document.write( "Question 1204507: Rachelle has
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Algebra.Com's Answer #840792 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
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\n" ); document.write( "\n" ); document.write( "let dime be \"d\" and nickel \"n\"\r
\n" ); document.write( "\n" ); document.write( "\"1d\"=$\"0.10\"
\n" ); document.write( "\"1n\"=$\"0.05\"\r
\n" ); document.write( "\n" ); document.write( "if Rachelle has $\"4.45\" in dimes and nickels, we have\r
\n" ); document.write( "\n" ); document.write( "\"0.10d%2B0.05n=4.45\".....eq.1\r
\n" ); document.write( "\n" ); document.write( "if the number of nickels is twenty more than the number of dime, we have\r
\n" ); document.write( "\n" ); document.write( "\"n=d%2B20\"...eq.2\r
\n" ); document.write( "\n" ); document.write( "go to\r
\n" ); document.write( "\n" ); document.write( "\"0.10d%2B0.05n=4.45\".....eq.1, substitute \"n\"\r
\n" ); document.write( "\n" ); document.write( "\"0.10d%2B0.05%28d%2B20%29=4.45\".....solve for \"d\"\r
\n" ); document.write( "\n" ); document.write( "\"0.10d%2B0.05d%2B0.05%2A20=4.45\"\r
\n" ); document.write( "\n" ); document.write( "\"0.15d%2B1=4.45\"\r
\n" ); document.write( "\n" ); document.write( "\"0.15d=4.45-1\"\r
\n" ); document.write( "\n" ); document.write( "\"0.15d=3.45\"\r
\n" ); document.write( "\n" ); document.write( "\"d=3.45%2F0.15\"\r
\n" ); document.write( "\n" ); document.write( "\"d=345%2F15\"\r
\n" ); document.write( "\n" ); document.write( "\"d=23\"\r
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\n" ); document.write( "\n" ); document.write( "go to\r
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\n" ); document.write( "\n" ); document.write( "\"n=d%2B20\"...eq.2, substitute \"d\"\r
\n" ); document.write( "\n" ); document.write( "\"n=23%2B20\"\r
\n" ); document.write( "\n" ); document.write( "\"n=43\"\r
\n" ); document.write( "\n" ); document.write( "Rachelle has \"23\" dimes and \"43+\"nickels.\r
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