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This is best done through an example. Consider this problem: 3/4 (divided by) 9/16

First, you "change" the problem into a multiplication problem by "flipping" (taking the reciprocal) of the (divisor or dividend I forget, the second fraction) and then changing the division symbol into a multiplication symbol. Thus, the problem becomes:

3/4 * 16/9

Now there are two ways to go from here, we'll take the long way first.

Multiply straight across.

3*16=48 < Top

4*9=36 < Bottom

48/36.

To reduce, you find the greatest common factor between the two numbers. To find this GCF, you "prime factorize" 48 and 36, that is, split it into a multiplication problem (again best explained by an example). Consider 28. Divide 28 by 2 and get 14. Divide 14 by 2 to get 7. Thus, 7*2*2 (More useful form, 7*22 but it means the same) is the prime factorization of 28. You do the same thing for 48 and 36.

PF of 48: 2*2*2*2*3*

PF of 36: 2*2*3*3

So now you look at all the common "numbers" of PF of 48 and PF of 36. Note that although they both have 2's, each 2 of the PF of 36 can only match with ONE 2 of 48.

48: 36:

2>>>>>2

2>>>>>2

2

2

3>>>>>3

Thus there are two 2's that matches, and one 3's that matches. So they "cancel" out.

Now you multiply all the numbers of the PF again, ignoring what you already canceled out.

For 48, there are two 2's, so it'd be 4.

For 36, there is one 3, so it'd be 3.

Now put them over each other, 4/3.

The process gets faster as you do it more and more and get faster at multiplication/division.

Here's a slightly shorter way:

3/4 * 16/9

See if you can reduce the numbers right here (This process is essentially the same thing, but you "prime factorize" first before multiplying). Prime factorize 3, 16, 4 and 9.

Top: 3*2*2*2*2

Bottom: 2*2*3*3

Look familiar?

If you could see that 16 cannot divide into 9, but can divide into 4, and that 3 cannot divide into 4, but can divide into 9, you could have simplified the problem right there:

3/4 * 16/9 => 3/4 * (4*4)/(3*3)=4/3

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15y ago

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