SOLUTION: Sam had a piggy bank full of nickels and dimes. He added up all the money and it was 123 coins worth $8.20. How many dimes did he have?

Question 963958: Sam had a piggy bank full of nickels and dimes. He added up all the money and it was 123 coins worth $8.20. How many dimes did he have?
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
N=nickels; D=dimes
N+D=123
D=123-N Use this to substitute for D.
$0.05N+$0.10D=$8.20 Substitute for D from above.
$0.05N+$0.10(123-N)=$8.20
$0.05N+$12.30-$0.10N=$8.20 Subtract $12.30 from each side.
-$0.05N=-$4.10 Divide each side by $0.05.
N=82 There were 82 nickels.
D=123-N=123-82=41 ANSWER: There were 41 dimes.
CHECK:
$0.05N+$0.10D=$8.20
$0.05(82)+$0.10(41)=$8.20
$4.10+$4.10=$8.20
$8.20=$8.20
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