Subtracting Mixed Fractions With BorrowingDate: 01/08/2004 at 20:46:02 From: Ethan Subject: Subtracting Fractions Hi Dr. Math - My name is Ethan and I'm in 7th grade. I am stumped when it comes to subtracting (mixed number) fractions. Then I found this website and I decided to ask. Say I have 5 6/8 to be subtracted from 10 3/8. Do I have to borrow because the 6 is bigger than the 3? I know that you can't subtract a smaller number from a bigger number. If you do have to borrow, then how do you do it? Date: 01/09/2004 at 10:50:52 From: Doctor Ian Subject: Re: Subtracting Fractions Hi Ethan, There are a couple of ways to approach a problem like this. One way-- which doesn't require borrowing--is to convert the mixed numbers to improper fractions, and then subtract those: 10 3/8 - 5 6/8 = 83/8 - 46/8 = 37/8 If you're not sure how to do that conversion, take a look at Converting Mixed Numbers to/from Improper Fractions http://mathforum.org/library/drmath/view/58876.html Another way to proceed is to "borrow" 1 from the larger integer. That is, 10 3/8 is the same as 10 + 3/8, right? So 10 3/8 = 10 + 3/8 = 9 + 1 + 3/8 = 9 + 8/8 + 3/8 = 9 + 11/8 And now we have 9 11/8 - 5 6/8 = ... which is easier to do, right? This is exactly the same thing we're doing in a subtraction problem, when we borrow from the next column over, e.g., . . . . . . . Here we exchanged one ten . for ten ones. 84 7 14 - 26 -> - 2 6 ---- ------ 5 8 We just exchanged one of our tens for ten ones, to make subtracting ones possible. The same idea works when subtracting times...except you have to remember that there are 24 (not 10) hours in a day, and 60 (not 100) minutes in an hour, or seconds in a minute: . . . . . . . . Here we exchanged one hour . for 60 minutes. . 7:15 6:75 - 5:41 -> 5:41 ------ ---- 1:34 Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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