Subtracting Mixed Numbers: A Cry for Help

March 26, 2008 · 17 Comments

Paraphrased from a homeschooling math discussion forum:

Help me teach fractions! My son can do long subtraction problems that involve borrowing, and he can handle basic fraction math, but problems like $9 - 5 \frac{2}{5}$ give him a brain freeze. To me, this is an easy problem, but he can’t grasp the concept of borrowing from the whole number. It is even worse when the math book moves on to $10 \frac{1}{4} - 2 \frac{3}{7}$ .

Several people replied to this question, offering advice about various fraction manipulatives that might be used to demonstrate the concept. I am not sure that manipulatives are needed or helpful in this case. The boy seems to have the basic concept of subtraction down, but he gets flustered and is unsure of what to do in the more complicated mixed-number problems.

Let’s take a quick look at what is involved in a calculation like this…

7 steps to subtract mixed numbers

$10 \frac{1}{4} - 2 \frac{3}{7} = ?$

Ignore the whole numbers for a moment, and focus on the fraction parts.
$\frac{1}{4} - \frac{3}{7}$
Convert both fractions to a common denominator. That in itself involves several steps:
- Look at the two denominators. Can you see a super-easy common denominator? For example, is one of the numbers a multiple of the other, or are they both divisors of some common number like 12?
- If not, then realize that the Least Common Denominator is an irrelevant sidetrack and don’t worry about it.
- Use the Easiest Common Denominator: multiply the denominators of your fractions. It always works.
- Multiply the numerators of your fractions by the same amount to create equivalent fractions.
- Cross out the original fractions and write your new fractions next to them.
$\frac{1}{4} - \frac{3}{7} = \frac{1 \times 7}{4\times 7} - \frac{3\times 4}{7\times 4} = \frac{7}{28} - \frac{12}{28}$
Compare the fractions. Notice that the one being subtracted (which is called the subtrahend) is larger than the one you are taking it away from (the minuend).
Borrow/rename from the whole number part of the minuend to make a bigger fraction. This involves several more steps:
- Subtract 1 from the whole number, crossing it out and writing your new number above.
- Mentally convert the “borrowed” 1 to an improper fraction with the common denominator.
- Mentally add that converted fraction to the fraction part of the minuend.
- Cross out the fraction part of your minuend (yes, again!) and write your newer value beside it.
$10 \frac{7}{28} - 2 \frac{12}{28} = 9 \frac{7 + 28}{28} - 2 \frac{12}{28} = 9 \frac{35}{28} - 2 \frac{12}{28}$
Subtract the fraction part of the subtrahend from the fraction part of the minuend. Write this answer down below.
$\frac{35}{28} - \frac{12}{28} = \frac{35 - 12}{28} = \frac{23}{28}$
Check to make sure the fraction part of your answer is in lowest terms. Whew! This time it is.
Now look at the whole number part of the calculation and follow the standard rules for whole number subtraction…

I can see how a student might get confused, can’t you? And transferring the problem to manipulatives will only add steps, making it even more complicated!

Photo by gotplaid?

Make it simpler, step 1

I would like to try a different approach. Sometimes, taking the long way around a problem actually leads to a shorter solution. As an added bonus, the round-about method may help a student make sense out of the calculation, rather than just trying to memorize and follow a complex recipe of steps.

Here is the basic principle:

Whenever you get stuck on a problem, think about anything you can do to make it simpler.

With that in mind, let’s look again at our mixed-number calculation:

$10 \frac{1}{4} - 2 \frac{3}{7} = ?$

How can we make it simpler? Ignore the fractions for now. If it was 10 - 2, that would be easy:

$10 - 2 = 8$

Now put the fractions back where they were:

$8 \frac{1}{4} - \frac{3}{7} = ?$

Can you see that this calculation will have exactly the same answer as the problem before?

This is a key principle of working with arithmetic — do whatever you can to make your calculation simpler, without changing the basic problem. Most of the first year of algebra is spent learning how to simplify equations in this way.

Make it simpler, step 2

Okay, now what else can we do? If we had a really short problem like $1 - \frac{3}{7}$ , that would be easy, right? I am sure you could do it in your head. So let’s “steal” 1 from the 8 and ignore everything else:

$1 - \frac{3}{7} = \frac{4}{7}$

Now we have to put that left-over $\frac{4}{7}$ back with the other part that we were ignoring. This is what our calculation really looked like:

$8 \frac{1}{4} - \frac{3}{7} = 7 \frac{1}{4} + [ 1 - \frac{3}{7} ] = 7 \frac{1}{4} + \frac{4}{7} = ?$

So far, we have made the problem simpler (only one mixed number instead of two), and we have taken away everything we were supposed to subtract. Our last step is a relatively straight-forward addition problem, putting the two fraction pieces together.

At this point, we are forced to find a common denominator — but at least we got the hard subtraction work out of the way first!

