document.write( "Question 903361: In changing a $5, Marge received 9 more dimes than nickels and 7 fewer quarters than dimes. How many of each coin did she receive?\r
\n" ); document.write( "\n" ); document.write( "nickels=x
\n" ); document.write( "dimes=9+x
\n" ); document.write( "quarters= x-7 = 9+x-7 = x-2\r
\n" ); document.write( "\n" ); document.write( "x+9+x+x-2=?
\n" ); document.write( "3x+7=?\r
\n" ); document.write( "\n" ); document.write( "Not sure what the number should be that the equation equals. Would it be 500?
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Algebra.Com's Answer #547931 by josgarithmetic(39625)\"\" \"About 
You can put this solution on YOUR website!
Use more variables, and do not restrict assigning to x or y or z.\r
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\n" ); document.write( "\n" ); document.write( "DIMES, NICKELS, QUARTERS -------
\n" ); document.write( "Try n, d, q for how many of each coin.\r
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\n" ); document.write( "\n" ); document.write( "\"9 more dimes than nickels\", \"d-n=9\".\r
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\n" ); document.write( "\n" ); document.write( "\"7 fewer quarters than dimes\", \"q=d-7\".\r
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\n" ); document.write( "\n" ); document.write( "You need to think carefully about forming those algebraic sentences; mistakes in there can easily happen.\r
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\n" ); document.write( "\n" ); document.write( "Another equation is for the \"In changing $5...\".
\n" ); document.write( "Account for the change:
\n" ); document.write( "\"0.05n%2B0.10d%2B0.25q=5\"
\n" ); document.write( "Simplify this money count equation.
\n" ); document.write( "\"n%2B2d%2B5q=100\"\r
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\n" ); document.write( "\n" ); document.write( "System of equations is
\n" ); document.write( "\"highlight_green%28system%28n%2B2d%2B5q=100%2Cd-n=9%2Cq=d-7%29%29\"\r
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\n" ); document.write( "\n" ); document.write( "More than one way to solve is possible. Try substituting for q in the other two equations. Use q=d-7. (Cannot do that to the d-n=9 because it has no q in it);
\n" ); document.write( "\"n%2B2d%2B5%28d-7%29=100\"
\n" ); document.write( "\"n%2B2d%2B5d-35=100\"
\n" ); document.write( "\"n%2B7d-35=100\"
\n" ); document.write( "Now use the d-n=9 equation to substitute for d, in the form of d=n+9:
\n" ); document.write( "\"n%2B7%28n%2B9%29-35=100\"
\n" ); document.write( "\"n%2B7n%2B63-35=100\"
\n" ); document.write( "\"8n=100-63%2B35\"
\n" ); document.write( "\"8n=100-28=72\"
\n" ); document.write( "\"n=72%2F8\"
\n" ); document.write( "\"highlight%28n=9%29\"-----ONE OF THE COIN, NINE NICKELS\r
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\n" ); document.write( "\n" ); document.write( "\"d=n%2B9\"
\n" ); document.write( "\"d=9%2B9\" because just found n=9,
\n" ); document.write( "\"highlight%28d=18%29\"-------HOW MANY DIMES\r
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\n" ); document.write( "\n" ); document.write( "\"q=d-7\"
\n" ); document.write( "\"q=18-7\"
\n" ); document.write( "\"highlight%28q=11%29\"-------THE COUNT OF QUARTERS
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