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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