document.write( "Question 1105533: I have 27 coins that total $3.30. I have 3 times as many quarters as dimes; how many nickels do I have? This is my question and I have tried several ways to set the problem up but none give me the correct answer. How do I begin to set this up?\r
\n" ); document.write( "\n" ); document.write( "Thank You,
\n" ); document.write( "Iron Heart
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Algebra.Com's Answer #720401 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "With the information given in this way, I can't tell whether substitution or elimination would be easier. Since the other tutor showed a solution using substitution, let me try elimination.

\n" ); document.write( "That means using two variables:
\n" ); document.write( "let x = number of dimes
\n" ); document.write( "then 3x = number of quarters
\n" ); document.write( "let y = number of nickels

\n" ); document.write( "\"x%2B3x%2By+=+27\" the total number of coins is 27
\n" ); document.write( "(1) \"4x%2By+=+27\"
\n" ); document.write( "\"10%28x%29%2B25%283x%29%2B5%28y%29+=+330\" the total value of the coins (in cents) is 330
\n" ); document.write( "\"85x%2B5y+=+330\"
\n" ); document.write( "(2)\"17x%2By+=+66\"

\n" ); document.write( "Subtracting (1) from (2):
\n" ); document.write( "\"13x+=+39\"
\n" ); document.write( "\"x+=+3\"

\n" ); document.write( "So the number of dimes is x=3; the number of quarters is 3x=9; and since the total number of coins is 27, the number of nickels is 15.

\n" ); document.write( "HMMM.... Looks very different than the solution using substitution... and both methods seem to require about the same a mount of work....
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