Question

Does anyone know how to simplify complex fractions or simplify the differences?

Answer 1

It would be better for you to post your actual question.
most people here will be able to help you on almost any topic in maths.

Answer 2

well it says simplify the difference: (x)over(3x+9) - (8)over(xsquared+3x)

Answer 3

so its:\[\frac{x}{3x+9}-\frac{8}{x^2+3x}\]

Answer 4

yep. How did you do that?

Answer 5

first thing you need to do is to factorise the denominators.
(you can use the equation editor to enter latex expressions here)

Answer 6

Oh..okay. thanks.
and im not quite sure how to factor 3x+9

Answer 7

can you notice anything common between the terms:
3x and 9

Answer 8

what divides into both these terms?

Answer 9

3

Answer 10

correct, therefore we can factorise it as follows:\[3x+9=3(x+3)\]try and do the same with the second denominator

Answer 11

oh. hah okay

Answer 12

BTW: if you don't know how to enter latex yet then you can enter your equation:

(x)over(3x+9) - (8)over(xsquared+3x)

as:

x/(3x+9) - 8/(x^2+3x)

to make it easier for others to read (and for you to enter)

Answer 13

so would the second one be x(x+3)
and thanks.

Answer 14

perfect!

Answer 15

so now we can rewrite the equation as follows:\[\frac{x}{3x+9}-\frac{8}{x^2+3x}=\frac{x}{3(x+3)}-\frac{8}{x(x+3)}\]

Answer 16

then would I multiply the opposite side by the factored denominator?

Answer 17

from here you now need to find the common denominator for both fractions which is the least common multiple (LCM) of both.

Answer 18

its both of them right? since there isnt more than one thats the same.

Answer 19

the two denominators are: 3(x+3) and x(x+3)
which term is common in both?

Answer 20

3 and x

Answer 21

the 3 and x are the ones that are not common :)

Answer 22

(x+3) occurs in both denominators, therefore (x+3) is the common term, therefore you need to include this just once

Answer 23

next you multiply it by all the other "uncommon" terms - which are 3 and x

Answer 24

so the LCM for the denominator would be 3x(x+3)

Answer 25

can you see how both denominators would divide exactly into this LCM?

Answer 26

so you multiply each side with the lcm?

Answer 27

No, i'm confused on where i go from writing down (x)/3(x+3) - (8)/x(x+3)

Answer 28

it might help to do an example with just numbers.
suppose we had:\[\frac{5}{3*4}+\frac{3}{2*4}\]

Answer 29

the LCM of the denominators here would be 3*2*4

Answer 30

so we would write:\[\frac{5}{3*4}=\frac{5*2}{3*4*2}\]and:\[\frac{3}{2*4}=\frac{3*3}{2*4*3}\]

Answer 31

this leads to both fractions having the "same" denominator, so we can then do the following:\[\frac{5*2}{3*4*2}=\frac{10}{24}\]\[\frac{3*3}{2*4*3}=\frac{9}{24}\]\[\frac{5}{3*4}+\frac{3}{2*4}=\frac{10}{24}+\frac{9}{24}=\frac{19}{24}\]

Answer 32

okay. So i multiply each side with the lcm?

Answer 33

that is why we first need to find a "common" denominator

Answer 34

ok, so lets summarise where we got to:\[\frac{x}{3x+9}-\frac{8}{x^2+3x}=\frac{x}{3(x+3)}-\frac{8}{x(x+3)}\]and we know LCM is 3x(x+3), therefore:\[\frac{x}{3x+9}-\frac{8}{x^2+3x}=\frac{x}{3(x+3)}-\frac{8}{x(x+3)}\]\[\qquad=\frac{x*x}{3x(x+3)}-\frac{8*3}{3x(x+3)}\]\[\qquad=\frac{x^2}{3x(x+3)}-\frac{24}{3x(x+3)}\]

Answer 35

now we have a common denominator so we can get:\[\qquad=\frac{x^2-24}{3x(x+3)}\]

Answer 36

where did you get \[x ^{2}-24\]?

Answer 37

from the working out above where we got to:\[\qquad=\frac{x^2}{3x(x+3)}-\frac{24}{3x(x+3)}\]

Answer 38

\[x*x=x^2\]\[8*3=24\]

Answer 39

so you multiplied each side by 3x(x+3)?

Answer 40

not quite, I multiplied each fraction with something that would turn its denominator into the LCM.
writing it like this might help:\[\frac{x}{3(x+3)}=\frac{x}{3(x+3)}*\frac{x}{x}=\frac{x^2}{3x(x+3)}\]

Answer 41

we can do this because multiplying any number by 1 doesn't change its value, i.e.\[\frac{x}{x}=1\]

Answer 42

similarly, we multiplied the second fraction by:\[\frac{3}{3}\]

Answer 43

sorry i had to get off for a minute.

Answer 44

did you follow the explanation above?

Answer 45

yes

Answer 46

ok - good :)