Please explain, in the simpliest most easy to understand way, how you reduce fractions?

Answer 1

well to reduce fraction you check to see if they are divisible by the same number

14/22

14 is divisible by 7 but 22 is not

14 is divisible by 2 and so is 22

7/11

its mostly trial and error when you reduce fractions...

15/30

you notice they are both divisible by 5 but they are divisible by 15 because 15 is half of 30

so you divide both sides by 15 and get 1/2

8/24

you see they are both divisible by 2 but they are also divisible by 8 and you get 1/3

sometimes if you divide them out you can reach the simplest form

10/20

divide both by 2
5/10
divide again by five this time and you get

1/2

that is how you reduce fractions so do trial and error if you're really unsure if they are divisible by a larger number (like the 15/30) check ya later ♥

Answer 2

A fraction is reduced when no number larger than one will divide evenly into both the numerator (top number) and denominator (bottom number).

Take the smaller of the two numbers (numerator and denominator) and attempt to divide it into both the numerator and denominator. If it divides evenly, you are finished!

If not, try the next smaller number to divide into both and continue each time trying the next smaller number until you find one that divides evenly into both numerator and denominator.

When you do find one that divides evenly, you now have a new fraction with smaller numerator and denominator. Start the process over again, trying the next smaller number each time. As you get "smaller" and "smaller" fractions, the original fraction is being reduced a little each time.

Eventually you have tried every number all the way down to one without finding another number that will divide into numerator and denominator. When you have reached this point, the fraction you have is the reduced form of the original fraction.

This is a tedious and unwieldy method of reducing fractions, but has the advantage that you do not have to use any reasoning ability at all, only the ability to divide and count backward down to one.

Good Luck!

Answer 3

To start, let's describe any fraction that you need to reduce can be written as X / Y.

What you need to do is find the largest common number (N) possible, which divides X and Y evenly.

In other words, this is what it generally looks like when you reduce a fraction:

X / N
-------- = ?
Y / N

Again, your task is to find the largest common number possible (N) that divides X and Y (evenly), to the smallest values possible (without leaving a remainder).

Example:

X = 15; Y = 20

Fraction = 15/20

If you mentally go through numbers (starting from 1 in your head), you will see that 5 is the largest common number that can divide both 15 and 20 without leaving a remainder.

Like this:

15 / 5 = .....3
--------- ----------
20 / 5 = .....4

You can see that the answer (3/4) cannot be reduced any further without leaving a remainder.

Answer 4

First, you have to find a common divider. If the number is large, and you aren't sure what the largest common divider is, then see if both numbers are even or odd. Let's say you've got 220/350. I don't know any large divisors, but I do know that 2 will go into both. So, let's divide each by 2. We get 110/175. Now, I don't know any large numbers that go into both 110 and 175, but I do know that both numbers can be divided by 5. So, divide them by 5 and we get 22/35. Now, that number is much easier to check, and we see that no number will go into both 22 and 35. Thus, the simplified form of 220/350 is 22/35.

Of course, you could have done that a lot easier, by noticing that both numbers ended in 0, which indicates that both numbers can be divided by 10, divide them both by 10 and we get 22/35, but I wanted to show how it can be broken down into smaller steps if you can't think of anything right off the top of your head.

Answer 5

while asked to cut back a fragment, you could continually think of of the appropriate elementary component.. very properly, as an occasion: sixteen/28 Which variety is divisible by sixteen and 28? .. Then initiate thinking of a lowest variety it is divisible by the two numbers, ok... so a million, yet then in case you divide the two numbers by a million, you get sixteen/28, it is not even interior the decreased variety. ok, attempt variety 2 b/c it is divisible by sixteen and 28, then you particularly get 8/14 8/14 <-- contemplate whether you could cut back further? particular. Now, think of of a variety it is divisible by 8 and 14, permit's attempt 2, then you particularly get 4/7 <-- then ask your self, can i nevertheless cut back this variety? in all hazard no longer answer: 4/7 is interior the decreased from of the fraction sixteen/28 while you're asked to discover the GCF(perfect elementary component).. what might it is? answer is 4. Why? sixteen/4 = 4 and 28/7 = 4

Answer 6

OK. Basically you find a number that both the numbers can be divided EQUALLY in to. And it must be the lowest one. Example:

6/12 They both can be divided equally by 6, but also 2. so we go with 2...
6 [divided by] 2 = 3
12 [divided by] 2 = 6

We have 3/6. Now. they bolth can be divided equal by 3, so...

3 [divided by] 3 = 1
6 [divided by] 3 = 2

and we send up with 1/2! OK? Email me if ya need More help.

Answer 7

divide both numerator and denominator which are top and bottom numbers with the GCF= Greatest Common Factor.\

Example: 100 and 125 is 25 because 25 is the biggest number that divides evenly in both. so 100/125= 4/5.

easy as that done.

Answer 8

well the simplest way to do it is when you get a fraction..divide the top and the bottom by 2 until you can't divde it anymore..or if you have a fraction like this
6/3.. you automatically know its going to be two..because the fraction bar is a symbol for division

Answer 9

If you know how to divide numbers by 2 or 3, you know how to reduce fractions.

eg. 18/32

divide top by 2 = 9
divide bottom by 2 = 16

and you get 9/16

allways divide by the same number until top and bottom can't be divided by the same number.

Hope this helps.

Answer 10

break up top and bottom into prime multipliers, then cancel out identical multipliers from top and bottom.

there are tricks to speed it up - if it ends in 0, it is divisible by 10; ends in 5 - by 5, if ends with a even number or zero - divide by 2. If sum of digits is divisible by 3, then the number is divisible by 3.

e.g. 6/10=(2*3)/(2*5)=3/5