$7 \frac{1}{4} + \frac{4}{7} = 7 \frac{1 \times 7}{4 \times 7} + \frac{4 \times 4}{7 \times 4} = 7 \frac{7}{28} + \frac{16}{28} = 7 \frac{23}{28}$

Show your work

How would this work in the “real life” of math homework? Start by crossing out the 10 - 2 and writing in the 8. It is easy to figure 8 - 1 = 7 mentally. I think most students can do $1 - \frac{3}{7} = \frac{4}{7}$ in their heads, but teachers will probably want them to show that step — especially at first.

The main thing students would need to write down would be that last line of addition:

$7 \frac{1}{4} + \frac{4}{7} = 7 \frac{7}{28} + \frac{16}{28} = 7 \frac{23}{28}$

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Categories: Grades 5+up · Math monsters
Tagged: Teaching, Subtraction, Arithmetic, Fractions, PUFM, Middle school, Math monsters

17 responses so far ↓

Jackie // March 27, 2008 at 12:04 am

Would converting them both to improper fractions with a common denominator help?
Roger // March 27, 2008 at 6:52 am

If basic fraction math isn’t a problem, then rearranging the terms might help, too.

$10 \frac{1}{4} - 2 \frac{3}{7} =$
$(10 + \frac{1}{4}) - (2 + \frac{3}{7}) =$
$10 + \frac{1}{4} - 2 - \frac{3}{7} =$
$(10 - 2) + (\frac{1}{4} - \frac{3}{7})$

Which is basically exactly what you’ve described, dealing with the whole number part separately from the fraction part. This is just another way of looking at it.
Denise // March 27, 2008 at 8:03 am

@Jackie—
I did offer the improper fraction suggestion to the forum poster, in addition to the method above. But I doubt she uses that idea. I don’t think many students want to deal with fractions involving such big numbers.
$10\frac{1}{4} - 2\frac{3}{7} = \frac{41}{4} - \frac{17}{7} = \frac{287}{28} - \frac{68}{28} =?$

@Roger—
What you have described is very similar to my suggestion. The biggest difference is that I keep the $\frac{1}{4}$ separated from the $\frac {3}{7}$ . Your method will still require “borrowing” from the whole number, converting the borrowed 1 to 28ths, and then doing the subtraction.

By ignoring the $\frac{1}{4}$ until the end, I am able to use the much easier calculation $1 - \frac{3}{7} = \frac{4}{7}$ plus an addition problem. Most of my students strongly prefer addition over subtraction.
Alane Tentoni // March 27, 2008 at 10:51 am

When I taught this to 7th graders, I used making change as an example.
First we would all agree that a quarter is 1/4 of a dollar, a dime is 1/10, and a nickel is 1/20. Then the example went like this:
“Suppose you have 3 dollar bills and 2 quarters in your pocket, and I bump into you outside the grocery store and ask if I can borrow 5 quarters — I need quarters, a dollar bill won’t help. So you go in the store and get the clerk to change one of your bills into quarters. How many quarters will she give you?”
While I’m talking them through it, I show them on the board how I’m “borrowing” from the dollars’ column and adding 4 to the fractions (quarters).
We do the same thing with nickels and dimes, and it always seemed to help a lot. It gave them something concrete to hang on to (a real-life example of a time when “borrowing” worked differently), without adding a manipulative for them to think about.
Denise // March 27, 2008 at 12:56 pm

Great idea! From reading the discussion, this student seems to think that “Whole numbers are whole, and fractions are parts, and never the twain shall meet.” If he could get his mind around the analogy of making change, everything would probably come clear for him.
jd2718 // March 29, 2008 at 1:19 pm

Denise,

It’s what I do, and when asked, how I help little kids.

There is a nice focus maintained throughout on the “big picture” - I mean, look how you never get lost in details to such an extent that we forget about the 10 - 2 part.

Very nice.

Jonathan
Maria Miller // March 29, 2008 at 6:11 pm

One more idea:
Subtract whole numbers separately: 10 - 2 = 8

Subtract the fractions separately:

1/4 - 3/7 = 7/28 - 12/28 = -5/28

It’s negative but all that means is you need to take it away from the other result you got… the 8.

All in all you have 8 and -5/28 and so the total is
8 - 5/28 = 7 23/28
Denise // March 29, 2008 at 7:13 pm

Thank you, Jonathan.

Maria, I have used that idea with whole number subtraction, for a student who struggled with borrowing. He enjoyed playing with an “advanced” topic like negative numbers. I don’t think I’d want to try it on a student who was struggling with fractions, however — sounds like a recipe for brain overload.

On the other hand, I can think of one student in my elementary math club who would probably enjoy doing all his homework that way…
Ms. Longhorn // March 31, 2008 at 7:08 pm

The way I teach my students who have trouble with this concept is what we call “consumation of differences”.

