document.write( "Question 87873: Number problems. Jill has #3.50 in nickles and dimes. If she has 50 coins, how many of each type of coins does she have?\r
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Algebra.Com's Answer #63773 by ankor@dixie-net.com(22740)\"\" \"About 
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n = number of nickels; d = number of dime
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\n" ); document.write( "Write a simple equation for each statement
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\n" ); document.write( "\"Jill has #3.50 in nickels and dimes.\"
\n" ); document.write( " .05n + .10d = 3.50
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\n" ); document.write( "\"If she has 50 coins,\"
\n" ); document.write( " n + d = 50
\n" ); document.write( "Rearrange for substitution; subtract d from both sides:
\n" ); document.write( " n = (50 - d)
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\n" ); document.write( " how many of each type of coins does she have?
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\n" ); document.write( "Substitute (50-d) for n in the 1st equation, solve for d:
\n" ); document.write( ".05n + .10n = 3.50
\n" ); document.write( ".05(50-d) + .10d = 3.50
\n" ); document.write( "2.5 - .05d + .10d = 3.50
\n" ); document.write( "-.05d + .10d = 3.50 - 2.50
\n" ); document.write( ".05d = 1.00
\n" ); document.write( "d = 1/.05
\n" ); document.write( "d = 20 dimes
\n" ); document.write( "Then
\n" ); document.write( "50 - 20 = 30 nickels:
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\n" ); document.write( "Check solutions:
\n" ); document.write( ".05(30) + .10(20) = 3.50
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\n" ); document.write( "That's not hard, is it? \n" ); document.write( "
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