document.write( "Question 4912: A collection of nickels, dimes and quarters has a value of $7.50. If there are 3 fewer dimes than nickels and twice as many quarters as nickels, how many dimes are there? \n" ); document.write( "
Algebra.Com's Answer #2415 by Abbey(339)\"\" \"About 
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Let x=number of nickels
\n" ); document.write( "let x-3=number of dimes (3 fewer than nickels)
\n" ); document.write( "let 2x=number of quarters (twice as many as nickels)\r
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\n" ); document.write( "\n" ); document.write( "Let 5x=the value of all nickels (5 cents for each nickel)
\n" ); document.write( "Let 10(x-3)= the value of all dimes (10 cents for each dime)
\n" ); document.write( "Let 25(2x)= the value of all quarters (25 cents for each quarter)\r
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\n" ); document.write( "\n" ); document.write( "Then adding all values together we know:
\n" ); document.write( "$7.50= total value of all coins
\n" ); document.write( "We are working in cents, so convert$7.50 to 750 cents and write the equation:
\n" ); document.write( "750=5x+10(x-3)+25(2x)\r
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\n" ); document.write( "\n" ); document.write( "Clear the parentheses:
\n" ); document.write( "750=5x+10x-30+50x
\n" ); document.write( "Combine like terms:
\n" ); document.write( "750=65x-30
\n" ); document.write( "Add 30 to both sides:
\n" ); document.write( "780=65x
\n" ); document.write( "Divide both sides by 65:
\n" ); document.write( "12=x\r
\n" ); document.write( "\n" ); document.write( "Using our original statments we know their are 12 nickels, 9 dimes (3 fewer than the nickels) and 24 quarters (twice as many as the nickels).\r
\n" ); document.write( "\n" ); document.write( "Check your answer:
\n" ); document.write( "12 * $.05 = $.60
\n" ); document.write( " 9 * $.10 = $.90
\n" ); document.write( "24 * $.25 =$6.00\r
\n" ); document.write( "\n" ); document.write( "$6.00+.90+.60=$7.50, So this is correct.
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