document.write( "Question 4082: Add the following fractions, showing the steps you used:\r
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document.write( "(a) 2/3 + 3/4 =
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document.write( "(b) (3/5 + 4/7) + 2/5 =
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document.write( "(c) 5/6 + 5/8 = \n" );
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Algebra.Com's Answer #1802 by carl(7)![]() ![]() ![]() You can put this solution on YOUR website! (note.. remember when adding or subtracting you need the same denominator \n" ); document.write( " for the fractions).....\r \n" ); document.write( "\n" ); document.write( "so... \r \n" ); document.write( "\n" ); document.write( "a. 2/3 + 3/4 = (think 12) \n" ); document.write( "2/3 becomes 8/12 (3 divided by 12 = 4 x 2 = 8) and 3/4 becomes 9/12, then add em..... 8 + 9 = 17 and drop the denominator of 12 and you get 17/12, now reduce the imporper faction to 1 and 5/12. \r \n" ); document.write( "\n" ); document.write( "(the improper faction is... divide 12 by 17 and get 1, then subtract 12 from 17 and get 5, place it over the denominator)\r \n" ); document.write( "\n" ); document.write( "b. (3/5 + 4/7) + 2/5 = first the ( ) then the other... \n" ); document.write( "here you go.... \n" ); document.write( "with 3/5 and 4/7 think 35... and you should get 1 6/35. BUT.. we are not done, now we must add 1 6/35 + 2/5 (ouch)! ok.. lets keep the 35, because 5 will go into 35.... so....\r \n" ); document.write( "\n" ); document.write( "1 6/35 remains and (2/5 changes to) 14/35 then lets add the top, 20/35 and drop the whole 1... and we get 1 20/35. lets reduce because we can take 5 from each... 1 4/7... right???\r \n" ); document.write( "\n" ); document.write( "c. 5/6 + 5/8 = ok.. because we are using 6 and 8... we can see they go into 24 very nicely.... \r \n" ); document.write( "\n" ); document.write( "5/6 becomes 20/24 (remember the divide 24 / 6 = 4 x 5 = 20) and then the 5/8 becomes 15/24.... so lets add the 20/24 + 15/24 and we get 35/24.... remember the improper issue, we get 1 11/24.\r \n" ); document.write( "\n" ); document.write( "Hope that helps. \n" ); document.write( "Carl \n" ); document.write( " |