document.write( "Question 179941: Joe has a collection of nickels and dimes that is worth $6.05. If the number of dimes is doubled and the number of nickels was decreased by 10, the value of the coins would be $9.85. How many dimes do we have? \n" ); document.write( "
Algebra.Com's Answer #134850 by MathTherapy(10552)\"\" \"About 
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Let number of dimes be d, and the number of nickels n
\n" ); document.write( "Then, d(.10) + n(.05) = 6.05, and
\n" ); document.write( "2d(.10) + (n – 10).05 = 9.85, or .2d + .05n - .5 = 9.85\r
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\n" ); document.write( "\n" ); document.write( " .1d + .05n = 6.05
\n" ); document.write( " .2d + .05n = 10.35\r
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\n" ); document.write( "\n" ); document.write( "Subtract eq (ii) from eq (i) - .1d = - 4.3
\n" ); document.write( " d = 43 \r
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\n" ); document.write( "\n" ); document.write( "Therefore, we have 43 dimes.\r
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\n" ); document.write( "\n" ); document.write( "To find how many nickels there are, we just substitute 43 for d in eq (i).
\n" ); document.write( "This gives us: 4.3 + .05n = 6.05
\n" ); document.write( " .05n = 1.75
\n" ); document.write( " n = 35 \r
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\n" ); document.write( "\n" ); document.write( " 43 dimes = $4.30
\n" ); document.write( " 35 nickels = $1.75
\n" ); document.write( " $4.30 + $1.75 = $6.05\r
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\n" ); document.write( "\n" ); document.write( "Doubling dimes = 2(43) = 86 = $8.60
\n" ); document.write( "Decreasing nickels by 10 = (35 – 10) = 25 = $1.25
\n" ); document.write( " $8.60 + $1.25 = $9.85
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