Juan has 15 pennies
subtract 15 coins from 51
51-15=36 remaining coins
8 more nickles than dimes
dimes=x
nickles=x+8
the sum of the number of dimes and the number of nickles= 36
x+(x+8)=36
2x+8=36 <--subtract 8 from both sides
(2x+8)-8=36-8
2x=28 <--divide both sides by 2
=
x=14
if dimes=x and x=14, he has 14 dimes
nickles=x+8=14+8=22 nickles
he has 14 dimes
Check: 14+22+15=51
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