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Borrowing and Subtracting with Mixed NumbersDate: 09/03/2008 at 16:40:09 From: Emerald Subject: 3 1/3 - 2 2/5 How do I make the first numerator larger than the second numerator when I'm subtracting mixed fractions? On one website it says: Make the first numerator larger than the second, and it gives this example: 5 1/3 = 4 4/3. I don't understand how 5 1/3 became 4 4/3. I looked for a step by step method on how this was done but I couldn't find one. I would normally start solving this problem: 3 1/3 - 2 2/5 like this: Starting with 3 1/3, I would multiply the denominator (3) with the integer (3) then add the numerator (1) and end up with 10/3 (3x3=9+1=10) Then I would do the same thing with 2 2/5 (5x2=10+2=12) which leaves me with 12/5. Now I'm at 10/3 - 12/5. This is where I get totally lost. I think I'm supposed to multiply the two denominators together (3x5=15). And then subtract the two numerators which leaves me with something like an improper 8 (10- 12=8) leaving me with 8/15. Now I strongly feel like this is incorrect, because I believe my answer should either be a mixed fraction or just a smaller proper fraction. My brain is quite frozen on the subject. Please Help...
Date: 09/03/2008 at 18:47:01
From: Doctor Ian
Subject: Re: 3 1/3 - 2 2/5
Hi Emerald,
You wrote:
>How do I make the first numerator larger than the second numerator
>when I'm subtracting mixed fractions?
The same way you would do it if making change. Suppose you have 4
dollar bills, and 3 dimes, and you want to give someone 1 dollar bill
and 7 dimes. You could trade 1 dollar bill for 10 dimes, giving you 3
dollar bills and 13 dimes. Then it's easy, right?
Same thing here. Suppose I want to subtract
4 3/10 - 1 7/10
I can change the initial number this way:
4 3/10 = 4 + 3/10
= 3 + 1 + 3/10
= 3 + 10/10 + 3/10
= 3 + 13/10
= 3 13/10
So now I have
3 13/10 - 1 7/10 = ...
which I can do easily.
It's really the same thing we do when subtracting:
43
- 17
----
I trade 1 group of 10, for 10 "groups" of 1,
3 [13]
- 1 7
--------
and now I can proceed.
So you see, it's all the same idea, over and over again--just
breaking up groups, so we can use the pieces.
>On one website it says: Make the first numerator larger than the
>second, and it gives this example: 5 1/3 = 4 4/3. I don't
>understand how 5 1/3 became 4 4/3. I looked for a step by step
>method on how this was done but I couldn't find one.
Does it make more sense now? One key to this is remembering that a
mixed number is really an implied addition:
5 1/3 = 5 + 1/3
We just leave out the "+" as a convenience.
>I would normally start solving this problem: 3 1/3 - 2 2/5 like
>this: Starting with 3 1/3, I would multiply the denominator (3) with
>the integer (3) then add the numerator (1) and end up with 10/3
>(3x3=9+1=10) Then I would do the same thing with 2 2/5 (5x2=10+2=12)
>which leaves me with 12/5. Now I'm at 10/3 - 12/5. This is where I
>get totally lost.
This is a perfectly valid way to do it. You just need to find a
common denominator--one that is divisible by both the denominators
you have.
The smallest number divisible by 3 and 5 is 15, so you'll multiply the
first fraction by 5/5 (which is the same as multiplying by 1, so the
appearance changes, but the value doesn't),
10 5 50
-- * - = --
3 5 15
And you'll multiply the other fraction by 3/3,
12 3 36
-- * - = --
5 3 15
And now you just subtract the numerators:
10 12 50 36 50 - 36 14
-- - -- = -- - -- = ------- = --
3 5 15 15 15 15
As a check, we can think about whether that's in the right ballpark.
We're subtracting 2 and something from 3 and something, so we should
get an answer around 1. Which we did.
>I think I'm supposed to multiply the two
>denominators together (3x5=15). And then subtract the two
>numerators which leaves me with something like an improper 8 (10-
>12=8) leaving me with 8/15. Now I strongly feel like this is
>incorrect, because I believe my answer should either be a mixed
>fraction or just a smaller proper fraction. My brain is quite frozen
>on the subject. Please Help...
Note that you could have also just changed the fractions to have
common denominators,
3 1/3 - 2 2/5
= 3 5/15 - 2 6/15
= 2 20/15 - 2 6/15
= (2 - 2) + (20/15 - 6/15)
= 0 + 14/15
= 14/15
Does this help?
- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
Date: 09/18/2008 at 16:49:28 From: Emerald Subject: Thank you (3 1/3 - 2 2/5) Thanks man. That was awesome! I used this knowledge when I took my placement exam and got a high enough score so that I didn't have to pay for a remedial math class. I'll definitely be coming back asking for more advice. Thanks again. Your new fan, Emerald |
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