document.write( "Question 517095: There are 223 coins total of quarters, dimes, and nickels. The total value of the coins is $20.10. How many coins do I have of each?
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Algebra.Com's Answer #344673 by JBarnum(2146)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "a quarter is .25 dime .10 nickel .05 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "oh boy elimination and substitution fun \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "subtrct to eliminate dimes \n" ); document.write( "42.40=.15Q-.5N \n" ); document.write( ". \n" ); document.write( "the prob above is that i dont have a 3rd equation and cant do much without one, like if they said there was 100 more dimes than nickels then id have a third equation and could solve this better(d=100+n). \n" ); document.write( ". \n" ); document.write( "yeh im gonna do guess n check \n" ); document.write( "im sure someone can do all that math but guess n check works to: \n" ); document.write( "right off bat 80 quarters is $20 so you know there will be very few quarters \n" ); document.write( "200 dimes is $20 so there will be alot of dimes \n" ); document.write( "so i started by taking 50 from the dimes to make 150 coins then 70 nickles to make 220 coins and then started swaping coins around to stay in the 220s till i got the right combo, i tried to keep the total at a constant 20.10 and realized i either needed 1 extra dime or 2 extra nickels 2 make that 10 cents \n" ); document.write( ". \n" ); document.write( ". \n" ); document.write( "6q=1.50 150d=15 72n=3.60 228=>20.10 nope \n" ); document.write( "8q=2 150d=15 62n=3.10 220=>20.10 nope \n" ); document.write( "5q=1.25 157d=15.70 63n=3.15 225=>20.10 nope \n" ); document.write( "5q=1.25 159d=15.90 59n=2.95 223=>20.10 YES \n" ); document.write( "Q=5, D=159, N=59\r \n" ); document.write( "\n" ); document.write( "apparently theres more than 1 solution lol \n" ); document.write( " |