document.write( "Question 955348: 70. Nickels, dimes, and quarters. Bernard has 41 coins consisting of nickels, dimes, and quarters, and they are worth a total of $4.00. If the number of dimes plus the number of quarters is one more than the number of nickels,
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Algebra.Com's Answer #583633 by macston(5194)\"\" \"About 
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N=nickels; D=dimes; Q=quarters
\n" ); document.write( "D+Q=N+1
\n" ); document.write( "N+N+1=41
\n" ); document.write( "2N=40
\n" ); document.write( "N=20 ANSWER 1 there are 20 nickels worth $1.00 so there 21 quarters and dimes worth $3.00
\n" ); document.write( "D+Q=21
\n" ); document.write( "D=21-Q
\n" ); document.write( "$0.10D+$0.25Q=$3.00 Substitute for D.
\n" ); document.write( "$0.10(21-Q)+$0.25Q=$3.00
\n" ); document.write( "$2.10-$0.10q+$0.25Q=$3.00 Subtract $2.10 from each side.
\n" ); document.write( "$0.15Q=$0.90 Divide each side by $0.15.
\n" ); document.write( "Q=6 ANSWER 2: There are 6 quarters.
\n" ); document.write( "D=21-Q=21-6=15 ANSWER 3: There are 15 dimes.
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\n" ); document.write( "$0.05N+$0.10D+$0.25Q=$4.00
\n" ); document.write( "$0.05(20)+$0.10(15)+$0.25(6)=$4.00
\n" ); document.write( "$1.00+$1.50+$1.50=$4.00
\n" ); document.write( "$4.00=$4.00\r
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