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.
\n" ); document.write( "5n+10(n+3)+25(30-2n+3)=355
\n" ); document.write( "

Algebra.Com's Answer #579562 by Alan3354(69443)\"\" \"About 
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( "
\n" );