document.write( "Question 181998This question is from textbook
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document.write( ": Rita has a collection of 105 coins consisting of nickels, dimes, and quarters. The number of dimes is 5 more than one-third the number of nickels, and the number of quarters is twice the number of dimes. how many coins of each kind does she have????\r
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document.write( "I would like help I don't know how to start. \n" );
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Algebra.Com's Answer #136611 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! Rita has a collection of 105 coins consisting of nickels, dimes, and quarters. The number of dimes is 5 more than one-third the number of nickels, and the number of quarters is twice the number of dimes. how many coins of each kind does she have???? \n" ); document.write( ". \n" ); document.write( "Let x = number of nickels \n" ); document.write( "then \n" ); document.write( "from:\"number of dimes is 5 more than one-third the number of nickels\" \n" ); document.write( "(1/3)x + 5 = number of dimes \n" ); document.write( "x/3 + 5 = number of dimes \n" ); document.write( "from:\"number of quarters is twice the number of dimes\" \n" ); document.write( "2((1/3)x + 5) = number of quarters \n" ); document.write( "2x/3 + 10 = number of quarters \n" ); document.write( ". \n" ); document.write( "Our equation comes from the fact that there are 105 total coins: \n" ); document.write( "x + (x/3 + 5) + (2x/3 + 10) = 105 \n" ); document.write( "Now, we solve for 'x'. Start by multiplying both sides by 3: \n" ); document.write( "3x + x + 15 + 2x + 30 = 315 \n" ); document.write( "6x + 45 = 315 \n" ); document.write( "6x = 270 \n" ); document.write( "x = 45 (number of nickels) \n" ); document.write( ". \n" ); document.write( "dimes: \n" ); document.write( "x/3 + 5 = 45/3 + 5 = 15 + 5 = 20 (number of dimes) \n" ); document.write( ". \n" ); document.write( "quarters: \n" ); document.write( "2x/3 + 10 = 2(45)/3 + 10 = 90/3 + 10 = 30 + 10 = 40 (number of quarters) \n" ); document.write( " \n" ); document.write( " |