Results 1 to 6 of 6

Math Help - Not getting the subtraction part neither

  1. #1
    Junior Member
    Joined
    Sep 2012
    From
    Florida
    Posts
    32

    Not getting the subtraction part neither

    Subtract:

    Box 1: Enter your answer as a reduced mixed number or as a whole number. Example: 2 1/2 = `2 1/2`, or 2_1/2 = `2 1/2`
    Enter DNE for Does Not Exist, oo for Infinity


    ^.. Walk me through the steps please.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    146

    Re: Not getting the subtraction part neither

    There are many ways to do this, but the systematic way to to get common demoninators everywhere. In this case, that means converting the mixed fractions into simple fractions - each with demoninator 8. Then combine them (subtract them). Then, if necessary, write the result as a reduced mixed number (do the division and write the remainder in fraction form).

    I'll do one step to get you started:

    5 \frac{1}{8} = 5 + \frac{1}{8} = \frac{5}{1} + \frac{1}{8} = \frac{(5)(8)}{(1)(8)} + \frac{1}{8} = \frac{40}{8} + \frac{1}{8} = \frac{40 + 1}{8} = \frac{41}{8}

    Now do the same thing for 4 \frac{7}{8}, and then you'll be ready to subtract them.
    Last edited by johnsomeone; October 12th 2012 at 03:30 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2012
    From
    Florida
    Posts
    32

    Re: Not getting the subtraction part neither

    4 7/8 = 4 + 7/8 = (4)(8)= (7)(8) +7/8 = 32/8 +7/8=32+8/8=40/8


    5?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,963
    Thanks
    1631

    Re: Not getting the subtraction part neither

    Another way to do it is this:

    (First, suppose the problem were 51- 47. Since 7 is larger than 1, you have to "borrow" from the tens place: 11- 7 is 4 and since we have borrowed from the tens place, similarly, we have 4- 4= 0 so 51- 47= 04= 4.)

    To subtract 5\frac{1}{8}- 4\frac{7}{8} note that you cannot subtract 7 from 8 we have to "borrow" from the ones place: 1+ \frac{1}{8}= \frac{8}{8}+ \frac{1}{8}= \frac{9}{8} and \frac{9}{8}- \frac{7}{8}= \frac{2}{8}= \frac{2}{2(4)}= \frac{1}{4}. And since we have borrowed from the 5 we have 4- 4= 0. 5\frac{1}{8}- 4\frac{7}{8}= \frac{1}{4}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,963
    Thanks
    1631

    Re: Not getting the subtraction part neither

    Quote Originally Posted by Rosestar View Post
    4 7/8 = 4 + 7/8
    Yes, that's true.

    = (4)(8)= (7)(8) +7/8
    What? how did this happen? You are writing "=" between things that are not equal! Don't do that!
    = 32/8 +7/8
    Now you are back on the right track

    =32+8/8=40/8
    It would be better to use parentheses: (32+ 8)/8= 40/8. But now "7" suddenly became "8". How did that happen?

    5?
    Last edited by HallsofIvy; October 12th 2012 at 04:20 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,791
    Thanks
    1687
    Awards
    1

    Re: Not getting the subtraction part neither

    Quote Originally Posted by Rosestar View Post
    Subtract:
    This is a pure and simple rant.
    The mathematics education community should 'outlaw' mixed fractions.
    That community pushes calculators and/or computer algebra systems neither of which handles those. I do not understand how you have both ways. It has to be the force of habit.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 25
    Last Post: January 15th 2010, 03:53 AM
  2. Stucked @ part ii after comleting part i
    Posted in the Geometry Forum
    Replies: 5
    Last Post: December 31st 2009, 06:16 PM
  3. Sum of a part arithmatic and part geometric sequence
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: October 25th 2009, 10:53 PM
  4. Replies: 14
    Last Post: February 13th 2009, 04:27 AM
  5. subtraction
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 9th 2008, 07:33 AM

Search Tags


/mathhelpforum @mathhelpforum