Anyone good with fractions? - The Horse Forum

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post #1 of 15 Old 08-29-2008, 11:07 AM Thread Starter
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Anyone good with fractions?

I have taken a couple year break from school, and now in my algebra class, the rules are escaping me as to what to do when adding fractions with variables(among other things).
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post #2 of 15 Old 08-29-2008, 11:18 AM
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Yay math!
What are they giving you? Could you post a question?
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post #3 of 15 Old 08-29-2008, 11:34 AM Thread Starter
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Ooh a taker. Okay, this looks really simple to me, I just have forgotten how to do it.

Solve for n.

N/2 + n/5 = 2/3

Or....

n + n = 2
2 .. 5 .. 3
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post #4 of 15 Old 08-29-2008, 01:23 PM
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n/2 + n/5 = 2/3

To solve for n we need a common denominator on the left side with "#n" on top. To do this we need to use the "clever form of 1" and do some multiplication.
The clever for of 1 is like 3/3 or 5/5, three divided by three is one, etc. I'm going to put my multiplications in brackets.
n(5) + n(2) = 2
2(5) .. 5(2) .. 3
So we perform our multiplications and add the n terms ontop...
7n = 2
10 .. 3
Cross multiply
7n(3) = 10(2)
21n = 20
And solve
N = 20/21

:) Hope I helped (and got the right answer! Wouldn't that be embarassing if I did it wrong!)
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post #5 of 15 Old 08-29-2008, 02:05 PM
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Okay, I'll give it a go too :) I'm going to describe it without cross-multiplying. There are many ways to get the answer, and while at first you learn rote methods, gradually you come to understand what you're doing and apply your principles rather than following a set series of instructions blindly :)

You know that the top of the fraction is the numerator and the bottom is the denominator?

If the denominators of two fractions are the same and you are adding or subtracting them, you add/subtract the numberators and put them on top of the denominator, e.g.

3/7 + 2/7 = (3 + 2)/7 = 5/7

4/13 - 1/13 = (4 - 1)/13 = 3/13

So what you need to do to add or subtract any kind of fraction is get them with the same denominator.

In your sum, the three denominators are 2, 5 and 3. So you need to get them all with the same denominator.

Now, Anabel mentioned 'the clever form of 1'. I've never heard it described like that, but as you know, any number multiplied by one is the same number, and 3/3 or 5/5/ or 134/134 = 1. So you can multiply both sides of the equation by 1 and it will stay the same.

That means that 2/3 is the same as 10/15, as I'm sure you've learned when simplifying fractions. It's just the same process in reverse.

What's the 'lowest common multiple' of 2, 5 and 3? In other words, what's the lowest number that has 2, 5 and 3 as fractions? In this case, it's 2 x 3 x 5 = 30, but sometimes it might be lower - but multiplying them together will always work, you just may need to do some more simplifying/cancellnig at the end :)

Now we know that the lowest common multiple is 30, so they all need a denominator of 30. The point of using that number is that you end up with whole numbers on the top half of the fraction, as is correct and appropriate.

To change n/2 to something/30, we multiply by 15/15, which is multiplying by 1 so we don't change anything at all really:

N/2 x 15/15 = 15n/30

Then you need to do the same for n/5 (multiply by 6/6) and 2/3 (multiply by 10/10). That leaves us with:

15n/30 + 6n/30 = 20/30

Note that what we've done here is multiply both sides of the equation by 1, rather than multiplying each part by something different - cos you're not allowed to do that!

Now, everything on either side of the equation has a denominator of thirty. So if you multiplied both sides by thirty, that denominator would disappear and we'd be left with:

15n + 6n = 20
21n = 20
N = 20/21

This method may seem more difficult or longer than Anabel's, but it's the one I instinctively go for, because it's not How To Solve Sums With Fractions, it's tackling it as if it's any maths problem whatsoever :)

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post #6 of 15 Old 08-29-2008, 02:11 PM
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Lol thanks for the great explaination! Basic math is not really my thing anymore haha I'm taking third year calculus this year
I just tend to do things the quickest way possible and rarely show work. 8)
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post #7 of 15 Old 08-29-2008, 02:19 PM
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I did A level maths and I'm studying chemical engineering at uni - I absolutely adore maths, I'm the kind of person who solves problems for fun and relaxation. I'm also really, really big on understanding the principles, not following a set pattern. I used to argue with my teachers in secondary school about the way they taught things in GCSE Maths - I understood that less able students needed the rote methods but I thought they were a bad idea xD

I always, always showed my working, though. For one thing, it looks prettier, for another, there was a guy in my class who always tried to cram every action into one line. Guess who found it harder to untangle his mistakes ;)

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post #8 of 15 Old 08-30-2008, 07:00 PM Thread Starter
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Oh guys you so totally rock. I am glad to know I have a resource here! Thank you Thank you Thank you!

I guess it was stuck with the getting the denominators the same because of the whole "do the same thing on both sides".
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post #9 of 15 Old 08-30-2008, 07:49 PM
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Yeah, you have to do the same thing to each side. But you're not really doing anything when you're making the denominators the same - you're just making the fractions more complicated.

There's nothing different between

1/2 + 1/4 + 1/8 = 7/8

And

4/8 + 2/8 + 1/8 = 7/8

You can simplify fractions and you can make them more complicated :) It's like writing 2 x 6 x n and 12n, or breaking apart 8n to give 4 x 2n or something like that.

If you believe everything you read, better not read.
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post #10 of 15 Old 09-01-2008, 12:01 AM Thread Starter
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You guys explain this much better than my instructor. Basically, she writes the definition and is halfway through writing out an example before people(I) have time to finish the notes, then all I can do is copy the example without much comprehension before she moves on to the next thing.

Gah, it is so frustrating.
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