Subtracting Mixed Numbers with Borrowing

Date: 03/31/2009 at 22:29:05
From: Bejoy
Subject: Simplify 6 (1/8) - 3 (4/12)

For 6(1/8)-3(4/12) my teacher got the answer 2(19/24) but i'm not
getting that answer.  i'm getting 3(5/24)

First i multiplied the denominator 8 by 3 then i multiplied the 
second denominator 12 by 2.  And i got the common denominator 24.  Now
my answer is 6(3/24)-3(8/24).  After this i subtracted those 
fractions and got 3(5/24) but my teacher got a totally different 
answer.  i have no idea how did she get that answer.  Can you please 
help me to get the answer that my teacher got?

Date: 04/01/2009 at 11:19:41
From: Doctor Ian
Subject: Re: Simplify 6 (1/8) - 3 (4/12)

Hi Bejoy -

When we subtract two mixed numbers, we subtract the whole number
parts, and we subtract the fractional parts, and put them together to
get an answer, right?

You have

     6  3/24
  -  3  5/24
  ----------

Now, you can subtract 3 from 6, 

     6  3/24
  -  3  5/24
  ----------
     3   ?

But can you subtract 5/24 from 3/24?  No, because 5/24 is larger!

Let's think about money for a moment.  Suppose we invent a coin that
is worth 1/24 of a dollar.  (It would be worth a little bit less than
a nickel.) 

I have 6 dollars and 3 of those coins.  I need to give you 3 dollars
and 5 of those coins.  Can I do it?  No, I don't have enough coins. 
What I _can_ do is find someone to "make change" for a dollar, giving
me 24 of those coins.  

Then I have 5 dollars and 27 coins.  So I can give you 3 of the 
dollars, leaving me with 2; and I can give you 5 of the coins, leaving
me with 22. 

In terms of your problem, this looks like

     6  3/24   trade 1 for 24/24       5 27/24
  -  3  5/24   ----------------->   -  3  5/24
  ----------                        ----------
                                       2 22/24  or 2 11/12

Does it make more sense now? 

Note that this is what you already learned to do when you subtract
whole numbers, and have to "borrow" or "regroup".  For example, if I
want to subtract

    543
  - 287
  -----

I can think of subtracting the various groups separately,

    500  40  3           5 hundreds  4 tens  3 ones
  - 200  80  7    or   - 2 hundreds  8 tens  7 ones
  ------------         ----------------------------

I can't subtract 7 from 3, so I have to "borrow" 10 from the second
column:

    500  30  13
  - 200  80   7
  -------------
              6

I can't subtract 80 from 30, so I have to "borrow" 100 from the third
column:

    400 130  13
  - 200  80   7
  -------------
         50   6

And now I can finish up:

    400 130  13
  - 200  80   7
  -------------
    200  50   6   ->  256

It's just the same idea, except sometimes you're using whole numbers,
and sometimes you're using fractions.  The same idea shows up when
using units like gallons and quarts,

    5 gal  1 qt          4 gal  5 qt
  - 2 gal  3 qt   ->   - 2 gal  3 qt
  -------------        -------------
                         2 gal  2 qt
and when using time,

    5 hr  15 min         4 hr  75 min
  - 1 hr  45 min  ->   - 1 hr  45 min
  --------------       --------------

This shouldn't be surprising, since we _could_ write those last two
problems in terms of mixed numbers!  A quart is 1/4 of a gallon, and 1
minute is 1/60 of an hour:

    5  1/4 gal          4  5/4 gal
  - 2  3/4 gal   ->   - 2  3/4 gal
  ------------        -------------
                        2  2/4 gal


    5  15/60 hr         4  75/60 hr
  - 1  45/60 hr  ->   - 1  45/60 hr
  -------------       -------------
                        3  30/60 hr

The only thing that differs among these cases is the group size.  With
the numbers you've learned about, each group (thousands, hundreds,
tens, ones) is always 10 times larger than the previous one.  But in
other systems, we have other group sizes:

  1 gal = 4 qt
  1 qt  = 2 pt
  1 pt  = 2 cup

  1 hr  = 60 min
  1 min = 60 sec

and so on. 

One final thing you might find helpful is that you can avoid this kind
of borrowing altogether.  How?  Well, suppose I have 10 dollars, and
you have 6 dollars.  I have 4 dollars more than you, right?

Now, suppose someone gives us EACH 4 more dollars.  Now I have 14, and
you have 10.  But I still have 4 dollars more than you.  That hasn't
changed.  Same as if someone gives each of us 329 dollars, or a
million dollars.  Adding the same amount to each number doesn't
change the difference between them.  Do you see why? 

How can I use this?  Well, suppose I have something like

     6  3/24
  -  3  5/24
  ----------

I can add 19/24 to each of those numbers, and the numbers change, but
the difference doesn't:

     6  3/24        6  22/24        6  22/24
  -  3  5/24  ->  - 3  24/24  ->  - 4 
  ----------      ----------      ----------

And now I have an easy subtraction!  Again, I can do this with
something as basic as

    504            505             555
  - 249  ----->  - 250  ------>  - 300
  -----  add 1   -----  add 50   -----
    
So really, I _never_ have to borrow to do a subtraction, if I don't
want to.  I can just keep adding until the number I'm subtracting is a
nice, round number--then I can do the subtraction easily. 

Sometimes one way is easier, sometimes the other way.  It's good to
know about both, so you have a choice.  And, if you solve a problem in
two different ways, getting the same answer both ways can give you
confidence that you haven't made a mistake. 

Anyway, does this help clear things up?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/