document.write( "Question 186830: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes was doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes does he have? \n" ); document.write( "
Algebra.Com's Answer #140052 by feliz1965(151)![]() ![]() ![]() You can put this solution on YOUR website! n + d = 6.05......... Equation A \n" ); document.write( "(n - 10) + 2d = 9.85...Equation B\r \n" ); document.write( "\n" ); document.write( "Solve for n in Equation A.\r \n" ); document.write( "\n" ); document.write( "n = 6.05 - d\r \n" ); document.write( "\n" ); document.write( "We plug that value for n into Equation B.\r \n" ); document.write( "\n" ); document.write( "6.05 - d + 2d = 9.85\r \n" ); document.write( "\n" ); document.write( "6.05 + d = 9.85\r \n" ); document.write( "\n" ); document.write( "d = 9.85 - 6.05\r \n" ); document.write( "\n" ); document.write( "d = 3.80....This is the total value of the dimes Joe has. However and as you know, each dime is 10 cents.\r \n" ); document.write( "\n" ); document.write( "How many dimes are in $3.80? There are 38 dimes.\r \n" ); document.write( "\n" ); document.write( "To find the value of n, we select EITHER Equation A or B, plug 3.80 for d and simplify to find n. If you look carefully, you will notice that Equation A is easier to use.\r \n" ); document.write( "\n" ); document.write( "I will use Equation A.\r \n" ); document.write( "\n" ); document.write( "n + d = 6.05...Equation A\r \n" ); document.write( "\n" ); document.write( "n + 3.80 = 6.05\r \n" ); document.write( "\n" ); document.write( "n = 6.05 - 3.80\r \n" ); document.write( "\n" ); document.write( "n = 2.25\r \n" ); document.write( "\n" ); document.write( "How many nickels in $2.25? \r \n" ); document.write( "\n" ); document.write( "Joe has 45 nickels.\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |