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Mathematics 21 Online
OpenStudy (breann):

what is the easiest way to find the least common denominator? i'm struggling with this.

OpenStudy (anonymous):

prime-factorisation

OpenStudy (saifoo.khan):

Ladder table.

OpenStudy (amistre64):

whats the issue?

OpenStudy (amistre64):

if its just adding and subtracting fractions, i neer bother with a LCD, i let the math do it for me

OpenStudy (breann):

in my textbook it says find the LCD of 11/12 and 3/20 then it just tells me that the LCD is 60. it doesn't really explain how it got that... then it has LCD= 2*2*3*5=60

OpenStudy (amistre64):

lets see what we can find: im gonna add them up just to get them cosy; lets assign them to some number N as a place holder. \[\frac{11}{12}+\frac{3}{20}=N\] now lets start clearing fractions, its prolly best to start with the smaller one: \[\frac{11(12)}{12}+\frac{3(12)}{20}=(12)N\] and reduce \[11+\frac{3(3)}{5}=12N\] \[11(5)+\frac{3(3)(5)}{5}=12(5)N\] \[11(5)+3(3)=60N\] and the number by the N is your lowest common denom

OpenStudy (amistre64):

the addition itself is unimportant since all we want to find is that 60 to begin with

OpenStudy (amistre64):

as far as how your book tells you to do it? i dunno

OpenStudy (breann):

well, then it says to write equivalent fractions. then subtract fractions with common denominators. then simplify. i just don't get all these steps and where they get certain numbers from.

OpenStudy (amistre64):

so in essense, they want you to do a whole bunch of work before you actually do the math in order to do the math ....

OpenStudy (breann):

pretty much. lol.seems ridiculous to me

OpenStudy (amistre64):

me too :) they figure if you find the smallest number that the bottoms have in common, that you can alter both fractions by a useful form of the number "1". then just combine the tops and keep the common bottom. but the math can do all that work in alot less time and effort

OpenStudy (amistre64):

lets forget subtracting at the moment and see if we can just follow the steps. \[\frac{11}{12}+\frac{3}{20}; cd=60\] \[\frac{11}{12}*\frac{5}{5}+\frac{3}{20}*\frac{3}{3}\] \[\frac{55}{60}+\frac{9}{60}\] \[\frac{55+9}{60}\] is how they want it to go; but i get that at the start without having to go thru alot of worry

OpenStudy (amistre64):

my last line up top was: \[11(5)+3(3)=60N\] \[55+9=60N\] \[\frac{55+9}{60}=N\] same outcome

OpenStudy (breann):

ok, where does the 5s and the 3s come in? that is whats confusing to me...

OpenStudy (amistre64):

ok, do you see that 5/5 = 1 and 3/3 = 1 tho? and that anytime we mucliply by "1" the value doesnt change

OpenStudy (breann):

ooooh. ok. yeah.

OpenStudy (amistre64):

12 * [ ] = 60 12 * [5] = 60 this is why we choose 5 ----------- 20 * [ ] = 60 20 * [3] = 60 and this is why we choose 3 they are the numbers that change the bottoms into 60

OpenStudy (breann):

OHH! wow. i feel like i was making it way more difficult for myself than it is.

OpenStudy (amistre64):

it happens :)

OpenStudy (breann):

Thank you! :)

OpenStudy (amistre64):

youre welcome :) and good luck

OpenStudy (anonymous):

For finding LCD you can use Prime Factorisation Ladder method as shown in the attachment that follows. Here you will notice that we divide by a factor even if only one of the numbers is divisible by it . We divide till you get 1 for all numbers at the end.... For your example of 12 and 20 we do as follows:

OpenStudy (anonymous):

|dw:1319540850997:dw|

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