document.write( "Question 1105659: Hi
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document.write( "bob 3/7 of his money.john spent 1/5 of his and alan spent 1/3 of his.all of them had the same amount of money left. If the 3 of them had spent $144 , how much did each have at first.\r
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document.write( "thanks \n" );
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Algebra.Com's Answer #720522 by math_helper(2461) You can put this solution on YOUR website! \n" ); document.write( "Bob starts with B dollars: (4/7)B = k (1) (Bob spent 3/7 so he has 4/7 left ) \n" ); document.write( "John starts with J dollars: (4/5)J = k (2) \n" ); document.write( "Alan starts with A dollars: (2/3)A = k (3)\r \n" ); document.write( "\n" ); document.write( "(3/7)B + (1/5)J + (1/3)A = 144 (4)\r \n" ); document.write( "\n" ); document.write( "From (1) and (2): (4/7)B = k = (4/5)J > J = (5/7)B \n" ); document.write( "From (1) and (3): (4/7)B = k = (2/3)A > A = (6/7)B\r \n" ); document.write( "\n" ); document.write( "Now substitute these last two equations for J and A in (4) to get one equation in one unknown:\r \n" ); document.write( "\n" ); document.write( " (3/7)B + (1/5)(5/7)B + (1/3)(6/7)B = 144 \n" ); document.write( " (3/7)B + (1/7)B + (2/7)B = 144 \n" ); document.write( " (6/7)B = 144 \n" ); document.write( " B = 144*7 / 6 = 168 > A = 144, J = 120\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " Ans: Bob started with $168, Alan with $144, and John with $120 \n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "Check: \n" ); document.write( " (3/7)(168) + (1/5)(120) + (1/3)(144) \n" ); document.write( " = 72 + 24 + 48 \n" ); document.write( " = 144 (ok)\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |