document.write( "Question 954026: Mary has $4.00 in nickels, dimes, and quarters. If she has twice as many dimes as quarters and twenty-five more nickels than dimes, how many coins of each type does she have? \n" ); document.write( "
Algebra.Com's Answer #582644 by macston(5194)![]() ![]() You can put this solution on YOUR website! Q=quarters; D=dimes=2Q; N=nickels=D+25=2Q+25 \n" ); document.write( "$0.25Q+$0.10D+$0.05N=$4.00 Substitute for D and N. \n" ); document.write( "$0.25Q+$0.10(2Q)+$0.05(2Q+25)=$4.00 \n" ); document.write( "$0.25Q+$0.20Q+$0.10Q+$1.25=$4.00 Subtract $1.25 from each side. \n" ); document.write( "$0.55Q=$2.75 Divide each side by $0.55 \n" ); document.write( "Q=5 ANSWER 1:There were 5 quarters \n" ); document.write( "D=2Q=10 ANSWER 2: There were 10 dimes. \n" ); document.write( "N=D+25=35 ANSWER 3:There were 35 nickels. \n" ); document.write( "CHECK: \n" ); document.write( "$0.25Q+$0.10D+$0.05N=$4.00 \n" ); document.write( "$0.25(5)+$0.10(10)+$0.05(35)=$4.00 \n" ); document.write( "$1.25+$1.00+$1.75=$4.00 \n" ); document.write( "$4.00=$4.00 \n" ); document.write( " |