document.write( "Question 41648: I am having trouble setting up this problem:\r
\n" ); document.write( "\n" ); document.write( "*John found a box of coins. The box contained nickels,dimes, and quarters. He added up the change and there were 85 coins totaling $6.25 in the box. If there were 3 times as many nickels as dimes, how many quarters were there?\r
\n" ); document.write( "\n" ); document.write( "-I came up with the equation 3x+y=85 (x=nickels, y=dimes),but this equation alone doesn't help so I think that I may need to use substitution, but I can't come up with another equation.\r
\n" ); document.write( "\n" ); document.write( "Please Help! Thank YOu \n" ); document.write( "

Algebra.Com's Answer #46036 by 303795(602) $\"\"$ $\"About$

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There are 3 nickels for each dime so (each dime + its associated 3 nickels) is worth 25 cents.
\n" ); document.write( "Use x for the number of dimes and y for the number of quarters
\n" ); document.write( "Using the value of the coins form equation 1
\n" ); document.write( ".25x +.25y=6.25
\n" ); document.write( "Multiply this equation by 4 to remove the decimals
\n" ); document.write( "x + y = 25
\n" ); document.write( "Using the number of coins form equation 2
\n" ); document.write( "4x+y=85
\n" ); document.write( "Subtract the first equation from the second to get
\n" ); document.write( "3x=60
\n" ); document.write( "so x = 20
\n" ); document.write( "so y (the number of quarters must be 5)
\n" ); document.write( "(There were 60 nickels, 20 dimes and 5 quarters (85 coins) with a value of $3.00+$2.00+$1.25=$6.25) \n" ); document.write( "

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