Question

Complex fraction help

Answer 1

I did it where the method of getting the same denominator for each side (numerator and denominator, then switch the 2nd one and multiply

Answer 2

what is the least common multiple of the denominators of the fractions there
That is what is the least common multiply of x-2,x+2,x^2-4?

Answer 3

If you could show your work that method maybe I can pinpoint my mistake..

Answer 4

I multiplied 3/x-2 and 6/x^2-4 by x+2 only for 3/x-2

Answer 5

First of all do you know x^2-4=(x-2)(x+2)?

Answer 6

yes

Answer 7

And so therefore the least common multiple of x-2,x+2,x^2-4 is x^2-4
or you can just say (x-2)(x+2)
Least common multiple is (x-2)(x+2)
So multiply the least common multiple on top and bottom.

Answer 8

thats why 3/x-2 is just missing x+2

Answer 9

This will get rid of the compound fraction.

Answer 10

the numerator once i got a common deniminator and subtracted I got 3x/x^2-4

Answer 11

is that right?

Answer 12

\[\frac{\frac{3}{x-2}-\frac{6}{(x-2)(x+2)}}{\frac{3}{x+2}+\frac{1}{x-2}} \ \cdot \frac{(x+2)(x-2)}{(x+2)(x-2)} \\ \frac{\frac{3}{x-2}(x+2)(x-2)-\frac{6}{(x-2)(x+2)}(x+2)(x-2)}{\frac{3}{x+2}(x+2)(x-2)+\frac{1}{x+2}(x+2)(x-2)}\]

Answer 13

\[ \frac{\frac{3}{x-2}-\frac{6}{(x-2)(x+2)}}{\frac{3}{x+2}+\frac{1}{x-2}} \ \cdot \frac{(x+2)(x-2)}{(x+2)(x-2)} \\ \frac{\frac{3}{\cancel{x-2}}(x+2)(\cancel{x-2})-\frac{6}{\cancel{(x-2)}\cancel{(x+2)}}\cancel{(x+2)}\cancel{(x-2)}}{\frac{3}{\cancel{x+2}}(\cancel{x+2})(x-2)+\frac{1}{\cancel{x+2}}(\cancel{x+2})(x-2)} \]

Answer 14

so you should have on top 3(x+2)-6
and on bottom you should have 3(x-2)+1(x-2)

Answer 15

and yes the top does give you 3x

Answer 16

but I think you need to play with your bottom

Answer 17

because 3(x-2)+1(x-2) shouldn't simplify to x^2-4

Answer 18

oh wait are you saying you didn't get rid of the compound fraction

Answer 19

you just combine the top fractions?

Answer 20

well and I wrote the problem down wrong

Answer 21

but i think you get the idea now

Answer 22

I put 1/(x+2) instead of 1/(x-2)

Answer 23

\[\frac{ 3 }{ x-2 }\times \frac{ x+2 }{ x+2 }=\frac{ 3x+6 }{ x^2-4 }\]

Answer 24

you subtract the 6 to get 3x/x^2-4?

Answer 25

\[\frac{\frac{3}{x-2}-\frac{6}{(x-2)(x+2)}}{\frac{3}{x+2}+\frac{1}{x-2}} \ \cdot \frac{(x+2)(x-2)}{(x+2)(x-2)} \\ \frac{\frac{3}{\cancel{x-2}}(x+2)(\cancel{x-2})-\frac{6}{\cancel{(x-2)}\cancel{(x+2)}}\cancel{(x+2)}\cancel{(x-2)}}{\frac{3}{\cancel{x+2}}(\cancel{x+2})(x-2)+\frac{1}{\cancel{x-2}}(x+2)(\cancel{x-2})} \\ \]
there so on top you should have 3(x+2)-6
and on bottom you should have 3(x-2)+1(x+2)

Answer 26

I mean 3x/x^2-4 just for the numerator

Answer 27

and yeah if you want you can combine the top fractions
and then the bottom fractions and then get rid of the compound fraction

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or you could did what i did in the first step which is get rid of the compound fractions by multiplying by the lcm of the denominators

Answer 28

Then for this part I multiplied the first part by x-2 and the other by x+2

Answer 29

so if i use my method would it be \[\frac{ 3x }{ x^2-4 }\times \frac{ x^2-4 }{ 4x-4 }\]

Answer 30

yep which is the same thing as 3(x+2)-6 on top
and 3(x-2)+1(x+2) on bottom

simplifying gives you 3x on top
and 3x-6+x+2=4x-4 on bottom

Answer 31

oh i see, instead of just crossing out x^-4 as both numerator and denominator, i actually multiplied it out...that's why I was confused

Answer 32

oh lol

Answer 33

yeah those (x^2-4) cancel

Answer 34

you did it!

Answer 35

you can breathe! :)

Answer 36

lol yes. final answer is 3x/4(x-1)