Instead of subtracting the fraction, you add to the number being subtracted, whatever fractional part needs to make it a whole number. So for your example of 10 1/4 - 2 3/7, you would take 4/7 and add it to create a whole number of 3, but to keep the problem balanced, you also add it to 10 1/4. (this also helps them with their continuity between equations and fractions)

By creating common denominators to add 10 1/4 and now 4/7, they end up with 10 7/28 + 16/28, which gives them 10 23/28 to work with, but now they are only subtracting 3 as a whole number instead of 2 3/7. So as a final step they are only subtracting the whole numbers instead of borrowing from the whole, plus working with adding to the numerator AND changing common denominators. It works really well!
Denise // April 1, 2008 at 7:48 am

I like this method! It is shorter than mine by one subtraction step — and as I said, my students have always hated subtraction.

This is also a good mental math trick for whole number subtraction problems: Add to both numbers whatever it takes to make the subtraction simpler. Consider the example…
$362 - 187 = ?$
Adding 13 to both numbers changes this to…
$375 - 200 = easy \; !$
Freehold 2 // April 1, 2008 at 10:37 am

[...] Carnival of Homeschooling Posted on Tuesday, 1 April, 2008 by ruby3881 The April Fool’s Edition of the Carnival of Homeschooling is up at Why Homeschool. Henry Cate offers a roundup of information on - what else could be more appropriate today - the history of April Fool’s Day. If you enjoy pranks you can read a few of those too. And of course, as usual there are dozens of wonderful posts from folks who write about homeschooling. One that caught my eye was a post to help parents cope with subtraction of mixed numbers. [...]
Brotz // May 15, 2008 at 5:55 pm

so can you give me an answer for this one please post it here or email it to me i got this off a web site

2 and 5/2 minus 2 and 3/2 what is the answer

its mixed numbers by the way i just dont know how to put up mixed numbers on here
Denise // May 15, 2008 at 6:54 pm

Is this the calculation you mean?
$2 \frac{5}{2} - 2 \frac{3}{2} = ?$
First, look at the whole parts. You have 2 whole things, and then you take away 2, so there are no whole parts left. Next, look at the cut pieces (fractions). Since they are all the same size (halves), you do not have to worry about the denominators. You have 5 pieces, and then you take away 3 of them, which means you have 2 halves left.
$Answer = \frac{2}{2}$
[Which, of course, is the same as 1 whole thing.]
caitlin // May 18, 2008 at 8:25 pm

what is the answer to 6 - 1 3/8? and how do you get it?
Denise // May 19, 2008 at 12:32 pm

It may help to think of pizzas, Caitlin. The fractions might really represent anything, but pizzas make an easy-to-imagine model to help you figure the problem out.
$6 - 1 \frac{3}{8} = ?$
Imagine starting with a pizza, and then eating $1 \frac{3}{8}$ . How much would you have left? Well, first you would eat a whole pizza, which would leave 5 pizzas left. Then, you take some Tums and come back for more pizza. You eat $\frac{3}{8}$ of the next pizza. What is left after that? In addition to a horrible stomach ache, you would have 4 whole pizzas left and part of the 5th pizza.

To make it precise: $6 - 1 \frac{3}{8} = 4 \frac{5}{8}$ .
jepons // June 25, 2008 at 11:14 am

subtracting mixed number from mixed number and finding the least common denominator.
77 and 65/90
-
58 and40/75.
Denise // June 26, 2008 at 8:35 am

Let’s start with the fractions this time. For a common denominator, you need to make a multiple of 90 and 75. That means, you need to find three numbers:

N = our common denominator.
x = what we will multiply 90 by.
y = what we will multiply 75 by.

Which gives us:
$N = 90 \times x$
and:
$N = 75 \times y$

The easiest common denominator can always be found by simply multiplying the two denominators that you have in your original fractions, like this:

$x = 75$
and:
$y = 90$
which gives us:
$N = 90 \times 75 = 6750$

But these numbers are so big that they are a mess to work with. To find a smaller common denominator, we break our numbers into factors and see whether there are any duplicates:

$90 = 2 \times 3 \times 3 \times 5$
so:
$N = 2 \times 3 \times 3 \times 5 \times x$
and:
$75 = 3 \times 5 \times 5$
so:
$N = 3 \times 5 \times 5 \times y$

Aha! Both denominators have a factor of 3×5, so we can pull that out:

$N = \left( 3 \times 5 \right) \times 2 \times 3 \times x$
and:
$N = \left( 3 \times 5 \right) \times 5 \times y$

Comparing these equations, can you see how to make them equal? Let x=5, and let y=2×3=6, which will give you:

$N = 90 \times 5 = 75 \times 6 = 450$

Finally, going back to your original equation, you need to make equivalent fractions:

$77 \frac {65 \times 5}{90 \times 5} - 58 \frac {40 \times 6}{75 \times 6} = 77 \frac {325}{450} - 58 \frac {240}{450} = ?$

I think you can finish it from there.

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