Question

x-1/x over x+1/x eliminate the complex fractions

Answer 1

\[\frac{x-1}{x}\times \frac{x}{x+1}\] is a start

Answer 2

i.e. divide by "multiplying by the reciprocal" just like with number
then cancel the \(x\) top and bottom

Answer 3

as my hero @Hero would write
\[\frac{x-1}{\cancel{x}}\times \frac{\cancel{x}}{x+1}=\frac{x-1}{x+1}\]

Answer 4

^hey i think you misread the question

Answer 5

@Gabysolis49, did you check your previous question?

Answer 6

x-1/x over x+1/x ..isnt it (x-1/x)/(x+1/x) ?

Answer 7

Hum actually he didn't the problem I have is written differently

Answer 8

okay i will help you

Answer 9

hey no that's different

Answer 10

(x-1/x)/(x+1/x) ..this is her problem

Answer 11

you would get ( x^2-1)/(x^2+1)

Answer 12

no you dont..you are not interpreting it right

Answer 13

jesus you still doing this?

Answer 14

wait what? authority/power > correct solution?

Answer 15

i know but thats what she meant as she said

Answer 16

@PrincetonTiger, the correct way to post it linearly if we were doing it your way would be

(x - (1/x))/(x + (1/x))

To post it the way I did it:
((x - 1)/x)/((x+1)/x)

@Gabysolis49 posted neither so it is open to interpretation.

Answer 17

perhaps it is
\[\large \frac{x-\frac{1}{x}}{x+\frac{1}{x}}\] equation editor is our friend

Answer 18

Yes like that^

Answer 19

@Hero please. She told you that the way you are interpreting is wrong. Humans can make mistakes and so did you. Just accept it. No point pushing your authority. It is simply rude

Answer 20

kk
multiply top and bottom by \(x\) and finish in one step right?

Answer 21

let me know what you get

Answer 22

x-1 over x+1

Answer 23

^ nope that isnt right

Answer 24

i would do one step for you ..((x^2-1)/x)/((x^2+1)/x))