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Mathematics 7 Online
OpenStudy (anonymous):

I need help on my fraction homework.

OpenStudy (sandra):

hey nya was helping you in your other one heh

OpenStudy (sandra):

your other question

OpenStudy (anonymous):

I don't understand what you were talking about :(

OpenStudy (sandra):

well, let's do another practice problem

OpenStudy (anonymous):

No. I need help on my actual problem!!!

OpenStudy (sandra):

well, you wanted to compare 5 1/3 and 6.7

OpenStudy (sandra):

to do this, you want to convert 6.7 into a fraction too

OpenStudy (anonymous):

YES!

OpenStudy (sandra):

so the first rule of numbers is, if you multiple *any* number by 1, it is the same number

OpenStudy (sandra):

so to convert 6.7 into a fraction, you want to multiply it by some version of a fraction by multiplying by some version of 1 - in this case, you can multiple it by 10/10 (which is one)

OpenStudy (sandra):

so 6.7 * 10/10 = (6.7 * 10)/10 = 67/10

OpenStudy (sandra):

now we want to convert 5 1/3 to a non-mixed fraction

OpenStudy (sandra):

to do that, why don't we convert it into "1/3's"

OpenStudy (sandra):

thirds

OpenStudy (sandra):

so 5 1/3 = 5 + 1/3 5 = 5 * 3/3 = 15/3 so 5 1/3 = 15/3 + 1/3 = 16/3

OpenStudy (sandra):

so really, you're comparing 16/3 to 67/10

OpenStudy (sandra):

it's kind of hard to tell from just that right?

OpenStudy (sandra):

and so what you have to do, is make sure the bottom number in both of those fractions is the same - that way you can just compare the top numbers to tell which is bigger

OpenStudy (sandra):

that process is called finding a "common denominator"

OpenStudy (sandra):

a common bottom number for both fractions

OpenStudy (sandra):

now this is where our first rule comes back into play - you can multiply any number by "1" and you have the same number

OpenStudy (sandra):

So we want to make it so 16/3 to 67/10 both have the same bottom number. One way to make sure of that is find some number both of those bottom numbers - 3 and 10 - divide into evenly

OpenStudy (anonymous):

Ok, I think I am going 2 find somebody else!

OpenStudy (sandra):

ok good luck =)

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