document.write( "Question 232563: Could you help me with this problem. I can't seem to figure it out correctly. \r
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document.write( "Amy has two more nickels than dimes and five more dimes than quarters. Her nickels, dimes, and quarters total $3.25. How many of each kind of coin has she? \n" );
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Algebra.Com's Answer #171930 by rjd41(9) ![]() You can put this solution on YOUR website! First, I recommend changing everything to cents.\r \n" ); document.write( "\n" ); document.write( "Dimes is listed on both sides of the problem, this is what you will focus on then. \n" ); document.write( "You want to show all coins listed in the problem:\r \n" ); document.write( "\n" ); document.write( "x - will represent dimes.\r \n" ); document.write( "\n" ); document.write( "10(x) + 5(x+2) + 25(x-5), notice the three coins and how they follow the problem.\r \n" ); document.write( "\n" ); document.write( "10x + 5x+10 + 25x-125 = 325\r \n" ); document.write( "\n" ); document.write( "so you get 40x + 10 - 125 = 325\r \n" ); document.write( "\n" ); document.write( "add 115 to each side to get 40x alone \r \n" ); document.write( "\n" ); document.write( "40x + 115 - 115 = 325 + 115 \r \n" ); document.write( "\n" ); document.write( "40x = 440\r \n" ); document.write( "\n" ); document.write( "divide each side by 40 to get x alone\r \n" ); document.write( "\n" ); document.write( "x =11, the 11 is how many dimes you have. 11(10) = 110\r \n" ); document.write( "\n" ); document.write( "Two more nickels than dimes, 11 + 2 = 13, 13(5)=65\r \n" ); document.write( "\n" ); document.write( "five more dimes than quarters so 11 - 5 = 6, 6(25)= 150\r \n" ); document.write( "\n" ); document.write( "check: 110 + 65 + 150 = 325 \n" ); document.write( " \n" ); document.write( " |