document.write( "Question 742235: John has a collection of nickels,dimes, and quarters. Altogether, there are 241 coins worth $26.20. There are 4 more nickels than dimes. How many of each type of coin are in the collection. \n" ); document.write( "
Algebra.Com's Answer #452381 by tommyt3rd(5050)![]() ![]() You can put this solution on YOUR website! # of Coins: N-nickels, D-dimes, Q-quarters\r \n" ); document.write( "\n" ); document.write( "there are 241 coins - means \n" ); document.write( "\n" ); document.write( "worth $26.20 - means \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are 4 more nickels than dimes means \n" ); document.write( "\n" ); document.write( "So we have 3 equations and 3 unknowns:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "So we begin...\r \n" ); document.write( "\n" ); document.write( "5 times the second equation minus the first equation will eliminate Q:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since N=D+4 we can write \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "This means that there D+4=95+4=99 nickels\r \n" ); document.write( "\n" ); document.write( "Finally we can see that there must be 241-99-95=47 quarters\r \n" ); document.write( "\n" ); document.write( "(so there are 99 nickels, 95 dimes and 47 quarters) \n" ); document.write( " |