document.write( "Question 887682: how many coins of each are there if there are 9 more dimes than nickles and 21 more quarters than dimes and the total is $10.00 \n" ); document.write( "

Algebra.Com's Answer #536798 by algebrapro18(249) $\"\"$ $\"About$

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Lets let n be the number of nickles we have, q be the number of quarters we have, and d be the number of dimes we have. Now we know that we have 9 more dimes than nickles. We can write his as the following equation:
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\n" ); document.write( "\n" ); document.write( "n = 9+d
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\n" ); document.write( "\n" ); document.write( "Now we also know that we have 21 more quarters than dimes. This can be written as the following equation:
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\n" ); document.write( "\n" ); document.write( "d = 21+q
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\n" ); document.write( "\n" ); document.write( "Now we also know that the number of nickles, dimes, and quarters together equals $10 in change. This can be written as:
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\n" ); document.write( "\n" ); document.write( "0.05n+0.1d+0.25q=10
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\n" ); document.write( "\n" ); document.write( "Now we have 3 equations with 3 variables each. We can now solve the system of equations. I'd suggest starting by plugging the equation for d into the equation for n and then solving the third equation for q. I'll let you do the algebra but you should get n = 30+q and then plug that and d=21+q into the first equation and solve for q. You should get q = 16. Then you can plug that back into the equations for n and d and get n = 46 and d = 37. Let me know if you need to see the work. \n" ); document.write( "

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