document.write( "Question 1010925: In a collection box , there are dimes and quarters whose total value is $28. If there were as many quarters as there are dimes , and as many dimes as there are quarters , the total value would be $36.40. How many coins of each kind are in the collection box ? \n" ); document.write( "
Algebra.Com's Answer #626449 by Edwin McCravy(20060)\"\" \"About 
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document.write( "Let D = the number of dimes\r\n" );
document.write( "Let Q = the number of quarters \r\n" );
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document.write( "$0.10D + $0.25Q = $28.00   <--There are D dimes and Q quarters.\r\n" );
document.write( "$0.25D + $0.10Q = $36.40   <--if there were D quarters and Q dimes\r\n" );
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document.write( "Divide the second equation through by 5\r\n" );
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document.write( "10D + 25Q = 2800\r\n" );
document.write( " 5D +  2Q =  728\r\n" );
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document.write( "Multiply the second equation through by -2\r\n" );
document.write( "And add them to eliminate D\r\n" );
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document.write( " 10D + 25Q =  2800\r\n" );
document.write( "-10D -  4Q = -1456\r\n" );
document.write( "------------------\r\n" );
document.write( "       21Q =  1344\r\n" );
document.write( "         Q =  64\r\n" );
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document.write( "Substitute in \r\n" );
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document.write( "  5D +  2Q =  728\r\n" );
document.write( "5D + 2(64) = 728\r\n" );
document.write( "  5D + 128 = 728\r\n" );
document.write( "        5D = 600\r\n" );
document.write( "         D = 120\r\n" );
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document.write( "64 Quarters and 120 Dimes\r\n" );
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document.write( "Checking:\r\n" );
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document.write( "64 Quarters is worth 64($0.25) = $16.00\r\n" );
document.write( "120 Dimes is worth 120($0.10)  = $12.00\r\n" );
document.write( "----------------------------------------\r\n" );
document.write( "                         TOTAL = $28.00\r\n" );
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document.write( "120 Quarters is worth 120($0.25) = $30.00\r\n" );
document.write( "64 Dimes is worth 64($0.10)      = $ 6.40\r\n" );
document.write( "-----------------------------------------\r\n" );
document.write( "                           TOTAL = $36.40\r\n" );
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document.write( "Edwin
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