1. ## dividing fractions

i got lost in complex fractions. i don't know how to input equations here..

2. {1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?

3. You could let $1=\frac{m-1}{m-1}$ see if that helps.

4. ## Fractions

Hello princess_21
Originally Posted by princess_21
i got lost in complex fractions. i don't know how to input equations here..
Sorry, I can't understand your attached document. But have another look at the one you've already visited

http://www.mathhelpforum.com/math-he...fractions.html

If you're still really stuck, can you write one out neatly by hand, and either scan it or take a digital photo, and attach that instead?

5. Originally Posted by princess_21
{1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?
is this it?

$\displaystyle \frac {1 - \frac 1{m - 1}}{1 + \frac 1{m - 1}} + \frac {1 - \frac 1{1 + \frac 1{m - 1}}}{1 - \frac 1{1 - \frac 1{m-1}}}$

6. Originally Posted by princess_21
{1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?
$\frac{1-\frac{1}{m-1}}{1+\frac{1}{m-1}}+\frac{1-\frac{1}{1+\frac{1}{m-1}}}{1-\frac{1}{1-\frac{1}{m-1}}}$

Is this the correct equation?

7. Originally Posted by ronaldo_07
$\frac{1-\frac{1}{m-1}}{1+\frac{1}{m-1}}+\frac{1-\frac{1}{1+\frac{1}{m-1}}}{1-\frac{1}{1-\frac{1}{m-1}}}$

Is this the correct equation?
haha, well, both you and i interpreted it the same way. chances are we're correct...unless there was a typo.

would you like to proceed?

8. Originally Posted by ronaldo_07
You could let $1=\frac{m-1}{m-1}$ see if that helps.
I would let $1-\frac{1}{m-1}=\frac{m-1}{m-1}-\frac{1}{m-1}$

This simplyfies to (m-2)

Do this also for $1+\frac{1}{m-1}$ and it should come in a nice form for you to easily work out m

9. here's another attachment..

10. yeah. that's the question. thanks

11. Originally Posted by princess_21
here's another attachment..
We have interpreted correctly look above for help I have provided

12. Originally Posted by Jhevon
haha, well, both you and i interpreted it the same way. chances are we're correct...unless there was a typo.

would you like to proceed?

yeah that's the question..

13. i solved it this way.
let m= m-1

and then i got (2m-2)/(m+1)... is this correct?

14. Originally Posted by ronaldo_07
I would let $1-\frac{1}{m-1}=\frac{m-1}{m-1}-\frac{1}{m-1}$

This simplyfies to (m-2)

Do this also for $1+\frac{1}{m-1}$ and it should come in a nice form for you to easily work out m
indeed. just to flesh it out a bit

$\displaystyle \frac {1 - \frac 1{m - 1}}{1 + \frac 1{m - 1}} \cdot {\color{red} \frac {m - 1}{m - 1}} + \frac {1 - \frac 1{1 + \frac 1{m - 1}} \cdot {\color{red} \frac {m - 1}{m - 1}}}{1 - \frac 1{1 - \frac 1{m-1}} \cdot {\color{red} \frac {m - 1}{m - 1}}}$

$= \frac {m - 1 - 1}{m - 1 + 1} + \frac {1 - \frac {m - 1}{m - 1 + 1}}{1 - \frac {m - 1}{m - 1 - 1}}$

$= \frac {m - 2}m + \frac {1 - \frac {m - 1}m}{1 - \frac {m - 1}{m - 2}}$

now perform a similar trick on the second fraction as i did before and continue

15. Originally Posted by princess_21
i solved it this way.
let m= m-1

and then i got (2m-2)/(m+1)... is this correct?
you cannot solve it that way. you would be changing the value of the fraction. you would have to use another variable and then make the back-substitution afterwards. just follow the trick you have been shown. it is the easiest way