document.write( "Question 1184047: A fruit seller sold some apples and oranges.
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document.write( "He sold 3/7 of the apples in the morning and 3/8 of the remainder in the afternoon.
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document.write( "He sold 1/3 of the oranges in the morning and 1/4 of the remainder in the afternoon.
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document.write( "He had sole sold the same number of apples and oranges at the end of the day.
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document.write( "If there were 8 more oranges than apples left unsold, how many fruits were there at first? \n" );
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Algebra.Com's Answer #817905 by Octo-pie7(11)![]() ![]() You can put this solution on YOUR website! let a = # of apples he started with \n" ); document.write( "let o = # of oranges he started with \n" ); document.write( "3/7a were sold in the morning, so 4/7a remained. \n" ); document.write( "3/8(4/7a) were sold in the afternoon, so 5/8(4/7a) remained. \n" ); document.write( "1/3 a were sold in the morning, so 2/3 o remained. \n" ); document.write( "1/4(2/3o) were sold in the afternoon, so 3/4(2/3o) remained. \n" ); document.write( "3/8(4/7a) = 1/4(2/3o) \n" ); document.write( "5/8(4/7a) + 8 = 3/4(2/3o) \n" ); document.write( "12/36a = 2/12o => 3/14a = 1/6o \n" ); document.write( "20/36a + 8 = 6/12o => 5/14a + 8 = 1/2o \n" ); document.write( "3/14a = 1/6o \n" ); document.write( "5/14a + 8 = 1/2o \n" ); document.write( "(6)(3/14a) = (1/6o)(6) => 18/14a = o => 9/7a = o \n" ); document.write( "(-2)[(5/14a) + 8] = [1/2o] (-2) => -10/14a - 16 = -o => -5/7a - 16 = -o \n" ); document.write( "9/7a = o \n" ); document.write( "-5/7a - 16 = -o \n" ); document.write( "———————- \n" ); document.write( "4/7a - 16 = o \n" ); document.write( " + 16 +16 \n" ); document.write( "4/7a = 16 \n" ); document.write( "7/4 * 4/7a = 16 * 7/4 \n" ); document.write( "a = 28 \n" ); document.write( "3/14 (28) = 1/6o \n" ); document.write( "6 = 1/6o \n" ); document.write( "6 * 6 = 1/6o * 6 \n" ); document.write( "36 = o \n" ); document.write( " |