document.write( "Question 136799: BETSY HAS A JAR CONTAINING 80 COINS, ALL OF WHICH ARE EITHER QUARTERS OR NICKLES. THE TOTAL VALUE OF THE COINS IS $ 14.60. HOW MANY OF EACH TYPE OF COIN DOES SHE HAVE?\r
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document.write( "CAN YOU PLEASE SHOW ME A SIMPLE FORMAT TO FOLLOW TO SOLVE MY PROBLEM AND HOW TO COME OUT WITH THE CORRECT ANSWER?
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document.write( "THANK,
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document.write( "lj23kfuller@yahoo.com \n" );
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Algebra.Com's Answer #100125 by jojo14344(1513)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Let \"x\" = # of coins for quarters \n" ); document.write( " \"y\" = # of coins for nickels \n" ); document.write( "then, x + y = 80 coins, right? This will be equation 1. \n" ); document.write( "Next, when we multiply 0.25(quarter) by \"x\" (# of coins) + \n" ); document.write( "nickel (.05) multiply by \"y\" (# of coins) equals $14.60 isn't it?\r \n" ); document.write( "\n" ); document.write( "Equating that fact, \n" ); document.write( "0.25x+0.05y=$14.60 and this is equation 2. \n" ); document.write( "Going back to equation 1, we get \"x = $80-y\" . Then substitue this value to x in equation 2, \n" ); document.write( "0.25($80-y)+0.05y=$14.60 \n" ); document.write( "$20-0.25y+0.05y=$14.60 \n" ); document.write( "$20-$14.60=0.20y \n" ); document.write( "y=27 coins (nickels) \n" ); document.write( "For x = 80-y= 80-27 \n" ); document.write( " x= 53 coins (quarter) \n" ); document.write( "To check: \n" ); document.write( "53quarters+27nickels = 80 coins \n" ); document.write( "also, \n" ); document.write( "0.25(53)+0.05(27)=$14.60 \n" ); document.write( "$13.25+$1.35=$14.60 \n" ); document.write( "$14.60=$14.60\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |