document.write( "Question 167378: Juan has a jar containing eighty coins, all of which are either quarters or nickels. The total value of the coins is $14.60. How many of each type of coin does he have?\r
\n" ); document.write( "\n" ); document.write( "Arbitrarily I tried 60 quarters, but that ended up $15.00; then I dropped it to 55 quarters, which ended up $13.75, but the remainng number of coins had to be 25 which would have brought the total to 80 coins (correct answer), but it equals $15.85 which is too much.(You see, I've been working at it.) Question: is there a formula for this stuff? Thank you very much. \n" ); document.write( "
Algebra.Com's Answer #123288 by ankor@dixie-net.com(22740)\"\" \"About 
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Juan has a jar containing eighty coins, all of which are either quarters or nickels. The total value of the coins is $14.60. How many of each type of coin does he have?
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\n" ); document.write( "Actually it is quite simple using two equations:
\n" ); document.write( "Let n = no. of nickels
\n" ); document.write( "let q = no. of quarters
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\n" ); document.write( "Total coin equation
\n" ); document.write( "n + q = 80
\n" ); document.write( "or
\n" ); document.write( "n = (80-q)
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\n" ); document.write( "Total value equation:
\n" ); document.write( ".05n + .25q = 14.60
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\n" ); document.write( "From the 1st equation, substitute (80-q) for n
\n" ); document.write( ".05(80-q) + .25q = 14.60
\n" ); document.write( "4 - .05q + .25q = 14.60
\n" ); document.write( "-.05q + .25q = 14.60 - 4
\n" ); document.write( ".20q = 10.60
\n" ); document.write( "q = \"10.60%2F.20\"
\n" ); document.write( "q = 53 quarters
\n" ); document.write( ":
\n" ); document.write( "I'll let you find the no. of nickels \n" ); document.write( "
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