document.write( "Question 949308: I have 30 total coins that total $3.55. There are 3 more dimes than nickels. How many quarters do I have.
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document.write( "5n+10(n+3)+25(30-2n+3)=355 \n" );
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Algebra.Com's Answer #579562 by Alan3354(69443) You can put this solution on YOUR website! I have 30 total coins that total $3.55. There are 3 more dimes than nickels. How many quarters do I have. \n" ); document.write( "5n+10(n+3)+25(30-2n+3)=355 \n" ); document.write( "5n+10(n+3)+25(30-(2n+3))=355 \n" ); document.write( "5n+10(n+3)+25(30-2n-3)=355 **************** The -1 applies to the 3\r \n" ); document.write( "\n" ); document.write( "-------------- \n" ); document.write( "n = # of nickels \n" ); document.write( "Solve for n \n" ); document.write( "5n+10(n+3)+25(30-2n-3)=355 \n" ); document.write( "5n + 10n+30 + 750-50n-75 = 355 \n" ); document.write( "-35n + 705 = 355 \n" ); document.write( "-35n = -350 \n" ); document.write( "n = 10 nickels \n" ); document.write( "----- \n" ); document.write( "Solve for q \n" ); document.write( " |