document.write( "Question 952962: There were 12 more nickles than dimes, and the total value of the coins was $5.10. How many nickles and dimes were there?
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Algebra.Com's Answer #581953 by macston(5194)![]() ![]() You can put this solution on YOUR website! d=number of dimes; n=number of nickels=d+12 \n" ); document.write( "$0.10d+$0.05n=$5.10 Substitute for n. \n" ); document.write( "$0.10d+$0.05(d+12)=$5.10 \n" ); document.write( "$0.10d+$0.05(d)+$0.60=$5.10 Subtract $0.60 from each side. \n" ); document.write( "$0.15d=$4.50 Divide each side by $0.15 \n" ); document.write( "d=30 ANSWER 1: There were 30 dimes. \n" ); document.write( "n=d+12=30+12=42 ANSWER 2: There were 42 nickels. \n" ); document.write( "CHECK: \n" ); document.write( "$0.10(30)+$0.05(42)=$5.10 \n" ); document.write( "$3.00+$2.10=$5.10 \n" ); document.write( "$5.10=$5.10 \n" ); document.write( " \n" ); document.write( " |