document.write( "Question 24108: What I have is a money mixture - I can't remember how to figure it. The problem is John has a change jar which he empties occasionally. This time he has all nickels and dimes. He has nine times as many dimes as nickels, and the value of the dimes equals $5.10 more than the value of the nickels. How many dimes and nickels does John have. I have two equations, 9d=n or d=n+$5.10 we are supposed to use a method of substitution or similar to solve. Please help! Three people here, and no clues. - Nicky \n" ); document.write( "
Algebra.Com's Answer #12804 by rapaljer(4671)\"\" \"About 
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I think you need to say
\n" ); document.write( "Let n = number of nickels
\n" ); document.write( "d = number of dimes.\r
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\n" ); document.write( "\n" ); document.write( "He has nine times as many dimes as nickes, so the number of dimes is 9 times the number of nickels. This means that d = 9n \r
\n" ); document.write( "\n" ); document.write( "Value of the dimes = 10d
\n" ); document.write( "Value of the nickels = 5n\r
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\n" ); document.write( "\n" ); document.write( "Value of the dimes equals 510 cents more than the value of the nickels
\n" ); document.write( "10d = 5n + 510\r
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\n" ); document.write( "\n" ); document.write( "This is a good place to substitute, since d=9n\r
\n" ); document.write( "\n" ); document.write( " 10d = 5n + 510
\n" ); document.write( "10*9n = 5n+ 510
\n" ); document.write( "90n= 5n + 510
\n" ); document.write( "85n= 510
\n" ); document.write( "n = 6 Nickels
\n" ); document.write( "9n = 54 Dimes\r
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\n" ); document.write( "\n" ); document.write( "Check. Value of 6 nickels = $.30
\n" ); document.write( "Value of 54 dimes = $5.40\r
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\n" ); document.write( "\n" ); document.write( "Dimes are worth $5.10 more than the nickels so it checks!\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC\r
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