document.write( "Question 150508This question is from textbook
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document.write( ": there were 3/4 as many women as there were men in a train yesterday. at the next station, 12 women and 6 men got in. as a result, there were 6/7 as many women as men on the train. how many men and women were in the train at first? \n" );
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Algebra.Com's Answer #110473 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! there were 3/4 as many women as there were men in a train yesterday. at the next station, 12 women and 6 men got in. as a result, there were 6/7 as many women as men on the train. how many men and women were in the train at first? \n" ); document.write( ". \n" ); document.write( "Let M = # of men \n" ); document.write( "(3/4)M = # of women \n" ); document.write( ". \n" ); document.write( "(6/7)(M+6) = (3/4)M + 12 \n" ); document.write( "To get rid of all the denominators, multiply both sides by 28: \n" ); document.write( "24(M+6) = 21M + 336 \n" ); document.write( "24M + 144 = 21M + 336 \n" ); document.write( "24M-21M = 336-144 \n" ); document.write( "3M = 192 \n" ); document.write( "M = 64 (Number of men in the train at first) \n" ); document.write( ". \n" ); document.write( "(3/4)64 = 3(16) = 48 (Number of women in the train at first) \n" ); document.write( " \n" ); document.write( " |